A /«(lH^<'4/^ I'-^-L'VL/Vi..^^ V. \ TREATISE . OF .ALGEBRA: WHEREIN THE PRINCIPLES ARE DEMONSTRATED, AND APPLIED IN MANY USEFUL AND INTERESTING INQUIRIES, AND IN THE RESOLUTION OF A GREAT VARIETY OF PROBLEMS OF DIFFERENT KINDS. TO WHICH IS ADDED, ^ THE GEOMETRICAL CONSTRUCTION OF A GREAT NUMBER OF LINEAR AND PLANE PROBLEMS, WITH THE METHOD OF RESOLVING THE SAME NUMERICALLY, BY THOMAS SIMPSON, F. R. S. FIRST AMERICAN, FROM THE EIGHTH LONDON EDITION. PHILADELPHIA : PRINTED FOR MATHEW CAREY. SOLD BY C. 8c A. CONRAD & CO.; BRADFORD Sc INSKEEP; HOPKINS Sc EARLE; JOHNSON 8c WARNER; KIMBER & CONRAD; BIRCH Sc SMALL; AND EVERT DUYCKINCK. PRINTED BY T. ^ G. PALMER. March 17, 1809. THE AUTHOR'S PREFACE TO THE SECOND EDITION. The motives that first gave birth to the ensuing work, were not so much any extravagant hopes the author could form to himself of greatly extending the subject by the addition of a large variety of new improvements (though the reader will find many things here that are nowhere else to be met with), as an earnest desire to see a subject of such general importance established on a clear and rational foun- dation, and treated as a science, capable of demon- stration, and not a myterious art, as some authors, themselves, have thought proper to term it. How well the design has been executed, must be left for others to determine. It is possible that the pains here taken, to reduce the fundamental princi- pies, as well as the more difficult parts of the sub- ject to a demonstration, may be looked upon, by some, as rather tending to throw new difficulties in the way of a learner, than to the facilitating of his progress. In order to gratify, as far as might be, the inclination of this class of readers, the demon- strations are now given by themselves, in the man- ner of notes (so as to be taken or omitted at plea- sure) ; though the author cannot by any mieans be induced to think that time lost to a learner which PREFACE. is taken up in comprehending the grounds whereon he is to raise his superstructure : his progress may, indeed, at first, be a little retarded; but the real knowledge he thence acquires will abundantly com- pensate his trouble, and enable him to proceed, af- terwards, with certainty and success, in matters of greater difFculty, where audiors, and their rules, can yield him no assistance, and he has nothing to de- pend upon but his own observation and judgment. This second edition has many advantages over the former, as well with respect to a number of new subjects and improvements, interspersed throughout the whole, as in the order and disposition of the ele- mentary parts : in which particular regard has been had to the capacities of young beginners. The work, as it now stands, will, the author flatters him- self, be found equally plain and comprehensive, so as to answer, alike, the purpose of the lower and of the more experienced class of readers. P. S. The great reputation of Mx. Simpson's Treatise o/'Algebra, and the favourable recep^ tion it has universally met with since its first publi- cation^ and which testifies it to be the best elementary work upon the subject^ has induced the proprietor to have this edition carefully revised and corrected throughout by a very eminent mathematician: he therefore trusts it will be found as worthy the appro- bation of the public as any former edition^ or as if revised by the author himself THE CONTENTS. SECTION L NOTATION SECTION IL Page 1 ADDITION SECTION III. 8 SUBTRACTION SECTION IV. 11 MULTIPLICATION 13 SECTION V. DIVISION SECTION VI. 28 INVOLUTION SECTION vn. 36 EVOLUTION - 42 SECTION VIIL THE REDUCTION OF FRACTIONAL AND RA- DICAL QUANTITIES - 45 SECTION IX. OF EQUATIONS - - 57 1 . The reduction of single Equations ibid. 2. The extermination of unknown Quantities, or the reduction of two or more equations to a single one 63 SECTION X. OF ARITHMETICAL AND GEOMETRICAL PROPORTIONS - 69 SECTION XL THE SOLUTION OF ARITHMETICAL PROB- LEMS - "7^ SECTION XII. THE RESOLUTION OF EQUATIONS OF SE- VERAL DIMENSIONS - 131 1. Of the origin and composition of equations ibid. 2. How to know whether some, or all the roots of an Equation be rational, and, if so, what they are 1 34 CONTENTS. Page 3. Another way of discovering the same thing, by- means of Sir Isaac Newton's method of divisors ; with the grounds and explanation of that method 136 4. Of the solution of cubic Equations according to Cardan ' - 143 5. The same method extended toother, higher Equa- tions - 145 6. Of the solution of biquadratic. Equations according to Des Cartes - 148 7. The solution of biquadratics by a new method, with- out the trouble of exterminating the second term 150 8. Cases of biquadratic Equations that may be re- duced to quadratic ones - 153 9. The resolution of literal Equations, wherein the given, and the unknown quantity are alike af- fected - 156 10. The resolution of Equations by the common me- thod of converging series - 158 11. Another way, more exact 162 12. A third method ^ - 170 13. The method of converging series extended to surd Equations - - 174 14. A method of solving high Equations, when two, or more unknown quantities are concerned in each 177 SECTION XIII. OF INDETERMINATE PROBLEMS 180 SECTION XIV. THE INVESTIGATION OF THE SUMS OF POWERS - -201 SECTION XV. OF FIGURATE NUMBERS - 213 1 . The Sums of Series, consisting of the reciprocals of figurate numbers, with others of the like nature 2 1 5 *2, The sums of compound Progressions, arising from a series of powers drawn into the terms of a geo- metrical progression 219 3. The combinations of Quantities ^ - 225 4. A demonstration of Sir Isaac Newton's Binomial theorem - - 227 SECTION XVI. OF INTEREST AND ANNUITIES - 229 1. Annuities and Pensions in Arrear, computed at simple interest - 231 CONTENTS. Page 2. The investigation of Theorems for the solution of the various cases in compound interest and annuities 234 SECTION XVII. OF PLANE TRIGONOMETRY - 241 SECTION XVIII. THE APPLICATION OF ALGEBRA TO THE SOLUTION OF GEOMETRICAL PROB- LEMS . - 254 1 . An easy way of constructing, or finding the roots of a quadratic equation, geometrically 267 2. A demonstration why a problem is impossible when the square root of a negative quantity is concerned 272 3. A method for discovering whether the root of a ra- dical quantity can be extracted - 284 4. The manner of taking away radical quantities from the denominator of a fraction, and transferring them to the numerator - 288 5. A method for determining the roots of certain high Equations, by means of the secti(Ai of an angle 30 1 AN APPENDIX, Containing the geometrical construction of a large varie- ty of linear and plane Problems; with the man- ner of resolving the same numerically 3 1 5 TREATISE OF A L G E B R A- SECTION I. Of Notation. XjlLGEBRA is that science which teaches, in a general manner, the relation and comparison of abstract quanti- ties : by means whereof silch questions are resolved whose solutions would be sought in vain from common arithme- tic. In algebra, otherwise called specious arithmetic^ num- bers are not expressed as in the common notation, but every quantity, whether given or required, is commonly represented by sqme letter of the alphabet; the given ones, for distinction sake, being usually denoted by the initial letters a, ^, r, d^ &c. ; and the unknown or required ones by the final letters 7/, w, x, ?/, &c. There are, moreover, in algebra, certain signs or notes made use of to show the relation and dependence of quantities one upon another, whose signification the learner ought, first of all, to be made acquainted with. The sign 4- signijies that the quantity to which it is prefixed is to be added. Thus a + b shows that the B 2 Of Notation. number re^presented by b is to be added to that repre^ sented by «, and expresses the sum of those numbers ;' so that if a was 5, and b 3, then would a + b ht 5 +3^ or 8. In like manner, a + b + c denotes the number arising by adding all the three numbers a, b^ and c toge- ther. Note. A quantity which has no prefixed sign (as the leading quantity a in the above examples) is always under- stood to have the sign -f- before it : so that a signifies the same as + ^ ; and a-^-b the same as +a + b. The sign — signifies that the quantity which it precedes is to be subtracted. Thus a — b shows that the quan- tity represented by b is to be subtracted from that repre- sented by a, and expresses the difference of a and bi so that if a was 5, and b 3, then would a — b be 5 — 3, or 2. In like manner, a + b — c — d represents the quantity which arises by taking the numbers c and d from the sum of the other two numbers a and b : as, if a was 7, <^ 6, c 5, and d 3, then would a + b — c — d be 7 + 6 — 5 — 3, or 5. The notes + and — are usually expressed by the words plus (or 7nore)^ and minus (or less). Thus, we read a + b^ a plus b \ and a — b^a minus b. Moreover, those quantities to which the sign + is pre- fixed are called positive (or ajfirmative) \ and those to which the sign — is prefixed, negative. The sign X signifies that the quantities betzveen which it stands are to be ynultiplied together. Thus axb denotes that the quantity a is to be multiplied by the quantity b^ and expresses the product of the quantities so multi- plied ; and ax b X c expresses the product arising by multiplying the quantities a, b^ and c, continually to- gether : thus, likewise, a-^-b X c denotes the product of the compound quantity a + b by the simple quantity c ; and a + b + c X a — b +c X a + c represents the product which arises by multiplying the three com- pound quantities a + b + c^ a — b + c^ and a + c con- tinually together ; so that if a was 5, b 4, and c 3, then OfNotatmu would a + /^+cX« — ^H-cx« + cbe 12x4x8, which is 384. But when quantities denoted by single letters are to be multiplied together, the sign x is generally- omitted, or only understood ; and so ab is made to signify the same as aXb \ and ahc the same 2& axh X c. It is likewise to be observed, that when a quantity is to be multiplied by itself, or raised to any power, the usual method of notation is to draw a line over the given quantity, and at the end thereof place the exponent of the power. Thus a + b\^ denotes the same as a -f- ^ X a -f ^, viz. the second power (or square) oi a + b con- sidered as one q uantity : thus, also, ab + be'] ^ denotes the same as ab + bc X cib -{-be x ab + bc^ viz. the third power (or cube) of the quantity ab + be. But, in expressing the powers of quantities repre- sented by single letters, the line over the top is com- monly omitted; and so a^ comes to signify the same as au or ax a, and b ^ the same as bbb or b X b X b : whence also it appears that a^ b^ will signify the same as aabbb ; and a^e^ the same as aaaaaee ; and so of others. The note . (or a full point), and the word into^ are likewise used instead of x, or as marks of multiplica- tion. Thus a + b . a -f- c and a-^b into a + e^ both signify the same thing as a + b x a + c^ namely, the product of a + bhy a + c. The sign -r- is used to signify that the quantity pre^ ceding it is to be divided by the quantity which comes after it : thus c -— ^ signifies that c is to be divided by b ; and a + b-rra — c, that a-^b is to be divided by a — c. Also the mark ) is sometimes used as a note of divi- sion ; thus, a + b^ ab^ denotes that the quantity ab is to be divided by the quantity a-^-b \ and so of others. jBut the division of algebraic quantities is most com- 4 Of Notation. monly expressed by writing down the divisor under the dividend with a line between them, in the manner of c a vulgar fraction. Thus ~-j- represents the quantity arising by dividing c by ^ ; and denotes the quan- tity arising by dividing a + bhy a — c. Quantities thus expressed are called algebraic fractions ; whereof the upper part is called the numerator, and the lower the de- nominator, as in vulgar fractions. The sign V is used to express the square root of any quantity to which it is prefixed: thus V25 signi- fies the square root of 25 (which is 5, because 5 X 5 is 2 5) : th us also \^cib denotes the square root of ab \ and W — Z— « denotes the square root of — , or of ^ a a the quantity which arises by dividing ab + be by d -, but — — ^t — (because the line which separates the d numerator from the denominator is drawn below V ) signifies that the square root of ab + be is to be first taken, and afterwards divided by d : so that if a was 2, 7 n A A ^ c^ 4\. ij ^'oT+bc , V36 6 . 6) 6, c 4, and a 9, then would ■ be or — ; ' _J d 9 9 but y 2— Z — is s^—-i or ^4, which is 2. The same mark -^z, with a figure over it, is also used to express the cube, or biquadratic root, &fc. of any quantity : thus v^64 represents the cube root of 64 (which is 4, because 4x4x4 is 64), and Vab + ed the cube root of ab -{- cd ; also Vl6 denotes the biquadratic root of 16 (which is 2, because 2 x 2 X 2 x 2 is 16); and Vab + cd denotes the biquadratic root of ab-^cd; and so of others. Quantities thus ex- pressed are called radical quantities, or surds ; where- Of Notation. 5 of those consisting of one term only, as Va and \^aby are called simple surds ; and those consisting of several terms, or members, as Va^ — b^ and Va^ — b^ + bc^ com- pound surds. Besides this way of expressing radical quantities, (which is chiefly followed) there are other methods made use of by different authors ; but the most com- modious of all, and best suited to practice, is that, where the root is designed by a vulgar fraction, placed at the end of a line drawn over the quantity given. Ac- cording to this notation, the square root is designed by the fraction |, the cube root by |, and the biquadratic root by l, &?c. Thus a p expresses the same thing with Va^ viz. the square root of a ; and a^ + ab'\z the same as Va^ + ab^ that is, the cube root of a^ ^ab: also a\i denotes the square of the cube root of a ; and <2-f z")* the seventh power of the biquadratic root of a-\-z', and so of others. But it is to be observed, that, when the root of a quantity represented by a single letter is to be expressed, the line over it may be ne- glected ; and so a^ will signify the same as a"]^, and b\ the same as ^ (3 or Vb. The number, or fraction, by which the power, or root of any quantity is thus designed is called its index, or exponent. The mark r^ {called the sign of equality) is used to signify that the quantities standing on each side of it are equal. Thus 2 -f 3 = 5, shows that 2 more 3 is equal to 5 ; and x=.a-^bj shows that x is equal to the difference of a and b. The note : : signifies that the quantities between which it stands are proportional : as, a : b : i c '. d^ denotes that a is in the same proportion to ^, as c is to ^; or that if a be twice, thrice, or four times, ^c. as great as b^ then accordingly c is twice, thrice, or four times, i^c. as great as d. t> Of Notation* To what has been thus far laid down on the signifi- cation of the signs and characters used in the algebraic notation, we may add what follows, which is equally ne- cessary to be understood. When any quantity is to be taken more than once, the number is to be prefixed, which shows how many times it is to be taken : thus 5a denotes that the quan- tity a is to be taken five times ; and Zhc stands for three times hc^ or the quantity which arises by multiplying he by 3 : also 7Va^ -f h^ signifies that >/a^ ■^h'^ is to be taken seven times ; and so of others. The numbers thus prefixed are called coefficients ^ and that quantity which stands without a coefficient is always understood to have a unit prefixed, or to be taken once, and no more. Those quantities are said to be like that are expressed' by the same letters under the same powers, or which differ only in their coefficients : thus 3^c, 5hc^ and ^hc are like quantities ; and the same is to be understood of the radicals 2v/ and *ts} . But unlike quan- ^ a ^ a tities are those which are expressed by different letters, or by the same letters tinder different powers : thus 2aby 2abc^ 5ab^^ and oba^ are all unlike* When a quantity is expressed by a single letter, or b\' s'everal single letters joined together in multiplication without any sign between them, as a, or 2ab^ it is called a simple quantity. But that quantity which consists of two or more such simple quantities, connected by the signs + or — , is called a compound quantity : thus a — 2ab -f- 5abc is a compound quantity ; whereof the simple quantities a, 2<3^, and 5abc are called the terms or members. The letters by which any simple quantity is expressed may be ranged according to any order at pleasure, and yet the signification continue the same : thus ab may be written ba ; for ab denotes the product of a by ^, and ba the product of ^ by a ; but it is well known, that when two numbers are to be multiplied together, it matters not which of them is made the multiplicand. Of Notation* t nor which the multiplier, the product, either way, coming out the same. In like manner it will appear that abc^ acb^ bac^ bca^ cab^ and cba^ all express the same thing, and may be used indifferently for each other, as will l3e demonstrated further on ; but it will be some- times found convenient, in long operations, to place the several letters according to the order which they have in the alphabet. Likewise the several members or terms of which any quantity is composed may be disposed according to any order at pleasure, and yet the signification be nowise affected thereby. Thus a — 2ab -j- Sa^b may be written a + 5a^b — 2ab, or — 2ab + a + 5a^b, kc. ; for all these represent the same thing, viz. the quantity which remains, when, from the sum oi a and 5a^b^ the quantity 2ab is de- ducted. Here follow some examples wherein the several forms of notation hitherto explained are promiscuously con- cerned, and where the signification of each is expressed in numbers. Suppose a = 6, ^ = 5, and c = 4 ; then will a2 +sab — c2 = 36 + 90 — 16 = 110, 2a^ — 3a^ b +c^ = 432 ~ 540 + 64 = — 44, ^2 xa + b — 2abc=S6x H —240= 156, a^ 216 + c2 = + 16 = 12 + 16 = 28, V2ac + c^(or 2ac + c^ H) = V64 = 8 (for 8 X 8 = 64), 2hc ^ 40 a2 ^S/l)2 ^^^ _ 36 — 1 ___ 35 _ 2 a'--\/b^ +ac ""l2 — 7^T"" ' V^ 2 ^ac^ x/2ac + c^ = 1+8 = 9, ^b^ — ac + V2ac + c2* = V25 — 24 + 8 = 3. This method of explaining the signification of quan- tities I have found to be of good use to young begin- ners ; and would recommend it to such as are desirous of making proficiency in the subject, to get a clear idea 8 Of Addition. of what has been thus far delivered, before they proceed farther. SECTION IL Of Addition. ADDITION, in algebra, is performed by connecting the quantities by their proper signs, and joining into one sum such as can be united : for the more ready effecting of which, observe the following rules. 1°. If in the quantities to be added^ there be terms that are like^ and have all the same sig^n^ add the coefficients of those terms together^ and to their sum adjoin the letters common to each term^ prefixing the common sign. Thus Sa And 5a + 7b Also 5a — 7b added to 3a added to 7a -f- 3^ added to 7a — 3<^ makes Sa.^makes 12a + 10^. makes 12a — 10b. Hence {''^Vab + 7Vbc * , ^, f^^ Sd I ^ , — ,— And the j — — — likewise <^Wab+ 2Vb^ sum of 1 ^ ^ the sum of ^\S>Vab -f WTc l^A _ ^ will be 11 Vab+ 1 8 VTc a .„ , 7b lOd will be — — — The reasons on which the preceding operation^ are grounded will readily appear by reflecting a little on the nature and signification of the quantities to be added : for, with regard to the first example, where Sa is to be added to 5a, it is plain that three times any quantity whatever, added to five times the same quantity, must make eight times that quantity : therefore 3a, or thi*ee times the quantity denoted by a, being added to 5a^ or five times the same quantity, the sum must consequently Of Addition. 9 2"*. JV/ieTij 271 the quantities to be added^ there are like terms ^ whereof some are affrmative and others negative^ add together the affirmative terms (if there be more than one^^ and do the same by the negative ones; then take the difference of the two sums {not regarding the signs) by subtracting the coefficient of the lesser from that of the greater^ and adjoin the letters common to each; to which difference prefix the sign of the greater* Examples of this rule may be as follows. I2a — 5b 2. — Sab +5 be — Sa +2b ^7ab — 9ac Sum 9a — 3^ Sum 4«Z> — 4^bc 6ab + 12bc — Scd 4. 5Vab — 7Vbc + Sd — Tab— 9bc+ 3cd SVab + SVbc — 12d — 2ab^ 5bc+ 12cd 7\'ab 4- 3v/;c 4- 9d Sum — 2,ab — 2bc + 7cd Sum 15V ab + 4V/^c + 5d 5. 1 2abc — 1 ^abd + 25acd— 72bcd 1 ^abc + 1 2abd -f 20acd'^ 1 Sbcd — 1 Sabc — 26abd— 1 5acd +12bcd 82abc+ ISabd — 1 Oacd — 1 6bcd Sum 47abc — 12abd -f- 20acd — 94:bcd make 8a, or eight times that quantity. From whence, as the sum of any two quantities is equal to the sum of all their parts, the reason of the second case, or example, is likewise obvious. But as to the third, where the given quantities are 5a — 7b and 7a — Sbj we are to consider, that, if the two quantities to be added together had been exactly 5a and 7a (which are the two leading terms), the sum would then have been just 12a; but, since the for- mer quantity wants 7b of 5a, and the latter Sb of 7a^ their sum must, it is evident, want both 7b and 3^ of 12a ; and therefore be equal to 12a — 106, that is, equal to what remains, when the sum of the defects is deducted. And, by the very same way of arguing, it is easy to con- ceive that the sum which arises by adding any number 10 Of Addition. 6. oa '^cc ^ i^^^q \^E^^ b a ^ a ^ a ha ^ a ^ a ^ 13a . 4cc ^ fbc ^ lab A- Sum -r- + S\} 3\/ — JL h n ^ n ^ n CC a tn the last example, and all others where fractional and radical quantities are concerned, every such Quan- tity, exclusive of its coefficient, is to be treated in all respects like a simple quantity expressed by a singlc letter. 3°. When^ in the quantities to be added^ there are terms without others like to them^ write them down with their proper signs* Thus a + 2<^ And aa-^-bh added to 3c -f- 3 + ^3 _ ^. 3 ^20 — bi:7 SECTION III. Of Subtraction. SUBTRACTION, in algebra, is performed by change ing all the signs of the subtrahend {or conceiving them to be changed), and then connecting the quantities, as in addi- tion. Ex. 1. From 8« + 5b Ex. 2. From 85 + 5b take 5a + ^b take 5a — Zb Rem. 3a + 2b. Rem. ':ia + 8b. Ex. 3. From 8a — 5b Ex. 4. From Sa — 5b take 5a -f 3/? take 5a — 3b Rem. 3a— 8^. Rem. 3a — 2b. In the second example, conceiving the signs of the subtrahend to be changed |o their contrary, that of 3b becomes -f ; and so the signs of 3b and 5b being alike, the coefficients 3 and 5 are to be added together, by case 1 of addition. The same thing happens in the third example ; since the sign of 3b, when changed, is — , and therefore the same with that of 5b. But, in the fourth example, the signs o£ 3b and 5b, after that of 3b is chang- ed, being unlike, the difference of the coefficients must be taken, conformable to case 2 in addition. 12 Of Subtraction, Other examples in subtraction may be as follows : From 20ax + 5bc — 7aa From 7Vax + 9>/b y take I'^ax — Zhc — Saa take — 5Vflx + 12v'^i!/ Rem. ^ax + %bc — %aa. Rem. 12\ «^— Z*^Vy. From 6 V<2« — ;cx + loVo^ — :v^ — 7v 22 take 9\/«<:/ — xX'\-\ Sy/a^ — x"^ — ^si— ^ c ^ „ s ,-- faa Rem. — SVaa — xx +25Va^ ~x^ + 2>J— From Ta^^^-^eJ^ + d c ^ c 1 c> . Sa fax , , take a^ -\ V — + ^ Rem. 6a^>^ — + 7sr^ + d—b. c ^ c In this last example, the quantity cf in the subtrahend being without a coefficient, a unit is to be understood ; for la^ and a^ mean the same thing. The like is to be ob- served in all other similar cases. The grounds of the general rule for the subtraction of algebraic quantities may be explained thus: let it be here required to subtract 5a — Sb from Sa + 5b (as in ex. 2). It is plain, in the first place, that if the affirma- tive part 5a were alone to be subtracted, the remainder would then be 8a + 5b <-^ 5a; but, as the quantity actu- ally proposed to be subtracted is less than 5«, by 3^, too much has been taken away by 3^ ; and therefore the true remainder will be greater than 8a + 5b — 5«, by Sb ; and so will be truly expressed by 8a + 5b — 5a-j-Sb: where- in the signs of the two last terms are both contrary to what they were given in the subtrahend ; and where the whole, by uniting the like terms, is reduced to 3a -{- 8b^ as in the example. Of Multiplication. f^ SECTION IV. Of Multiplication* BEFORE I proceed to lay down the necessary rules for multiplying quantities one by another, it may be pro- per to premise the following particulars, in order to give the learner a clear idea of the reason and certainty of such rules. Firsts then^ it is to he observed^ that when several quantities are to be multiplied continually together^ the re- sult^ or product^ will come out exactly the same^ midtiply them according to what order you will. Thus a X b x (\ a X c X b^ b X c X a^ &fc. have all the same value, and may be used indifferently : to illustrate which we maj'^ suppose c = 2, ^ = 3, and c = 4; then will a X b x c-=z 2X3X4 = 24; axcx^ = 2x4x3 = 24; and bxc X<^ = 3X4X2 = 24. Secondly. If any number of quantities be multiplied continually together^ and any other number of quantities he also multiplied continually together^ and then the two products one into the other ^ the quantity thence arising will be equal to the quantity that arises by multiplying all the proposed quantities continually together. Thus will abc X de z= a X b X c X d X e \ so that, if a were = 2, ^ = 3, c = 4, d=5^e=z6^ then would abc X de = 24x S0= 720, and flX<^Xcx^Xe = 2x3x4x5x6 = 720. The ge- neral demonstration of these observations is given below in the notes. The following demonstrations depend on this prin- ciple, that if two quantities^ whereof the one is n times as great as the other (n being any number at pleasure)^ be multiplied by one and the same quantity,, the product,, in the one case,, will also be n times as great as in the other. The greater quantity may be conceived to be divided into n parts, equal, each, to the lesser quantity; and the product of each part (by the given multiplier) will 14 OJ Multiplication. The multiplication of algebraic quantities may be con- sidered in the seven following cases. be equal to that of the said lesser quantity ; therefore the sum of the products of all the parts which make up the whole greater product, must necessarily be n times as great as the lesser product, or the product of one single part, alone. This being premised, it will readily appear, in the first place, that b x ci and a x b are equal to each other : for, b X CL being b times as great as 1 x « {because the niultiplicajid is b times as great^ it must therefore be equal to 1 X ci (or a) repeated b times, that is, equal to ax b^ by the definition of multiplication* In the same manner, the equality of all the variations, or products, abc^ bac^ acb^ cab^ bca^ cba (where the num- ber of factors is 3) may be inferred : for those that have the last factors the same (xvhich I call of the same class^ are manifestly equal, being produced of equal quantities multiplied by the same quantity : and, to be satisfied that those of different classes^ as abc and acb^ are like- wise equal, we need only consider, that, since ac x b is c times as great as « X <^ (because the multiplicand is c times as great) it must therefore be equal Xo aX b taken c times, that is, equal to a x ^ X c, by the definition of mid- tiplication. Universally. If' all the products, when the number of factors is 7z, be equal, all the products, when the number of factors is ?2 + 1, will likewise be equal : for those of the same class are equal, being produced of equal quantities multiplied by the same quantity: and to show that those of different classes are equal also, we need only take two products which differ in their tv/o last factors, and have all the preceding ones acGording to the same order, and prove them to be equal. These two factors we will suppose to be repre- sented by r and 5, and the product of all the preceding ones by /; ; then the two products themselves will be represented by prs and psr^ which are eqTial, by case 2. Of Multiplication. 15 1*. Simple quantities are multiplied together by uiulti- plying the coefficients one into the other^ and to the pro- duct aniiexing the quantity which^ according to the ynt- thod of notation^ expresses the product of the species; pre- fixing the sign + or — , according as the signs of the given quantities are like or unlike* Thus 2a Also %ah And 11^^ mult, by 2>h mult, by Sc mult, by 7ah makes ^ah. makes 30abc. makes 77aabdf Thus, by way of illustration, abcde will appear to be = abced^ &c. For, the former of these being equal to every other product of the class, or termination e (by hypothesis and equal multiplication), and the latter equal to every other product of the class, or termination d ; it is evident, therefore, that all the products of different classes, as well as of the same class, are mutually equal to each other. So far relates to the first general observation : it re- mains to prove that abed x pqrst is =a X b X c X d X p X q X r X s X t. In order to which, let abed be denoted by :> ;, then will abed x pqrst be denoted by :v X pqrst ^ or pqrst X X (by case l), that is, hy pxqXrXsXtXx-, which 1!^ equal to x X p X q X r X s X ty or a xbxcxdxpXq X ^ X s X tj by the preceding demonstratioiu The reason of rule 1 depends on tjiese two general observations : for it is evident from hence, that 2a x ^b (in the first example) is =2x«XoX^ = 2x 3X«X^=6xaX^ = ^ab : and, in the same manner, 11 adf x Tab (in the third example) appeals to be =: 11 X « X dxj X 7 X a X b = 11 x 7 X a x axbxdxf^77x aabdf = 77 aabdf But the grounds of the method of proceeding may be other- wise explained, thus; it ha3 been observed that ab (according to the method of notation) defines the pro- duct of the species a, b {in the first example), therefore the product of a by 3^, which must be three times as great (because the multiplier is here three times as great^. 10 &f Multiplication. In the preceding examples^ all the products are q^r- mative^ the quantities given to be muhiplied being so ; but, in those that follow, some are affirmative^ and others negative^ according to the different cases specified in the fetter part of the rule ; whereof the reasons will be ex- plained hereafter. Mult. -f 5a Mult. — 5a Mult. — Sa by — e^b by 4.. p^b by ^ 6b Prod. — ^Oab. Prod. -^ ^oab. Prod. + 30abl Mult. -f- 7'v ax Mult. — 7a'^ au f xx ^y — 5\ cy by — 6b\ ac — vy Prod. — 35 X ^'ax x Vcz/. Prod.-f-42aZ>x V aa^f :>f^xV«a-z/y In the two last examples, and all others where radi- cal quantities are concerned, every such quantity may be considered, and treated in all respects as a simple quantity, expressed by a single letter ; since it is not the form of the expres3ion, but the value of the quantity that is here regarded. 2°. A fraction is multiplied by multiplying the nume- rator thereof by the given multiplier^ and making the pro- duct a numerator to the given denominator. rr» a , ac . Sac ^ , , 6aacd ihus -7- X c makes -7- ; also --- x 2ad makes ; b b b b will be truly defined by 3<2^, or ab taken three times : bu^ since the product of a by 2>b appears to be Zab^ it is plain that the product of 2rt by Zb must be twice as great as that of a by 3^, and therefore will be truly expressed by 6a^. Thus, also, the product of the species ah and c, in the second example, being abc by bare notation, it is evi^ dent that the product of 6ab by c will be truly defined by 6abc^ or acb six times taken, and consequently the product of 6ab and 5c ^ by SOabc^ or 6abc taken five times, the mul- tiplier here being five times as great. The reason of rule 2° may be thus demonstrated : let the numeratgr of any proposed fraction be denoted by A» Of Multiplication. 17 likewise -^ X 7Vax makes ^ ; lastly, ^ X/2ab makes Vaa + XX 3°. Fractions are multiplied i?ito 07ie another by multi- plying the numerators together for a nexu numerator^ and the denominators together for a nexv denominator. rjy. a c ac 2ab Sad 10a%d 21xy SaVa __ eSaxyV g — 5aVx — 2a _ Vaz S^ 8bV^ ' Sbc "" b ^ lOaW^ . and ^^^^ 5bVaa + xx _ 3bbc ' v^ a -{-"z "^ 15ab X ^xy X *>/aa 4. xx a+zx S/ab the denominator by B, and the given multiplicator by C : then, I say, that -— - is equal to -- x C. For, since -— denotes the quantity which arises by dividing AC by B, and ~ the quantity which arises by dividing A by B, it is evident that the former of these two quantities must be C times as great as the latter, because the dividual is C times as great iu the one case as in the other ; and there- fore must be equal to the latter C times taken, that is, AC A -J- must be equal to -^ X C, as was to be shown. JtS B The reason of rule 3° will appear evident from the preceding demonstration of rule 2"^. For, it be- A AC ing there proved that -- x C is equal to ----, it is ob- t%. B B A C AC vious that ~ x r— can be only the D part of — ; be- D ,18 Of Multiplication. 4°. Surd quantities under the same radical sign are mul- tiplied like rational quantities^ only the product must stand under the same radical sign. Thus, VF X VF = V35 \^Va X J^b = Vab ; \/7fc X i^5ad z=l25abcd; Wab X 5Vc = \S\^abc ; 2aV2cy X 3b V 5 ax (z=: eab X V2cy X S^Sax) ^ J , , "Tab Isx ^ 5c JlSd = eabVlOacxy ; and V — X -% V— r = ^ 5x ^ 3a 9d^ 2b 35abc hoAdx 45dx ^ 6ab C cause --, the multiplier here, is but the D part of the AC former multiplier C : but r^^ is also equal to the D AC part of the same ~— - ; because its divisor is D times B AC as great as that of -— - : therefore these two quantities, 13 A C AC -- X -rr and -—--, being the same part of one and the sam^ r> D BD quantity, must necessarily be equal to each other ; which zvas to be proved. ^ As to rule 4° for the multiplication of similar ra- dical quantities, it may be explained thus : suppose VA and VB to represent the two given quantities to be multiplied together ; let the former of them be de- noted by a^ and the latter by ^, that is, let the quan- tities represented by a and b be such, that aa may be = A, and ^/^ = B ; then the product of VA by VB^ or of a by ^, will be expressed by ab^ and its square by ab X abi but ab x ab i^ -=1 a X b X a X b -= aa X bb (by the general pbservations premised at the beginning of this section) ; whence the square of the product is like- wise truly expressed by aa X bb^ or its equal A X B ; and consequently the product itself by VA X B, that is, by the quantity which, being multiplied into itself, produces A X B. Of Multiplication. 1 9 4^°. Powers^ or roots ^ of the same quantity are 77iultiplied together by adding their exponents: but the exponents here understood are those defined in p. 5, where roots are represented as fractional powers. Thus, x^ X ^Ms = ^* ; a + zf X a + z^ = a + z]' i x'^ X x^ ^ X = a:^ ; and x^ X x =: x'^ z=: x ; also aa -f 22*]^ X «a -f zz*]^ is = aa 4- zz^ ^ = aa + 22 ; and In the same manner, the product of >/A X V'B will appear to be VAB : for if VA be denoted by cr, and \/B by b ; or, which is the same thing, if aaa = A, and W^ = B ; then will VA X v'B^= a x <^ (or ab) and its cube z=z ab X ab X ab = aaa X bbb = AB (by the aforesaid observations) ; whence the product itself will evidently be expressed by VAB. ^ The grounds of these operations may be thus explained. First, when the exponents are whole num- bers, as in example 1, the demonstration is obvious, from the general observations premised at the begin- ning of the section: for, by what is there shown, x^ X x^y or XX X XXX is = xX^XxX^X^^^^ (by notation). But in the last example, where the expo- nents are fractions, let c + z/"j^ be represented by x j that is, let the quantity x be such, that :v X ^ X -v x a: X ^ X ^, or ^^ may be equal to c +y; so shall c -f z/ 1 be expressed by :i^^ ; because, by what has been already shown, x^ X x^ is = x^ : and in the same manner will c + 2/"] 3 be expressed by x^ ; because x^ x x^ X ^ is likewise = x^. Therefore c + y"] X c -f- z/"]^ is = x^ X x^ = x^ = the fifth power of c -|- t/]^^ which is c + y'\^^ by notation. 20 Of Multiplication. 6°. A compound quantity is multiplied by a simple 07ie^ by multiplying every term of the multiplicand by the mul- tiplier. Thus a + 2b"^Zc Also a2 — 5aVx'\-7b mult, by 3(2 mult, by 8c makes Sa^-f-Grt^ — 9ac; makes 8a^c — 40acV x '\- 5^bc ; And 5a^ ^-^8ab + 6aC'—7bc+12b^ — 9c^ ~ mult, by Sabc makes 15a^bc — 24a^b^C'\'18a^bc^ — 21 ab^c^+36ab^C'''27abc^. To explain the reason of the two last rules, let it be first proposed to multiply any compound quantity, as a + b — c — dy by any simple quantity f; and, I say, the product will be ctf + bf —- cf — df For the pro- duct of the affirmative terms, a + b^ will he of + bf because to multiply one quantity by another is to take the multiplicand as many times as there are units in the multiplier, and to take the whole multiplicand (a + /;) any number of times (f)^ is the same as to take all its parts (a, b) the same number of times, and add them together. Moreover, seeing a + b — c ^-^ d denotes the excess of the affirmative terms (a and b) above the negative ones (c and d)y therefore, to multiply a + b — c — d by f is only to take the said excess f times ; but f times the excess of any quantity above another is, manifestly, equal to f times the former quantity, minus f times the latter ; but f times the former is, here, equal to qf + bf (by what has been already shown), and/times the latter, for the same reason, will be equal to cf + df and therefore the product oi a + b — c — d by f is equal to af + bf — cf — df; as was to be proved. Hence it appears, that a compound quantity is multiplied by a simple affirmative quantity, by multi- plying every term of the former by the latter, and con- necting the terms thence arising with the signs of the mul- tiplicand. Of Multiplicatmu 21 7°, Compound quaiitities are multiplied into one another by multiplying every term of the multiplicand by each term of the multiplier^ successively^ and connecting- the several products thus arising with the signs of the multiplicand^ if the multiplying term be afirmative^ but with contrary signs ^ if negative. Thus the product of 5a + Sx multiplied by Sa + 2x .„ , r 13aa -f 9ax 1 ^^^^1^^ 1 +10ax + 6xx S which contracted by unit- f i^aa + 19ax + 6xx. ing the like terms, is (. But to prove that the method also holds good when both the quantities are compound ones, let it be now pro- posed to multiply A — B by C — D ; then, I say, the pro- duct will be truly expressed by AC — BC — AD+ BD, For, it has been already observed, that to multiply one quantity by another is to take the multiplicand as many times as there are units in the multiplier ; and, therefore, to mutiply A — B by C — D is only to take A — B as many times as there are units in C — D : Now (according to the method of multiplying com- pound quantities) I first take A — B, C times (or multiply by C), and the quantity thence arising will be AC — BC, by what is demonstrated above* But I was to have taken A — B only C — D times ; therefore, by this first operation, I have taken it D times too much ; whence, to have the true product, I ought to deduct D times A — B from AC — BC, the quantity thus found ; but D times A — B, by what is already proved^ is equal to AD — BD ; which subtracted from AC — BC, or written down with its signs changed, gives the true product, AC — BC — AD-f-BD, as was to be demonstrated. And, universally^ if the sign of any proposed term of the multiplier, in any case what- ever, be affirmative, it is easy to conceive that the re- quired product will be greater than it would be if there Of Multiplication*, Likewise the product of c^ + a^h + alP' + ly^ by a^h fa^+a^l? + a^^+aP \ ^s \ ^a^b — aW — ab^^b^S Which, by striking out the terms that destroy one another, becomes a'* — b^* were no such term, by the product of that term into the whole multiplicand -, and therefore it is, that this product is to be added, or written down with its proper signs, which are proved above to be those of the multi- plicand. But if, on the contrary, the sign of the term by which you multiply be negative ; then, as the required product must be less than it would be if there were no such term, by the product of that term into the whole multiplicand, this product, it is manifest, ought to be subtracted, or written down with contrary signs. Hence is derived the common rule, that like sig?is pro- duce +, and unlike signs^ — . For, first, if the signs of both the quantities, or terms, to be multiplied be affirmative, and therefore like^ it is plain that the sign of the product must likewise be affir- mative. Secondly, also, if the signs of both quantities be nega- tive, and therefore still like^ that of the product will be affirmative, because contrary to that of the multiplicand^ by rvhat has been just now proved. Thirdly, but if the sign of the multiplicand be affirma- tive, and that of the multiplier negative, and therefore unlike^ the sign of the product will be negative, because co7itrary to that of the multiplicand. Lastly, if the sign of the multiplicand be negative, and that of the multiplier affirmative, and therefore still unlike^ the sign of the product will be negative, because the same zvith that of the multiplicand. And these four are all the cases that can possibly hap'- pen with regard to the variation of signs. Of Multiplicatio7i. 2Z Other examples in multiplication, for the learner's ex- ercise, may be as follows ; from which he may, if he please, proceed directly to division, by passing over the interven- ing scholium. 1. Multiply by product Multiply by product Multiply by product Multiply by product x^+xy +y^ x^ — xy +y^ x^+x'y + x^y^ — x^y — x^y^ — xy^ + x^y^ + xy^ + y^ 2. 2^2 — Sax + 4x^ 5a^ — ^ax — 2x^ — 4a'x^ + 6ax^ — 8;c^ 3. lOa'^ — ^ra^x + 34aV — \Mx^ — S^v^ 2a — 2b + 2c 2a — Ab + 5c 6aa — 4ab -f- 4ac — 12ab+ Sbb— Sbc + 15ac — lObc + lOcc 4. 6aa — Idab -f- 19ac -f- Sbb--18bc + lOcc. o? — Sd^b+ Sab^^b^ a^ — 2ab + b^ aS — Sa^b+ 3a^^~ aH^ — 2a*^+ ^aH^— 6a^'- + 2ab^ + ^3^2 _ 3^2/,3 ^ 3^^4 _ 1,5 a^ — . sa^'b + lOa^b^ — lOd'b^ + Sab"" — b\ SCHOLIUM. The manner of proceeding, in referring the reasons of the different cases of the signs to the multiplication of compound quantities, may perhaps be looked upon as indirect, and contrary to good method ; according to which, it may be thought tibat these reasons ought to 24 Of Multiplication^ have been given before, along with the rules for simple quantities, as this is the way that almost all authors on the subject have followed. But, however indirect the method here pursued may seem, it appears to me the most clear and rational ; and I believe it will be found very difficult, if not impossible, without explaining the rules for compound quantities first, to give a learner a distinct idea how the product of two simple quantities with negative signs, such as — h and — c, ought to be expressed, when they stand alone, independent of all other quantities : and I can- not help thinking farther, that the difficulties about the signs, so generally complained of by beginners, have been much more owing to the manner of explaining them, this way, than to any real intricacy in the sub- ject itself; nor will this opinion, perhaps, appear ill grounded, if it be considered that both — a and — ^, as they stand here independently, are as much im- possible in one sense, as the imaginary surd quantities V — b and s/ — c ; since the sign — , according to the established rules of notation, shows that the quan- tity to which it is prefixed is to be subtracted ; but, to subtract something from nothing is impossible, and the notion or supposition of a quantity less than nothing, absurd and shocking to the imagination : and, cer- tainly, if the matter be viewed in this light, it would be very ridiculous to pretend to prove, by any show of reasoning, what the product of — h hy — c, or of v' — b by S/ — c, must be, when we can have no idea of the value of the quantities to be multiplied. If, indeed, v/e were to look upon — h and — c as real quantities, the same as represented to the mind by b and c (which cannot be done consistently, in pure alge- bra, where magnitude only is regarded), we might then attempt to explain the matter in the same manner that some others have done ; from the consideration, that, as the sign — is opposite in its nature to the sign +, it ought therefore to liave, in all operations, an oppo- site effect ; and, conseqn.ently, that as the product, when Of Multiplication. ' 25 the sign + is prefixed to the multiplier, is to be added, so, on the contrary, the product, when the sign — is prefixed, ought to be subtracted. But this way of arguing, however reasonable it may appear, seems to carry but very little of science in it, and to fall gready short of the evidence and conviction of a demonstration : nay, it even clashes widi first principles, and the more established rules of notation ; according to which the signs + and — are relative only to the magnitudes of quantities, as composed of diffe- rent terms or members, and not to any future opera- tions to be performed by them : besides, when we are told that the product arising from a negative multiplier is to be subtracted, we are not told what it is to be sub- tracted from ; nor is there any thing from whence it can be subtracted, when negative quantities are independent- ly considered. And, farther, to reason about opposite effects, and recur to sensible objects and popular consi- derations, such as debtor and creditor, tPc. in order to demonstrate the principles of a science whose object is abstract number, appears to me not well suited to the nature of science, and to derogate from the dignity of the subject. It must be allowed, that, in the application of alge- bra to different branches of mixed mathematics, where the consideration of opposite qualities, effects, or posi- tions can have place, the usual methods have a better foundation ; and the conception of a quantity abso- lutely negative becomes less difficult. Thus, for ex- ample, a line may be conceived to be produced out, both ways, from any point assigned ; and the part on the one side of that point being taken as positive^ the other will be negative. But the case is not the same in abstract number, whereof the beginning is fixed in the nature of things, from whence we can proceed only one way. There can, therefore, be no such things as nega- tive numbers, or quantities absolutely negative, in pure algebra, whose object is number, and where every muhiplication, division, ^c. is a multiplication, divi- E 26 Of Multiplication. sion, &fc. of numbers, even in the application thereof; for, when we reason upon the quantities themselveSj and not upon the numbers expressing the measures of them, the process becomes purely geoinetrical^ whatever symbols may be used therein, from the algebraic notation ; which can be of no other use here than to abbreviate the work. However, after all, it may be necessary to show upon what kind of evidence the multiplication of negative and imaginary quantities is grounded, as these some- times occur, in the resolution of problems : in order to which it will be requisite to observe, that, as all our reasoning regards real^ positive quantities, so the alge- braic expressions, whereby such quantities are exhi- bited, nuist likewise be real and positive. But, when the problem is brought to an equation, the ease may indeed be otherwise ; for, in ordering the equation, so much may be taken away from both sides thereof, as to leave the remaining quantities negative ; and then it is, chiefly, that the multiplication by quantities absolutely negative takes place. Thus, if there were given the equation a = r, in order to find x ; then, by subtracting the quantity a from each side thereof, we shall have — -- = c — a; b which multiplied by — ^, according to the general rule^ gives X :=z — cb + ab ; that is, — -- by — b will give + X ; c by — b, — cb ; and — a by — b, -j- ab ; which appear to be true ; because the products being thus ex- pressed, the same conclusion is derived, as if both sides of the original equation had been first increased by c, and then multiplied by b ; where both the mul- b tiplier and multiplicand are real, affirmative quantities, and where the whole operation is, therefore, capable of a clear and strict demonstration: but then it is not in consequence of any reasoning I am capable of forming Of Multiplication. 27 about — - and •— b^ or about + c and — b^ considered b independently, that I can be certain that their product ought to be expressed in that manner. So, likewise, if there were given the equation a — — = c ; by transposing a, and taking the square root b _ f • ^2 on both sides, we shall have y — ^ = Vc — a; and this multiplied by V — by will give V x^ (or x) =; V — cb + ab: which also appears to be true, because the result, this way, comes out exactly the same as if the operations for finding x had been performed alto- gether by real quantities : but, notwithstanding this, it is not from any reasoning that I can form about the j ^» multiplication of the imaginary quantities y f and V — by &c. considered independently, that I can prove their product ought to be so expressed ; for it would be very absurd to pretend to demonstrate what the pro- duct of two expressions must be, which are impossible in themselves, and of whose values we can form no idea. It indeed seems reasonable, that the known rules for the signs, as they are proved to hold good in all cases what- ever, where it is possible to form a demonstration, should also answer here : but the strongest evidence we can have of the truth and certainty of conclusions derived by means of negative and imaginary quantities, is the exact and con- stant agreement of such conclusions with those determin- ed from more demonstrable methods, wherein no such quantities have place. In the foregoing considerations, the negative quali- ties — by — c, &c. have been represented, in some cases, as a kind of imaginary or impossible quantities ; it may not, therefore, be improper to remark here, that such imaginary quantities serve, many times, as much to discover the impossibility of a problem, as imaginar}- 28 Of Division* surd quantities : for it is plain, that, in all questions re- lating to abstract numbers, or such wherein magnitude only is regarded, and where no consideration of posi- tion, or contrary values, can have place ; I say, in all such cases, it is plain that the solution will be altogether as impossible, when the conclusion comes out a negative quantity, as if it were actually affected with an imagi- nary surd ; since, in the one case, it is required that a number should be actually less than nothing ; and, in the other, that the double-rectangle of two numbers should be greater than the sum of their squares ; both which are equally impossible: but, as an instance of the impossibility of some sort of questions, when the conclusion comes out negative, let there be given, in a right-angled triangle, the sum of the hypothenuse and perpendicular = ^, and the base = h^ to find the perpendicular ; then, by what shall hereafter be shown in its proper place, the answer will come out , and is possible, or impossible, according as the quantity — is affirmative or negative, or as a is greater or less than h ; which will manifestly appear from a bare contemplation of the problem : and the same thing might be instanced in a variety of other examples. SECTION V. Of Division. DIVISION in species, as in numbers, is the converse of multiplication, and is comprehended in the seven fol- lowing cases. 1°. JVhen one simple quantity is to be divided by a?i- other^ and all the factors of the divisor are also found iii the dividend^ let those factors be all cast off or expunged^ and then the remaining factors of the dividend^ joined together^ will express the quotient sought. But it is to be observed, OfDivisioJU 29 that, both here and in the succeeding cases, the same rule is to be regarded, in relation to the signs, as in multipli- cation, 'Diz. that like signs give +, and unlike^ — . It may also be proper to observe, that, when any quantity is to be divided by itself, or an equal quantity, the quotient will be expressed by a unit, or 1. Thus ^ -4- « gives 1 ; and 2ab -—- 2ab gives 1 ; moreover, Sabcd-r- ac gives Sbd ; and 16bc -^ 8b gives 2(? : for here the dividend, by resolving its coefficient into two factors, becomes 2x8 X b X c ; from whence casting off 8 and ^, those common to the divisor, we have 2 x c, or 2c. In the same manner, by resolving or dividing the coefficient of the dividend by that of the divisor, the quotient will be had in other cases : Thus, 20abc^ divided by 4c, gives Sab ; and — Slab \^xif X ^xx -f i/if^ divided by — 1 TaVxi/j gives + Sb \/xx + yy» 2°. But if all the factors of the divisor be not found in the dividend^ cast off those {if any such there be) that are common to both^ and write down the reinaining factors of the divisor^ joined together^ as a deno7ninator to those of the dividend ; so shall the fraction thus arising express the quotient sought. But if, by proceeding thus, all the factors in the dividend should happen to go off, or va- nish, then a unit will be the numerator of the fraction re- quired. Thus, abCy divided by bcd^ gives -: ^ax And 16a^bx^^ divided by Sabcx^^ gives ^ — : The first rule, given above, being exactly the con* verse of rule 1° m the preceding section, requires no .other demonstration than is there given. The second >'ule (as well as those that follow hereafter upon frac- tions) depends on this principle, that, as many times as any one proposed quantity is contained in another, just so many times is the half, third, fourth, or any other assigned part of the former, contained in the half, third, fourth, or other corresponding part of the latter ; and 30 Of Division. Likewise, QTabVxy^ divided by 9aWxy^ gives £- : a And SabVai/^ divided by 16a^bVa^^ gives — .. 2a 3°. One fraction is divided by another ^ by multiplying the denominator of the divisor into the numerator of the dividend for a new numerator^ and the numerator of the divisor into the denominator of the dividend for a new de- nominator. 1 hus, y, divided by -, gives ~~ : d be * , 5ax ,. . 1 1 , ^bc . 35adx Also,— , divided by—, gives ^^: . , 6a^b ,. . , , , 5ab^ . ISa^bx And , divided by , ogives —-. 5x' ^ 2x'^ 25ab^x just so niany times, likewise, is the double, triple, qua- druple, or any other assigned multiple of the former, con- tained in the double, triple, quadruple, or other corre- sponding multiple of the latter. The demonstration of this principle (though it may be thought too obvious to need one) may be thus : let A and B represent any two proposed quantities, and AC and BC their equimul- tiples (or, let AC and BC be the two quantities, and A AC A and B their like parts) : I say, then, that —77 = -^ : BC B AC for the multiple of-——- by BC is manifestly = AC; BC A A and -~- X BC, the multiple of ~ by the same BC, is = o (^!/ ^^^^^ ^ ^^^ tnultiplication^j = — rj— (yid, p. 14 and 15) = AC : therefore, seeing the equimultiples of the two proposed quantities are the same, the quantities them- selves must necessarily be equal. The second rule, given above, is nothing more than a bare application of the principle here demonstrated : Of Division. 31 But, in cases like this last, where the two numerators, or the denominators, have factors common to both, the conclusion will become more neat by first casting oif such common factors. Thus casting away ab out of the two numerators, and X out of both the denominators, we have — to be divided by — ; whereof the quotient is — - : in the 12ac^ 4acx Sc^ x . Q>c^d same manner, — tt -^ Tr-, or —-. -4 — ., gives — --; ' lObb 5bd 2b d ^ 2bx J QiaUxy 7a^\/xy Q ^ 7a . 12b and — -i-— ^ -4 ^V-^9 or — r- — -, crives — . 5c lObc ' 1 2b' ^ 7a When either the divisor or the dividend is a whole quantity^ instead of a fraction, it may be reduced to the form of a fraction, by writing a unit or 1 under it. since, by casting off the factors common to the dividend and divisor (as directed in the rule), it is plain that we take like parts of those quantities : therefore the quotient arising by dividing the one part by the other will be the same as that arising by dividing one whole by the other. As to rule 3°, wherein it is asserted that -— -^ — - = rrjrr^ it is evident that AD and BC are equimultiples of the AC A . given quantities -- and -- ; because -- X BD is (by ride 2"* in rmdtiplication) = — jt— = AD, and -- X BD = crd = CB : whence it follows that the quotient of A C -. divided by -- will be the same with that of AD di- B D AD videdL by BC ; which, by notation, is -^tti ^s was to BC be shown. The grounds of the note subjoined to this rule are these: by casting away all factors common 32 Of Division, Thus, — -, divided by 7^ (or ~- ), giv ves ■ 35cy/ ^ , o; , 5a2|^\ J. ., ,1 O^t-^ . ISa^bif And 5a^£> (or ), divided by -^^^ gives r-^. 1 / o2y \yOCOC 4°. iS'z/r^ quantities^ under the same radical sign^ are divided by one another like rational quantities^ only the quotient must stand under the given radical sig?L Thus, the quotient of Vab by Vb is Va : That of s/l6xxy by VSxy is V2x : T-u . f hOabb , I'sab . llOabbc ' l2b 1 hat ot v/ by \/ — is v r—> oi^ V — • ^ Sc ^ ^ c ^ \5abc' > 3 And that of 6abVlOacxy by 2aV2cy is SbVjax. 5°> Different poxvers^ or roots^ of the same quantity are divided one by another^ by subtracting the exponent of the divisor from that of the dividend^ and placing the remainder as an exponent to the quantity given. But it must be ob- served, that the exponents here understood are those de- fined in p. 5 ; where all roots are represented as fractional powers. It will likewise be proper to remark further, that, when the exponent of the divisor is greater than that of the dividend, the quotient will have a negative exponent, or, which comes to the same thing, the result will be a fraction, whereof the numerator is a unit, and the deno- minator the same quantity with its exponent changed to an affirmative one. Thus, x% divided by a:^, gives x^ : And « + zl^, divided by a -f z"]^, gives a -f- zY '• Likewise, x^^ divided by ;c*, gives x^ : to the two numerators, we take equal parts of the quan- tities ; and, by throwing off the factors common to both denominators, we take equimultiples of those parts. The two preceding rules, being nothing more than the converse of the 4th and 5th rules in multiplication, are demonstrated in them: though, perhaps, the case in rule 5, where the exponent comes out negative, may stand in need of a more particular explanation. Accord^ Of Divisioiu 33. Moreover, c + yl , divided by c + i/1 , gives c + 2/ r * Lastly, x^^ divided by ^*, gives x , or --. x^ 6°. A compound quantity is divided by a simple one ku dividing every term thereof by the given divisor. Thus, 3a^) Zabc + l^abx — 9aab (c + 4;^ — 3a : Also,— 5ac)l5a^fc— 12acy^+5ad^ {—3ab 4— i^--,— : 5 c, and so of others. 7°. But if the divisor J as well as the dividend^ be a com- pound quantity y let the terms of both quantities be disposed in order y according to the dime?isions of some letter in them^ as shall be judged most expedient ^ so that those terms may stand first wherein the highest pozver of that letter is involved^ and those next where the next highest power is involved^ and so on: this being done^ seek how many times the first term of the divisor is contained in the first term of the dividend^ xvhichy when founds place in tA^ quotient^ as i2i division in- vulgar arithmetic^ and then multiply the xvhole divisor thereby^ subtracting the product from the respective terms of the dividend; to the remainder bring doxvn^xvith their pro- per signs J as many of the nextfolloxving terms of the divi- dend as are requisite for the next operation; seeking again how often the first terin of the divisor is contained in the first term of the remainder^ which also write down in your quotient^ and proceed as before^ repeating the operation till all the terms of the dividend are exhausted^ and you have nothing remaining. ing to the said rule, the quotient of x^ divided by x^ was — 2 1 asserted to be a: , or — , Now, that this is the true x^ value is evident; because 1 and x^ being like parts of x^ and x^ (which arise by dividing by a:^), their quotient will consequently be the same with that pf the quantities themselves. F 34 Of Division. Thus, if it were required to divide a^ -f 5a^x + Sax^^ + x^ by a 4- X (where the several terms are disposed ac- cording to the dimensions of the letter d)jl first write down the divisor and dividend, in the manner below, with a crooked line between them, as in the division of whole numbers ; then, I say, how often is a con- tained in <2^, or what is the quotient of a^ by a; the answer is «^, which I write down in the quotient, and multiply the whole divisor, a + at, thereby, and there arises a? -{^a^x ; which subtracted from the two first terms of the dividend, It^ves 4a^x ; to this remainder I bring down -f 5ax^y the next term of the dividend, and then seek again how many times a is contained in 4a^x ; the answer is 4aXy which I also put down in the quotient, and by it multiply the whole divisor, and there arises 4a^x + 4ax^y which subtracted from 4a^x + 5ax^ leaves ax^j to which I bring down x^^ the last term of the dividend, and seek how many times a is contained in ax^y which I find to be x^ ; this I therefore also write down in the quotient, and by it multiply the whole divisor ; and then, having subtracted the product from ax^ +x^y find there is nothing remains ; whence I conclude, that the re- quired quotient is truly expressed by ^^ + 4a^ + x"^. See the operation. a + x) a^ + Sa^x -f- 5ax^ + x^ {or -j- 4ax + ,%^ 4a^x 4- 5ax^ 4a^x •j-4ax^ ax^ + x^ ax^ -f- .y' In the same manner, if it be proposed to divide a^ — ^a^x-^ lOa^x^ — lOa^x'^ + 5ax*~x^ by a^ —2ax+x^y the quotient will come out a^ -^^ 3a^x -f 3ax^ — > x\ as the quotient will come out will appear from the process! O/Dhismi. 35 ^Sa^x + Sax^^x^ a^ — 2ax + x^) a^ — 5a^x + lOa^x- — lO^^^^ ^ 5^^.4 _ ;^5 (^^^s — Za^x-^- %a^x^ — 3a- A?^ '"°" + Q^a^x^—' Ta^x^ + Sax"^ +^S^x^_~_6fx^ +^'^1 O O So, likewise, if a* ►— x^ be divided by « — - :>:, the quo- tient will he a^+a^x + a^x^ +ax^ +x^ ; as by the work will appear. a _ ^t") fl* — x^ (a^ -{^a^x -}- «V + o^f^ + .y* a*^- c**-- — a^« ax*- ax"- -x' -x' Moreover, if it were required to divide a^ — • Sci^x^ 4 Sa^x^ — x^ by a 3 — ^a^x + 3ax^ — x^^ the process wil! stand thus : ^3 _ Sa^x +\a^--^3a^::^ + 3a^x^ — x^(a^+Sa^x+Sax^+rc' Sax- — X 3 )a^ — 3a^x +Za'^x^ — a^x^ 4. 2,a^x — 6«^;c2 4- a^x^ + Za^x"^ ^:^a5x — 9a^x^ +9a^x^—3 a^x'^ ^ + 3a'^x'^ — 8a^;f3+6a^4 — x^ -j-a^AT^ — 3^/^A"* + 3«;c^ — x^ 4- «3 -^3 — Sa^x"^ +3ax^-^x^ O 36 Of Involution. But it is to be observed, that it is not always that the work will terminate without leaving a remainder; and then this method is of little use ; and in all these cases it will be most commodious to express the quotient in the manner of a fraction, by writing the divisor under the dividend, with a line between them, as has been shown in the method of notation. It would be needless to offer any thing by way of de- monstration to the two last rules, the grounds thereof being already sufficiently clear from what has been de- livered in the last section, and the rules themselves no- thing more than the converse of those there demonstrated. I shall here show the reason why, in division, as well as in multiplication, like signs produce +, and unlike — . in order thereto, it must first be observed, that, according to the nature of division, every quotient whatever mul- tiplied by the given divisor, ought to produce the given dividend ; whence it is evident, 1. That +u) +ab {+b ; because -f a^ mult, by + ^, gives + ab ; 2. That -f ^ 'a~+~x]i X a + ^l^^ : but a^ x cT (by rule 5° in multiplication) is = a^ = a, and a + x~\'^ X a + x^'^ = a + x^^ ; there- lore our square, or its equal product, is likewise expressed ,2 by a X a + x\^. The 4th rule, or case, for the involution of fractions is grounded on rule 3° in multiplication, and requires nc> other 4emonstration than is there given. 40 Of Involution* a^ 4- ^a^b + ^ab^ + b^ the cube, or third power. a -^ b eC^ ^AfO^b + 6a^b^ -f 4Vthe fourth power," ^ +^ c* + 4a^^ + 6a'/^^ + 4«2Z>3 + ab"" 4- Q-^^ -f- 4^3/^^ -f 6fl^^^ + 4g^^ + b' a^ + 5a^^ + XOcv'b'^ + lOa^^^ + ^ab"^ + ^S the 5th power. a +b A« + 5a^b + 10a'' b^ + lOa^^ + SaH" -f «(^* 4- a^b -f- 5a^^^ -f lOa^b^ + 10a^Z>^ + 5^^^ + b^ a^ 4- 6a*^ 4- ISa^b^ + 20a^^ + 15aH^ + 6ab^ + ^% the 6th or required power of a 4- ^. So, likewise, if it be required to involve or raise a — ^ to the sixfh power, tjie process will stand thus : c —b c? — 2a^ 4- ^^, second pow ear. a ^^b a^ _ 20.^ + ab' — a^b+ 2ab^ -^ b^ a^^3a^b^+ Sab^ — W, third power. a — b ^4_4^3<^^ 6«V;2_ 4a^3 4- /^'^ fourth power. n — b as — 4>a^b+ 6a^^— 4aV 4. ab^ fl* _ Sa^b 4- 10<^^^^ — \OaH^ + Sab'' — ^% fifth power. ^ or — b Of Involution. 41 — a^b+ a''b^ — \0a^b^-\'\0aH^ — 5a¥'hh^ a^ _ e>a^b + ISa-^b^ — ^Oa^b^ + ISaF-b"^ — (5ab^ + b^^ the sixth power oi a-^b ; and so of any other. But there is a rule, or theorem, given by Sir Isaac Newton^ demonstrated hereafter, whereby any power of a binomial a + b^ or a — ^, may be expressed in simple terms, without the trouble of those tedious mul- tiplications required in the preceding operations ; which is thus : Let n denote any number at pleasure ; then the 72th power of a + 3 will be a^ + ncT^ b + — — HI — . «-3,„ 72.72— -1.72 2 n-3,» . 