N We send this as a sample of Part Third, or the advanced course of Prof. Olney's "Special or Elementary Geometry." This book will be bound up with Parts First and Second, under the title of Olney's Elements of Geometry and Trigonometry-UNIVERSITY EDITION. SHELDON & COMPANY, New York. STODDARD'S 1VTMATHEIATICAL SERIES. A TREATISE ON SPECIAL OR ELE MENTARY GE OMETRY. IN FOUR PARTS. UNIVERSITY EDITION. INCLUDING PLANE, SOLID, AND SPHERICAL GEOMETRY, AND PLANE AND SPHERICAL TRIGONOMETRY, ~WITH THE NECESSARY TABLES. WHICH DISTINGUISHES THE UNIVERSITY FROM THE SCHOOL EDITION, COMPRISES 160 PROPOSITIONS AND PROBLEMS FOR EXERCISE IN GEOMETRICAL INVENTION, A TREATISE ON APPLICATIONS OF ALGEBRA TO GEOMETRY, AND AN INTRODUCTION TO THE MODERN GEOMETRY. BY EDWARD OLNEY, PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIGAN. NEW YORK: SHELDON & COMPANY, 677 BROADWAY.,1872. Stoddlard's [athematical Series. STODDARD'S JUVENILE MENTAL ARITHMETIC - - $ 25 STODDARD'S INTELLECTUAL ARITHMETIC - - - - 4) STODDARD'S RUDIMENTS OF ARITHMETIC - - - 50 STODDARD'S NEW PRACTICAL ARITHMETIC - - 1 00 SHORT AND FULL COURSE FOR GRADED SCHOOLS. STODDARD'S PICTORIAL PRIMARY ARITHMETIC 30 STODDARD'S COMBINATION ARITHMETIC - 75 STODDARD'S COMPLETE ARITHMETIC -— 1 25 The Combination School Arithmetic being Mental and Written Arithmetic in one book, will alone serve for District Schools. For Academies a full high course is obtained by the Complete Aritnhmietic and Intellectual Arithmetic. HIGHER MATHEMATICS OF TIlE SERIES, BY EDWARD OLNEY, PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIGAN. A COMPLETE SCHOOL ALGEBRA, in one vol, 410 pages, $1.50. Designed for Elementary and higher classes in Schools and Academies. A GEOMETRY AND TRIGONOMETRY, in one vol., 8vo, - Price $2.50 A GEOMETRY AND TRIGONOMETRY, UNIVERSITY EDITION, in one vol., 8vo, - - - - Price $83.00 A GENERAL GEOMETRY AND CALCULUS in one vol. - Price 2.50 The other books of this Series will be published as rapidly as possible. Entered according to Act of Congress, in the year 1872, by SHELDON & COMPANY, In the Office of the Librarian of Congress, at Washington. PART III. AN ADVANCED COURSE IN GEOMETRY. CHAPTER I.:EXERCISES IN GEOMETRICAL INVENTION. SECTION A THEOREMS IN SPECIAL OR ELEMENTARY GEOMETRY. 623. This chapter will afford a review of Parts I. and II., while it will greatly extend the student's knowledge of geometrical facts. Great pains should be taken to secure good habits as to neatness of execution in the construction of figures, orderly and proper arrangement of thought, and in style of expression. The practice of constructing every figure upon geometrical principles-guessing at nothing —cannot be too strongly commended. As to the formgn of a geometrical argument, observe the following order: 1st. The enunciation of the theorem or problem in general terms. 2d. The elucidation of the general statement, by reference to the particular figure which it is proposed to use. 3d. A description of the figure, with reference to any auxiliary construction which is used in the demonstration or solution. 4th. The demonstration proper, 624. If two adjacent sides of a quadrilateral are equal each to each, and the other two adjacent sides equal each to each, the diagonals intersect at right angles. SuG's.-Ist. Draw a quadrilateral having such sides as the data require, and draw its diagonals. 2d. State the proposition with reference to the figure. 244 EXERCISES IN GEOMETRICAL INVENTION. 3d. [In this case the regular third step is not required, as no auxiliary lines are necessary.] 4th. Prove that the diagonals are at right angles to each other. The demonstration is based upon a corollary in Section I., Part II., Chapter I. 625. CoR.-One of the diagonals is bisected. [State which one, and show why.] 626. If a parallelogram has one oblique angle, all its angles are oblique; and if it has one right angle, all its angles are right angles. Suc's.-Let the student be careful to follow the order as heretofore given. No auxiliary construction is needed. The demonstration is based upon the doctrine of parallels. 62T. The sum of three straight lines drawn from any point within a triangle to the vertices is less than the sum, and greater than the half sum of the three sides of the triangle. SuG's.-The first statement is proved from (276) and the second from (274.) 628. A line drawn from any angle of a triangle to the middle of the opposite side, is less than the half sum of the B adjacent sides, and greater than the difference between this half sum and half the third side. /I~C SUG's. —lst. Draw a triangle, as ABC, bisect one side, as AC, /A D / and draw BD. 2d. Make the statement with reference to the figure. 8d. Produce BD until DE = BD, and draw AE and EC. 4th. The first step in the proof is to show the triangle ADE equal to CBD, and ADB equal to DCE; whence AE = BC, and EC = AB. FIG. 356. 629. If lines be drawn from the extremities of either of the oblique sides of a trapezoid to the middle of the opposite side, the triangle thus formed B. CGQ- G is half the trapezoid. /"E,.,/\' SUG's.-The third step, or construction, consists in F D drawing HE parallel to AD and hence to BC (?), and FIG. 357. FG through E parallel to AB. 630. Any line drawn through the centre of the diagonal of a parallelogram bisects the figure. THEOREMS IN SPECIAL GEOMETRY. 