72.72 1.72 2.72 3 a b^ ■\ ; . a b^ -^ 1.2.3 1,2. 3. 4 «-*;a . 72. 72 — -1.72 2. 72'— ^3. 72 4 n-5 , . ^ '^ '' + 1.2.3.4.5 '^ *'' "^^ And the 72th power of « — b will be expressed in the very same manner, only the signs of the second, fourth, sixth j> &?c. terms, where the odd powers of b are involved, must be negative. An example or two will show the use of this general theorem. First, then, let it be required to raise <2 + ^ to the third power. Here 72, the index of the proposed power, be- ing 3, the first term, a^^ of the general expression, is equal to c? ; the second 72a b = Zc^b ; the third ^lliLZZi, a^-^B^ = 3a^^; the fourth n.n-l.n-ab^ + b'^. Again, let it be required to raise a + ^ to the sixth power; in which case the index, 72, being 6, we shall, by proceeding as in the last example, have a** =r a^, '^ Fot one of the factors, ini ==0. Ep 4 G 42 Of Evoluttoji. na-^b = em% IlILilL, /-^^2 ^ 15^4^2 g.^. and 1.2 ' consequently, « + bY = «^^ + 6a*^ + 15a^^a + 20a^<^^ + ISaH"^ + 6a3* + b^ ; being the very sanie as was above determined by continual multiplication. Lastly, let it be proposed to involve cc + xy to the fourth power. Here a must stand for cc^ b for xy^ and n for 4 ; then, by substituting these values, instead of a^ b^ and 7Z, the general expression will become c^ + Ac^xy + Qcf^x^y^^ + 4c^x^y^ + ^^y^y the true value sought. From the preceding operations it may be observed, that the uncise, or coefficients, increase till the indices of the two letters a and b become equal, or change values, and then return, or decrease again, according to the same order : therefore we need only find the coefficients of the first half of the terms in this manner, since from these the fest are given* SECTION VIL Of Evolution* EVOLUTION^ or the extraction ofroots^ being directly the contrary of involution^ or raising of powers y is per-* formed by converse operations^ viz. by the division of indi- ces^ as involution rvas by their multiplication. Thus the square root of ^^, by dividing the exponent by, 2, is found to be x^ ; and the cube root of x^^ by di- viding the exponent by 3, appears to be x^: moreover, the biquadratic root of a -f at") ^ will be a + x^ l and the cube root of aa + xxy will ht aa + xx\^* "^. In the same manner, if the quantity given be a fraction ♦ or consist of several factors multiplied together, its root will be extracted, by extracting the root of each particular factor. Of Evolution, 43 Thus the square root of d^b^ wil l be ab i that ot will be — : and that of --.-1,.^_L. will be — ~ 2 ■ . : moreover, the square root of aa — xx |- 4x CL — 'X will be aa — '^:v V ; its cube root aa — xx\^' ; apd its biquadratic root aa — xoc\^ ; and so of others. AIL which being nothing more than the converse of the ope- rations in the preceding section, require no other demon- stration than what is there given. Evolution of compound quantities is performed by the following rule. Firsts place the several terms ^ whereof the given quan^ tity is composed^ in order^ according to the dimensions of some letter therein^ as shall be jitdged most commodious ; then let the root of the first term be founds and placed in the quotient ; which term being subtracted^ let the first term of the remainder be brought down^ and divided by twice the, first term of the quotient^ or by three times its square^ or four times its cube^ &c. according as the root to be ex- tracted is a square^ cubicj cr biquadratic one^ &c. and let the quantity thence arising be also rvritten doxvn in the quo- tient^ and the whole be raised to the second^ third^ or fourth^ &c. power ^ according to the aforesaid cases^ respectively y and subtracted from the given quantity ; and^ if^^y thing 7'emai?i^ let the operation be repeated^ by always dividing the first term of the remainder by the same divisor^ found as above. Suppose, for example, it were required to extract the square root of the compound quantity 2ax + a^ + x^ : then, having ranged the terms in order, according to the dimensions of the letter «, the given quantity will stand thus, a^ + 2ax + x^y and the root of its first term will be « ; by the double of which I divide 2aXy the first of the remaining terms, and add + x^ the quantity thence arising to a^ already found, and so have a + xm the quotient ; which being raised to the second power, and subtracted from the given quantity, nothing 44 Of Evolution. remains ; tHei!?efoi'e a + at is the square root required* Sec the operation. a^-^^ax-^-x^ (a + x 2a) 2ax c^ -*> "lax + x^^ second power of a + at. o~o oT In like manner, if the quantity a^ — 2a^x + 3a^;c^ — 2ax^ + x^ be proposed, to extract the square root thereof ; the answer will come out a^ — (2^: + x^^ as ap- pears by the process. a^ — "2.0^ X + oc^x^ — 2ax^ + x^ {a^ — ax + a;^ 2a^)—2a^x a"^ — 2a^x -f a^Y^, second power of a^ — ax. 2d^) 2a^x'^^ first term of the remainder. g ^ — 2a^x 4- Sa^x^ — 2ax^ 4- x"^^ square of rt^ •— a:c + a:^ rr ~ - - Agahi, let it be required to extract the cube root of a^ — Q>a'^x + 12ax^ — 8^^, and the w^ork will stand thus : a^ — ea-x + 12ax^ ~Sx^ (^a — 2x 3a^) — 6a^x a^ — 6a^x '{-\2 ax'^ — 8^^, cube of a — 2x. 'o O 0~ Lastly, let it be required to extract the biquadratic root of 16^^"^ — 96x^2/ + 216::c2z/3_216;cz/« +8lz/% and the process will stand as follows : 16x^ — 9Q)X^y + 216.x'22^^ — 216xy^ + SU/ (2x — Sy - 16x^ — 96;c3z/-f 216.%'^ — 2iexy^ -f SJy^ O And, in the same manner, the root may be deter mined in any other case, where it is possible to be ex- tracted ; but, if that cannot be done, or if, after all, there be a remainder, then the root is to be expressed in the maimer of a surd, according to what has been already shown. As to the truth of the preceding rule, it is too obvious to need a formal demonstration, every ope- Of Evolutipn. 45 ration being a proof of itself. I shall only add here, that there are other rules beside^ this, for extracting the roots of compound quantities, which sometimes bring out the conclusions rather more expeditiously; but as these are confined to particular cases, and would take up a great deal of room to explain in a manner suf- ficiently clear and intelligible, it seemed more eligible to lay down the whole in one easy general method, than to discourage and retard the learner by a multiplicity of rules. However, as the extraction of the square root is much more necessary and useful than the rest, I shall here put down one single example thereof, wrought ac- cording to the comnlon method of extracting the square root, in numbers : which I suppose the reader to be ac- quainted with, and which he will find more expeditious than the general rule explained above. Examp. a^ ^ 4^3^^. ^ ^^^% ^ 4^^3 ^ ^a ^^^2 ^ 2<7.v -f jc^ 2a^ + 2ax) +A:a^x + ^a^x^ ' + A^a^x -f 4*a^x^ 2a^ +4ax+x^) +2a^x^ +4ax^ + x-^ '^2a^x^ '}-4ax^ +x' ^ O SECTION VIII. Of the Reduction of Fractional and Radical Quan- tities. THE reduction of fractional and radical quantities is of use in changing an expression to the most simple and commodious form it is capable of; and that, either by bringing it to its least terms, or all the members thereofj if it be compounded, to the same denomination. 45 Reduction of A fraction is reduced to its least terms by dimding^ both the numerator and denominator by the greatest com- mon divisor » nh Thus, --, by dividing by b^ is reduced to - : be c And — 77-> by dividing by ab^ is reduced to ~ : abb 0^7 ^ Moreover, will he reduced to — , or 4^di Sab 1 ' And M=. will be reduced to — 72a^xWxy 6a Thus, also, — ^ — ^- , by dividing every term of the numerator and denominator by 2a, is reduced to 6a — ^b 2a ' ^^^ 6^^V:iTalF^ ^ ^^ ^^^^^^^S every terni by 2ax. is reduced to : Sa-^2x Lastly, !-~ 1- —T — , by dividm^ both the •^' a^+3ab + 2b^ ^ ^ ^ numerator and denominator by tUe compound divisor , . . .^a^ +2ab+bb a + b.is reduced to • -I — . ^ ' a + 2b But the compound divisors, whereby a fraction can sometimes be reduced to lower terms, are not so easily discovered as its simple ones ; for which reason it may not be improper to lay down a rule for finding such divisors. Firsts divide both the numerator and denominator by their greatest simple divisors^ and then the quotients one by the other (as is taught in case 7, section 5), always ob- serving to make that the divisor which is of the least di- "tnensions ; and^ \f(^^y thing remain^ divide it by its great- est sim.ple divisor^ and then divide the last compound di-^ Fractional ^antities. 47 visor by the quantity thence arising ; and if any thing yet remain^ divide it likewise by its greatest simple divisor^ and the last compound divisor by the quantity thence arising; proceed on in this manner till nothing remains; so shall the last divisor exactly divide both the numerator and denomi- nator^ without leaving any remainder. Note. If, after you have divided any remainder by its simple divisor, you can discover a compound one which will likewise measure the same, and is prime to the divi- sor from whence that remainder arose, it will be conve- nient to divide also thereby. And, if in any case it should happen that the first term of the divisor does not exactly measure that of the dividend, the whole dividend may be multiplied by any quantity that shall be necessary to make the operation succeed. Ex. 1. Let it be required to reduce the fraction Sa' + lOa^'b-^SaH^ . v i . . . r j to Its lowest terms, or to find a^b + 2d'b'^ + 2ab^ + ^^ the greatest common measure of its numerator and de- nominator. Here, dividing first by the greatest simple divisors, 5a^ and ^, we have a^ + 2ab + b^^ and a^ + 2a^b -f 2a^2 ^ ^3 . ^^^j ^f ^^ latter of these be divided by the former, the work will stand thus : a^ + 2a^b+ ab^ where the remainder is + ab^ + P ; which be- ing divided by ^^, its greatest simple divisor, gives a + b', by this divide a^ + 2ab + b^^ and the quotient will come out a + b^ exactly ; therefore the last divisor, a + ^, will exactly measure both quantities, as may b^ proved thus : a+b) 5a^ + lOa^b + 5a^o^ {Sa'^ + 5a^ 5a^+ Sa'^b Sa'^b+Sa^b^ 5^b + SaH^ 48 Reduction of a + F) a^b + 2aH^ + 2ab^ + b^ (a^ +ab^+b'^ aH^+2ab^ a%^+ ab^ ab^ -f b"^ ab^ 4- b'^ O O In both which cases nothing remains ; therefore the fraction eiven will he reduced to — -. -. ^ a^b-\-ab'^ A-b^ Ex. 2. Let it be proposed to reduce the fraction — - to its lowest terms: and then the €? — c^x — ax^ -f x^ work will stand as follows : . ax^ 4- a:^) a^ '+ O + -f O — :\;* (« + x a^ — a^x — ' a^: \'^ + ax^ a^x + a*A:^-— a^^— x^ ^3^_ (fx^'-^ax'^ + x^ + 2a^x^ +0—2;^ a^ + o— ..\^) a^ — a^x — ax^ +x^ (a-^x a^ , — o — ax^ — a^x -f- + .x^ — d^x 4- -f x^ From whence it appears that a^ ^ q _^ ^,2^ ^^ ^2 _ ^2^ will measure both a*^ — x^ and a^ — a^x -^ax^ + x^ ; and, by dividing thereby, the fraction proposed is reduced to ^2 ^ ^2 ^ These operations are founded on this principle, that xvhatever quantity measures the whole^ and one part of another^ must do the like by the remaining part. For that quantity, whatever it be, which measures both the divisor and dividend, in the first example, must evidently measure a* + 2a^b + ab^ (being a multiple of Fractional ^antities. 49 Example 3. In the same manner the fraction = ; —5 — , ^ 3 Will be reduced to . L- — • See the process. X — Sa x^^^ax^ — ^a^x + 6a^)x^'--2>ax^ — Sa^^s-f 1 %a^x—^a\x — 2a — 2ax^-{' +12a^x — Sa'* remainder — ^^(fi^jt^ — 4^^^;^' — : ^a'^ ; which divided by — 2a^^ gives x"^ + 2(7a; — 2a2 for the next divisor. x^ + 2ax — 2a2) a:^ — ax^ — ^a^x + 6«3 (^ — 3a a:^ + 2ax^ — 2a^x — tiax'^ — Qa^x -f 6a^ — ^ax'^ — 6a^x 4- 6a^ o o' ;^2+2a;c— 2a2)Ar4— SaAT^ — %a^x'^+\^a^x — ^a^x"^ — SaX'\Aa^ x^-\-2ax^ — 2r^x^ ^^5ax^ — 6a^x^ 4. 1 8a^x *-^5ax^ — lOa^x'^ + XOa^x + 4a^v2 4. 8c^x — 8a* 4- 4a^x^+ 8a^x — 8a^ Now if, by proceeding in this manner, no compound divisor can be found, that is, if the last remainder be only a simple quantity, we may conclude the case pro- posed does not admit of any, but is already in its lowest the former) : whence, by the principle above quoted, the same quantity, as it measures the whole dividend, must also measure the remaining part of it, ab^ -f- b^ : but, the divisor, we are in quest of being a compound one, we may cast off the simple divisor ^^, as not for our purpose : whence a + b appears to be the only com- pound divisor the case admits of: which, therefore, must be the common measure required, if the example pro- posed admits of any such. II 50 Reduction of terms. Thus, for instance, if the fraction proposed if^ere to be — ; it is plain by or + ax + x^ inspection, that it is not reducible by any simple divisor ; but to know whether it may not, by a compound one, I proceed as above, and find the last remainder to be the simple quantity 7xx: whence I conclude that the frac- tion is already in its lowest terms. Another observation may be here made, in relation to fractions that have in them more than two different letters. When one of the letters rises only to a single dimension, either in the numerator or in the denomi- nator, it will be best to divide the said numerator or de- nominator (whichever it is) into two parts, so that the said letter may be found in every term of the one part, and be totally excluded out of the other ; this being done, let the greatest common divisor of these two parts be found ; which will, evidently, be a divisor to the whole, and by which the division of the other quantity is to be tried; as in the following example, where the P . . . x^ + ax^ 4- bx^ — 2a^x + bax — 2ba^ traction eiven is ■ ■ ■ ; • ^ XX — bx + 2ax — 2ab Here the denominator being the least compounded, and b rising therein to a single dimension only, I divide the same into the parts x^ -f- 2ax^ and — bx — 2ab ; which, by inspection, appear to be equal to x + 2a x ^-i and X -f 2a X — b. Therefore .v -f- 2a is a divisor to both the parts, and likewise to the whole, expressed by X + 2a X X — b ; so that one of these two factors, if the fraction given can be reduced to lower terms, must also measure the numerator : but the former will be found to succeed, the quotient coming out v2 — ax -j. bx — ab^ exactly: whence the fraction it- -^ . , , x^ — ax 4- bx — ab i . i . self is reduced to ; ; which is not re- X ducible farther, by x — b^ since the division does not terminate without a remainder, as upon trial will be found. Fjractional ^antities* 51 Having insisted largely on the reduction of fractions to their least terms,- we now come to consider their reduc- tion to the same denominator. Fractions are reduced to the same denominator by multi- plying the numerator of each into all the deno7ninators^ ex- cept its own^ for a nexu corresponding numerator^ and all the denominators continually together^ for a common deno- 7ninator. Thus, ---and —-will be reduced to -— and —-; b d bd bd a c ^ e adf cbf , bde _, _, and -, to ^, ^, and _; J 2ax J 5bx 6a^x , Sbxcd , and — -, and , to -, .and r ; and cd 3a 2acd Sacd so of others. But when the denominators have a common divisor, the operation will be more simple, and the conclusion neater, if, instead of multiplying the terms of each fraction by the denominator of the other, you only multiply by that part which arises by dividing by the common divisor. As, if there were proposed the frac- tions — r and — -; then, the denominators havincr the ad cd ' ° factor d common to both, I multiply by the remaining factors a -^ X X ^ +x « 2a:^^ t^y/Z:. 10a — 5x = 2 ^ ; and ilf^+--=rr = ^^ ^' a Vxij + a 5x1/ + 5aV*\y 4- lOrt^ — Sax , . , — " s^= ; and so m other cases. aVxy + a^ The reason of the two kinds of reduction hitherto explained, is grounded on this obvious principle, that the equimultiples, or like parts of quantities, are in the same ratio to each other, as the quantities themselves ; or, that the quotient which arises by dividing one quantity by another, is the same that arises by dividing any part or multiple of the former, by the like part or multiple of the Fractional ^lantities. 53 Besides these, there are yet two other sorts of reduc- tion which authors have treated of under the head of fractions ; which are, the reducing of a whole quantity to an equivalent fraction of a given denomination^ and a com* pound fraction to a simple one of the same value. Nei- ther of these, indeed, are of any great use in the sokition of problems ; however, it might be improper to leave them entirely untouched. 1°. A whole quantity is reduced to an equivalent frac- tion by multiplying it by the given denominator^ and writ- ing the midtiplier underneath the product^ with a line be- tween them. Thus the quantity tf, reduced to the denominator by will be — , and 'the quantity c + d^ to the denominator 7 Mil a-j-b X c + d ac -{-be -^ad-^bd a + h. will be — -n^—t or — • ' a-^b ' a + b 2°. A compound fraction is reduced to a simple one of the same value by multiplying the numerators together for a new numerator ^ and the denominators together for a nexu denominator. . • But by a compound fraction here, we are not to un- latter: for, in reducing to the lowest terms, it is plain, that, instead of the whole numerator and denominator, we only take that part of each which is defined by the greatest common measure ; whereas, in reduction to the same denominator, we, on the contrary, make use of equimultiples of those quantities ; since, in multiplying any numerator into all the denominators, except its own, we multiply it by the very same quantities by which its denominator is multiplied. The rule for reducing a compound fraction to a simple one, may be explained thus. It is plain c 1 that the part of — defined by -~, which arises by a b dividing by ^, will be equal to ~ (the divisor here bd o4 Reduction of derstand one consisting of several tenns, connected toge- ther by the signs + and — (which i^ the general definition of a compound quantity), but such a one as expresses a given part of some other fraction. Thus ^ of -^ will be equal to ^, and the ^ o 5 15 b part of ~- will be = 7-/ a bd Of the Reduction qf Radical Quantities. The reduction of surd quantities, like that of fractions^ may be either to the least terms, or to the same denomi- nation. A radical quantity is reduced to its least terms by resolv- ing it into factors^ and extracting the root of that which is rationaU Thus, V28 is reduced to V4 x VT"; which, by extracting the square root of 4, becomes 2\/ 7 ; also \^c^b is reduced to Va^ X V <^ ; which, by extracting the root of a^, becomes aV b : likewise, K/a^b'^c^^ or 1"^ is reduced to s-cv'b^ x V^c^ or obK^bc^ : 1 4a^x — 4a^x^ . - , J 4a^ '-> \ 7~i ^^5 reduced to y— ^ x • onH ▲ / ^_ Slb''c—'162b'x a''o'*c moreover. fax — x^ 2a lax — x^ ,'^ll6a^x'^+ 16ttV c being b times as great) ; therefore the part of -— • de- fined by --, being a times as great as that defined by -7-, c ac must be truly expressed by — X «, or its equal ~ ; as was to be shown. Radical ^lantities. 55 reduced to ^J—- x Sj-TZZZ a^x^ ^ax is reduced to ^__-- x V 2 ' ../. "> ^^' ":;r X 4^/f? -i- ^ ^ ^^^ gQ q£ ^^y other: all which is V c^ — 2bx \ evident from case 4 of multiplication, and case 3 of invo- lution. But it is to be observed, that, in resolving any expression in this manner, the factor out of which the root is to be extracted, is always to be taken the greatest the case will admit of. It also may be proper to take no- tice, that this kind of reduction is chiefly useful in the addition and subtraction of surd quantities, and in uniting the terms of compound expressions that are commensurable to each other, where the irrational part, or factor, after reduction, is the same in each term. Thus, VI8 -f^Vsi is red uced to_3V/2 + 4^2", or rV2'; and Vsf + \^50f — ^72^ is reduced to 2aV"2" + 5«VT — 6aV 2 = a>/ 2. Moreover, by re- . . }l2a^x , flScFx , /48a^ duction, V + \] — - — becomes = \J 4. ' ^ 5 > 4 ^ 20 ^ hsa^x . Hx . ^ f^ ^ I3x ^ 20 ^20 ^ ^ 20 ^20 An d SaS/4 a^x^ + Sx"^ -f 3xV9a -f- ISa^x^ becomes eaxVa^ + 2;c2 ^ 9axVa^ + 2x^ == XSaxVa" -f 2a,^ Surd quantities^ under different radical signs ^ arc reduced to the same radical sigiiy by reducing- their indices to the least coinnion denominator. 3^, reduced to the same radical sign, will become a^ and a^ (for the indices are here i and |-, and these are equivalent to f and |, where both have the same denominator). In the same manner, 2"j^ and 3J3 will become 26" and 3|^, or 8|« and 9|«. And It JL ' ' universally^ A"* and B^ will, when their exponents o6 jReduction of^ £sf*c. are reduced to the same denomination, become *A\^ andll mp The principal use of this sort of reduction is, when quan- tities under different radical signs are to be multiplied or divided by each other. That the reduction of a radical quantity to another of a different denomination, by an equal multiplication of the terms of its exponent, makes no alteration in the value of the quantity, may be tlius demonstrated. m Let A " be any quantity of this kind ; then, the terms of its exponent being equally multiplied by any number mr r, I say, the quantity A ^^ hence arising, is equal to the given one A**. For, if X be assumed == A"'* , or, which is the same thing, if the value of x be such, that x'*'' = A ; then the nth root of x^"^ being x'^ (by case 2 of section 6) and the nth root of A being A" (by notation)^ these two quantities JL x** and A ^ must, likewise, be equal to each other : and, if they be both raised to the 72th power, the equa- lity will still continue ; but the 72th power of the former (x^) is = x'^'^ (by case 2 of involution) ; and the mth. 1 m power of the latter (A " ) is A"« (^by notation) ; there- fore ;c"^^ is = K'' . But, x being = A«^, we have mr mr m ^mr^^nr^ ^^ notation; and consequently A"''= A'* ; which xvas to be proved. Of Equations. 57 Thus, VT, multiplied by a/10, or T^, into lo]^^^ will give 125]^ X lOu"!^, or 12500l^ : also, Vax into \/a^Xy or ax\^ into 6?^;^1^, will give 02:^^1^ x a^^v*"]^, or a';v^^^ : and Vax^ divided by Va^x will give fl^vV^ 1^ ;^ J^ ... - ZZZTT-i o^ — 1 • Lastly, 2x^ multiplied into VZax^ a4;rf]6 ^ ^ will give Vl^ X Va^x*, or V\2ax^. SECTION IX. Of Equations. AN equation is, when two equal quantities, differently expressed, are compared together, by means of the sign = placed between them. Thus, 8 — 2 = 6 is an equation, expressing the equa- lity of the quantities 8 — 2, and 6 : and ^ = a + 6 is an equation, showing that the quantity represented by x is equal to the sum of the two quantities represented by a and b. Equations are the means whereby we come at such conclusions as answer the conditions of a problem ; where- in, from the quantities given, the unknown ones are deter- mined ; and this is called the resolution or reduction of equations. Reduction pf Single Equations. Single equations are such as contain only one un- known quantity; which, before that quantity can be discovered, must be so ordered and transformed, by the addition, subtraction, multiplication, or division, ^c* of equal quantities, that a just equality between the two parts thereof may be still preserved, and that there may result, at last, an equation, wherein the unknown quaiv I 58 Of Equatiorui. tity stands alone on one side, and all the known ones on the other. But, though this method of ordering an equa- tion is grounded upon self-evident principles, yet the ope- rations are sometimes a little difficult to manage in the best manner -, for which reason the following rules are subjoined. 1 °. Any term of an equation may be transposed to the coritrary side^ if its sign be changed^. Thus, if a: + 6 = 16 ; then will .v = 16 — 6, that is. And, li X — 4 = 8; then will .v = 8 -f 4, or x = 12 ; Also, if 3y = 2;c -f 24 ; then will Zx — 2x :=. 24, that is, AT = 24 : Again, \^ 5x — 8 = 3x + 20 ; then wall Sx '—* ^x ■=. 20 + 8, or 2x = 28 : Lastly, if ax + bx — c •\- d — ex =f — g -^hx — kx ; then, by transposition, ^ax + bx — ex — hx + kx z=zf-^ g + c — d ; where all the terms affected by x (the un- known quantity) stand now on the same side of the equa- tion. 2°. If there be a?iy quantity by which all the terms of the equation are multiplied^ let them all be divided by that quantity ; but if all of them be divided by any quantity^ let the common divisor be cast axvay. Thus, the equation yza: zzl ab is reduced to x = b\ also, lO.v = 70 (or 10 X -v = 10 X /) is reduced to X = 7 ; and x^ = ax^ + bx^ is reduced x -=. a -{• b'. ^ The reason of this rule is extremely evident ; since transposing a quantity thus is nothing more than sub- tracting or adding it on both sides of the equation, ac- cording as the sign thereof is positive or negative. Thus, in the equation ^ + 6 = 16 (which, by transposition, be- comes Ar = 16 — ^6==10), the number 6 is subtracted from both sides ; and in the equation x — 4 = 8 (which, by transposition, becomes a; = 8 + 4 = 12), the number 4 is added on both sides. Of Equations* 59 X h , Moreover (bij the latter part of the rule)^ — = — is re- ax^ abx^ ' ac\ duced to X = d i and = — , to ax^ = abx^ c c — acx^ ; which, if the whole be divided by ax^^ will be farther reduced to a: = ^ — c. 3°. If there be irreducible fractions^ let the whole equa- tion be multiplied by the product of all their denominators ^ or^ which is the same things let the 7iumerator of every terin in the equation be multiplied by all the denominators^ except its own^ supposing such terms (if any there be^ that stand 7vithout a denominator^ to have a unit subscribed* / XX Thus, the equation x + — + — =11, is re^- duced to e>x + Zx +2x = 66 ; and x + ^^-t- = 12 + ^ ~ ^ , to 40x + Sa;- + 16 = 480 + 5x — 15 : so, o T1 • X X + b , * ^ , likewise, a = is reduced to a^c — ex =z ax a c ax ex + ab ; and f- a = —- to cd^x + a^b + obx = acx a + x b + cx^. ■ 4"^. If ill your equation^ there be an irreducible surd^ wherein the unknown quantity enters^ let all the other terms be transposed to the contrary side (by rule 1) ; and then^ if both sides be involved to the power denominated by the surd^ an equation xvill arise free from radical quantities ; un- less there happen to be more surds than one^ in which case the operation is to he repeated. Thus, \^ X + 6 = 10, by transposition, becomes V x (=10 — 6) = 4 ; which, by squaring both sides, gives X =16. - So, likewise, \^aa -^ xx — c ■=. x^ becomes Vaa + xx ■==z c + X 'y which, squared, ,gives aa + xx = cc + 2cx f wv, o]* aa* — cc = 9.cx {by rule 1). The reasons of 60 OfEquattons^ this, as well as of the two preceding rules, depend on self- evident principles : for, when the equal quantities, on each side of an equation, are multiplied or divided by the same, or by equal quantities, or raised to equal powers, the quan- tities resulting must necessarily be equal. 5°. Having'^ by the preceding rules {if there he occasion)^ cleared your equation of fractional and radical quantities^ and so ordered it^ by transposition^ that all the terms^ where* in the unkno-wn quantity is founds may stand on the same side thereof let the whole be divided by the coefficient^ or the sum of the coefficients^ of the highest power of the saidwi- known quantity : and then, if your equation be a simple one (that is, if the first power, or the quantity itself, be only concerned), the work is at an end ; but if it be a qua- dratic or cubic one, &Pc. something further remains to be done ; and recourse must be had to the particular me- thods for resolving these kinds of equations, hereafter to be considered in their proper place. I shall here subjoin a few examples for the learner's exercise, wherein all the foregoing rules occur promise cuously. Ex. 1. Let 5x — 16 = 3;^ + 12: then (% rule 1) 5x — 3x = 12 + 16, or 2x = 28 : whence {by rule 5) .V = — = 14. 2 Ex. 2. Let 20 — 3;^ — 8 = 60 — 7x: then, — 3;^ -f- Tx = 60 — 20 -f- 8, that is, 4a: = 48 ; and consequently x z= 4 Ex. 3. Let ax — b :=zcx + d; then ax **^ ex :=: d + b^ or a— ^ XX = d + b ; and therefore x = (by rule 5*). Ex. 4. Let 6x^ — 20x :=zl6x +2x^t then, dividing by 2x {according to rule 2) we have 3x — 10 = 8 -f- a: ; whence 3;^- -— . a; = 8 + 10, that is, 2;^ = 18 : and therefore x =^- o Of Equations,. 61 Ex* 5. Let oax^ — - ahx^ = ax^ ^f- "lacx^ : here, dividing the whole by ax^^ we have Zx -^b = .v + 2c ; therefore 7 1 2c + ^ 2j^ = 2c + ^, and .r = — - — . ^ -^ ^a;. 6. Let ^ +— = 21 : then (% rw/e 3) 4.v + 3;^- 3 4 252 ' = 252 ; and therefore x =— — = 36. ^;,. 7. Let ^±i + ^i±^ = 16 ^ ^±1 : then 12:^ + 12 2 3 4 + 8:v + 16 = 384 — ^x — 18 : whence 26.v = 338, and 338 ,^ A^ = = 13. 26 Ex. 8. Let a = c : then ax "--^bb =. ex ; whence X ax *— c^* = bb^ and x =: ■ a — c JE:v. 9. Let h -7- H = ^- ^^^^5 ^^-^ + ^^^ + a b c abx = rt/!?c^, or be + ac + ab x x: = abed; and consequent- ly ;v = -■ (bt/ rule 5). •^ bc + ac + ab^ ^ ' axP' 4. rtc^ i::v. 10. Let ax + b^ ^ "^ : then, ax + h^ x a+ X n + xz= ax^ + ac^j that is, a^x + ab^ + ^a:^ + b^v = r/;c2 + ^r^ ; whence a^x + <2X^ + b^x — ax^ = ac^ -— a^2, or rt^A" + b^x = ^c^ — ab^ ; and therefore a: =:: ac^ — ab^ aa + bb* Ex. 11. Let 1 = 1 : then ax + ab + bx a + ;f X •=. ax -^ XX ; whence — xx •\- bx, = ■ — abi which, by changing all the signs (in order that the highest power of X may be positive), gives xx — bx z=l ab. But the same conclusion may be otherwise brought out, by first changing the sides of the equation ax + ab + bx = nx ^^ OfEquatiom. + XX ; which thereby becoming ax + xx=:ax + ab4. l>x we thence get xx ~i>x = ab, the same as before. Ex. 12. Let :^ + 12 = ir: then ^ - ^ and V/5X = 15; whence {by rule 4), 5;. = 225, and therefore x = — £ = 45. 5 iix. 13. Let VlTJ^ = 2 +V/^ : then (^y rw/e 4) 12+^_=4 + 4Vx + x; whence,_by transposition, 8= 4V^ ; and, by division, 2 = V:^ ; consequently, E.. 14. Let.v + V^^-+.- = ;;;-==. Here {by rulej>)_x X Va^+x^ X a^ + x ^ = 2a^ ; whence x x V'a^ + x^ = a^ — x^^ and x^ X a^ + x^ =z a"^ 2aV + x"^ (by rule 4), that is, a^x^ + x"^ =z a^ q^x^ + x"^ • therefore Za'x^ = a\ and ^c^ = -fl = 2!. Ex, 15. Let V X + Va + X = — . Then Va + X Vax + XX + a + X z=z 2a^ or Vax +xxz:za — x; whence ax + XX := a^ — 2ax + x^. and a; = — = ~ 3a 3* Ex. 16. Let v'^s — a^ =z X — c: then, by cubing both sides, x^ — a^ = x^ — Sc;^^ j. 3c^x — c^ ; whence Scx^ — Sc^x =za^ — c\ and x^ — ex z:z — ^tL^hy divid- oc 3 "^ ing the whole by oc, Ex. 17. Let v^«a 4- AT^ =;= \/^4 _j^ ^4 . ^hen, by raising both sides to the fourth power we have aa + xx"]^ = ^^ -f a'% that is, .^^ + 2aV -f at^ =: Z^'* + at^ ; and consequently 1^/2 2^2 :2aa ^' Of Equations* 63 Ex.. 1 8. Let X = \a^ + x Vbb + xx -r- a. Here .v + a =r \a^ + X s/bb -f XX ; which, squared, gives^ jv^ + 2a.v + fl2 --. ^ ^ ^ \/}jl) j^ xx^ or AT ^ + 2c7Ar = :^V/>^ + ivA; ; di- vide by ^, so shall x + 2a = \/bb + x:^ ; this squared again gives a:^ + 4ax + 4a^ = bb + xx ; whence 4^wr = ^^ — 4a^, and therefore x =. a. 4a Of the Extermination of Unknown Qtcantities, or the reduction of two or more equations to a single one. It has been shown above, how to manage a single equation ; but it often happens, that, in the solution of the same problem, two, three, or more equations are con- cerned, and as many unknovvn quantities occur pro- miscuously in each of them ; which equations, before any one of those quantities can be known, must be re- duced into one, or so ordered and connected, that, from thence, a new equation may at length arise, affected with only one unknown quantity. This, in most cases, may be performed various ways, but the following are the most general. 1°. Observe xvhich of all your tmknoxvn quantities is the least involved^ and let the value of that quantity be found in each equation {by the methods already explained)^ looking' upon all the rest as knoxun ; let the values thus found be put equal to each other {for they are equal^ because they all ex- press the same thing) ; whence fiexv equations -w'llL arise ^ out of which that quantity will be totally excluded; xvith these new equations the operation may be repeated^ and the unknown quantities exterminated^ one by one^ till^ at last^ you come to an equation containing only one unknoxvn quan- tity. ^ ' 2 . Or, let the value of the unknoxvn quantity^ xvhich you would first exterminate^ he found in that equation xvherem it is the least involved^ considering all the other quantities as known; and let this value^ and its powers^ be substituted for that quantity audits respective powers in the 6i Of Equattom* other equations ; andy -with the nexv equations thus arising^ repeat the operation^ till you have only ane unknown quan- tity and one equation. 3°. Or, lastly^ let the given equations be multiplied or divided by such numbers or qua7itities^ whether known or unknown^ that the term which involves the highest power of the unknown quantity to be exterminated^ may he the same in each equation ; and then by adding^ or subtracting the equations^ as occasion shall require^ that term will vanish^ and a new equation emerge^ wherein the number of dimen- sions ( if not the number of unknown quantities^ will be di- minished. But the use of the difFereixt methods here laid down will be more clearly understood by help of a few exam- ples. EXAMPLE I. Let there be given the equations x + y -=. 12, and 5x + Sy ■= 50 ; to findx and y. According' to the frst method^ by transposing 2/ and 3y, we get .v= 12 — z/, and 5x =^ 50 — 3z/ ; from the last of which equations, x = — Zl^. Now, by equating , r 1_ ... 50—32/ these two values 01 x^ we have 12 — ' ^ = — -l and therefore 60 — 52/ = 50 — 3z/ : from which y is given = — = 5 ; and :\^ (= 12 — z/ = 12 — 5) = 7- According to the second method^ x being, by the first equation, = 12 — z/, this value must therefore be substi- tuted in the second, that is, 60 — 5y must be written in the room of its equal 5x ; whence will be had 60 — 5y + 2y = 50 : and from thence z/ = — = 5^as before. But^ according to the third method^ having multiplied the first equation by 5, it will stand thus, 5:r + 5z/ = 60 ^ from whence subtracting the 2d equation, 5x + 3z/ = 50, there remains - - ?^ ^ ^ 9 > whence y = 5? still the same as before^ Of Equations, 65 The first of these three ways is much used by some authors ; but the last of them is, for the general part, the most easy and expeditious in practice, and is, for that reason, chiefly regarded in the subsequent exam- ples. EXAMPLE IL Let i'^^-^^y- ^l^ Here the second equation being multiplied by 4, in or- , der that the coefficients of y in both equations may be the same, we have 12x — 8y = 80. Let this equation and the first be now added to- gether ; whence y will be exterminated, there coming 204 out 17x :=. 204 ; from which x = - — = 12 : therefore, 17 , , ^ . , 124^ — 5x 124 — 60 64. by the first equation, y (= ^ — = — . -> = — ) = 8. EXAMPLE in. ^. i 5x — 3v= 90 Given. < ^ ^ J' .^^ (^ 2:v + 5z/ = 160. Here multiplying the first equation by 2, and the second by 5, in order that the coefficient of x may be the same in both, there arises \0x — 6z/= 180 10^ + 25z/ = 800. By subtracting the former of which from the latter we havfc 312/ = 620 : hence y = = 20 j and so, by the first oX , 90 4-3Z/ 90 4-60. equation, x (= ^— ^ = — ) = 30. But the value of x may be otherwise found, inde- pendent of the value of y ; for, by multiplying the first equation by 5, and the second by 3, and then adding them together, y will be exterminated, and you will get 25a; -f- 930 ^x = 450 -I- 480 ; whence x = -;;-- = 30, the same as be- o 1 Jbre, K 66 Of Equations. EXAMPLE IV. [-+^ = 16 Given. ; — 96 = 2?/ + 64, and 12a; + 122/ + 20.r — 480 = 30z/ ~ l^x + 1620; which, contracted, become Ax — 2z/ = 160, and 4i7x — 182/ = 2100: from the last of which subtract 9 times the former; so shall 11a: = 2100 — 1440 = 4fX — - 160 660 ; therefore x = 60, and y (= z=z2x — 80} = 40. EXAMPLE VI. r x + 2/ = i3^ Let < a; + 2 = 14 > ; to find ^, «/, and ; t 2/ +2 = 15 J - y ■ By subtracting the first equation from the second, in order to exterminate ac, we have z -— 2/ = 1 ; to which the third equation being added, 2/ will likewise be exter- minated, there coming out 22 = 16, or 2 = 8 ; whence y (^z^l)=i7'r and:v(=13 — j/) = 6. Qf Equations. EXAMPLE VIL Let 2 ^ 3 ^ 4 ^ y ^ A^ S 4^ 5 . — + ^ + — = 38. L4 ^ 5 6 Here the given equations, cleared of fractions, become 12x + Sy + 62 = 1488 20;^ +15y + 12zz=z 2820 30;^ + 24z/ + 202 = 4560. Now, to exterminate 2, let the second of these equations be subtracted from the double of the first ; and also the triple of the third from the quintuple of the second ; ^hence is had 4fX + y =: 156 10x + 3y=z420: from which 12^; — 10;\r = 468 — 420, and at = — =24. 2 Therefore y (= 156— 4;c) = 60 ; andz (=il?iz:2^Ili2f ) = 120. EXAMPLE VIIL r^ + ioo= y + io exterminate y. Multiply the first equation by f and the second by ^, and subtract the latter product from the former ; whence you will have bfx — agx + cfy — ahy = df — ak -, which, , ,. . . . df- — ak4-as^x — bfx by transposition and division, gives y^-^ 7 i^— . Let this value of y be now substituted in the first equa- tion, and there will arise adfx — a^kx -f- a^gx"^ — abfx'^ -f cdf — cak + cagx — cbfx bx -rz d\ which, multiplied by cf — ah^ and contracted, gives ag — - bfx !^^ + df — ■ ak + eg >-^ bh X x =■ ck '— hd* Of Equations. 69 EXAMPLE XII. Supposing ax^ +bx-\'C'= 0, and fx^ +gx + A = ; to exterminate x. Proceeding here as in the last example, we have fbx ^fc — agx — <^/i = O , and, from thence, x = 7- J—. ... ah — /?T b X ah — fc Whence, by substitution, a X — - H Ti ' ^ Jb — agJ Jb — ag 4. c = 0, This, by uniting the two last terms, and divid- 3 111 • <2/2 — fcY bh — eg ^ mg the whole by a, gives ^ ■- + -7- = O ; con- fb — a g j f^ — «.§' sequently, ah — fc\^ +fb — ag X bh — c§* = 0. After the same manner x may be expunged out of the equations ax^ + bx"^ -f rx -f- <^ = O, and fx^ -f gx + A = 0, &c. But, to show the use of the above example, sup- pose there were given the equations x^ -{-yx ^-^y^ =: 0, and x^ + 3xy — 10 = O : then, by comparing the terms of these equations with those of the general ones, ax^ + bx + c =:0y and fx^ + gx + h =z ; we have « = 1, b =z y^ c = — 2/% / =: 1, g z=: Sy^ and h = — 10 ; which values being substituted in the equation ah—Jc]^ +fb — ag X bh — c^ = 0, it thence becomes — 10 + yyy 4 y — 2>y X — lOz/ -f 3i/2 = 0, that is, 100 — 202/^ + y^ + 20?/^ — 62/"* = ; or, 100 = 5//^; whence y may be found, and from thence the value of x also. SECTION X. Of Proportion. QUANTITIES of the same kind may be compared together, either with regard to their differences, or ac- cording to the part or parts that one is of the other, call- ed their ratio. The comparison of quantities according 70 Of Proportimu ^ to their difFerences, is called arithmetical; but, according to their ratios, geometrical. When, of four quantities, 2, 6, 12, 16, the difference of the first and second is equal to the difference of the third and fourth, those quantities are said to be in arithmetical proportion. But, when the ratio of the first and second is the same with that of the third and fourth (as in 2, 6, 10, 30), then the quantities are said to be in geometrical proportion. Moreover, when the difference, or the ratio, of every two adjacent terms (as well of the second and third, as of the first and second, ££?c.) is the same, then the proportion is said to be continued: thus, 2, 4, 6, 8, &fc. is a continued arithmetical proportion ; and 2, 4, 8, 16, ^c. a continued geometrical one. These kinds of proportions are also called progressions, being carried on according to the same law throughout. Arithmetical Proportion. THEOREM I. Of any four quantities^ a, Z», c, and "so of the rest. Whence the truth of the propo- sition is mnnitest. THEOREM VIII. The sum of any number of quantities^ in continued geome- trical proportion^ is equal to the difference of the rectangle of the second and last terms ^ and the square of the first ^ di- vided by the difference of the first and second terms. For, let the first term of the proportion be denoted by rt, the common ratio by r, the number of terms by w, and the sum of the whole progression by x\ then it is manifest that the second term will be expressed by a X r^ or ar ; the third by ar x r, or ar'^ ; the fourth by ar"^ X ^, or «r^, and the ;2th or last term by ar^"'^ ; and there- fore the proportion will stand thus : a + ar + ar^ -f ar^ . . . • -f ar'^"^ + ar^"'^ = x ; which equation, mul- Of Proportion. 75 tiplied by r, gives ar + ar^ -f <2r^ + ar"^ ......+ «r^^-^ 4- ar^ = rx ; from which the first equation being sub- tracted there will remain — a + ar'^ =^ rx ---^ x ; whence /ar^ — a r X ar^"^ — a\ ar X ar^^"^ — aa X = { = I == ; as \ r — 1 r — 1 / ar — a was to be demonstrated. SECTION XL The Application of Algebra to the Resolutioji of Numerical Problems, WHEN a problem is proposed to be solved algebrai- cally, its true design and signification ought, in the first place, to be perfectly understood, so that, if needful, it may be abstracted from all ambiguous and unnecessary phrases, and the conditions thereof exhibited in the clearest light possible. This being done, and the several quantities therein concerned being denoted by proper symbols, let the true sense and meaning of the question be ti'anslated from the verbal to a symbolical form of expression; and the conditions, thus expressed in algebraic terais, will, if it be properly limited, give as many equations as are ne- cessary to its solution. But, if such equations cannot be derived without some previous operations, which fre- quently happens to be the case, then let the learner ob- serve this rule, viz, let him consider what method or pro- cess he would use to prove, or satisfy himself in, the truth of the solution, were the numbers that answer the condi- tions of the question to be given or affirmed to be so and so ; and then, by following the very same steps, only using unknown symbols instead of kilown numbers, the question will be brought to an equation. Thus, if the question were to find a number, which being multiplied by 5, and 8 subtracted from the pro- ax — h 2axb + bK 7b The Application of Algebra duct, the square of the remainder shall be 144 ; then, hav- ing put a = 5, ^ = 8, and c = 144, suppose the niimber sought to be - - - 4 (or) then 5, or a times that number 1 ^o will be J from which 8, or b^ being sub- 1 . tracted, there remains J which squared is - - 144 Therefore a^x^ — 2axb + b^ is =z c (or 144), according to the conditions of the question. In the same manner may a question be brought to an equation when two or more quantities are required. After the conditions of a problem are noted down in algebraic terms, the next thing to be done is to consider whether it be properly limited, or admits of an indefinite number of answers ; in order to discover which, observe the following rules, RULE r. When the nu7nber of quantities sought exceeds the num- ber of equations given^ the question^ for the ^ general party 7S capable of innumerable answers. Thus, if it be required to find two numbers (x and y) with this one single condition, that their sum shall be 100, we shall have only one equation, viz. x +y= 100, but two unknown quantities, x and 2/, to be determined ; therefore it may be concluded that the question will admit of innu- merable answers, RULE IL But if the number of equations^ given from the conditions of the question^ be just the same as the number of quantities s ought y then js the question truly limited. As, if the question were to find two numbers, whose sum is 100, and whose difference is 20 ; then, x being put for the greater n^timber, and y for the less, we shall have a; -f z/ = 100, and x — z/ = 20: therefore, there being here two equations and two unknown quantities, to the Re&olutnon of Problems. 77 the question is truly limited ; 60 and 40 being the only two numbers that can answer the conditions thereof. RULE III. When the number of equations exceeds the number of quantities sou^ht^ either the conditions of the problem are inconsistent one with another^ or what is proposed iii gene-- ral terms can only be possible in certain particular cases. But it is to be observed, that the equations understood here, as well as in the preceding rules, are supposed to he no ways dependent upon, or consequences of one ano- ther. If this be not the case, the question may be either unlimited or absurd, or perhaps both, at the same time that it seems truly limited \ as will appear by the follow- ing example : Wherein it is required to find three numbers, under these conditions, that the sum of once the first, twice the second, and three times the third, may be equal to a given number b ; that the sum of four times the first, five times the second, and six times the third, may be equal to a given number c ; and that the sum of seven times the first, eight times the second, and nine times the third, may be equal to a third given number d. Now, the three numbers sought being respectively denoted by X, y, and 2, the question, in algebraic terms, will stand thus : ;^^ + 2z/ -f- 32 = Z» 4:?^ + 5z/ -f- 62 = f 7x + Sy + 9z = d. Here, there being three equations, and just the same number of unknown quantities, one might conclude the question to be truly limited: but, by reflecting a little upon the nature and form of these equations, the con- trary will soon appear ; because the last of them includes no new condition but what is comprised in and may be derived from the other two ; for if from the double of the second the first equation be taken away, the value of 7x + 8y -f 92 will from thence be given = 2c -^ ^, ■ Hence it is manifest, that giving the value of 7x + Si/ + 92, in the third equation, contributes noticing to- 7^ The Application tf Algebra wards limiting the problem ; and that the problem itself is not only unlimited, but also impossible, except when d is given equal to 2c -— b. Having laid down the necessary rules for bringing pro- blems to equations, and for discovering when they are truly limited, it remains that we illustrate what is hither- to delivered by proper examples. Arithmetical Problems. PROBLEM I. To find that number^ to which 7 S being added^ the sum .shall be the quadruple of the said required number* Let the number sought be represented by - a" ; then will its quadruple be denoted by - Ax ; whence, by the conditions of the question, a; + 75 = 4^^ ; this equation, by transposing a:, becomes 75 = 3x : 75 from whence, dividing by 3, we have at = — =25, o which is the number that was to be found j for it is plain that 25 + 25 X 3 = 25 X 4 = 100. PROBLEM IL What number is thatj which being added to 4, and also multiplied by 4, the prb duct shall be the triple of the sum 7 Let the number sought be denoted by - x\ so shall the sum be denoted by - a: + 4, and the product by - - - Ax : whence, by the conditions of the question, 4x =z x -^ 4 X 3 ; that is, 4x z=z 3x + 12 ; from which, by transposi- tion, a; = 12. PROBLEM in. To find txvo numbers such^ that their sum shall be 30, and their difference 12. If X be taken to denote the lesser of the two numbers ; then, by adding the difference 12, the greater number will be denoted by -%• -f 12 ; and so we shall have 2Ar + 12 = 30, by the question. From which equation, 2.r = 30 — 12 = 18; and con- io the Resolution of Problems, 79 18 sequently, ::c = — = 9 ; whence the greater number A* {x + 12) is also given = 21. PROBLEM IV. To divide the number 60 into three such parts^ that the first may exceed the second by 8, and the third by 16. Let the first part be denoted by x ; then the second will be X — 8, and the third x — 16: the aggregate of all which, or ^x — 24 is r= 60, by the question* 84 Hence 3x = 60 + 24 = 84, and ^ = — = 28 : so that 3 28, 20, and 12 are the three parts required. PROBLEM V. The sum of 660/. xvas raised^ for a certain purpose^ by four persons^ A, B, C, and D : whereof B advanced twice as much as A i C as much as A and B ; a?id D as much as B and C : what did eachpersoyi contribute? Let the sum or number of pounds advanced by 1 A be called ' J ^ ' then will the number of B's pounds be denoted by ^x ; that of C's by - - - Zx \ and that of D's by - - 5x y the sum of all which is given equal to 660/. that is^ \\x = 660 : from whence x = = 60. Therefore 60, 120, 11 ' ' 180, and 300/. are the respective sums that were to be de- termined, ^* PROBLEM VL A certain simi of money was shared among five persons^ A, B, C, D, and E ; whereof B received 10/. less than A ; C 16/. more than B ; D 5/. less than C ; E 15/. more than D : moreover it appeared that the shares of the txvo last together xvere equcd to the sum of the shares of the other three : what xvas the whole sw7i shared^ and how mxich did each receive ? , >will be the share of <^ j^ I ' LE; 80 77ie Application of Algebra Let X denote the share of A : then^'"+ f and therefore 2x -f-17 = 3x — 4, ^z/ the question: from whence, by transposition, 21 = at ; so that 21, 11, 27, 22, and 37/. are the several required shares j amounting, in the whole, to 118/. PROBLEM VIL To find three numbers on these conditions^ that the stint of the first and second shall be 15 ; of the first and third 16 ; and of the second and third 1 7. If the first number be denoted by a; ; then it is plain; by the question, that the second will be represented by 15 — x^ and the third by 16 — x. But the sum of these two last is given equal to 17; that is, 31 — 2.v =17; whence, by transposition, 14 = 2^; ; and consequently x = 14 — = 7. Hence 15 — :v = 8, and 16 — .v = 9 ; which are 2 the other two numbers required. PROBLEM VIIL To find that number y -which being doubled^ and 16 sub- tracted from the product^ the remainder shall as much ex- ceed 100 as the required number itself is less than 100. The number sought being denoted by x^ the double thereof will b/x represented by 2x ; from which subtracts- ing 16, the remainder will be 2x — 16; and its excess above 100 equal to 2x — 16 — 100 : therefore 2x — ? 16 — too = 100 — X, bij the question; whence 3.v = 216 ; 1 216 and consequently x = -^ = 72. 3 PROBLEM IX. 7o divide the number 75 into two such parts that three times the greater may exceed seven times the lesser by 15. Let the greater part be = a: ; then will the lesser io the Resolution of Problems. 81 part = 75 — .r, and we shall have Sx — 15 = 7S*-^x X T ; or, which is the same thing, Sx — 15 = 525 — 7x : from whence 10.v = 540, and conseqilently :v = 54. PROBLEM X. Two persons^ A and B, having received equal sums of money ^ A out of his paid.away 251. aiid B of his 60/. and then it appeared that A had just truice as much money as B : what money did each receive ? Suppose .V to denote the sum received by each per- son ; then A, after paying away 25/. had x — 25 ; and B, after paying away 60/. had x — 60 : hence x — 25 = 2x — 120, by the question; and therefore 120 — 25 = 2x — .v^ that is, 95 = X. PROBLEM XL To find that number whose \ part exceeds its \ part by 12. Let the number sought be represented by x ; then will XX = 12, by the conditions of the problem ; which equation, by multiplying every numerator into all the de- nominators except its o\vn, gives Ax — 3^ = 144, that is, X = 144. PROBLEM XIL What sum of motley is that xuhose ^ party i fia^'t^ and -|- party added together ^ shall amount to 94 pounds ? If X be the number of pounds required, then will v X X 1 — I f- ^ — = 94 : from whence, by reduction, 20;\: -f 15^ + 12;f = 94 X 60, that is, ATx = 94 X 60 ; and there.- fore..v = 2 X 60= 120. PROBLEM XIIL In a mixture of copper j tin^ and lead ^ one half of the whole — 16lb. was copper ; one third of the whole — 12lb. tin ; and one fourth of the xvhole + 4lb. lead : -what quantity of each 7vas there in the composition P ' M 82 7 lie Application of Algebra Litt oc denote the weight of the whole : then will V I i <| — — 12 ^bethe weight of the y 3x '3jc = 5z ; and from thence 2 = — , and ?/ = — ; which values being substituted in the first equation, we have tjX k)X J. KjX . 1 t f 1 ax = — -I- f*. z/, or M = J but, by the fourth 4 5 20 ^ ^ 13x equation, w = x r— a; therefore x ' — a z=, -*-, an^ X = = 40 : consequently y (— ) = 30, z ( — ) = 24, and u (x — 14) = 26 ; and the whole sum (^x + y +z + u)=z 120A PROBLEM XXXIV. Tofmdfour iiumberSi such that the first together with half the second^ may be 357 («)^ the secoiid zvith ^ of the third equal to 476 (^), the tiiird with \ of the fourth equal to 595 (c), and the fourth w%th\ of t^e first equal to 714 (^). The required numbers being denoted by x, y, z, and w, and the conditions of the question expressed in algebraic terms, we have the four following equations : X -J- 2 + a. y + Z T = b. 2 + u T = c. M + 5 = d. to the Resolution of Probkms* 95 From the first whereof we get x = aj — ^ ^ ; and from the 4th, x^Sd — 5ic \ whence a — —^sd^-^Su^ and y ^2q — lOd + lOii ; but, by the second, y = b — — ; therefore 2a — 10^'+ lOw = ^ , and z = Si? 3 3 — 6a + sod — SOii ; but, by the third, 2: = c ; whence Sb — 6a + SOd -^ SOu = c ~ — , and 123 — 24a + 120^ — 120m = 4c — ic ; consequently u = 126 — 24a -f 120^/ —4c ^^^ , . u — —- = 676 ; whence z (c — — >i lly ^ 4 ^ .= 426, y(z=b — ^)=: 334, and .v (= a ~ JL) = 190. Otherxvise* Let the first of the required numbers be denoted by ^ (as above) ; then, the sum of the first and ^ the second being given equal to a, it is manifest that \ the second must be equal to a, minus the first, that is =i a — .v, and therefore the second number = 2a — 2x : moreover, the sum of the second, and \ of the third, being given = b ; It is likewise evident, that -J of the third must be equal td b^ mimis the second, that is = 3 — %a + 2x, and conse- quently the third number itself = Sb — 6a -f- 6.v : in the same manner it will appear that \ of the fourth number = c — Sb + 6a — 6x ', and consequently the fourth number itself = 4c = 123 + 24a — 24.T : whence, by tlw question^ 4c — 123 + 24a — 24,r -j = o^, and therefore 5 — 5t/ + 20c — 603 4. 120a , ^^ , ^ = ^ — -^pj^ 1: = 190 ; as above. ^ PROBLEM XXXV. To divide the number 90 (a) into four such parts ^ that if the first be increased by 5 (3), the second decreased by 4 (c), the third multiplied by 3 (/), and the fourth dz- 96 The Application of Algebra vided by 2 (^), the result^ in each case^ shall be exactly the same. Let X, 2/, 2, and u be the parts required ; then, by the question^ we shall have these equations, viz. x + y + z+u=:a^ and X -{- b =: y — c =: dz = — . e Whence, by comparing dz with each of the three other equal values, successively, x =t dz — b^ y ■=: dz -f c, and It = dez'y all which being substituted for their equals, in the first equation, we thence get dz — b-\-dz+c + z+ dez = a ; whence dez -f 2dz + z z=z a 4- b — c, and z = — = -—, = 7. There- ^ ' de + 2d+l fore X (^z=: dz — b) = 16; y (= dz + c) = 25 ; and ?^ ( = dez ) = 42. PROBLEM XXXVL If A a7id B together^ can perform a piece of tvork in 8 (a) days; A aiid C together in 9 (J?) days^ and B and C i/z 10 (c) days; how many days will it take each person^ alone^ to perform the same work ? Let the three numbers sought be represented by x^ 2/, and 2, respectively: then it will be, as x (days): a (days) : : 1, the whole work, : — , the part thereof performed by A in a days ; and, as z/ : a : : 1 : — , the y part performed by B, in the same time ; whence, by the question^ f- — - == 1 (the whole work). And, by X ^ y proceeding in the very same manner, we shall have , • , . . b b ^ these two other equations, viz. -— + — = 1> and c c 1 = 1 : let the first of these three equations be y ^ divided by <2, the second by bj and the third by c^ andl vou will then have to the Resolution of Problems ». 9r L + L-JL X y a JL i- — -1 X X b"* y z c which added all together, and the sum divided by 2, give 1 1 = -r-H — r-l ; fro«i whence each of X y z ^a^ 2b^2c the three last equations being successively subtracted, we get JL — — JLo-i. J. 1 _ — be + ac + ab _£_.J L-lJ — ^^~ ^c+^ y 2a 2b 2c 2«^c '^ 1 1 1 1 be + ac — ab ^r — =: — J- — — — — : ; . Hence X 2a 2b 2c '^abc 2abc 1440 ^^ , ^ be -{• ac + ab — 90 + 80 + 72 ^'' ^ _ 2abe _ 1440 —1723 •^ "" be — ac + ai^ ~ 90 — 80 + 72 "■ ^"^' 2<^^c 1440 ^^,^ /?>c- + ac — ab 90 + 80-/2 ^ ^' Otherwise* Let the work performed by A in one day be de- noted by X : then his work in a days will be ax^ and in b days it will be 7»x ; therefore the work of B in a days will be 1 — ax y and that of C, in b days, 1 — bx^ by the conditions of the problem ; whence it follows that the work of B, in one day, will be expressed by 1 ' '— ax ' 1 — — bx "— ^, and that of C, in one day, by ; but the sum of these two last is, by the question^ equal to — 1 1 1 part of the whole work, that is, (- — -— 2^^ = — ; a b c O 98 The Application of Algeh. ra , 1.1 1 be 4- ac — ab - ' whence xz=z^+ — — — =z _-- , equal to the 2a 2b 2c 2abc . ^ work done by A in one day ; by which divide 1 (the 2abc whole), and the quotient, :, will o-ive the re- bc + ac — ab ° quired number of days in which he can finish the whole. PROBLEM XXXVII. 7b fi7id three numbers on these conditions^ that a times the first ^ b times the second^ and c times the third^ shall be equal to a given member p ; that d thnes the first^ e times the second^ and f times the thirds shall be equal to another given number q ; and that g times the ftrst^ h times the second^ and k times the third^ shall be equal to a third given 7iumber r. Let the three required nun\bers be denoted by .r, y, and z-i and then we shall have ax + bi/ + cz = j&, dx + eij -}-fz = q^ gx + hi/ + y^2 = r. From d times the first of which subtract a times the se- cond, and from g times the first subtract a times the third, and you will have these two new equations, {bdy — aey + cdz — afz -^^ dp — aq^ bgy — ahij + cgz — akz ^=^gp''^ ar ; or, which are the same, bd — ae X y + cd — afx z =: dp — aqy and, bg — ah X y + cg — ak X z ^gp — ^^'• Multiply the first of these two equations by the coeffi- cient of y in the second, and vice versa^ and let the last of the two products be subtracted from the former, and }'0u \yill next have en — qf- bg — ah X z — bd — ae X eg — a^ X Z =: bg — ah "■■ op — aq — bd — ae X gp — ar; and therefore z = ^.y- -' x W^^^ -bd-ae xJF^^ . ca — afi X bg — ah — bd — ae X eg — ak whence .r and y may also be found* to the Resolution of Problems* 9^9 ^ocample. Let the given equations be X •\' y + 2 = 12, Sat + 3z/ + "^^ = 38, 3;^- + 6z/ + 102 = 83 ; Or, which is the same thing, leta=:l,^ = l,c=:l,/7=12, ^= 2, e = 3, /= 4, y = 38, ^ = 3, A = 6, i = 10, and r = 83 : then these values being substituted above in that . 3 — 6 X 24 — 38 — 2 — 3 X 36 — 83 of 2, It will become === — ----- - 2_4X 3 — 6 —2 — 3X 3 — 10 ::::: HT = z=. 5 X whence, also, we find 6 — 7—1 dp — aq — cd — afxz ^ _ 24 — 38 — 2 — 4X5 ^^"~ dd—ae ^ "^ 2 — 3 12 — 4. = Zlf = 4, and ;. C= LZi^-^^ — 1 ' a . 1 = 3. Having exhibited a variety of examples of the use and application of algebra, in the resolution of problems producing simple equations, I shall now pi'oceed to give some instances thereof in such as rise to quadratic equa- tions ; but, first of all, it will be necessary to premise something, in general, with regard to these kinds of equa- tions. It has been already observed, that quadratic equations are such wherein the highest power of the unknown quantity rises to two dimensions ; of which there are two sorts^ viz* simple quadratics, and adfected ones, A simple quadratic equation is that wherein the square only of the unknown quantity is concerned, as xx = ab ; but an adfected one is, when both the square and its toot are found involved in different terms of the same equation, as in the equation x^ + 2ax = bh. The re- solution of the first of these is performed by barely extracting the sqviare root, on both sides thereof: thus, in the equation x^ = ah^ the value of x is given = \ ab (for, if two quantities be equal, their square roots must necessarily be equal). The method of solution, when the equation is adfected, is likewise by extracting the 100 The Application of Algebra square root ; but, first of all, so much is to be added to both sides thereof as to make that where the unknown quantity is a perfect square ; this is usually called com- pleting the square^ and is always done by taking half the coefficient of the single power of the unknown quantity in the second term, and squaring it, and then adding that square to both sides of the equation. Thus, in the equation x$c + 2ax = bb^ the coefficient of X in the second term being 2a, its half will be a^ which, squared and added to both sides, gives x"^ + 2ax -}- a^ z= b^ + a^ ; whereof the former part is now a perfect square. The square being thus completed, its root is next to be extracted ; in order to which, it is to be observed that the root, on the left-hand side, where the unknown quantity stands, is composed of two terms or members ; whereof the former is always the square root of the first term of the equation, and the latter the half of the coefficient of the second term : thus, in the equation, x^ + 2ax -f- a^ == b^ -f- a^, beforf us, the square root of the left-hand side, x'^ -f 2ax + cr.^ will be expressed by x -}- a (for x + a X x -j- a =i x^ + 2ax -f a^). Hence it is manifest that x + a =: S/b^ + a^, and therefore x = V/6^ -f- c^ ' — a; from which X is known. These kinds of equations, it is also to be observed^ are commonly divided into three forms, according to the different variations of the signs : thus x^ -f- 2ax = Z>^ is called an equation of the first fori^^i ; ^2 — 2ax = b^ one of the second form ; and x^ — 2ax zzi — b^ one of the third form ; but the method of extracting the root, or finding the value of .t, is the same in all three, except that, in the last of them, the root of the known part, on the right-hand side, is to be expressed with the double sign ± before it, x having two different affirmative values in this case. The reason of which, as well as of what has been said in general, in relation to these kinds of equations, will plainly ap- pear, by considering, that any square, as x^ — 2ax + ^r^ raised from a binomial root, x — a (or a — - a;) is composed of three members ; whereof the first is the square of the first term of the root ; the second, a rect- to the^ Resolution of Problems > 101 angle of the first into twice the second ; and the third, the square of the second : from whence it is manifest, that, if the first and second terms of the square be give'n or expressed, not only the remaining term, but the root itself, will be found by the method above delivered. But now, as to the ambiguity taken notice of in the third form, where x^ — 2ax = — ^^, or x^ — %ax + a^ = a^ — ^2 . xhQ square root of the left-hand side may be either x — a, or a — x (for either of these, squared, produces the same quantity) ; therefore, in the former cose, xz=La+ Va^ — /y^, ?jid, in the latter, x = a — s/d^ — h^ ; both which values answer the conditions of the equation. The same ambiguity would also take place in the other forms, were not the root (at) confined to a positive va- lue. When the highest power of the unknown quantity hap- pens to be affected by a coefficient, the whole equation must be divided by that coefficient ; and if the sign of that power be negative, all the signs must be changed before you set about to complete the square. All equations whatever, in which * there enter only- two different dimensions of the unknown quantity, where- of the index of the one is just double to that of the other, are solved like quadratics, by completing the square : thus, the equation x^ + 2ax'^ = b^ by completing the square, will become x^ + 2ax'^ •{- a^ = b + a^ -, whence, by extracting the root on both sides, x^ + a = Vb -{- cr ; th erefore y ^ =z V b + a^ — «, and consequently x = ^Vb + a^ — a. These things being premised, we now proceed to the re- solution of problems. PROBLEM. XXXVIII. To find that number^ to which 20 being addcd^ and from which 10 being subtracted^ the square of the sum^ added to txvice the square of the remainder^ shall be 17475. Let the number sought be denoted by x ; then, by the 102 The Application of Algebra co nditions of the question, we shall have x + 20*]^ + 2 X oc — lOl^ = 17475 ; that is, x^ + 40a: -f 400 + 2a:^ .— , 40x + 200 = 17475; which, contracted, gives %x^ = 168 75. Hence x^ = 5625 ; and, consequently, x ^ V5625 = 75. PROBLEM XXXIX. To divide 100 into txvo such parts ^ that^ if they be mul- tiplied together y the product shall be 2100. Let the excess of the greater part above (50) half the number given, be denoted by x; then 50 + a: will be the greater part, and 50 — x the lesser ; therefore, by the question^ 50 + x X 50 — x, or 2500 — x^ = 2100 ; whence x^ = 400, and consequently x = V400 3= 20; therefore 50 + x = 70 = the greater part, and 50 — x = 30 = the less. PROBLEM XL. What two 'numbers are those^ which are to one another in the ratio of^ (a) to 5 (<^), and whose squares^ added toge- ther, make 1666 (c).^ Let the lesser of the two required numbers be x ; then, a : b : : X : — = the greater; therefore, by a the qxiestion, x^ -{ = c ; whence a^x^ + b'^x^ = aV, a^c , / X and x^ = -~ — --; consequently .v — ^ „ a^ + b^ ^ ^ ^ a^ + b^ f c bx ^ K^'iT' — r = 21 = lesser number, and — = 35 :=: ^ a^ ^ 6^ a the greater. PROBLEM XLL To find two numbers^ whose difference is 8, and product 240. If the lesser number be denoted by .v, the greater will be A' + 8 ; and so, by the question, we shall have x'^ + 8.t tQ the Resolution of Problems* 103 = 240. Now, by completing the square, x^ + %x + 16 (= 240 -f 16) = 256; and, by extracting the root, X + 4< = V256 = 16 : whence x = 16 — 4 = 12 ; and AT + 8 = 20 ; which are the two nijmbers that were t*o be found. PROBLEM XLII. Tc find two numbers whose difference shall be 12, and the sum of their squares 1424. Let the lesser be a:, and th en the gr eater will be .r + 12 ; therefore, by the problem^ x + 12*)^ + x'^ = 1424, or 2x^ + 24.V + 144 = 1424; this, ordered, gives x^ -f \2x = 640 ; which, by completing the square, becomes x^ + 12;v + 36 (= 640 + 36) = 676 ; whence, extract- ing the root on botli sides, we have .v + 6 = (V676) 26 ; therefore x = 20, and ;^^ + 12 = 32, are the two numbers required. For 1-2-20 =12, ^"^ 132^+ 202=1424. PROBLEM XLIIL To divide 36 into three such pai'ts that the second maif exceed the first bij 4, a7id that the sum of all their squares may be 464. Let X be the first part, then the second will be x + 4 ; and, the sum of these two being taken from (36) the whole, we have 32 — 2x^ for the thircl, or remaining part ; and so, by the question^ x'^ + x + ^ + 32 — 2xY = 464, that is, e>x^ — 120;^- + 1040 = 464 ; whence 6^ _ 120:v = — 576, and x^ — 20x = — 96. Now, by completing the square, x^ — 20a: + 100 (= 100 — 96) = 4 ; and, by extracting the root, x - — 10 = ^ 2. Therefore .r = 10 q: 2, that is, ;c = 8, or ;\r = 12 ; so that 8, 12, and 16 are the three numbers require^. PROBLEM XLIV. To divide the number 100 (a) into two such parts that their product and the difference of their squares 7nay be equal to each other. 104 The Application of Algebra L^t the lesser part be denoted b y x\ t hen the greater will be a — ^oc^ and we shall have a — at x a. = a — ~x^ — x^^ that is, ax^^x^ :=: (^ ' — 2«.r ; whence x^ •— 2>ax =: — a^ . and, by completing the square, x^ — Zcpc + — = ( — "^ "^-^ "dT ' ^ which the root being extracted, there comes out x ~ ± \/-_^, and 2 ^4 therefore ;^ = — ± W_^. But at, by the nature of the problem, being less than «, the upper sign (+) gives X too great ; so that a; = -^ y— ^ = 38,19658, fcPc. must be the true value required. PROBLEM XLV. The sum^ and the sum of the squares^ of two numbers be- ing given ; to find the numbers. Let half the sum of the two numbers be denoted by «, half the sum of their squares by ^, and half the difference of the numbers by x ; then will the numbers themselves be represented by a — at, and a •\- x^ and their squares l^y ^2 — 2flfx + x^^ and c^ + 'Hax + a:^ ; and so we have c^ — 2<2Ar + x^ + c^ + "^ax + x^ =z 2^, bi/ the ques- tion. Which equation, contracted and divided by 2, gives a^ ^ x^ ^ b ; whence x^ z= b — a-, and consequently x = V^ — a^- Therefore the numbers sought are a — Vb — a^, and a + Vb — a^» PROBLEM XLVL The sumy and the sum of the cubes of two numbers being given ; to find the numbers* Let the two numbers be expressed as in the preceding problem, and let the sum of their cubes be denoted by c. I'herefore will a — .r]'' + a + xf = c, that is, by involution and reduction, 2a^ + 6ax^ = cj whence to the Resolution of Problems • 105 iQUX^ ■=• c — 2«^, vT^ = "T — = , and .v ==: 6a 6a 3' PROBLEM XLVII. The sum^ and the sum of the biquads'ates {or 4thfowers) of two numbers being given ; to find the numbers. The numbers being denoted as above, we shall liere have a — x\ + a -f a;]^ = ^, that is, Sa^ + I'la^x^ + 2;c^ = d\ from which, by transposition and division, x^ + ^(j^x^ = i^ — «"* ; and, b}'- completing the square, x^ -f. 6a2x2 -f- 9a4 = |^ + Sa^ ; w hence x '^ -f Sa^ = V^^ + 8a^ ; and, consequently, x = \ — Sa^ + \^\d + 8a^ PROBLEM XLVIII. The swn^ and the sum of the 5th powers of two numbers bei7ig given ; to find the numbers* The notation in the preceding problems being still retained, we shall have 2a^ + SOa^;^^ + \Oax'^-=.e\ and e o^ therefore x^ -f ^c^x^ = ; and x'^ -f- ci^ = 10a 5j_ ^ v/ 1 J whence ^ = \f V f- a^. ^\Qa S ^ lOa^ 5 PROBLEM XLIX. What two numbers are those^ whose product is 120 {a)^ and if the greater be increased by 8 {b\ and the lesser by 5 (c), the product of the txvo numbers thence arising shall be 300 (d) P If the greater number be denoted by ^, and the lesser by t/, we shall have xy = a, and X -j-b X y -i- c = //, by the conditions of the question- Subtract the first of these equations from the second, and you will have x + b x y -{- ^ — xy = d — a, that isy ex + by + be = d — a; where both sides being multi- P 106 The Appltc-^ xx + a — x ■=. bb -, whence, by transposition and division, \^ax — xx ^=' — — — : therefore, by squaring again, ax — xx = L or x^ — ax z=: ( L-) = L b^a a" , a ^ \ 2ah' ~ b'' a ^ — — , and .r = -— - + v = — 4- 2 4' 2^>4 2^ -— V2a — <^2 __ 54 _. |.|^^ greater part ; whence a — .r ==: 36 = the lesser part* PROBLEM LII. A grazier purchased as mayiij sheep as cost hrni 60/. bul: ofzvliich he reserved 15^ and sold the remainder for 54/. and gained tiw) shillings a-head by them : the question is, hoxv jnamj sheep did he buy^ and xvhat did they cost him a- liead? Let the number of sheep be x\ then if 1200, the number of shillings which they all cost, be divided by at, the quotient, , will, it is evident, be the number X of shillings which they cost him a-piece ; and so the number of shillings they were sold at per head will be [- 2, by the question ; and therefore this, mul- tiplied by x — 15, the number of sheep so sold, will giye 1200 + 2.V 30, equal to the whole number X @f shillings which they were all sold for; that is, 11/0 + 2:^ — .1522- = 1080: hence we have liro.r + 2a^ X — 18000 = 1080;i\ 2x'^ + 90.V = 18000, x^ + 45.r =r 9000, and x = V9506.25 — 22.5 ■=. 75, the number of 1200 sheep ; and conseqnentlv rr^ 1 6 shillings, the price ©f / 5 each. 108 The Application of Algebra PROBLEM LIIL Trvo country 'Xvomen^ A and B, betwixt them^ brought 100 (c) eggs to market; they both received the same sum for their eggs ; but A, who had the largest and best^ says to B, Had I brought as many eggs as you^ I should have received 18 (a) pence for them; butj replies B, had I brought no more than you^ I should have received only 8 (hi) pence for mine: the question is^ to fnd how many eggs each person had. If the number of eggs which A had be = x^ the number of B's eggs will be = c — x ; therefore, by the problem^ it will be, c — x i a i x x \ ^ = the num- ber of pence which A received ; and as .r : ^ : : c — x i b X c — X X the number of pence which B received : ax b ^ c I X whence, again, by the problem^ — '■ — = ; and c •— — X X therefore ax'^ = b x c — xY = bc^ — 2bcx + bx^ ; ^bcx bc^ which equation, ordered, gives x^ •\ — ^- — ■, = • ; a — o a —— b from whence x comes out ( = v > + ^ ^ a — b .^ ) ^ . — = 40. But the value of x may a — b a — b be otherwise more readily derived from the equation ax^ = b X c — xY', without the trouble of completing the square ; for the square root being extracted on both sides thereof, we have xV a =c — x X \^ b ; whence x\/a + x\/b=zcx/b^ and consequently x _ c Vi \/ a + '\/ h 100^/8 100 V 4 .^ , r : = = 40, as before* V 18 + V8 \/9 +\/4 "^ to the Resolution of Problems. 109 PROBLEM LIV. One bought \20 pounds of pepper^ and as inaJiy ofgingei^ and had one pound of ginger more for a crown than of pep- per; and the whole price of the pepper exceeded that of the ginger by six crowns : how many pounds of pepper had he for a croxvuj and how many of ginger ? Let the number of pounds of pepper which he had for a crown be x^ and the number of pounds of ginger will be ;^ + 1 ; moreover, the whole price of the pep- .„ , 120 ,1 r i_ • 120 per will be — crowns, and that ot the gmger ; '^ X X + X therefore, by the question^ — — = 6 ; whence ' ^ ^ ^ X X '\-l 120a: + 120 — 120.r =: 6:^^ + 6^, and therefore x^ + x = 20 j which, solved, gives ;r = 4 = the pounds of pep- per, and X + 1 z=z 5 = those of ginger, PROBLEM LV. To find three numbers in arithmetical progressio7i^xvhere- of the sum of the squares shall be 1232 (a), and the square of the mean greater than the product of the two extremes by 16 (^). Let the mean be denoted by x^ and the common dif- ference by y ; then the numbers themselves will hQ x — i/, AT, and X + y \ and so, by the problem^ we shall have these two equations : X — yl -f y ^ + X 4- 2/^ = a^ and x^ •=^ X — y X X + y + b: these, contracted, become Sx^ + 2y^ = a, and x^ = x^ — y^ + b ; from the latter whereof we get y^ = b = 16; and consequently y = V ^ = 4 ; which, substituted for y in the former, gives Zx^ •}- 2b =z a; whence x^ = ^ , and therefore F .r = y — — = 20 ; so that the three required numbers 3 are 16, 20, and 24. For / ^^^ + ^^^ + 24^ = 1232. "'"^'^ t202--16 X24 =16. 1 1 .0 7 '// e Application of Algebra PROBLEM LVL To fmd two numbers whose difference shall he 10 (a), and if ^00 (Ji) he divided by each ofthem^ the difference of the quotients shall also be equal to 10 («). The lesser number being represented by :c^ the greater will be represented by ^ + a; and therefore^ by th-: pro- blem^ = a ; which, freed from fractions, X oc -^ a gives hx + ha -^ bx •=. ax^ + a^x^ that is, ba = ax^ + a^x y whence, dividing by a, and completing the square^ we have x^ + ax •}- \a^ :=. b + \a^ ; therefore x + ^a:=: \/h ^ i-a^, and consequently x = \^b + \(j^ — i^ = 20, the lesser number : whence x •\-a^=. 30, the greater num- ben PROBLEM LVIL To find two numbers whose sum is 80 («), and if they bt divided alternately by each other ^ the sum of the quotients shall be ^ (b). If one of the numbers be .r, the other will be « — Xj X a t X and we shall therefore have -f- = b : which a — X X equation, brought out of fractions, becomes x'^ + a^ — ^ax + x"^ = abx — bx^ ; and this, by transposition, gives 2x'^ + bx^ — 2ax — abx = — a^, that is, 2 + 6 X c^'^ — 2 -f. 6 X cix = — a^ ; whereof both sides being divided by 2 + 6, we have x^ — ax = ■ — a^ ; whence, by completing the square, x^ — ax -f. a^ a^ ft2 p ^— — = , ; hence x — la = ± V j, 4 4 2 -\- b ^ 4< a + b^ if m fl ft and X ^ — ± V— — = 60, or = 20 ; which two 2 > 4 2 + /^ ' are the numbers that were to be found. PROBLEM LVIIL , To divide the number 134 {a) into three such parts ^ that once the ^first^ txvice the second^ and three times the to the Resolution of Problems. Ill r/iird^ added together^ may be = 278 (^), arid that the su?7z of the squares of all the three parts may be = 6036 (c). Let the three parts be denoted by x^ i/, and 2, respec- tively ; then, from the conditions of the problem, we shall have these three equations : X + y + z = a^ X +2y + 3z = b^ x^ -{- y^ + z^ =z c. Let the first of these equations be subtracted from the second, whence y +2z = b -^ a^ or y = b — cr — Sz; also, if the double of the first be subtracted from the second, there will come out z — x z=z b — 2dz, or x ^z -f 2a — b: wherefore, iifho^ put =z b — a (= 144), g = A — 2a (= 10), and for y and .r, their equals/ — 2z and z — gj be substituted, our third equation, x^ + y^ -[- 2;^ = c, will become zz — 2^2 + gg + ff — 4/2 4- 422 + 22 = c ; which, ordered, gives 2^ — ^^ "^'^ . X 2 = ^' ^- ; whence, by putting h = cyf t p. 298 ^ "^ ^ ( = . — ), and completing the s.quare, &f c. 2 is h Ic — p — ^ ^2 149 1 >0 : therefore y {=■ f — 22) = 44, and .v ( = 2 — g) ^ '0. PROBLEM LIX. A traveller sets out from one city B, to go to another C, at the same time that another traveller sets out from Qfor B ; they both travel uniformly^ and in such proportion that the former^ four hours after their meetings arrives at C, and the latter at B, in nine hours cfter : now^ the question is to find 'n how many hours each person performed the journey. D B 1 . C Let D be the place of meeting, and put a = 4, /i = 9, and X = the number of hours they travel before they meet : then, the distances gone over, with the same uni- form motion, being always to each other as the times in 112 The Application of Algebra which they are described, we therefore have, BD : DC I I X (the time in which the first traveller goes the distance BD) : « (the time in which he goes the distance DC) ; and, for the same reason, BD : DC : \ b (the time in which the second goes the distance BD) : x (the time in which he goes the distance DC) : wherefore, since it appears that x is to a in the ratio of BD to DC, and b to X in the same ratio, it follows that x i ax : b \ x; whence oc^ = ab^ and x =z Vab (= 6) ; therefore a + Vab =10, and b + \'^ab =15, are the two numbers required. PROBLEM LX. There are four numbers in arithmetical progression^ xvhereofthe product of the extremes is 3250 («), and that . of the means 3300 (Ji) : what are the numbers? Let the lesser extreme be represented by z/, and the common difference by x ; then the four required numbers will be expressed hy y^y + x, y + 2x^ and z/ + 3x : there- fore, by the question j we have these two equations, viz. y X y + 3a% or 2/2 + 3xy = a, and y ^ X X y + 2x^ or 2/2 + Sxy -f 2x^ = b ; whereof the former being taken from the latter, we get 2x^ = b — a: and from thence x = j^ _II1~. = 5 But, to find y from hence, we have given y^ + 3xy = a (by the first step) ; therefore, by completing the square, £5fc. y = a 4- — . = 50 : and so the four numbers are 50, 4 2 35^ 60, and 65. PROBLEM LXL 4 The sum (30) and the sum of the squares (308) of three numbers in arithmetical progression being given ; to find the numbers. Let the sum of the numbers be represented by 3^, .the sum of their squares by c, and the common diffe- rence by X : tlien, since the middle term, or number, to the Resolution of Problems* 1 1<3 from the nature of the progression, is = b^ or | of the whole sum, the least term, it is evident, will be ex- pressed hj b — :r, and the greatest by b -^ x \ and therefore, by the queMwn^ we have this equation, b — v"]2 + b^ + b + x\^ = c; which, contracted, gives 3b^ + 2:>c^ = c ; whence 2x^ = c — ob^^ and x=: JiZZ^ = ^' Therefore 8, 10, and 12, are the three numbers sought. PROBLEM LXII. Hmnng giveii the sitm (^), and the sum of the square^ (c), of any giueri number af terms in arithmetical progreS' sion; to find the progression. Let the common difference be ^, the first term x -f e, and the number of terms n : then, by the question^ we shall have X '{' € + X -\- Ze -\- X '\' :^e X A- ne ^ b^ and X 4- e'Y^ X -f '^e^-^ x -|- Se]^ x -f- ne]^ = c. But (by section 10, theo. 4) the sum of the first of these progressions is nx -j — 1— '- — : and the sum of the second (as will be shown further on) is = 7ix^ + 7i»n+ 1 . xe + -^- :: ■ : therefore oirt^ two equations will become nx -f . i- = b, and 2 71 -4- 1 . 2n -f- 1 . e?^ nx^ + 21 • n -{- i . xe + :•- = o o Let the former whereof be squared, and the latter muK ti plied by 7Z, and we shall thence have nV + 7z2 .'^T+T.xe + ^.^'•^^ + 0''"' ^ ^2^ and 4 '2 W2 -f- 1 . 2'i -h 1 • C n^X^ H> 72^ . n + 1 . ;c^ + ^\'^^^^'r!.J^ =: ^^C 114 7 lie Applkatioji of Algebra let the first of these be subtracte d from the second, so shall — T • = nc—b^* But ^^^ ' ^ + ^ ' ^^ + 1 __ y^^ » y^ + 1] M s =?i^ ,¥Ti X 6 4 2;z -f- 1 72 + 1 2 r-T V 8^i + 4^ — 67Z — 6 6 4 ^ 24 72 + 1 . 2/2 2 __ ^^^ • ^ + 1 • ^2 — 1 ^^^ • 72^ — ^ 1 : 24 "^" "^12 12 Therefore ^i-l-^^^ •-^~ = ?2c — b\ and ^ = lt:^nc —12¥ (b 72 +1 . g \ \ -J . ; whence .r ( -; — — / is known- Example : Let the given number of terms be 6, their sum 33^ and the sum of their squares 199 ; then, by writing these numbers, respectively, for 72, ^, and c, we shall have e = 1 ; whence .v = 2, and the required numbers 3, 4, 5, 6, 7, and 8. PROBLEM LXIIL Two post-boys^ A ^tz^ B, set oiit^ at the same time^from tivo cities 500 7niles asunder^ in order to 7neet each other : A rides 60 7niles the first daij^ 55 the second^ 50 the third^ and so on^ decreasing 5 miles every day : bjit B goes 40 miles the first day^ 45 the second^ 50 the third^ &c. increasing 5 miles every day ; 7ioxv it is required to find in xvhat number of days they xvill meet. In order to have a general solution of this problem, let the first day's distance of the post A be put = m^ and the distance which he falls short each day of the preceding = ^y also the first day's distance of the pqst B = /;, and the distance which he gains each day :=. e ; and let x be the required number of days in which they meet : then the whole distance travelled by A will be expressed by the following arithmetical progression : m + m — d + m — 2d + m — 3d^ &c. and that of B by to the Resolution of Problems. 1 1 5 p+p + e+p + 2e+p + 3^, &c. where each pro- gression is to be continvied to x terms. But the sum of the first of these progressions {by Sect. 10, ^/zeor. 4) is = '^ix — ^iLlZliil-., and that of the second = px + ^' ^ ^ -— : therefore these two last expressions, add- ed together, must, by the conditions of the question, be equal to 500 miles, the whole given distance ; which we will call ^, and then we shall have p -^ r n X x + XXX — IX e — d , ^ ^ g-x X x—1 , , 5 = ^ orfx + ^ = b, b) writing y= p + w, and ^ = . y^, o r the dis- tance travelled by B, will appear to be ^ — — liJt— miles which A travels, will be x X ni + therefore, by the questio7ij we have 6 \ ?: X -^'f" \ X e z=, rnx + — ; which, divided by at, and con- . .A • 2;c2 4- 3^ -^ 1 ^ ex -h e , tracted, gives = m -f ; whence , , 3x Sex „ . 3^ 1 J I :ir + — — = Sm + — ; and, by com- 2 2 2 2 ' -^ pletmg the square, ^^+-~- ^T+T^ — "7^ + 2 2 lo Id 9e^ , Se 1 9 18^ 9^2 - (= 3.^ + ^ - — + ^ --^ + l6 = 48m 4- 1 4- 6c' -f 9e^\ _ 48m + 1 -^ Se f , 16 / "" 16 ' w enc^ tD the Resolution of Problems. 1 1 f /i + -. — -, = -I— L, and X = 4 * "* required. PROBLEM LXV. The sum of the squares (a), and the continual product (b) of four numbers in arithmetical progression being given ; to find the numbers. Let the common difference be denoted by 2^, and the lesser extreme by 2/ — 3x ; then, it is plain, the other three terms of the progression will be expressed by 2/ — x^ y + x^ and y + 3x^ respectively ; and so, by the question^ we have y — Z'Y + y — x' Y 4- fj -f X p 4> y -f- 3>x'Y = «^, ani y — ctx X y — xxy + xxy + ^x=:b^ that is, by reduction, 4z/2 + 20:^2 = a, and y^ — XOy^:^^ + 9x'^ = b ; from the former of which y^ =. \a — Sx"^ : and there- fore y^ = \a^ — I ax^ + 25;t^^ : these values being sub- stituted in the latter, we have -^^a^ — ^ax^ + 25^;^ 5ax^ — lax^ + 50;^^* + 9x^ = b^ and therefore x"^ r= ; whence, by completint? the square, 84 16 X 84 ^ ^ 16 ^5 ^4 _Sa^ 25£ r = A 4. _J!L_>) =. 84 4 X 84 X 84 ^ 84 84 X 84'' -~ ; therefore x^ = —^ ■ — , and x =r 84 X 84 2 X 84 84 ' 4. 5a' 2\/8U 4. a^ 168 known. ; whence y (= V^a — S.r^) is also 118 The Application of Algebra PROBLEM LXVI. The difference of the means {a)^.and the difference of^he extremes {F) of four numbers in continued geometrical pro- portion being given ; to find the numbers. Let the sum of the means be denoted by x ; then the gi'eater of them will be denoted by ^ "^ ^ , and the lesser by — — - : whence, by the nature of proportionals, it At .„ - X •\' a X — a X — a x — clV will be — -— : — -- — : : — — ~ : -L., the lesser 2 2 2 2x + 2a ' X — a .r 4- a x ■\' a x -^^ a^ , extreme, and : — • — : : — ! — : — — — L , the '22 2 2;v— 2a' greater extreme : therefore, by the problem^ we have X -^ (lY ^ — «T 7 1 ^1 T — 13 ! L =r h ; and consequently at -f a I — 2x — la 2x - f 2a ^ ^ ' X — aY =^2b X X --^a X x + a^ that is, 6x^a + 2a^ = — — — 1)0^ -1- a^ 2b X ^^ — a^ ; whence x^ = -7 — -~p-> and consequently '3a .^' = aJ ^+ a ■ Sa PROBLEM LXVIL The sum^ and the sum of the squares of three numbers in geomet7'ical proportion being given ; to find the numbers. Let the sum of the three numbers be denoted by a, and the sum of their squares by ^, and let the numbers themselves be denoted by x^y^ and 2 : then we shall have X +y +z =:a, x'^ + y^ +z^=zby and xz = z/^. Transpose .2/ in the first equation, and square both sides, so shall x"^ + 2xz + 2^ = a^ — 2ay + y^ ; from whei>ce» subtracting the second equation, we have 2^2 — 2^2 ^ ^^2 — 2ay + y^ — b: but, by the third, 2xz ^ 2?f ; therefore y- z=^ e^ — 2ay + z/^ — b\ and conse- to the Resohilion of Problems* 119 quently y = -^- = ~ - ^. ^ow, to find -x and 2, t/ may be looked upon as known ; and so, by the second eqviation, we have given x" + z^ = b — z/^ ; from whkh subtracting 2xz = 2z/*^, there arises x^ — 2xz + z^ =:b — 3y^ ; where, the square root being extracted, we have cc — 2 = Vb — 5y^ : but, by the first equation, we have x + z = a — y ; whence, by adding and sub- tracting these last equations, ther e results 2 x :=z a — y + Vb •— oz/z/, and 22 = a — t/ — Vb — Si/y. PROBLEM LXVIIL The sum Qs)^ and the product (p)^ of any two nujyibers being given; to find the sum of tlie squares^ cubes^ biqua- drates^ &c. of those nu?nbers. If thfe two numbers be denoted by x and y ; then Avill .V + z/ = 5 j , the problem, and xifc:ip j ^ ^ The former of which, squared, gives xx + 2xy + yy^=i s^ ; from whence subtracting the double of the latter, we have x^ + 2/2 = ^^ — 2/?, the sum of the sqitares. Let this equation be multiplied by x + z/ = 5 ; so shall x^ +xy X X + y + y^ = s^ — 2sp^ that is, x^ +p x s + y^ =: s^ — 2sp (because xvj = /?, and x + y =z s) ; and there- fore x^ + 2/2 _- ^3 — 3^^^ f/i^ ^^;;^ of the cubes. Multiply, again, by .v + ^ = ^ ; then will x"^ + xy X x^ + y'^ + 2/4 3= ^4 — 3^2^^ ^^ x^ + p X A>2 — 2p + yA __ ^^4 — 3^2^ (because x"^ + 2/^ = ^2 — 2p). Conse- quently AT'* + 2/* == 5^ — 45'2j& + 2/j^, tlifC sum of the biqua- drates. Hence the law of continuation is manifest, being such, that the sum of the next superior powers w^ill be always obtained by multiplying the sum of the powers last found by 5, and subtracting, from the product, the sum of the preceding ones multiplied by/?. And thus the sum of the ;ith powers, expressed in a general i^anner^ wfll be S"" '^ 72.S ^-2^ + n • ^i^^ . s'^"'^ p^ — n . ^-^^^. The Appficatioii of Algebra PROBLEM LXIX. The sum of the squares (a), and the excess (^) of the product above the sum of two numbers being given ; to find the numbers* Let the sum of the numbers be denoted by ^, and their product by r ; then the sum of their squares will be 5.2 . — 2r {by the last problem)^ and we shall have r — « = b^ and 5^ — « 2r = a, whence, by adding the double of the former equation to the latter, s^ — 2^ = a+2b-y and consequently s-=:\/a-\-^b'^\ +!• From which r (z=zb + s) is likewise known ; and from thence the numbers themselves. PROBLEM LXX. The sum {a) ^ and the sum of the squares (b) of four numbers^ in geometrical progression^ being given ; to find the nu7nbers. If X and y be taken to denote the two middle numbers, the two extreme ones, by the nature of progressionals, x^ Xp" will be truly represented by and -^, y X Put the sum of the two means = s^ and their rect- angle = r; so shall the sum of the two extremes (XX yy\ + ^ i be = a — 5, and their rectangle also = r (by the nature of the question). But (^by problem 68) the sum of the squares of any two numbers whose sum is s^ and rectangle r, will be = ss — 2r ; and, for the same reason, the sum of the squares of our other two numbers (whose sum is a — ^, and rectangle r) will be = rt — 5 p — 2r. Therefore, by adding these aggre- gates of the squares of the means and extremes together^ We get this equation, viz. s^ + « — ^ «"] ^ — 4r t=r <^. to the Resolution of Problems. 121 Moreover, from the equation, — + liL ^ a — s. y ^ we get x^ -{-y^ = xy x o, — s -=1 r x (i — s\ but {by the same pr oh. just now quoted) x'^ +if z=zs'^ — Zsr ; therefore ^3 — ^8r ■=. ar — *r, or r = ; which vaUie be- ing substituted for r, in the preceding equation, we have 5^ -f a — ^1 = h. This, solved, gives ' 2^ + a ' 7 t> ^ = v/ — ^^^^^^^^ — I — — '- whence every thino; else is > 2 ^4aa 2a ^ b readily found. PROBLEM LXXI. The sum (a) and the sum of the squares, (J)) of five iiwur bers^ in geometrical progression^ being given ; to find thr numbers. Let the three middle numbers be denoted by x^ ?/, XX 22 and 2 : then the two extreme ones will be — and — ; and therefore we shall have XX 22 ^ 4 ^ r* % ^^^^ question. ^+^ + 2/^+2^ + -,= ^,/ •' ^ . XX Z^ Put X + z = u; then, by the first equation, ( ^ y y r= a — u — y. Wherefore, seeing the sum of the two extremes is expressed by a — u — y, and their rectangle by y^ (see theor. 7, sect. 10), the sum of their squares will (by problem 68) be = « — u — yY — Sz/^ ; and, in the very same manner, the sum of the squares of the two terms {^x and 2) adjacent to the middle one (tf) will be = u^ — 2^/2. Whence, by substituting these values, our equations become -^ + u + y •=^ a^ and y R 1 22 The Application ofAlgebm a — u — yY — 2z/2 + iv" ~2y'' + y^-by which, by re^ duction, are changed to aa ~ 2«w — 2ay + 2iiu + 2uy — 2yy = ^, and ay '^-^ uu — ^^ + 2/^ = ^* To the former of which add the double of the latter ; so shall aa — 2au = b -, and therefore w = — — . 2 2a From whence, and yy + a — u X y =^ uUj the value of ^ (= yuu -I J ) is likewise given. PROBLEM LXXII. The sum («), the swn of the squares (J?)^ and the sum of tlie cub^s (c), of any four numbers in geometrical proportion being given ; to find the numbers* Let half the sum of the two means be x^ and half their difference y ; also let half the sum of the two extremes be 2;, and half their difference v^ and then the numbers themselves will be expressed thus, z — v^ x — z/, x +y^ z -^ v: whence, by the conditions of the problem, we have z — V + X — y - f X + 2/ + z ^ v = a^ z — v Y + x—y'Y^ x — y ^ •\- Z'\-v' Yz:z b^ z — V X z -^v =ix — y X ^ + y (theor. 1, p. 72) ; which, contracted, are 2z +2x z=a^ 2z^ + 2v^ + 2x^ + 2z/2 = b^ 2z^ + ezv"^ + 2x^ + 6xy^ = c, z^ — IT =z x^ — y^, liCt x^ — z^ + v^j the value of z/-, in the last of these equations, be substituted instead of 2/^, in the two preced- ing ones, and wc shall have 2z^ + 2v'^ -f- 2x^ + 2x^ ~ 22- + 2v^ = b, and 22^ + 621^- + 2x^ + 6x^ — 6xz^ + 6xv^ = c ; which, abbreviated, become 4x^ + 4v^ = ^, and Slz^ + Sx^ —6xz^ + 6x + 62 X v^ = <•. to the Resolution of Problems* 123 Let \b — AT^, the value of v^^ in the former of these equations, be substituted, for its equal, in the latt er^ and we shall next have 2z^ + ^x^ — ^xz^ + 6x + 62 X :^f) — -2 __. ^ . moreover, if for 2, in the last equation, its equal ^a — a: be substituted, there will come out 2 x |a — xY + 8^ — ^x X ^CL — xf + Sa X \b — XX z=ici that is, 6ax^ — Sa^x -| 1 = c ; therefore x'^ — 4 4 ax c b a^ . ^, a — = — * — ; and, consequently, ,%" = — '— 2 6a 8 24* > 1 ^' 4 — — — + — * whence, 2, ^, and y are likewise 6a 8 48' ^ ' ' :7 known, T/i^ *awe otherwise. ^/ Let the sum of the two means = ^, and their rect> angle = r ; so shall the sum of the two extremes = a — 5, and their rectangle also = r (by the questtoii) : from whence, and prob, 68, it is evident, that the sum of the squares of the means will be = 5^ — 2r, and the sum of the squares of the extremes = a — ^j" — - 2r; also, that the sum of the cubes of th e mea ns will be = s^ — or5, and that of the extremes = a — ^"j* — Zr X a — 5: by means whereof, and the conditions of the problem, we have given the two following equa- tions : viz* s^ + a — s\ — • 4r = ^, or, 2*^ — 2as ^^^r:=^b — aa'f and s^ + a — ^1^ — Zra = c, or Zas^ — od^s — Zar = c — » a? : divide the former by 2, and the latter by 3a, and then subtract the one from the other, so shall r = — -— f- 6 2 c a -— , whence the value of 5 (= 3a' ^ 2 4'- -J. 2r + — ^ bij the first equation) is also 1 24 The Application of Algebra given, being (when substitution is made) = ~ .— I aa b 2c 12 "i" 3«' PROBLEM LXXIIL Having given the sum («), and the sum of the squared (^h)j of any number of quantities in geometrical progression; to determine the progression* Let the first term be denoted by x^ the common ratio by 2, and the given number of terms by n : then, by the conditions of the problem, we shall have X + xz + xz^ + xz^ + xz"^ . . • + xz^^'^^ = «, .v^ + c^z^ + xH'^ + x^z^ +xH^ . , . + x^z^"""^ =3 b. Multiply the first equation by 1 — i 2, and the second by 1 — z^ J so shall .V — xz''^ = a X 1 — 2;, and Divide the latter of these by the former ; whence will be had x + xz^ = — X 1 + 2 : let this equation and a the first be now multiplied cross-wise, into each oth er, in order to exterminate x ; so shall a X 1 + 2" = — X 1 -f 2 X 1+2+2^+2^ . . . 2«-i. a If n be an even number^ put 2w = n \ then our last* equation, when multiplication by 1 + 2 is actually aa made, will stand thus, ^ x 1 + 2^^"^ = 1 +22+22* b . . . . + 2z^'-^ + 222^^-1 + z^"^', which, divid' ... aa \ 12 cd by .2-, becomes -^+-^^+2- = ^ + ^^;;^::^ + ~ + ~+- + 2+22 +22^ .... +22^^^ to the Resolution of Problem^ 12>5 + 22>»-i + 2W», Let s be now put (s= h ^) = the sum of the halves of the two terms of the series adjacent to (2) the middle one; then, the rectangle of these quantities being 1, the sum of their squares (or half the sum of the two terms of the series next to those) will be = 5^ — 2 (by problem 68) ; and the sum (— + 2^) of half the' two next terms to these last = s^ -— Zsy &c. &c. Hence, by making r/ = — — ,— , and putting the value of — + 2"* (as expressed in the said problem 68) = Q, and then substituting above, &fc. our equa- tion becomes dOi =1+5+5^ — 2+^^ — 3^ + ^4 — 4^.2 ^ 2, &c. continued to m terms ; whence the va- lue of s may be determined. Thus, let 72, the number of terms given, be four; then m being = 2, Q (= ~ + 2^) will be s^ — 2; and our equation will, here, ht d x 6'^ — 2 = 1 -{• s. If n be = 6, Q ( = — + 2^) will be = ^^ — 3^ ; and we shall have d x s^ — 3,y = 1 +5+^2 — 2 c= 52 ^ ^ — J . 2ccA so in other cases, where n is an even numben ^ Ifnbe an odd number^ put 2m = ?z ■— 1 ; and let both sides of the equation c?Xl+2« = — X 1+2 X 1 + 2 + 2^ . . . 2»-i a be divided by 1 + 2 ; so shall a X 1 -2 +2^-2^ . . . -2"-2^2^-^ = -- X 1 + 2 + 2^ . . + 3«-i a (because 1 +2X 1 — 2 + 2^ — 2^ + 2'*...-- z'^"^ + 2"-- t26 The Application of Algebra {1 _ 2; + 2^ — 2^ + Z"^ .. .— 2:«-2 ^ 2»-i * > _ 1 ^ 2^^ : whence, by transposition, and substituting m, « — A X 1 + 22 ^ 2^ . . . + 22m _ a + 1. X a a 2 + 2^ + z^ . . . 2^^! ; put 22-±J^ =, c J and let the whole equation be divided by a X 2^' ; then will 1 2^ ' ^'^"■^ + 1 cx 2;m-.l + 1 ~i • • + 2'n-3 ^ 2*, Now, if m be an even number, the powers of 2 in the former part of the equation will be the even ones, and those in the latter the odd ones : but if m be an odd number, then vice versa. In the first case, our equation may be written thus : 4 + "i:i'- + -T + 4- + 1 +2' +2' • • • + 2"^' + 2- = ^- X ^ + 4=3 • • • + 4 + — + ^ + ^' • • • ^""' + 2""' * Where, since 1- % = 5, -^ + z^ __. ^^,2 — ^ 2, -- +2^ 2 2" 2^ ^ ^.3 — n^^ j. 2;4 = 54 — 4^2 ^ 2^ fe'c. we shall, by sub- 2^^ stituting these values in each series (p roceeding fro m the middle both ways) ha ve 1 + s^TZTi + ^^ _ 4^^ ^ 2 + £sfc. == c into s + s^ — 2s + fcV. But, in the second case, where M is an odd num- ber, and the even powers of 2 come into the second series, we shall, by the very s ame met hod, have s + s^ — 3s + s ^ — 5s^ + 5s + ^c. = c into 1 +s^~ 2s + ^4 _ 4^^2 j^,2 + ^c. to the Resolution of Problems. 12? In both which cases the terms are be so far conti- nued, that the exponent of 5, in the highest of them, 71 — 1 may be = — > — . Thus, if n^ the given number of terms, be 3, then m {— ~- — ) being = 1, the equa- V tion belongs to case 2, and will be * = c, barely. If n =. 5j then m = 2 ; and therefore 1 + s^ — 2 = C5, or s^ — 1 = C5, by case 1. If n be 7, m will be 3 ; and so ^ 4- i'^ — 3^ = c X 1 + s^ — 2, or s^ — 2^ = c X s-^ — 1) ^^ ca^g 2 . Lastly, if tz = 9, then vi = 4, and therefore 1 +s^ — 2 -f- 6-^ -- 4a-^ + 2 = c Xs + s^ — 3^, or s^ — 3s^ + 1 = c X s^ — 2^, btf case 1. PROBLEM LXXIV. Having given the sum (a), and the sum of the cubes (Ji)^ of any number of terms in geometrical progression ; to de^ termine the progression. By retaining the notation in the last problem, and pro- ceeding in the same manner, we here have a=.x + xz + xz^ . . , -L ATZ"-! = ^' — , and 2 — 1 ' h:=zx'' + X^Z^ + X^Z^ . . . + X^Z^ «-3 _ tz L. (l^u Z^ 1 ^ -^ theorem 8, sect. 10). Divide the last of these equations by the former, so shall b _ Z 1X2^^ 1 - Z^n 4. 2;n ^ 1 — = x^ X =:x^ X -z (be- a 2^ — 1 X 2'^ — 1 z^ + z + 1 ^ cause = 2^ I. 2 -f 1, and = z^"- -f 2"* + 1 ). Let this equation, and the square of the first, «2 =: ^2 ^ __ 2! Z — ^ be now multiplied, cross- 2 — 22 -f- 1 wise, in order to exterminate x; whence will be had ^ ^ 2^" _ 22" + 1 , 2^" + 2;n + 1 , . , — X -T- ^— = a^ X ■ ^■^- : which, -<2 2^ ™ 22 + 1 , ^2 ^ :^ + 1 ' 128 The Application of Algebra the numerators being divided by 2", and the denomina- tors by x^ will stand thus, z'^ z^ h X : = a^ X -^. Put (as 2;_2+JL 21 + 1 + 2^ Z Z before) the sum of z and — = 5 ; then, their rectan- gle being 1, the sum of their 72th powers (z^ + — ) will be had in terms of s (from problem 68), which sum let be denoted by S ; so shall our equation become h X — — — = a^ X "^ — '. whence the value of s s — 2 s + 1 may, in any case, be determined. Thus if (ji) the given number of terms be 3 ; then S (the sum of the cubes of z and — ) being =5^ — 3*, we have b X = a^ x ; that s — 2 s -{- 1 ———————— s^ - ^o J. 1 is, by division, ^ X 6^ + 2^ 4- 1 = a^ X '^^^ If the number of terms be 4; then will S = ^^ — 45" + 2 ; and therelore b x = ^"^ X : s — 2 5 -f 1 which, by an actual division of the numerators, is res- duced to ^ X 'S^ + 2s^ = a^ X ^^ — ^'^ — 36* -f 3. Again, taking ii = 5, we have S = 6^ — 5s^ -f- Ss 5 and therefore b X '■ = «"* X .9 4-1 which, by division, is reduced to ^ X 6-^ + 2^^ — s^ — 2.s -f 1 = a^ X "S"* — s^ — 4^^ -f 4a^ + 1 : and so of others ; where it may be observed, that the values of S — 2, and S + 1, will be always divisible by their respective denominators, except the latter, when n is either 3,, or a multiple of 3. io the Resolution of Problems^ 129 PROBLEM LXXV. The sum of any rank of quantities (a + b + c + d + e + &c.) being given = P, the sum of all their rectan- gles (ab + ac + ad^ &c. + be -{- bd^ &c. + cd^ &c.) = Q, the sum of all their solids {abc + abd + abe^ &c. + acd + ace^ &c. + bcd^ &c.) = R, &c. &c. it is pro- posed to determine the sum of the squares^ cubeSj biqiia- drateSj &c. of those quantities. rp =: b + c + d^ is?c, = sum of all the quan- tities after the first (a), , q z=.bc +bd -\- be^ &c. +cd + ce^ &c. = the sum Put*^ of their rectangles, r = bed + bce^ &c. + cde^ &c. = the sum of their solids. Then will P = a +/^, Gi = pa + q, R = ya + r, S = r« + *, T =.sa + t^ &c. By squaring the first of these equations, we have P2 = (2^ -|- 'jiap + p^ y from whence the double of the second being subtracted (in order to exterminate 2a/?), there results P^ — 2Q = a^ + p^ — 2q. Where P^ — 2Q expresses the true sum of all the proposed squares (f- j^ b^ ^ c^ ^ d^ &c. ; because, all the quantities a, b^ c, d^ &c. being concerned exactly alike in the original, or given equations, they must neces- sarily be alike concerned in the conclusions thence de- rived ; so that if substitution for p and q were to be ac- tually made in the equation P^ — 2Q =. a^ + p^ — 2q^ here brought out, it is evident that no other dimensions of b^ c, d^ e^ &c. besides the squares, can remain there- in, as no dimensions of a^ besides its square, have place in this equation. In order to find the sum of all the cubes, put A( = P)=:a-}-^ = sum of the roots, and B ( = P2 _ 2Q) zn a^ + p'^ — 2q =z sum of the. squares ; then, by multiplying the two equations together, we have PB = a^ + pa^ yf- //a — 2qa + // — 2pq. 1 30 The Application of Algebra From whence (to exterminate pd^^ the next inferior power of a after the highest, a^) let QA = pa^ + p^a + qa + pq (the product of the equations Q and A) be subducted ; and there will remain PB — QA = a^ — 3^<7 + p^ — Zpq. To this last equation (in or- der to take away the next inferior power of a) add three times the equation R = ^a -f- r, so shall PB — QA. + 3R = a^ + p^ — Spq + 3r. From whence it is evident that PB — QA + 3R must be the re- quired sum of all the cubes a\ + b^ + c^ + d^ &c* for reasons already specified with respect to the preceding case. To determine the sum of the biquadrates, put C = (3^ + p^ — Spq + 3r = the sum of all the cubes ; tht^n multiplying by the equation V =. a + p (as be- fore)^ we get PC = «^ = pa^ -f- p'^a — opqa -^ 3ra + ^4 — 3^2^ ^ r^p^,^ From which (to exterminate pa^^ subtract QB =: pa^ + p^a — 2pqa + qa^ + p^q — 2f (the product of the equations Q and B) : so shall PC — QB = «'* — qa^ — pqa + 3ra + p"^ — 4^p^q + Zpr + 2^2. tQ ti^-g ^^^ j^^ _ ^^2 ^ p^fj^ ^ ^.^ ^ ^^. then will PC ~ QB + RA = «^ + ^ra + p^ — 4/^ -f 4pr + 2q^ : lastly, subtract 4S = 4ra + 4^, so shall PC — QB + RA — 4S = «^ + /?4 — 4p^q + 4pr + 2^2 — 4^^ _. j)^ ^j^g g^j^ q£- ^y[ the biquadrates. In like m.anner (the last equation being, again, mul- tiplied by P = « -f- /'i the preceding one by Q = j&a -f- ^, &c. &c.) the sum of the fifth powers will be found = PD — QC + RB — SA + 5T: from whence, and the preceding cases, the law of continua- tion is manifest; the sum (F) of the sixth powers being PE — QD H- RC — SB -{- TA — 6U ; and the sum (G) of the seventh powers = PF — QE + RD — SC + TB ~ UA + rW, &>. £ifc. But, if you would have the several values of B, C, D, E, £sPc. independent of one another, in terms of the given quantities P, Q, R, S, T, &fc. then will B = P2 _ 2Q, C = P3 _ 3PQ + 3R, D = P-* — 4P2Q + 4PR + 2Q2 — 4S, to the Resolution of Problems. 121 K = P^ — 5P3Q + 5P2R + 5PQ2 — 5PS — 5QR + oT, £s?c. £?c. which values may be continued on, at pleasure, by multiplying the last by P, the last but one by — Q, the last but two by R, the last but three by — S, 8Pc. and then adding all the products toge- ther; as is evident from the equations above derived. — ■ These conclusions are of use in finding the limits of equations, and contain a demonstration of a rule, given for that purpose, by Sir Isaac Newton^ in his Universal Arithmetic. SECTION XIL Of the Resolution of Equations of several Dimensions. BEFORE we proceed to explain the methods of re- solving cubic, biquadratic, and other higher equations, it will be requisite, in order to render that subject more clear and intelligible, to premise something concerning the ori- gin and composition of equations. Mr. Harriot has shown how equations are derived by the continued multiplication of binomial factors into each other; according to which method, supposing X' — a^ X — b^ X — c, X — d^ &c. to denote any number of such factors, the value of x is to be so taken, that some one of those factors may be equal to nothing : then, if they be multiplied continually together, their product must also be equal to nothing, that is, x — -a X x — b X ^ — " c X X — <:/, &c. = : in which equation x may, it is plain, be equal to any one of the quantities, «, ^, c, d^ &c. since any one of these being substituted instead of x^ the whole ex- pression vanishes. Hence it appears, that an' equation may have as many roots as it has dimensions, or as are expressed by the number of the factors, whereof it is sup- posed to be produced. Thus the quadratic equation a X X — ^ = or x^ .[ x -f- ab = 0, has 1§2 The Resolution of Equations two roots, a and b ; the cubic equation x — a X oc — b X a: — c = 0, or — al ab'\ .^3 _[ z?* V a;^ + (7C V ;^ — abc = 0, has three roots, a, b^ and c ; a nd the biquadratic equation, x — a X ^ — b X X i — c X ^^ — d =0^ or -dj ^^ + ^'^J -bed} has four roots, a^ b^ c, and d. From these equations it is observable, that the coefficient of the second term is al- ways equal to the sum of all the roots, with contrary- signs ; that the coefficient of the third term is always equal to the sum of their rectangles, or of all the pro- ducts that can possibly arise by combining them, two and two ; that the coefficient of the fourth is equal to the sum of all their solids, or of all the products which can possibly arise, by combining them three and three ; and that the. last term of all is produced by multiply- ing all the roots continually together. And all this, it is evident, must equally hold good, when some of the roots are positive and the rest negative, due regard being had to the signs. Thus, in the cubic equation :v — a X oc — /9 X ^ -f c = O, or .\^ -I- — bYX^ + 4- ab" + ab'\ -^ ac> X r^bc) 4- abc =R O (where two of the roots, a, ^, are positive, and the c^her, — - c, is negative, the coefficient of the second term appears to be — a — b + c^ and that of the third, ab ^ — ac — bc^ or ab -i- a x — c -}- b X — c, con- formable to the preceding observations. . Hence it follows, that, if one of the roots of an equation be given, the sum of all the rest will likewise be given ; and that, in every equation where the second term is wanting, the sum of all the negative roots is exactly equal to that of all the posi- tive ones ; because, in this case, they mutually destroy of several Dtmensiom» 133 each other. But when the coefEcient of the second term is positive, then the negative roots, Jaken together, ex- ceed the positive ones. But the negative roots, in any equation, may be jchahged to positive ones, and the posi- tive to negative, by changing the signs of the second, fourth, and sixth terms, tol iso on, alternately. Thus, the fore- going equation _ +ab^^ {x — a X x^^b X ^ + (? = ) x^ + — al + ab't — l?> x^ —^ ac> X - + c} — be} abc = Oj by changing the signs of the second and fourth + «1 + «^1 terms, becomes x^ + +bYx'^ — ac>x — abc = 0, or — ^ J —bcS X -^-a X oc + b X ■^' — <-' == 0; where the roots, from -f- «, + 6, and ■;-— c, are now become — «, — ^, and + c*. Moreover, the negative roots may be changed to positive ones, or the positive to negative, by increasing or di- minishing each, by some known quantity. Thus, in the quadratic equation x^ + 8x + 15 = 0, where the two roots are — 3 and — 5 (and therefore both negative) if z — 7 be substituted for x^ or, which is the same thing, if each of the roots be increased by 7, the equ^ion will become z — 7^ -*- 8X2 — r -f 15 = ; that is, z^ — 6z4-8=:0, orz — 2 X 2 — 4 = 0; where the roots are 2 and 4, and therefore both positive. This method of augmenting, or diminishing the roots of an equation is sometimes of use in preparing it for a solution by taking away its second term ; which is always perform- ed by adding or subtracting ^, A^ or |. part, qs?c. of the coefficient of the said term, according as the proposed equation rises to two, three, or four, fcPc. dimensions. Thus, in th^ quadratic equation x^ — • Sx -f- 15 = 0, let the roots be diminished by 4, that is, let x — 4 be put = 2, or .V = 4 -f 2 : then, this value being substituted for X, the equation will become z + 4]^ - — S X z -^-4 + 15 = 0, or ;o^ — 1=0; in which the second term is wanting. 1 34 The Resolution of Equations Likewise, the cubic equation 2' -^ az^ + bz — c =0, by writing a; = 4- ^^ and proceeding as abovis:, o will become x^^ 1 2r^ + — ^ ^=0; and so of Others. Hence it appears, how any adfected quadratic may be reduced to a simple quadratic, and so resolved without completing the square ; but this by the bye. I now pro- ceed to the matter proposed, viz, the resolution of cubic, biquadratic, and other higher equations ; and shall begin with showing Ilorv to determine -whether some^ or all the roots of an equa^ tio?i be rational^ and^ f^^j "i^hat they are. Find all the divisors of the last term, and let them be substituted, one by one, for x in the given equation ; and then, if the positive and negative terms destroy each other, the divisor so substituted is manifestly a root of the equa- tion ; but if none of the divisors succeed, then the roots, for the general part, are either irrational or impossible : for the last term, as is shown above, being always a multi- ple of all tfie roots,' those roots, when ratioiial, must neces- sarily be in the number of its divisors. Examp. 1. Let the equation x^ — 4:^;^ — 7x + \0 =zO^ be proposed ; then the divisors of (10) the last term being + 1^ — 1^ ^ 2, — 2, -f. 5, — 5, + 10, — 10, let these ^ quantities be successively substituted instead of ;c', and we shall have, 1 — 4 — , 7-f-10= 0, therefore 1 is a root ; — 1 — 4 -|» 7 + 10 = 12, therefore — 1 is no root ; 8 — 16 — 14 -f 10 = — 12, therefore 2 is no root ,• — 8 — 16 -f. 14 -f 10 = O, therefore — 2 is anodier root ; 125 — 100 — 35 -f- 10 = O, therefore 5 is the third root. It sometimes happens that the divisors of the last term are very numerous ; in which case, to avoid trou- ble, it will be convenient to transform the equation to another, wherein the divisors ave fewer ; and this is best of several Dimensions. ,135 effected by increasing or diminishing the roots by a unit, or some other known quantity. Examp* 2. Let the equation propounded be y^ — 4z/^ — . 8z/ + 32 = O; and, in order to change it to another, whose last term admits of fewer divisors, let ;i^ + 1 be substituted therein for y^ and it will become oc"^ — Gat^ — ♦ 16a: + 21 = 0; where the divisors of the last term are 1, — 1,3, — 3, 7, — 7, 21, and — ^ 21 ; which being suc- cessively substituted for x^ as before, we have, 1 — 6 — 16 + 21=0, therefore 1 is one of the roots ; 1 — 6 + 16 + 21 = 32, therefore — 1 is not a root ; 81 — 54 — 48 + 21 = 0, therefore 3 is another root. But the other two roots, without proceeding further, will appear to be impossible ; for, their sum being equal to - — 4, the sum of the two positive roots (already found), -with a contrary sign (as the second term of the equation is here wanting), their product, therefore, cannot be equal to (7) the last term divided by the product of the other roots, as it would, if all the roots w^ere possible. However, to get an expression for these imaginary roots, let either of them be denoted by t;, and the other will be denoted by — 4 — Vy which, multiplied toge- ther, give — 4^v — ^- = 7 ; whence v = — 2 + V — 3, and consequently — 4 — v = — 2-— V — 3. Now let each of the four roots, found above, be increased by unity, and you will have all the roots of the equation proposed. When the equation given is a literal oiiCj you may still proceed in the same manner, neglecting the known quantity and its powers, till you find what divisors suc- eeed ; for each of these, multiplied by the said quantity, will be a root of the equation. Thus, in the literal equa- tion, x^ + 3rt;c^ — 4sic )ns, whose terms differ by unity ; whereof the terms corre- sponding to the term 0, in the first progression, are 3, 4, 3, and 5 : therefore the two former progressions being ascending ones, and the two latter descending, I try the quantities + 3, + 4, — 3, — 5, one by one, and find that they all succeed. And after the same manner we may proceed in other cases; but, in order to try whether any quantity thus found be a true root, we may, instead of substituting for .V, divide the whole equation by that quantity coi>* i 58 The Resolutmi of Equations^ nected to x^ with a contrary sign ; for, if the division ter- minate without a remainder, the said quantity is, mani- festly, a root of the equation. Thus, in the last example^ where the equation is x^ + x^ — 29x^ — 9a: + 180 = O, the numbers to be tried being + 3, + 4, — 3, and — 5, I first take — 3 and join it to at, and then divide the whole equation, x"^ + x^ — 29x^ — 9x + 180 (= 0), by a: — 3, the quantity thence arising, and find the quotient to come out x^ + 4a:^ — 17 x — 60, exactly. Therefore + 3 is on€ of the roots. Again, in order to try + 4, the second number^ I divide the quotient thus found, by a; — 4, and there comes out x^ + Sx +15; therefore + 4 is another root : lastly, I try — 3, by dividing the last quotient by ■%■ + 3, and find It also to succeed, the quotient being x + 5* See the ope- ration at large. ,v„3) x^ + x^ — 29;^^— 9:v + 180(;c^+4Ar* — 17x — 60 x"^! — Sx^ ^4x^ — 29a;2 4- 4x^.^ 12;c^ — 17;%2-- 9x ^17x^ -hSlx — eOx -f 180 ' — 60x + 180 O O . 4) .v^ + 4x^ — 17x — 60 (a:^ + Sat + 15 x^>^4x^ + 8^:^— 17x 4- Sx^ — 32a;' + 15a:- -4-15A:- -60 -60 o gJ several Dimensions* 139 X + S) x^ + 8x -{- 15 {x + 5 x^ + Sx + 5X + ^5x^ 15 15 As another instance hereof, let there be proposed the equation 2x^ ~ 3;^^ + 16:\: > — 24 = ; then, expounding X by 2, 1, 0, and — • 1, successively, and proceeding as in the foregoing examples, we have 2 + 12 1.2.3.4. 6 . 12 + 2 — 1 1 — 9 1.3.9 + 3 + 1 — 24 1.2.3.4. 6 . 8 &c. + 4 + 3^ 1 — 45 1,3,5.9. 15 . 45 + 5 + 5 Therefore, the quantities to be tried being 4 and |, I first attempt the division by at •— 4 ; which does not answer : but, trying -x* — ^, or (its double) 2:v — - 3, I find it to succeed, the quotient being x^ + 8, ex- actly. The reason why the divisors, thus found, do not always succeed, is, because the first progression, 2, 1 , 0, -— 1 is not continued far enough to know whether the corres- ponding progression may pot break off, after a certain number of terms ; which it neyei: can do when the bu- siness succeeds. Thus, in the last example, where we had two different progressions resulting^ had the operation, or series, 2, 1, 0, — 1, been continued only two terms farther, you would have found the first of those pro- gressions to fail ; whereas, on the contrary, the last (by which the business succeeds) will hold, carry on the pro- gression, 2, 1, 0, -r- 1, as far as you will. The grounds of which, as well as of the whole method, upon which thq foregoing observations are founded, may be explained in the following manner. Let, there be assumed any equation, as ax^ + bx^ -f- cx^ + dx -{^ e =z % wherein «, ^, c, d^ and ^, represent any whole numbers, positive or negative, and let px + q denote any binomial divisor by which the said expression 140 The Resolution of Equations ax^ + bx^ + cx^ + dx + e \^ divisible, and let the quo- tient thence arising be represented by rx^ -f- sx^ ^ tx -{- v, or, which is the same in effect, let ax^ -f- bx^ + cx'^ + dx -{- e ^^ px -\- q X r..^ -f sx^ -f- tx -^ v. This being premised, suppose x to be now, successively, expounded by the terms of the arithmetical progression 2, 1, 0, '^— 1, «— 2 {as abcvey-, and then the corresponding values of our divisor px + q^ will, it is manifest, be expounded by 2p -f- q^ p + q^ q^ — P + 9f ^^^ ' — ^P + ^i respectively ; which also constitute an arithmetical progression, whose common difference is/?; which common difference (/?) must be some divisor of the coefficient (a) of the first term, otherwise the division could not succeed, that is, p could not be had in a without a remainder.. Hence it appears that the binomial divisor, by which an expression of several dimensions is divisible, must always vary as x varies, so as to be, successively, expressed by the terms of an arithmetical progression, whose common dif- ference is some divisor of the first, or highest term of that expression. It also appears that the said common difference is al- ways the coefficient of the first term of the general divi- sor ; and that the term {q) of the progression, which arises by taking :v = 0, is the second term. Therefore, when- ever, by proceeding according to the method aboye pre- scribed, a progression is found, answering to the conditions here specified, the terms of that progression are to be con- sidered only as so many successive values of some general divisor, as px -f- q. Whence the reason of the whole pro- cess is manifest. After the same manner we may proceed to the in- vention of trinomial divisors, or divisors of two dimen- sions ; for, let mx^ + px + q, be any quantity of this kind, wherein 772, /?, and q represent whole numbers positive or negative, and let the terms of the progres- sion 3, 2, 1, 0, . — 1, - — '2, ~ 3, be written therein, one by one, instead of x ; whence it will become 9m -f 3p ^ q^ 4m + 2p + q, m + p + q, q, m ^p + q, 4m — , 2p 4- ^, and 9m — Sp + ^, respectively ; where m must be ^ome divisor of the coefficient of the first term of the of several Dimensions^ 141 given expression ; otherwise, the division could not suc- ceed. Hence it appears, 1°, That the coefficient (jn) of the first term of the divisor must always be some numeral divisor of the co- efficient of the first term of the proposed expression, 2°, That the product of that coefficient by the square of each of the terms of the assumed progression, 3, 2, 1, Oj — 1, —- * 2, •— 3, being subtracted from the corre- sponding value of the general divisor, the remainders {Zp + q, 2p + q, p+ q, q, —p + q, ~2p + q, ~ Sp -f- q) will be a series of quantities in arithmetical pro- gression, whose common difference is the coefficient of the second term of the divisor. 3°. And that the term (^) of this progression, which arises by taking ^t- = 0, will always be the third, or last term of the said divisor. From whence we have the following rule. Instead of x in the quantity proposed^ substitute^ successivelij^ four or more adjacent ter?ns of the progression 3, 2, 1, 0, — - 1, — 2, — - 3 ; and from all the several divisors of each of the numbers thus residting^ subtract the squares of the corresponding terms of that pro- gression multiplied by some numeral divisor of the highest term of the quantity proposed^ and set doxvn the remainders right against the corresponding terms of the progression 3, 2, 1, 0, — 1, — 2, — 3; and then seek out a colla- teral progression which runs- through these remainders ; which being found^ let a trinornial be assumed^ whereof the coefficie7it of the first term is the foresaid numeral divisor ; that of the second term^ the common difference of this col- lateral progression ; and xvhereof the third term is equal to that term of the said progression which arises by taking Oi -=.0 ; and the expression so assumed will be the divisor to be tried. But it is to be observed^ that the second term raust have a negative or positive sign^ according as the pro- gression^ found among the divisors^ is an increasing or a decreasing one. Thus, let the quantity proposed be x'*' — x'^ — 5x^ -f- 12^ — 6; and then, by substituting 3, 2, 1, 0, — 1, — 2, successively, instead of x, the numbers resulting will be 39, 6, 1, — 6, — 21, and — 26, respectively; which, together with all their divisors, both positive and 142 The Resolution of Equations negative, I place right against the corresponding terms of the progression 3, 2, 1, 0, - — 1, ' — • 2, in the following manner : 1 — 1 — 2 39, 6. 1 .' 6. 21 . 26. 13.3 - 1 3.2 7.S 13.2 — 1 — 1 — 1 — 2 . — 3 • — 3 . ~ 7, , , 2 . 13 39 6 6 21 26 Then, from each of these divisors I subtract the square of the corresponding term of the first progression muhi- plied by unity (as being the only numeral divisor of the first term), and the work stands thus : 3 30. 4.-6—8 10.^12.-22—48 + 4 — 6 2 2._1.^2 3.~ 5.— 6 7.— 10 + 2 — 3 1 + 2 + 6. 3. 2. 1. — 1.— 2. — 3 — . 6 -1 20. 6. 2. 0.— T 2.— 4 8.-22 — 4 + 6 -2 22. 9.-2 — 3 r 5 - 6.— 17 — 30 — 6 + 9 Here I discover, among the remainders, two colla- teral progressions, viz. 4, 2, 0, — 2, — 4, —^ 6, and — 6, — 3, 0, -f- 3, + 6, -f 9 ; therefore the quantity to be tried is either x^ + 2x — 2, or Ai-^ — Sx + 3 ; by both of which the business succeeds. This invention of trinomial divisors is sometimes of use in finding out the roots of an equation when they are irrational, or imaginary. Thus, let the equation given be x"^ — 4^^' -f 5x^ — 4.v — 1=0; and let x be successively expounded by the terms of the progression 3, 2, 1, 0, and the numbers resulting will be 7, — 3, — 1, and 1 ; which, together with their divisors, being ordered according to the preceding directions, the ope-r ration will stand as follows : 5 7 . 1 — 1 .—7 — 2 .—8 .—10—16 2 3 . 1 .— 1 3 _1 ._3 ._ 5— 7 1 1 .—1 # # —2 ^ ^^ 1 .—1 # # 1 —1 ^ ^ ,2 — 1 + 1 — 5 — 2 + 1# Here we have two progressions, — 2, — 1, 0, 1 ; and — 8, — 5, — 2, 1; therefore the quantity to be tried is either x^ — ;c + 1, or ^^ — ?^x -f 1 ; but I take the tf several Dimensions. 143. first, and having divided x"^ ^— 4x^ + 5x^ — 4x + 1 thereby, find it to succeed, the quotient coming out x^.^3x + 1, exactly. Therefore x"^ — 4x^ + 5x^ — 4;v + 1 being universally equal to x^ — x + 1 X ;^2 — Sx + 1, let ;<:2 — X + 1 be taken = 0, and also x^ — 3;c -f- 1 = ; from the former of which equations we have x =z | ±V — | ; and from the latter x =: ^ ± VK^ Therefore the four roots of the given equation ^rU+Vtrj^ 1 _ V""::^^, I + V^andf — V|l whereof the two last are irrational and the two first imaginary. And in the same manner, the roots of a li- teral equation, as z^ — 4az^ + 5a^z^ — 4a^z + a^ = 0, where the terms are homogeneous, may be derived ; for, let the roots be divided by «, that is, let x be put = ^ ^ or ax = z ; and then, this value being substituted for a z, the equation will become x"^ -— 4^;^ + 5x^ — 4^ + 1 == ; from which x will be found, as above j whence z ( = ax^ is also known. Having treated largely of the manner of managing such equations as can be resolved into rational factors, whether binomials or trinomials, I come now to ex- plain the more general methods, by which the roots of equations, of several dimensions, are determined ; and shall begin with The Resolution of cubic Equations^ according to Cardan. If the given equation have all its terms, the second term must be taken away, as has been taught at the be- ginning of this section; and then the equation will be reduced to this form : viz. x^ + ax z=: b \ where a and b represent given quantities. Put x =: y +z ; and then, this value being substituted for x^ our equation becomes i/ + 3t/^2jf%2;2 ^ z^ +a X y '+ z = ^, or y^ + z^ + Syz X y -i-z+axy+z = b. Assume, now, 3yz = — a ; so shall the terms Syz X y +z and a X y + z destroy each other, and our equation will be reduced to y^ +Z^ =: b. From the square of which, let four times the cube of the equation yz =z — ^a be subtracted, and 144 The Resolution of Equations we shall have t/« — Si/^z^ 4. :^« -= Z)« 4. __ ; and there- fore, by extractin g the square root, on both sides, y^ — "i? from 2/3 + 2^ = ^, gives 2t/^ = ^ + y Z^^ -f- - — , and / 46/^ = \^^ + -^ ; which added to, and subtracted 2z3 = b -J^±t: hence y^l + JH + t} ; and consequently x (^z=z y b jo^ a^ and 2 = — « — V 1 2 ^ 4 ^ 27 + -) = T + V4 +27I -^T-slj + TrV which is Cardaii!^ theorem : but the same thing may be exhibited in a manner rather more commodious for practice by substituting for the second term its equal — \^ b 1 6^ a^l'^ (= — -^ = 2;, because uz =z — T + VT+i?! ^ |a). And, this being done, our theorem stands thus : "^^^4 '27! b lo- — +V h 2 ' > 4 ^' Example 1. Let the equation ^' + Sz/^ + Qy = 13 be propounded ; and, in order to destroy the second term thereof, let ;v .— 1 b e put := s: y; so shall x ' — Ip + 3 X ^' — iT + 9 X ^^ — 1 = 13, or x'^ + 6x = 20 ; therefore, in this case, a being = 6, and b = 20, we j + ^^+Tri of several Dimensions^ 145 10^+ VlooTsl^ — ^ -TY = 20,3923]^ 10 + V 100 + 8 b 2 - — = 2.732 — .732 = 2 ; and consequently y 20,3923]^ ( = :^— .1)=:1. Exam. 2. If the equation given be y^ — 32/^* — • 2y'^ . — 8 = 0; then, by writing x + 1 f or z/^, it will be- come T+T]' — 3 X ^ + iT — 2X^ + 1 — 8 = 0, or' ::c' — 5x ■=• 12 : therefore, ^ being = — 5 , and ^ = 12, .V will here be equal to 6 + \/36 — ^i^\^ — — 4 1 1 1 ,6666, y c. — -^ == ==^ = 6 + 5,6009> + — ^ 6 + \/36 — VV l"^ ^ + 5,6009 1 3 = 2,26376 + ,73624 = 3 ; and consequently y^ (=.v + l)=4: which is the only possible value of y^ in the given equation. And it will be proper to take notice here, that this method is only of use in cases where two of the three roots are impossible (except when b^ a^ . , they are equal) ; for 1 being, in all other cases, a negative quantity, its square root is manifestly impos- sible. I shall now give the investigation of the same ge- neral theorem, for the solution of cubics, by a different method ; which is also applicable to other higher equa- tions. Supposing, then, the sum of two numbers, z and z/, to be denoted by 5, and their product (zy^ by /?, it will appear (^from prob. 68, p* 119) that the sum of their cubes (z^ + z/^) will be truly expressed by 6*^ — 3j&5. If, therefore, z^ + y^ be assumed = ^, we shall also have s^ — ops = b : but, zy being = /?, or y = — , our z /i3 lirst equation z^ '\- if =. b^ will become z^ + ^ = Z' ; from which, by completing the square, &Pc. z is found U 146 I' fie Resolution of Equations '] z=.\h '\- \^\ob — p^\^\ whence j/ ( = ~) is given = P — ~ ~ ; and consequently « ( = 2; + J/) = i|^ ^\\t)h — /^Js ^ _P_ — . whicl> is, evidently, the true root of the equation 6^ < — '^ps = h* From whence the root of the equation o<^ + ax = b^ wherein the second term is positive, will be given, by writing x for .9, and ^a for — p ; whence x is found r= U + J— 4. rL — =====p^^=E=rr, the same as ' ^4 ^271 ,, . bd , a'b before. In like manner, if things be supposed as above, and there be now given z^ + y'^ = b ; then, by the problem there reforred to^ we likewise have s^ — 5ps^ + Sp^s = b. But the first equation, by substituting — for its equal y, becomes 2^+1!=: b \ whence z^^ — ^2^ == — /% z^ = :^b + s/^bb — "7"% and 2 = |/^ + v 1/;^ — j^l^ ; and consequently s (= 2 + 2/ = 2+-^) = 2 1^ + \/\bb —"7^1^ + — ^ — ^ = the true root of the equation s^ — 5/?5^ + 5p^s =z b. Which by substituting x for 6', and — — for /?, gives x = 1^ + V kbb + STp — ==^^^FlT> fo^ the true root of the equation x^ + a>.^ + ^a^x ~ ^% of several Dhnensions. 14r Generally^ supposing z^ + y^ = b^ or z" + — = ^ (because ^ = ^), we have z^'^ — <^^ = — /?« ; whence z^ -z^^b + s/^bb - — Ji''\ and z = ;^b + V^bb — /^J« : therefore ^ (^ + t/j = 2 + -^ =: ^3 + Vic?^; — pn\^ + --^ , ; which ^b + V ^bb + y'U is the true root of the equation 6** — npt^^"^ + n . 2 ^ 2 3 ^ 72 — 5 n — 6 n — 7 ^o r^^ \ -^— . _^— . -^— . /.^6--« — ^c. (= 2« + r) This equation, by writing x for 6*, and — for — /?, becomes .\« + aA;^^^ ^ n ~ S ^ ^2, n--4 ^ 7^ — 4 27^ 272 37^ 272 372 472 7 1 •. . ^ . M' . ^" = (> ; and its root a; = — + v/ — 4- — 2 ^4 /^^^ - ■ 7 — :- nx* Wherein the two precedincr b , jb^ a" l« r , o 2 ^ 4 72' theorems are included, with innumerable others of the same kind ; but as every one of them, except the first, requires a particular relation of the coefficients, seldom 148 The Resolution of Equatiolis occurring in the resolution of problems, I shall take no further notice of them here, but proceed to The Resolution of Biquadratic Equations^ according to Des Cartes. Here the second term is to be destroyed as in the so- lution of cubics ; which being done, the given equation \yill be reduced to this form, x^ + ax^ + &x + c = ; wherein a^ h^ and c may represent any quantities whatever, positive or negative. Assume ^^ + /?jv 4- ^ X a:^ -f r;c + 5 = a:^ + ^^^ -^ hx -^^ c ; or, which is the same thing, let the biquadratic be considered as produced by the mul- tiplication of the two quadratics x^ + px + y = 0, and x^ + rx + s = O: then, these last being actually mul- tiplied into each other, we shall have x"^ + ax^ + hx + c = .:v'^ + ^ }► a;^ + qK x^ -{• P^ \ X '\' qs \ whence, ^ prj ^^ ^ by equating the homologous terms (in order to deter- mine the value of the assumed coefficients /^, ^, r, and ^) we have p -j- r = 0^ s + q + pr =: a^ ps + qr =z by and qs =z c ; from the first of which r = — p ; from the second s +q ( = a — pr) = a + p^ ; and from the third s — q z=z — . Now, by subtracting the square of the last of these from that of the precedent, we have 4qs =z a^ + 2ap^ + p^ -" , that is, 4c = fl^ -f- 2ap- -}. j^4 _^ — (because qs = c) ; and therefore // + 2ap^ _ 4. X P^ =^^^ i fr<^n^ which p will be determined, as in example the second, of the solution of cubics. Whence s {= ^a + ^p^ 4 ), and 5^ ( = ha + ip^ — 2p ' — ) are also known. And, by extracting the roots ol the two assumed quadratics .^^ -f- px -f ^ == O, and of several Dimensions. 149 P x^ + rx + s z=z O, we have x^ in the one, = — — ± ^lE — q ; and, in the other, = ■ ± y II — s = ^ ± v/ — — s^ because r = — p. Therefore the four roots of the b iquadratic , x^ -f ax^ + bx + c z=z Q^ T+VT '2 V4 ^'~ o + V4 1^ are 2 " p.^JptZ ^ 4, and ■ 2 EXAMPLE. Let the equation propounded be y^ — 42/^ -— . 8y ^ 3^ = 0; then, to take away the second term thereof, let X + 1 = y 'y whence, by substitution, x^ ^ — 6x^ — - 16:^ + 21 = 0; which being compared with the ge- neral equation, x^ ^ + axr^ + bx + c z=z 0^ we here have a =. — 6, ^ = — 16, and c = 21 ; and conse- quently /^« — 12/^ — 48/^2 (^_-. ^6 ^ 2ap^ it 4^ I P"^) = 256 ( = b^). Now, to destroy the second term of this last equation also, make 2; + 4 = />^ ; and then, this value being substituted, you will have z^ — 962 = 576 ; whence, by the method above explained, z will be found ( = 288 + \J2SsY — 32]^ ^ + 32 = ^3 ■ ^- = =:=.j) = 12. Therefore p C- 288 + Vz88T — 32l^ P a p^ b Vz + 4) is = 4, 5 (= _ + ^ + — ) = 3, and 5r(=:|.+^^~-)=:7; consequently L + ^S^s :. 3, A _ JpTZZ ^U^L ^ ^4 2^4 2 ^ 150 The Resolution of Equations ^7=-. + V—, and -p--4inr,^ ■—2 — V — 3 ; which are the four roots of the equation x"^ — ^x^ ---, 16^ + 21 ; to each of which let unity be added, and you will have 4, 2, — ^ "% '^ — ^5 ^^^ — ^ — -V— 3, for the four roots of the equation proposed j whereof the two last are impossible. And that these roots are truly assigned, may be easily proved by multiplyinjj^ the equations, y — 4 = 0, ^ — 2 = 0, z/ + 1 — V — 3 = 0, and 2/ 4. 1 + V — 3 = 0, thus arising, continually together ; for, from thence, the very equation given will be produced. The Resolution of Biquadratics by another Method* In the method of Des Cartes^ above explained, all biqua- dratic equations are supposed to be generated from the multiplication of two quadratic ones : but, according to the way which I am now going to lay down, every such equation is conceived to arise by taking the difference of two complete squares. Here, the general equation x^ -f- ax^ 4- bx^ ^ ex -^ d = being proposed, we are to assume x^ -f- ^ax + A"J^ — Ba; + CY = X* + ax^ + bx^ + ex + d : in which A, B, and C represent unknown quantities, to be deter- mined. Then, x^ -f- iax + A, and Bx + C being actually involv- ed, we shall have ^ ^ + ^a^x^ + a Ax -f A^Y =z x'^ + ax^ + bx^ -f ^ # „ B2jc2_2BC;c— C^J ex +d: from whence, by equating the homologous terms, will be given, 1. 