245 631. Prove that the sum of the angles of a triangle is two right angles, by producing two of the sides about an angle and through this angle drawing a line parallel to the third side. Prove the same by producing one side of the triangle and drawing a line through the exterior angle parallel to the noh-adjacent side. 632. If any point, not the centre, be taken in a diameter of a circle, of all tie chords FIG. 358. which can pass through that point, that one is the least which is at right angles to the diameter. 633. If from any point there extend two lines tangent to a circumference, the angle contained by the tangents is double the angle contained by the line joining the points of tangency and the radius extending to one of them. 634. The angle included by two lines drawn from any angle of a triangle, the one bisecting the angle and the other perpendicular to the opposite'side, is B half the difference of the other two angles of the triangle. E D FIG. 359. SUG'S. ABD = 90~ - A, whence ABD - EBD = 90~ - A - EBD. Also, DBC = 90 ~- C, whence EBC = 90~ - C + EBD -- 90~ -A- EBD, etc. 635. If three lines be drawn from the acute angles of a right angled triangle-two bisecting these angles, and a third a perpendicular to one of the c bisecting lines-the triangle included by these lines will be isosceles. D SUG's.-It is to be proved that OD = CD. COD = OAC + ACO = 45~, etc. A F B 636. If one circumference be described Fr. 360. on the radius of another as a diameter, any straight line extending from their point of contact to the outer circumference is bisected by the inner. SuG. —The demonstration is based upon (159, 211). 246 EXERCISES IN GEOMETRICAL INVENTION. 637. Prove that the sum of the angles of a regular five point star (Fig. 101) is two right angles. Show, also, that the figure formed by the intercepted portions of the lines is a regular pentagon. 638. If the sides of a regular hexagon are produced till they meet, show that the exterior figures will be equilateral triangles. P' 639. If from two given points on the same e i side of a given line, two lines be drawn meeting in the line, their sum is least when they A 3 make equal angles with the line. FIG. 361. FIG. 361. 640. If from two given points without a circumference, two lines be drawn meeting in the circumference, their sum is least when they make equal angles with a t.angent at tlhe common point. 641. The side of an equilateral triangle inscribed in a circle is equal to the diagonal of a rhombus, whose other diagonal and each of whose sides are equal to the radius. 642. If two circumferences intersect each other, and from either point of intersection a diameter be drawn in each, the other extremities of these diameters and the other point of intersection are in the same straight line. 643. If any straight line joining two parallels be bisected, any other line through the point of bisection and included by the parallels, is bisected at the same point. 644. If the sides of any quadrilateral are bisected, the quadrilateral formed by joining the adjacent points of bisection is a parallelogram. SuG's.-lst. Draw a quadrilateral, bisect its sides, and join the adjacent points of bisection. E F ~2cld. State the proposition, with reference to the A -- figure. 3d. Draw the diagonals. 4th. Give the proof. It is based on the similarity of triangles. 645. COR. 1.-The parallelogram is onehalf the trapezium. Prove it. What figure is formed by joining the centres of EF, FG, FIG. 362. and FC, HC, etc.? THEOREMS IN SPECIAL GEOMETRY. 247 646. CORn. 2.-Lines joining the middle points of the opposite sides of any trapezium bisect each other (?). 64i. If two straight lines join the alternate ends of two parallels, the line joining their centres is half the difference of the parallels. A B SUG'S.-We are to prove that EF = i (CD - AB). E CH = EF = 2 (CD - AB). C H D FIG. 363. 648. In ally right-angled triangle the line drawn from the right angle to the middle of the hypotenuse is equal to one-half the hypotenuse. 649. The perpendiculars which bisect the three sides of a triangle meet in a common point. SUG's. —First show that the intersection of two of the perpendiculars is equally distant from the three vertices of the triangle. Then that a line drawn from this point to the middle of the third side is perpendicular to it. 650. The three perpendiculars drawn from the angles of a triangle upon the opposite sides intersect in a common point. P B SuG's.-Draw through the vertices of the F j/ / triangle lines parallel to the opposite sides. The proposition may then be brought under A' the preceding.. 651. CoR. —The following triangles are similar —viz., BOE, BDC, AOD, and 9 AEC, each to each; also BOF, BDA, DOC, FIG. 364. and CFA. Prove it. 652. If from a point without a circle two secants be drawn, making equal angles with a third secant passing through the centre of the circle, the intercepted chords of the first two are equal. SUG.-Prove by revolving one part of the figure. 248 EXERCISES IN GEOMETRICAL INVENTION. 653. The sam of the alternate angles of any hexagon inscribed in a circle is four right angles. M c - 654. If two circles intersect in A and B, and from P, any point in one circumference, the chords PA and PB be drawn to cut the other in c and D, CD is parallel to a tangent at P, FIa. 365. 655. If two lines intersect, two lines which bisect the opposite angles are per_pendicular to each other. 