2A -f. ia^ _ B2 = ^, or 2A + ia^ — b — B^ ; 2. a A — 2B€ = c, or aA — c = 2BC ; 3. A^— C2 =^,orA2 — ^ =C\ Let now the first and last of these equations be multi- plied together^ and the product will, evidently, be of several Dimensions. 151 equal to i of the square of the second, that is, 2A^ + ^aa — b X A."" — 2d\ — d X ^aa — T ( = B^C^) = i. X a^^ — ^«^A + c^ ( = B^C^). Whenc e, denot- ing the given quantities \ac — d^ and \& + d^ X ^aa — b by k and /, respectively, tliere arises this cubic equation, A^ — ibA.^ + kA — 4/ = : by means whereof the value of A may be determined (as hath been already taught) ; from which, and the preceding equations, both B and C will be known, B being given from thence =r V2 A 4- ^aa — 6, an.d C = II~. 2r> The several val^es of A, B, and C, bein g thus foun d, that of X will be readily obtained : for x^ -f hax + A")^ — Bjv -f Cl]^ being universally, in all circumstances of x^ equal to x"^ + ax^ -f bx^ -f- ex + d^ it is evident, that when the value of x is taken such, that the latter of these compressions becomes equal to nothing, the for- mer must likewise be = ; and consequently x^ -f ^ax + A'Y = Ba; 4- Cp: whence by extracting the square root on both sides, x^ + iax -f- A = -t B;t^ ± C ; which, solved, gives x-^-^ .|B — \a tH V^^^ ^ ib]"^. C — A = ± |B — J« ± S/-^^ -f iah -f ib^ di C — "A; exhibit- ing all the tour different roots of the given equation, ac- cording to the variation of the signs. This method will be found to have some advantages over that explained above. In the first place, there is no necessity here of being at the trouble of exterminating the second term of the equation, in order to prepare it for a solution : secondly, the equation A^ — i^A^ + kA — ^l = 0, here brought out, is of a more simple kind than that derived by the former method: and, thirdly (which ad- vantage is the most considerable), the value of A, in this equation, will be cominenswate and rational (and therefore the easier to be discovered), not only when all the roots of the given equation are commensurate^ but when they are irrational 2ind even impossible; as will appear from the examples subjoined. 1 52 The Resolution of Equations Exajn* 1. Let there he given the equation x^ + 12.x'' — 17 = 0. Which being compared with the general equation ^4 ^ ^;^3 ^ ijyp. -|- cr + ^ = 0, Ave have a = O, Z> = 0, c = 12, and ^ = — 17 : therefore >^ ( \ac — d\ = 17, / (ic^ + dx iaa — ^) = 36; and consequently A3 — ii^'A^ + kA — ^l z= A3 + 17A — 18 = 0; where it is evident, by bare inspection, that A = 1. Hence B (= V^A + iaa -1) = VY, C (= ^^-^) = — 12 12 _ ^^ n ^^ = == — 3V2 ; and ;^ = ± -1^2 ± \/— if 3^2 2\/2 ^ 2 = ± ^V 2 q: Y q= 3 V 2 . Therefore the four roots of the equation are |V 2 + W ~ 3V 2 — ---, 3\/2 — — , 2' iV2 _^— 3V2 — |, — iV/2 + sj and — -^VT — y 3V'T — --- ; whereof the first and se- cond are impossible. Exam. 2. Let the equation given be x^ — 6:>^^ — 58^^ _.114x — 11=0. Here a = — 6, ^ = — 58, c = — 114, an d d = — 11 ^ whence k (iac — d) = 182, / (ice + d X kaa — b) =. 2512; and therefore A^ + 29A2 + 182A — 1256 = 0. Where, trying the divisors 1, 2, 4, 157, ^c, of the last term (according to the method delivered on p. 134) the third is found to succeed ; the value of A being, therefore, = 4. Whence there is given B = V75 = 5V3, 90 — C = — ^ ^"^ ^y 2Q[\A X (= ± iB — ia ± 10\/ 3 ' ^ of several Dimensions. 153 Exam* 3, Z^^ M^r^ ^e ^loti; proposed the literal equation z^ + 2a2' — 3r«V — 38a^2 + a^ = 0. This equation, by dividing the whole by at*, and writing x = — , is reduced to the following numeral one, x^ + 2x^ — 37^^ — 38^ +1=0. If, therefore, tf, b^ c, and d be now expounded by 2, — ^7^ — 38, and 1, respectively, we shall here have k (^ac — ^) = — 20^ I (K^ + d X ^aa — ^) = 399 ; and, therefore, by sub- stituting these values, A3 + y A2 — 20A — 3|« = O. or, 2A3 + 37A2 — . 40A — 399 = 0. Which equation, by the preceding methods, will be found to have three commensurable roots, |, — 3, and — 19: and any one of these may be used, the result, take which you will, coming out exactly the same. Thus, by tak- ing — 3 , for A, we shall have x^ + x — 3= ±V2x 4a; + 2 : bu t, if A b e taken = J, then will x'^ + x + I = ± VT X 3:^ + 1 ; lastly, if A be taken = — 19, then xr^ + X — 19 = -f- 6\/lO. All which are, in effect, but one and the same equation, as will readily appear by squaring both sides of each, and properly transposing; whence the given equation oc^ + 2x^ — 37 :x^ — 38:^ + 1 = 0, will, in every case, emerge. And the same ob- servation extends to all other cases, where there are more roots than one, it being indifferent which value we use ; unless that some are to be preferred, as being the most simple and commodious. Having given the general solution of biquadratic equations by the means of cubic ones, I shall now point out two or three particular cases, where every thmg may be performed by the resolution of a quadra* tic only. X iS4f The Resolution of Eqiuttlons These are discovered from the preceding equations^ 2 A + i«2 _ ^ _ B^ aA— c = 2BC andA^ — ^= C^: wherein, if A be supposed = 0, it is plain that ia^ — ^ = B2, — c = 2BC, and — . ^ = C2 : whence B =r \/^aa — ^, C = ^ ' — ■= » = V — c^, and consequent- ly rf = ^ ; by making/= 6 — \aa. 4/ Therefore, in this case (wherein d = — r), the general equation o^ 4- j^ a: + A = ± B:^ ± C, will become :»c* + ifl;c = ± x\/ — -y -f V— .^. ^ But, z/ B ^^ supposed = ; then will 2 A + i«^ — 6 » 0, and also c A — c = ; whence A = i^ — |a^ 3=if=:-l--, anddiereforeC( = VA2 — flr) = Vi//— rf: so that in this case (where c = ^) the generd equation becomes x^ + hax + ^ == 4- V^/^/ — '"^ ; which, solved, gives x:=z — ^a± ^^TaY ~ hf± V\ff—d. Lastly, if Q be supposed = O, then will a A — c = 0, sold A* — t/ = ; consequently A = — = V 6^, and B ( = V 2 A + 4^^ — b) = y — — /: therefore, in thisi case (where d = — ), we shall have :^ + iax H = ± aa ci »VfZ7. From the whole of which it appears, that if c be » ^, or d. either, equal to ^, or to ~ (/ being c= 2 4/ era of several Dimensions* 155 b — yia) ; then, the roots of the given equation, o^ + ax^ + bo(^ + ex + d z* O^ may be obtained by the reso- lution of a quadratic, only. Exam. 1. Let there be given x^'^ — • 25x^ + 60x •— 36 3=0. Here « = 0, * = •— 25, c = 60, and a? = — 36 ; cc therefore, / (= — 25) being = _ (= — 25), 4^ we have [by case 1) x^ + ia:v = ± rvV — / =F V — d^ that is, AT^ = ± 5x If 6 : which, solved, gives ^ = ± | A VY r6, that is, :v = I ± |, or A* = — f =t ^ : so that 3, 2, 1, and — 6 are the four roots of the equation pro- pounded. Exam. 2. Let there be now given x^ + 2qx^ + Zq^x^ + 2q^x — r* = 0. Then, a being = 2qy b = 3^^, c = 2^', and d =s — r^, thence will / ( = b — iaa) = 2q^^ and 2l ( =e 2q^) = c ; and so (the exan^ple belonging to ca se 2) we have y ( = — jg + \f|^ l' ~ i/ ± V i// — 6^j = — i? ± V — lyy ± V ^^ + r\ Exam. 3. Lastly^ suppose there tu ere given the equation Here, a being =i — 9, 6=15, c = — 27, and c^ = 9, it is evident that — ( = 9) = ^ ( = 9) : therefore, bif c /2c cas^ 3, we have at* + iax + — c= ± x\j — + ^aa — b 5 156 Tha Resolution of Equations that IS, x^ ^^ Aix -^ 3 ( = ± xS/ ^ + »-ji — 15) = ± ^x\/ 5 I which, solved, chives 9 ± -^^ T '»- 'J7>> ± vl-vl The Resolution of Literal Equations^ toherein the given and the unknown quantity are alike af- fected. Equations of this kind, in which the given and the unknown quantities can be substituted, alternately, for each other, without producing a new equation, are al- ways capable of being reduced to others of lower dimen- sions. In order to such a reduction^ let the equation^ if it he of an even dimension^ he first divided by the equal pow- ers of its two quantities in the middle term ; then assume a new equation^ by putting some quantity {or letter^ equal to the Sinn of the two quotients that arise by dividing tHoSe quantities one by the other ^ alternately ; by means of -which equation^ let the said quantities be exterminated ; whence a numeral equation will emerge^ of half the dimensions with the given literal one* But^ if the equation propounded be of an odd dimensioiz^ let it be^ first,, divided by the sum of its ttvo quantities^ so will it become of an even dimension^ and its resolution will therefore depend upon the preceding rule* Exam. 1. Let there be given the equation x^-^Aiax^ +. Sa^x"^ — 4rt'x + a^ =.0. » . ". XX 4i%' Here, dividing by a^x^^ we have — f- 5 aa a 4a ^ aa ^ . xx ^ aa , ' >^ , ^ , ^ 1 =0 (or 1 4x f- h^ = .V XX . aa XX a x 0, by joining tiie corresponding terms) ; and, by mak- X a ing 2 = — + — , and squaring both sides, we have of several Dimensions. 157 also, 22 = — + 2 + — , or 2» — 2 = ^ + fl!« :v::c' aa :>jx Therefore, by substituting these vakies, our equa- tion becomes 2^ — 2 — 4z + 5 = 0, or 2* — 42 = X d — 3 ; whence 2 =c 3. But — + --- being = 2, we ax have AT^ — • 'zax = — «^ ; and consequently x = \za ± Vl^2^^^~aa = 4« X 2 ± v 22 — 4 = ^a X 3 ± V"5^ in the present case. Exam. 2. i^if ^A^re he given x^ + 4a;v* ^- 12(2^ a:^ ~ 12aV + 4a^x + a^ = 0. In this case, we must first divide by x + cr, and the quotient will come out x^ -f 2>ax^ — ISa^x^ -j- ^a^x + fx* = 0: whence, by proceeding as in the former ex- , , XX aa X a ^ ^ ^ ample, we have — +.^+3X 1 ^— 15 = 0, aa XX ax or 22-— 2 + 32 — 15 = O, and from thence 2 = x^lr — 3 Exam. 3. Suppose there were given 7x^ — 2&ax^ *— 26a^x + ra^ = 0. This, divided by a^x^^ becomes 7 X — + -r — a? x^ 0^ c^ X a 26 X -r + -5 = 0. Now, making, as before, 2 = 1 , cr x^ " a , X x^ a^ we have 2^ — • 2 = — + — ; and, multiplying again X a , x^ by 2 = — + — , we likewise have 2^ — 22 = — 4. ax a^ X a * x^ a^ ' ^ * x^ vV C* iA. C* + — +-7= -r- + z -\ r; and, therefore, a x^ a^ x^ X a- ~ 32 = --- + -T • which values being substi^ a^ x^ tuted as above , our equation becomes 7 X z^ — 32 — 26 X 2:^ ~ 2 = 0, or 72» ~ 262^ — 2I2 + 52 = O, 158 The Resolution of Equations Where^ trying the divisors of the last term, which are 1, 2, 4, 13, £sPc. the third is found to answer; 2, conse* quently, being = 4. Exam. 4. Wherein let there be given 2a:^ — . IZa^x^ — * 13a^x2 + 2a* =0. Here, dividing, first, by at + a, the quotient will be 2x^ — 2ax^ — lia^x^ + llaV — lla'^x^ — 2a* a; + ^Q^ = Q ; whi ch, divid ed again by a^x^^ gives x^ , a^ ^ x^ , a^ ^^ X a cr x^ or x^ ax = 0, that is. 2 X z^ — Iz — 2 X 2^ — 2 — llz + 11 = 0, or 2z^ — 22* — 172 + 15 = (yid. p. 119) : whence 2 = 3. A literal equation may be made to correspond with a numeral one^ by substituting a unit in the room of the given quantity (or letter) : and equations that do not seem, at first, to belong to the preceding class may some- times be reduced to such, by a proper substitution ; that is, by putting the quotient of the first term divided by the last, equal to some new unknown quantity (or letter) raised to the power expressing the dimension of the equa- tion. Thus, if the equation given be 2x^ + 24^;^ — • 315;c^ 2x^ +216:>c + 162 = ; by putting = y\ we have x =.3yi whence, after substitution, the given equation becomes 162jy^ + 6482/^ — 28352/2 + 648t/ -f- 162 = 0: which now answers to the rule, and may be reduced down to 2j/4 + 82/^ — 35Z/2 + 8z/ + 2 = 0. Of the Resolution of Equations by Approximation and Converging Series. The methods hitherto given, for finding the roots of equations, are either very troublesome and labo- rious, or else confined to pruticular cases ; but that by converging series, which we are here going to explain, is universal, extending to all kinds of equations; and^ by Appro^matmi. 15S though not accurately true, gives the value sought, with little trouble, to a very great degree of exactness. When an equation is proposed to be solved by this method, the root thereof must, first of all, be nearly estimated (which, from the nature of the problem, and a few trials, may, in most cases, be very easily done) ; and some letter, or unknown quantity (as 2) must be assumed, to express the difference between that value, which we will call r, and the ti'ue value (x) ; then, instead of x^ in the given equation, you are to substitute its equal, r ± 2, and there will emerge a new equation, affected only with z and known quantities ; wherein all the terms having two or more dimensions of 2, may be rejected, as in- considerable in respect of the rest ; which being done, the value of z will be found, by the resolution of a simple equation ; from whence that oi x ( = r ± 2) will also be known. But, if this value should not be thought suffi- ciently near the truth, the operation may be repeated, by substituting the said value instead of r, in the equation exhibiting the value of z ; which will give a second cor- rection for the value of x. As an example hereof, let the equation x^ + 10:^^ -f- 50x = 2600, be proposed : thence, since it appears that X must, in this case, be somewhat greater than 10, let r be put = 10, and r + z =z x ; which value being substituted for x^ in the given equation, we have r^ + Zr'^z + 3r2^ + 2' + lOr* + 20r2 + 102^ + 50r + 502 = 2600: this, by rejecting all the terms where- in two or more dimensions of 2 are concerned, is re- duced to r^ + 2,r^z + lOr* + 20r2 + 50r + 5O2 = car^ 1 ^ 2600 — 7-3 — lOr* — 50r 2600 ; whence 2 comes out = — — 1 3r2 + 20r + 50 = 0.18, nearly : which, added to 10 (= r), gives 10,18 for the value of x. But, in order to repeat the opera- tion, let this value be substitufed for r, in the last equa- tion, and you will have 2 = — ,000534r ; which, added to 10,1"8, gives 10,1794653, for the value of x^ a second time corrected. And, if this last value be again sub- stituted for r, you will have a third correction of x ; from whence a fourth may, in like manner, be found; and 160 The Resolutmi ofEquatiom so on, until you arrive to any degree of exactness you please. But, in order to get the general equation from whence these successive corrections are derived, with as little trouble as possible, you may neglect all these terms, which, in substituting for x and its powers, would rise to two or . more dimensions o\ the converging quantity : for as they, by the rule, are to be omitted, it is better entirely to ex- clude them than to take them in, and afterwards reject them. Thus, in the equation jjct^ + at^ + ;^ = 90, let r + z be put := X ; and then, by omitting all the powers of 2 above the first, we shall have r^ + ^rz = x^^ and ^,3^ 3^.2^ ~ x^^ nearly; which, substituted above, give yZ ^ 3^2^ ^ ^2 ^. 2r2 + ^ + 2 = 90 ; whence % is found = 90 r^ r^ _ r . Therefore, if r be now taken equal 3r^ 4_ 2r + 1 ' ^ to 4 (which, it is easy to perceive, is nearly the true value c \ u 11 u ( 90 — 64—16 — 4 6. of X) we shall have 2 ( = = — 1 = ^ ^ 48 + 8 + 1 S7 0.10, &?c. which, added to 4, gives 4.1, for the value of x^ once corrected : and, if this value of x be now substituted for r, we shall have z { = ) == ,00283 ; which, added to 4.1, gives 4.10283, for the value of x^ a second time corrected. In the same manner, a general theorem may be derived, for equations of any number of dimensions. Let ax^ + <$>^ '»-i + ^;v'»-^ + <3^^"~' + e^»-^, ^c. = Q, be such an equation, where 72, a, ^, c, d^ &:c. represent any given quantities, positive or negative ; then, putting r + 2 = .r, we have, by the theorem in p. 41, x^ =2 r^ + 7?r"~^2:, &c. Ar**~^ = r""^ + 72 — 1 X r'*"'^2, &c. ^n~2 __. ^n-2 ^jj^ — 2 X r^'^^z^ &c. &c. Which values being substituted in the proposed equations, hy Approximation. 16 J it beco mes ar"^ + nar^-^z + hr''-^ ->r n — 1 X b^'^^H + ^^n-2 ^ j^ — ^ ^ cr'^-s^ + ^r"*-3 + ;z — 3 X dr^'^'^z^ &c. = ' Q. From which z is found = Q — ar"" — br^"^ — cr^-^ — r/r"-^ — er«-^, &c. nar^''^+ /z- 1 x br^'^+n-2 x cr^'^+ ;z-3 X dr^'^ -f- /z- ^ ,< ^r"-*, i^c^ As an instance of the use of this theorem, let the equation — x^ + 300r = 1000 Be propounded. Here n being = 3, « .= — 1, 3 = 0, r = 300, and Q = 1000, we shall, by substituting these values above, have 1000 4- r^ — 300r . r • , / ^ u z = — : m which (as it appears, by inspection, that one of the values of x must be greater than 3, but less than 4) let r be taken = 3 ; and z will 127 become = -^j- = 0.5, and consequently jjc ( = r jit ( o + z) -=1 Z.S^ nearly. Therefore, to repeat the opera- tion, let 3.5 be now written instead of r, and z will 7.125 come out = '- = — 0,027 ; which, added to 263.>^5 3.5, gives 3.473, for the value of x^ twice corrected. And, by repeating the operation once more, x will be found = 3,47296351 ; which is true to the last fi- gure. When the root of a pure power is to be extracted, or, which is the same thing, if the proposed equation be x^ = Q ; then, a being = 1, and b^ c, J, &c. each = ; z, in Q •— r" this case, will be barely = ~ — 3— ; which may serve as a general theorem for extracting the roots of pure powers. Thus, if it were required to extract the cube root of 10 ; then, n being =; 3, and Q = 10, z will 10 ■ y^ be = — ; in which let r be taken = 2, and we 3r2 ' ' $hall have z = — = 0,16: therefore ;c = 2,16; from 12 ' Y 162 The Resolution of Equations whence, by repeating the operation, the next value of x will be found = 2,1 544. The manner of approximating hitherto explained, as all the powers of the converging quantity after the first are rejected, only doubles the number of figures at every operation. But I shall now give the investigation of other rules, or formulae^ whereby the number of places may be tripled, quadrupled, and even quintupled, at every operation. Let there be assumed the general equation az + h^^ tf- CT? -f- t/z^, &fc. = /? ; 2, as above, being the converg- ing quantity, and «, ^, c, d^ &c. such known numbers as arise by substituting in the original equation, after the va- lue of the required root is nearly estimated. Then, by transposition and division, we shall have p hz^ CZ^ dz* cji r 1 u • . 2 = -i , crc. irom whence, by reject- a a a a p ing all the terms after the first, and writing q z=z~^ there will be give z ^= q i which value, taking in only one term of the given series, I call an approximation of the first de- gree, or order. To obtain an approximation of the second degree, or such a one as shall include two terms of the series, let the value of 2, found as above, be now substituted in the bz^ second term — , rejecting all the following ones ; so shall 2=^ 2. :=: q ^, which triples the number of a a ^ a figures at every operation. For an approximation of the third degree, let this last value of z be now substituted in the second and third terms, neglecting every where all such quantities as have more than three dimensions of q : whence z will be had (= a ^H ^ ^) = a — — ^^ + 2bb — ac _ by Approximation* 163 The manner of continuing these approximations is sufficiently evident: but there are others, of the same degrees, differing in form, which are rather more com- modious, and whereof the investigation is also some- what different. It is evident from the given equation, that z = . 7 — ^ . , 3 . — If, therefore, the first a + bz + cz^ + dz-^^ &c. value of z, found as above, be substituted in the denomi- nator, and all the terms after the second be rejected, we shall have 2 = — 2-r- = 2--. ; which is an approxi- a + bq aa + bp mation of the second degree. But, if for z you write its second value, q ^-^ you will then have x ( = ^ , - • ) ?r a + bq — — i. + cq^ a ^ ; being an approximation of the third bb a + bq^-'^^'-^c.q- degree. A . t . • b ^ ''Xbb '-^ ac , . , A gam, by writmg q q^ + . q^ m the ^ a '^ aa room of 2, and neglecting every where all such terms as have more than three dimensions of q^ you will have ^(= ^ _ ) a + bq- q^ + q^+cX q^ q^ + «? : which is , J bb ^ ^Zb^ 2>bc a + bq — c . q^ ^ Ld . q^ ^ a ^ aa a ^ an approximation of the fourth degree. It is observable, that the powers of the converging quantity q^ in the former approximations, stand, all of them, in the numerator ; but here^ in the denominator : 164 The Resolution of Equations but there is an artifice for bringing them alike, into both^ and thereby lessening the number of dimensions, without taking away from the the rate of convergency* To begin with the approximation z = — ^ , which is of the third degree, ^ a '■ he put ^ = = the coefficient of the last term of the * a b denominator divided by that of the last but one \ so shall 2: = -— /- ; whereof the numerator and the a '\' bq — bscf" denominator, being equally multiplied by 1 + sq^ it becomes z = r-4 — j — z tt-\ • a A^ bq — bsq^ + asq + bsq^ — bs^q-^ but, the approximation being only of the third degree, hs^cf may be rejected, and so we have p + pqs a 4- sp . p Z = — ■ -^ ' = — ■■ . a -^-h -^-as . q aa -\- b '\- as • p In the same manner, in order to exterminate the third dimension of q out of the equation, ^ = P a + bq c . q^ + — + d . q^ ^ a ^ aa a put ty = — -I — — = the coefficient of the last term i a bb — ac of the denominator dwided bv that of the last but one ; P then will z =: a +hq^^-^ _c . / + ~ a ^ bq •— bsq^ + bswq ^ tb bb ^ . bb . a (because s -=. — — -7-J ; whereof the terms being equally multi plied by 1 -f- wq^ &c. we thence have z == — -—7 — —, — 2"— — - — , . , o a + bq — bsq^ + axvq + bwtf by Approximation^ .165 pXH +t oq' == : which is an approxi- i^ -f ^ -f aiv . q -^7V — s • bg ap X a -i- wp a X aa + b + axi^ .p-^w — s . pp mation ot the fourth degree^ and quintuples the number of figures at every operation. By pursuing the same method, other equations might be determined, to include five or more terms of the given series ; but, then, they w ould be found more tedious, and perplexed in proportion; so that no real advantage, in practice, could be reaped therefrom. I shall, therefore, proceed now to illustrate what is laid down above by a few examples. Exam* 1. Let the equation given be x^ + 20:^ = 100. Here, x appearing, by inspection, to be something greater than 4, make 4 + 2 = x i then the given equa- tion, by substitution, becomes 282 + z^ -^ 4, There- fore, in this case, a = 28, ^ = 1, c = 0, &c. and A A' .1 ^P f 112 28. f; = 4 ; and consequently i^ ( = = ) = ^ ^ ^ aa -^ bp ^ 7%S 19r 0.14213 ; which is one approximation of the value of z* b c But, if greater exactness be required, then s ( j-) being here — , and w (^ — f- ,, — ) = — , we shall, ^ 28 ^a ^ bb~ac^ I4' ' according to our two l2LStformiilce^ have Z (— a + sp .p _ 28 4-1 X 4 _ aa + b + ds . /; 28 X 28 + 2 X 4 ~ 90 I X. IQ/ 28 X 7 + 2 = 1386_= 0'14213564, nearly; and ^ r ap X a-^ Tvp _. 28 X 4 X 28 4- I axaa.^£^a^.p+^I^^.pp 28 X 784 + l¥-f-|- ~ 28 X 28 f ^ _ 28 X 198 _ 5544 _ 7 X 796 + ^ "" 49 X 796 + 1 "^ 39005 "" 166 The Resolution of Equations 0.1421356236, more nearly ; which value is true to the last figure. Exam, 2. Suppose the given equation^ when prepared for a solution^ to be 7682 + 482^ + 2^ = — 96. In this case, a = 768, b = 48, c =;: 1, c^ = 0, /? = 1 . ' 2b. ^ ad — be 1 48 = — , and T(; (= —) + 24 ^ a bb — ac 8 48 X 48 — 768 1 13 ^- ^ /I 4- fyns = — . rherelore 8 48— 16 32 a-\^b-^as.q — — 96 — 96 X — I X aV _ — 95 4- | _ — 191 _ "~ 768 + 48 -I- 3;i X — j ^^^ — 6—4"" 1516 "" — 0.1259894, nearly ; or z = p '\-pn w_ a + b + aw .q-^-w — s . bqq 96 — 96 X — i V ^\ — 96 -f 768+48 + 7:^X— i-f/^X n 768 — 6 — 9 + ^f^ ^ -96X128 + 9X16 ^ _ 12144 ^ _ o,l259894802, 753 X 128 + 5 96389 ' • more nearly. In the same manner the roots of other equations may be approached : but, to avoid trouble in preparing the equation for a solution, you may every where neglect all such powers of the converging quantity, 2, as would rise higher than the degree or order of the approxima- tion you intend to work b}^ And, further to facilitate the labour of such a transformation, the following ge- neral equations of the values of />, «, b^ c, d, &c. may be used. pz=zk-- ar —-* /3r^ — yr^ — ^r^, csPc. a=zcc + i£(ir + Syr^ -f 4^r^^ ^c. A = ^ + 3rr + 6h^ + lOer^ fcPc* c = y + 4i'r -f lOfr^ + uPc. d=S' -^oer -f &c. The original equation being ax + ^x^ + yx^ + h.^ + tx*^ E^fc. = k : from whence, by making r + 2 = ;^, the above values are deduced- by Approximation. 16ir The better to illustrate the use of what is here laid down^ I shall subjoin another example ; wherein let there be given x'^ + 2:c^ + Sx^ + 4x^ -f 5x (or 5x -f 4^:^ + 3;^^ + ^x* + A*) = 54321 J to find Xj by an approximation of the second degree. In this case, k being = 54321, « = 5, j3 = 4, y = 3, J^ = 2, and e = 1, we have p =z 54321 — 5r — 4r^ — 3r^ — 2r^ — r% a=z5 + 8r+ 9r^ + 8r^ + 5r^, and ^ = 4 + 9r + 12r2 + lOr^. Which values, by assuming r = 8, will become p = 11529, a = 25221, and b = 5964: whence <^ ( = ^1 = 0,45, and 2 ( = _A^) = 11^2. =0,41; a^ \ ^ a + bq^ 25^21+2683 and therefore ;v (= r + z) = 8,41, nearly. To repeat the operation, let 8,41 be now substituted for r; so shall p = 135,92, a = 30479, b = 6876, ^ ( = t) = 0.00445, and 2 (= — ^) = 1^?^ = a^ ^ a + bq' 30479 + 30 0,004455: which, added to 8,41, gives 8,414455, for the next value of x. The formulce^ or approximations determined in the pre- ceding pages, are general, answering to equations of all degrees howsoever affected ; but in the extraction of the roots oi pure powers, the process will be more simple, and the theorems themselves very much abbreviated. For let x^ :=. k be the equation whereof the root x is to be extracted ; then, by assuming r nearly equal to x^ and making r x 1 + z = ^, our equation will become r^ X 1 + z]" = ^, or 1 + zY = — ,' that is, 1 + nz n — 1 2 . n — 1 n — 2 „ , n — 1 2 2 3 ^ 2 • — - — . . 2*, £sPc. = — : from whence, by trans- 3 4 r'* position and division, % + ^ . z^ + ^ ■- . ^^ . . 2} 168 The Resolution of Equations 2 3 4 nr"^ Here, by a comparison with the general equation, az + bz^ + cz^ + dz\ ^c. — p, we have ^ = 1, /. __ ^ — 1 ^ w — 1 n — ^ 72 — 1 w — 2 = — - — , c = . ^ = . . 2 2 3' 2 3 — -— ^, £s?c. and p = ^ : whence q f-^) =z p ; s ,^ c. n — 1 n — 2 72-4-1 ' 2i^ /- J ~ — ^ _- . and w ( — ^a b^ 2 3 6 ^ a h _ 72 1 yV.72 2.7 2 3 I- .72 1 .72 — 2 h — ^'l^ 1 rT^zrt'i:^: ^ _ 72 1 72 2 . 72 — 3 2^2 2 . 72 — 2 ___ " 1 2 . n + 1 "^ 72 — 1 72 — -2 72 1 — — - + -r- x/2— -3 — 272 + 2= :; + 1 ^ 2 . 72 + 1 ; : ^ 1 ■ n — 2 n — 1 72 — 2 72 ^, .=. X— n + l=: — j 5— = — . There- 2 . ;z + 1 . .. . . fore, for an approximatiSh of the third degree, we have a X s fi . p ___ t 4- \t> . n -^ 1 . p __ aa + b + as . p ^ — 1 22 4-1 "^ 2 "^ 6 ' ^ p + n ' » fi — , ^^^ £^^ ^^ approximation of the 1 + 2;/ — 1 . ^y fourth degree, z = ^ = a + b 4- aw . q -{- w - — s • bq^ 22 — 1 n n // 4 1 n — 1 ^^ ^2 ^2^2 6 2 ^ by Approximationi 169 p + hip^ XT : — r t 2 r ^ HeilCe it IS evi- 272 1 2n 1 72 — 1 1+ .p + • .p^ dent that the root x {r x 1 + z) of the given equation a-" = L will be equal to ;* -(- ^ ■- ' hT nearly ; 1 + 2n—l. ^p and equal to r + —.Jl2Ll + ^^^P 2?? — 1 ^ . 272 — 1 , n — 1 . p- ^ 2 ^^ .12 more riearhj. But both these theorems will be rendered a little more 727*^ commodious, by putting v = —- -^ and substitut- ing — , in the place of its equal, /;, whence, after V 7' X 6^ J f2 -4- 1 proper reduction, x will be had = r -f V X 6 v -f 4?^ — 2 V , , , r X 2t; + 72 nearly; and equal to r +- — ===== - ^ 'yX2Z;+272 l+^'U 1.272 — 1 more nearly* I shall now put down an example or two, to show the use and great exactness of these last expressions. 1., Let the equation given be .v^ = 2, or, which is the same thing, let the square root of 2 be required. Then, assuming r = 1.4, we have ?i = 2^ A = 2, , 71Y'' 2 X 1,96. „^ , , p V (^ = ^ — 1 — ) — 98; and therefore r 121^^4- n-fi ^ ^ MJ<__591 ^ ^ ^ ^ X 6t; + 4?2 — 2 9« X 594 197 197 —• = 1,4 + = 1,41421356; which is 70 X 198 ' ^ 13860 ' the value of vT according to the former approximation ; Z irO The Resolution of Equationa but, according to the latter, the answer will come out 5544 ^••^ + c^^r.. = 1.41421356236; which is true to the last figure ; and, if with this number the operation be re- peated, you will have the answer true to nearly 60 places of decimals* 2. Let it be required to extract the cube root of 1728. Here, taking r = 11, we shall have v (■— - — ^) i — r« 3993 ;= —1- = 16.05793 ; and therefore r 4- 397 ^ "^'"^-'^ =11,99998, 2v + 2n — 1 X-i) + \ X 71' — 1 X 2?2 — 1 which differs from truth by only part of a ^ ^ 50000 ^ unit. 3. Let it be proposed to extract the cube root of 500. Here, the required root appearing to be less than 8, but nearer to 8 than 7, let r be taken = 8, and 3 V 512 we shall have v ( =: ) = — 128; and there- ^ —12 ^ r . r X "2v 4-71 fore r + 2v -i- 2n — 1 X V + } X n — 1 X 2n — 1 = 7.937005259936 ; which number is true 96389 to the last place. 4. Lastly, let it be proposed to extract the first sursolid root of 125000. In which case, k being = 125000, n = 5^ r = 10, and v =. 20, the required root will be found =;= 10,456389. Besides the different approximations hitherto delivered^ there are various other ways whereby the roots of equa- tions may be approached ; but, of these, none more gene- ral, and easy in practice, than the following : by Approximation. 171 Let the general equation, az + bz^ + cz^ + dz^ + ez*^ cifc. = /?, be here resumed; which, by division, becomes z = 7 -^ . If, P P P P P tjhierefore, we make A == — j and neglect all the terms P after the first, we shall have z = --- ; being an approxima- tion into of the first degree. And if this value of z be now substituted in the se- cond term, and all the following ones be rejected, we shall then have z = , — = = — - a b \ CL K b Q — + — X-T- _ A + — p p A p p (by making B = ) ; which is an approximation of the second degree. In order now to get an approximation of th^ third de- gree, let this last value be substituted in the second term, neglecting all the terms after the third f so shall z = — J T : but here, in the room of a b A . c ^ ' ' ^ p p ^ P z^^ either ,of the squares of the two preceding values of 2, or their rectangle may be sub3tituted, that is, either -T" X -ri -T7 X -77> or —- X T7 ; but the last of these A A B B A B (= — ) is the most commodious ; whence we have z = B B . ^ aB + bA-hc r=7r; supposmg C = ^b + Aa+1- ^ ^ p p p Again, for an approximation of the fourth degree, we . b ^Bc^cBAcA have --2=;~x— ; ~^=— X-itX-^= — X— ; P P ^ P p L B p C . d . d B A 1 d 1 ,. - anci — z^ = ~x -prX i^X --:■ = — X —; which va- fl p C B A p L 172 The Resolution of Equations lues being substituted in the general equation, and all the terms after the four first rejected, there now comes out rr: --_ ; by makmg D = ---. -i— . i) p In like manner, for an approximation of the fifth de- gree, we shall have — ^=:— x— -, — 2^ = — Xtt-X t- p p Y> p p B C cB d ^ d C B A dA , e ^ = /7d> " == 7"" D "^ CT^ B =^' ^"^ 7 ' C B A 1 e ^ D^Z^li^A^pD'' consequently z Z' P P p P h, = — -' • Whence the law oi con- tinuation is manifest ; whereby it appears, that, if there be - ^ a ^ ciA •\- h ^ aB -f bA 4- c taken A = — , B = 3. C = -:_ P P P ^ <2C + ^B -f- cA 4- ^ ^ oD -^ bC + cB -^ dA -^ e U z=. — , H. = ^^ t P P ,. aK+hB+cC+dB+cA+f ^ a¥+bE+cD+dC+eB+fA+g p p ,^ ^ ... 1 A B C D E F ^ . - J-fc. then will ^, _, ^, — , -g., ^, — , £ifc. be so many successive approximations to the value of 2, ascending gra- dually from the lowest to the superior orders. An example will help to explain the use of what is above delivered ; wherein we will suppose the equation given to be 122 + 62^ + 23 :=2. Here a = 12, ^ = 6, c = 1, <^ = 0, e = O, &fc. and /; = 2 ; whence A ( = ^) = 6, B (= — ^-J = c,o r /- ^ ^^ + ^^ + S - 12 X 39 4-6 X 6 4-1 _ I by ApprQximatton. 173 505 j^ , flrC 4- ^B + cA -f d ^ __ 6X 505 4- 6 X 39 + 6 = 1635, &Pc. A 2 Therefore, -- = — = 2, nearly. r> 13 B 78 ; C 505 -^ C 101 .,; = — = 2, Still nearer. D 654 ' From the same equations the general values of B, C, D, ^c. may be easily found, in known terms, independent of each other. 1 hus B ( = ) = -r -f (because A = — j ; ^ P P P\ P ^ P^ , ^ , «B hK. c ^ «^ . 2ab . c alsoC(= _ + _+_) = — + _. + ^; p p p p^ p^ p , ^ . aC bB cA d . fl4 3^2^ ^ac + ^(^ and D(=: — 4 1 J = 1 -—^ P P P P P^ P^ P" •\ , &c. Therefore A _ op B '^'^^p' B _ p xa^ + bp C "" r/^ + '■J.abp -\-cp^ '' C __ p X a^ -^ ^abp + cp^ ^ a"^ + Za^bp +~2ac -^Jb . p^ + dp^ D _ px a"^ + Sci^bp -f- 2ac + bb .p- +dp^ ^ a^ + A^ctbp + 3ac + 2>bb . ap^ + be + ar/ . 2/>^ + ^/)^' &Pc. which are so many different approximations to the va* lue of z. Thus far regard has been had to equations which consist of the simple powers of one unknown quantity, and are no ways aft'ected, either by surds or fractions. If either of these kinds of quantities be concerned m an equation, the usual way is to exterminate them by multiplication, or involution (as has been taught i» 174 The Resolution of Equations sect. IX.) But as this method is, in many cases, very laborious, and in others altogether impracticable, espe- cially where several surds are concerned in the same equation, it may not be amiss to show how the method of converging series may be also extended to these cases, without any such previous reduction. In order to which it will be necessary to premise, that if A + B represent a compound quantity, consisting of two terms, and the latter (B) be but small in comparison of the for- mer ; then will B 1 A + B 1 A A^^^A 2°. A + B]2 = A2 + 1 1 B B A2i ^ A*B or A^ + ^A or- B A+B]! Ai 2AI Ak 2AxAi 4°. A + B]i = A3" + —i or Ai + 3A3 1 Ai" B r-. A + B]i A + B"|^ = A^ + 1 1 or- AJB B j>, nearly. SAt At 3A X As — Ai . Ba' 3 or A4 -f 4A4 B 4A B A + Jb]! Ai 4AI Ai 4AxAij All which will appear evident from the general theo- rem at p. 41 : from whence these particular equations, or theorems, may be continued at pleasure ; the values here exhibited being nothing more than the two first terms of the series there given. But now, to apply them to the purpose above mentioned, let there be given V 1 4- a:^ 4- V2 4- x'^-\- v' 3 -f ;f 2 = 10, as an exam- ple, where, x being about 3, let 3 -f ^ be therefore sub- stituted for »v, rejecting all the powers of e above the by Approximation* 1 '/5 first, as inconsiderable, and then the given equation will stand thus, V 10 + 6^ + Vll + 6g + v" 12 + "6t* .-= 10:_jDut, by theorem 2, V 10 + 67" will be = VlO 3V/ 10 X ^ -j ^ nearly; for, in this case, A = 10, and B = 6^^ and therefore A^ + = V 10 4- 2A ^ oVTo X e lO in like manner isVll +6^ = Vll + ?i:^iiJi-£, &c. and consequently VlO + ^"^^^ ^ ^ 4. 11 _ ^ ^ •^_ ^ 10 ^ Vn + ^^V^^ + Vli + £i:ili5-i = 10; which. 11 12 ' contracted, gives 9.944 + 2.5^18^ = 10; whence 2.718e = .056, and e = .0205; consequently x = 3.0205, nearly. Wherefore, to repeat the operation, let 3.0205 + e be n ow substituted f or x ; then will V 10.12342 -L 6.Q41 g + V 11.12342 + 6.041^ + V 12.12342 + 6.041t^ = 10; whence, ^z/ theorem 2, 6.041^ , V 10.12342 + — 7= + V 11.12342 + 2V 10.12342 6.041^ 6.041^ — + V 12.12342 + — , = 10, or 2N/11. 12342 - ." -• 2a/ 12.12342 ' 9,9987814 + 2.7224^ = 10: from which e comes out = .000447, and therefore x = 3.020947 ; which is true to the last place. Again, let it be proposed to find the root of the equa- 20^ W ^v -i- ^2 tion --; + ^^ ^ "^ = 34. Put 20 + V lb -^ 5x + x^ 25 e = X ; then, by proceeding as before, we shall havd 400 4- 20e 20 -f e X V^405 -4- 40, and the latter by B, and then vsubtracting the one from the other, we shall have bAs — Bas = bOi — B7; and therefore 5 = /; ^ : . Ab — ao whence x (=f+ s) is given. Again, by multiplying the former equation by a, and the latter by A, £s?f. we shall have aBt — Abt *= cQ -- Ay, artd therefore t = ^Bl"^ = A^ - aQ . ^^^^^^^ Ba ~ bA Ab — aB !/ (= ,^ -f i^ l^l^ewise given. 2 A 178 The Resolution of Equations It is easy to see that this method is also applicable, in cases of three or four equations, and as many unknown quantities ; but as these are cases that seldom occur in the resolution of problems^ and, when they do, are reducible to those already considered, it will be needless to take further notice of them here : I shall, therefore, content myself with giving an example or two, of the use of what is above laid down^ le Let there be given x^ + y^ = 10000, and x^ — z/* =3 25000 ; to find x and y. Then, by writing f + s = x^ g- -}- 1 =z 2/, and proceeding according to the afore- going directions, we shall have f"^ -f- 4f^s -f- g^ + A^g^t = lOOOO, and/^ + Sf'^s — ^^ _ 5^^ -. 25000, or 4f^s ^- 4g^t = 10000 — /^ — g^, and Sf'^s — 5g*t = 25000 + ^^ — /^ : therefore, in this case, A = 4/^, B = 4^'^ Q =. 10000 — /^ — g-^ a =z 5f\ d = — 5g^^ and q = 25000 + ^^ — /^^ But it appears, from the first of the two given equations, that x must be something less than 10, and from the second that y must be less than x : I thei'efore take y^= 9, and ^ = 8 ,* arid then A becomes = 2916, B = 2048, Q = — 65/, a = 32805, ^ = — 20480, ^ = — 1281 ; and therefore * cfJ^p = _o.,=,.„,,(^=^,=-o.u. hence x = 8.87, and y = 7.86, nearly. Therefore, in order to repeat the operation, let f be now taken = 8.S7, and g^ = 7.86 ; then will A = 2791, B = 1942, Q = _ 6.76, a z= 30950, b = ^ 19083, and ^ = 94 ; consequently ^ ( = -^ ^) = .00047, and t (= ^f ~ ^) = — .00415 ; whence x == 8.87047, Ab — . an and y = 7.85585; both which values are true to the last figure. Example 2. Let there be given 20;^ + xy^[^ + ^x\^ xy -=12, and >/x^ + y^ + , ' =13. Here the -^ Vx^ — y^ given equations, by writing f + s for x, and g -^ ^ by Approximation. 1 7-9 ior t/, will become 20f -f 20^ -f fg^ -f 2fgt + ^p^]"^ + ^/ Sf + Hs = 12, and V/^ + ^ + 2/^^ + 2^'^ 4- > ,>^ ' o /> = 13 ; but 2QA + f^^ + 2^"^ + 2/^^ + ^'^'1^5 t)y what is shown in — . .2 p. 174, will be transformed t^WTW]^ + ^^ ' -= 3 X 20/ +fg^ X 20s + 2fgt + g^s (supposing all the terms that have^ more than one dimension of s and t^ to be rejected, as inconsiderable) ; also V/^ ^^^^2/3+ 2gt^ is trans- fi+gt ^ 1 formed to V/^ 4. 0-2 1 , .. • _. ^^^ — ^\//2+^2' s/f2_^^2_^,2fs—2gt 1 fs — P-t to —=====■ — ===== '; — ; therefore our equations will stand thus, 3 X 20/+/^^ Vsf + -7= = 12, and VP + g^ + /, ^ + ^f\ — t r — g'x vp — ^ = 13: which equations, ify be assumed = 5, and g = 4, will be reduced to 5.6462 + .01045 X 36^* + 40^ + 6 .3245 + .6324 ^ = 12, and 6.4031 + 781^ + .625^ + 20 + 5^ + 4^ X .3333 -^ .1852^^ + .1482? = 13 5 whence 1.008^ -f- .418if = .0293, and 1.59^ — 5.255^ = .0698: therefore, in this case, A = 1.008, B = 0.418, Q == .0293, a = 1.59, b = — 5.255, and q IjQ B(7 = .0698 : consequently s (= -^ ^) = 0.305, and t Ao — ao - Ao' — — dQ (= -— ^) = — *0040j therefore x = 5.0305 and At? — aB"^ ' y = 3.9960, mo The Resolution of SECTION XIIL Of Indeterminate or Unlimited Problems. A PROBLEM is said to be indeterminate, or unlimited^ when the equations expressing the conditions thereof are fewer in number than the unknown quantities to be determined -, such kinds of problems, strictly speaking, being capable of innumerable answers : but the answers in whole numbers, to which the question is commonly re- strained, are, for the general part, limited to a determinate number ; for the more ready discovering of which, I shall premise the following LEMMA. Supposing to be an algebraic fraction, in its lowest terms, x being indeterminate, and «, ^, and c given whole numbers ; then, I say that the least integer^ cix db b for the value of x^ that will ^.Iso give the value of — C an integer, will be found by the following method of cal° culation. Divide the denominator (c) by the coefficient (a) of the indeterminate quantity ; also divide the divisor by the re- mainder^ and the last divisor again by the last reinainder i and so o?iy till a unit only remains. Write down all the quotients in a line ^ as they follow^ under the first of "which write a unit^ and under the second zvrite the first ; then multiply these tivo together ^ and hav* ing added the first term of the loxver line {or a unit^ to the product^ place the sum under the third term of thq upper line : multiply^ in like mamier^ the next two corres- ponding terms of the tzvo lines together^ and^ having added the second term of the lower to the product^ put down the result under the fourth term of the upper one: proceed on^ in this way^ till you have multiplied by every number in the upper line^ Indeterminate Problems. 181 Then multiply the last number thus found by the absolute quantity (^) in the numerator of the given fraction^ and divide the product by the denominator ; so shall the re- mainder be the true value of x^ required ; provided the number of terms in the upper line be even^ and the sign ofb negative^ or if that number be odd and the sign of b af- firmative; but^ if the number of terms be even^ and the sign ofb crffirmative^ or vice versa, then the difference between the said remainder and the denominator of the fraction xoill be the true answer. In the general method here laid down, a is supposed less than c, and that these two numbers are prime to each other : for, were they to admit of a common measure, whereby b is not divisible, the thing would be impossi- ble, that is, no integer could be assigned for x^ so as aoc i b » '* to give the value of an integer : the reason of which, as well as of the lemma itself, will be explained a little farther on : here it will be proper to put down an example or two, to illustrate the use of what has been al- ready delivered. Examp. 1. Let the given quantity be '^ — • Then the operation will stand as follows : 87 )256( 2 82)87(1 2, 1, 16, 2 5)82(16 1, 2, 3, 50, 103 2)5(2 50 71X + 10 1 25 6)5150( 20 lo = Ar, Examp. 2/ Given ri)89(i 18)71(3 1, 3, 1 17)18(1 1, 1, 4, 5 1 10 50": 1 82 The Resolution of ^xanip. 3. Gwen • ^ . ^ 450 3rr)450(l 1, 5, 6 73 )3/7( 5 1, 1, 6 37 12)73(6 250 1 1850 74 Examp* 4. Given 450)9250(20 250 450 200 = AT. 987; 1, 2, 3 9_ ^ 8)9(1 18 r 26 PROBLEM II. Supposing 9x -J~ loz/ = 2000, it is required to find all the possible values ofx and y in xvhole positive numbers. By transposing 13z/, and dividing the whole equation u r. 1 2000— 13v ^^^ 2 — % by 9, we have x = ^ = 222 — y -^ ^ ; which, as X is a whole positive number^ ' by the question, must also be a whole positive number, and so likewise -^ J from which the least value of y^ in whole num- bers, will come out = 5 ; and consequently the corres- ponding value of X = 215. From whence the rest of the answers, which are 16 in number, will be found, by adding 9 continually to the last value of y, and subtracting 13 from that of x^ as in the annexed table, which exhibits all the possible answers in whole num- bers. x=215;202|189ll76|1631l50ll37|124|lll|98|85| 72| 59| 46| 33! 20/ 7 ^=51 14| 23| 32| 4l| 501 59\ 68J 77l86195|1041113|122113lll40ll49 In the same manner, the least value of y, and the greatest of x being found, in any other case, the rest of the answers will be obtained, by only adding the co- efficient of x^ in the given equation, to the last value of ^/, continually, and subtracting the coefficient of y from tlie 184 The Resolutkn t)f corresponding Value of $c\ Hence it follows, that, if tli€ greatest value of x be divided by the coefficient of y, the remainder will be the least value of x^ and that the quo- tient + 1 will give the number of all the Answers. But it is to be observed, that the equations here spoken of are such wherein the said coefficients are prime to each other ; if this should not be the case, let the equation given be, first of all, reduced to one of this form, by dividing by the greatest common measure. PROBLEM IIL To find hoxv many different ways it is possible to pay \00L in guineas and pistoles only; reckoning guineas at 21 shillings each^ and pistoles at If. Let AT represent the number of guineas, and y that of the pistoles ; then the number of shillings in the guineas being 21;c, and in the pistoles 17y, we shall therefore have 21 at + 17z/ = 2000, and consequently x = ^ = 95 4- — -^ ^ : which beinj^ a whole 21 ^21 ^ number, by the question, it is manifest that — 2— must also be an integer : now the least value of z/, in whole numbers, to answer this condition, will be found = 4, and the expression itself = 3 ; the corresponding, or greatest value of X being = 92 ; which being divided by 1 7, the coefficient of y (according to the preceding note^^ the quo- tient comes out 5, and the remainder 7 ; therefore the least value of X is 7, and the number of answers ( = 5 + 1) ;= 6 ; and these are as follows : X = 92 75 58 41 24 7 7/=: 4 25 PRO 46 >BLI 67 88 iM IV. IOC), To determine whether it be possible to pay 100/. m gui- neas and moidores only; the former being reckoned at 2\. shillings each^ and the latter at 27* Here, by proceeding as in the last question, we have Indeterminate Problems. 185 2000 — 27y 21 ^2\x + 27y = 2000 ; and consequently x' \ = 95 — J/ —* ■ *^ — : where, the fraction being in its least terms, and the numbers 6 and 21, at the same time, admitting of a common measure, a solution in whole numbers {hy the note to the preceding lemma) is im- possible. The reason of which depends on these two considerations : that, whatsoever number is . divisible by a given number, must be divisible also by all the divi- sors thereof; and that any quantity which exactly mea- sures the whole and one part of another, must do the like by the remaining part. Thus, in the present case, the quantity 6?/ ■— 5, to have the result a whole num- ber, ought to be divisible by 21, and therefore divisible by 3, likewise, which is here a common measure of a and c : but 6z/, the former part of 6z/ — 5, is divisible by 3 ; there^- fore die latter part — 5 ought also to be divisible by 3 ; which is not the case, and shows the thing proposed to be impossible. PROBLEM V. A butcher bought a certain number of sheep and oxen^for which hepaidlOOL ; for the sheep hepaidl7 shillings apiece^ and for the oxen^ 07iewith another^ he paid 7 pounds apiece; it is required to find hoiv many he had of each soj't. Let X be the number of sheep, and y that of the oxen; then, the conditions of the question being ex- pressed in algebraic terms, we shall have this equation, viz, 17 X + 140z/ = 2000; and consequently x = 2000 — 140v ^,^ „ 4z/— 11 1 . , , . ^ = 117 — Sy --^ ; which bemg a whole number, — must therefore be a whole 17 number likewise : whence, by proceeding as above, we find 2/ = 7, and x ■=. ^0 ; and this is the only answer the question will admit of; for the greatest value of x cannot in this case be divided by the coefficient of 2/, that is, 140 cannot be had in 60 ; and, therefore, ac- 2B 186 The Resolution of^ cording to the preceding note, the question c^n have only one answer, in whole numbers. PROBLEM VI. A certain number of men and women being merry -making together^ the reckoning came to SS shillings^ toxvards the discharging ofwhich^ each man paid 2>s. €d. and each wo- man \s. 4^d, : the question is^ to find how many persons of both sexes the company consisted of Let X represent the number of men, and y that of the women ; so shall 42x + 162/ = 396, or 2\x -f- 8z/ 193 ■ 21^ = 198; and consequently y = 1- = 24 — 2x 8 5^1%* — " 6 5^4%* ' 6 ' : whence, y being a whole number, t- 8 8 must likewise be a whole number ; nnd the value of x^ answering this condition, will be found = 6 ; and consequently that of z/ (= 24 — 12 — 3) = 9 ; which two will appear to be the only numbers that can an- swer the conditions of the question ; because 21, the co- efficient of .T, is here greater than 9, the greatest value of 2/. PROBLEM VIL One bought 12 loaves for 12 pence ^ whereof some were tivO'penny ones ^others penny ones^ and the rest farthing ones : xvhat number xvere there of each sort ? Put X = the number of the first sort, y = that of the second, and z = that of the third ; and then, by the con- ditions of the question, we have these two equations, viz> X + y + zz=: 12, and 8x +4y = 15410 ; to determine the least value ofx^ and the greatest ofy^ in whole positive numbers. By transposition and division we have 15410—87^ ^^ 87^ — 50 , , . y z=. — ^ =60 ^ : where the Irac- ^ 256 256 tion being the same with that in examp. 1 to the pre- mised lemma^ the required value of x will be given from thence = 30 ; from thence that of y will likewise be known. But I shall in this place show the manner of deducing these values, independent of all previous consi- derations, by a method on which the demonstration of the lemma itself depends. In order to this, it is evident, as the quantity ^7x — b (supposing b = 50) is divisible by 256, that its double 174a;' — 2b must be likewise divisible by 256. But 256.V is plainly divisible by 256 ; and if from this the quantity in the preceding line be subtracted, the re- mainder, 82^ -f- 2^, will be likewise divisible by the same number ; since xvhatsoever number measures the xvhole^ and one part of another^ must do the like by the re- marning part : for which reason, if the quantity last found be subtracted from the first, the remainder 5x — Zb will also be divisible by 256 : and, if this new remain- der, multiplied by 16, be subtracted from the preceding one (in order to farther diminish the coefficient of :r), the difference 2x + 50b must be still divisible by the same number. In like manner, the double of the last line, or remainder, being subtracted from the preceding one, we have x — 103^, a quantity still divisible by 256 : but — ^ = 20 H ; therefore x — 30 must 256 256 be divisible by 256 ; and consequently x must be either equal to 30, or to 30 increased by some multiple of 256 ; but 30, being the least value, is that required. It may not be amiss to add here another example, to il- lustrate the way of proceeding by this last method: where - ... 987Ar 4-651 ;n let us suppose the quantity given to be — - — ^-^ Indeterminate Problems^r 1 89 Then, making i = 651, the whole process will stand as follows: From - - 1235a: sub. - - - 9^7X'^b l.rem. « - 24<^x~b 1. rem. X 3 - 744^x~Zb 2. rem. - - 243:v + 4/^ 3. rem. - - Sx-^sb 3. rem. X 48 - 240.y — 2406^ 4. rem. - - 3x -f 244/^ 5. rem. - - - 2x — 249^ 6. rem. - - x + 493b ; vvhere, x being without a coefficient, let 493 5, or its equal 320943, be now divided by 1235, the common measure to all those quantities, and the remainder will be found 1078 ; therefore x + 1078 is likewise divi- sible by 1235 ; and consequently the least value of x (= 1235 — 1078) = 157. The manner of working, according to this method, may be a little varied ; it being to the same effect, whether the last remainder, or a multiple of it, be subtracted from the preceding one, or the preceding one from some greater multiple of the last. Thus, in the example before us, the quantity 248x — ^, in the third line, might have been multiplied b}^ 4, and the preceding one subtracted from the product ; which would have given 5x — 5b (as in the sixth line) by one step less. If the manner of proceeding in these two examples be compared with the process for find- ing the same values, according to the lemma^ the gi'ounds ••f this will appear obvious. PROBLEM X. Supposing e^f^ andg to denote given integers ; to deter- X "~" e X - f viine the value of x^ such that the quaiitities , — -^, and - ^ , maij all of t herd be integers* 1 90 The Resolution of By making -~ — = z/, we have x = 28z/ + e ; which vaKie being substituted in our second expression, it becomes — — ^ ; which, as well as y. is to be 19 ^ a whole number : but — , ^ T, ~J ^ by making b z=z e — f, will be = 2/ H — ^ — ^-— ; and therefore 19// and 18// + 2(^ being both divisible by 19, their difference, 7j — 2^, must also be divisible by the same number ; whence it is evident, that one value of y is 2b ; and that 2b +192 (supposing z a whole number) will be a general value of y ; and consequently that x (= 28z/ + e) = 5322 + 56b + (? is a general value of Xy an- swering the two first conditions. Let this, therefore, be substituted in the remaining expression ^ ; which, by that means, becomes i ± — ZZK = 352 ^ ' 15 ^3^+. ZUi (supposing <3 = 11^ +> — ^ = 12e — liy^ — ^). Here 152 and 142 + 2/3 being both di- visible by 15, their difference 2 — 2/3 must likewise be divisible by the same number ; and therefore one value of 2 will be 2/3, and the general value of 2 = 2/3 + 15w : from whence the general value of x (= 5322 + 56b + e) is given = 79S0io + 1064/3 + 56^ + e; which, by restoring the values of b and /3, becomes ,7980w + 12825^— 11 reo/' — 1064^. Now, to have all the terms affirmative, and their co- efficients the least possible, let w be taken == — e + 2f + g', whence there results 4845^ + 4200/' + 6916^, for a new value of x : from wl>ich, by expounding e, f and^, by their given values, and dividing the whole by 7980, the least value of x^ which is the remainder of the division, will be known. Indeterminate Problems^ 19> PROBLEM XL If5x + 7z/ + 112 = 224 ; it is required to find all the possible values ofx^ y^ and 2, in whole numbers. In this, and other questions of the same kind, where you have three or more indeterminate quantities, and only one equation, it will be proper, first of all, to find the limits of those quantities. Thus, in the pre- , . 224 — 7y — ll2 , , sent case, because ^ is = --^ , and because the least values of y and z cannot, by the question, be less than unity, it is plain that x cannot be greater than 224 7 11 *^*^ . . -, or 41 : and, in the same manner, it will appear that y cannot be greater than 29, nor z greater than 19 ; which, therefore, are the required limits in this T\/r • • 224 — 7z/ — ll2 . Moreover, smce :v is = — ^ = 45 case. — y — 2z — — i — ^— I! — = ^ whole number, it is 5 manifest that ^ — "^ — must also be a whole num- 5 ber : let 2 + 1 be therefore considered as a known quantity, and let the same be represented by b^ and then the last expression will become ■ — ; from which, by proceeding as above, we shall get y = 2b = 22 + 2 ; whence the corresponding value of x comes out = 42 — 52. Let z be now taken = 1 ; then will a;* = 37 and y = 4 ; from the former of which values let the coeffi- cient of y be continually subtracted, and to the latter, let that of X be continually added, and we shall thence have 37, 30, 23, 16, 9, and 2 for the successive values of x-^ and 4, 9, 14, 19, 24, and 29 for the correspond- ing values of y ; which are all the possible answers when 2 = L 192 The Resolution of Let z be now taken = 2, then x = 32, and j/ = 6 ; let the former of these values be increased or decreased by the multiples of 7, and the latter by those of 5, as far as possible, till they become negative ; so shall we have 39, 32, 25, 18, 11, and 4, for the successive values of X in this case, and 1, 6, 11, 16, 21, and 26, for the respective values of t/ : which are all the answers when 2 = 2.^ . Again, let z be taken = 3 ; then, by proceeding as above, the corresponding values of x and y will be found equal to 34, 27,20, 13,6; and 3, 8, 13, 18, 23, respec- tively ; and so of the rest : whence we have the follow* ing answers, being 60 in numbers. z - y .oc 4. 9.14.19.24.29. 37.30.23. 16. 9.2. 2 1. 6.11.16.21.26. 39.32.25.18. 11.4. 3 3. 8.13.18.23. 34.57.20.13. 6. 4 5.10.15.20.25. 29.22.15. 8. 1. 5 2. 7.12.17.22. 31.24.17.10. 3. 6 4. 9.14.19. 26.19.12. 5. 7 1. 6.11.16. 28.21.14. 7. 8 3. 8.13.18. 23.16. 9. 2. 9 5.10.15. 18.11. 4. 10 2. 7.12. 20.13. 6. 11 4. 9.14. 15* 8. 1. 12 1. 6.11. 17.10. 3. 13 3. 8. 12. 5. 14 5.10, 14. 7. 15 2. 7. 9. 2. 16 4 4. 171. 6. 183. 1 . Indeterminate Problems. 193 PROBLEM XII. If \7x + 19j/ +2I2 = 400; it is proposed to jind all the possible values of x^ ^/, and 2, in whole positive numbers. When the coefficients of the indeterminate quanti- ties x^ 2/, and z are nearly equal, as in this equation, it will be convenient to substitute for the sum of those quantities. Thus, let :v + ^ + 2 be put = m ; then, by subtracting 17 times this last equation from the pre- ceding one, we shall have 22/ + 42 = 400 — 1 7w ; and by subtracting the given equation from 21 times the assumed one, .r + 2/ + 2 = 7??, there will remain 4'X + 2y=i2\m — 400. Therefore, since y and 2 can have no values less than unity, it is plain, from the first of these two equations, that 400 — 1 7m cannot be less than 6, and therefore m not greater than , or 23 : also, because, by the second of the two last equa- tions, 21??z — 400 cannot be less than 6, it is obvious that m cannot be less than IL^^ or 19-. therefore 21 19 and 23 are the limits of m in this case. These be- ing determined, let 4x be transposed in the last equation, and the whole be divided by 2, and we shall have m y z=z 10?n — ' 200 — 2x + — ; which being a whole number, by the question, — must likewise be a whole number, and consequently 7n an even numbeir ; which, as the limits of in are 19 and 23, can only be 20, or 22 : let, therefore, m be first taken = 20, then y will be- come = 10 " — 2x and 2 (m — x — 2/) = 10 + x ; wherein x being taken equal to 1, 2, 3, and 4, suc- cessively, we shall have y equal to 8, 6, 4, 2, and 2 equal to 11, 12, 13, 14, respectively, v.hich are four of the answers required. Again, let m be taken = 22 ; then will 2/ = 31 — 2x^ and 2 = ;c — 9 ; wherein let X be interpreted by 10, 11, 12, 13, 14, and 15, suc- cess! velv, whence z/ will come out 11. 9, 7, 5, 3. and 1 ; 2C 19^4 The Resotution of and X equal to 1, 2, 3, 4, 5, and 6, respectively. 'J'here- lore we have the ten following answers ; which are all the question admits of. .V = 1 2 3 4 10 11 12 13 y — 8 6 4 2 11 9 7 5 11 12 13 14 1 2 3 4 14 3 15 1 6 PROBLEM XIII. Supposing 7x +9y + 23z = 9999 ; it is required to determine the number of all the answers^ in positive inte- gers* In cases like this, where the answers are very many, and the liumber of them only is required, the following method may be used. In the general equation ax -f- by -^ cz ■= k (where a and b are supposed prime to each other) let z be assumed = ; and find the greatest value of x^ and the least of ?;', in the equation ax -\- by z:^ k^ thence arising ; de- noting them by g and / : find, moreover, the least posi- 'tive value of ;z, in whole numbers, from the equation am '\'bn-=L c^ together with the corresponding value of ;;?, whether positive or negative ; then, supposing q tG represent an integer, the general value of x may be ex- pressed by ^ — bq — 7722;, and that of 2/ by I -}- aq ^-^nz ^ as will appear by substituting in the general expression ax + by -\- cz^ which thereby becomes ag — abq - — arnz -\- bl -{- abq — bnz -{- cz -=1 k^ as it ought to be, because ag + bl =: i, and all the rest of the terms destroy one another. And it may be observed farther, by the bye, and is evident from hence,, that any two corresponding values of 7n and 72, determined from the equation am -f- bn = c, will equally fulfil the conditions of the general equation ; but the least are to be used, as being the most commodious. As to the limits of z and ^, these are easily determined ; the former from the original equation, and the latter from the general va- lue o£ X ; by which it appears that q cannot exceed ' > ~ ' ■ ; w^herein the greatest, or the least value of z is to be used, according as the second term, after substitu- Indeterminate Problems. 195 tion for m^ is positive or negative. But, besides this, there is another limit, or particular value of q to be determined, which is of great use in finding the number of answers. It is evident, from the given equations, that the values of x will begin to be negative, when z is so increased as to exceed ^ ? ; and that those of m y will, in like manner, become negative, when z is taken greater than —^ — 2 : therefore, as long as ©• — hq . . I -\- ag . , _ ^ contmues c-reater than (supposmQ- the m n ^ ^^ ^ value of q to be varied) so long will x admit of a greater assumption for z than y will admit of, without producing negative values ; and vice versa, By^ mak- ing, therefore, these two expressions equal to each other, the value of a will be cfiven C= -^ -) = — : ^ ^ ^ a7n+ nh^ c expressing the circumstance wherein both the values of X and ?/, by increasing 2, become negative together. But this holds true only when m is a positive quantity ; for, in the other case, the last term ( — mz) in the ge- neral value of X being positive, the particular values do not become negative by increasing, but by diminishing the value of 2 ; it being evident, that no such can re- sult from any assumption for 2, but when y is greater than 4-. b To apply these observations to the equation, 7x -f 9y + 232 = 9999, proposed, we shall, in the first place, by taking 2 = 0, have x = 1428 — y : whence the least value of y is given = 5 ; and the greatest o.f x = 1422. Again, from the equation am -f- hn = c, or 77n ■+• 9n = 23, we have m = 3 — 7i — ^ — - — ; in which the least positive value of 7i is given = 1 ; and the corresponding value of 771 = 2 ; and so the general values of x and ?/ do here become 14St2 -^ 196 The Resolution of 9q — 22, and 5 + 7^ — 2, respectively. From the former of which the greater limit of q is given = 1422 — 2 ^^^_ ,p ng" — ml . , , ~— , or 157| ; and irom -^ , expressmg the les- ser limit, we have 61, for the value of ^, when the least value of x becomes equal to that of y. These limits being assigned/ let q be now interpreted by 0, 1, 2, 3, 4, 5, ^c. successively, up to 61, inclusive: whence the number of answers, or variations of y cor- responding to every interpretation, will be found as in the margin. From whence it appears that the arith- metical progression 4-fll -j-18-f"25 -f 32, £sPc. con- tinued to 62 terms, will truly ex- press the number of all the an- swers when q is less than 62 : which number is therefore given = 4 + 61 X r -f 4 X 31 = 13485. In all which answers it is evi- dent, that x^ as well as z/, will be positive (as it ought to be) : because it has been proved that the least value of x^ till q be- comes ( = '-^ '^) = 61|, will be greater than that of y ; which is positive, so far. But now, to find the answers when q is upwards of 61, w^e must have re- course to the general value of x ; which, in these cases, by the different interpretations of 2, becomes negative before that of y. Here, by beginning with the greatest limit, and writing 157, 156, 155, 154, ^c. successively, in the room of ^, it will ap- pear, that the number of answers wiil be truly ex- pressed by the series 4 -f 8 + 13 + 17 + 22, £sPc. con- tinued to 157 — 61 terms: which terms being united in pairs (because, in every two tenns, the same fraction in the limit of 2 occurs) the 7 y = N. Ans. 5 — 2 4 1 12 — 2 11 2 19—2 18 3 26—2 25 4 33 — 2 32 is?c. ^c. &fc. nq- — mL 9 X = 2JD N. Ans. 157 9—22 4^^ 4 156 18 — 22 9 8 155 27—22 13^ 13 154 36 — 22 18 17 153 45 — 22 22| 22 &c. ^c. ^c. ^c. Indeterminate Problems* 197 series 12 + 30 + 48 + &fc. thence arising, will be a true arithmetical progression j whereof the common difference being 1 8, and the number of terms = -1- = 48, the sum will therefore be given = 20880 : to which adding 13485, the number of answers when 5^ was less than 62, the aggregate 34365 will be the whole number of all the answers required. PROBLEM XIV. To determine how many different ways it is possible to pay 1000/. xvithout using any other coin than croxvns^ gui- neas^ and moidores. By the conditions of the problem we have 5.r + 21z/ + 272 = 20000 ; where taking 2: = 0, ;c is found := 4000 — 42/ — — , and from thence the least value of 2/ = (0 being necessarily included here by the question) : whence the greatest value of x is given = 4000. More- over, from the equation 5m -j- 21n =• 27, we have n — 2 m = 5 — 4;^ — ^~ ; from which ?2 = 2, and w ::^ — 3: so that the general values of x and z/, given in the preceding problem, will here become 4000 — 21^ + 3z, and 5q — 22. Moreover, from the given equa- .• .u . 1' ' r . u 20000 tion, the greatest limit or z appears to be = = jor _ mz __ 4000 + 3 X 740 740 ; whence we also have 21 «^^ 1 V . r 1 ^ 4000 = 296 = the; greatest limit ot q ; and ^ = = ^ ^ 21 190, expressing the lesser limit of y, when the value of X, answering to some interpretations of 2, will become negative, while those of y still continue affirmative. To find the number of all these affirmative values, up to the greatest limit of ^, let 0, 1, 2, 3, 4, 5, qI?c. be now written in the room of f/ (as in the margin). Whence it is evident that the said number is composed of the 198 The Resolution of <1 ^ = Quot. N. Ans. — 22 1 1 5 — 22 2^ 3 2 10 — 22 5 6 3 15 — 22 n 8 4 20—22 10 11 5 25 — 22 124 13 &?c. tfc. £57^C. ^c. series 1+3 + 6 + 8 + 11+13, £sfc. continued to 29/ terms ; which terms (setting aside the first) being united in pairs, we shall have the arithmetical pro- gression 9 + 19 + 29, &?c. where the number of terms to be taken being 148, and common difference 10, the last term will therefore be 1479, and the sum of the whole progression 110112: to which adding (1) the term omitted, we have 110113, for the number of all the answers, including those wherein the value of x is negative ; which last must therefore be found and deducted. In order to this we have already found, that these ne- gative values do not begin to have place till q is greater than 190: let, therefore, 191, 192, 193, t^c. be sub- stituted, successively, for q ; from whence it will appear that the number of all the said negative values is truly exhibited by the arithmeti- cal progression 4 + 11 + 18 + 25, ^c. continued to 296 — -190 terms ; where- of the sum is 393/9 ; which subtracted from 110113, found above, leaves /0/34, for the number of answers required. After the manner of these two examples (which il- lustrate the two different cases of the general solution, given in the preceding problem) the number of answers may be found in other equations,' wherein there are three indeterminate quantities. But, in summing up the numbers arising from the different interpretations of ^, due regard must be had to the fractions exhibited in the third column expressing the limits of 2 ; because, to have a regular progression, the terms of the series in t\\Q fourth column, exhibiting the number of answers^ 7 X. Quot. N. Ans. 191 32—11 H 4 192 32 — 32 10| 11 193 32 — 53 i/| 18 194 32-/4 24| 25 ^c. £5?c. £5fc. isPc. Indetermbiate Problems. 