656. The angle included by two lines drawn from a point within a triangle to the vertices of two of the angles, is greater than the third angle.'Suc's. —The demonstration may be founded on (219) or (231). 657. In a triangle whose angles are 90~, 60~, and 30~, show that the longest side is twice the shortest. 658. Lines which bisect the adjacent angles of a parallelogram are mutually perpendicular. 659. If from any point in the base of an isosceles triangle lines are drawn parallel to the sides, a parallelogram is formed whose perimeter is constant and equal to the sum of the two equal sides of the triangle. 660. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides of the triangle, their sum is constant and equal to the perpendicular from one of the equal angles of the triangle FIG. 366. upon the opposite side.. 661. If from any point within an equilateral triangle, three perpendiculars be let fall upon the sides, their sum is constant and equal to the altitude of the FIG. 367. triangle. THEOREMS IN SPECIAL GEOMETRY. 249. 662. If from a fixed point without a circle two o tangents be drawn terminating in the circumference, the triangle formed by them and any tan- c gent to the included arc has a constant perimeter c/ equal to the sum of the first two tangents. A B FIG. 368. 663. The sum of two opposite sides of a quadrilateral circumscribed about a circle, is equal to the sum of the other two. 664. If two opposite angles of a quadrilateral are supplementary, it may be circumscribed by a circumference. 665. The square described on the sum of two lines is equivalent to the sum of the squares on a b b the lines, plus twice the rectangle of the lines. a a b SUGo'.-Be careful to give the construction fully, and M show that the parts are rectangles, etc. a i T. FIG. 369. 666. The square described on the difference of b7two lines is equivalent to the sum of the squares on the lines, minzius twice the rectangle of the lines. a FIG. 370. 66T. The rectangle of the sum and difference b9.of two lines is equivalent to the difference of the squares described on the lines. Sce. —The three preceding propositions are but geo- a-b b b metrical conceptions and demonstrations of the algebraic formulc, (a + b)2 = a2 + 2ab + b2, (a - b)2 = a2 - 2ab + b2, and (a + b) (a - b) = a2 - b2. FIG. 371. 250 EXERCISES IN GEOMETRICAL INVENTION. VARIOUS DEMONSTRATIONS OF THE PYTHAGOREAN PROPOSITION. 668. The square described on the hypotenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides. 1st METHOD.-Let ABC be the given triangle, and B ACED the square described on the hypotenuse. Complete the construction. Show that the four triangles are equal. J H _ \ The square HF is (AB - BC)2. The student can complete A the demonstration. FIG. 372. \t H \ 2d METHOD.-Let ACED be the square on the hypotenuse. Let fall the perpendiculars EF, DC, etc. Show that the three triangles are equal, and L XC that FD and LB are the squares of the two sides AB and BC. G A F B FIG. 373. G/' i" 3d METHOD.-Let BL and BH be the 1 E squares on the sides. Produce FL and HG till they meet in K. Draw DA and EC perpendicular to AC, and draw DE and KB. Prove that ACED is a square, and also B + \ that the triangles ABC, CLE, BFK, KBC, L DKE, and AHD are all equal to each other. Af _/.'"'_ \~ / The demonstration is then readily made. FIG. 374. G H 4th METHOD.-This is the demonstration i\ L usually given in our text-books. Drawing the \B ss \ \ squares on the three sides, let fall BI perpen\ dicular to AC and produce it to K. Draw BD, BE, HC and AF. Show that the triangle HAC I/ I! = BAD, and that the former is half the square / It AG, and the latter half the rectangle AK.,'.. Hence AG = AK. In lilke manner show that.'.LC _ CK. O C K E FIGo. 375. THEOREMS IN SPECIAL GEOMETRY. 251 We will now give a few other figures by means of which the demonstration can be effected, and leave the student to his own resources in effecting it. I\ h (a4 b) 22a b a 5th METHOD. 6th MIETHOD. 7th METHOD. 8th METHOD. FIG. 376. 9th METHOD.-The truth of the theorem appears also as a direct consequence of (360).:. FIG. 377. 669. In an oblique angled triangle the square of a side opposite an acute angle is equivalent to the sum of the squares of the other two sides diminished by twice the rectangle of the base, and the distance from the acute angle to the foot of the perpendicular let fall upon the base from the angle opposite. SUG'S.-lt is to be shown that AB = -BC2 + AC2 - 2AC x DC. B Observe that AD2 = (AC - DC)2 = AC2 + DC2 - 2AC x DC. Whence, by a simple application of the preceding theorem, the truth of this becomes apparent. A D FIG. 378. 670. In an obtuse angled triangle the square of the side opposite the obtuse angle is equivalent to the sum of the squares on the other two sides, increased by twice the rectangle contained by the base and the distance from the obtuse angle to the foot of the perpendicular let fall from the angle opposite upon the base produced. SuG. —The demonstration is analogous to the preceding, C being made obtuse in this case; whence AD = AC + DC, etc. 252 EXERCISES IN GEOMETRICAL INVENTION. 6T1. The following is an outline of a general demonstration covering the three preceding propositions: E. Letting AE, BD, and CF be the three perpendiculars fiom / the angles upon the opposite sides, and observing that a A F -B circumference described on any side as a diameter passes C through the feet of two of the perpendiculars, (356) and D.:. (355) readily give the following: AB x AF = AC x AD = AC2 AC x CD, A F B and AB x BF = BC x BE-BC2 B x CE; FIG. 379. adding, AB2 = AC2 + BC2 ~ 2AC x CD (or 2BC x CE), the + sign being taken when C is obtuse, and the - sign when C is acute. If C is right CE and CD become O, whence AB2 = AC2 + BC2. 6T2. DEF.-The line drawn from any angle of a triangle to the middle of the opposite side is called a medial line. 6''3. The sum of the squares of any two sides of a triangle is equivalent to twice the square of the medial line drawn from their included angle, plus twice the square of half the third side. SUG.-Proved by applying (669, 670). 6T4. The three medial lines of a triangle mutually trisect each other, and hence intersect in a commnon point. SuG's. —To prove that OE = ABE, draw FC parallel to AD until it meets BE produced. Then the triangles AEO and FEC are equal (?); whence EF = OE. Also, A El BO = OF(?). F Having shown that OE = — BE, by a similar construcFIG. 380. tion we can show that OD = ~AD. Finally, we may show that the medial line from C to AB cuts off L of BE, and hence cuts BE at the same point as does AD. ANOTHER DEM.-Lines through 0 parallel to the sides trisect the sides, etc. 675. In anyquadrilateral the sum of the squares of the sides is equivalent to the sum of the squares of the diagonals, plus four times the square of the line joining the centres of the diagonals. X 6r6. CoR.-The sum of the squares of the sides of a parallelogram is equivalent to the sum of the FIG. 381. squares of the diagonals. THEOREMS IN SPECIAL GEOMETRY. 253 67'. In any quadrilateral which may be inscribed in a circle, the product of the diagonals is equal to the sum of the products of the opposite sides. 6T8. In any triangle the rectangle of two sides is equivalent to the rectangle of the perpendicular let fall from their included angle upon the third side, into the diameter of the circumnscribed circle. SUG.-This proposition is an immediate consequence of the similarity of two triangles in the figure. FIG. 382. 6T.9. CoR — The area of a triangle is equivalent to the product of its sides divided by twice the diameter of the circumscribed circle. 680. If there be an isosceles and an equilateral triangle on the same base, and if the vertex of the inner triangle is equally distant from the vertex of the outer one and from the ends of the base, then, according as the isosceles triangle is the inner or the outer one, its base angle will be I of, or 24 times the vertical angle. 681. Of all triangles on the same base, and having their vertices in the same line parallel to the base, the isosceles has the greatest vertical angle. SuG's.-Circumscribe a circle about the isosceles triangle. By what is the vertical angle mneasured When the triangle is isosceles? By what, when it is not? 682o. Two triangles are similar, when two sides of one are pro-. portional to two sides of the other, and the angle opposite to that side which is equal to or greater than the other given side in one, is equal to the homologous angle in the other. ALGEBRAIC DEMONSTRATIONS. 683. The difference of the squares on any side of a regular pentagon and any side of regular decagon, inscribed in the same circle, is equivalent to the square of the radius. SuG's.-We will give the outline of what may be termed an Algebraic Demonstration of this proposition. This method is often thle most convenient and ex 254 EXERCISES IN GEOMETRICAL INVENTION. peditious. Letting p represent a side of the pentagon, d a side of the decagon, and r the radius, the student should be able to discover the following relations: (1) r:d::d:r-d, or -2 - dr = d; — (2) V/d,2 _ 4p3 +r/ - _-Lp~ = r. From (2), 2rVra/ - ~4p2 = 2re _ d2 = r2 + dr, by substituting for d2 its value fiom (1). Hence 4r' - p2 = re + 2dr + d2, or 32 -2dr- =p2 + d2. In this, substituting the value of dr as found in (1), we readily obtain r2 = p2 _ 2'. Q. E. D. 684. Demonstrate algebraically that the square on the sum of two lines, together with the square on the difference, is double the sum of the squares on the lines separately. 685. The sum of the squares of the three medial lines of a triangle is three-fourths of the sum of the squares of the sides. 686. The square of any side of a triangle is equivalent to twice Ithe sum of the squares of the segments of the medial lines adjacent to its extremities, minus the square of the non-adjacent segment of the third medial line. Deduced algebraically from the preceding. 68'. The sum of the squares of the three greater segments of the medial lines of a triangle is equivalent to one-third the sum of the squares of the sides of a triangle. Deduced algebraically from the preceding. 688. The lines from the vertices of a triangle to the points of tangency of the inscribed circle intersect in a common point. SuG's. DC is parallel to AF, BD = BF = b, CF = CP A de a/a =, AD = AP = a, DC = d, FCG = e. OF= AF x b ab be b d = e- e. OF = AF x ~C ~~ a + b' a b' a(b + c) + be' C a e a? How if B is obtuse?'79T. To pass a circumference through two given FIG. 4. points, having its centre in a given line. 18 274 EXERCISES IN GEOMETRICAL INVENTION.'98. To draw a line parallel to a given line and tangent to a given circuimference. SuG.-Draw a diameter perpendicular to the given line., N'> 7'99. To draw a common /! tangent to two given circles. I P o ) SUG.- IST METHOD.