199 itvust be united by twos, threes, or fours, &Pc. according as one and the same fraction occurs every second, third, or fourth, &fc. term (the odd terms, when there happen any thing over, being always to be set aside, at the beginning of the series). And it may be observed farther, that, to determine the sum of the progression thus arising, it will be sufficient to find the first term only, by an actual addi- tion : since not only the number of terms, but the com- mon difference also, will be known ; being always equal to the common difference of the limits of z (or of the quo- tients in the said third column) multiplied by the square of the number of terms united into one ; whereof the reason k evident. But all this relates to the cases wherein the coefficients of the indeterminate quantities, in the given equations, are (tw o of them at least) prime to each other : I shall add one example mqre, to show the way of proceed- ing when those coefficients admit of a common measure. PROBLEM XV. Supposing \%x -^^ I5y + 20z = 100001 ; it is required to find the number of all the answers in positive integers. It is evident, by transposing 202 and dividing by (3) the greatest common measure of x and z/, that Ax + 5z/, 22; 2 and consequently its equal 33333 — 62 ■ — , must be an integer, and therefore 22 — 2 divisible by 3 : but 32 is divisible by 3, and so the difference of these two, which is 2 -f 2, must be likewise divisible by the same number, and consequently 2 = 1+ some multiple of 3. Make, therefore, 1 + 32^ = 2 (ti be- ing an integer) ; then the given equation, by substitut- ing this value, will become 12x + \5y + 60ic + 20 = 100001 ; which, by division, fcPcv is reduced to 4x + 5y + 20u = 33327 : wherein the coefficients of .v and y are now prime to each other, and we are to find the number of iill the variations, answering to the dif- ferent interpretations of ?/, from to the greatest lim^l inclusive. 200 The Resolution^ fcf'c. By proceeding, therefore, as in the aforegoing cases^ we have x = 8331 — y — -■ whence the least va- lue of z/ is given = 3, and the greatest oi x =z 8328. Moreover, from the equation 4;?z + Sii = 20, we have mz=z5' • — ; whence n = 0, and m z= 3. 4 Therefore the general values of x and y (given in problem 1 3) do here become 8328 — 5q — 5w, and 3 + 4^' ; from the for- 8328 mer of which the greatest limit of q is given = = 1 665. Now, since the value of y will here continue posi- tive, in all substitutions for q and w, as no negative quanr tity enters therein, the whole numbers of answers will be determined by the values of x alone. In order to this, let q be successively expounded by 1665, 1664, 1663, ^c. and it will thence appear that the said number will be truly defined by 1666 terms of the arithmetical progression 1 -f 2 -f 3 + 4 + 5, £s?c. whereof the sum is found to be 1388611. When there are four indeterminate quantities in the given equation, the number of all the answers may be de- termined by the same methods ; for any one of those quan- tities may be interpreted by all the integers, successively," up to its greatest limit, which is easily determined ; and the number of answers corresponding to each of these in- terpretations may be found as above ; the aggregate of all which will consequently be the whole number of answers required : which sum or aggregate may, in many cases, be derived by the methods given in section 14, for sum- ming of series by means of a known relation of their terms. But this being a matter of more speculation than real use, I shall now pass on to other subjects. ? X Quot. N. Ans. 1665 1664 1663 3 — 5ii 8 — 5u 13 — 5u H 1 2 3 The Jmestigation^ ifc. ^Oi SECTION XIV, The Investigation of the Sums of the Powers ofNwn- bers in Arithmetical Progression^ BESIDES the two sorts of progressions treated of in section 10, there are infinite varieties of other kinds ; but the most useful, and the best known, are those consisting of the powers of numbers in arithmetical progression ; such as 12 + 2^ + 32 + 4^ • . . . w% and 1^ + 2^ + 3* + 4^ • • • • n^, &fc, where n denotes the number of terms to which each progression is to be continued* In order to investigate the sum of any such progression, which is the design of this section, it will be requisite^ first of all, to premise the following LEMMA. If any expression, or series, as ers of an indeterminate quantity 72, be universally equal to nothing, whatsoever be the value of n ; then, I say, the sum of the coefficients A — a, B — ^, C — • c, &c. of each rank of homologous terms, or of the same powers of ;2, will also be equal to nothing. For, in the first place, let the whole equation An + B/z^ + Cii" &c. 1 r. u V 'A ^ u 1 _ an ^bn^ - en' &c. j = ^' ^^ ^^^^d^d by n, and this being universally so,*" be the value of n what it will: let, therefore, n be taken = 0, and it will become 4 1=0; which being rejected as such, out of the last equation, we shall next have + Bn + Cn^ + D723 &C.1 ^ , ,. .,. ^bn ^cn^~ dn' Scc.J = ^' whence, dmdmg 2D iO^ The Investtgation of again by n, aiid proceeding in the very same manner, B .— ^ is also proved to be = ; and from thence C »— . c, D — d^ he. Q. E. D. Now, to apply what is here demonstrated to the pur- pose above specified, it will be proper to observe, first, that as the value of any progression, 1^ + 2^ + 3^ + 4* 72^, varies according as 72, the number of its terms, varies, it must (if it can be expressed in a ge- neral manner) be explicable by 7z, and^ its powers with determinate coefficients ; secondly, it is obvious that those powers, in the cases above proposed, must be ra- tional, or such whose indices are whole positive num- bers ; because the progression, being an aggregate of whole numbers, cannot admit of surd quantities ; lastly, it will appear that the greatest of the said indices can- not exceed the common index of the progression by more than unity ; for, otherwise, when n is taken indefinitely great, the highest power of ii would be indefinitely greater than all the rest of the terms put together. Thus, the highest power of n, in an expression univer- sally exhibiting the value of 1^ + 2^ -f- 3^ . . . . • w^, cannot be greater than n^ ; for 1^ -f 2^ + 3^ . . . . . n^ is manifestly less than n^ (or n^ + n^ + n^ + &c. con- tinued to n terms) ; but 7i^, when n is indefinitely great, is indefinitely greater than n^^ or any other inferior power of 72, and therefore cannot enter into the equation. This being prertiised, the method of investigation may be as follows. Case 1°. To jind the sum of the progression 1 -f 2; -f- 3 -f- 4 .... 72. Let Atz^ -|- B72 be assumed, according to the foregoing- observations, as a universal expression for the value ofl+2 + 3-f4. . . . . 72; where A and B represent unknown, but determinate quantities. Therefore, since the equation is supposed to hold universally true, whatso- ever be the number of terms, it is evident, that, if the num- ber of terms be increased by unity, or, which is the same the Sians of Progressio7is, 3K)3 tiling, if 72 + 1 be written therein, instead of if^ the e quaT iity will still subsist, and we shall have A X ^ + l]^ + B X « + 1 = 1+2 + 3+4 n X n + 1. From which the*first equation being subtracted, there re- mains A xn + if — An^ + B xn + \ — Bn = /z + 1 : this contracted will be 2 An + A + B = 72 + 1: whence we have 2 A — Ix^ + A + B — 1=0: wherefore, by taking 2 A — 1=0, and A + B — 1 = {accord- ing to the lemma) ^ we have A = I, and B = | ; and consequently 1+2 + 3+4 72(= An^ + B^z) = 72^ n 72 X w -f- 1^ - — I , or . 2^2' 2 Case 2*'« To find the sum of the progression 1^ + 2^ + 3^ 72% or 1 + 4 + 9 + 16 n^. Let A72^ + B/2^ + Cw, according to the aforesaid ob* servations, be assumed = 1^ + 2^ + 3^ + 4^ . . . . 72^ : then, by reasoning as in the preceding case, we shall have A xlT+lT + B xT+TT + C Xn +\ = 12 + 22 + 32+4^ . . . . 72^ + 72 + 1]^ ; that is, by involving 72 + 1 to its several powers, A72^ + 3A722 + 3A72 + A + ^r^ + 2B)2 + B + C72 + C = 1^ + 22 + 32 + 42 . • . . 72^ + 72 + l"]^: from which, subtracting the former equa- ^ In this investigation, it is taken for granted, that the sum of the progression is capable of being exhibited by means of the powers of 72, with proper coefficients : which assumption is verified by the process itself : for it is evident from thence, that the quantities A72^ + B;2, and 1 + 2 + 3 + 4 ... 72, under the values of A and B there determined, are always increased equally, by taking the value of 72 greater by a unit: if, therefore, they be equal to each other, when 72 is = (as they actually are), they must also be equal when 72 is 1 ; and so, likewise, when 72 is 2, £s?c. £s?c. And the same reasoning holds good in all the following cases. U04f The Investigation, of tion , we g et ZKii^ + 3Aw + A + 2^n + B + C ( = > -f Ip) = n^. + 2)1 + 1; and consequently ?A — 1 X ^^^ + 3A + 2B — 2 X ?2 + A + B + C — 1 = 0; whence (^bif the lemmd)^ 3A — r 1 = 0, 3A + 2B ~ 2 = 0, and A + B + C — 1 = ; therefore A =:i- B = - — ^^ ^i_ c = l — A — B=i-- 3' 3 ^' 6 ' and consequently 1+4 + 9 + 16 n2=:~ + , n 72 . n 4- 1 . 2?2 + 1 6' 6 3 ' 2 Case 3°. To determine the mm of the progression P +2- + 33 + 43 ^ 71% or 1 + 8 + 27 + 64. . . . . , n'. By putting A/z'* + Bn^ + Ctz^ + D;z = 1 + 8 + 27 + 64 .•,... 72^, and proceeding as above, we shall have 4A723 + eAn^ + 4A72_+ A + 3B?22 + 3B/z + B + 2C;? + C + D (= n _+ lY) = n^ + 3y;^ + 37Z + 1 ; and therefore 4A — "1 x n^ + 6A + 3B — 3 X ^^^ + 4A + 3B + 2C1II1 X 7i + A + B+ C +D — 1=0: 1 r> / 3—- 6A. 1 ^, 3— 4A — 3B. hence A = -, B (= _^) = _,C (= ^ ) • = — , D (=1 — A — B — C) = 0; and therefore n^ 71^ or 1^ + 23 + 3^ + 43 72^=: f- _ + — , ^^ 424 . Z — L. In the verv same manner it will be found 4 that ^A r^A , A ^^^ W'* ^^^ ^^ ^ 5 2 ' 3 30 r ^r ^ n^ n^ 5n^ n^ ^ ^ 6 2 ^ 12 12^ 72^ 72^ 72* 72^ 72 :^6 + 2^ + 3^ . 4 . • 72^ = — + — + + ^^^ 7^22 6^ 42 the Sums of Progressions. 205 In order to exemplify what has been thus far deliver- ed, let it, in the first place, be required to find the sum of the series of squares 1+4 + 9 + 16, &fc. continued to 10 terms : then, by subst ituting 1 for h, in the ge- n . w + 1 . 271 + 1 , n^ , 72^ 71 ^ neral expression (ox ~ + _ + ), found by case 2°, there will come out 385, for the re- quired sum of the progression : which, the number of terms being here small, may be easily confirmed, by ac- tually adding the 10 terms together. Secondly, let it be required to find the number of cannon-shot in a square pile, whose side is 50. Here, by writing 50 for 7i 7Z . 72 + 1 . 2/1 + 1 • 1 11 t m the same expression, — , we shall have ^50 X 51 X 101>^ 42925, expressing the number of shot in such a pile. Lastly, suppose a pyramid composed of 100 stones of a cubical fi.gure ; w^hereof the length of the side of the highest is one inch ; of the second two inches ; of the third three inches, i^c. Here, by writing 100 instead of tz, in the third general expression, we have 25502500, for the number of solid inches in such a pyra- mid. Hitherto, regard has been had to such progressions as have unity for their first term, and likewise for the common difference ; but the same equations, or theo- rems, with very little trouble, may be also extended to those cases where the first term, Knd the common difference, are any given numbers, provided the for- mer of them be any multiple of the latter. Thus, sup- pose it were required to find the sum of the progression 6^ + 8^ + 10* £ffc. (or ^,Q> + 64 + 100 Ssfc.) conti- nued to eight terms : then, by making (4) the square of the common difference a general multiplicator, the given expression will be reduced to 4 X 3^ + 4^ + 5^ .... 10' : but the sum of the progression P + 2* + 3* + 4^ . . . . 10^ is found, by the second theorem, to be 385 ; from which, if (5) the sum of the two first terms (which the 206 The Iwoestigatiqn of series 3^+4^ + 52 lo^ wants) be taken away, the remainder will be 380 ; and this, multiplied by 4, gives 1 520, for the true sum of the proposed progressirn : and so of others. But if the first term be not divisible by the common dif- ference, as in the progression, 5^ + 7^ + 9^ &Pc. the specu- lation is a little more difficult ; nevertheless, the sum of the series, in any such case, may still be found, from the same theorems. Let the series jn + el^ + m + "2eY + m + ZeY . . • . • m -f iiey be proposed, where m and e denote any quan- tities whatever, and where n represents the number of terms. Then, by actually raising each root to its second power, and placing the terms in order, the given expres- sion will stand thus : m^ -f 77i' + 772^ ... . m^ 2me + 4me + 6me .... 2?ime \^ . Now, it is evident e^ + 4^2 ^ 9(?2 7i2 that the sum of the first rank, or series, is 7i x in^ - also the sum of the second, or 2me X 1 -f 2 -f 3 -f 4 . . . . yz, Tl X 71 ~ I 1 appears (<^^/ case 1) to be 2me X ? an(d that of '} the third, or e^ x 1 + 4 -f 9 - f- 16 ti^ (^bi/ case 2) = e^ X — ^ i— : therefore the sum of the 6 whole progression, m + eY + m -{- 2ef -f m + SeY' • . . . w -f neY is =: ?i , 7n^ + n . ?i + 1 > me -f n . 71 -f 1 . 272 -f- 1 . e^ 6 In li ke manner, if the series proposed be m + eY + ~mr+2eY + w^ + ^ef ^ + ^^1^ J then may it be resolved into 1 + 1-f- 1 1 X 1+24- 3. . , . . n 1 4-4-f 9. . , ..7Z2 .1 X ^nie .1+8+27 n^ X ^ J X 3772^^ f , 1 ^ ^ : whose sum, by the Slims of Progressions. SOT' the Ibrementioned theorems, will appear to be 7z . /z -f 1 • Sm^e . n . 7Z -f 1 . 2/2 -f- 1 . me^ «. ^3 + _ + _ + !i_liLZ — !_£-, And, by following the same method, 4 . . . the sums of other series may be determined, not only of powers, but likewise of rectangles, and solids, fc^c. provided their sides, or factors, be in arithmetical pro- gression. Thus, for example, let there be proposed the series of rectangles m + e . p -^ e + 7?z + 2^ . p + 2e -f- m '{' oe . p -{- Se . • , . + m + ne . p + ne. Then, the factors being actually multiplied together, and the terms placed in order, the given series will be resolved into the three following ones : 77ip + 7np + mp + 7Tip . . • . + inp m -{-p.e -\'m-{-p»2e + 7n'{-p*3e +771 -f-/?.4e'....-f-7/z-fj&.?2r ef2 + 4^2 + 9e^ + 1 6^2 f- 7i^e^. Whereof the respective sums {by case 1 and 2) are 71 . 71 + 1 , „ n , 71 + 1 . 271 4^ 1 mp X n^Tu +p . e X 1 •, and e^ X ■ — : • — and the aggregate of all these, or n + 1 . 7z + 1 . 2?2 4- 1 , . 71 X mp H — . m +p . e -J ■ ■ — • e^, is con- 2 6 sequently the true sum of the series of rectangles pro- posed. From this last general expression, the number of can- non-shot in an oblong pile, whether whole or broken, will be known. For, supposing ^ = 1, our series of rectangles becomes 7n + 1 . p + 1 +m+2.j&-t-2-f m + 3 . p + 3 + m -\-7i . p + n ; and the sum thereof = n X mp -\ -i-— • 7n +p -j -I- -11. = 2 6 the number sought: where m + 1 and p + ^ repre- sent ther length and breadth of the uppermost rank, or tire: 7i being the number of ranks one above another. 208 The Investigation of But the expression here brought out may be reduced to n -f- 1 .72 — 1 .--. X 2/?2 + 7i -f 1 * 2p + 71 + 1 -{ — ', which 4 3 is better adapted to practice, and which, expressed in words^ gives the following rule. To twice the length, and to twice the breadth of the uppermost rank, add the number of ranks less one, and multiply the two sums together ; also multiply the number of ranks less one, by that number more one, and add -^ of this product to the former ; then -| of the sum multiplied by the number of ranks will be the an- swer. As a rule of this sort is of frequent use to persons con-^ cerned in artiller}', it may not be improper to add an ex- ample or two, by way of illustration. 1. Suppose a complete pile, consisting of 15 tires, or ranks, and suppose the number of shot in the upper- most (which, in this case, is a single row) to be 32* Then the first product mentioned in the rule will be 64 + 14 X 2 -f- 14 = 78 X 16 = 1248 ', and the se- cond = 14 X 16 = 224; \ whereof is r4|, and this, added to 1248, gives 1322|; whereof ^ part is 330f 5 which, multiplied by 15, gives 4960, for the whole num- ber of shot in such a pile. 2. Let the pile be a broken one, such, that the length and breadth of the uppermost tire may be 25 and 16, and the number of tires 11. Here, we have 50 -f- 10 X 32 -f 10 = 60 X 42 = 2520j for the first product ; and 12 X 10 = 120, for the second : therefore X 1 1 = 640 x 1 1 = 7040, is the true an- 4 swer. Having exemplified the use of the theorem, for find- ing the sum of a series of rectangles, I shall here subjoin one instance of that preceding it, for determining the sum of a series of cubes ; wherein the value of the first 10 terms of the progression 2 -f- v 2^ + 3 + 2 V ^^ 4. 4 -f- 3 s/^y + 5+4 \/TY ^c. is required. Here, the Sums of Progressions, 209 e being 1 + y/ 2, m will be = 1 ; therefore, by writing 10, 1, and 1 +\/2 for 7^, w, and e^ respectively, in the Ml 1 .^ 10.11.3.1 -f- ^2 general expression, it will become 10 + — - 10. 11 .21 . 1 + V/^T , 100 . 121 . 1 + V 2]^ _ + _ - + - _ 24815 + ireOO \/2, the value sought. If any one be desirous to see this speculation carried further, so as to extend to series of powers, whose in- dices are fractions ; such as square roots, cube roots, &fc. I must beg leave to refer to my Essays^ where it is treated in a general manner. Here, I must desire the reader to observe, once for all, that the theorems above found will hold equally true, in case of a descending series, such as m — ef + ^ "~- ^^ j ^^* ^^ ^ — ^1 + ^'^ "~ ^^ I &Pc. provided the signs of the second, fourth, &fc. terms be changed ; as is evident from the investigation. Although the subject of this section has, already, been pretty largely insisted on, yet it may not be im- proper to add a different method, whereby the same conclusions will, in many cases, be more easily derived : in order to w^hic^ it is necessary to premise the subse- quent LEMMA. lfa + b + c + d + e + &fc. be a series, whereof the terms, g, b^ c, <^, &:c. are so related to each other, that the sum, or value thereof, can be universally ex- pounded by an expression of this form, viz. An -f B x ^ ^ ^ — ' ^ + ^ X ^ X n — 1 X n — 2 + D X ?i X 72 1 Xn — 2 X n — 3 £sPc. n being the number of terms to which the series is to be continued, and A, B, C, D, ^l\ determinate coefficients ; then, I say, the values of those coefficients will be as hereunder specified, viz. A = a B=:-^ + ^ 2 2E 210 The Investigation of ■"" 2.3.4 ' p, __a — 4^ + 6c — 4a~9b + c + d+e. Moreover, by taking the quadruple of the first of thes^ from the second, £s?c. we get ^^24D + — « + 3^ — 3c + ^, and 120D + 120E=:— 4a + ll<^ — 9c + <^ + ^j the Swm of Progressions. 211 from the latter of which subtract the quintuple of the for- mer, and there will remain =^120E =a — 4^ + 6c — 4^ + ^. Now, divide each of the equations marked thus ('^), by the coefficient of its first term, and there will come out the very values of A, B, C, D, csPc. above exhibited^ Q. E. D. COROLLARY. If every term of the proposed series, a^ h^ c, d^ 8icc» be subtracted from the next following, the first of the remainders, — a + ^, — ^ + c-> — c + ^y — ^ + ^9 &c. divided by 2, gives the value of B, the coefficient of the second term of the assumed series. And if each of the quantities thus arising be subtracted from its suc- ceeding one, the first of the new remainders, a — 2^ + c, b — 2c + d^ c — 2d + e^ &c. divided by 6, will be equal to C, the coefficient of the third term of the same series. In like manner, if each of these last re- mainders be again subtracted from its succeeding one, the next remainders will be — a + Sb — 3c + c/, — b + 3c — 3d + e^ &:c. whereof the first, divided by 24, gives the coefficient of the fourth term, ££?c. &fc. There- fore, if the first remainder of the first order be denoted by P, the first of the second order by Q, the first of the third by R, the first of the fourth by S, ^c. then ^ being = B, -^ =1 C, = D, ^ 2 ° "2.3 '2.3.4 '243.4.5 = E, &fc. it is manifest that the sum of the series a + b ^.c + d+e-i-f &c. will be truly expressed by 1 . 2 ^ __ 1.2.3 ^ R X ^^^ — ^ ^ ^ — 2 X7i — ^ 1.2.3.4 S X ^ ^ ^ — lXy^~2x?^ — 3 X n — 4 np 1.2.3.4.5 ' Example 1. Let the sum of the series of squares t 4. 4 -f- 9 + 16 4 . . * + 71* be required; Then, ti«k- 1 + 4 + 9 + 16 + 25 . . . , > • Sn X n — 1 J n X n — Ixn- -2 2 . ■' 3 ZIa. The hvocsti^ation of ing the differences of the several orders, according to the preceding corollary, we have 1,4,9, 16,25, 36, &c. 3,5, r, 9, 11, &c, 2, 2, 2, 2, &c. G, G, G, &c. Therefore a in this case being = 1, P = 3, Q =c 2, and R, S, ££?c, each = G, the sum of the whole series, w^, is found z=z n + — ^^^ + 3n^ + n _ ~ 6 ~ n X ^ + 1 X 2n + 1 6 Example 2. Let it be required to find the sum of n terms of the following series of cubes, viz. 27 + 64 + 125 + 216 + 343 -f 512, csfc. Proceeding here as in the last example, we have 27, 64, 125, 216, 343, 512, &c. 37, 61, 91, 127, 169, &c. 24, 3G, 36, 42, he. 6, 6, 6, &c. G, O, &c. Therefore, by substituting 27 for a^ S7 for P, 24 for C, and 6 for D, we thence get ^ 37/z X 71 — 1 . 24n X n — 1 X n — 2 , 27n + ~ . + g-^ ~ + 6^Xn-lX ;z-2x^^-3 ^^j^.^^ abbreviated, be- 1.2.3.4 ' comes 1 1 1- 15;z, the sum, or value, re- 4^24^ ' quired • Example 3. Let the series propounded be 2 + 6 + 12 + 2G + 3G, &fc. In this case, we have 2, 6, 12, 20, 30, &c, 4, 6, 8, 10, &c. 2, 2, 2, &c. the Sums of Progressions. 213 Hfence, a being = 2, P = 4, Q = 2, and R, S, ^c. each = 0, the sum of the series will therefore be 4.n X n — 1 2nXn — 1 Xn — 2 ii^ + 3n^ +2n %n + + — ^ = T nxn-flx^i-f^ And, in the same manner, the sum of the series may be truly found, in all cases where the differences of any or- der become equal among themselves : and even in other cases, where the differences do not terminate, a near ap- proximation may be obtained, by carrying on the process to a sufficient length. SECTION XV. Of Figurate Numbers^ their sums^ and the sums of their reciprocals^ xvith other matters of the like nature. 2d- 3d ^ii1^th I 6th Lrthj 1 • 1 4. 5 10 . 15 20. Z5 35 . 70 * THAT series which arises by adding together a rank 'Units (called fig. No. of the 1st ord.) Figurate numbers of the 2d order p Figurate numbers of the 3d order ] Figurate numbers of the 4th order ! Figurate numbers of the 5th order LFigurate numbers of the 6th order Therefore the figurate numbers "Ist order'l f 1 . 1 . 1 2d order 1 | 1 . 2 . 3 of the < 3d order ^ are <{ 1.3. 6 4th order I 1 . 4 . 10 L5thorderJ Ll . 5 . 15 Hence it is manifest, that, to find a general expression for a fig\irate number of any order, is the same thing as to find the sum of all the figurate numbers of the preceding order, so far. Let n be put to denote the &c. &c. &c. &c. &c. 214 The Sums of Figurate Numbers y distance of any such number from the beginning of its respective order, or the number of terms in the pre- ceding order whereof it is composed : then it is evident^ by inspection, that the sum of the first order, or the ?7th term of the second, will be truly expressed by 72, the number of terms from the beginning. It is also evident, from sect. 14, p. 203, that the sum of the second order, 1+24-3+4 11, will be !^ + ii (= ii X !LdLi) ^ 2^2^12^ which, according to the preceding observation, is also the value of the nth term of the third order. Hence, if the numbers, 1,2, 3, 4, 5, fcPc. be successively written instead of ;2, in the general expression 1 , we shall thence have ^ + i,| +|,| +|, i_6 + 1^ 2^^ + 1^ ^,^ foj. the values ot the first, second, third, fourth, fifth, Sfc. terms of this order, respectively ; whence it appears, that the series 1+3 + 6+10 + 15 + 21, &fc. may be re- solved into these two others, viz. i +1 + 1 +l + V+¥,&c. and l+l+.l+l+l + |,&c. The former of which being a series of squares, its sUm ■will therefore be = 1 1 {by case 2, p. 203), and that of the latter series (<^2/ case 1, p. 203) appears to be 1 : and the aggregate of both, which is I 1 (or — X X ) will be the true 62^3^ 1 2 3^ value of the proposed series 1 + 3+6 + 10 + 15 l^c. continued to ii terms, and therefore equal, like- wise, to the 72th term of the next superior order, 1+4 + 10 + 20 + 35, £ffc. Let, therefore, 1, 2, 3, 4, 5, &?c. (as above) be successively written for 72 in Q''^ 7?^ 72 this general expression, -^ -| + "T^ ^^^ ^^ '^^ ^" come 1 4. 1 +1, l+l +1, + V +1+1, V + y t> ^c. for the values of the first, second, third, fourth, Isc, and their Reciprocals. 215 terms of the fourth order, respectively; whence it ap- pears that the series 1 + 4 + 10 + 20 + :^5^ ^c. may be resolved into these three others, viz. 1 + 8 + 2r + 64 + 1 25+216 . . . . n^ ^6 ' 1 -f 4 + 9 + 16 + 25 + 36 n^ 1 +2 + 3+4 + 5<^+ 6 n ^ 3 ^ , 72^ 71^ 71^ 71^ 71^ n whereof the sums are -^ + ^ + -^, _ + -_ + _^^ and — + ^ (by p. 202 and 203) the aggregate of 6 6 , . , n^ TV' lln^ . ^ /- ^^ 7Z + 1 Zl— X — -^^ — ) will consequently be the true value of 3 4 the whole series. Aftet the same manner the sum of the ^ ^ , 1 -n I ^^ 72 + 1 72 + 2 71+5 fifth order will appear to be — X — ^- — X —^ — X n +4 12 3 4 from whence the law of continuation is ma- 5 ' nifest. And it may not be amiss to observe here, that though the conclusions thus brought out are derived by means of the sums of powers determined in the preced- ing section, yet the same values may be otherwise obtain- ed, by a direct investigation, from either of the two gene- ral methods there laid down. In order now to find the sum of the reciprocals of any series of figurate numbers ; suppose 1 + <^ + ^c + bed + hcde + hcdef + &c. to be a series whose terms con- tinually decrease, from the first to the last, so that the last may vanish, or become indefinitely small : then, by taking the excess of every term above the next Ibllow- ing one, we shall have 1 — h^h X 1 — c^ he X 1 — d^ bed X 1 — e^ hcde x 1 — /, &c. The sum of all which 216 The Sums ofFtgurate Numbers. is evidently equal to the excess of the first term above the last, or equal to the first term, barely; because the last is supposed to vanish, or to be indefinitely small in respect of the first. Hence it appears, that 1 — b -{- b X 1 -— c +bc X 1 — d + bed X 1 — e + bode X 1 — /, &c. = 1. Let b be now taken = -— , c = —^^ d ■=. — i-l a a + p ^ -f ^ ' c = — lL_, f = , &c. Then, 1 — b being = a + r a + 5 a — m ^ a — m , a — m . a — 7n -, 1 — c =: , 1 — d= , 1 — e: a a-j-p a + q^ a -^ r^ he. we shall, by substituting these several values in the , a — ?n m a — m m above equation, have -| x 1 X ^ a a a+p a m 4-p a — m ^ m vi -{- p m + (/ a — w.o — 31^ X 1 X ^ X —^-^ X h &c. a+p ci + f] ci f^+/? a + q a -^-r 1 1 ^ . ^ . w ^ -{- P \ = 1 ; and consequently 1 ^ X + ,.. X — 3^ X — -^^ + &c. = ; by dividmg a+p a + ^ a + r a — m the whole by . a Hence, if q be taken = 2/j, r ~ 3p^ s z=: 4/?, &c. wi and /3 be put =z a + p^ we shall have 1 + — + m.m+p 7n . m+ p . m + 2p m . m + p . m + 2p ..m + Sp jn+]) '^(i.^+p.^+2p'^(i.^+p.^ + 2p.i + 3p J- &c. ad infinitum^ = ■ ; which, when ^ ^ |3 p 771 7)1 m . 771 + 1 771 , 7n + 1 . m + 2, ^ = 1, becomes 1 +-+ ^-^- + ^-^-j:-^-^^^ , 2^^^ ^ lZZ : this, by taking m = 1, and 1 1.2 1.2.3 s = n, gives 1 H 1 " -f == , ^ "^ and their Reciprocals. 217 2.3.4 . p 7^ — 1 , . + 6CC. = ; exhi- n . /i 4- 1 . /z 4-2 . n -f- 3 n — 2 biting the general value of a series of the reciprocals of figurate numbers, infinitely continued ; whereof the or- der is represented by n ' from whence as many parti- cular values as you please may be determined. Thus, . by expounding nhy 3, 4, 5, £s?c. successively, it appears that 14 h — + f- — , &c. = 2, 1 + 1 4. i. 4. 1 + 1, &c. = ^. ^ 4 10 20 35' 2 14- — + — + — + — ,&c. = ±, 5 15 35 70 3 and so on, for any higher order ; but the sums of the two first, or lowest orders, cannot be determined, these being infinite. By interpreting /3 and 7n by different values, the sums of various other series may be deduced from the same general equation. Thus, in the first place, let jS = 711 m 4- 2 ; so shall the said equation , become 1 4 ^ 4- 77Z . m 4- 1 , 711 , m 4- \ . in . w 4- 1 « + — — > &c. w4-2.;?2 4-3 7^4-3.^72-1-4 /;z4-4.?/z4-5 = m 4- 1 ; which, divided by 7?2 . m 4- 1, gives 1.1.1.1 m. 711 + 1 m + l •711 + 2 Tn + 2%7n + S m + 3 .7n+4> &c. = — . m Again, by taking- 3 2= w 4- 3, and dividing the whole equation by 771 . m + 1 • ni + 2^ we have -__L__ + !_ — , + 771 . m + 1 , m + 2 7n + 1 . m +k, , m + 3 OT + 2 . w + 3 . OT + 4. m.?n + 1 .2 2 F 2:18 The Sums of In like manner we shall have — = -,&c.=:. ^ w -f- 1 . w -4- 2 . m -f 3 . w + 4 m.m + \ .m + 2 .^ From whence the law for continuing the sums of these last kinds of series is manifest ; by which it appears, that if, instead of the last factor, in the denominator of the first term, the excess thereof above the first factor be sub- stituted, the fraction thence arising will truly express the value of the whole infinite series. A few other particular cases will further show the use of the general equations above exhibited. Let the sum of the series 1+ — J--1 x — + 2 4 62 4 6 8o ^ - j: -. u — X — X X — X — X — n Sec. ad znfimtum. be 5 7 9 5 7 9 11 -^ required. Here, by comparing the proposed series with 1 -^ - P 4- =~- + &c. (= — - — ^ — ), we have m = 2, ^.^+p ^ —p — m 3 = 5, and /? = 2 ; and consequently — — = 3 = ^ ' ^ ^ / f3— / — m the true value of the series. Let the sum of an infinite series of this form, viz. ^ 1 — -f &c. be de- 1 . 2 . 3, &c. 2 . 3 . 4, &c. 3 . 4 . 5, &c. manded. Here (according to the preceding rule) we have + ^ o + t; :^ ^c- = -. : = ^ 5 .22.33.4' 1.1 + :; — : — r-) ^c. 1.2.32.3.4 3.4.5 1.2.2 4 11 1__ ^^ ^ 1 _ 1, 1.2. 3. 4*^2.3.4. 5 3.4.5.6' ^' 1.2.3.3 is' £5fc. i^c. Compound Progressions. 219 If, instead of the whole infinite series, you want the sum of a given number of the leading terms only ; then let the value of the remaining part be found, as above, and subtracted from the whole, and you will have your desire. Thus, for instance, let it be required to find the sum of the ten first terms of the series -\ ^ -f , i^c. Then the remaining part, -f- 1 1 . £s?c. beincj = — (hf the 12 . 13 ^ 13 . 14 ^ 14 -15' ^ 11 ^ ^ rule above) ^ and the whole series =1, the value here souffht will therefore be 1 = — • The like of ^ 11 11 others. The sums of series arising from the multiplication of the terms of a rank of figurate numbers into those of a decreasing geometrical progression, are deduced in the following manner. By the theorem for involving a binomial (given at p. 40, and demonstrated hereafter) it is known that n - (or 1 — .v]-'") is = 1 + mx + m • ^^i^li . x"" 1 — X I 2 ^ m + 1 m + 2 „ . m + 1 711-^2 m + 3 ^23 ^234^ &c. In which equation let in be expounded by 1 , 2, 3, 4, 5, &c. successively, so shall 1°. = 1 + ;^ + :^2 ^ -^^,3 ^ ^,4 ^ ^5 ^ ^^^ 1 X 2^ ^ = 1 + 2a; + 3x^ + 4a,3 + 5.v^ + 6.r% &c« 1 — x\ 3°. = 1 + 3.V + Q>x^ + 10;^"' + 15x^+ 21a:% &(^. 1 — x\ 4^°- = 1 + 4.V +10v- + 20a- 4- o5.v^ 4- 56.^^^c; 226 The Sums of 5\ ^ , = 1 +5Ar + 15x^ + 35x^ + YOx^ + 126;^^Scc* 1 — x\ 6^ , ' '•' '. = 1 + 6;tf + 21.^2 + 5&x^ + 126;^'* + 252;^*, &c» 1 — ;c I All which series (whereof the sums are thus given) are ranks of the different orders of figurate numbers, multi- plied by the terms of the geometrical progression 1, x, x^^ a:', x*^ &c. From these equations the sums of series composed of the terms of a rank of powers, drawn into those of a geometrical progression, such as 1 + 4x + 9x^ -f 16;c^, &€• and 1 + 8x -j- 27x^ -f 64^^, &c. may also be de- rived ; there being, as appears from the former part of this section, a certain relation between the terms of a series of powers and those of figurate numbers ; the lat- ter being there determined by means of the former. To find here the converse relation, or to determine the former from the latter, it will be expedient to multiply the several equations above brought out, by a certain num- ber of terms of an assumed series 1 -f- Ax + Bx^ -|- Cx^^ &c. in order that the coefficients of the powers of x may, by regulating the values A, B, C, D, &^c. become the same as in the series given. Thus, if the series given be 1 -f 4x + 9x^ + 16.^^ -f 25a;'*, &c. ; then, by multiplying our third equation 1 -f- Aa: by 1 + Ax, we shall have =====^ =z 1 -{• 3 + A X x^ 1 • — ' X j + 6 + 3A X x^ 4- 10 -f- 6 A X x^ + &c. which series, it is tivident, by inspection, will be exactly the same, in every term, with the proposed one, if the quantity A be taken = 1. The sum of the said series, infinitely continued, is 1 -4- iV therefore truly represented by ~ -, -f In lite manner, if the fourth equation 1 + 4.V + 10^' + 20.v3 ^ g5_^4^ gjc. be multiplied by Compound Progressio7is» 221 A . T> 9 u Ml • 1 4- A^c- 4- Bx2 ^ , 1 + A.r + B:v^5 there will arise — — — = 1 + 1 — x\ 4-f A X ^ + 10 -f. 4A -f-B X ^^ + 20 + 10 A -f 4B X x^j &c. where, the several terms of the series being com- pared with those of the series 1 + 8.r + 27 x^ + 64.\^, &c. we have 4 4- A = 8, and 10 4- 4A 4- B = 27 ; whence A = 4, and B = 1 ; and consequently, by substituting these values, ■■ . ■ - — - = 1 + 8jc 4- 27^ 4- 64^^ + i — xy 125jv^&c. Again, by multiplying the fifth equation, = 1 — x^ \^5x + 15 x^ 4- 35;c3, &c. by 1 + A^t- 4- B;^^ ^ ^^3^ . - 1 4- A;c 4- B;c2 4- Cx^ ^ ~ it becomes — i ■ = 14-54-Axa;' + 1 — x] 15 4.5A +B X ^^ + ^5 + 15A 4. 5B 4- C X a:^ &c. And, by comparing the several terms of the series with those of 1 4. 16x + SlJt^ 4. 25^x^, &c. we get 5 4- A = 16, 15 4- 5A 4- B = 81, and 35 4- 15A 4- 5B 4- C = 256: whence A = 11, B (=81 — 15 — 55) = 11, and C ( = 256 — 35 — 220) = 1 ; and consequently 1 4. Hat 4- 11.^^ -f .Y^ --I — ■ = 1 4- 16^ 4. Ux^ + 256x\ &c. 1 — xy By proceeding the same way, it will be found, that 1 4- 26x 4- 66;c2 4- v6;c^ -h^^^ . -^ 7; =^^— = 1 +2^^ + 3V 4. 4^Ar^ + 1 — X \ &c. &c. And, itniversally^ putting a •=. vi^ h ^=:. m * ^ "^ .^ m 4- 1 7?i 4- 2 p J 1 . , . ci=zm^ — - — . - — - — , &c. and multiplying the gene- ral equation -- — 1 + ax + bx^ + cx^ 4. dx\ &c. 1— ;vT" -r -r , by 1 4- A.Y 4- Bx^ 4. C;v3 ^ j^^^ g^^^ ^^^j.^ ^^.jg^^ 1 4- Ay -L Ba:# , &c. " T^^^T^ =l4-«-f-Ax^ + ^22 The Sums of b + aA + B X ^^ + c + bA + aB + C X oc^^ & z _ . ^^^ ^l^g^^ ^£ ^^ ^^^^ _ 1 — x]2 — -^- *- l3— ^ by ivhat has betn above determined ; and 1 — xY consequently the sum of all the three equal to _£_ ^f^^fj+lP + Pl_:J^ ^ ,he whole infi. nite series y — Z' • ^ — q *^^ +f — ^P * g — 2^ • 2''+^ +v &Pc. But the sum of the t first terms only is wanted ; therefore the sum of all the remaining terms, after the t first, must be found in like manner, and be deducted from the sum of the whole, here given. Now, to do this, we are first to get the leading term of the said re- maining ones I which, according to the law of the series, will be expressed by f — p — tp . g — q — tq . 2'+^ : whence, if we make f — tp z=i h^ g — tq •=. k^ and r + tv :=. s^ It is evident, that the series to be deducted will be h — p . k — q • 2^ + A — 2/? . k — 2q. 2*+^, &c. which, having the very same form with that first pro- posed, its sum will therefore be had by barely writing h for f^ k for g^ and s for r, in the value above deter- mined : which, thereby, becomes 224 The Sums, ^e. In t he same wanne r, suppo si ng the t first terms of the series a — . d — p , c — / ? . d — p . &c. x z** + a — ^p . Ij — zp . c — ..p . d — -/? . &c. X z^'+S &c. were to be required ; by putting the continual product of all the quantities a, h, c, d, &c. = P ; the sum of all the P P P products ( f- -- ^ , &c.) that arise by omitting one letter in each, = Q; the sum of all those P P (-7 + —1 &^c.), by omitting two letters, = R, £sr*c. we shall here have ,s L 1 — ^1 1 — ^1 &c. for the sum of the whole infinite series : and if / we make « = a -^ tp, b r=z b — tp^ r =z r + tv^ &c. it is evident that the sum of the remaining terms, after the t first, will be truly expressed by &c. where x = 2«, and P, Q, R, S, cs^c. are the same in / / / y relation to a^ Z>, c, ^, &c. as P, Q, R, S, £^c. in respect to a b^ c, d^ &€• A multitude of other cases and examples might be given, there not being, in the whole scope of the mathematical sciences, a subject of greater variety and intricacy than this business of series : but to pursue it farther here would be inconsistent with the general plan of this work. Such, therefore, as are desirous of a greater insight into the matter, may, if they please, turn to my Miscellanies^ where it is carried to a greater length. Of Combinations » 22S' , From the series for figurate numbers, derived in the former part of this section, the investigation of a general theorem for determining how many dif- ferent combinations any number of things will admit of, when taken two by two, three by three, &fc, may be very easily deduced. Let the number of things in each combination be^ jirst^ supposed txvo only ; and let n hcy universally^ put to represent the xvhole number of things or letters^ «, ^, c, d^ &c. to be combined. When the num- ber of things is only two, as a and b^ it is evident that there can be only one combination («/?) ; but, if n be increased by 1, or the letters to be combined be three, as a, ^, c, then it is plain that the number of combina- tions will be increased by 2, the number of the preced- ing letters a and b ; since, with each of those, the new letter c may be joined ; and therefore the whole num- ber of combinations, in this case, will be truly ex- pressed by 1 -f- 2. Again, if n be increased by one more, or the whole number of letters be four, as a, b^ o, d; then it will appear that the number of combina- tions must be increased by 3, since 3 is the number of the preceding letters, with which the new letter d can be combined, and therefore will here be truly ex- pounded, by 1 -f- 2 -f- 3. And, by reasoning in the same manner, it will appear, that the whole number of combinations of two, in five things, will be 1 -f- 2 -f 3 -f 4 ; in six things, l+2 + 3-f-4-|-5; and in seven, l-f2 + 3-f-4-f5-f6, &Pc. Whence, uni- versally, the number of combinations of ?z things, taken two by two, is = 1 + 2 -}- 3 -f 4 -f- . . • , . n — 1 : which being a series of figurate numbers of the second order, where the number of terms is n — 1, the sum thereof, by case 1, p. 203, will tlierefore be truly defined . n — 1 n n — 1 Let 210ZV the iiumber of quantities iri each covibiiiation be supposed to be three. Tf- h plain, that, in three things, rv, b, (\ there rrm ^ G 226 Of Combinations. l)e only one combination ; but, if n be increased by 1, or the number of things be 4, as a, b^ c, d^ then will the number of combinations be increased by (3) the number of all the combinations of two, in the preceding letters rt, 3, c ; since with each two of those the new letter d may be combined ; therefore the number of combina- tions, ia this case, is 1 + 3. Again, if n be supposed to be increased by 1 more, or the number of letters to become five, as a, ^, c, d^ e ; then the number of combinations will be increased by six more (=14-2 4- 3), that is, by all the combinations of two, in the four preceding letters, «, b^ c^ d: since, as before, with each two of those, the new letter e may be combined; Hence the number of combinations of n things, taken three by three, appears to be 1 + 3 -f- 6 + 10, &c. continued be ?2 — 2 terms ; which being a series of figurate numbers of the third order, the value thereof, by what is before determined (p. 214) will be tiTily expressed by n — 2 71 ~ 1 n . ^ n n — 1 n — 2 X X — , or Its equal, — X X • 1 2 3 ^ ' 1 2 3 And universally, since it appears that increasing the number of letters by 1 always increases the number of combinations by all the combinations of the next in- ferior order with the preceding letters (for this obvious reason, that to each of these last combinations, the new letter may be joined), it is manifest, that the combina- tions, of any order, observe the same law, and are ge- nerated in the very same manner as figurate numbers, and therefore may be exhibited by the same general expressions ; only, as there are 2, 3, 4, 5, £s?c. things necessary to form the first, or one single combination, according to the different cases, it is plain that the number of terms must be less by 1, 2, 3, £sPc. respec- lively, than (n) the number of things ; and, therefore,^ instead of 7Z, in the aforesaid general expressions, we must substitute n-. — 1, ?2 — 2, or ;z — 3, £s?c. respectively, to have the true value here. Hence, the number of combinations of two things, in n things, will be a — 1 n n n — 1 n. n — 2 n — 1^ ^2 — - ^ -TT' ^^' Y ^ --^ ' of three, -.-^ x —^ X y. Of the Binomial Theorem. 227 n 7^ — 1 n — 2 r,r, n — 3 n — 2 n — 1 «r_X-^ X -^; offour,__x-^X -^ n n n — 1 n — 2 n — 3 , . , ^^ ^^ X — , or — X X — :; — X — (vid. p. 215) : 41 2 3 4^^^ whence, universally, the number of combinations in the number, tz, of things, taken two by two, three by three, cSTc. will be expressed by — X X — ; — X , ^c. continued to as many factoids as there are things in each combination. From this last general expression, showing the com- binations which any number of quantities will admit of, the known theorem for raising a binomial to any given power, is very easily and naturally derived. For it is plain that <^« [T >a + bc=.a-\'bxa + C', which, multiplied by a + d^ gives a^ + cV a^ ^ bdy a + O cd) + d] bcdz=. a + b X a-\-c X a -{-d; and this, again, multiplied by a + e, gives + b^ + c ^'++d + e. + dej + bd\ + brd^ + c^ I + cdej a + bXa-^-cxa + dxa + e. Whence it appears, that the coefficient of a, in the second term, is always the sum of all the other quantities ^, c, J, &c. added together ; and that the coefficient of the third term is the sum of all the products of those quantities, or of all their possible combinations, taken two by two ; since, from the nature of multiplication, they must be all concerned alike, in every term : whence it is also manifest, that the coefficient of the fourth term must be the sum of all the solids of the same quantities, or of all their possible combinations, taken three by three, 228 Of the Binomial Theorem. Hence, if the number of the quantities b^ c, d^ «, ox a •\- b raised to the power 7z, is truly ex- 71 ' 1 71 ' 1 pressed by a^ + 7iba^~'^ + 7iX b^a^"^ + w X X ^^~' b^ ci""-^-, i^c. or a^ + 72rt«~^ b + ?i X ^^ "~ a'^'-^S^ 72 — — 1 71 - 2 + 71 X — - — X — :; — a^'^^b^^ ^c. as was to be shown. Of Interest ayid Ammities. ^29 SECTION XVL Of Interest and Annuities. INTEREST may be either simple or compound: simple interest is that which is paid for the loan of any principal, or sum of money, lent out for some li- mited time, at a certain rate per cent, agreed upon be- tween the borrower and the lender, and is always propor- tional to the time. Thus, if the rate agreed upon be 4 per cent, per ann. or, which is the same thing, if the interest of 100/. for one year be 4/. then the simple interest of the same sum for two years will be 8/. ; for three years 12/. ; and for four years 16/. ; and so on for any other time in proportion. Compound interest is that which arises by leaving the simple interest of any sum of money, after it becomes due, together with the principal, in the hands of the borrower, and thereby converting the whole into a ne^7 principal. Thus, he who lets out 100/. for one year, at the rate of 4 per cent, has a right to receive 104/. at the year's end ; which sum he may leave in the borr rower's hands, a second year, as a new principal, in order to receive interest for the whole ; and this inte- rest (which will be found 4/. Ss. 2|^.), together with 4/. the interest of the first principal, for the first year, will be the compound interest of 100/. for two years ; and so on, for any greater number of years. But I shall first give the investigation of the theorems for simple interest. Let the rate per cent, or the interest of 100/. for one year = r ; the months, weeks, or days in one year = t ; the months, weeks, or days which any sum, a^ is lent out for = n ; and the amount of that sum, in the said time, viz. principal and interest, = b. Then it will be, as 100 is to r (the interest of 100/.) so is the proposed sum (a) to — , the interest of that sum, for the same time. Again, as f, the time in which 230 Qf Interest and Animifies* the said interest is produced, is to n (the time proposed), so is , the interest in the former of these times, to • 100 ' 100^* that in the latter ; which, added to «, the principal, gives a -I = b. the whole amount: from whence we 100^ - , 100^? 100? X b — a J also have a = , r = and n =: 100? + nr an ' : the use of which equations, or theorems, ar will appear by the following examples : Examp, 1. What is the amount of 550/. at 4 per cent. in seven months ? In this case we have a = 550, r = 4, ? = 12, 7z = 7 : , ^1 . ^^^ ^^^ . 550X7X4 J and consequently a -j = 550 -4 = 562|-/. ^ ^ toot 100 X 12 ^ or 562/. 16^. 8d, the true value sought. Examp. 2. What is the interest of 1/. for one day, at the rate of 5 per cent. ? Here r being z= 5^ t = 365, a = 1, and /z = 1, we have ^IlL = £ = 1 = 0.0001369863, £s?c. = 100? 100 X 365 100 X 73 the decimal parts of a pound required. Examp. 3. What sum, in ready money, is equivalent to 600/. due 9 months hence, allowing 5 per cent, dis- count ? Here r being =: 5, ? = 12, 72 = 9, and b = 600, we have a (by theorem 2) = i22iL^22iLil = 578,313/. or 578/.. ^^ ^ 100X12 + 9X5 6s. 3\d, which is the value required. Examp. 4. At what rate of interest will 300/. in fifteen months amount to, or raise a stock of, 330/. ? In this case, we have given ?=12, n =15, a = 300, and ^ = 330 ; whence \by theorem 3) r will come out 100 y 12 V 30 = — — = 8 ; therefore 8 per cent, is the rate 300 X 15 ^ required. Of Interest and Annuities. 231 Examp* 5. In how many days will 365/. at the rate of 4! per cent, amount to, or raise a stock of, 400/. t ri 1 .N u 100 X 365 X ^^ Here (by theorem 4) we nave n = ^ . ^ ^ ^ 365 X 4 :=5 ^7S = the number of days required. Of Annuities or Pensions in Arrear^ computed at Simple Interest. Annuities or pensions in arrear are such as, being- payable, or becoming due, yearly, remain unpaid any number of years : and we are to compute what all those payments will amount to, allowing simple interest for their forbearance, from the time each particular payment be- comes due : in order to which, TA =.the annuity, pension, or yearly rent. Y J 7Z = the time, or number of years, it is forborne. j r = the interest of 1/. for one year. L. 7n = the amount of the annuity and its interest. Then, as 1 : r : : A : r A, the interest of the proposed sum or pension A, for one year ; which, as the last y earl's rent but one is forborne only one year, will express the -whole interest of that rent, or payment : moreover, since the last year's rent but two is forborne two years, its in- terest will be 2r A : and, in the same manner, that of the last year's rent but three, will appear to be 3rA, ^c. &?c. whence it is manifest that the sum total of all these, or the whole interest to be received at the expiration of n years, for the forbearance of the proposed annuity or pension, will be truly defined by the arithmetical pro- gression r A + 2r A + 3r A + 4r A + 5rA, fcfc. continued to n — 1 terms, that is, to as many terms as there are years, excepting the last. But the sum of this pro- n -— 1 gression is equal to n X X rA {by theor.4,sect. 10). Therefore, if to this the aggregate of all the rents, or ?zA, be added, we shall have nA -f '^^^^ X ^A = m : 232 Of Inter ent and AnnuUie^^ whence we also have A = n — 1 , r =^ n 4- n X X r 2 2wz — 27iA , J2m 1 . . == , and 72 r= W~~ + />2 — p .^ supposmg /' nxn — ixA ^ rA _ 1 _ 1 ""7" '2" Examp* !• If 600/. yearly rent, or pension, be for- borne five years, what will it amount to, allowing 4 per cent, interest for each payment from the time it becomes due? Here we have given A = 600, n = 5, and r = .04 (for as 100 : 4 : : 1 : .04)^ which values substituted, in theorem 1, give m = (72 A •\- n x Ar = 3000 + 240) = 3240/. for the value that was to be found. Exavip. 2. What annuity, or yearly pension, being for borne five years, will, in that time, amount to, or raise u stock of, 3240/. at Af per cent, interest ? In this case we have given n -=. 5^ r z=z .04, and m = 3240, and therefore, by theorem 2, A (= • n'\-^nxn — \Xr — — — -) = 600 ; vv^hich is the annuitv required. 5 + 4/ ' ^ Examp. 3. At what rate of interest will an annuity oi 560/. in seven years, raise a stock of 4508/. ? In th^s case we have given A = 560, n = 7, and m = 1 ,^, 1 r.^ ^ / 2;7^ — 272 A \ 4508 ; whence (by theor. 3) we have r (= ■■■ ) nxn — 1 X A/ ^ 9016 — 7840 ^ ^^^ _ ^^^^ interest of 1/. for one year ; 42 X 560 therefore it will be, as 1 : ,05 : : 100 : 5 per cent, the rate required. Examp. A:. How long must an annuity of 560/. be for- borne, to raise a r.tock of 4^08/. supposing interest to be ^ per cent. P Of Interest and Annuities^ 233 Here, we have given A = 560, r = .05, m = 4508 ; whence, by theorem 4, we also have /?(=-' ) =19.5 ; r At and consequently n (= y ^ + /?2 — /?) = 7 ; which is the number of years required. Note* If the rent or pension be payable half-yearly, or quarterly, the method of proceeding will be still the same, provided n be always taken to express the num- ber of payments, and r the interest of 1/. for the time in which the first payment becomes due. Thus, if it were required to find what 300/. half-yearly pension would amount to in five years at 4 per cent, interest : then the simple interest of 1/. for half a year being = ,02, and the number of payments = 10, we, in this case, have A = 300, r = ,02, and 72 = 10 ; and consequently m {by theo-^ n — • 1 rem 1) = r A -f- ?z x X r A = 3270/. which is the value sought. And the like is to be observed in what fol- lows hereafter. Of the Present Values of Annuities, or Pensions^ computed at Simple Interest. "A = the annuity, pension, or yearly rente y r = the interest of lA for one year. , [ /z = the number of years. ^ V =. the present value of the annuity. Then, because the amount of the annuity, in n years, is found above to be nA. -^ ^n . n — 1 . r A, and sincfc 1/. present money, is equivalent to 1 + nr to be re- ceived at the end of the time n, we therefore have 1 +nr ; 1 ,: : wA + ^n . n — 1 . rA (the said amount) nK -\-^n . n — 1 . rA . . , , : ^ , Its required value, m present money. But it may be observed, that this method, given by authors for determining the values of annuities, ac- cording to simple interest^ is, in reality, a particular sort, 2 H 234 Of Interest and Annuities. or species of compound interest ; since the allowing of in- terest upon the annuity, as it becomes due, is nothing less than allowing interest upon interest ; the annuity it- self being, properly, the simple interest, and the capital, from whence it arises, the principal. It is true, the sum 1 -f- nr^ expressing the amount of !/• is given, strictly speaking, according to simple interest: but the conclu- sion (as a late author ^ very justly observes) would be more congruous, and answer better, were the same al- lowances to be made therein as are made in finding the amount of the annuity ; that is, were interest upon in- terest to be taken once and no more. Agreeable to this assumption, r, the interest of 1/. being considered as an annuity, its amount in n years (by writing r for A, in the general formula above) will be given = nr -|- |?2 . ;? •^- 1 . r" : to which the principal 1/. being add- ed, the aggregate 1 + nr + ^n . n — 1 . r^ will there- fore be the whole amount of 1/. in the time 7i; and so we shall have 1 + nr -{- ^n . n — 1 . r^ : 1 : : ?zA -|- ^n • n — 1 . r A : — = ^, the true 2 + 2nr + n ^ 7i — 1 . r^ value of the annuity, according- to the said hypothesis. From which equation others may be derived, by means w^hereof the different values of A, w, and r, may be successively determined. But, as this method of allow- ing interest upon interest, once and no more, is arbitrary, and the valuation of annuities, according to simple inte- rest, a matter of more speculation than real use, it being not only customary, but also most equitable to allow com- pound interest in these cases, I shall not stay to exemplify it, but proceed to The Resolution of the various Cases of Compound In- teresty and of Annuities^ as depending thereon, fp __ f the amount of 1/. in one year, iyzz. princi- Let ^ "" \ pal and interest. [V = any sum put out at interest. * Mr. Hardyy in his Annuities. Let <^ OJ Interest and Annuities. 235 ■ n = the number of years it is lent for. a = its amount in that time. A = any annuity forborne n years. m =z its amount. ___ ( the present value of the annuity for the same ^ ^ "" I time. Therefore, since one pound, put out at interest, in the first year is increased to R, it will be, as 1 to R, so is R, the sum forborne the second year, to R^, the amount of one pound in two years ; and therefore as 1 to R, so is R^, the sum forborne the third year, to R^, the amount in three years : whence it appears that R'*, or R raised to the power whose exponent is the number of years, will be the amount of one pound in those years. But as 1/. is to its amount R**, so is P to (a) its amount, in the same time ; whence we have P x R'* = «. More- over, because the amount of one pound, in n years, is R», its increase in that time will be R" — - 1 ; but its interest for one single year, or the annuity answering to that increase, is R — 1 ; therefore, as R — 1 to R'* — 1, A X R" 1 so is A to m. Hence we get — rr = w. Fur- K — 1 thermore, since it appears that one pound, ready money, is equivalent to R'*, to be received at the expiration of A X R" 1 n years, we have, as Rn to 1, so is — (the sum in arrear) to ^, its worth m ready money ; and therefore Axl — i- R« R-l ="• From which three original equations others may be derived, by help whereof the various questions relating to compound interest, annuities in arrear, and the present values of annuities^ may be resolved. 235 Of Interest and Annuities. Thus, because PR''' is = a, there will come out P = =p1v — , and R = -^ I " , &c. or by exhibiting the same equations in logarithms (which is the most easy for practice) we shall have 1°. Log. a r=z log. P 4- 72 X log. R. 2°. Log. P = log. a — n X log. R. 3°. Log.R=^°S'"-^°g-P. ?l o ^ _ log. 6? — log. P ^•^"- i^R • Which four theorems, or equations, serve for the four cases in compound interest. Again, since m is = : HI-., we shall have 1°. Log. m = log. A + log. R« — 1 — log. R — 1. 2°. Log. A = log. ?n — log. R" — 1 + log. R — 1. o __ log* ^^1^ — m -f A — log. A ^" ^~— iS^TR 4°. Rn_^ + ^_1=0. A ^A To which the various questions relating to annuities in arrear are referred. Moreover, seeing A x -~ is = v^ we thence have R — 1 1 I"". Log. V = log. A + log. 1 — — — log. R — 1. 2°. Log. A = log. V + log. R — 1 — log. 1 — ^• o ^ _ log. A — log. A -f u — vR %j • Tl — — —————— —————^ — • log. R 4% R«+^^i: + ixR'» + ^ = a V V Of Interest and jAnnuities* 237 The use of which theorems, respecting the present va- lues of annuities, as well as of the preceding ones, for com- pound interest and annuities in arrear, will fully appear from the following examples. Examp. 1. To find the amount of 575L in seven years, at four j&^r cent, per annum^ compound interest. n^ In this case we have given P = 575^ R = 1,04, and n-=i7 \ therefore, hy theorem 1, log.~ a = log. S7S + 7 log. 1,04 = 2,8789011 ; and consequently a = 756,66, or 7S^L 13^. ^\d. the value required. Examp. 2. What principal, put to interest, will raise a stock of 1000/. in fifteen years, at 5 per cent. P Here we have given R = 1,05, tz = 15, and a = 1000 ; therefore, by theorem 2, log. P = log. 1000 — 15 log. 1,05 = 2,6821605 ; and consequently P = 481,02 or 481/, 0^. 4|c/. the value sought. Examp. 3. In how long time will 57 SL raise a stock of 756/. 13^. 2\d. at 4< per cent.P In this case we have R = 1,04, P = 575^ and a:= 756,66; whence, bytheor.4, ^^^og- 756,^6 -log- 575 log. 1,04 = 7, the number of years required. Examp. 4. To find at what rate of interest 481/. in fif- teen years, will raise a stock of 1000/. Here we have given P = 481, a = 1000, and w = 15 ; therefore, by theorem 3, log. R = —^^ ^^' = . 0211903, whence R = 1,05 ; consequently 5 per cent. is the rate required. The four last examples relate to the cases in compound interest ; the four next are upon the forbearance of an- nuities. Examp. 1. If 50/. yearly rent, or annuity, be forborne seven years, what will it amount to, at 4 per cent, per an- num^ compound interest ? Here we have R = 1,04, A = 50, and tz = 7 5 and 238 Of Interest and Annuities* therefore, by theor, 1 , log. m (= log. A + log. R" — 1 — log. R— 1) = log. 50 + log. Ifiiy — 1, — log. ,04 = 2,596597 ; and consequently 7n = 395/. the value that was to be found. Examp. 2. What annuity, forborne seven years, will amount to, or raise a stock of 395/. at 4 per cent, com- pound interest ? In this case we have given R = 1,04, n = 7, and m = 395 ; whence, hij theorem 2,1 og. A ( = log. m — log. R*«— 1 + log. R — 1) = log, 395 — log. 1^] 7 — 1 + log. ,04 = 1,6989700; and consequently A = 50/. which is the annuity required. Examp. 3. In how long time will 50/. annuity raise a stock of 395/. at 4 per cent, per annum^ compound inte- rest? Here we have R = 1,04, A = 50, w = 395 ; and therefore, hii theor. 3, n (= -Sli^! — l^^ "^^ ^^' ) ^ ^ log. R ^ :=:: 1_ = 7 the number of vears required. ,0170333 Exam A 4. If 120/. annuity, forborne eight years, amount to, or raise a stock of 1200/. what is the rate of interest ? In this case we have given n = 8, A = 120, and m ■=. 1200, to find R ; therefore, by theorem 4, we have R» — loR + 9 = 0, from which, by any of the methods in sect. 13, the required value of R will be found = 1,06287; therefore the rate is 6,287, or 6/. 5s. 9d. per cent, per annum. The solution of the last case, where the rate is re- quired, being a little troublesome, I shall here put down an approximation (derived from the third general for- mula^ at p. 165) which will be found to answer very near the truth, provided the number of years is not very great. Of Interest and Annuities » 239 Let Q = ; then will 2 . w^ — nA. 3000Q + 2n—l. 400 be the rate 6Q.5Q + 37Z — 4 + I-. n — 2 . tin— 13 per cent, required. Thus, for example, let n = 8, A = 120, and m = 1200 ; m will Q = 42000 + 6000 then will Q = '- =14, and the rate itself 2 . 240 ' 84 X 90 + 75 = 6,287, as above. The preceding examples explain the different cases 6i annuities in arrear ; in the following ones the rules for the valuation of annuities are illustrated. Examp. 1. To find the present value of 100/. annuity, to continue seven years, allowing 4 per cent, per annmri^ compound interest. Here we have given R = 1,04, A = 100, and n = 7 ; and therefore, by theorem 1, log. t; ( = log. A + log. 1 — ~ — log. R — 1 ) = log. 100 + log. 1 — log, ,04 = 2,778296 j and consequently 1,04^ t; = 600,2 = 600/. 4sy. which is the value that was to be found. Examp. 2. What annuity, or yearly income, to con- tinue 20 years, may be purchased for 1000/. at 3^ per cent. P In this case, R = 1,035, ;z = 20, v = 1000; whence, by theorem 2, we have log. A ( = log. v + log. R — 1 _ log. 1 _ ) = 1,847336; and con- sequently A = 70,36, or 70/. 7s. %d. 240 Of Interest and Annuhies. Examp. 3. For how long time may one, with 600/. pur- chase an annuity of 100/. at 4/?^r cent. ? In this example, we have R = 1,04, A = 100, and V = 600 ; and therefore, by theorem 3, n ( = log. A — log. A 4- ^ — ^^Rn ^ ,, u c —2 ^ — .^ A ~ 7 the number of years log. K required* Examp. 4. To determine at what rate of interest an annuity of 50/. to continue ten years, may be purchased, for 400/. Here A = 50, n = 10, and v = 400; whence, by A A theorem 4, R'^+i (-lxR"H being = 0, we have R" — 1,125R^° + ,125 = 0; which equation resolved, gives the required value of R = 1,042775 ; and consequently the rate of interest 4,2775/. per an-^ num. The solution of this last case being somewhat tedious, the following approximation (which will be found to an- swer very near the truth, when the number of years is not very large) may be of use. Assume Q = — 1— Jt — '- — : so shall 2n K — 2v 3000Q — 2/1 -f- 1 X 400 express the 6Q. 5Q— 3?2 — 4 + 1 . n + 2 . ll/z + lS rate per cent, very nearly. Thus, for example, let A (as above) be = 50, n = 10, andi; = 400; then, Q being =i ^^ ^^ ^/^ = 27.5, ' ^ 1000 — 800 , 82500 — 8400 . ^^^^ r ^-u ^ ^.^ we have , or 4,2775, for the rate per 165 X 103,5 + 246' ' ,^ ' cent, the same as before. Of Plane Trigonometry. 241 SECTION XVII. Of Plane Trigonometry* DEFINITIONS. 1. PLANE trigonometry is the art whereby, hav- ing given any three parts of a plane triangle (except the three angles), the rest are determined. In order to which, it is not only requisite that the peripheries of cir- cles, but also that certain right lines, in and about the circle, be supposed divided into some assigned number of equal parts. 2. The periphery of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minutes into 60 equal parts, called seconds, or second- minutes, £sfc. Any part of the periphery is called an arch, and is measured by the number of degrees and minutes, c£?c. it contains. 3. The difference of any arch from 90 degi'ees, or a , quadrant, is called its complement, and its difference from 180 degrees, or a semicircle, its supplement. 4. A chord, or sub- tense, is a right line drawn from one ex- tremity of an arch to the other ; thus BE is the chord or sub- tense of the arch BAE or BDE. 5. The sine (or right-sine) of an arch is a right line drawn from one extremity of the arch perpen- dicular to the diame- ter passing through the other extremity: thus, BF is the sine of the arch AB, or BD. 21 242 Of Plane Trigonometry. 6. The versed- sine of an arch is the part of the diame- ter intercepted between the arch and its sine : so AF is the versed-sine of AB, and DF of DB. 7. The co-sine of an arch is the part of the diameter intercepted between the center and the sine ; and is equal to the sine of the complement of that arch. Thus, CF is the co-sine of the arch AB, and is equal to BI, the sine of its complement HB. 8. The tangent of an arch is a right line touching the circle in one extremity of that arch, continued from thence, to meet a line drawn from the center through the other extremity ; which line is called the secant of the same arch : thus AG is the tangent, and CG the secant of the arch AB. 9. The co-tangent and co-secant of an arch are the tangent and secant of the complement of that arch : thus HK and CK are the co-tangent and co-secant of the arch AB. 10. A trigonometrical canon is a table exhibiting the lengths of the sine, tangent, csfc. to every degree and mi- nute of the quadrant, with respect to the radius, which is supposed unity, and conceived to be divided into 10000000 or more decimal parts. Upon this table the numerical solution of the several cases in trigonometry depends ; it will therefore be proper to begin with its construction. PROPOSITION I. The number of degrees and minutes^ &c. in an arch be-^ ing given ; to find both its sine and co-sine. This problem is resolved, by having the ratio of the circumference to the diameter, and by means of the known series for the sine and co-sine (hereafter de- monstrated). For, the semi-circumference of the circle, whose radius is unity, being 3,141592653589793, £5?r. it will therefore be, as the number of degrees or mi- nutes in the whole semicircle is to the degrees or minutes in the arch proposed, so is 3,14159265358, ^c. to the length of the said arch ; which let be denoted by a ; then, by the series above quoted^ its sine will be ex- Of Plane Trigonometry* 245 pxessed by a - ^ + ^-^^ - ^ . 3 . /'o . 6 . 7 ' &fc* and its co-sine by 1 — 2 ^4 ^6 2 2.3,4 2.3.4.5.6 ^ 2 . 3 . 4 . 5 . 6 . r . 8 Thus, for example, let it be required to find the sine of one minute: then, as 10800 (the minutes in 180 de- grees) : 1 : : 3,14159265358, i^c. : .000290888208665 = the length of an arch of one minute : therefore, in this case, a = .000290888208665, and -^ ( = ^) ^ • o O = .000000000004102, &fc. And, consequently, .000290888204563 = the required sine of one minute* Again, let it be required to find the sine and co-sine of five degrees, each true to seven places of decimals. Here ,0002908882, the length of an arch of 1 minute (fpund above), being multiplied by 300, the number of minutes in 5 degrees, the product .08726646 will be the length of an arch of 5 degrees : therefore, in this case we have a = ,08726646, —|.= —, 0001 1076, o a^ + = + ,00000004, ^120 ' i^c> and consequently ,08715574 = the sine of 5 degrees. Also, — =,00380771, 2 a* — = ,00000241 ; 24 ' ^and consequently ,9961947 = th^ co-sine of 5 degrees. After the same manner, the sine and co-sine of any other arch may be derived ; but the greater the arch is, the slower the series will converge, and therefore a greater number of terms must be taken to bring out the conclusion to the same degree qi exactness. 244 Of Plane Trigonometry. But there is another method of constructing the trigo- nometrical canon ; which, though less direct, is more geo- metrical ; and that is by determini^ng the sines and tan- gents of different arches, one from another, as in the en- suing propositions. PROPOSITION IL The sine of an arch being given; to find its co-sine^ tan- gent^ co-tangent^ secant ^ and co-secant. Let AE be the proposed arch, EF its sine, CF its co-sme, AT its tangent, DH its co-tangent, CT its secant, and CH its co-secant : then {by Euc. 47. 