-There are two sets of tangents, AC, BD, and "vs~ t~ ~,a' that if PE =AP - OC, OE is parallel'- /._;, - to AC, etc. FIG. 423. P }?:::.... Pt 2D METHOD. p0O, p'O' ------. --- being parallel to each other,. pp'T gives the intersection of the tangent with the line passing through the centres, FroIG. 424. since pO: p'O':: OT: O'T, or pO -p'O': p'O':: 00': O'T. Also, PO - P'O': P'O':: 00': 0'T. Hence O'T is constant for all positions of the parallel radii. Prove that if the parallel radii are on different sides of the line joining the centres, T' is the point where the internal tangent cuts 00'. QUERIES.-How many tangents can be drawn-ist. When the circles are external one to the other; 2d. When they are tangent externally; 3d. When they intersect each other; 4th. When tangent internally; 5th. When one lies within the other? D" 800. To describe a circle tangent to a given circumference!\ x\ and also to a given line at a given point. O. \0' 4 \ / SuG's. —There may be two casesBst. VWhen the given circle is exterior \ \, "\/\,- \ to the one sought; and 2d. When it./A is interior. In either case the centre of the required circle is in the perpendicular AO'. In the former case, C, the centre of the required circle, is at FIG. 425. r + r' from C; and in the latter O' is at r - r'fiom C. AO = r, and CD =AB B =AB' = r'. PROBLEMS IN SPECIAL OR ELEMENTARY GEOMETRY. 275 801. To construct a trapezoid when the E A B four sikdes are given. o / -\ SUG. —Knowing the differen'ce between the two parallel sides, we may construct the triangle AEC, c D and hence the trapezoid. FIG. 426. 802. On a given line, to construct a polygon similar to a given polygon. SUG's.-One method may be learned from A (90). Ex. 8, page 152, furnishes anotller ( --- method. The following is an elegant metllod: To Construct on A' homo!ogous with A, a polygon similar to P. Place A' parallel to A, FGc. 427. and the figure will suggest the construction. 803. To pass a plane through a given line and tangent to a given sphere. SUG's. —Pass a plane through the centre of the sphere and perpendicular to the given line. Thrllough the point of intersection and in this secant plane draw tangents to the great circle in which the secant plane intersects the surface of the sphere. The points of tangency will be the points of tangency of the requiled planes (?), of which there are thus seen to be two. 804. DEF.-A Tangent Plane to a cylindrical or conical surface is a plane which contains an element of the surface, but does not cut the surface. The element which is common to the surface and the plane is called the Eleiment of Contact. 805. To pass a plane through a given point and tangent to a given cylinder of revolution. SuG's.-lst. When the point is in the surface of the cylinder. Through the point draw an element of the cylinder, by passing a line parallel to the axis, or to any given element. Through the same point pass a plane perpendicular to this element, making a right section (a circle). To this circle draw a tangent. The plane of the element and tangent is the tangent plane required. [The student should prove that any point in the plane affirmed to be tangent, not in the element passing through the given point, is without the cylinder.] 2d.'When the given point is without the cylinder. Pass a plane through the given point perpendicular to the axis of the cylinder, thus making a right section of the cylinder (a circle). In this secant plane draw tangents to the section. Through the points of contact of these tangents draw elements of the cylinder. These elements are the elements of contact of the tangent planes. Hence planes passing through them and the given point are the tan 276 EXERCISES IN GEOMETRICAL INVENTION. gent planes required. [The student should remember that this is but an outline, and be careful to fill it up, giving the proof.] 806. To pass a plane through a given point and tangent to a conical surface of revolution. 80'. To find, with the compasses and ruler, the radius of a material sphere whose centre is inaccessible. A SUG's.-With one point of the compasses at any point in the surface, as A, trace a circle of the sphere, as acb. a j --- The chord Aa is measured by the distance between the compass points. In like manner measure three other chords, as ac, ab, and bc. Draw a plain triangle having these chords for its sides, and circumscribe a circle about it. Thus aD is found. Knowing aA, and aD, and rememij i / bering that AaB is right angled at a, the triangle B AaB can be drawn in a plane (?), whence AO becomes Fro. 428. known. SECTION II. APPLICATIONS OF ALGEBRA TO GEOMETRY. 808. The mathematical method which is called technically A.pplications of Algebra to Geometry consists in finding, by means of equations, the numerical values of the unknown parts of a geometrical figure, when a sufficient number of the parts are given numerically. 809. By reference to the COMPLETE SCHOOL ALGEBRA, page 238, it will be seen that the algebraic solution of a problem consists of two parts: 1st. The Statement, which is the expressing by one or more equations of the conditions of the problem, i. e., the relations between the known and unknown quantities (parts of the figure) to be compared; and 2d. The Solution of these equations, so as to find the values of the unknown quantities in known ones. 810. In applying the equation for the solution of such problems as are now proposed, we have to depend upon our previously acquired knowledge of the properties of geometrical figures for the relations between the known anid unknown quantities, which will enable us to form the necessary equations, i. e., to make the State APPLICATIONS OF ALGEBRA TO GEOMETRY. 277 ment. The resolution of the equations thus arising is effected in the ordinary ways. [See NOTE, page 239 of THE COMPLETE SCHOOL ALGEBRA.] 811. The details of this method will be most readily obtained from a careful study of examples. EXAMPLES. 812. In a right angled triangle, given the hypotenuse and the sum of the other two sides, to find these sides separately. SOLUTION.-Let ABC be a triangle, right angled A at B. Let the known hypotenuse be h, the unknown base, y; the unknown altitude, x; and the known sum of the base and altitude, s. We have here two unknown quantities, and hence must have two equations, in order to find their Y values. One of these equations is furnished directly FIG. 429. by the statement of the problem, which says that the sum of the base and perpendicular is to be given. HenceEquation 1 is x + y=s. A second relation between x and y and the known quantity h is furnished by the relation given in PART II. (346). WhenceEquation 2 is X2 + y2 = h'. Solving these equations we findy = is ~ 4-/2h2 - s2, and x = —s:F ~i/24 - 82. If h 10 and s = 14, wefind x = 6, andy = 8; or x = 8, andy = 6. GEOMETRICAL SOLUTION —It is exceedingly interesting and instructive to compare the algebraic solution of such problems with their geometrical solution, when the problem can be solved in both ways. The geometrical solution of this problem is as follows: Take DC = s, the sum of the two sides, 0 and make ODC = 45'. From C as a centre, A, with a radius h, the hypotenuse, describe an arc cutting DO, as in A and A'. Draw AC and the perpendicular AB, also A'C and the per- A pcndicular A'B'. Both the triangles ABC and A'B'C fulfil the conditions. For AB = DB (?), whence AB + BC = s, and AC = -, by construction. So, also, A'B'= DB' (?), whence A'B' D + B'C = s, and A'C = h, by construction. s _ COMPARISONS OF THESE SOLUTIONS.-ist. We find in the algebraic solution, that, in FIG. 430. 278 EXERCISES IN GEOMETRICAL INVENTION. general, y may have two values —viz., -s + V/2h2 -, and ~s - -- 82; and that when y = Is + 5/2-2 - 82, x = -s - 4-/W --.2; but, when y = As - ~-/Vh2 - s2, X = ~s + _/2h-2 - s2. Correspondingly, we find in the geometrical solution that the base (y) may have two values-viz., BC, and B'C; and that when the base is BC, the altitude (x) is AB; but, when the base is B'C, the altitude is A'B'. 2d. From the algebraic solution, we observe that the base y = -s:~ ~Z/22- s-, may be considered as made up of two parts-viz., a rational part, is, and a radical part, -/2 _ — s2; and that the altitude, x = -s FT V2h2 s2, is made up of the same parts, only observing that, if the base is considered as the sum of these parts —viz., is + IV/2-2 s2, the altitude is their difference- viz. 4S -- 4~V22 - s2. If, however, the base is 4s - - /21- s I2, the altitude is is + ~/V/2h2 -_ s. Now, we can discover exactly the same things geometrically, and can show exactly what is the geometrical meaning of each of the parts of the values of y and x. To.do this, draw Cf* bisecting AA'; let fall the perpendicularf e, and draw Ak and fg parallel to DC. Cf is perpendicular to DO (?), and hence equal to Df (?). Also, De = eC = fe = is (?). From the right angled 1 s isosceles triangle DfC,fC =- -- DC = (?). Hence, from AfC, Af = 2~_ __7- 21 AC -JC -2= 2h2, the quantity under the radical sign is negative, and the radical becomes imaginary. This means, that no triangle can be formed under these circumstances. This case appears in the geometrical solution also, for then h < -_.s, or less than fC, and consequently the arc struck from C as a centre, with radius h, will not touch DO, and we get no triangle. * This part of the construction should not be allowed on the figure till it is wanted-i.e., till this stage of the discussion. APPLICATIONS OF ALGEBRA TO GEOMETRY. 279 813. ScmH —This problem is discussed thus at length as an illustration of what may be done by such methods. Of course, all problems are not equally fruitful; but the student should not rest satisfied.with a mere determination of the values of the unknown parts in known terms, when anything farther is revealed either by the process or result of the algebraic solution. Especially should he desire to become expert in seeing what geometrical relations are indicated by thefocrm of the answer obtained. 814. Given the lengths of the medial lines from the acute angles of a right angled triangle, to determine the triangle, i. e., to find the base and perpendicular. SuG's.-Let AD - a, CE = b, AB = 2x, and CB = 2y; then o 4x + y2 = a, and 4y2 + 2 =b(?)..2 = AB = -,/ 2 and 2y = CB = 1 4b - a The form of these results indicates that CB sustains the A E B same relation to CE and AD that AB does to AD and CE-a FIG. 431. fact which is evident from the nature of the case. Again, if 4a2 < b6, 2a is imaginary; and if 4b- < a2, 2y is imaginary. In either case the triangle cannot exist. So also if 4a2 = b2 2x = 0; and if 4b2 = a2, 2y = 0, and there can be no triangle. This may be seen from the figure by conceiving AB, for example, to diminish. As A approaches B, AD approaches equality with DB, and CE with CB. Hence the limit is AD = ICE. Thus we see that either medial line must be more than half the other,-a proposition which is proved by this solution. 