1.) we shall have CF = VCE^ — EF^ ; from whence the co-sine will be known ; and then, by reason of the similar triangles, CFE, CAT, and CDH, it will be, 1. CF : FE : : CA : AT; whence the tangent is known. 2. CF : CE : : CA : CT ; whence the secant is known. 3. EF : CF : : CD : DH ; whence the co-tangent is known. CD : CH ; whence the co-secant is also c 4. EF known. Hence it appears, 1. That the tangent is a fourth proportional to the co- sine, the sine, and the radius. 2. That the secant is a third proportional to the co-sine and the radius. 3. That the co-tangent is a fourth proportional to the sine, the co-sine, and the radius. 4. That the co-secant is a third proportional to the sine and the radius. 5. And that the rectangle of the tangent and co-tangent is equal to the square of the radius. Of Plane Trigonometry* 245 PROPOSITION III. The co-sine CF of an arch AE being given ; to find the ine and co-sine of half that arch* From the two extremities of the diameter AB draw the subtenses AE and BE ; and let CQ bisect the arch AE in Q and its chord (perpendicularly) in D ; then, since the angle BE A is a right one- (% Euc. 31. 3.), the triangles ABE and ^^^^ ^^JE ADC are similar ; and, therefore, AC being = ^ AB, AD must be = i AE, and CD = \ BE : b ut AE is = VA B X AF, an d BE = VAB X BF ; therefore AD = iVAB_ __ CD = Iv^AB xW = Vp^ X~BF = the co-sine Hence it is evident, that the sine of the half of any arch is a mean proportional between the half radius and the versed-sine of the whole arch ; and its co-sine a, mean proportional between half the radius and the versed-sine of the supplement of the same arch. PROPOSITION IV. The sine AD, and co-sine CD, of an arch AQ being given ; to find EF, the sine of the double of that arch. {See the preceding fgure)* Since AE = 2AD, and BE = 2CD, and the triangles ABE and AEF are alike {bif Euc* 8. 6.), we have, as AB (2AC) : AE (2AD) : : BE (2CD) : EF ; whence it appears, that the sine of double any arch is a fourth proportional to the radius, the sine, and the double-co- sine of the same arch. PROPOSITION V. The sine CD, and tangent BE, of a very small arch^are nearly in the ratio of equality* For, the triangles ADC and ABE being similar. 246 OfPlmc Trigonometry. thence will AD : AB : : DC : BE : but as the point C approaches to B, the difference of AB and AD will become indefinitely small in respect of AB, and there- fore the difference of BE and DC will liker wise become indefinitely small with respect to BE or DC. Corollary. Because any arch BC is greater than its sine and less than its tangent ; and since the sine and tangent of a very small arch are proved to be nearly equal, it is manifest that a very small arch and its sine are also nearly in the ratio of equality. PROPOSITION yi. To find the sine of an arch of one minute* The sine of 30 degrees is known, being half the chord of 60 degrees, or half the radius ; therefore, by prop. 2. and 3. the sine of 15 degrees will be known ; and, the sine of 15 degrees being known, the sine of 7"" 30' will be found {by the same propositions)^ and from thence the sine of 3° 45' ; and so likewise the sine of half this ; and ^o on, till 12 bisections being made, we come, at last, to .the sine of an arch of 52", 44'^^ 03';", 45'""; which sine (by corol. to the preceding prop.) will, as the co-sine is nearly equal to the radius, be nearly equal to the arch itself. Therefore we have, as 52", 44"', 03"", 45""', is to 1', so is the length of the former of these arches (found as above) to the length of an arch of one minute, or t!iat of its sine, very nearly. If it be taken for granted, that 3,1415926535, ^c. is the length of half the periphery of the circle whose radius is unity, we shall have, as 10800, the number of minutes in 180°, or the whole semicircle, is to one mi- nute, so is 3,1415926535, &fc. the whole semicircle, to 0,000290888208, the length of an arch of one mijnute, or that of its sine, very nearly. 1^ Of Plane Trigonometry, 24>7 PROPOSITION VIL If there he three equidifferent arches AB, BC, and AD, it will be^ as the radius is to the co-sine of their common difference BC or CD, so is the sine CF, of the mean^ to half the sum of the sines BE + DG, of the two extremes ; and as the radius is to the sine of the common difference^ so is the co-sine FO of the mean^ to half the difference of the sines of the two extremes. For, let BD be drawn, cutting the radius OC in 7n ; also draw mn parallel to CF, meeting AO in 7Z, and BH and 77iv parallel to AO, meeting DG in H and v : then, because the arches BC and CD are equal tQ each A E other, OC is not only perpendicular to BD, but also bisects it (^Euc. 3. 3.) ; whence it is evident that Bm, or Dw, will be the sine of BC or CD, and Om its co- sine ; and that mn^ being an arithmetical mean between the sines BE and DG, of the two extremes, is equal to half their sum, and T^v equal to half their difference. Moreover, by reason of the similarity of the triangles OCF, OwiTz, and Dmv, it will be as OC : Om \ \ C¥ , mn \ q, ^ j^ and as OC : Dw : : FO : D^ / ^' ^- ^' COROL. 1. Since, from the foregoing proportions, mn is =: Om X CF Dm X FO . . . , ^ - , and U'o (= z;H) = — ~^^ , it is evident OC OC 248 Of Plane Trigonometry. that DG (= mil + D.) will be = Q>»xCF + DmxFO and BE (= mn — va) = -— : from whence it appears that the sine (DG) of the sum (AD) of any two arches (AC and CD) is equal to the sum of the rectangles of the sine or the one into the co-sine of the other, alternately, divided by the radius ; and that the sine (BE) of their difference (AB) is equal to the difference of the same rectangles, divided also by the radius, COROL. 2. 20w X OF Moreover, seeing DG + BE (2w?2) is = OC and DG — BE (= DH = 2D^) = !H^-jJ!2, from the former of these, we have DG = — ^^-~ BE, and from the latter, DG = — ^^ + BE ; which equa- tions, expressed in words, give the following theorems, Theor. 1. If the sine of the mean of three equidtfferent arches (supposing the radius unity) be multiplied by twice the cosine of the common difference^ and the sine of either extreme be subtracted from the product^ the remainder will be the sine of the other extreme. Theor. 2. Or, if the co-sine of the mean be multiplied by ttvice the sine of the common differ ence^ and the product be added to or subtracted from the sine of one of the extremes^ the sum or remainder xvill be the sine of the other extreme. These two theorems are of excellent use in the con* struction of the trigonometrical canon; for, supposing the sine and co-sine of an arch of 1 minute to be found, by prop. 6 and 1 , and to be denoted by p and y, respec- tively ; then, the sine of 2 minutes being given from prop. 4, the sine of 3 minutes will from hence be known, being == 2^ X sine 2' — sine 1' (by theor. l) or = 2p X co-sine of 2' + sine of I' (by theor. 2). After the same Of Plane Trigonometry » 249 inanner the sine of 4' will be found, being = 2§^ X sine of 3' — sine of 2', or = 2/? X co-sine of 3' + sine of 2'. And thus the sines of 5, 6, 7, &fc. minutes may be successively derived by either of the theorems ; but the former is the most commodious. If the mean arch be 45°, then, its co-sine "being = V|^, it follows (yr(?m theorem 2) that the sine of the ex- cess of any arch above 45°, multiplied by 2V|. or \/2, gives the excess of the sine of this arch above that of another arch as much below 45° ; thus, VS x sine of 10° = sine of SS"" ■— sine of ZS"" ; and Vi x sine of 15° = sine of 60° — sine of 30° ; and so of others : which is useful in finding the sines of arches greater than 45°. But, if the mean arch be 60 degrees, then its co-sine being i, it is evident, from the same theorem^ that the sine of the excess of any arch above 60°, added to the sine of another arch as much below 60°, will give the sine of the first arch, or greater extreme : thus, the sine of 10° + sine 50° = sine 70°, and sine 15° -f- sine 45° = sine IS"^ ; from whence the sines of all arches above 60 degrees, those of the inferior arches being known, are had by addition only* PROPOSITION VIII. In dny right-angled plane triangle ABC, it will fe, as the base AB is to the perpendicular BC, 50 is the radius {of the tables) to the tangent of the angle at the base. Let DA be the radius to which the table of sinea and tangents is a- dapted, and DE the tangent of the an- gle A ; then, by rea- son of the similarity of the triangles ABC and ADE, it will be, as AB : BC : : AD : DE. % E. D. 250 Of Plane Trigonometry, PROPOSITION IX. In every plane triangle^ it will he^ as any one side is to the sme of the opposite angle^ so is any other side to the sine of its opposite angle. For, let ABC be the proposed triangle; take CF = AB, and upon AC let fall the perpendiculars BD and EF ; which will be the sines of the angles A and C, to the equal ra- dii AB and CF. But the trian- gles CBD and CFE are simi- lar, and there- fore CB : BD : : CF (AB) : FE ; that is, as CB is to the sine of A, so is AB to the sine of C. S>.£.D. PROPOSITION X. In every pla7ie triangle^ it will be^ as the sum of any two sides is to their difference^ so is the tangent of the complex ment of half the angle included by those sideSj to the tan- gent of the difference of either of the other two angles and the said complement. For, let ABC be the triangle, and AB and AC the two proposed sides ; and upon A, as a center, with the radius AB, let a semicircle be described, cutting CA produced, in D and F ; so that CF may express the sum, and CD the difference of the sides AC and AB ; join F, B, and B, D, and draw P AD ^ DE parallel to FB, meeting BC in E ; then the angle FBD being aright one {by Euc. 31. 3.) ADB will be the comple- ment of the angle F, which is equal to half the pro- posed angle A (by Euc. 20. 3.). Moreover, seeing the Of Plane Trigonormtry. 251 angles FBD and EDB are both right ones, for EDB is = FBD (= a right angle), because DE is parallel to FB, it is plain, that, if BD be made the radius, BF will be the tangent of BDF, and DE the tangent of DBE : but, because of the similar ti*iangies CFB and CDE, CF : CD : : BF : DE ; that is, as the sum of the sides AC and AB is to their difference, so is the tan- gent of BDF to the tangent of DBC \ which angle is ma- nifestly the excess of ABC above BDF, or ABD ; and also the excess of ADB above ACB. ^E. D. PROPOSITION XL As the base of any plane triangle is to the sum of the two sides^ so is the difference of the sides to the difference of the segments of the base^ made by a perpendicular falling from the vertical angle • < For, let ABC be the proposed triangle, and BD the perpendicular ; from B as a center, with the interval BC, let the circumference of a circle be described, cut- ting the base AC in G, and the side AB, produced, in F and E : then wall AE be the sum of the sides, AF their differ- ence, and AG the difference of the segments of the base AD and DC: but (^by Eiic. 36. 3.) AE X AF = AC X AG ; and therefore AC : AE AP: AG. %E.D. Z52 Of Plane Trigonometry, The solution of the cases of right-angled plane trimigks. Given. Sought. Proportion. 1 2 Thehypothe- nuse AC and the angles. The leg BC. As the ! adius (or the sine of B) is to the hyp. AC ; so is the sine of A, to its opposite side BC. (^by firo/i. 9 ) The hypoth. AC and one leg AB. The angles. As AC : rad. : : AB : sine of C ; whose complement gives the an- gle A. 3 The hypoth. AC and one le^ AB. Theother leg BC. i^et the angles be found by case 2 ; then, as rad. : AC : : sme of A : BC {by firo/i. 9,) 4 5 6 The angles and one leg AB. The hy- pothenuse AC. As sine of C : AB : : rad. (sine of B): AC (iby prop. 9, ) The angles and one leg AB. The other leg BC. As sine of C : AB : : sine of A : . BC {by prop. 9.) Or, rad. : tang, of A : : AB : BC (byfirofi 8.) The two legs AB and BC. The angles. As AB : BC : : rad. : tang, of A {by prop, 8) ; whose complement gives the angle C. 7 The two legs AB and BC. The hy- pothenuse AC. Find the angles by case 6, and from thence the hyp. AC, by case 4. Of Plane Trigonometry. 253 The solution of the cases of oblique plane triangles. Given. Sought. Proportion. 1 rhe angles and one side AB Either of the other sides, suppose BC. As sine of C : AB : : sine of A : BC (dy firo/i. 9.) 2 Two sides AB, BC and the angle C opposite to one of them. The other angles A and ABC As AB : sine of C : : BC : sine of A ; which added to C, and the sum subtracted from 180°, gives the angle ABC. 3 Tsvo sides AB, BC and the angle C opposite to one of them The other side AC. Find the angle ABC by case 2 ; then, as sine of A : BC : : sine of ABC : AC. 4 5 Two sides AB, AC and the included angle A. The other angle C and ABC. As AB+ AC : AB— AC : : tang, of the comp.ofl A : tang, of an ang. which added to the said com. gives the greater ang. C ; and subtracted leaves the lesser ABC (firo/i, 10.) Two sides AB, AC and the included angle A. The other side BC. Find the angles by case 4; and then BC, by case 1 . 6 All the sides. An angle, suppose A Let fall a perpendicular BD, opposite the required angle, and suppose DG = AD ; then (by prop. 11.) AC : BC + B A : : BC — BA : CG, which subtracted from AC, and the re- mainder divided by 2, gives AD ; whence A will be found, by case 2, of right angles. 254 The Application of-Aigehr SECTION XVIII. The Application of Algebra to the Solution of Geo- metrical Problems. WHEN a geometrical problem is proposed to bfe re- solvecUDy algebra, you are, in the first place, to describe a figure that shall represent or exhibit the several parts or conditions thereof, and look upon that figure as the true one J then, having considered attentively the nature of the problem, you are next to prepare the figure for a so- lution (if need be), by producing and drawing such lines therein as appear most conducive to that end. This done, let the unknown line, or lines, which you think will be the easiest found (whether required or not), together with the known ones (or as many of them as are requisite), be denoted by proper symbols ; then proceed to the opera- tion, by observing the relation that the several parts of the figure have to each other ; in order to which, a com- petent knowledge in the elements of geometry is abso- lutely necessary. As no general rule can be given for the drawing of lines*, and electing the most proper quantities to substitute for, so as always to bring out the most simple conclusions (be- cause different problems require different methods of so- lution), the best way, therefore, to gain experience in this matter is to attempt the solution of the same problem se- veral ways, and then apply that which succeeds best to other cases of the same kind, when they afterwards occur. I shall, however, subjoin a few general directions, w^hich will be found of use. 1°. In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles ; and if an angle be given, let the perpendicular be opposite to that angle-, and also fall from the end of a given line, if possible. 2°. In electing proper quantities to substitute for, let thoge be chosen (whether required or not) which lie p ro Geometrical Problems. 25S neatest the known or given parts of the figure, and by help whereof the next adjacent parts may be expressed, without the intervention of' j surds, by addition and sub- traction only. Thus, if the prbblem were to find the perpendicular of a plane triangle, from the three sides given, it will be much better to substitute for one of the segments of the base, than for the perpendicular, though the quantity required ; because the whole bas^ being giveuj the other segment will be given, or expresseel, ^y subtraction only, and so the final equation come out a simple one ; from whence the segments being known, the perpendicular is easily found by common arithmetic : whereas, if the perpendicular were to be first sought, both the segments would be surd quantities, and the final equa- tion an ugly quadratic one. 3°. When, in any problem, there are two lines or quantities alike related to other parts of the figure, or problem, the best way is to make use of neither of them, but to substitute, for their sum, their rectangle, or the sum of their alternate quotients, or for some line or lines in the figure, to which they have both the same relation. This rule is exemplified in prob. 22, 23, 24, and 27. 4°. If the area, or the perimeter of a figure be givefi^ or such parts thereof as have but a remote relation to the parts required, it will sometimes be of use to assume another figure similar to the proposed one, whereof one side is unity, or some other known quantity; from whence the other parts of this figure, by the known proportions of the homologous sides, or parts, may be found, and an equation obtained, as is exemplified in prot. 25 and 32. These are the most general observations I have been able to collect ; which I shall now proceed to illustrate thy proper examples. PROBLEM I. The base (^), and the sum of the hypothenuse and per- pendicular («), of a right-angled triangle^ ABC ^ being given ; to find the perpendicnlar. 256 The Applicatmi of Algebra Let the perpendicular BC be denoted by at; then p the hypothenuse AC will ^ be expressed by a — x : but {by Euc. 47. 1.) AB^ + BC2 = AC^ ; that is, b^ + x^ •=■ c^ — %ax_ + XX ; whence x = = the 2a B perpendicular required- PROBLEM IL 'The diagonal and the perimeter of a rectangle^ A BCD, being given; to find the sides. Put the diagonal BD = «, half the perimeter (DA + AB) = b^ and AB = x\ then will AD = <^ — X ; and, therefore, AB^ + KW being = BD% we have x^ + b^ — 2bx + x^ z=z a^ ; which, solved, gives \/2d^ — b^ ^b PROBLEM IIL The area of a right-angled triangle ABC, and the sides of a rectangle^ Y.BDF J inscribed therein^ being given; to deter ?nine the sides of the triangle. Put DF = Uy DE = ^, BC = Xy and the measure Q of the given area ABC = d: then, by similar tri- angles, we shall have x — b (CF) : «(DF): :x (BC) : AB = ^^^. rr.1 r ax ^ X Therefore > X - = X — b 2 dy and consequently ax'^ == 2dx — 2hd, or ;^^ — ~ = — — : which, solved, ^ a a to Geometrical Problems. 257 d ^ives X z=, — ± a ~ /7/Tr 2bd , from whence AB and AC will likewise be known. PROBLEM IV. Having the lengths of the three pcypendiculars PF, PG, PH, drawn from a certain point P xvithin an equilateral- triangle ABC, to the three sides thereof; from thence to determine the sides* L,tt lines be drawn from P to the three angles of the triangle ; and let CD be perpendicular to AB : call PF «; PG ^; PH c; and AD = oc : then will AC ( = AB) = 2x, and CD (== \/AC2 _ ^£)2j _ V^3^^. _ x\^ 3 ; and consequently the area of the whole triangle ABQ^(=: CD X AD) = xxs/ S. But this triangle is composed of the three trian- gles APB, BPC, and APC ; whereof the respective areas are ax^ bx^ and ex. Therefore we have xxV 3 = fi^ + hx -f ex ; and from thence, by division, x = PROBLEM V. a + b^c Having the area of a rectangle DEFG inscribed in a given triangle ABC ; to determine the sides of the rectangle^ Let CI be perpendi- cular to AB, cutting DG in H ; and let CI = a, AB = ^, DG = X, and DX H the given area = cc: then it will be, as ^ : ax ^,^- - : :«:--- = CHj which, taken from CI, 2og The Applicatmi of Algebra leaves «•*—--- = IH ; and this, multiplied by x^ givers. ax 7~ :=z cc •=: the area of the rectangle ; whence we have abx — ax^ = hcc^ x^ ^^hx ■=. — ^4 a 2^4 a have abx — ax^ = hcc. x^ — ^:v = , :^ — — = ± PROBLEM VL Through a given point P, xvithin a given circle^ so t& draw a right line^ that the two parts thereof PR, PQ, m- tercepted between that point and the circumference of the circle^ may have a given difference Let the diameter APB be drawn; and let AP and BP, the two parts thereof (which are supposed given) be denoted by a and b ; mak- ing PR = x^ and PQ = x ■\- d{d being the given dif- ference). Then, by the na- ture of the circle, PQ X PR being = PA x PB, we have X + d X X •=• ah ^ or XX -^-^ dx-=^ah \ whence x is found = V^F+Tdl— \d. PROBLEM VIL Fro7n a given point P, without a given circle^ so to draw a right line PQ, that the part thereof RQ, inter- cepted by the circle^ shall be to the external part PR, in a given ratio • to Geometrical Problems. 259 Through the center O, draw PAB; put PA = tf, PB = /^, PR = X, and let the given ratio of PR to RQ be that of m to n ; then it will be, as m : n : : — = RQ; therefore X nx m PQ = PA X PB, or ^ nx , , X X X '\ =: ab; there- m fore mx^ -f nx^ = maby and X = \}- mab m + n PROBLEM VIII. The sum of the two sides of an isosceles triangle ABC Z?f- ing equal to the sum of the base and perpendicular^ and the area of the triangle being given ; to determine the sides. Put the semi-base AD = x^ the perpendicular CD = y, and the given area ABC =1 a^ I so shall xy = a^, and 2\^xx -j- ifi/ = 2x + y {by EL 47, 1, and the conditions of the problem). Now, squaring both sides 6f the last equation, we have 4*xx + 4z/z/ = A^xx -f- 4!xy + yy ; whence Syy = 4x2/, and consequently y = -— : which value, substituted , ^ AfXX in the former equation, gives = a^ ; from whence ha" . .— (= -) = faV 3 ; and AC %m (= V;vx + yt/ The Application of Algebra Jsaa , ^aa hzSaa^ aV\ :=z\aV^. PROBLEM IX. The segments of the base AD and BD, and the ratio of the sides AC and BC, of any plane triangle ABC being given ; to find the sides* Put AD = «, BD == b, AC = .y ; and let the given ratio of AC to BC be as m nx to n, so shall BC = — • in But AC2 _ AD^ (= DCO = BC^ — BD% that is, in Species, ;c^ »— a^ 'b^. mm Hence we have 7r?x^ A and — n^x^ -=. rn^ X aa-^ bb^ Jaa — X = m\j ■ ^ mm — ; bb nn PROBLEM X. The base AB (a), the perpendicular CD = b^ and the difference (d) of the sides AC — BC, of any plane trian- gle ABC, being given ; to determine the triangle. (See the preceding figure,) Let the sum of the sides AC + BC be denoted by x\ dx then {by prop* 11, sect. 18) we shall have a\ x \ \ d \ — = the difference of the segments of the base ; therefore a dx the greater segment AD will be = -^ + aa -f- ^^ But AD2 + DC2 2 AC^ a" •\' 2a^dx -f d^x'^ 4aa + _ x^ + 2dx + dd ^ ' 4 2a that is, whence to Geometrical Problems. 261 a^ + %a^dx + d^x^ + 4a^^^ = a^x^ + 2 aV;v + a^i^ ; which, , , . \aa 4- 4bb — dd solved, gives x = ^V aa-dd. ' PROBLEM XL The base AB, the sum of the sides AC + BC, and the length of the line CD drawn from the vertex to the middle of the base^ being given; to determine the triangle. Make AD (= BD) = a, DC = b, AC + BC = c, and AC = :v : so shall p BC = c — ^, But -^^ AC^ + BC^ is = 2AD2 + 2DC2 (by El. 1 2, 2) ; that is, o<^ + c — x'Y — 2a2 + 2^2 ; which, by reduction, becomes x^ — ex :=z d^ + l^ — \c^ ; whence x is found = |c ± Vaa + PROBLEM XIL The two sides AC, BC, and the line CD bisecting the vertical angle of a plane triangle ABC, being given ; to find the base AB. Call AC a; BC ^ ; CD c ; and AB.;\:: then a + b '. PC : I a '. AD = bi DB = and a + b bx But (by a + b El. 20, 3) AC X CB -, AD X DB = CD2, that is, - abx^ ab — =r~ = c^; from axbj whence x wiU be found = a^b . ^* ab^ ab 162 The Application of Algebra PROBLEM XIIL The perimeter AB + BC + CA, and the perpendicular BT^^ falling from the right angle B, to the hypothenuse AC, being given ; to determine the triangle* Let BD = «, AB = x, BC = y, AC = z, and AB + BC + CA = ^: then, by reason of the similar tri- angles ACB and ABD, it will be, as 2 : 2/ : \x \ a:, and therefore xy=iazi more- over, x^+y'^ =2^ (by Euc, 47^ 1), and x -j- y -{- z =: b (by the question^. Trans- pose 2 in the last equation^ and square both sides, and you will have x"^ + 2xy -j- y^=zb^ — 2bz + 2^ ; from which take x^ + y"^ =. 2^, and there will remain 2xy •=. b^ — 2^2 ; but, by the first equation, 2Arz/ is = %az ; therefore 2a2 "=. IP' — 2^2 and 2 = 2a -f 2/^ ' a: and y from hence, put whence 2 is known. bb But to find = c, and let this value 2a + 2^ of 2 be substituted in the two foregoing equations, X + y ■=: b — 2, and zy = 02, and they will become X + y =. b — c, and xy :=: ac : from the square of the former of which subtract the quadruple of the latter, so shall x^ — 2xy + y^ =z b — c^ — ^cic ; and conse- quently X — z/ = \ ^ — cY — " 4<2C. This equation be- ing added to, and subtracted from x + y = b — c, gives 2x =z b' — c + \b ■ _ \f^ _ cf — 4ac. . cl^ — 4aCy and 22/ = ^ — c PROBLEM XIV. Having the perimeter of a right-angled triangle ABC, and the radius DF of its inscribed circle; to determine all the sides of the triangle* to Geometrical Problems. 263 From the center D, to the angular points A, B, C, and the points of contact E, F, G, let lines DA, DB, DC, DE, DF, DG be drawn; making DE, DF, or DG — a, AB = x, BC c= 2/, AC = and It is evident that — 4- 2 ^ x + y + z-b. ab 2 BDC, and ADC) will be = ay az or its equal — (expressing the sum of the areas ADB, xt/ "2 = the area of the whole triangle ABC ; and consequently 2x1/ = 2ab : moreover (^z/ Euc. 47, 1), x^ + y^ = z^ . to which if 2xy = 2ab be added, we shall have x^ + 2xy -f- y^^ or .V + y1^ = 2^ + 2ab ; but, by the first step, x +y']^ is = b — 2 I 2 = ^2 __ 2bz + 7? ; therefore, by making these two values of x + y\^ equal to each other, we get z^ + 2ab ■=: b^ — 2bz + z^ ; whence 2a =: b — 22, and z = ^b — a. But, to find x and y, from hence, we have now given x + y (= b — 2) = -^^ + «, and xy = ab : the former of these equations, multiplied by x^ gives :x^ + xy =: — + ax ; from which the latter xy = ab being subtracted, we have x^ = \bx + ax - — ab^ or x^ — • — - — X X =z — < ai> : whence, by completing. 2G4 The Application of Algebra . . 2a + b ± V4a^ — 12ab 4- b^ the square, ksrc, x = : so that the three sides of the triangle are, ^b — a^ 2a-\^b + V4a^ — 12ab + b^ ^ ^^^ 2a + b—\^4a^~ 1 2ab+b^ 4 ' 4 Otherwise. The right-angled triangles ADE, ADG, having the sides DE, DG equal and AD common, have also AE equal to AG : and, for the like reason, is CE = CF J and consequently AC (AE + CE) = AG + CF. Whence it appears that the hypothenuse is less than the sum of the two legs AB + BC, by the diameter of the inscribed circle, and therefore less than half the pe- rimeter by the semi-diameter of the same circle. Hence we have AC = |(^ — a, and AB -|- BC = |^ + a. Put, therefore, |^ — /2 = c, ^b + a = d^ and half the dif- ference of AB and BC = x ; then will AB =z d + x^ and BC =zd—x; and consequently 2d^ + 2x^ (A B^ + BC^) = c^ ( AC^), whence x is found = V^ c^ — d^ ; therefore ABis^l^' + ^ic^ — ^, andBC = |^— V'p'irS^- PROBLEM XV. All the three sides of a triangle ABC being give7i ; to find the perpendicidary the segments of the base^ the area^ and the angles. Put AC = (7, AB = ^, BC = c, and the segment AD = X ; then BD being = ^ — x^ we have c^ — b — x'\ 2 (= CD2) = a^ — x^, that is, c^ — b"^ -f- 2bx P A D B x^ = a" — x'^ '-, whence 2bx = aa + bb -^ cCj and to Ge07netrical Problems* '^^S 2^ AC + AD X AC — AD = a + aa -^ bb > — cc 2b aa A- bb — cc 2ab + aa 4- bb — cc a — ■' = ' 1 V 2b 2b ^ 2ab — aa — bb 4- cc _ « -f b ^ — c^ c^ — a — b^ hence CD = — x Ja + bf — c^ x c^ — ^^ZZ^Y ; and the area ( ) = ^ 2 ^ \sj^T~bY — ? X c^ — ^=Tp- But, because the difference of the squares of any two lines, or numbers, is equal to a rectangle under their sum and difference, the factor a + bl^^ — c^ will be = a 4- b -{- c X a '\- b — c>, and the remaining factor c^ — a — b']^:=zc + a*-^bxc*---a + b: and so the area will be likewise truly expressed by i\ g + b + c X a + b — c X c + a-^b X c — a^ b la -{- b -^ c a 4- b — c c 4- a* — *b c- — a 4- b = V -32j-^-V-^ — i— ^— T^ = /y 5 .s — c . s — b . s — a-yhy making s = — ^It . In order to determine the angles, which yet remain to be considered, we may proceed according to prop. 11, in trigonometry, by first finding the segments of the base: but there is another proportion frequently used in practice, which is thus derived : let B A be produced to F, so that AF may be = AC ; and then, FC being joined, it is plain that the angle F will be the half of the angle A 5 and DF ( = AC + AD) wili be given 2 M 266 The Application of Algebra {from above) = 2b iff — ^ + ^ -f g « + b — c ~ c: but DF (: ^s X s — ) is to DC C—\77s — c . 5 — - ^ . * — . a), so is the radius to the tan- gent of F ; and, consequently, s X s — c:**— ^X* — a : : sq. rad. : sq. tang, of F ; diat is, in words, as the rect- angle under half the sum of the three sides, and the excess of that half sum above the sid^ opposite the required angle, is to the rectangle under the differences between the other two sides and the said half sum, so is the square of the radius to the square of the tangent of half the an- gle sought. PROBLEM XVI. Having given the base AB, the vertical angle ACB, and the right line CD, which bisects the vertical angle ^ and is terminated by the base ; to find the sides and angles of the triangle. Conceive a circle to be described about the triangle, and let EG be a diameter of that circle, cutting the base AB perpendicularly in F ; also, from the cen- ter O suppose OA and OB to be drawn, and let CD be produced to E (for it will meet the pe- riphery in that point, be- cause the angles A CD and BCD, being equal, must stand upon equal arches EA and EB). Now, because the angle AOB at the center, standing upon the arch AEB, is double to the angle ACB at the periphery, standing upon the same arch {Euc* 20. 3.), to Geometrical Problems • 267 that angle, as well as ACB, is given ; and, therefore, in the isosceles triangle AOB, there are given all the angles and the base AB ; whence AO and FO will be both given, by plane trigonometry, and, consequently, EF (AO — FO) and EG ( = 2AO), Call, therefore, EF = a, EG = bj CD = c, and DE = x; and sup- pose CG to be drawn; then, the angle ECG being a right one (Euc. 31. 3.), the triangles EDF and EGC will be similar ; whence x : a : :b : x + c; therefore, by multiplying extremes and means, we have x^ + ex = abj and consequently x = \/ab + ^cc — ^c ; from which DF (VED^ — EF^), half the difference of the segments of the base, will be found, and from thence all the rest, by plane trigonometry. Before I proceed further in the solution of problems, it may not be improper, in order to render such solutionis more general, to say something here, with regard tp the geometrical construction of the three forms of adiected quadratic equations. mz. Cx^ + aAT = be, < x^ ^— a;c = be, {^ax-^x^' = be. Construction of the first and second forms. With a radius equal to |a, let a circle OAF be de- scribed; in which, from any point A in the periphery, apply AB equal to b — c (b being supposed greater than c), and produce the same till BC becomes = c ; and from C, through -p the.center O, draw CDE ^ cutting the periphery in D and E ; then will the value of X be expounded by CD, in the first case, and by CE, in the second. 268 The Application of Algebra For, since (by construction^ DE is = a, it is plain, it' CD^be called x, that CE will be x + a-, but if CE be called a;, then CD will be x — a\ but (by Euc. 37. 3.) CE X CD = AC X BC, that is, ■%■ -f a x x (x^ + ax) is = bcj in the first case ; and x X x '— a (x^ -^ ax) = bCy m the second : which two are the very equations above exhibited. When b and c are equal, the construction will be ra- ther more simple ; for, AB vanishing, AC will then coin- cide with the tangent CF ; therefore, if a right-angled triangle OFC be constituted, whose two legs, OF and FC, are equal, respectively, to the given quantities -Ja and ^, then will CD ( = CD — OF) be the true value of x in the former case, and CE ( = CD + OF) its true value in the latter. Co7istruction of the third form. With a radius equal to la, let a circle be de- scribed (as in the two preceding forms), in which apply AB, equal to the sum of the two given quantities ^ + c, and take therein AC equal to either of them ; through C draw the diameter DCE ; then either DC, or EC, will be the root of the equation. For, the whole diameter ED being = a, it is evident B that, if either part thereof (DC or EC) be denoted by x^ the remaining part will be a — x: but DC X EC = AC x CB (Eiic. 35. 3.), that is, ax — ♦ x^ = bc^ as was to be shown. The method of construction, when b and c are equal, is no ways different ; except that it will be unnecessary to describe the whole circle; for, AC being, here, perpen- dicular to the diameter ED, if a right-angled triangle OCA be formed, whose hypothenuse is ^a, and one of its legs (AC) = b, it is evident that the sum (EC) and the differ- ence (DC) of the hypothenuse and the other leg will be the two values of x required. to Geometrical Problems. 269 Note* If b and c be given so unequal, that b -—* c^ in the ' two first fortns, or b +^c, in the last, exceeds (^) the whole diameter ; then, instead of those quantities, you may make use of any others, as ^b and 2t:, or ^b and 3Cj whose rect- angle or product is the same ; or you may find a mean proportional between them, and then proceed according to the latter method. PROBLEM XVII. The bas-e AB, the vertical angle ACB, and the right line CD, drawn from the vertical angle .^ to bisect the base^ being given; to find the sides and perpendicular • Suppose a circle to be described about the triangle; and let CQ be perpendicular to AB, and ED equal and parallel to CQ : moreover, from the center F, let FA«, FB, and FC be drawn ; also !et CE be drawn (parallel to AB). Put the sine of the given angle ACB, to the radius 1, = m, its co-sine = n^ the xSemi-base BD = ^, the bisecting line, CD = b, ^d the perpendicular CQ (DE) = X', then, since (by Euc. 20. 3.) the angle BFD is equal to ACB, it will (by plane tri- gonometry) be, as m (sine of BFD) : a (DB) : t n (sine Tia of PBF) : — = DF ; and, as m (sine of BFD) : a (DB) : : 1 (sine of BDF) : iL = the radius BF, or m FC; whence EF (ED -DF) =;^-'if, or ^^^TTJ!?. m m . But {by Euc. 12. 2.) DF» + FC« + 2DF X FE = DC« ; 270 The Application of Algebra that IS, m species, — -- -j -| x = 6^^ nr m^ m m cP- T?c^ . 'inax ,, i ^ . , or — r H — Ir : but, since the sum of the iir rrr m squares of the sine and co-sine of any angle whatever is equal to the square of the radius, or, in the present case, w^ + ^^* = 1 , therefore is 1 — n^ = wi^, and conse- ^2 ^^2^2 ^2 ^^2 quently — ~ (or —- X 1 — n^) = — X m^ = a^ ; m^ rrr nr m^ 2?2CIX whence our equation becomes a^ + = b^ ; which, m ordered, gives x ^ — — = — X n AB m DC -^ DB X DC — DB . m — X — tt; > where — expresses the n AB n tangent of the angle ACB : therefore, in any plane trian- gle, it will be, as the base is to the sum of the semi-base and the line bisecting the base, so is their difference to a fourth proportional ; and as the radius is to the tangent of the vertical angle, so is that fourth proportional to the perpendicular height of the triangle: whence the sides are easily found. The same otherwise. Let the tangent of the angle ACB, or BFD, be re- presented by j&, and the rest as above ; then it will be {by trigonometry) as /? : 1 (the radius) : : a (BD) : — P == DF ; therefore FE (DE — DF) = .r ~ ~, and FC^ P (= FB^ = DB^ + DF^) = ^2 ^ fl ; and consequently ^ + «' + 5 + 7X^-y(I^F2 + I'C' + 2DFxFE) = hb (== DC2), that is, a^ + — -b^-, whence x ^pX ^ ~ — , the same as before. to Geometrical Problems. 271 PROBLEM XVIII. The areaj the perimeter^ and one of the angles of any plane triangle ABC being given^ to determine the triangle. Suppose a circle to be inscribed in die triangle, touch- ing the sides thereof in the points D, E, and F ; ^Iso from the center O, suppose OA, OD, OC, OF, OB, and OE to be drawn: and upon BC let fall the perpendicular AG ; put- ting AB + BC + AC = 6, the given area = a^, the sine of the angle ACB (the radius being 1) = w, the co-tangent of half that angle (or the tangent of DOC)=n,^dAC = :c. Therefore, since the area of the triangle is equal to ^ABxOE+iBCxOF + ^AC X OD, that is, equal to a rectangle under half the perimeter and the radius of the inscribed circle, we have — X OE = aa \ and therefore OE = 2aa But AD being = AE, and BF = BE ; it is manifest that thei sum of the sides, CA + CB, exceeds the base AB, h\f the sum of the two equal segments CD and CF ; and S0 is greater than half the perimeter by one of those equf il segments CD; that is, CA + CB z=z \b + CD: but {by trigonometry^ as 1 (radius) : n (the tangent of DOC) : : ^ (OD) : DC = ^; whence CA + CB(=i^ + CD) = l^ + 2na^ which, taken from (b) 2na^ the whole perimeter, leaves ^b — = the base i IB Make, now, ^b -f- 2na» = c ; then will BC = c = a* ; also, (by trigonometry^ it will be, as 1 (radius) : m (the sine 272 The Application of Algebra of ACG) : : X (AC) : mx = AG ; half whereof, mul- tiplied hy c~x (BC), gives mcx »— wr/v2 • = a^, the area of the tr iangle: from whence x Comes out = |c ± 4 m PROBLEM XIX. The hypothenuse^ A,C^of a right-ajigled triangle ABC, aiyithe side of the inscribed square BEDF, being given; to determine the other tivo sides of the triangle. Let DE, or DF = «r, AC = b, AB == x, and BC = 2/ ; p then it will be, isls x : y -, i x ^ — a (AF):«(FD); whence we have ax'=.yx — t/a, and consequently xy •=. ax '\- ay. Moreover, xx + yy = bbi to which equation let the double of the former be add- ed, and there arises x^ + 2xy + y^ :=.b^ + 2ax that is, X + yf = b^ + 2az/; ^2a Xoc-\-y^orX'\- ip^ — 2a X .^ + ^ = ^^ ; where, by con- sidering :v -f- ?/ as one quantity, and completing the square, we have x -f y Y — 2a X ^ + 2/ -}- a^ = ^^ -|- «2 . whence x -^-y — a = s/b^ + a^, and ^ + z/ = Va^ -f- b^ -f <2 ; which put = c : then, by substituting c — x instead of its equal (?/) in the equation xy ■=: ax -^^ ay^ there will arise ex • — x^ = ac V^c — ac, and y =z-tc - whence x will be found = |c -{- • V-^cc — ac. It appears from hence that c, or its equal Vaa + ^^^ + a, cannot be less than 4a, and therefore ^^ not less than 8a^ ; because the quantity ^cc — ac^ under the radical sign, would be negative, and its square root im- possible ; it being known that all squares, whether from positive or negative roots, are positive ; so that there cannot arise any such thipgs as negative squares, to Geometrical Problems. 273 F E imless the conditions of the problem under consideration be inconsistent and impossible. And this may be demon- strated, from geometrical principles, by means of the fol- lowing LEMMA. The sum of the squares of any two quantities is greater than a double rectangle under those quantitieSy by the square of the difference of the same quantities. For, let the greater of the two quantities be represented by AB, and the lesser by BC (both taken in the same right line). Upon AB and BC let the squares AK and CE be constitut- -^ ed; take AP = H I K BC, and complete the rectangles PH and CF. There- fore, because AB = AH, and AP = BC, it is plain that PH and PD are equal to two rectangles under the proposed quan- tities AB and BC ; but these two rectangles are less than the two squares AK and CE, which make up the whole figure, by the square FK, that is, by the square of PB, the difference of the two quantities giv^n ; as was to be proved. Now, to apply this to the matter proposed, let there be given the quadratic equation :>c^ -f ^^ = 2ax^ or x =: a ± s/aa — bb : then, I say, this equation (and consequently any problem wherein it arises) will be impossible, when Ofl — ^6 is negative, or b greater than a. For, since b is supposed greater than a, 2bx will likewise be greater than 2ax ; but 2ax is given :=z xx + bb^ therefore 2bx will be greater than xx -f- bb^ that is, the double rectangle of two quantities will be greater than the sum of their squares, tvhich is proved to be impossible. 2N 274 The Application of Algebra PROBLEM XX. The base AB (a) and the perpendicular BC {F) of a right-angled triangle ABC, being given; it is proposed to find a point D in the perpendicular^ such that^ if two right lines be drawn from thence^ one to the angular point A, and the other (DE) perpendicular thereto^ the triangles DEC, ABD, cut off by those lines ^ shall be to one another in a given ratio. Let AB be produced to F, so that the angle BFD may be equal to the angle BC A ; putting AC = c, CD = a% T ^ \e\ ^y^U F J 3 A and the given ratio of the triangle DEC to ABD as ;w to ;?. Then, by reason of the similar triangles ABC, DBF^ it will be, a (AB) : b (BC) x i b — x (BD) : BF == h^—hx , . „ h^ — hx a" + l^ — bx ; whence AF = a -| = a a a = ^Izi^ (because a^ + b^ - c^\ Also, as ADE is a a ^ ^ right angle, the angles FAD, EDC will be equal ; there- fore, the angles C and F being equal {by con.)^ the tri- angles AFD, DCE must be similar ; and co nsequendy ^p c^ — ^>a1\ . ^^o^..<,^.. AF X BD ,b—xxc^~bx -):CD2(^2).. •(- 2a ). b — — X X ax^ the area of the triangle AFD : ( ==-), the area 2X6-2 — bx of the triangle DEC : wherefore, the area of the tri- to Geometrical Problems. 275 1 ATiTM. • BD X AB b — X X ct 1,111 angle ABD being , or , we shall have which, reduced, gives x = vf — H — ■ b — X X a-^ b — X X a . mini ; -■ : • {by the question) ; and ^ X c ' - 9 , -^ « , mbx mc^ consequently nx^ =: m x c^ — f^^^ or x^ + = — ^ •^ 71 n mb n ' 4n^ 2n The geometrical construction of this problem, from TYtbx j^f^^ the equation x'^ 4 = — , may be as follows; In 71 n CB let there be taken CH : CB : : lyi i n, and let HK be drawn parallel to BA ; then CH being = — , and CK 71 Vie = — , our equation will be changed to x^ + x X CH = AC X CK, or to CD x CD + CH = AC x CK. Upon CH as a diameter let the circle CTHQ be describ- ed, in which inscribe CG = AK ; and in CG, produced, take CS = CA ; and from S, through the center O, draw the right line STOQ, cutting the circumference in T and Q, and make CD = ST ; then will D be the point required. For CG being = AK, and CS = CA ; therefore will AC x CK = CS x GS = ST x SQ {Euc. 37. 3.) = ST X ST + TQ = CD X CD + CH 5 the very same as above. PROBLEM XXI. Having the perimeter of a right-angled triangle ABC, and three perpendiculars DE, DF, and DG^ falling front a point within the triangle upon the three sides thereof; t — > ap ; also, if, from the same equation, b times x + y + z = /) be subtracted, there will remain ax — bx + p + c — b X z = ^p^ — bp ; which two last equations, by putting d z=z b — a^ e = p + c — ci', f = I ^^ — ctpy g z^ p + c — ^, and h = ^p^ — bp^ will stand thus : dy + ez = fj and — dx + gz = h ; whence y = /*— ez , ^z — h *J . — ^ and X = ^5 . — . a a Let these values of x and y be substituted in :\^ -f p — 2€fz -^ (^Z^ g^^^ — 2^, and we shall have ^rhz4-h^ //2 d^ = 2^, or e^ -!- g^ — d^ XZ^ — 2/4- ^gh X z = — /2 — A2 : put e^ + ^2 _ ^2 ^ k, ef ■\- gh =z I, and f^-{-h^z=zm; so shall J^z^ ~ 2/2 = — m; whence 2^ — --~ = ^ ~, and 2 = 7- k k k to Geometrical Problems. ^77 , //2 m I * V IS ~ T = - ± V/^ — km from which x (^z:z ^' — ) and y ( =—7 — ) will also be known. a d If a, b^ and c be all equal to each other, the point D will be the center, and each of the given perpendiculars a radius of the inscribed circle ; and the value of 2, in this case, will be barely equal Xo\p*-^a\ for the equation, by — ay + p + c — a X'Zz=. \p^ — ap^ above found, here becomes pz = \p^ — ap. But, if only a and b ( or DE a nd DF) be equal, then the equation will become /? + c — aX2: = \p'^ — ap \ and therefore z = -^ ^ ; in which, if c be taken = 0* p ^c — a ^ 2 will be = 1^- -^ ; where a is the side of the inscrib- p — a ed square. PROBLEM XXir. The perpendicular CD, the difference of the sides^ AD — BD, and the vertical angle D, of any plane triangle ABD, being given; to determine the sides* From B,upon AD (produced, if need be), let fall the perpendicular BE : let the sine of the angle BDE = 5, its co-sine = c (the radius being unity) \ also let the perpendicular CD = /?, the lesser side BD = .v, and the greater DA = a: -f ^; then {by prop. 9, in trigonometry^ 2& \ ', s \ \ X \ sx z=z BE ; and as 1 : c : : a; : c^ = ED. Now AB2, being = AD^ + DB2_ADx2DE {Euc. 13. 2. ), will be ex- po unded b y x '\- ([f ^ x^ k c R — X •\' d X 2cx^ or 2x^ -f- 2dx + d^ .^ 2cx^ — 2cdx ; whence, by reason of the 278 The Application of Algebra similar triangles ABE and ADC, it will be, as Sat^ ^ ^dx ^ d^ — 2cx^ — 2cdx (AB^) : s^x^ (BE^) : : cc^ + 2xd + dd (AD2) \ p^ (DC2); and, consequently, by multiplying extremes and means, s^x^ -f 2s^dx^ + ^d^x^ = 2^2 .2 ^ 2^2^;^ ^ ^2^2 _ 2/^2c;c2 — 2p^cdx ; from whence, by transposition and division, we have x"^ + X — ^— ^ = 0. Which equation answering the condi- tions of the second case of biquadratics, explained at p. 154, we shall therefore have x^ + dx -^^ p^c — p^ 12 //,2^2 "i^x — p^ r \'— J- + ^ -^ — i- ; and consequently x = — |^ + Supposing ^, c, and j& to be the same as before, put half the given difference of the sides = a, and half their sum = X ; then the greater side AD will be = a: -f a;, and the lesser BD ■= x — a ; wherefore (by trigonometry) 1 \ s I I X — a \ s X ^ — ^ = BE ; and 1 : c : : .r — - a : c X X — a = DE : but AB ^ is = A D^ -f D B^ — 2D E X AD z=zx 4-al2 + X — oY — 2c X oc — ax x + a=i 2x^ 4- "^a^ — 2cr^ + 2ca^ ; whence, by reason of the similar triangles ABE, A DC, it will be 2x^ -f 2^^ ~ 2c .2 + 2ca2 (AB2) : s^ x x — a]^ (BE^) : : x + a]^ (AD^) : /?2 (D C^); and consequently s^ x x — af X X 4- ^1^ = S^Y^ ^ 2(^2 — 2cx^ -f 2C(22 X p^, or ^^^ 4 _ 252^^x2 ^ ^2^4 ^ 2/?2;c2 — 2cp^x^ + 2/?2flr2 ^ 2cp^a^ ; whence, by transposition and division, x"^ — 2a^x^ — 2^ + !f^ = ?^! + ^ ^ a^. Substitute A" to Geometrical Problems. 279 /= fl2 +Pl-%, and 5- = «^ X ^UrJfL - a^. then the equation will stand thus, x^ — 2fx^ = g : whence i;^ is found = Sf± V/2 -f ^. If, instead of the diiference, the sum of the sides had been given, in order to find the difference, the naethod of operation would have been the very same ; only, instead of finding the value of x in terms of a^ by means of the equa- tion s^x^ — 2s^a^x^ + s'^a^ = 2fx^ — 2cp^x^ + 2p^d^ + 2cp^a^^ that of a must have been found, in terms of ►r, from the same equation. PROBLEM XXIII. Having one leg AB of a right-angled triangle ABC ^ to find the other leg BC, such that the rectangle under their difference {^Q, — AB) and the hypothenuse AC may be equal to the area of the triangle. Put AB = a, and BC = ^t- ; so shall AC =.Vaa + xx\ and —- = .V — a . Vaa + xx^ by the conditions of the problem. By squaring both sides of this e quation, we have ^a^x^ = ;^ — 2ax + a^ x aa -f xx\ in which the quantities x and a being concerned exactly alike, the solu- tion will therefore be brought out from the general method for ex- tracting the roots of these kinds of equations ^delivered at p. 156): according to which, having di- vided the whole by c^xP^^ we get — = -21 ,— , 2 + -^ 4 a X X — H ; which, by making 2 = -^ + — , will be X a ax reduced down to l = ^ ~ 2x2, or z^ — 2z = 1 : -^hence z is given =; 1 + V f . But since — + — =: 2, a ' X ' 280 The Application of Algebra we have jc* — azx = . — a^ \ and therefore ;v = — + 2 ^ y — «2 = _X2: + V22 — 4: which, by sub- stituting the value of 2, becomes :v = — X At i + Vf + Vv'i'— J. PROBLEM XXIV. To ^mw a right line DF yro;?z one angle D of a given rhombus ABCD, so that the part thereof ¥G^ intercepted by one of the sides including the opposite angle and the other side produced^ may be of a given length. Let DE be perpendicular to AB ; and let AB ( = AD) C = «, AE = b, FG = c, and AF = x : then DF2 (= AF^ +AD2— 2AExAF) :=z XX + aa — 2bx ; and, b}^ similar tri- _ -^ —^ angles, xx -^ aa • — A E B F 2bx(DF^):xx(AF^) : cc fFG2) : x — n o'] ^ TBF^ ) ; and consequently ^x 4- aa — 2bx X xx — 2ax 4- aa z=: ccxx. Make ma = ^, and na = c ; so sh'i;l ou r equati on becon>e XX + aa — 2max x ^^ — 2a -J-aa^ n^ a^x^ : w hich^ di* X fi X a vided by a^x^. jrives f- 2w x 1 2 = •^ ° a AT ax J XX X a n'^ : this, by making 2 = f > becomes z — 2m X Z — 2 = 72* : theref ore z^ — 2m -f- 2 X 2 == n^ — 4w, and 2 = 1+ y yz 4- sjn^ -J- 1 — wz")^ = ^ + ^4- Vc^ + a - /^? by restoring the values a " m and 72. From whence the value of x will be also J| to Geometrical Problems., ^l known \ for 1 being = ^, we have, by recluc- a X tion, x^ — azx = — aw, and therefore a: = — x 2 + V'2^ 4. PROBLEM XXV. TA^ diagonals AC, BD, and all the cmgles DAB, ABC, BCD, and CD A, of a trapezium ABCD, being giv^n; to determine the sides. Let PQRS be another trapezium similar to ABCD, whose side PQ is unity j and let QP and RS be produced till they meet in T : also let PR and QS be drawn, and make Rz; and Sw perpendicular to TQ. Let the (natural) sine of the given angle STP, to the radius 1, be put = wz ; that of TSP, or PSR, = n ; that of TRQ =/; ; the CO sine of SPQ = r ; that of RQv = s ; AC = a; BD ::::: ^ ; and PT = X. Then (bij plane tm- gonometry) n : m -. -. x x V^ = — -, and 1 : ^ f PS) 71 n .iiriFwzzz^IHf^; whence (by Euc. 13. 2.) QS^ (= QP^ + ps^ ^ 2PQ X Pzt.) = 1 +^!:fi! ~ £!:^. nn n Again {by trigonometry)^ p : m : \ 1 + x (TQ) : QR = • mx and 1 : s m 4- mx (QR) : Qy P ' P 7ns -|- TUSX T And therefore PR^ ( ~ PQ^ + QR2 _ 20 282 The Application of Algebra 2PQ X Qt>) = 1 + ^ElHI _ ^J!1L±3^. But Z'/ P because of the similar figures ABCD, PQRS, it will be, A O : B pg : : PR* : QS% that is, a^ : b^ : : m -i- /.,^j 2??2^ 4- 2m,sx m^x^ 2rmx 1 -f- , _ ; 1 -J- — — - • pp p nn 71 - - « a^/ V 2a^rmx ... Z>^/?i^ and consequently a^ ^ * z=z b^ -] rm n pp 2b^mP-x^ b^m^x^ 2b^ms 2l^msx 4 1 -— — • : whence, writ- PP PP P P ^ a^m^ b'^m^ bbms bbmP' aarm , mc: / = , 9" = , and ^ ^ nn pp p pp n ' uTi^ ^i/bbtyis h-=.h^ — c^ A , we have fx^ + 2^-;^ = h : pp p ' J -r ^ which, solved, gives x = V -77 + -% — -^ • from whence SQ will also be known : and again, by reason of the simi- lar figures, it will be as QS : QP (unity) : : BD : AB ; which, therefore, is likewise known : from whence the rest of the sides BC, CD, and DA will all be found by plane trigonometry. The last problem is indeterminate in that particular case, where the trapezium may be inscribed in a circle, or where the sum of the two opposite angles is equal to two right ones ; for then there can but one diagonal be given in the question, because the value of the other depends entirely upon that. PROBLEM XXVI. Supposing BOD to be a quadrant of a given circle ; to fnd the scmi'diameter CE, or CL, of the circle CEGL, inscribed therein ; and likewise the semi-diameter of the little circle n¥mV^ touching both the other circles DLB, LME, and also the right line OB. Let BQ, P/z, and CE be perpendicular to BO ; join C, 7iy and O, n ; and draw OC meeting BQ in Q, and nr parallel to BO, meeting CE in r : put OB io Geometrical Problems. 283 ( = BQ) = 1, OQ (= V2, /^j/ Euc. 47. 1.) = h, and Pn C = nni) = a:. Then, by reason of the similar tri-^ angles OBQ, OEC, it will be, OQ : BQ : ; OC : B P E CE : whence, hy composition^ OQ + BQ (OC + CE) : CE ; that is, ^ + 1 : 1 BQ ; 1 : : : QL CE = = i^-^l = ^2 X b—V ~ ^2 — 1 — -* 1 ; which let be denoted by a, then we shall have Ctz + Cr = 2<2, and Cn — Cr = 2^; ; and therefore nr ( = V^^^ + Cr X C7i — . Cr) = 2>/a;c. Moreover, On + P/z being = 1 , and On — P/i = 1 — *2.x \ thence will OP^ V 1 — 2:v ; which, also, b eing = P E + OE (2^6 a: + «), we therefore have V 1 — • 2:v = "ilS^ax + <^ ; whereof both sides being squared, there arises 1 — 2a; = A^ax -f- Aas/ax + a^, or 1 — c^ — 2a: — Aax = '\a>/ax ; which, because 1 — aa is 2^7, will be « — 1 + 2a X -^^ = "la's/ ax : this, squared, gives a^ — 1 -f- 2a X 2aa" + 1 + 2a1^ X ^^ = ^a^-^ j whence 1 + 2a}^ x x- — 1 + 2a X 2a;v — 4«^.^ = — ^« ; "which, by writing 2B4 The Application of Algebra b — 1 instead of its equal a^ becomes 2^ • — \\^ X ^^ — ^ Yb 9 7i) — 9 X 2x z= 2b — 3 ; therefore x^ — — X 2b — 1 Y 2x = =z==z-=r- ; from whence x is found = 2u — i7 II 7 b — 9 '^ sJ'yJj — ;f. x ^o —. i]^ + 7b — ^Y 'tzz2^..JE£' ^ 7b- V 2 — 1 , . , , . , or to — : rzr^ ; which last is the root required, the 2V' 2 — 1 I other being manifestly too large : but this value will be "49" reduced to ^ . Therefore OP ( = V 1 — 2x) *, . /5I — 10.2 5V'^ — 1 ^^ _ IS given = y = = 7Vn ; and con- ^ 4-9 7 sequently BH = ^-^Q 5 from whence we have the follow- ing construction. In the tangent BQ take BH = -|.BO ; draw HO, cut- ting the circumference BDL in F, and make the angle OFP = iOHB, and draw PN parallel to BQ, meeting OH in 72, the center of the lesser circle required. SCHOLIUM. in the preceding solution it was required, not only to extract the square root of the radical quantities 136 -^ to Geometrical Problems. 285 96 y 2 and 51 — lOv/2, but likewise to take away the radical quantity from the denominator of the fraction v^T ■' 1 — -^ , and confine it, wholly, to the numerator: 2V 2 — 1 I all which being somewhat difficult (and, for that reason, omitted in the introduction, as too discouraging to a young beginner), I shall therefore take the opportunity to ex- plain here the manner of proceeding in such like cases, when they happen to occur. First, then, with regard to the extraction of roots out of radical quantities, let there be proposed A ± v B^ A be- ing the rational, and VB the irrational part thereof; and let the root required be represented by y/~± vTT- the square of which will be x + y ± 2v'^^ or x + y ± V4:xy ; therefore we have x + y ± V4xy = A rh VBI Let the irrational, as well as the rational parts of these two equal quantities be now compared together ; so shall X + y z=: Aj and V4:xy = VB : from the square of the former of which equations subtract that of the latter ; whence will be had xx — 2xy + yy = A^ B ; and, by taking the square root, x — y = v A^ b"; which added to, and subt racted from x + y ~ A, ^c. A + VA^ — B , A — VA^TZb" gives X = ^ , and y = -_ r. In the first of the two cases above specified, the quan- tity whose square root is to be extracted being 136 96V 2, we have A = 136, and B = 18432 (= 96]^ X 2} ; whence we have x {— A + V A^ — B^ _ ^^^ A a/ A 2 P and y {=: ) = 64; and consequently \/x~V^ =z V72 — V64 = 6V2"— 8, the required square root of 136 — - 96V 2. After the very same man- ner, die square root of the other radical quantity 51 10V2, or 51 — V200 will be found to be 5^2— 1 : 286 The Application of Algebra for, A being here = 51, and B = 200, we have .v = 50^ and 2/ = 1 ; and consequently V x — V^ «/ = 5V 2 — 1. What has been said, thus far, relates to the extraction of the square root only ; but the same method is easily extended to the cube, biquadratic, or any other root. Let us take an instance thereof in the cube root ; where we will suppose the given quantity, out of which the root is to be extracted, to be represented by A ± VB, as before. Then, if the rational part of the root be de- noted by x^ and the irrational part by V y ; the root itself will be expressed by ^ ± V y ; and its cube by x^ ± 3x^ V y -^ 3xfj ± y V 1/ : from whence, by proceeding as in the extraction of the square root, we have x^ + 2xy *: A, and 3x^ \/ y + y V y =: VB, Let the sum and the difference of these two equations be taken, and there will come out x^ -f 3x^ ^ y + ^^y + ^/ V ^ = A + v'B, and x^ — 3x^ V~y -f 3xy — y V'y = A — VB ; where- of the cube root being extracted on both sides, we thence have x +\/y = A -fv'B]^, and x^^\/y = A — v B]^ - let the two last equations be added together, and the sum be divided by 2 : so shall x = — i-!^ — ! — ^ — *■ ; and, by multiplying the same equations together, wc get ^:^ ' — '2/ = A^ — ;> B") s", and consequently y :=i x^ — ^ A^ — B"J^ ; whence y is likewise known. Universally^ let the index of the root to be extracted be denoted by /2, and let the root itself be represented by x±.s/ y (as above). Then this expression raised to 11 - 1 the nih power, will be x"" ± nx "~V y + n X — r — ^"^1/ ± n X — -— X : x^-^y V 2/, &c. from whence, 2 3 still following the same method, we shall here get 72 — 1 x^ ^nx ^''^"^y-i Sec. = A, and to Geometrical Problems. 287 v2.Y«-V"^ + n X ^~ X ^ x^-^y V"y, &c. = VB : let, therefore, the root of the sum, and also of the dif- ference of these two last equations, be taken, and you will have a:- + V z/ = A + y/B^ ^, and x en V if =z A — v/ B I '^ ; which two equations being added to- gether, X will be found = i— ^^ ! ^ — L A -f v'B1« _^ A^ — B]"^ , ., , ::=: -1 — 1- ± y *. and, ir the same ^ 2 X A + v^l~' equations be multiplied together, you will have L ,L x^ CO 1/ = A2 — B J « ; whence ^2/ = .r^ ± A^ — B | « : the use of which conclusions will appear by the following ex- amples. First, let it be proposed to extract the cube root of the radical quantity 26 + 15VT, or 26 -f- V675. Here, A being = 26, B = 675, and 72 = 3, we have .. _ 26^s/6 7f\^ i ^ — _ 3,732 ± ,268 '' 2 2>c 26-f V675T' 2 ) == 2 ; and tj ( = 4 — 676 — 675")^) = 3 ; and conse- quently x + V ?/ = 2 -f- V 3 = the value required for - + V~3^ X 2 -f v/3 X T+v"3 = 26 -f. \/675. Again, let it be required to extract the biquadratic root of 161 -f- V25920. In this case, A being =161, B = 25920, and 71 = 4, we have x (= ^^^^^^^^^^r - ^ - l!!£ijL£i5) == o, and 2 X 321,99, Ssfc.]^ 288 The Application of Algebra t/ (= 4 + 25921 — 25920]*j = 5 j therefore the root sought is, here, = 2 -f V 5. Lastly, if it were r equired to find the first sursolid root of 76 + V5808 ; then, by proceeding in the same manner, x will be found (= - — ^ ''- ) = 1, and «/(=!—' 5776 — 5HOs]^) = 3: and so of others. But it is to be observed, that the second part of the Value of x^ to which both the signs + and — are pre- fixed, is to be taken afilrmative or negative, according as that or this shall be found requisite to make the value of X come out a whole, or rational number : and that, if neither of the signs give such a value of x^ then this method is of no use, and we may safely conclude that the quantity proposed does not admit of such a root as we would find. It may also be proper to remark here, that, if the upper sign in the value of x be taken, the tipper sign in that of y must be taken accordingly ; and that the application of logarithms will be of use to facili- tate the extraction of the root A + VB f «, as being suiH- ciently exact to determine whether a: be a whole number, and, if so, what it is. Thus much in relation to the extraction of the roots af radical quantities ; it remains now to explain the man- ner of taking away radical quantities out of the deno- minator of a fraction, and transplanting them into the nu- merator. In order to which, supposing r to denote a whole num- ber, it is evident, in the first place, that X'' — y'' =~.ir^ X ^^-1 + x^'^y + x^^y^_ . • • + y'"'^ ; since, by an actual multiplication, the product appears to be i ^^ + ^'""'^ + ^"''^' "^ ^'''^'' ^'' \ where all the terms, except the first and last, destroy one another. Hence, by equal division, we have to Geometrical Problems. 289 _J_^._r-+^^^+^-g^_+y-_ And, in the X — y x^ — y^ very same manner, it will appear that X +y x^ ± y^ the sign — or +, in the denominator, takes place, accord- ing as the number r is even or odd. Let now x = A***, and y = B^ ; then our 'equations will become 'j^^ni, Qn J^rm _ gj-ji ^ J Arm—tn ___ ^rm— 2mJ^n i J^r}n—5mT^2n , , , ^ -^ grn— ri From which theorems, or general formulce^ the mat- ter proposed to be done may be effected with great far cility: for, supposing —-——^, or ^,, ^ ^^ to be a fraction having radical quantities A"", B" in the denomi- nator, it is plain, that its equal value, given by the said equations, will have its denominator entirely free from ra- dical quantities, if r be so assumed that both rwz and rn may be integers. To exemplify which, let the fraction V2 — 1 ^=n be propounded ; then, A being = 2,6 = 1, m = ^, and n = 1, we shall, by taking r = 2, have (^from theorem t) r=-^^ = ^ *' '^^ = VT+ 1. 2P 290 The Appiication of Algebra 1 Again, let the given fraction be -- _ ^.- or : _^ ~7- In which case., A being = cx^ cx\- + <:"* + x""]^ B 1= c^ + jf'*, w = |, and 7i = 1, we shall, by taking r = 4, have =, c If the numerator be not a unit, you may proceed in the same manner, and multiply afttrwards by the numerator given. Thus, in the case mentioned at the v "2"-— 1 beginning of this scholium^ w^e had given ^ — ^^, 2V 2 — 11 1 which may be reduced to V 2 — 1 X -, or to ^2 — 1 X 2V 2 — 1 1 1 2V 2 — 1 Sl — 1 8l — 1 2V^ -fl but is found (by theorem Vi to be = — 8| — 1 ^ ^ ^ 8—1 2^ 2 -4- 1 2 V 2 -f 1 whence our expression becomes V 2 — 1 : which, by multiplication, £5?c. 7 7 _ 5\/2~l is reduced, at length, to ' 49 PROBLEM XXVIL Having one leg^ AC, of a right-angled triangle ABC, to find the other leg BC, so that the hijpothenuse AB shall be a mean proportional between the perpendicular Cli fall- ing thereon^ arid the perimeter of the triangle. to Geometrical Problems* 291 Put AC = a^ and BC = ;c ; then will AB = AC X BCT ax Vxx + aa^ and CD ( = therefore, by the ques* tion^ X -{- a + Vxx -f aa : \^xx + aa', : Vxx + aa ax : — ; and conse- \\\x + aa ax^ 4- a^x quently + ax js^ V XX -J^ aa = XX -h aa : whence a^x'^ + 2a^x^ + a'^x^ =; XX + aa — axY X ^^ + ^^* Divide by a^x^ (according X fl to the rule at page 156), so shall — + 2 -| = a ^ X a ' P X ^ a - + --M X - + -z\ which, by making z = 1 is reduced to :s + 2 = z — l"|^ x z, or z^ a X — 22^ = 2. This, solved (by the rule for cubics), gives 2; = £ + -Lx 35+ v/TlGl 1^ + — X 4=r — ^ ^ ^ 35+\/116ir^ = 2,359304 : whence .v ( = — X 2 ± Vz^. — 4) will like wise be known. PROBLEM XXVIII. The base AB, and the perpendicular BE, of a right- angled triangle ABE, being given ; it is proposed to find a point (C) in the perpendicular^ from whence txvo i'ight lines C A and CD being drawn^ at right angles to each other ^^ the tzvo triangles ACD and K^C^ formed from thence^ shall be equal. Suppose DF parallel to AC, and let AF be drawn : putting AB = a, BE ^b, BC = v, and AC {Va^ + x''-) 292 The Application of Algebra ^ y. Then, since FD is parallel to AC, the triangle ACF will be equal to ACD, or ABC ; and therefore CF = BC =^5 whence we have EF (=£6 — BF) :=:b — 2x, and EC(=EB — BC) = ^ '— •%• : more- over, by reason of the si- milar triangles ABC and CDF, we have y (AC) : .V (BC) : : X (CF) : FD XX Whence, because of the parallel lines AC and FD, (EF) hx^ — _^ it wiU be, ~ (FD) : b — 2x A y ^ y (AC) '. b — X (CE) ; and consequently = b — 2x X y-, or bx^ ■=. b — 2x X y^ which equation, by writing a^ + a;^, instead of its equal, y^, becomes bx^ — x^ = ba^ + bx^ — ^a^'x — 2x^ ; whence we have x^ + 2a^x = ba^^ and therefore x r= llb^~W~^ J ^ — — . PROBLEM XXIX. Three lines^ AO, BO, CO, drawn from the angular points of a triangle to the center of the inscribed circle^ being given ; to find the radius of the circle^ and the sides of the triangle. If upon CQ produced, the perpendicular BQ be let fall, and the radii OE, OF be drawn to the points of contact, the triangles BOQ and A OF will appear to be equiangular ; because all the angles of the triangle ABC being equal to two right ones, the sum of all their to Geometrical Problems. 293 halves, OCB + OBC + OAF, will be equal to one right angle ; but the two former of these, OCB + OBC, is equal to the external angle QOB ; therefore QOB + OAF = a right angle = QOB + OBQ, and consequently OAF = OBQ. Putnow AO=«, BO = ^, CO = c, and OF = xi then, because of the similar trian- gles, we have a : X : : b : OQ, = — ; whence BQ2 a bbxx ■=z b^ --^ , aa and BC^ (= CO^ + BO^ + 2CO x OQ) =z b^ + t" + '^—. But BC2 : BQ2 : : OC^ : OE^ ; that is, a j:. [ — : : : c^ : x^. lLl^ + vr8o5 + )^im.2xy) = xy^ — 4xy^ + 2xy ; sine of 7 A (= sine of 6A x y — sine of 5 A = xy^ — 4xy^ + 3xy^ -^ xy^ -f 2xy^ — . :v) = xy^ — • 5xy^ + 6xy^ — x : whence, universally ^ the sine of the multiple arch 72 A, where n denotes anj^ whole positive number whatever, will be truly expressed by ;c x into this series : „ - n— 2 ^» . 72-^3 n — 4 ^ . n — 4 n — 5 n — 6 " ^ 2^-1 — ^Xy^'+'-^x-Y-xy—^^'^r^'^T'^^ &?c. Moreover, from the second theorem, we have co-sine of 2 A (c= co-sine of A x y — co-sine of = y'^t^^y" — ^. r2 SOO The Application of Algebra Co-sine of 3 A ( = co-sine of 2 A X y — co-sine of A = y^ ^.j—Ia^ y' — ^y . T ^ 2^""^ 2 ' Co-sine of 4a ( = co-sine of 3 A X y -^ co-sine of 2 A = 2 "~"2 '^■^ 2 whence, universally^ the co-sine of the multiple arch ^^A will be truly represented by — •— -2 — -j •^ >i ^ 7Z' — 3 ^. n n — 4 w — 5 ^ ^ n X —^xy--^ — -x—^X—^ xy^'^ + -X X X - — — X v"""^, ^c. which series, as well 2 3 4 as that for the sine, is to be continued till the indices of y become nothing, or negative. But, if you would have the sine expressed in terms of X only^ then, because the square of the sine + the square of the co-sine is always equal to the square of the radius, and, therefore, in this case, x^ + %y^ = 1, it is manifest that the sines of all the odd multiples of the given arch A, wherein only the even powers of y enter, may be exhibited in terms of x only, without surd quantities : so that 4 — 4x^ being substituted for its equal 2/^, in the sines of the aforementioned arches, we shall have 1st. Sine of 3 A = 3a: — 4x^ ; 2d. Sine of 5A=: 5x — 20^2 + 16x^ ; 3d. Sine of 7A=z7x — 56x^ + 112;^^ — 64^^ ; 4th. Sine of 9A = 9a? — 120^:^ +432^^^ — 576x^ +256x^; And, generally^ if the multiple arch be denoted by 7iA, then the sine thereof will be truly represented by n n^ — 1 n r? — 1 n^ — 9 . n n^ — 1 n^ — 9 n^ — 25 ^^ 7 , ^ n^ ~ 1 7i2 _ 9 71^^25 n* — 49 ^ ... , ^ X ^ ^ X ■ ^ Q X ^S ^c. 4.5 6.7 8.9 to Geometrical Problems. 301 •From this series, the sine of the sub-multiple of any arch, where the number of parts is odd, may also be found, supposing {s) the sine of the whole arch to be give;i : for let x be the required sine of the sub-multiple, and 71 the number of equal parts into which the whole arch is divided; then, by what has been already shown, , „ _ n 11^ — 1 o . w n^ — 1 we shall have nx X -r — -- X ^^ + , X X 1 2.3 1 2.3 X ^^, ^c, = 5 : from the solution of which equa- tion the value of x will be known. Hence, also, we have an equation for finding the side of a regular po- lygon inscribed in a circle : for, seeing the sine of any arch is equal to half the chord of double that arch, let ^v and Iw be written above for x and ^, respectively, and TIV JZ 71^ , 1 ^3 then our equation will become —. ■ — X — X — ^ 2 12.38 71 77.2 - -1 92 ► — 9 v' ,&fc.= 71 + — X — - X X T^l :|t^, < or Tiv X 1 2 . 3 4 .5 32 1 V^' — 1 v^ n 7?,^- -1 ^^^ — 9 v' X — + X X X —: ,&fc.: = ru; ex- 2 . 3 4 1 2 . 3 4. 5 16' pressing the relation of chords, whose coiTesponding arches are in the ratio of 1 to n. But, when the greater of the two arches becomes equal to the whole peri- phery, its chord (w) will be nothing, and then the equa- tion, by dividing the whole by nv, yfill be reduced to 2.3^4'*"2.3^4.5^16~ "TTT ^ n2 — 9 72^ — 25 Z)« ^. ^ , . , X rr, ere. = ; where 72 is the num- 4.5 5.6 64 ber of sides, and v the side of the polygon. From the foregoing series, that given by sir Isaac NewtOHy in Ph'iL Trans, mentioned in p. 242 of this Trea- tise, may also be easily derived. For, if the arch A and its sine x be taken indefinitely small, they will be to one another in the ratio of equality, indefinitely near, by .302 The Application of Algebra what has been proved at p. 246 ; in which case, the ge- neral expression, by writing A instead of x^ will become uK X X A^ + — X X 1 2.3 1 2.34.5 Therefore, if tz be now supposed indefinitely great, so that the njultiple arch nK. may be equal to any given arch 2, the squares of the odd numbers, 1, 3, 5, £sPc. in the factors ?i^ — 1, n^ — 9, r^ — 25, &Pc. may be rejected as nothing, or inconsiderable, in respect of r^ ; and then 72^ A' n^A^ the foreoxjinc; series will become tzA — 1 ^ ^ 2. 3^2.3.4.5 , £s?c. wherein, if for nA, its equal, 2.3.4.5.6.7 e, be substituted, we shall then have % - — 2.3 , l^c. which is the sine 2.3.4.5 2.3.4.5.6.7 of the arch z, and the same with that before given. Moreover, the aforegoing general expressions may be applied, with advantage, in the solution of cubic, and certain other higher equations, included -in this form, n — 3 o«4 n — 4 n — 5 VIZ. 2" — 02"-* H X c^z"^-^ X X 'In 2n Sn For, if z be put = y \}' — , the equation will be transform- — I— cd to — X into this series 71 I » 71 71' — 3 ^ 71 n--^^ n*-^S „__^ yn^ny^-^+j X -^ X y'^' + j X -^ X -^ X ^"^ TyTi ^ 7t 7t ^c. =/, and consequently ^ — — X y^"^ H X - 2 2 '^ 2 2 to Gaometrical Problems. 203 X y""^— ~ X ^^-^-^ X ^■=^X^/'^-^£sPc.===^l4" -. from n I whence, as it is proved above, that the former part of the equation (and therefore its equal) represents the co-sine of n times the arch whose co-sine is |z/, we have the follow- ing rule : Find^from the tables^ the arch whose natural cosine is JL jf , 71 a ~=r- {or its log", co-sine = log* if*"^ — ^og: — ) the radius fjj 2 n n\ being unity ; take the 72th part of that arch^ and Jlnd its cO'siney which multiply by 2y — , and the product xvill be n -2 the true value of z^ in the proposed equation z" — az^"'- A X d^z^-^ .— X X a^z"^-^^ ks?c\ 2n 2n 2n Thus, let it be required to find the value of z, in the cubic equation 2^ — 4322 = 1 728 ; then, we shall have n = 3, a = 432, and f = 1728; consequently . ~J^ - (= ) = ,5, and the arch corresponding aj^ 144jir 71 I thereto == 60° ; whence the co-sine of (20°) ^ thereof will be found ,9396926; and this, multiplied by 24 = 2\/^)'Si (= 2\} — ), gives 22,55262 for one value of z. But ^ n besides this, the equation has two other roots, both of which may be found after the very same manner; for, since 0,5 is not only the co-sine of 60°, but also of 60° + 360°, and 60° -f- 2 x 360°, let the co-sine of (140°) ^ of the former of these arches be now taken; which is — • ,7660444, and must be expressed with a negative sign, because the arch corresponding is greater than one right angle, and less than three, Then^ the value 304 The Application of Algebra thus found being, in like manner, multiplied by 24 ( = 2y — ), we shall thence get — 18,38506 for ano- ther of the roots ; whence the third, or remaining root will also be known ; for, seeing the equation wants the second term, the positive and negative roots do here mu- tually destroy each other; and therefore the remaining root must be — 4,16756, the difference of the two for- mer, with a negative sign. PROBLEM XXXV. From a given circle ABCH it is proposed to cut off a segment ABC, such^ that aright line DE drawn from the middle of the chord ^ AC, to make a given angle therewith^ shall divide the arch BC of the semi-segment into two equal farts BE and^C. Let the chord BC be drawn, and upon the diameter HDB let fall the perpendicular EF : put the radius OB of the circle = r, and the tangent of the given angle CDE (answering to that radius) = f, and let OF = 2 ; then will EF = V rr -^2, and BC ( = 2EF) = 2V/rr — 22, and consequently BD ( = BC2 . 4r^ — 42^ BH^ ^ . r -j-z 2r from r-f-22X r- which taking BF = r — 2, we have DF = But, by trigonometr y, EF : DF : : rad. : tang. DEF, that is, v/TT^ii : ^-f ^^'^ — ^ : : r : t. Whence we have r + 2zY X r ~ 2? = /^ X r — z^ ; where td Geometrical Problems. 305 the whole being divi ded b y r — z, there results r + 22"]^ X r*— 2 = ^^ X r + z: which, ordered, gives z^ — • ^rr — tt r X z^-^ X rr — tt. 4 4 Zrr — tt Put = «, and — X rr — tt = fi then it 4 4 *^ will be 2^ — az =/. Therefore find, from the tables, the arch whose co-sine is — ^ — (the radius being unity) ; take f thereof, and find its co-sine ; which, mul- tiplied by 2V^a^ gives the true value of z, (See the last problem.) Thus, for example, let the radius OB (= r) = 1, and the given angle CDE = 25°, whose tangent (?) is therefore = ,4663 ; whence ^a = ,23188, and if = ,09782. Now, by logarithms^ it will be log, if — ■■ log. ^a — • |log. ia = — 1.9425328 = log. — ~=i = log. co- sine of 28° 50' ; whereof the third part is 9° 36f ', whose log. co-sine (to the radius 1) is ^ — 1.9938609; which added to the ilog. of ^a ( = — 1.6826316) gives — 1.6764925 = log. of 0,47478, whose double ,94956, is the true value of 2, or FO : whence the correspond- ing arch BE = 18° 16^', and consequently BC (= 2BE) = 36° 33'. By means of this problem, that portion of a spherical surface representing the apparent figure of the sky is determined. PROBLEM XXXVI. The base AB, and the difference of the angles at the baic being given^ while the angles themselves vary; to find the locus of the vertex E of the triangle. Let the base AB be bisected in O, and the angle BOD so constituted as to exceed its supplement AOD by the given difference of EAB and EBA; and let ED, APQ, BSF be perpendicular, and EPF parallel to OD : then, since the angle BCE (BOD) as much exceeds 2R S06 The Application of Algebra ACE, as CAE exceeds CBE, it is evident that the sum of the two angles BCE, CBE, of the triangle BCE, is equal to the sum of the two angles ACE, CAE of the triangle ACE ; and consequently, that the remaining an- gles AEC and BE C, are equal the one to the other : there- fore, by reason of the similar triangles EFB, EAP, we have EF : EP : : BF : AP, that is, OD + OQ : OD — OQ : : QA + DE : QA — DE ; whence, by com- position and divi- sion, 20D : 20Q • : 2QA : 2DE; wherefore OD X DE is = OQ X QA ; which is the known property of an equilateral hyberbola with respect to its asymptote. PROBLEM XXXVII. To Jind the solidity of a conical iingula BFCB, cut off hy a plane BRFSB passing through one extremity of the base diameter. Let EPF be parallel to the base diameter BC, cut- ting AD the axis of the cone in P ; also, let An be per- pendicular to BF ; join P, 72, and let RS be the conju- gate axis of the elliptical section BRFSB : then the part ABF, above the said section, being an oblique el- liptical cone, its solidity will be expressed by ,7854 X SR X BF X — , that is, by the area of its base BRFSB to Geometrical Problems* 507 drawn into one-third of the perpendicular height. But the triangles BCF and AP/z will appear to be equi-an- gular ; for, APF and A?zF being both right angles, the circumference of a circle, described on the diame- ter AF, will pass through P and w; and so the angles AF?z (BFC) and AP;2, as well as AFP (FCB) and AnV, insisting on the same arcli, are, respectively, equal. Hence we have BC:BF::A/z:AP; and therefore BF X A?2 = BC X AP: this value being substituted above, the content of the part ABF becomes SR x BC x AP x ,2618 : which, because SR is known to be = VBC x l^F, is farther reduced to BC X AP X VBC x KF X ,2618. This, subtracted from BC^ X AD x ,2618, the content of the whole cone ABC, leaves BC^ X AD — BC X aFx VBC x H X ,2618 for the required solidity of the singula BCF ; which, because DP X BC , .p DP X EF .,, , r, and AP = -^-^^ _-_, will be re- AD = BC — EF' BC — EF' duced to BC^ — EFx VBCxEF BC — EF X ,2618 DP X BC. 508 The Application of Algebra PROBLEM XXXVIII. Let A and B be two equal weights^ made fast to the ends of a thready or perfectly fiexihle line pVnGiq^ supported by two pins ^ or tacks^ P, Q, in the same horizontal plane ; over which pins the line can freely slide either way ; and let C be another weight^^ fastened to the threhd^ in the middle^ be- tween P and Q : nozu the question is^ to find the position of the weight C, or its distance below the horizontal line PQ, to retain the other ttvo weights A and B in equilibrio. Let PR ( = iPQ) be denoted by a, and Rn (the distance sought) by x ; and then Pn, or Qn, will be re- presented by Vrt^ + x^. Therefore, by the resolution of forces^ it will be, as V a^ -f x^ (P?z) : x (R/z) : : the whole force of the weight A in the direction Pn, : A.X " ' • " - , its force in the direction tzR, whereby it en- Va^ + x^ deavours to raise the weight C ; which quantity also ex- presses the force of the weight B in the same direction : but the sum of these two forces, since the weights are sup- posed to rest in equilibrio, must be equal to that of the 2Av weight C ; that is, — = C ; wheiice w^e have 4A^x^ :=: C^fl^ + C^;^^, and consequently x \ aZ V4A2. to Geometrical Problems^ 309 PROBLEM XXXIX. To determine the position of an inclined plane AE, along which a heavy body^ descending by the force of its own gra- vity from a given point A, shall reach a right line BP, given by position^ in the least time possible* Through the given point A, perpendicular to the horizon, let there be drawn the right line RB, meeting BP in B ; also conceive the semicircle AER to be de- scribed, touching BP in E ; then let AE be drawn, which will be the position required ; because the time of descent a- long the chord AE being equal to that along any other chord A/z, it will consequently be less than the time of the descent along A^, whereof An is only a part: therefore, if AQ and OE be now made perpendi- cular to BP, we shall have (by reason of the similar triangles) AB : AQ : : AB + AO : (OE) AO ; whence, by multiplying extremes and means, AB X AO = AQ X AB + AQ x AO ; therefore AB X AO — AQ X AO = AQ X AB, and AO (OE) = -^--r -r-r^ ; from which BE and AE are also given. xV.JlS — AQ The geometrical construction of this problem is ex- tremely easy ; for, if AQ (as above) be drawn perpen- dicular to BP, and the angle O AQ be bisected by AE, the thing is done : because, OE being drawn parallel to AQ, the angle OE A is = QAE = EAO ; and so, AO being = OE, the semicircle that touches BP will pass through A. 310 llie Application of Algebra PROBLEM XL. A ray of lights from a lucid point P, hi the axis AP, of a concave spherical surface^ is reflected at a given point E tn that surface ; tofnd the point D w here the refected ran Tneets the axis. Draw EQ perpendicular to AP, and from the center C let CE be drawn; also make CE r= a^ CQ = b^ CP = c, CD = X ; -R \mA(byEucA2.2>.) P E will be r= VV + c2 + 2bci wherefore, the an- gles of incidence andreflection CEP D Q A and CED being equal, we have, as PC (c : CD {x) : : PE (Va^ + c? + ^Z^^) : ED = cc\/ a^ -^ c^ -^ 2bc 1 r .1 — ; also, for the same reason, wc have PE X ED — PC X CD = EC% that Is, --2 ■ ■ ex = a^ ; which, reduced, gives ca^ X = ^ -— showino; how far from the center the rav cuts the axis. But if the lucid point P be supposed in- finitely remote, so that the ray PE may be considered as parallel to the axis AP, the expression will be more sim- ple ; for then (2^, in the divisor, may be rejected as nothing in comparison of 23c ; that being done, CD, or x^ be- comes = — - : which, therefore, if E be taken near the 2b ' ' vertex A, will be = -ia, very nearly. PROBLEM XLL To fnd the magnitude and position of an image formed hij refraction at a given lens. Let MN be the given lens, DOBCF the axis thereof^ to Geometrical Problems^ 311 and D;z-the object whose image FH ^ve would find ; also let CB be the radius of that surface of the lens MBN, which v^ nearest the object, and OB that of the other surface: make RC^ perpendicular to DF, and from 72, to any point Eih the surface of the lens, draw the incident ray 7zE; and let the continuation thereof be El, and let the direction of the same ray, after the first refraction at E, be E2 ; and, after it is refracted a second time, at ^, let its direction be e3H ; draw CE and O^R, and make ndv parallel to DF, calling O/;, b ; CB, c; BD, d\ Dn^ p ; and the distance of the point E from the axis DF, :x^ ; and let the sine .of incidence be to the sine of rviiVaction, out of air into glass, as m to 71. Then, the thickness of the lens being looked upon as inconsiderable in respect of the focal distance Fb^ we shall have, as d : x — p (E^) '• i d + c. (nv)- : dx -\-cx ' — dp — cp ^ , . , , , , ^ . \ . -7—^- ^ = ^'1 ; which added to Cv (/?) gives dx -^ ex -—' cp ^ ^ , ^ dx -l- ex — cp 7 = CI : theretore w : n : : i- : d d 72 X dx -j- ex *— cf dm = C2. Moreover, b (Ob) : x (BE) : : b + c (OC) : ^ ^ ^X"" ^ CR; whence R2 ( = CR C2)r= b -^ c X X •n X dx 4- CX' — cp ^ R2 : R3 = 7nx X b -^c cp ' In "^ dm ex '-^ dx but n ; 771 : and therefore 312 The Application of Algebra C3 ('—■RS RCi — ^ — nx XX l^ 4'C cp^-^cx^-^dx . ^"^ ^ '^ hn d Let X be now taken = 0, and then C3 will become cp -~, which let be represented by C3, and draw B^, pro- ducing the same till it meets e3, produced, in H : then ^1. r r- r>n\ d -^ C X X m — H X b '^- c X X ^3 bemg (= C^^ — C3) = ^ ^ , and the triangles H3^ and HEB equiangular, it will be, as (EB ~^3) -^ — : (B^-) c : : nbcd (EB) X : BH = -=== — ===- — ; = the requir- m — n X b + c X d^-^ nbc cd distance of the image from the lens ; and as c (BC) : ^ (CG) : : BH (or BF) : FH = ^^^ = nbcp ~ ~ (= HF) the magnitude or m — n X b + c X d -^ 7ibc the image, or its linear amplification. COROL. 1. Because the values of BH and HF are alike affected by b and c, it follows that both the distance and magni- tude of the image will remain unaltered, if the place of the lens be the same, let which side you will be turned towards the object. COROL. 2. If d be made infinite, or the distance of the object from the lens be supposed indefinitely great, BF will be- nbc , . . r 1 . come === , ; which is the principal focal dis- m — n X b •}- c tance, at which the parallel rays unite, and this dis- tance, when both sides of the lens have the same con- vexity, or ^ is = c, will become = : but in •^' ' 2m ^-271 a planO'ConveXj where b is infinite, it will be = m • to Geometrical Problems, and, in a meniscus^ where b is negative, or one surface con- cave, and the other convex, it will be === . m — 71 X b — c The same ansxvered othertvise^ alloxving also for the thick- ness of the lens* Supposing, as before, that F is the place of the image of an object at D, let FR and DS be supposed perpen- dicular to the axis FBQ, intersecting the continuation R of Ee (the intercepted part of the ray* DEe'F) in r and Sj and meeting the radii O^, CE (produced) in R and S; likewise let Ea and ec be perpendicular to QF, and Ei; and Ew parallel thereto : then, because the ray is supposed to be indefinitely near the axis, ac may be taken for the thickness of the lens, which let be de- noted by t ; putting ^F = 2, ce -=.1/^ aY, = x^ Ob ,= ^, CB = c, and BD = d (as before). By similar tri- angles, Ca (c) : aE (x) : : CD (c -f- ^) : DS = ^i2SS±J; and, by the law of refraction, m : n : : DS : Ss *= whence D^ (= DS — Ss) = 71 X X c -\- d — X m c n ^ X X c -\~ d _ , ^ T^. 1 X — , and vs ( = aE — D^) = 7n a ■ 2S 314 The Application of Algebra^ is' c X X r — — , by making r = — , and a' = 1 — r. c m Now, vs : Ev (BD) : : «E t aQ (BQ) ; that is. a: X ^ ~ — : ^ : ^ .v : ■■ — ^ = BQ ; which is given c er — qd from hence. Again, in the very same manner, Oc (b) : ce (t/) : : OF (^ + z) • FH = y X b -^ X ^ ^^^ ^ . ^ . . FR : Rr = -^ X ^ , ^ — : whence Fr = 1 X ^ , VI b mo = ?2!! and wr (Fr — ce\ -=. y x ^5 a^^cl b b therefore cQ ( = 1 = : from which sub- wr qz — br tracting the value of BQ, found above, we get this equation, viz. < , = t : whence the va- qz — * br cq — qd cd lue of 2, by making the given quantity t + = ^5 re — qd vb^ comes out = :!— -. But, if you had rather have the qg — b same in original terms, it is but substitutin g for ^ ; w hence, f. J , rbcd 4- rbt X re ' — qd alter reduction, z = ' ^ : qd X b + c — rbc + qt X re — - qd ^vhich, by restoring ?7i and 72, becomes mnbed -f 7ipt x nc — m^-'n x d m — nxmdxb+c — mnbc+m — nxtXnc — m — nxd where, if t be taken = O, we shall have nbcd , r 1 z = - ', ■ zr==z , the very same as lound m-^72X^X^4-c — nbc by the preceding method. APPENDIX: CONTAINING THE CONSTRUCTION OF GEOMETRICAL PROBLEMS, WITH THE MANNER OF RESOLVING THE SAME NUMERICALLY. PROBLEM L The base^ the sum of the two sides^ and the angle at :he vertex of any plane triangle being given; to describe the triangle. CONSTRUCTION. DRAW the indefinite right line AE, in which take AB equal to the sum of the sides, and make the angle ABC equal to half the given angle at the ver- tex, and upon the point A, as a center, with a radius equal to the given base, let a circle ?zCw be described, cutting BC in C ; join A, G, and make the angle BCD = CBD, and let ^„ CD cut AB in D ; then will , n ACD be the triangle that was to be constructed. 316 The Construction of DEMONSTRATION. Because the angles BCD and CBD are equal, there- fore is CD = DB (^Euc. 6. 1.), and consequently AD + DC = AB : likewise, for the same reason, the angle AD.C = BCD + CBD (Euc. 32. 1.) is equal to 2CBD. ^. E. D. Method of Calculation. In the triangle ABC are given the two sides AB, AC, and the angle ABC, whence the angle A is known ; then in the triangle ADC will be given all the angles, and the base AC ; whence the sides AD and DC will also be known. PROBLEM II. The angle at the vertex^ the base^ and the difference of the sides ^ being given ; to determine the triangle* CONSTRUCTION. Draw AC at pleasure, in which take AD equal to the difference of the sides, and make the angle CDB equal to the complement ^ of half the given angle to a right angle ; then from the point A draw AB eqitfil to the given base, so as to meet DB in B, and make the angle DBC = CDB, then will ABC be the triangle required. DEMONSTRATION. Since [by construction^ the angles CDB and DBC are equal, CB is equal to CD, and therefore CA — CB = AD : moreover, each of those equal angles being equal to the complement of half the given angle, their sum, which is the supplement of the angle C, must therefore be equal to two right angles, — the (whole) given angle, and con- sequently C = the given angle. ^. E, D, Method of Calculation, In the triangle ABD are given the sides AB, AD, Geometrical Problems. 5ir and the angle ADB ; whence the angle A will be given, and consequently BC and AC. PROBLEM III. The angle at the vertex^ the ratio of the including sides ^ and either the base^ the perpendicular^ or difference of the segments of the base^ being given ; to describe the triangle. CONSTRUCTION. Draw CA at pleasure, and make the angle ACB equal to the angle given ; take CB to C A in the given ratio of the sides, and join A, B: then, if the base be given, let AM be taken equal thereto, and draw ME parallel to CA meeting CB in E, and make ED parallel to AB ; but, if the perpendicular be given, let fall CF perpendicular tg AB, in which take CH equal to the given perpendicular, and draw DHE parallel to AB ; ^IM N lastly, if the difference of the segments of the base be given, take FG = AF, and join C, G, and take GN equal to the difference of the segments given, drawing NE parallel t^ CG, and ED to BA (as before) ; then will CDE be^^e triangle which was to be constructed. DEMONSTRATION. Because of the parallel lines AB, DE ; M^, AC ; and NE, GC ; thence is DE = AM, and EI =^ NG ; and also CD : CE : : CA : CB (iiwc. 4. 6.). ^, E. D. 318 The Construction of Method of Calculation. Let AC be assumed at pleasure ; then the ratio of AC to BC being given, BC will become known ; and there- fore in the xriangle ACB will be given two sides and the included angle ; whence the angles B and A, or E and D^ will be found ; then in the triangle EDC, EHC, or EIC. according as the base, perpendicular, or the difference of the segments of the base is given, you will have one side and all the angles ; whence the other sides will be known. PROBLEM IV. The angle at the vertex^ and the segments of the base^ made by a perpendicular falling from the said angle^ being given; to describe the triangle. CONSTRUCTION. Let the given segments of the base be AD and DB ; ■iisect AB by the perpendicular EF, and make the angle EBO equal to the difference between the given angle and a right one, and let BO meet EF in O ; from O, as a cen- ter, with the radius OB, de- scribe the circle BGAQ, and draw DC perpendicular to AB, meeting the periphery of the circle in C j join A, C, and C, B, then will ACB be the. triangle that was to be constructed. DEMONSTRATION. The angle ACB, at the periphery, standing upon the arch AQB, is equal to EOB, half the angle at the cen- ter, standing upon the same arch ; but EBO is equal to th» difference of the given angle and a right one {by con- strutiion) ; therefore ACB (EOB) is equal to the angle given. ^ E. D. Method of Cafculation, Draw CFG parallel to AB ; then it will be, as the base AB. : the difference of segments CG ( : : EB : CF) : : the sine oi the given angle at the vertex (EOB) : Peometrkal Problems^ 519 iRe sine of (COF = CBG) the diiference of the angles at tlie base ; whence the angles themselves are given. After the same manner, a segment of a circle may be described to contain a given angle, when that angle is greater than a right one, if, instead-of BO being drawn above AB, it be taken on the contrary side, PROBLEM V. Hamng given the base ^ the perpendicular ^and the tingle a^ the vertex of any plane triangle^ to construct the triangle. CONSTRUCTION. Upon AB, the given base {see the preceding figure)^ \<:^\ the segment ACGB of a circle be described to contain the given angle, as in the last problem ; take EF equal to the given perpendicular, and draw FC parallel to AB, cutting the periphery of the circle in C ; join A, C, and B, C, and the thing is clone ; the demonstration whereof is e\^ident from the last problem. Method of Calculation* In the triangle EBO are given all the angles and the side EB ; whence EO will be known, and consequently OF (= DC — EO) ; then it will be as EB : OF : ': the sine of EOB (the given angle at the vertex) : the sine of OCF, the complement of (COF or CBG) the difference of the angles at the base; whence these angles themselves are likewise given. This calculation is adapted to the logarithmic canon ; but, by means of a table of na- tural sines, the same result may be brought out by one proportion only : for BE being the sine of BOE, and OE and OF co-sines of BOE and COF (answering to the equal radii OB and OC), it will therefore be, BE : EF : : sine BOE ( ACB) : co-sine BOE + co-sine COF ; from which, by subtracting the co-sine of BOE, the co-sin^, of COF (= CBG) is fouhd. PROBLEM VI. The angle at the vertex^ the sum cf the txvo including sides ^ and the difference of the segments of the base being given^ to describe the triangle. 02O The Construction of CONSTRUCTION. Draw the right line AC at pleasure, in which take AB equal to the difference of the segments of the base, and make the angle CBE equal to half the supplement of the given angle ; and from A to BE apply AE equal to ' the given sum of the sides ; make the an- gle EBD = BED, and letBD meet AE in D, and from the center D, with the radius DB, describe the circle DEC, cut- ting AC in C, and join D, C ; then will ACD be the triangle required. DEMONSTRATION. The angle EBD being -= BED, therefore is DE = DB = DC, and consequently AD + DC = AE. More- over, the angle CDE, at the center, is double to the angle CBE, at the periphery, both standing upon the same arch CE ; which last (by construction) is equal to half the sup- plement of the given angle ; therefore CDE is equal to the whole supplement, and consequently ADC equal to the given angle itself. ^. E. D* Method of Calculation* In the triangle ABE are given the two sides AB, AE, and the angle ABE ; whence the angle A will be given ; then in the triangle ABD will be given all the angles and the side AB ; whence AD and DC (DB) will be also given. PROBLEM VII. The angle at the vertex^ the sum of the including sides ^ and the ratio of the segments of the base being given ; to deter* mine the triangle. CONSTRUCTION. Let AG be to GB in the given ratio of the segments of the base, and, upon the right line AB, let a segment Geometrical Problems. 321 of a 6ircle be described, capable of containing the given angle ; draw GC per- pendicular to AB, meeting the periphe- ry in G ; join A, C, afnd C, B, and in AC, produced, take CH = CB ; jom B, H, and in HA take HD equal to the given sum of the sidts, draw DE parallel to AB, and EF to BC ; then will DEF be the triangle required. DEMONSTRATION. Let Ftz be perpendicular to DE. Whereas {by con- struction) CH is equal to CB, and FE parallel to CB, therefore is FE = FH {Euc* 4. 6.) and consequently FE ^- FD = HD: also, because FE is parallel to CB, therefore is the angle DFE == AGB : moreover, the triangles ABC, DEF, being equiangular, it will be, as AG : GB : : D;i : ?zE. ^E. B. Method of Calculation. From the center O, conceive AO and OC to be drawn ; supposing KOI perpendicular, and CI parallel to AB : then it will be, as AK is to CI (KG) so is the sine of AOK (= ACB, see prob. 4.) to the sine of COI, the difference of the angles ABC and BAC ; which are both given "from hence, because their sum is given by the question : therefore in the triangle DHE are given all the angles and the side HD, whence the base DE will be known. PROBLEM VIIL Having the angle at the vertex^ the difference of the in- cluding sides ^ arid the difference of the segments of the bascj to describe the triangle. CONSTRUCTION. Take AB equal to the difference of the segments of the base, and make tjie angle KUn equal to half the 2T ' 322 The ConstructioJi of given angle; from A to Btz apply A E = the difFer- ence of the sides ; produce AE, and make the angle EBO = BEO, and let BO meet AE, produced in O, and from the center O, at the dis^ tance of OB, de- scribe the circumfe- rence of a circle, cut- ting AB produced in C J join O, C ; then is AOC the triangle sought. DEMONSTRATION- Because the angle EBO is = BEO {by construction) ; therefore is EO = BO = CO, and consequently AO ~ OC = AE, Furthermore, because the angle AOC is double to ADC, and ADC = ABE {Euc. corol. 22. 3.), therefore is AOC also double to ABE. % E. D. Method of Calculation. The two sides AB, AE, and the angle ABE being given, the angle A will from thence be found; then in the triangle ABO will be given all the angles and the side AB, whence OB (OC) and OA will be known. PROBLEM IX. The angle at the vertex^ the difference of the including sides ^ and the ratio of the segments of the base^ being gvoen^ ■to determine the triangle. CONSTRUCTION. Let AG be to GB in the given ratio of the seg- ments of the base, and up- on the right line AB let a segment of a circle ACB be described {by prob. 4.} _ _^ capable of the given an- A 6 H gle ; draw GC perpendi- cular to AB, meeting the periphery in C, and join A, C, Geometrical Problems* 323 and B, C ; in AC take AP = BC, and draw BP ; also, in AC, take CQ equal to the given difference of the sides, drawing QE parallel to PB, and ED to BA ; then will CDE be the triangle which was to be described. DEMONSTRATION. The angle DCE is equal to the given angle by con» struction; also EQ being parallel to BP, DE to AB> and AP = BC, therefore must DQ = EC {Euc. 4. 6.), and consequently DC — EC = CQ. Moreover, it CG be supposed to cut DE in tz, then D/z : E/z : : AG : GB. ^. E. D. Method of Calculation, Let Cm be equal to CE, and let Em be drawn. It will be, as AB is to AG — BG, so is the sine of ACB to the sine of the difference of CBA and CAB {by prod. 4.) ; then in the triangle DEm will be given all the angles and the side Dwz, whence DE will be given. PROBLEM X. The angle at the vertex^ the perpendicular^ and the dif- ference of the segments of the base^ being given^ to construct the triangle. CONSTRUCTION. Draw RS at pleasure, in which take DE equal to half the difference of the segments of the base, and make EC perpendicular to RS and equal to the given per- pendicular, and the angle DEn equal to the difference between the given angle and a right one; join D, C, and draw D;iO parallel to CE, and in DC take the 324 The Construction of point /, so that np (when drawn) may be equal to 7iE ; draw CO parallel to n/?, meeting DnO in O ; and upon O as a center, with the radius OC, describe the circle BCA, cutting RS m B and A ; join A, C, and B, C, and the thing is done. DEMONSTRATION. Join O, B, and O, A : since OC is parallel to pn^ therefore is OC : DO \ \ pn -. tzD, or OB : DO : : nE : nD ; and consequently the triangle OBD similar to the triangle ^zED (hy Eiic. 7. 6.). Therefore, seeing the angle DE/z is {by construction) equal to the excess of the given angle above a right one, ACB must be equal to the angle given [by prob. 4.). Moreover, since AD is = DB, AE — BE w^ill be equal to 2DE, which is the given difference of the segments (by construction^, ^E.D. 3Iethod of Calculation, In the triangle CDE, right-angled at E, are given both the legs DE and EC, whence the angle EDC will be known, and consequently ODC ; then, as the radius is to the sine of DBO (: : OB : DO : : OC : DO) so is the sine of ODC to the sine of OCD ; whence DOC, the differ- ence of the angles ABC, BAC (see prob. 4.), is also given, and from thence the angles themselves. PROBLEM XL The angle at the vertex^ the perpendicular^ and the ratio of the segments of the base^ being given^ to construct the triangle* CONSTRUCTION. Take AF to FB in the given ratio of the seg- ments of the base, and upon the right line AB describe a segment of a circle ACB capable of the given angle ; make FC perpendicular to AB, ineeting the circumference of the circle in C, in which Geometrical Problems. 325 take CG equal to the given perpendicular ; draw DGE parallel to AB, meeting AC and CB in D and E ; and then DCE will be the triangle required. DEMONSTRATION. Because of the parallel lines DE and AB, it will be as AF : DO (: : CF : CG) : : FB : GE, or AF : FB : : DG : GE ; whence it appears, that DG and GE are in the ratio given. Also the angle DCE and the perpendicular CG ar.e respectively equal to the given angle and perpendi- cular, by construction. ^. E. D. Method of Calculation. As AB is to AF — BF {see proh. 4.j, so is the sine of AC B to the sine of the difference of A and B; whence^ both A and B will be given, because their sum, or the an- gle at the vertex, is given: then, in the triangles DGC, EGC, will be given all the angles and the perpendicular CG, whence the sides will also be known. PROBLEM XII. The hase^ the sum of the sides, and the difference of the angles at the base^ being given^ to describe the triangle. CONSTRUCTrON. At the extremity of the base AB, erect the perpen- dicular BE, and make the angle EBC equal to half the given dif- ference of the an- gles at the base ; from the point A, to BC, apply AC equal to the sum of the sides ; and make the angle CBD = BCA; then will ABD be the triangle required. DEMONSTRATION. From the centre D, with the radius CD, describe the 326 The Construction of semicircle CHF, and join F, B. Then, whereas by con- struction the angle CBD is = BCD, therefore is I3B = DC ; whence it appears that AD -f Y>1^ is = AC, and that the semicircle must pass through the point B : there- fore the angle CBF, standing in a semicircle, being a right angle, and therefore = ABE, let FBE, which is common, be taken away, and there will remain ABF = EBC ; but DF being equal to DB, it is manifest that ABF (EBC) is equal to half the difference of the angles ABD and DAB. % E. D. Method of Calculation. As the sum of the sides (AC) is to the base (AB), so is the sine of ABC, or of the complement of half the given difference, to the sine of (C) half the angle at the vertex \ whence the other angles BAD and ABD are also given. PROBLEM XIIL The hase^ the difference of the sides^ and the difference oj the angles at the base^ being given^ to determine the triangle* CONSTRUCTION. At the extremity B of the given base AB, make the angle ABD equal to half the given difference of the angles at the base ; and from A to BD apply AD = the difference of the sides ; draw ADC, B A and make the an- gle DBC = BDC, and ABC will be the triangle re- quired. DEMONSTRATION. Because the angle DBC is = BDC, CD will be = CB, and AC will exceed BC by AD. Moreover, since A + ABD = (CBD) CDB (Euc. 32. 1.), therefore is A -f 2ABD (= CBD 4- ABD) = ABC, and consequent- ly ABC — A = 2ABD, equal to the difference given. 4 ^. D. Geometrical Problems, >S27 Method of Calculation. In the triangle ABD are given the two sides, AB and AD, and the angle ABD,^ whence the angles A and ADB will be given, and from thence the angles CBA and ACB. PROBLEM XIV. The difference of the angles at the hase^ the ratio of the sides ^ and either the hase^ the perpendicular^ or the difference of the segments of the base^ being giveii^ to describe the ti'i- angle. CONSTRUCTION. Draw AC at pleasure, and make the angle ACD equal to the given difference of the angles at the base, and take CD to C A, in the given ratio of the sides ; draw ADE, upon which let fall the perpendicular CQ; take QE equal to QD, and join E, C ; then, if the base be given, let AB be taken equal thereto, and draw BF parallel to CA (meeting CE in F), and FG parallel to E A ; but if the perpendicular be given, let CP be taken equal thereto, and through P draw FPG parallel to AE ; lastly, if the difference of the segments of the base be given^ then let AR be taken equal to that difference, draw RH parallel to CA, and FHG to EA ; then will CFG be die triangle required. DEMONSTRATION. Since QE = QD, and the angle EQC = DQC, there- fore is CE = CD, and the angle E = QDC = A + ACD {Euc. 32. 1.), and therefore E — A = ACD ; whence, by- reason of the parallel hues AE, GF, £sPc. we have GFC — FGC == ACD, also FG = AB, GH = AR, and CF : CG : : CE (CD) : CA. .% E. D. 328 The Constnictton of Method of Calculatzoii, Let CA and CD be expressed by the numbers exhibit- ing the given ratio of the sides : then in the triangle ACD will be given two sides and the included angle A CD ; whence the angles CAE (CGF) and CEA (CFG) wiM be given, and from thence the sides CG and CF. PROBLEM XV. The hascy the perpendiciuar^ and the difference of the an- gles at the hasey being given^ to construct the triangle* CONSTRUCTION. Bisect the given base AB by the perpendicular DF, In which take DE equal to the given height of the triangle ; draw CEGH parallel to AB, and make the angle EDH equal P to the given dif*- ference of the an- gles at the base ; draw EAQ, and take Q therein^ so that QD 3= DH ; and, paral- lel to QD, draw AO, meeting DE in O ; upon O, as a center, with the radius OA, describe the circle AGFCB, and from the point G, where it cuts the right line CH, draw GA and GB ; then Will AGB be the triangle required. DEMONSTRATION. Let OG and BC be drawn. By reason of the paral- lel lines QD and AO, it will be QD (DH) : AO (OG) 3 : ED : EO ; therefore the two triangles EHD, EGO, having one angle, E, common, and the sides about the other angles D and O proportional, are equiangular ^EuG. 7. 6.), and consequently EOCt = EDH. More- over, because DOEF is perpendicular both to AB and GC, and AD equal to BD, it is evident that the circle passes through the point B, and that the arches FC, FCt, Geometrical Problems. 329 as well as the angles ABC, BAG, are equal ; and con- sequently that the angle GBC is the difference of the an- gles BAG, ABG : but this difference GBC is equal to EGG, or EDH (Euc. 20. 3.), that is, equal to the differ- ence given. ^. JE. D. Method of Calculation. First, in the right-angled triangle AED are given both the legs AD and DE, whence the angle DEA will be given : then it will be, as the radius is to the sine of the angle H, the complement of the given difference ( : : DH : DE : : DQ : DE), so is the sine of DEA to the sinejof Q ; whence AGE (QDE) will also be given ; from which take GGE, and there will remain AGG, equal to twice ABG, the lesser angle at the base. PROBLEM XVI. The sum of the sides^ the difference of the segments of the base^ and the diff'erence of the angles at the base^ being given^ to describe the triangle. CONSTRUCTION. Make AD equal to the sum of the sides, and the angle ADE equal to half the difference of the angles at the base ; from A to DE apply AE equal to the given difference of the segments of the base; make the angle CED = EDC, and from the point C, where EC cuts AD, with the radius EC, describe the semicircle FEB, cutting AE, produced in B ; join B, C, and the thing is done. DEMONSTRATION. Upon AB let fall the perpendicular CQ. Because EQ is = BQ {Etic. 3. 3.), therefore will AQ — BQ = AE : also, because the angles CED, EDC are equal {by construction)^ CD will be = CE = CB, and consequently AC -f CB = AD. Moreover, AJftC 2U 33a The Construction of (i:«c. 32. l.)=:2ADE w BAG =: BEC — BAG = AGE (Euc. 20. 3.) ^ E. D. Method of Calculation. In the triangle ADE are given the sides AD, AE, and the angle D, whence the angle A will be given ; then, in the triangle AGE are given all the angles, and the side AE, whence AG and GB (GE) will be likewise given.! PROBLEM XVII. The difference of the angles at the base^ the ratio of the segments of the base^ and either the sum of the sides ^ the difference of the sides^ or the perpendicular^ being given^ to construct the triangle. GONSTRUGTION. Let AG be to BG in the given ratio of the segments of the base ; and upon AB let a segment of a circle BPA be described (by problem 4), to contain an angle equal to the difference of the angles at the base ; raise GP per- pendicular to AG, cut- ting the periphery of the circle in P, and in AG produced, take GD = GB, and draw PA, PB and PD : then, if the perpendicular be given, take PF equal thereto, and through F draw EFG parallel to AD ; but if the sum or difference of the sides be given, let a fourth proportional PE, to AP ± PD, AP, and the said sum or difference be taken, and draw EFG as above ; then will PEG be the triangle required. DEMONSTRATION. Since GP is perpendicular to AD, and GD = GB, the angle D will be equal to DBP = A + BPA : whence, because EG is parallel to AD, PGE will be = PEG •Ar BPA {Euc. 29. lO^ and consequently PGE — PEG Geometrical Problems. 331 1=1 ABP, whicU, by construction, is equal to the given difference of the angles at the base. J Again, by reason of the parallel lines AD and EG, it will be, EF : FG : : AC : (BC) CD. Likewise, for the same reason, AP ± PD : PA : : PE ± PG : PE : : given sum or diff. of sides : PE (by constniction) and consequently PE ± PG = the said given sum or differ- ence, i^. E. Z). Meth od of Calculati o n . First, it will be, as AB is to AD, so is the sine of APB to .the sine of APD {by proh, 4) ; then, in the triangle PGE, will be given all the angles, and either the perpen- dicular, or the sum or difference of the sides, whence the sides themselves are readily determined. Note. The perpendicular cutting the circle in two points, indicates that this problem is capable of two differeixt so- lutions* PROBLEM XVIIL The difference of the sides: ^ the dijference of the sej^mcnts 'ofthe base^ and the difference of the angles at the basc^ be-' ing given^ to describe the triangle. CONSTRUCTION. Draw the indefinite line AQ, in which take AD equal to the given difference of the sides, and make the angle QDH e- qual to the com- plement of half the difference of the angles at the base ; from A to DH apply AC = the given dif- ference of the segments ; and having produced the same to L, make the angle DCE equal to CDE, and let CE meet AQ in E, and upon the center E, with the radius EC, describe an arch, cutting AL in B ; join E', B, so shall AEB be the triangle required. 332 The Construction of DEMONSTRATION. Upon AB let fall the perpendicular EP. Because the angle DCE = CDE, therefore is ED = EC, and consequently AE — EB (= AE - — EC = AE — ED) == AD. Also, since EB = EC, therefore will PB z=z PC, and consequently AP — BP (AP — PC) r= AC. Moreover, the angle EBC being = ECB {Euc. 5. 1.), and ECB — A == CE A {Euc. 32. 1.), it is plain that EBC — A = CEA, equal to the given difference, because the triangle EDC is isosceles, and the angle at the base equal to the complement of half the said difference, by construction. ^. E. JD. Method of Calculation. In the triangle ADC are given two sides, and the angle ADC, whence the angle A will be kno^vn ; then, in the triangle ACE, will be given all the angles and the side AC, whence AE and CE (BE) will also become known, PROBLEM XIX. The perpendicular^ the difference of the angles at the base^ (ind the difference of the segments of the hase^ being given ^ to construct the triangle, CONSTRUCTION. Upon AQ, equal to the given difference of the seg- ments of the' base, let a segment of a circle QCA be described, capable of the difference of the angles at the base ; bisect AQ with the perpendicular TL, in which let TE be taken equal to the given perpendicular ; draw EC parallel to AQ, cutting the pe- riphery of the circle in C \ also draw CP perpendicular to AQ, and in AQ produced take PB = PQ ; join Cj A|^nd C, B j then will ACB be the triangle required. Geometrical Problems. 335 DEMONSTRATION. Since {by construction) CP is perpendicular to QB, and PB equal to PQ, thence will the angle B = PQC, and B (PQC) — BAG = ACQ = difference of an- gles given : also, for the same reason, will CP = TE, and AP — BP = AP — PQ = AQ. 4 ^- ^• Method of Calculation. From the center O, conceive O A and OC to be drawn : then, in the triangle AOT, will be given all the angles, and the side AT, whence OT and OE will be given ; then it will be, as AT : OE : : sine of AOT (ACQ) : sine of OCE ; whence all the angles in the figure are given. PROBLEM XX. The segments of the base^ and the sum of the sides of any plane triangle^ being ^iven^ to determine the triangle. CONSTRUCTION. From the greater segment AQ, take QF equal to the lesser segment BQ ; make QL perpendicular to AB, and draw AI, mak- ing any angle with AB at pleasure, in which take AE e- qual to the given sum of the sides, and join B, E ; make the angle AFG = AEB,and bisect EG in H, and from B as a center, with the ra- dius EH, describe tjiC/z, cutting the perpendicular QL in C \ join C, A, and C, B, and the thing is done. DEMONSTRATION. From the center C, with the radius CB, let the circle BDLKF be described ; and let AC be produced to meet its periphery in D. By reason of the similar triangles AEB, AFG, it will be, as AE : AB : : AF : AG; whence AG X AE = AF X AB ; but {by Euc. 37. 3.) ,334 The Construction of AF X AB = AK X AD ; therefore is AG X AE = AK X AD : whence, as EG and DK are equal, by construc- tion, it is evident that AG and AK, as well as AE and AD, must be equal. ^ E. D. Method of Calculation, As AE : AB : : AF : AG ; which taken from AE, and the remainder divided by 2, gives BC (EH), the lesser side of the triangle. PROBLEM XXI. The segments of the base^ and the difference of the sides^ being' given J to describe the tj'iangle. CONSTRUCTION. Take AF equal to the difference of the given segments AQ, BQ (^see the preceding figure)^ and draw AI making any angle with AB at pleasure, in which take AG equal to the given difference of the sides ; join F, G, and make the angle ABE = AGF, and from the center B, at the distance of -jEG, describe nCm^ cutting the perpendicular QL in C ; join C, B, and C, A, then will ACB be the tri- angle that was to be constructed : the demonstration of which is so very little different from the foregoing, that it would be needless to give it here. LEMMA. If a given right line AB be divided in any given ratio^ €tt C, and the right line (ZT^O he taken to A.C i?i the ratio qfBC to AC — ^ BC ; andfroin O as a center^ at the distance oyTOC, a circle CPD be described^ and txvo right lilies AP, BP be drawn from A and B, to meet any -where in the pe- riphery thereof; I say these lines will be to one another {every where) in the given ratio of K^ to H'Si, For, [since CO : AC : : BC : AC — BC; therefore, by composition, CO : AO : : BC : AC ; and, by per- mutation, CO : BC : : AO : AC; whence, by di- vision, CO : BO : : AO : CO, or PO : BO : : AO : PO ; wherefore, seeing the sides of the triangles FOB, AOP, about the common angle O, are proportional. Geometrical Frobtems. 335 tllose triangles must be similar {^Euc. 6. 6.), and there- fore the other sides also proportional, thiat is, PO (CO) i AO : : BP : AP : whence {by the second step) BC : AC ::BP;AP. §>. E. D. PROBLEM XXII. The segments of the base^ and the ratio of the sides^ being given^ to determine the triangle. CONSTRUCTION. Let AQ and QB be the segments of the base ; and let the whole base AB be divided at C, in the given ratio of the sides ; take CO to AC, as BC to AC — BC, and with the radius CO de- scribe the circle CPD, and raise QP perpendicu- lar to AO, meeting the periphery in P ; join A, P, and B, P ; then will ABP be the triangle required : the demon- stration of which is manifest from the preceding lemma. Method ofCalculatioiu Since the ratio of AC to CB, and the length of the whole line AB are given, thence will AC and CB be given, and consequently OC (-— -— ] ; from 336 The Construction of O Q whence the perpendicular PQ ( = VCQ X DQ) is likewise given. PROBLEM XXIII. Having the hascy the perpendicular^ and the ratio of tht sides^ to describe the triangle* CONSTRUCTION. Let the base AB be divided at C, in the given ratio of the sides, and let the circle CPD be described as in the last problem ; ^ in OR, perpen- -^=^=J5^ dicular to AD, take On equal to the given perpendi- cular, and, through 72, draw PnP pa- \ ^ rallel to AD, cut- ting the periphery of the circle in P ; join P, A, and P, B, and, the thing is done. The truth of this is also evident from the preceding lemma. Method of Calculation. Upon AD let fall the perpendicular PQ, and join O, P : then PO ( = -— -^ --— ) will be dven ; there- \ AC — BC/ ^ fore, in the triangle OPQ, are given OP and PQ, from whence not only OQ, but AQ and BQ are also given. Note* The parallel P;zP, cutting the circle in two points, shows that this problem admits of two different solutions. PROBLEM XXIV. The difference of the segments of the base^ the perpendi- cular^ and the ratio of the sides^ being given^ to construct the triangle. CONSTRUCTION. Let AB be the difference of the segments of the base {see the last figure^ ^ and let every thing be done as in the preceding problem : take Q(b = QB, and join P, b ; then will A^P be the triangle required. The reasons of Geometrical Proplems. 337 which are obvious from what has been said already; and the numerical solution is also evident from the last problem, PROBLEM XXV. The ratio of the segments of the hase^ the perpendicular^ and the ratio of the sides being given ^ to construct the tri^ anq-le. CONSTRUCTION. Draw any right line ABC at pleasure, in which take AE to EB in the given ratio of the sides, and AF to Elf B 5: FB in the given ratio of the segments of the base, and make FQ perpendicular to AB, and equal to the given height of the triangle ; make also EC : AE : : BE : AE — BE, and with the radius CE describe the circle ERS, and from the point R, where it intersects the perpendicular FQ, draw RA and RB, and draw QP and QT parallel to RA and RB ; then will PQT be the triangle that was to be described, DEMONSTRATION. By the foregoing lemma, AR : BR : : AE : BE ; there- fore, by reason of the parallel lines, it will be QP : QT {: : RA : RB) : : AE : BE. And, for the same reason, PF : TF : : AF : BF. ^ E. D. Method of Calculation, Having assumed AB at pleasure, there will be given BE, AE, BF, and CE f= A|J1_||) ; whence RF \ AE — BE/ ( = \EF X CE + CF) is also given ; then, in the right-angled triangle BRF, will be given both the legs 2X 338 The Construction of BF and RF, whence the angle B is given ; lastly, in the triangle FQT will be given all the angles and the side FQ, whence QT and TF will be given, and consequently PQ and FP- PROBLEM XXVI. To divide a given angle ABC into twoparts^ CBF, ABF, so that their sines may obtain a given ratio. CONSTRUCTION. In BA, and CB produced, take BE and BD in the given ratio of the sine of CBF to the sine of ABF ; draw DE, and parallel thereto draw BF, and the thing is done. For, by trigonometry^ BE : BD : : the sine of D (= CBF) : the sine of BED (= ABF). Hence the numerical solution is also evident ; since it will be, as the sum of BE and BD is to their difference, so is the tangent of half the given angle ACB to the tangent of half the difference of the two required parts FBC and FBA. PROBLEM XXVIL To divide an angle given into two parts ^ so that their tan- gents may be to each other in a given ratio. CONSTRUCTION. Take any two right lines AD, BD, which are in the ratio given, and upon the whole compounded line A B let a seg- ment of a circle BCA be de- scribed, capable of the angle given J make DC perpendicu- lar to AB, meeting the peri- phery in C, and draw AC and BC, then will ACD and BCD be the two angles required. Geometrical Problems. 339 The reason of which is evident, at one view, from the con- struction. The method of solution is also very easy ; for it will be, as AB is to AD — DB, so is the sine of ACB to the sine of B — A {see problem 4) ; whence B and A, and also BCD and ACD, are given. PROBLEM XXVIII. To divide a given angle ABC into two parts^ so that their secants may obtain a given ratio. CONSTRUCTION. Take BE to BT in the given ratio of the secants ; jt)in T, E, and let BF be drawn perpendicular to ET, and the thing is done. The truth of which is manifest from the construction. Method of Calculation. The angle EBT and the ratio of the sides BE and BT being given, the angles E and T will also be given, and consequently their complements EBF and FBT. PROBLEM XXIX. From a given point O^ to drat^ a right line OF, to cnt two right lines AC, AB, given by position^ so that the parts thereof OE, OF, intercepted between that point and those lines^ may be to one another in a given ratio. CONSTRUCTION. From O, through A, the point of concourse of BA and CA, let OAD be drawn, in which take AD to AO 340 The Construction of in the given ratio of FE to EO, and draw DF parallel to Ac, cutting AB in F ; join F, O, and the thing is done ; as is manifest from Euc. 2. 6. Method of Calculation. Since the point O and the lines AC and AB are given by position, OA and all the angles at the point A are given; therefore, from the given ratio of AD and AG, AD will be likewise given ; then in the triangle DAF will be given AD and .all the angles (because FDA = C AO) ', whence AF is also given. PROBLEM XXX. To divide a given arch CD i7ito two such parts ^ that the rectangle under their sines may be of a given magni- tude* CONSTRUCTION. Upon the radius OC let fall the perpendicular DF, in which (pro- JD duced if need be) take FG = |OC, and thereon con- stitute a rectangle FIHG equal to the given rect- angle ; and, sup- posing HI to cut the circumfer- ence in E, draw OB to bisect DE j then will CB and DB be the parts required. Geometrical Problems. 341 DEMONSTRATION. Draw CM and DNE perpendicular to the radius OB, and Nti and E^ perpendicular to DF. It is evident, by construction, that the triangles OCM and DN^i are similar (because Nn is parallel to CO, and IS^D to CM) ; therefore OC : CM : : DN : N/z (= \Ee\ and consequently CM X DN = OC X |E^ = ^OC X E^ = FGxEe = the given rectangle by construction. ^. E. D. Method of Calculation. Dividing the measure of the given rectangle by half the radius, FI will be given, which, added to OF, the co-sine of CD, gives the co-sine (OI) of CE, the differ- ence of the two parts ; whence the parts themselves will be known. PROBLEM XXXI. Having the ratio of the sines ^ and the ratio of the tan- gents of two angles^ to determine the angles. CONSTRUCTION. Let AD be to ED in the given ratio of the sines, and AD to FD in the given ratio of the tangents j and about the center D, with the interval DE, let the semicircle ERK be described ; and, upon AF, describe another semicircle, cutting the former in H, and through H draw AR, and join H, D ; then will DHR and DAR be the two angles required. DEMONSTRATION. Join F, H, and draw DQ perpendicular to AR. The angle AHF, standing in a semicircle, being a right one, the lines FH and DQare parallel (l)yEiic.27. 1.) ; 342 The Construction of and therefore AD : FD : : AQ : HQ : : tang, of DHQ : tang. D AQ. Likewise DA : DE (DH) : : sine of DHQ : sine of D AQ, as was to be shown. Method of Calculation. If AR be supposed to meet the periphery in R, and RN be drawn parallel to HF, meeting AK in N ; then will DN = DF, and AN : AR : : AF : AH ; but {by Euc. 37. 3.) AR : AK : : AE : AH ; whence, by compounding the terms of these two proportions, ^c. AN : AK : : AF X AE : AH^ ; whence AH, as well as AD and DH, being known, the angles A and K will also be known. PROBLEM XXXIL To draw from a point A, in the circumference of a given circle^ two subtenses AB and AD, which shall be to one another in the given ratio ofm to 72, and cut ojftwo arches AB and ABD, in the ratio of\ to S, CONSTRUCTION. Draw the diameter AH, and take the subtense AQ, in proportion thereto, as n — m to 2m ; from the center O draw OB paral- lel to AQ, meeting the periphery in B ; join A, B, and make the subtenses BC and CD each equal to AB, and draw AD, and the thing is done. DEMONSTRATION. Join H, Q, and draw BE and CF perpendicular to AD. The angle AOB (QAH) at the center, standing up- on the arch AB, is equal to the angle BAD at the cir- cumference, standing upon double that arch ; therefore, AQH being equal to AEB, or a right angle {Euc* 31. 3.), the triangles AQH, AEB, must be equiangular, and consequently AB : AE : : AH : AQ ; but, by con- Geometrical Problems* 343 struction, AH : AQ i \2m : n — my whence AB : AE : : 2m: n — w, or AB : 2AE : : 2;?z : 2n -— 2?w ; there- fore (^by composition) AB : AB + 2AE (: : 2m : 2n) : : m : n. But AB being = BC = CD, EF is = BC = AB, DF = AE, and AD = 2AE + AB, Hence AB : AD ::m: n. % E. D. Method of Calculation. Let AP be perpendicular to OB ; then, because of the similar triangles OAP, AHQ, it will be as AO : OP (: : AH : AQ) : : 2m i n*— m (by construction) ; therefore OP n — m X AO '2m ' (BP = AO ~ OP) = Sm- n X AO 2m , and consequently AB (V2AO X BP) = 4 Sm — n X AO ; whence AD is also given. PROBLEM XXXIIL The area and hypothenuse of any right-angled plane tri- angle being given^ to describe the triangle. a CONSTRUCTION. Upon the given hypothenuse AB, as a diameter, let the semicircle ACB be described, and upon OB, equal to half AB (^by Euc. 41. 1.), constitute the rectangle OE equal to the given area of the triangle, and let the side thereof, EF, cut the periphery of the circle in C ; join A, C, and B, C, and the thing is done. o44 The Constructio7i of DEMONSTRATION. The triangle ABC, standing upon the whole diameter AB, is equal to the rectangle OE, of the same altitude, standing upon half AB {by Euc. 41. 1.), which last (by constructwri) is equal to the area given. 31ethod of Calculation* Join O, C, and let CD be perpendicular to AB ; then it will be, as A02 (AO x OC) : AO x DC ( : : OC : DC) : : radius : sine of DOC ; which, in words, gives this theorem. As the square of half the hypotlieniise of any right-an- gled plane triangle is to the area^ so is the radius to the sine of double the lesser of the two acute angles* N* B. Since no sine can be greater than the radius, it is plain, that, if the square of half the hypothenuse be not given greater than the area of the triangle, the problem will become impossible ; in which case the side EF, in- stead of cutting, will pass quite above the circle. PROBLEM XXXIV. To describe a right-angled triangle^ whose area shall be equal to a given square^ and the sum of its two legs equal to a given right line AB. CONSTRUCTION. Upon AB let a semicircle be described ; make ACD = half a right an- gle, and CD = twice (PQ) the side of the given square ; draw Iq DE parallel to AB, meeting the circum- ference in E, and EF perpendicular to AB, p intersecting AB in F, in which, produced, take FG ~ FB, and draw AG ; so shall AFG be the tri- anjrle required* DEMONSTRATION. It is evident that AF + FG = AB ; and also that the P Geometrical Problems* 3^45 area AFG = |AF x FG = |AF x FB = |FE^ ( = ^DH^) = 1CD2 = PQ2. ^ E. D. Method of Calculation. If the radius CE be drawn, in the right-angled trl- angle CEF, there will be given CE ( r= ^AB) a nd EF]* ( = 2PQ2 j whence CF ( = V^AB^ — 2PQ2) will be known, and, from thence, both AF and FG. LEMMA. The area of any right-angled triangle^ ABC, is equal to a rectangle under half its perimeter and the excess of that half perimeter above the hypothenuse^ or longest side. . DEMONSTRATION. In the proposed triangle let the circle EGF be in- scribed, and from the center D, to the angular points A, B, C, and the points of contact E, F, G, let the right lines DA, DB, DC, DE, DF, and DG be drawn. It is plain that the sum of the three triangles ADB, BDC, and ADC, is equal to the whole y/\(^ triangle ABC ; but the triangle ADB is equal to the rect- angle 1 AB X DG ; and so of the rest : therefore the sum of the rectangles ^AB X DG + |CB X DF + |AC X DE is equal to the whole triangle ABC ; but the sum of these rectangles {by Euc, 1. 2.) is equal to the rectangle under half the perimeter AB + BC + AC and the semi-diameter DG, which last rectangle is, therefore, equal to the triangle given. But the angles E and G being right ones {Eiic. 17. 3.), and the side AD common, and also DE equal to DG, thence will AE = AG {Euc. 47. 1.). And in the same manner will CE = CF ; consequently AC CAE + CE) will be = AG + CF; whence it 2Y 346 The Construction of appears that the hypothenuse is less than the sum of the two legs by BG -f BF, or twice the radius of the in- scribed circle, and therefore less than half the perimeter by once that radius, or DG : whence the proposition is manifest. PROBLEM XXXV. The perimeter and area of a right-angled triangle being given^ to describe the triangle. CONSTRUCTION. Make AB equal to the given perimeter, which bisect in C, and upon AC let the rectangle ACDE be consti- tuted equal to the given area j take CF = CD, and, from Tu F, through D, draw the indefinite line FH, to which, from B, apply BI = AF ; then, upon, AB let fall the perpen- dicular IK, so shall BIK be the triangle that was to be constructed. DEMONSTRATION. Sinc^ {bij construction^ CD is = CF, therefore is IK = FK, and consequently IK + IB + BK = FK + AF + BK = AB. Again, the excess of the half perimeter AC above the hypothenuse BI ( AF) being = CF = CD, it is evident ^from the premised lemma) that the area of the triangle will be = ACDE = the given area by con- struction. ^. E. D. Method of Calculation. Dividing the area by half the perimeter, CD ( = CF) will be given ; then, in the triangle BFI, will be given BF, BI, and the angle F ( = 45°) ; whence the angle B will also be known, and from thence BK and BI. Geometrical Problems. 347 PROBLEM XXXVL To make a right-angled triangle equal to a given square ABCD, whose sides shall be in arithmetical progression. CONSTRUCTION. sAB In AB, produced, take BF = — — , and upon AF describe the semicircle AEF, cutting BC, produced, in E ; take BQ = done. 4EB ; join E, Q, and the thing is DEMONSTRATION. Since, by construction, QB : BE : : 4 : 3, therefore will BQ2 :*EB2 : : 16 : 9, and BQ^ + BE* : BE^ : : 16 + 9 (25) : 9, that is, EQ^ : BE^ : : 25 : 9 {Euc. 47. 1.) ; whence EQ ; BE : : 5 . 3 {Euc. 22. 6.) ; therefore the sides BE, BQ, and EQ, being to one another in the ratio of the numbers 3, 4, and 5, are in arithmetical progression. And, because BQ is = 2EB2 2BF X AB 4EB thence will EB x_BQ "2 = AB2. ^ E. D. 3 3 Method of Calculation. Seeing BF is = -±^, (BE V AB x l^^) will be = AB V|; whence BQ (i|^) and EQ (^^ likewise given. \ \ will be us Whe Construction of PROBLEM XXXVII. In a g'iveJi circle CIHK^ to describe three equal circles E, F, and G, ivhich shall touch one another^ and also the periphery of the given circle* ^ CONSTRUCTION. From the center C let the right lines CH, CI, and CK be drawn, dividing the periphery into three equal parts, in the points H, I, and K ; join I, K, and in CK, produced, take KL = ^IK j draw IL, and, paral- lel thereto, draw KF meeting CI in F ; make HE and KG each == IF, and upon the centers F, E, and G, through the points I, H, and K, let the circles FrI, EwH, and G/zK be described, and the thing is done. DEMONSTRATION. Draw FE, FG, and EG. Because {by construction^ HE, IF, and KG are equal, CE, CF, and CG will likewise be equal, and FG pa- rallel to IK {by Euc. 2. 6.), and therefore, KF being parallel to IL {by construction^ the triangles IKL and FGK are equiangular ; whence, IK being == 2KL, FG is = 2GK (2Fr) {Euc. 4. 6.) : whence it is manifest that the circles F and G touch each other. Geometrical Problems. 349 Moreover, the angles ECF, ECG, and FCG, as well as the containing sides CE, CF, and CG being respectively- equal, EF, FG, and EG must also be equal (Z>z/ Hue. 4. 1.), and therefore EF or EG = 2FI or 2GK ; whence it is evident that the circles E, F, and G also touch one ano- ther. But all these circles touch the given circle, because they pass through given points H, I, K in its periphery, and have their centers in right lines joining the center C and the points of concourse. Method of Calculation, In the triangle FGK we have given the angle FGK (150°), and the ratio of the including sides (viz. as 2 to 1) ; whence the angle FKG will be given ; then, in the triangle FCK will be given all the angles and the side CK ; whence CF and also FI will be given. But, if you had rather have a general theorem for expressing the ratio of F I to CI, then let EC be produced to meet FG in r. Therefore the angle rFC being = 30% Cr wi ll be = |CF ; whence {bi/ 'Euc. 47. 1.) FI or Fr (V FC2 —_Cr^) is = FC X VT, and therefore CI = FC + FC V| ; consequently CI : FCj : 1 -f^Vj: 1 ; whence, by division, CI : FI ( : : 1 + V | : V~l) • • '^ I + 1:1. ^ PROBLEM XXXVIII. In a given circle CEHG to describe jive equal circles K, L, M, N, andOy which shall touch one another^ and the circle given. CONSTRUCTION. Let the whole periphery EGH be divided into five equal parts, at the points E, F, G, H, and I (by Euc. 11.2.), and draw CE, CF, CG, CH, and CI; join G, H, and in CH produced take HP = |.GH ; draw PG, and parallel thereto draw HM, meeting CG in M ; take FL, EK, lO, and HN, each equal to MG, and upon the centers K, L, M, N, and O, let circles be described through the points E, F, G, H, and I, and the thing is done. 350 The Construction of The demonstration whereof is evident from the last proposition : and in the same manner may 6, 8, or 10, £sPc. equal circles be inscribed in a given circle, to touch one another. The method of calculation in this, or in any other case, will also be the same as in the last problem ; for in the triangle MNH will be given the ratio of NM to NH (as 2 to 1), and the included angle MNH equal to 126% 120°, 11:3^°, or 108°, i^c. according as the number of circles is 5, 6, 8, or 10, ^c. from which the angle MHN will be given; then in the triangle CMH will be given all the angles, and the side CH, to find CM, PROBLEM XXXrX. The perimeter of a right-angled triangle^ whose sides are in geometrical progression^ being given^ to describe the triangle. CONSTRUCTION. Upon AC, equal to the giv.. n perimeter, describe the semicircle ABC, and let AC be divided in D, ac- cording to extreme and mean prcportion ? make DB perpendicular to AC, meeting the periphery of the Geometrical Problems. 351 circle in B, and, having joined A, B, and C, B, let AE and CE be drawn to bisect the angles BAC, BCA ; and, from point of intersec- tion E, let EF and EG be drawn pa- rallel to BA and BC, cutting AC in F and G ; then will EFG be the triangle that was to be constructed. DEMONSTRATION. Since (by constructioii) AC : AD : : AD : DC ; there- fore is AC^ : AC X AD :: AC X AD : AC X DC {by Etic. 1. 6.),«or ACy : ABq : : ABq : BC<7 (^by cor. to Euc. 8. 