815. The hypotenuse and radius of the inscribed circle of a right angled triangle being given, to determine the triangle. Results.-Calling the hypotenuse h, the radius r, the base x, and the per2r + h + 2 - 4hr - 4r2 pendicular y, we have, z = 2r h h2 - 4hr- 4rand 2r + h:: q/t2- 4hr - 4r2 2 The results being the same in other respects, the double sign before the radical indicates that the base and perpendicular are interchangeable — a fact which is evident fiom the nature of the case. If the radical is 0, i. e., if h2 - 4hr - 4r2 = O, x = r + th, and y = r + Ah, and the base and perpendicular are equal. Let the student show the same thing geometrically (from a figure). Also, if h2 - 4hr - 4r2 - O, h = 2r (1:~ V2). In this result the negative 280 EXERCISES IN GEOMETRICAL INVENTION. sign is to be rejected, since it would make h negative, as C a/2 > 1. The value h = 2r (1 + v/2) is readily seen from the figure when AB = CB. Thus AC = 2DB = 2 (DO + OB) = 2 (r + / E r V/2) = 2r (1 + V/2) (?). A F B FIG. 43-2. 816. A tree of known height standing perpendicular on a horizontal plane, breaks so that its top strikes the ground at a given distance from the foot, while the other end hangs on the stump. How high is the stump? That is, given the base and the sum of the perpendicular and hypotenuse of a right angled triangle, to determine the perpendicular. Result.-Let a be the height of the tree, b the distance from the foot to the point where the top strikes, and x the height of the stump; then x- a2 - b2 2a a'-_ b.2 b2 b2 Since — = ca - -, - is the distance below the middle, at which 2a 2a' 2a the tree breaks. 81;. In a rectangle, knowing the diagonal and perimeter, to find the sides. 818. Knowing the base, b, and altitude, a, of any triangle, to find the side of the inscribed square, x. ab Result, x - a + b 819. In an equilateral triangle, given the lengths, a, b, c, of the three perpendiculars from a point within upon the sides, to determine the sides. SUG'. —Find an expression for the altitude in terms of the sides; and then get two expressions for the area of the whole triangle. Equate these. Result, each side = 2 (a -- V/3 820. In a right angled triangle whose hypotenuse is h, and difference between the base and perpendicular d, to find these sides. - dt i V/4h2 _ d2 d =F _/4112 - d Results, x= 2 x + d - APPLICATIONS OF ALGEBRA TO GEOMETRY. 281 QUERIES. —What is the geometrical significance of the fact that the results are the same except as regards the signs? In the first result why must the negative value of the radical be rejected; and in the second the positive? 821. In an equilateral triangle given the lines a, b, c, drawn from a point within or without, to find the sides. Result.- Each side = 2 la'.]- 5~.~ c —,v/ (a~b2 6202 + c2a2) - 3 (a'4 + 6' q d') + The radical is + when the point is within, and - when it is without. 822. The perimeter of a right angled triangle and the perpendicular from the right angle upon the hypotenuse being given, to determine the triangle. SuG's.-Let s be the perimeter, p the perpendicular upon the hypotenuse, and x + y, x - y the two sides about the right angle. Then the hypotenuse = s - 2x, and we readily form the two equations p (s - 2x) = x2 - y2, and S(S ~ 2p) (X + y)2 + (x -y2 = (s - 2x)2 (?). Hence x and this value substi4(- + P), tuted in either equation will give y. 823. The base of a plane triangle is b and its altitude a, required the distance fiom the vertex at which a parallel to the base must cut the altitude in order to bisect the triangle. Result, - QUERY.-What does the fact that b does not appear in the result show? 824. Having given the area of a rectangle inscribed in a triangle, can the triangle be determined? Can it, if the rectangle is a square? If the rectangle is a square and the triangle right angled? If the rectangle is a square and the triangle equilateral? 825. The sides of a triangle being a, b, c, to find the perpendicular upon c from the opposite angle. Result, p = V2 (a2 + 2) + 2)a2b - a4 - -C SUG's.-Observe that a and b are similarly involved in the result, but c is differently involved from either. This is evidently as it should be, since a and b are the sides about the angle fiom which p is let fall; and are thus similarly related to p. But c, the side on which p falls, is differently related to p from 282 EXERCISES IN GEOMETRICAL INVENTION. either of the others. The student should be able to write the value of the perpendiculars upon each of the other sides, from this one. Thus, that on a is 2a, ('; ~ 2 + 02' - 4 a-. 826. The sides of a triangle are a, b, c, to find the side of an inscribed square one of whose sides falls in c. SvG's.-The altitude may be found from the preceding, hence may be assumed as known. Call it p. Then the side of the required square is Cp What is the side of the square standing on a? On b? QuERY. -Will the square be the same on whichever side it stands? Observe that though the values here found are apparently different, they smay not be so really, since p is different in each case. But let the student decide. 