6.), and consequently AC : AB : : AB : BC ; whence, the triangles ABC, FEG, being equiangular, FG : FE : : FE : EG. Also, EF is = AF, because the angle FEA (= EAB) = FAE ; and in the very same man- ner is EG = GC ; therefore EF + FG + EG (= AF + FG + GC) = AC. Moreover, the angle FEG (= ABC) is a right angle, by Euc. 31.3. ^ E. D. Method of Calculation, Becau se [by con struction) AD (= x/^ACy — |AC) =; AC X V4— 2. thence is AB ( \/ Ao x AD) = AC X \f V T — |. and BC (VAC x CD = AD) = AC X V I — I : but, by reason of the similar tri- angles ABC, FEG, it will be, as AC -f- AB + BC • (FG + FE + EG) AC : : A C : FG : : AB : FE : BC : EG ; or as \f v'J— \ + ^ + Vj : 1 : : AC FG : : AB : FE : ; BC : EG ; whence FG, FE,and EG are given. 3j2 The Construction of PROBLEM XL. To draw a right line PQ, to touch two circles C and O, given in magnitude and position. CONSTRUCTION. Upon the line CO, joining the centers of the given circles, describe the semicircle CDO, in which inscribe CD equal to the difference of the semi-diameters CF and OE ; and from the point B, where CD produced meets the periphery BF, draw PB perpendicular to C/B ; then will BP touch both the circles. DEMONSTRATION. Join O, D, and draw O A perpendicular to PQ. The angle CDO, standing in a semicircle, is right ; therefore the angles B and A being both right ones, by construction, the angle AOD must also be right, and the figure DOAB a rectangle, and consequently AO = BD = BC— CD = CF — CD = OE {by con- struction). Wherefore, seeing CB and OA are respec- tively equal to CF and OE, and both the angles A and B right ones, it is evident that the right line PQ touches both the circles. ^. E. D. The numerical solution of this problem is extremely easy ; for since the two sides CO and CD of the right- angled triangle CDO are both given, the angles DCO and AOC, determining the points of contact B and A, are from thence given, at one operation. But if it be required to draw a right line [ab) to touch both circles, and to pass between the centers C Geometrical Problems* 353 and O ; then, instead of taking CD equal to the differ- cnce of the semi-diameters CF, OE, let Cd be taken equal to their sum, and the rest of the process will be exactly the same, PROBLEM XLL To draw a right line AD through two circles GAEF, HCSR, given in magnitude and position^ so as to cut off* segments thereof^ AKBm, CTDn, equal respectively to two given segments EQFa, SPR^. CONSTRUCTION. Upon the subtenses EF, SR, from the centers G and H, let fall the perpendiculars GQ and HP ; and from /72r the same centers, at the distances GQ, HP, let two cir- cles GQK, HPT be described ; then draw a right line AD to touch both these circles, by the last proposition, and the thing is done ; for the lines FE, AB being at the same distance from the center G, the segments cut off by them must consequently be equal : and, in like manner, the segments SPR6, CTDw, are also equaL PROBLEM XLIL To describe the circumference of a circle through a given point P, to touch two right lines AB, AC^ given in position. CONSTRUCTION. Join A, P, and bisect the angle B AC, with the right line AK, and, from any point Q in that line, draw QT 2Z 354 The Construction of perpendicular to AC ; then, from Q to AP, draw QS = QT \ draw, likewise, PO parallel to SQ, meeting AK in O ; and from O, as a center, with the radius OP, de- scribe the circle PKF, and the thing is done. DEMONSTRATION. Let OH be perpendicular to AC, and OW to AB ; then, by reason of the parallel lines, it will be QS : OP (: : AQ : AO) : : QT : OH ; whence, as QT = QS^ OH will be = OP ; and therefore the circumference PKF will pass through the point H, and so, AHO be- ing a right angle, AC must touch the circle in that point. Moreover, the triangles AOH and AOW being equiangular, and having one side common, O W will there- fore be = OH, and the circle also touch AB in the point W. ^. E. D. Mf^thod of Calculation, Having assumed AQ at pleasure, there will be given, in the triangle AQT, all the angles and one side, whence QT (= QS) will be given: then, in the triangle AQS, will be given AQ, QS, and the angle QAS, whence the angle AQS (= AOP) will be given. Lastly, in the tri- angle AOP will be given all the angles and the side AP, whence AO and PO will be given. Othertvise. Say>asthe sine of OAH : radius (: ; OH : OA : : Geometrical Problems. ^5$ OP : OA) : : sine of OAP : sine of OP A ; then, in the* triangle AOP will be given all the angles and the side AP ; whence the other sides AO and OP will be found. PROBLEM XLIII. To describe the circumference of a circle through two giv- en points D, G, to touch a right line AB, given in position. CONSTRUCTION, Draw DG, and bisect the same by the perpendicular FC, meeting AB in C ; join C, D, and make FP perpendi- cular to AB ; and, from F to CD, pro- duced, draw FS = FP; make DH pa- rallel to FS, and from H, the inter- section of CF and j^ ^PT C DH, with the radius DH, describe the circle HDQ, and the thing is done. DEMONSTRATION. Join H, G, and draw HT parallel to FP, meeting AB in T ; then, because of the parallel lines, it will be, FS : HD (: : CF : CH) : : FP : HT ; wherefore, as the ante- cedents FS and FP are equal, the consequents HD and HT must likewise be equal ; and, therefore, since HT is perpendicular to AB, the circumference of the circle will touch AB in T ; and it will also pass through the point G, because the tw^o triangles DFH, GFH, having two sides and the included angles respectively equal, are equal in every respect. ^ E, D. Method of Calculation. The angle FCA, and the numbers expressing FC and DG being given, in the triangle CFD will be given (besides the right angle) both the legs CF and FD, whence CD and the angle FCD will be known ; 356 The Construction of then it will be, as the sine of FC A (TCH) : radius (: : TH : CH : : DH : CH) : : sine of HCD : the sine of CDH ; therefore in the triangle HCD there will be given all the angles and the side CD ; whence CH and HD will be known. PROBLEM XLIV. Having given AB, and also AD and ^G^ perpendicular to AB ; X.0 find a point T in AB, to which^ if txuo right lines DT, GT be drawn^ the angle DTG, formed by those lines^ shall be the greatest possible* CONSTRUCTION. Describe, by the last problem, a circle GDQ, that shall pass through G and D, and touch AB, and the point of contact T will be the point required. DEMONSTRATION. Join G, T, and D, T ; and from any other point R, in the line AB, draw RG and RD ; also, from the point Q, where GR cuts the circle, draw QD ; then, the angle GQD, being exter- nal with regard to the tri- angle DQR, will be greater than GRD ; therefore GTD, standing in the same segment with GQD, will be also great- er than GRD. ^ E. D. Method of Calculation. DfRw DE parallel to AB ; then in the triangle GDE will be given DE, EG ( = BG — AD), and the right angle DEG, whence the other angles EDG, EGD, and the side DG will be found; then in the triangle CFP, similar to GDE, will be given all the angles and the side FP /= AD4-BG\ ^^i^^j^^^ PC ^iii bg gi^en; from which, by proceeding as in the last problem, all the rest will be found. AK, PT CB Gesmetrical Problems* 357 PROBLEM XLV. To describe a circle^ which shall touch two right lines AB, AC, g-iven in position^ and also another circle O, g'iven in magnitude and position* CONSTRUCTION. Let the angle CAB, made by the concourse of the two lines, be bisected by AK ; and, from any point P in this line, let fall PQ perpendicular to, AB, which produce to R, so that QR may be equal to the semi-diameter of the given circle; and through R, parallel to AB, draw HM, meeting KA produced in H ; draw HO, to which, from P, draw Pv = PR, and draw OE parallel to Fv^ meeting AK in E, and cutting the periphery of the given circle in r ; lastly, from E, with the radius Er, describe the circle ErKN, and the thing is done. DEMONSTRATION. Draw EG pei-pendicular to HM, cutting AB in F; then, by reason of the parallel lines, PR : EG ( : : HP : HE) : : Fv : EO ; therefore, PR being = Pz; {by construction)^ EG and EO must likewise be equal ; from which the equal quantities FG and Or being taken away, the remainders EF and Er will be equal; and 358 "She Construction of therefore the circumference rKN passes through F ; but it also touches AB in that point, because EF {by construe- tioii) is perpendicular to AB ; it likewise touches AC, be- cause AE bisects the angle BAC ; lastly, it touches the circle O, because the right line OE, joining the centers O and E, passes through the point r, common to both pe- ripheries. Method of Calculation Supposing AO drawn, and AS perpendicular to HM, in the triangle AHS (besides the right angle) will be given AS (= rO) and the angle AHS (= EAF = |BAC), whence AH will be known; then in the tri- angle AHO will be given AH, AO, and the included angle, whence 'AHO and HO will also be given : then it will be, as the sine of EHG is to the radius ( ; : EG : EH : : EG : EH) so is the sine of EHG to the sine of EGH ; therefore in the triangle HEG will be given all the angles and the side HO ; whence EG and EH are known also. PROBLEM XLVI. To describe the circumference of a circle through a given point P, so as to have given parts cut ojfby two right lines AB, hXL given In position* CONSTRUCTION. Let the arcs to be cut off by AC and AB be similar respectively to the arcs ab^ be of any given circle abcq^ whose chords ab^ be subtend, at the center, any given angles aqb^ bqc. Let the angle ^Z>c be bisected by bdy take, in AB and AC, any two pointy E, D, equi- distant from A , and, having drawn DE, make the an- gle EDF = qbd, CDR = qba, and BFR = qbc ; then from the intersection R of the lines DR and FR, with the radius RD, describe an arch mS;z, cutting the line AP in S, draw RS and ARK, and also PQ, parallel to • RS, meeting AK in Q ; then from the center Q, with the radius PQ, describe the circle KPI, and the thing is done. Gejomctricdl Problefns* 359 DEMONSTRATION. Draw QH and QG parallel to RF and RD, meeting AB and AC in H and G. The angles BED and CDE being equal, BFD will exceed CDF by twice EDF, or by twice qbd^ that is, by as much as gbc exceeds qba^ or, lastly, by as much as BFR exceeds CDR ; therefore, seeing the whole angle BFD as much exceeds the whole angle CDF, as the part BFR of the former exceeds the part CDR of the latter, the remaining parts RFD and RDF must be equal, and consequently FR = RD = RS» But, by reason of the parallel lines, it will be, RF : QH : : RD : QG : : RS : QP ; whence the antecedents RF, RD, RS, being equal, the consequents QH, QG, QP, must be also equal, and the circumference must pass through the points H and G ; whence the solution is manifest. Method of Calculation. If two perpendiculars be conceived to fall from Q upon AB and AC, they will, it is plain, be in the given ratio of the sines of the angles QHI and QGL ; there- fore the position of the line AQK will be given {fro?n 360 The Construction of prob. 26) by saying, as the sum of the said sines is to their diffe rence, so is the t angent of half BAC to the tangent of halfBAQ— CAQ. Again, it will be as sine QAH : sine QH A (: : QH : QA : : QP : QA) : : sine QAP : sine QPA ; therefore, in the triangle AQP, are given all the angles and one side AP, whence AQ and PQ will be found. PROBLEM XLVIL Having the three perpendiculars^ let fall from the angles of a plane triangle on the opposite sides ^ equal to three given right lines K^, L/, and Mw, to describe the triangle* CONSTRUCTION. Draw the indefinite right line RS, in which take AB equal to O : find a fourth proportional to M;w, L/, and K^, with which, as a radius, from the center A, let an arch rCs be described; and from B, with "the ra- dius L/, let another arch be described intersecting the for- mer in C ; join A, C, and B, C, and upon RS let fall the perpendicular QC, in which, produced, take QP = L/, and draw PF parallel to RS, meeting AC, produced,^ in F, draw FG parallel to CB, and AFG will be the triangle required. DEMONSTRATION. Draw FE, Gjr, and Av perpendicular to the three sides ©f the triangle. Geometrical Problems. 361 The triangles ABC, AGF ; AFE, AG^ ; and GFE, AGz;, are equiangular, by construction; therefore Gq : FE : : AG : AF : : AB (K>^) : AC i^-^) - • Mm : L/ ; whence, as the consequents FE and hi are equal, by construction^ the antecedents Gq and Mm must be equal likewise. Again, BC (L/) : AB (Ky^j (: : FG : AG) : : FE (L/) : A^ ; and consequently Kk = Kv. 4 E. D. Method of Calculation. Since K/f, L/, and Mm afe given, AC f = — =^-- — I will be known ; then in the triangle ABC will be given all the three sides, whence the angles are known ; lastly, in the triangle AFG will be given all the angles and the perpendicular EF, whence the sides are also known. PROBLEM XLVIII. The position of three points^ in the same right line^ being given^ it is proposed to find a fourth^ where lines^ drawn from the former three^ shall make given angles with each other. CONSTRUCTION. Let the three given points be A, B, and C : make the angles ACE and C AE respectively equal to the given angles which the lines drawn from B, A, and B, C are to make ; and let AE and CE meet in E ; through A, C, and E, let the circumference of a circle AE'CD be described, and, through E and B, draw EBD, meeting it in D, then will D be the point re- quired. 3 A 36^ The Construction of DEMONSTRATION. Join A, D, and C, D. The angle EDA is equal to ACE, standing on the same segment j and for the like reason is EDC = CAE. 4 E. D. Method of Calculation. In the triangle ACE are given all the angles and the side AC, whence AE will be given ; then in the triangle ABE will be given the two sides AE, AB, and the in- cluded angle ; whence ABE and all the rest of the angles in the figure will be given. D A, PROBLEM XLIX. Three points^ A, B, C, being any how given ; to find a fourth^ where lines^ drawn from the former three^ shall make given angles with one another. CONSTRUCTION. Join the given points, and upon the right line AB describe a segment of a circle, capable of the given angle which that line is to subtend ; com- plete the circle, produce BA, and make the angle DAQ equal to the angle which BC is to subtend, and let AQ meet the pe- riphery in Q ; draw QC, cutting the same periphery in P ; join A, P, and B, P, and the thing is done. DEMONSTRATION. The angle ABP is equal to the given angle which AB was to subtend {by construction) ; and the angles QAB and QP A, standing upon the same segment, being equal to each other, their supplements DAQ and BPC must like- wise be equal, i^. E. D. Geometrical Problems* 363 Method of Calculation. Join B, Q ; then, in the triangle ABQ will be given all the angles and the side AB, whence BQ and ABQ will be known ; then in the triangle CBQ will be given two sides, and the included angle CBQ; whence the angle CQB, equal to BAP, will be known ; lastly, in the triangle APB will be given all the angles and the side AB, from which AP and BP will be found. PROBLEM L. To draw a right line EG through a circle O, given in magnitude and position^ which shall also cut a right line Q^C^ given in position^ in a given angle^ and have its parts EF, FG intercepted hy the circle and that right line^ in the given ratio of the two right lines ah and he. CONSTRUCTION. At any point B, in the right line QC, make the angle QBA equal to the given angle, and through the center O, perpendicular to BA, draw DQ meeting BA in R, and CG in Q ; bisect ah in d^ and in RB take R^ = bd^ and pq = ^c, and draw pm and qn paral- lel to DQ ; from the point 72, v/here qn intersects QC, draw TzL parallel to BA, meeting pm in m ; through the points Q and m draw QmF, cutting the periphery of the circle in F, and through F, pa- rallel to BA, draw EFG, and the thing is done. DEMONSTRATION. The lines GE, BA, and nL, being parallel, the an- 364 The Construction of gles QGE, QB A, &>V. will be equal, and likewise SF : FG : : Lm : mn ; but Lw {by construction') is ( = "Rp) = dh^ and mn (=pq) == be ; therefore SF : FG : i db : be, and consequently EF (2SF) : FG : : ab (2bd) : be. Method of Calculation* L,7i {dc^ : Lw {db) : : the tangent of LQ/z (the comple- ment of the given angle QBR) : the tangent of LQw ; therefore, in the triangle OQF will be given one angle OQF and two sides, QO, FO ; whence, not only the angle SOF, but also SO and SF will be known. PROBLEM LI. To apply, or inscribe^ a given right line AD betrveen the peripheries of two circles C and O, given in magnitude and position, so as to be inclined to the right line CO, joining the centers, in a given angle, CONSTRUCTION. Make OCB equal to the given angle, and let CB be taken equal to the given line ; upon the center B, with the radius of the circle C, let the arch nDm be describedy ^J>^ cutting the circle O in D : then draw BD, and, parallel thereto, draw CA, meeting the periphery in A ; join A, D, and the thing is done. DEMONSTRATION. Because {by construction^ CA and BD are equal and parallel, there iore will AD and CB be also equal and pa- rallel {by £uc. 33. 1.). ^ £. D. Geometrical Problems'^ 365 Method of Calculation* In the triangle CBO are given two sides, CO 'and CB, and the angle OCB ; whence OB and the angle COB will be known ; then in the triangle OBD will be given all the three sides, whence all the angles, and consequently DOC^ will also be known. PROBLEM LIL From a given rectangle ABCD, to cut off a gnomon ECG, zvhose breadth shall be every where the same^ and whose area shall be just half that of the rectangle* CONSTRUCTION. In BA take BH equal to BC, or AD ; and in DA, produced, take AP a mean propor- tional between B A and |AD (so that AP2 may = the given area AGFE). From P to the middle of AH draw PO ; make OE = OP, and DG = BE ; complete the rectangle EAGF, and the thing is done. DEMONSTRATION. If the semicircle EPQ, from the center 0,be described, it is plain that AQ = EH = BH — BE = AD — DG = AG ; and consequently that AE x AG = AE X AQ = AP2 {Euc. 13. 6.) ^ E. D. (= Method of Calculation, In the right-angled triangle AOP are eriven AO AB — B C\ -j and AP (= V^AB x BC) ; whence OP will be known, and from thence both AE and AG. 366 The Construction of PROBLEM LIIL Three points^ A, B, C, being given j it is proposed to find a fourth^ Y^from xvhence lines^ drawn to the three former^ shall obtain the ratio of three given lines a, b^ and c^ respec- tively. CONSTRUCTION. Having joined the given points, take AF, in AB, equal to «, and AI = Cj also make the angles AFG C^ and A IK equal, each, to ACB ; and from the cen- ters F and G, with the radii b and AK respec- tively, let two arcs be described intersecting in H; from which point draw HF and HA ; then draw BP to make the angle ABP = AHF, and it will meet AH (produced) in the point P, required. DEMONSTRATION. Let BP, CP, and GH be drawn. The triangles ABP, AHF being equiangular {by constructio7i\ it will be AP : BP : : AF (a) : FH (b) ; also AB : AP : : AH : AF ; and AB : AC : : AG : AF (because ABC and AGF are likewise equiangular) ; whence it is evident, since the extremes of the two last proportions are the same, that AP x AH = AC X AG, or AC : AP : : AH : AG ; therefore the triangles ACP, AHG being equian- gular (Euc. 6. 6.), we have AP : CP : : AG : GH (AK) : : AF (a) : AI (c). ^ E. D. Method of Calculation. In the triangles AFG, AIK are given all the angles and the sides AF and AI, whence AG, FG, and AK (GH) will be found ; then in the triangle FGH will be given all the sides, to find the angle HFG ; which, added to AFG, gives AFH (APB), from whence, and the two given sides AF and FH including it, every thing else is readily determined. Geometrical Problems. 367 PROBLEM LIV. To describe a triangle (ABC) similar to a given one AMN, such that three lines (AP, BP, CP) may be drawn from its angidar points to meet the same point (Pj, so as to be equal to three given lines AD, AF, and AK, respec- tivelij* CONSTRUCTION. Draw DE and KG, making the angles ADE and AKG, each, equal to the given angle N, and intersecting AN in E and G ; from the centers D and E, with the intervals AF and AG, let two arcs be described, intersecting in H ; draw AH, in which take AP = AD; andfrom.P, to AM and AN, apply PB A. and PC equal, respectively, to AF and AK, and let B, C, be joined ; so shall ABC be the triangle that was to be de- termined, DEMONSTRATION. The three lines AP, BP, CP are, respectively, equal to the three given lines AD, AF, AK, by con^aiction ; we therefore have only to prove that the triangle ABC is similar to the given one AMN. Now, supposing DH and EH to be drawn, it will be AP : PC (or AD : AK) : : AE : AG (EH) ; whence the triangles APC and AHE will be equiangular {Euc* 6. 6.), and consequently AC : AH : : AP (AD) : AE : : AN : AM {Euc. 5. 6.) ; but the triangles ABP and ADH (having AP = AD, PB = DH (^2/ construction)^ and the angle DAP common) are equal in all respects ; therefore, by substituting AB in the room of AH, our last proportion becomes AC : AB : ; AN : AM ; whence it is manifest that the triangles ABC and AMN are equiangular. ^. E. D. 368 The Construction of Method of Calculation* In the triangles ADE, AKG are given all the an- gles and the sides AD and AK, from which AE, DE» and AG will be known ; then in the triangle DHE will be given all the sides, to find the angle EDH> which added to ADE gives ADH ; from whence, and the two given sides including it, AH (= AB) will be known. PROBLEM LV. In the triangle ace^ besides the angle c, are given the seg- ments of the sides ab and de^ and the angles aeb anddbe sub- tended thereby; to describe the triangle* CONSTRUCTION. Upon AB, equal to ab^ let a segment of a circle be described to contain an angle equal to aeb : make the angle ABF = ace^ BA?2 = dbe^ and the line BF = ^f/; from the point n^ where A/z cuts the periphery of the circle, through F, draw nFE, meeting the periphery in E ; join A, E, and B, E, and draw EC parallel to BF, meeting AB, produced, in C ; and then the thing is done. DEMONSTRATION. Let BD be parallel to FE. Since the lines BD, EF, and ED, FB are parallel, therefore is ED = BF ( = ed)^ and the angle ACE also = ABF {ace) £uc. 28. 1. Moreover, the angle BEn Geometrical Problems. 369 (DBE) is equal to BAn (dbe\ both standing upon the same segment Bn. ^. E. I). Method of Calculation. Join B, n ; then in the triangle AB^i will be given all the angles and the side AB, whence B/z will be known ; then in the triangle nBF will be given B/z, BF, and the included angle /zBF, whence BF;2 (CDB) and all the rest of the angles in the figure will be known. PROBLEM LVI. To make a trapezium^ whose diagonals^ and two oppO' site sides ^ shall be all of given lengths^ and whereof the an- gle formed by the given sides ^ when produced till they meet^ shall also be given. CONSTRUCTION, Draw the indefinite right line AC, and take therein AB equal to one of the two given sides ; make the angle CBG equal to the given an- q^ C\ gle,andletBG t"^ be made equal \ to the other \ given side; up- *5 on the centers A and G, with intervals equal to the two dia- gonals, let two arches be de- scribed, cut- ting each other in D ; make DE equal and parallel to GB ; join D, B, and E, A : then ABDE will be the trapezium required. DEMONSTRATION. Draw DG, DA, and BE, and let BA and DE be pro- duced to meet each other in F. The lines BG and DE are equal and parallel, by- construction J therefore BE is = DG, which last (by 3 B ^fO The Construction of construction) is equal to one of the given diagofials, as AD is equal to the other ; moreover, the sides AB and ED (BG) are equal to the given sides, by construction ; and the angle F is equal to the given angle CBG, because DF is parallel to GB. ^ E. D. Method of Calculation, Suppose AG to be drawn ; then in the triangle ABG will be given the two sides BA and BG, and the included angle ABG ; whence the side AG and the other two angles will be known ; then in the triangle ADG will be given all the sides ; whence the angle AGD will be known, and from thence the whole angle BGD ; lastly, in the triangle BGD wuU be given the two sides BG and GD, and the included angle BGD ; whence the side BD wiU likewise be known. PROBLEM LVIL The segments of the base AD, DB, and the line DC, bisecting the vertical angle ACB of a plane triangle^ being given^ to describe the triangle* CONSTRUCTION. In AB, produced, take DO to AD, as DB to AD --^ r — ^^' ^^^ from the cen- ter O, with the radius OD, de- scribe the cir- cle DCQ ; also from the center D, at the given distance DC, describe the circle mCn^ and from C, the intersection of the two circles, draw CA and CB, and the thing is done. DEMONSTRATION. Since DO : AD : : DB : AD — DB ; therefore (bij the lemma in p. 334.) AC : CB : : AD : DB ; whence CD bisects the angle ACB {bij Euc. 3. 6. j. ^. E. D. Geometrical Problems, 371 Method of Calculation^ Draw CP perpendicular to AQ. Because, by construction, OD is = -r-— tt^; ' ^ AD — BD Aerefore will DQ = ^^ ; whence, by reason AU — — JljJL) of the similar triangles DCQ, D PC, it wil l be, as 2ADx BD DC AD — BD whence AC and CB are given DC : DP = A D — BD X D C^ 2AD X BD PROBLEM LVIIL Having given the base^ the angle at the vertex^ and the Une drawn from thence to bisect the base ; to construct the •triangle. CONSTRUCTION. Upon the given base AB describe {by prob. 4.) a seg- ment of a circle ADB ca- pable of the given angle ; and, from the point F, in which the perpendicular DF bisects AB, with a ra- dius FC equal to the bisect- ing line, describe nCm^ cut- ting the periphery ACB in C ; join A, C, and B, C, and the thmg is done. The demonstration of which is evident from the construction. Method of Calculation, From the center O let OA and OC be drawn; then in the triangle AOF will be given all the angles and the side AF ; whence FO and OC (OA) will be known, and in the triangle CFO will be given all the sides ; whence the angle FOC, and its supplement DOC, ex- pressing the difference of the angles at the base, will also be known. 372 The Construttion of PROBLEM LIX. The base J the difference of the angles at the base, and the line drawn from the vertical angle to bisect the base of any plane triangle^ being given ; to describe the triangle* CONSTRUCTION. Upon AB, equal to the given base, let a segment of a circle AHEB be described to contain an angle equal to the difference of the JS. ^\ angles at the base ; bisect AB in C, and take CD to AC in the duplicate ratio of AC to the given bisecting line KL ; make CS. and DI perpendicular to AB, cutting the circle in S and I ; draw A I, cutting CS in G ; and through G draw the chord EGH paral- lel to AB ; join A, E, and A, H, and in AI take AN equal to KL; draw MNP parallel to EH, meeting AE and AH in M and P ; then will AMP be the triangle which was to be constructed. DEMONSTRATION. Since (by construction) CG is parallel to DI, and KL^ : AC^ : \ AC : CD ; therefore {Eiic. 4. 6.) KL^ : ACq : : AG : GI : : AGy : GI X AG ; but GI x AG = EG X GH = EG^ {Euc. 35. 3. and 3. 3.) ; therefore KL^ : AC^ : ; AG^ : EG^ ; and consequently KL : AC : : AG : EG : : AN : NM ; but AN is {by con- struction) equal to KL, therefore NM is = AC, and consequently MP (2MN) = AB. Moreover, the dif- ference of the angles at the base, P — M, is ( = AHE — AEH) = AEB ; which {by construction^ is equal to the difference given. ^. E. D, Method of Calculation. From the center O draw OA and OI ; also draw li> parallel to EH, meeting OS in v : then it will be {by Geometrical Problems* 373 Construction) as KLy : AC^ ( : : AC : Iv) : : the sine of AOC or AEB, the given difference of the angles at the base, to the sine of SOI, which, added to AOS, gives AOI, whose supplement, divided by 2, will be OIG ; from whence OGI and its supplement OGA are given ; and consequendy ANM (equal to AGE) ; then in the triangle ANM will be given AN, NM, and the includ- ed angle ANM, whence the angles M, A, P will also be given. PROBLEM LX. The perpendicular^ the angle at the vertex^ and the sum of the three sides of a triangle being given ; to describe the triangle. CONSTRUCTION. Make AB equal to the sum of the sides, which bi- sect in P, making PO perpendicular to AB, and the angle PAO equal to half the given angle at the ver- tex ; from the center O, with the radius OA describe the cir- cle AHB,andinOP, produced, take PK equal to the given per- pendicular, and draw KH parallel to BA, cutting the circle in H; join A, H, and B, H, and make the angles BHF and AHE equal to HBF and HAE, respectively ; then will EHF be the triangle required. DEMONSTRATION. Join O, B, and O, H, and draw HQ perpendicular to AB. The angle EFH is = BHF + HBF == 2HBF {by construction) = HOA {Euc. 20. 3.) : and, in the same manner, is FEH = HOB ; hence it follows that EFH + FEH (= HOA + HOB) = AOB ; and, by taking each of these equal quantities from two right angles, 374 The Construction of we have EHF = OAB + OBA {Euc. 32. 1.) = 20AB ^ == the given angle {by construction^ • Moreover^ QH is = PK — the given perpendicular; and, EH being = AE, and FH = BF {Euc. 6- !.)> EH + HF + EF will therefore be = AB = the given sum of the sides* ^ ^- ^- . Method of Calculation. In the triangle AOP are given all the angles and the side AP, whence OP and AG (HO) are known : then in the triangle OHK will be given the sides OH and OK (OP + PK), whence HK w^ill be given ; next, in the triangle BQH will be given QH and BQ (BP — HK), whence QBH, and its double QFH, w^ill be given ; lastly, in the triangle EFH are given all the angles and the per- pendicular Qi-i, w^hen'ce the sides will also be given. But the answer may be more easily brought out, by first finding HOK, the difference of the angles ABH and BAH, as in the fifth problem. PROBLEM LXI. The sum of the three sides^ the difference of the angles at the base^ and the length of the line bisecting the vertical angle of any plane triangle being given ; to describe the tri- angle. CONSTRUCTION. Make AB equal to the sum of the sides, wliich bisect in E by the perpendicular DE^z, and make the angle yzE?" equal to half the given difference of the angles at the base, taking Er equal to the line bisecting the vertical angle : through r draw Cnr parallel to AB, cutting DE7Z in n ; draw tzA, to which draw Em = Er, and draw AD parallel to Ew, meeting nED in D ; and on the center D, at the distance Geometrical Problems* 375 of DA, describe the circle ACB, Cutting rnC in C ; join A, C, and B, C, and make the angle BCF = CBF ; also make ACG = CAG, and let CF and CO meet AB in F and G ; then will FCG be the triangle that was to be described. DEMONSTRATION. Upon AB let fall the perpendicular CP ; let CQ bisect the vertical angle GCF, and let DH be drawn parallel to Er, meeting Cr in H. Then, by reason of the paral- lel lines, it will be as Er : DH (: : En : D;i) : : Ew : DA 5 whence, Er being = Ew (by constructwii)^ DH and DA are also equal, and the point H falls in the periphery of the circle : therefore the angle ;?DH (nEr) at the cen- ter, standing upon half the arch HC, will be equal to the angle HAC, at the periphery, standing upon that whole arch, that is, equal to the difference of the angles ABC and BAG ; but the angle GFC being double to ABC, and FGC double to BAG {by construction)^ the differ^ ence of GFC and FGC will be double to the difference between ABC and BAG, and therefore equal to 2;zEr (2?zDH}, the difference given. Moreover, because GCQ = FCQ, 2PCQ will be the difference between PCG and PCF, which must likewise be equal to 2;2Er, the differ- ence of their complements PGC and PFC ; whence PCQ = ;zEr, and consequently CQ = Er. Furthermore, since the angle ACG = CAG, and BCF = CBF, thence will CG = AG, and CF = FB 5 and therefore CG + GF + FC = AB. ^ E. D. Method of Calculation. In the triangle E/zr are given all the angles and the side Er, whence E;z will be given ; next, in the triangle AE?i will be given (besides the right angle) both the legs E/z and EA, whence the angle EwA is given ; then it will be, as the radius to the sine of DH/z or Er;z (: : DH : D;z : : DA : D;z) so is the sine of D72A to the sine of DA?? ; whence AD;z, the supplement of ACB, is also given, from which all the rest of the angles in the figure are given by addition and fiubtractioa onl/. 276 The Construction of This method of solving the problem, it may be observ- ed, requires three operations by the sines and tangents ; but the same thing may be performed by two proportions only ; for as Er : AE : : the secant of rEn : the tangent of En A J whence all the rest will be found as above. 9 PROBLEM LXII. • To reduce a given triangle into the form of another^ or to make a triangle xvhich shall be similar to one triangle^ and equal to another. CONSTRUCTION. Upon the base AB of the triangle ABC, to which you would make another triangle equal, describe ADB similar to the trian- D gle required; draw CF pa- rallel to AB, meeting AD inF;takeAE a mean pro- portional be- tween AD and AF ; and, pa- rallel to DB, G- B draw EG ; then will AGE be the triangle that was to be constructed. DEMONSTRATION. Let FR and DQ be perpendicular to AB ; then the triang. ADB : triang. ACB : : DQ : FR (schoL Euc. 1. 6.) : : AD : AF {Euc. 4. 6.) : : AD^ : AD X AF (JEwc. 1. 6.) : : AD^ : AE^ (by construction) : : triang. ADB : triang. AEG {Euc. 19. 6.). Therefore the ante- cedents of the first and last of these equal ratios being the same, the consequents ACB and AEG must necessarily be equal. ^. E. D. Method of Calculation. In the triangle ADB are given all the angles and the side AB, whence AD will be given ; next, in the tri- angle AFR will be given all the angles and the side Geometrical Problems, 377 FR (=CH), whence AF will be given ; and then AD and AF being given, AE = V AD x AF will also be given, PROBLEM LXIIL To find a point in a given triangle ABC ^ from xvhence right lines drawn to the three angular points^ shall divide the whole triangle into parts (COA, AOB, BOC) having the same ratio one to another^ as three given right lines m^ n, and p^ respectively > CONSTRUCTION. In CA and AB produced, if need be, take CE and AF, each equal to m + n +/?, joining E, B, and F, C ; take C^ = w, Ac = n, and draw eh and cf parallel to EB and CF, meet- ing the sides of the given tri- angle in h and f\ draw also hOi and /? pa- rallel to AC and AB, and ^ ^ -n T"/^ a at O, the in- /^ ^ ^ ^ ; -^ tersection of these lines, will be the point required. E DEMONSTRATION. Let ^H and BD be perpendicular to AC. The trian- gles CBE, Che^ as also CBD, C^H, are similar; there- fore, m (C^) : m + 72 +/? (CE) : : C^ : CB : : <5»H : BD : : the triangle AOC : triangle ABC. In the very same manner it may be proved, that the part AOB is to the whole triangle ABC, as n tom-f-n-f/^> whence it fol- lows, that the remaining part BOC must be to the whole triangle, as /? to m -f- 72 -|- /? ; therefore these parts are to one another in the given ratio of ?n, n^ and /?. ^. £. D. 3C 378 The Construction of Method of Calculation. ^"^^^ lm + 7z+/^:n :: AC: A/(QO), both AQ and QO will be given from thence ; then, in the triangle AOQ, will be given two sides, and the in- cluded angle, from which every thing else will be known. PROBLEM LXIV^ To divide a given trapezium ABCD, -whose opposite sides AB, CD are parallel^ according to a given ratio^ by a right line QN, passing through a given point P, and fall- ing up07i the two parallel sides. CONSTRUCTION. Bisect AD in G, ■j^T W C X and draw GH pa- rallel to AB (or DC), meeting BC in H : then divide GH in M, accord- ing to the given ratio, and through M draw PQN, and the thing is done. DEMONSTRATION. Draw EMF and IHK parallel to AD, meeting DC and AB in E, I, K, and F. Because of the parallel lines, we have GD = ME = HI, and AG = FM = KH ; whence, as GD is = AG (by construction), ME will be = FM, and HI = HK ; and the triangle EMN will be = FMQ, and IHC = BHK (^Euc. 4. 1.) ; whence it appears that the trapezium AQND is also equal to the parallelogram DF, and the tra- pezium QBCN equal to the parallelogram FI ; but these parallelograms are to one another as their bases, or as GM to MH (^Euc. 1.6.); therefore GM : MH : : ANQD : QBCN. ^E.D. Method of Calcidatioiu Whereas AB and DC are parallel, GH is an arith- metical mean between them, and therefore equal to half Geometrical Problems* o79 their sum» Therefore, as the whole line GH and the ra- tio of its parts GM, MH are given, the parts themselves will also be given. PROBLEM LXV. To cut off from a given trapezium ABCD, -whose vp- posite sides AB^ CD are parallel^ a part AQND equal to a rectangle given^ by a right line passing through a given point P, andfallifig upon the two parallel sides, ij^ee the figure to the last problem. CONSTRUCTION. Bisect AD in G, and draw GH parallel to AB ; vipon AD {by Euc. 45. 1.) describe the parallelogram ADEF equal to the rectangle given, and through the intersec- tion of GH and EF draw PQN, and the thing is done. The demonstration whereof is manifest from the preceding problem. PROBLEM LXVL To divide a given trapeziu7n ABCD, whose sides AB and DC are parallel^ into tzvo equal parts^ by a rights line parallel to those sides. CONSTRUCTION. Produce AD ^. and BC till they 7 \ meet in H, and / \ make AG equal / \ . and perpendicular / ^\ toHD;drawHG, T^/ E/^ \ andbisect the same / "^'X/N^ with the perpendi- /p. ^x^'n \ cular PQ = HP ; //\ "^"-AX \ join H, Q, and in // J\ ni\' \ HA take HE, A^---x ^^LX A equal to HQ, and "^ *- , ^ "^ parallel to AB draw EF, and the thing is done. DEMONSTRATION. Since HE^ (= HQ^ ^ hP^ + PQ^ = 2HP2 t=r: HG^ HA2 + AG^ HA2 + HD^ . -^ = = 1 ) IS an anth- 380 The Construetion of metical mean between HA^ and HD% it is evident that the triangle HEF will also be an arithmetical mean be- tween the triangles H AB and HDC (or ABFE = EFCD) ; because those triangles, being similar, are to one another as (HE% HA^, HD^) the squares of their homologous sides. 4L' -^- ^* Method of Calculation. Since all the sides and angles of the trapezium are sup- posed given, the side CD and all the angles of the trian- gle HDC will be given; therefore HD and AH will be given. But the same thing may be had without the an- gles ; for, since DC is parallel to AB, we have AB — DC : AD : : DC : HD ; whence HE will be given, as be- fore. PROBLEM LXVII. known : whence HE, firiD^ 4- HA2 will also be To divide a given trapezium ABCD according to a given ratio ^ by a right line LH cutting the opposite sides AC, BD in given angles, CONSTRUCTION. Produce the said opposite sides till they meet in E ; draw AD, and BX CF parallel to it, meeting BE in F ; divide BF in G, accord- ing to the given ratio ; and, having made EAK equal to the given angle which LH is to make wi-h AC, take EH a mean proportional between EG and EK ; tiien draw HL parallel to AK, and the thing is done. DEMONSTRATION. By construction, EG : EH : : EH : EK : : EL : EA {Eiic. 5. 6.) ,• whence it follows that EG X EA = Geometrical Problems. 381 EH X EL^ and consequently that the triangles EHL and EAG are also equal to each other (£i(c. 15. 6.), from which taking away EDC, common, the remainders CDHL and CDGA will be likewise equal, and conse- quently ALHB = AGB, being the dift'erences between those remainders and ACDB. But the triangle ADF is = ACD, standing upon the same base AD and between the same parallels ; therefore (by adding AGD, common) AGF is also = CDGA (= CDHL) ; but AGF (CDHL) : AGB (ALHB) : : GF : GB {Euc. 1. 6.). ^ E. D. Method of Calculation. In the triangles ABE and ABK are given all the angles and the side AB, whence BE, BK, and EC will be known ; then in the triangle EFC will be given all the angles and the side CE, whence EF, and from thence FG, and EG, will be known; lastly, from the known values of EK, EG, and EF, the value of FH ( = VEG X E"K — EF) will be found. PROBLEM LXVIIL Two right li?ies AG a7id AH, 7neeti7ig in a point A, keing given by position ; it is required to draw a right line TzP to cut those lints in given angles^ so that the triangle KnV^ formed from thence^ mau be equal to a given square ABCD. CONSTRUCTION. Let the angle ABE be equal to the given angle APn, and let BE meet AG in E; drawEF perpendicu- lar to AH, make BQ equal 2EF, and upon AQdescribe the semicircle AmQ, cutting BC in rn ; draw mn parallel to AH, meeting AG in w, and n? parallel to EB, and A^P will be the triangle required. 382 The Construction of DEMONSTRATION. The triangles AEB arid A;zP, being similar, are to one another as the squares of their perpendicular heights EF and mB (/zS) : but mW is = BQ x AB = 2EF x AB therefore it will be, as the triangle AEB (EF x J-AB) the triangle A;zP : : EF^ : 2EF x AB : : EF : 2AB : EF X |AB : AB2 [Euc. 1. 6.); wherefore, the antece- dents being the same, the consequents must necessarily be equal, that is, An? = ABCD. ^. E. D. Method of Calculation. In the triangle ABE are given all the angles and the side AB, whence EF will be given, and consequently S;z (= VAB X 2EF) ; whence AP and An are also given. LEMMA. If from any point C, in one side of a plane angle KAL, a right line CB be draxvn^ cutting both sides AK, AL in equal angles (ACB, ABC); and from any other point D in the same side AK another right line be drawn^ to cut off an area ADE equal to the area ABC ; Isay^ that DE will be greater than CB. DEMONSTRATION. Complete the parallelogram DCBG, and join B, D, and in BG (produced if need be) take BF = BE, and draw FD. Since the triangles ABC, AED are equal, by suppo- sition, and have one angle. A, common, therefore will AD : AC : : AB (AC) : AE {Euc. 15. 6.), and consequently AD -f AE greater than AC -f. AB ( Euc. 25. 5. ) ; whence it is mani- fest that CD must be greater than EB,: or BG than BF. Moreover^ because the angle ABC Geometrical Problems. 383 ( = ACB = CDG) is =: CxBC, it will be greater than GBD, which is but a part of GBC ; and therefore ABD must, evidently, be greater than GBD ; wherefore, seeing BF and BE are equal, and that DB is common to both the triangles DBE, DBF, it is manifest that DE is greater than DF {Euc. 19. 1.) ; but DF is greater than jyO (by the same)^ because the angle DGF (DCBj, be- ing obtuse, is greater than GFD, which must be acute {Euc, 32. 1.) : consequently DE is greater than DG, or its equ^l CB. ^ E. B. PROBLEM LXIX. From a given polygon ABCDEF, to cut off* a given area AFEIK, by the shortest right line^ Kl^ possible. CONSTRUCTION. Let the given area to be cut off be represented by the rectangle LMNO ; and let the sides AB and DE, by which the dividing line is terminated, be produced till they meet in G ; make upon OL ^by Euc. 45. 1 .) a rectangle OQ equal to AFEG, and let a square GSTV be constituted (by Euc. 14. 2.) equal to the whole rectangle QN : bisect the angle BGD by the right line GH, and make GR perpendicular to GH ; ajid draw 384 The Construction of KI, by the last proMem, parallel to RG, so as to form the triangle KGI equal to the square GSTV, and the thing is done. DEMONSTRATION. Since, by construction, KGI ( = GSTV) = QN, let AFEG = OQ be taken away, and there will remain AFEIK = LN. Moreover, since the angle HGI is :^ HGK, and the angle IHG (HGR) a right one, the angles I and K are equal ; and therefore, by the preceding lemma, IK is the shortest right line that can possibly be drawn to cut off the same area. ^. -E, D. Method of Calculation* Let the area of the figure AFEG be found, by divid- ing it into triangles AFG, EFG, and let this area be add- ed to the given area to be cut off, and then, the square root of the sum being extracted, you will have GS the side of the square .GT ; from whence GI will be determined, as in the last problem. Note. In the same manner may a given area be cut off by a right line making any given angles with the opposite sides. PROBLEM LXX. Through a given point P, to draw a right line PED to cut txvo right lines AB, AC given in position^ so that the triangle AJ^^ ^formed from thence^ may be of a given mag^ nitude* CONSTRUCTION. Draw PFH parallel to AB, intersecting AC in F; DB Geometrical Problems. 385 and upon AF let a parallelogram AFHI be constituted equal to the given area of the triangle ; make IK perpen- dicular to A I, and equal to FP ; and, from the point K, to AB, apply KD = PH ; then draw DPE, and the thing is done. DEMONSTRATION. Supposing M to be the intersection of DF and IH, it is evident, because of the parallel lines, that all the three triangles PHM, PFE, and MDI are equiangular ; there- fore, all equiangular triangles being in proportion as the squares of their homologous sides, and the sum of the squares of PF (IK) and DI being equal to the square of PH (KD), by construction and Euc. 47. 1. it is evident that the sum of the triangles PFE and DMI is = the triangle PHM; to which equal quantities, vi Jig* 1, let AFPMI be added, so shall ADE be likewise equal to AFHI : but, infg. 2, let PFE be taken from PHM, and there will remain EFHM = DMI; to which adding AIME, .we have AFHI = ADE, as before. ^ E. D. Method of Calculation. By dividing the given area by the given height of the point P above AB, the base AI of the parallelogram AFHI will be known, and consequently PH ( = KD) ; whence DI (= VKD^ — PF^) will likewise become known. This problem, it may be observed, becomes impossible when KD (PH) is less than KI (PF) ; which can only happen, in case 1, when the given area is less than a pa- rallelogram under AF and FP. PROBLEM LXXI. To cut off from a given polygon BCIFGH, a part EDBHG equal to a given rectangle KL, by a right line ^D passing through a given point P. CONSTRUCTION. Let the sides of the polygon CB and FG, which the dividing Ime ED falls upon, be produced, till they meet in A ; upon ML (by Euc. 45. 1.) make the rectangle 3D 386 The Construction of MN equal to AGHB, and, by the last problem, let ED be so drawn through the given point P, that the triangle A ED, formed from thence, may be equal to the whole li s rectangle KN ; then will EDBHG be equal to KL : for since AED is = KN, let the equal quantities AGHB and MN be taken away, and there will remain EDBHG = KL. Method of Calculation* Let the area of the figure AGHB be found, by divid- mg it into triangles, and l^t this area be added to the area given, and the sum will be equal to the area AED, or the rectangle KN ; from whence AD will be found, as in the last problem. PROBLEM LXXIL HaviJig the base^ the vertical angle^ and the length of the line bisecting that angle and terminating in the base^ to describe the triangle. i CONSTRUCTION. Upon the given base AB let a segment of a circle ACB be described [by problem 4) to contain the given angle, and, having completed the whole circle, from O, the center thereof, perpendicular to AB, let the ra- dius OE be drawn j draw EB, and make BG perpeit- Geometrical Problems, 385 dicular thereto, and equal to half the given bisecting line, and from G, as ^ -.s^ r; a center, with the ra- dius GB, let a circle BHF be described, intersecting EG (when drawn) in F and H ; from E to ABdrawED = EF, and let the same be produced to meet the circumference in C ; join A, C, and B, C ; so shall ABC be the triangle required. DEMONSTRATION. The triangles CBE and BDE are similar, because the angle BEC is common to both, and the angles BCE and DBE stand upon equal arches BE and AE ; there- fore EC : EB : : EB : ED, and consequendy ED x EC = EB2: but {by Euc. 36. 3.) EB^ = EF X EH = ED X EH (b7j comtruction). Hence EDxEC = EDx EH, and consequently EC = EH ; from which taking away the equal quantities ED and EF, there remains DC = FH = the given line bisecting the vertical angle (^by construction) : and it is evident that DC bisects the angle ACB, since ACD and BCD stand upon equal arches AE and EB. ^ E. D. Method of Calculation. If BE be considered as a radius, BR Q AB) will be the co-sine of the angle EBR, and BG the tangent of BEG ; therefore BR : BG (or AB : DC) : : co-sine EBR (ACE) : tang. BEG, whose half complement EHB is likewise giv- en from hence : then the angle HB^ (supposing EB pro- duced to b^ being the complement of EHB, we shall have tang. EHB : rad. ( : : sine EHB : co-sine EHB : : BE : EH : : EB : EC) : : sine ECB : sine CBE = sine EDB = co-sine OED, half the difference of the angles (ABC and B AC) at the base. 388 The Construction of PROBLEM LXXIIL Having given the two opposite sides ab^ cd^ the two diago- nals aCj bd^ and also the angle aeb in which they intersect each other ; to describe the trapezium* CONSTRUCTION. In the indefinite line BP take BD equal to bd^ and make the angle DBF equal to the given angle aeb^ and BF = ere; also from the centers D and F, with the radii dc and ab^ let two arches mQ.n and r^s be described intersecting each other in C ; join D, C, and F, C, and make BA equal and parallel to FC ; then draV AD, AC, and BC, and the thing is done. DEMONSTRATION. Since {by construction^ AB is equal and parallel to CF, therefore will AC be equal and parallel to BF {Euc» 33. 1.), and consequently the angle AEB {Euc. 29. 1.) = DBF = aeb. ^. E. D, Method of Calculation, Join D, F ; then in the triangle DBF will be given two sides DB, BF, and the angle included -, whence the angle BFD and the side DF will be known ; then in the trian- gle DFC will be given all the three sides ; whence the angle DFC will be known, from which BFC (BFD — DFC) = BAC will also be known. Geometrical Problems, ' 389 PROBLEM LXXIV. Having given the two diagonals and all the angles^ to describe the trapezium, CONSTRUCTION. Assume AS at pleasure ; and, having produced the same both ways, make the angles QAC, RBC equal, respectively, to two opposite angles a and e of the tra- pezium ; moreover, make ACF equal to ace^ one of the remaining angles ; and from F, the intersection of AF and BQ, take FG = the given diagonal dc^ and q:f J) OP draw GH parallel to CB, meeting FB in H. Then from A and B {by the lemma^ p. 336) let two lines AE and BE be drawn to meet in FC, so as to be in the given ratio of tz^ to FH ; in AE take AN = ae^ and/ draw NM paral- lel to FC, meeting AC in M ; lastly, draw NP making the angle MNP == ced^ and meeting FB in P ; so shall A MNP be the true figure required. DEMONSTRATION. Let ED be parallel to NP, and let DC and PM be drawn. It is evident, by construction, that the diagonal AN, and all the angles of the trapezium, are equal to the respective given ones ; it therefore remains only to prove that PM is equal to the other given diagonal dc. Now the angle RBC being = CED {by coiist ruction)^ the circumference of a circle may be described through all the four angular points of the trapezium BCED ; and so the triangles FBE and FCD (as both the angles FBE and FCD stand upon the same chord ED) will 390 The Construction of be similar ; and consequently BE : DC ( : : FB : FC) : : FH : FG {dc). But (by construction) AE : BE : : ae : FH ; therefore, by compounding these two pro- portions, we have AE : DC : : ae : dc ; but (because of the similar figures ADEC, APNM) we also have AE : DC : : AN (ae) : PM ; and consequently PM = dc. ^. E. D. Method of Calculation. All the angles of the triangles ABC, FAC, and FBC being given, we shall have sine ACB x sine F : sine ABC X sine ACF : : AB : AF ; and sine FHG (FBC) : sine FGH (FCB) : : FG (^c) : FH ; whence AF and FH are known. Find AK = ^^^ and KO = ^I^ : which FH +ae h ii — ae last is equal to (OE) the radius of the circle determining the point E {see the aforesaid lemma). Therefore, in the triangle F'OE are given two sides FO and OE, besides the angle F ; whence the angle FOE will be given ; then in the triangle AOE will be given OA, OE, and the in- cluded angle ; whence the angle OAE, which the diago- nal AN makes with the side AP, will be known, and from thence every thing else required. This problem, as the circle described frOm O cuts FC in two points, admits of two different solutions (except, only, when FC touches the circle). If the circle neither puts nor touches that line, the problem will be impossible ; the limits of the ratio of AE to BE (and consequently of ae to dc) growing narrower and narrower, as AB be- comes less and less, with respect to AC, or according as the sum of the opposite angles (a -f e = QAC -f- RBC) approaches nearer and nearer to two right angles ; so that, at last (supposing AC and BC to coincide), AE and BE will be, every where, in the ratio of equality ; therefore cd can here have only one particular ratio to ae ; and the diagonal ANE may be drawn at pleasure, the problem being, in this case, indeterminate. Geometrical Problems. 391 PROBLEM LXXV. Supposing the right lines m^ n^p to represent the lengths of three staves erected perpendicular to the horizon^ in the given points A, B, C ; to find a point P, in the plane of the horizon ABC, equally remote from the top of each staff. CONSTRUCTION. Join A, B, and B, C, and make AE and BF perpen- dicular to AB ; also make BG and CH perpendicular to BC, and let AE be taken = m, CH =/?, and BF and BG each = w; draw EF and GH, which bisect by the -£ \ r 'ft G £.— ■ •-•••■" I perpendiculars LN and IK, cutting AB and CB in N and K ; make KP and NP perpendicular to BC and BA, and the intersection Y of those perpendiculars will be the point required. DEMONSTRATION. Conceive the planes AEFB arid BCHG to be turned up, so as to stand lerpendicular to the plane of the ho- rizon ABC and ircerser.t it in the right lines AB and BC ; then, becaup BF and BG arc equal to each other, and perpendicul}^ to the plane of the horizon, it is 392 The Construction of evident that the points F and G must coincide, and that AE, BG (BF), and CH will, represent the true position of the staves : suppose KG, KH, PG, PH, PE, and PF to be now drawn ; then, since (by construction^ GI = HI, and the angle GIK = HIK ; therefore is GK = HK [Euc. 4. 1.): moreover, since KP is (by con- struction) perpendicular to BC, it will also be perpendi- cular to the plane BCHG, and consequently the angles PKG and PKH both right angles: therefore, seeing the two triangles GKP, HKP have two sides and an included angle equal, the remaining sides PG and PH must likewise be equal (^Euc. 4. 1.). After the very same manner it is proved that PF (or PG) is equal t© EP. ^ E. D. Method bf Calculation* Draw Ir perpendicular, and H^ parallel to BC ; then, by reason of the similar triangles H^G and IrK, it will be as BC (%) : BG — CH (G^) : : 5£±£5 (I^) ^ BG — CH X BG + CH , . , , , : Kr = =rp; ; which subtracted 2BC from Br ( = |BC) gives BK : and in the same man- ner will BN be found ; then in the trapezium KBNP will be given all the angles and the two sides BK and BN ; from whence the remaining sides, &c. may be easily determined. PROBLEM LXXVI. The base^ the perpendicular^ and the difference of the sides being groen^ to determine the triangle. CONSTRUCTION. Bisect the base AB in C, and in it take CD a third pro- portional to 2AB ^nd the given difference of the sides MN ; erect DE equal to the giren perpendicular, and draw EK parallel to AB, and take therein EF = MN ; draw E AG, to which, from F, apply FG = AB ; draw AH parallel to FG, meeting EK In \ : then draw BH, and the thing is done. Geometrical Problems, 39G DEMONSTRATION. By reason of the parallel lines, FG (AB) : FE (MN) : : AH : EH (DP) ; therefore AB x DP = AH x MN, or 2AB X DP = 2AH X MN; to which last equal quantities adding 2ABxCD = MN%(^y construction) we have 2AB X CP = 2AH X MN X ^•- MN2 ; but 2AB X CP is = BH2 — AH^ (by a known property of triangles) ; therefore BH^ _ AH^ = 2AHxMN'+MN% or BH^ = KW + 2 AH x MN + MN^ = AH + U^Y {Euc. 4. 2.), consequently BH = AH + MN. ^ E. D. Method of Calculation. In the right-angled triangle ADE we have DE and / MN2\ AD f = |AB — j^) > whence the angle DAE (FEG) will be found ; then in the triangle EFG will be given two sides and one angle, from which the angle GFK ( = BAH) will also be known. PROBLEM LXXVIL The base^ the perpendicular^ and the sum of the two sides being' given^ to describe the triangle* CONSTRUCTION. Bisect the base AB in C, and in it produced take CD a third proportional to 2AB, and the sum of the sides, MN ; erect DE equal to the given perpendicular, and draw HE parallel to AB, and take therein EF = MN; draw EAG, to which, from F, apply FG = AB ; draw AH parallel to FG, meeting EF in H ; then draw BH. and the thing is done. / 3E m^ The Gonstructton of DEMONSTRATION. Because of the parallel lines, FG (AB) : FE (MN) 2 : AH : EH (DP) ; and therefore 2lMN x AH = 2AB X DP; which equal quantities being subtracted from MN^ = 2AB x CD, {by construction) there wiQ j^r " ' j\i remain MN^ — 2MN x AH = 2AB x CP = BH^ _ AH^ ; whence, by adding AH^ to each, we have MN^ — 2 MN X AH + AH2 = BHS that is, MN_AHT=: BH^ therefore MN — AH = BH, or MN = BH + AH. ^. E. D. Method of Calculation. In the triangle AED are given (besides the right an- gle) both the legs ; whence the angle DAE (= FEG) will be given ; then in the triangle FEG one angle and two sides will be known, from which the angle EFG ( = BAH) will be determined. PROBLEM LXXVIIL The difference of the two sides ^ the perpendicular^ and the vertical angle being groen^ to determine the triangle. CONSTRUCTION. Upon the indefinite line FEQ erect the given perpen- dicular DC, making the angle DCE = half the given angle; let EF, expressing the given difference of the sides, be bisected by the perpendicular GI, meeting EC in I ; also let EC be bisected in H, and make EK perpendicular to CE, and equal to EI; and, having Geometrical Problems. „„„ drawn HK taVe ht • rrr, from L to FQ appj^'il^B -^^ '"!f ^^ -^^"^ thereto ; i^i^7-'^«-EK, and join C,B; also draw EM, maHnrv +1,1^ "^o DEC, and cuui., CB rS? tt^M^EF apply CA = CM, so shall ACB be the tir^ >^q^^recl. Upon EM le^ ^-^ ^^^ perpendicular CN, and join L, M, p.id F. J» Now, LB^ = EK^ C^^y --- ^truction') ^MT Cm. X HK — f^ V^^^^- ^- ^0 == inriTcH N> HL — HK 0>«/ construction) = CL X EL; whencf- rf^^ -"^ • • ^^ • ^L 5 therefore the triangles ^J^ii and ELB must be equiangular [Euc. 6. 6.), and consequently LBM = LEB = CED = CEM {by con- struction). Therefore, since the external angle CEM of the trapezium LEMB, is equal to the opposite internal angle B, the circumference of a circle will pass through all the four angular points ; and consequently the an- gle LMB will be = LEB, both standing upon the same chord LB ; but it is prov^ed that LBM is = LEB j therefore LMB = LBM = FEI ; and so the triangles BLM and EIF, being isosceles, and having LMB = EFI, and also LB = EI (by construction)^ they will be equal in all respects, and consequently BM = EF ; whence BC — AC r = BC — CM ==: BM) = EF, the given 396 The Construction of difference {by construction^. Moreover, CEN being = CED {by construction^^ CN will be = CD; and so CM being = CA, ACD will be = MCN, to which add- ing DCM, common, we have ACB = DCN = 2DCE. Method of Calculation. Seeing EG and EH are the sine and tangent of EIG and EKH, to the equal radii EI and EK, it will therefore be EG : EH (or EF : EC) : : sine EIG (ECD) : tan- gent EKH. But EC : CD : : the radius : co-sine ECD ; whence, by compounding these proportions, EF : CD : : radius x sine ECD : co-sine ECD x tangent ^__^^ radius x sine ECD , ^ ^ t^^tw *. KH : : — ( = tangent ECDj : tangent ■^ co-sinc jlCD will u fj,oj^ ^hich EKL, half the complement of EKH ^'^^ ( : . (riven ; then it will be, as the radius : tangent : sine LBE .^EL : : LB : EL) : : sine LEB (CED) words, give the Fc&) ; which proportions, expressed in As the difference om% theorem : the tangent of half the veH^^^ ^^ ^^ the perpendicular^ so is gle ; and as the radius is to\ff'^S^^ ^^ ^^^^ tangent of an an- ment of this angle^ so is the con^^S^^^^ ^f ^^^^f ^^^^ comple- to the sine of ha f the diference oft^^^^lf^^^^ vertical angle r/z< ^f^c perpendicular, the differ ^g^^ence of the angles :t £^^^^ * mi^U^ dt the base* "■^o^rxM LXXIX. ''nine the triangle. (angles at the base heint, g^n, i^'iJt CJ)NSTRUCTI0N. l^tt a triangle ABC be constructed by the ^ast problem.yvhose per-- pendicular and diffe- rence of the sides shall be the same with those given, and whereof the vertical angle ACB is also equal to the given difference of angles; Geometrical Problems. 397 then upon C, as a center, with the radius CB, let an arch be described, intersecting AB, produced, in D ; join C, D, and ACD will be the triangle required. For CD be- ing = CB, the angle CDB will also be = CBD = A + BCA {Euc. 32. 1.}. The method of calculation is als© the same as in the preceding problem. PROBLEM LXXX. The perpendicular^ the. sum of the two sides ^ and the ver- tical angle being given^ to describe the triangle. CONSTRUCTION. Upon AB, the given sum of the two sides, erect AC equal to the given perpendicular ; and make the angle ACH equal to the complement of half the given an- gle ; upon AB (by prob. 72) let a triangle ABF be constituted, whose vertical angle, AFB, shall be equal to the given one, and whereof the bisecting line FE (ter- minating in the base) shall be = DC ; then daw CO and CH parallel to FB and FA, so shall GCH i)e the triangle required. DEMONSTRATION. It is evident that the angle HCG is = AFB = the given one. Moreover, if EM a^d EN be taken as perpendiculars to AF and BF, the/ will be equal to each other, and also equal to the given one AC, because' all the angles EFN, EFM, and ADC are equal, by construction, and EF is likewise = CD ; whence, as the angles AHC, AGC are respectively equal to EAM, 39B TJie Construction x>f EBN, it is evident that HG = EA, and GC = EB, and consequently that HC + GC ( = EA + EB) = AB* Method of Calculation. By the problem above referred to, AB : CD (EF) : : co-sine ADC (AFE) : tangent of an angle ; which let be denoted by Q. Now, CD : CA : : radius : sine ADC ; which pro- portion being compounded with the former, we have AB : CA : : co-sine ADC x radius : tangent Q x sine A -TN^ co-sine ADC \ radius . * t^^, -^^^ = -■ ^b^ADC (--tangent ADC) : tau- gent Q. Then, by the same problem, it will be as tangent ^Q : radius : : sine ADC : co-sine of the difference of the angles (G and H) at the base. The above proportions, given in words at length, exhibit the following theorem: As the sum of the sides is to the perpendicular^ so is the co-tangent of half the vertical angle to the tangent of an angle ; and^ as the tangent of half this angle is to the ra- dius^ so IS the sine of half the vertical angle to the cosine of half the difference of the angles at the base. PROBLEM LXXXI. To constitute a trapezium of a given ?nagnitude under four given lines* CONSTRUCTION. p/\^ Make a right P / ^^ angle b with two aL .^!"nn^ Qf ^\^Q given lines Ab^ be ; and with the other two complete the tra- pezium AbcT) : upon AD let fall the perpendicular cE, in which pro- duced (if neces- sary) take EF, so that the rectangle Geometrical Problems,. 399 under it, and AD, may be double the given area : more- over, take a fourth proportional to AD, A^, and hc^ with which, from the center F, let an arch be described, meet- ing another arch, dt^scribed from D with the radius Dc in C ; join D, C ; and from A and C draw the other two given lines AB, CB, so as to meet, and they will thereby form the trapezium A BCD, as required. DEMONSTRATION. Draw Ac, AC, and FC ; upon AD and AB let fall the perpendiculars CP, CQ ; and make FG perpendicular to PCG. Because AD^ + DC^ + 2AD x DP (= AC% Euc. 12. 2.) = AB2 + BC2 + 2AB X BQ, and AD^ -f. Dc^ + 2AD X DE (= Ac2) = A^^ ^ ^^.2 (^^^^ 47. 1.), it follows, by taking these last equal quan- tities from the former, that 2AD x DP — 2AD x DE (2AD -f- EP) = 2AB x BQ, and consequently that BQ ; EP (FG) : : AD : AB : : BC : FC Qnj construction) ; whence the triangles BCQ, FCG are similar, and so CQ : CG : : BC : FC : : AD : AB {bij construction)^ and therefore CQ X AB = CG X AD ; hence, by adding CP x AD to each, we have CP X AD -j. CQ X AB (= twice the area ABCD) = CP X AD + CG X AD = EF X AD = twice the given area (by construction). ^ E. Z). Method of Calculation. From DE ( = A^^ + ^c^ AD^ - Dc-n ^^^ ^^ \ 2AD / = -T-TT- ) the value of DF, and likewise that of the AD / angle ADF, will be found ; then, all the sides of the tri- angle DCF being known, the angle FDC will likewise be known ; which, added to ADF, gives (ADC) one of the angles of the trapezium. It may so happen tliat a trapeziui?i, liaving one right angle, cannot be constituted under the four given lines ; in which case it will be necessary (instead of form- ing the trapezium A^cD) to lay down AD first, and in it {produced if needful) to take DE equal to 400 The Construction^ &f c. ABT 4- BCT — Ti3T— iDCV ^ . , , L_ 2AB "" ^^ ^^^ ^^^ ^^ altitude of a rectangle, formed on the base 2 AD, whereof the contained area is equal to the difference of AB^^ -f BUy and ADT +"51:7 (which line DE is to be set off on the other side of D, when the latter of these two quan- tities is the greater) : this being done, the rest of the so- lution will remain the same, as is manifest from the first and second steps of the demonstration ; the process from thence to the end being nowise different. It may be further observed, that the problem itself be- comes impossible, when the two circles, described from the centers D and F, neither cut nor touch ; the greatest limit of the area, and consequently of EF, being when they touch each other ; in which case the sum of the radii DC, FC becoming = DF, the point C will fall in the line DF, and the angle DCF will become equal to two right angles ; but the sum of the opposite external angles CDP and CBQ is always equal to DCF ; because CDP (supposing Ctz parallel to AP) is = DC^z, and CBQ ( = CFG) = FC/z : hence it is evident that the limit, or the greatest area will be when the sum of the opposite angles is equal to two right angles, or when the trapezium may be inscribed in a circle. FINIS, Hi U DAY USE RH'l URN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subjea to immediate recall. prr/n I..D MAR 7 W? Rare Books . > ^ Special Collections LD 21A-50m-8,'61 ll^\^^f^nVc^Vnm\^ ( C1795S10 ) 476B ^^^^"jg.J^eg^*'''"^ {vi3S9750