82T. Having the area of a rectangle inscribed in a given triangle and standing on a specified side, to determine the sides of the rectangle. Result, b being the base on which the rectangle stands, 1 the altitude from this base, and s the given area, we have for the sides x = + / _ - sband y = F~ PV 4 pb SuG's. —The ~ and:F signs indicate that, in general, there can be two equal rectangles inscribed standing on the same base. The student will do well to illustrate it with definite numerical values, as p = 10, b = 6, s = 10. Again, b- must be greater than -, and P4 > -,. e., s must be less than jpb. That is, the greatest rectangle is half the area of the triangle, since I pb is lhe area of the triangle. 828. The Algebraic solution of a problem often enables us to effect a geometrical construction. We will give a few examples. Through a given point within a circle, to draw a A chord of a given length. V /0) J SOLUTION.-Let s be the length of the required chord, and >K' P the given point. Since P is a known point, call AP - a, PB = b, AB being the diameter through P. Let CD FIG. 433. represent the required chord, and calling CP, x, PD = s - x. Then sx - x2 = ab; whence x = s ~ V/is - ab. APPLICATIONS OF ALGEBRA TO GEOMETRY. 283 E To effect the geometrical construction, let s be the given D chord, and P the point in the given circle. Draw the diameter through P, and erect PE perpendicular to it. Make EH = s; then since PE = ab, PH - - V / - ab. Now take Pi = As, and from P as a centre, with a radius Pl -= As + Vis' — ab, strike the arc DI intersecting the circumference. DPC is the chord required. s FIG. 434. From the radical Vis2' - ab we see that, if ab > is2, X is imaginary, as we say in algebra. In such a case the problem is geometrically impossible, as will appear from the construction, for then PE is greater than EH, which makes HP, the representative of /t -- ab, impossible. If is2 = ab, z has but one value, and the segments are equal. 829. To find a point in a tangent to a circle from which, if a secant be drawn to the extremity of the diameter passing through the point of tangency, the external segment shall have a given length. SOLUTION.-Let AB = d be the diameter of the given circle, DX = a the external segment jf the required secant, and the whole secant BX = x. Then 2 -- ax = d9, and x = a~ V/d + A.', To effect the geometrical construction, construct B the radical by taking AC = la; whence BC =,/d2 + ~a. Now make CY = a, and with B as a centre, and BY as a radius, strike an arc cutting the tangent, as in X. Then is FIG.435. BX = x = la + V/d2 + ~a2. The negative value of the radical is inapplicable in this elementary, geometrical sense, since as V/d2 + jaA > la, this would make x a negative quantity. Again we see that no real value of a can render x imaginary. We can observe the same things from the geometrical construction. Thus, if the negative value of the radical were taken, x would be numerically less than BC, by pa, or AC. But BC - AC < BA. Hence an arc struck from B with the required radius would not cut the tangent. We see also that a may have any value between 0 and oo. 830. Given the hypotenuse and area of a right angled triangle, to construct the triangle. SuG's.-Let h be the hypotenuse, s2 the area, and x the perpendicular from the right angle upon the hypotenuse. Then hx = 2s2,,or:s: s:, andh: 2s:: 2s: 2x. -' The figure will suggest the construction. FIG. 436. 284 EXERCISES IN GEOMETRICAL INVENTION. 831. Through a point between two lines which intersect, to draw a line which shall cut off a triangle of given area. SuG's.-Let AY = x, and the required area = s2. We have h:H:: x -b:. H - hx ~~x 2s~x — b' A B DD y N 7 j ( A h A FIG. 437. 89 To construct this, find c = 7, i. e., construct a thlird proportional to h and s. Then construct V/c (c - 2b), i. e., find a mean proportional between c and c - 2b; let this be m. Whence x = c + m1. In general, there may be two solutions, if any, since there are two values of x. [This should also be observed fiom the figure.] But if 2b > - tlere is no solution. S2 If =- 2b, there is but one solution. In the latter case where is the given point s2 O? What is the geometrical difficulty when 2b > h? Can m be numerically greater than c? 832. To construct the four forms of the affected or complete quadratic equation, viz., (1.) x2 + px - q = o, (2.) x2 - px - q = o, (3.) x2 - x + q = o, (4.) x2 + px + q = o, without solving the equations. FIRST FORM. a2 + px - q = o.-Dlraw any,L two lines as LM, NP, intersecting in some point 0. Resolve q of the equation into two factors, as r and q', so that we have x2 + -r X q' = N E o. TakeOA=p, OB = r,OC = q'. Bisect CB and AO by perpendiculars, and from their inOF tersection F as a centre, with a radius FB, draw a circle. Then DO, or AE, is x, the positive root. For x (x + p) = rq', or x2 + px - rq' = o. C\ The negative root is OE. Thus, let OE = (- x). Al Then DO = AE = (- ax- p). Hence (-x) FIG. 438. ( — - p) = 2 f p = rq, orX2 + px - r = 0. This construction is evidently always possible irrespective of the relative magnitudes of p, r, q'; a fact which agrees with tlie statement in algebra that this form always has real roots. SECOND FORM. x2 — p - rq' = o.-The construction is the same as for the first form; only, in this case OE is the positive, and DO the negative root. Thus for OE = x (positive), we have DO x OE = (x - p) x = rq', or x2 - px - APPLICATIONS OF ALGEBRA TO GEOMETRY. 285 rq' =o. For DO = (- x), we have DO x OE = DO(OA + AE) = DO (OA + DO) = (-) (p- x) = rq', or 2 - - rq'= o. Observe that in the first case the negative root is numerically greater than the positive; while it is the reverse in this form. This agrees with the conclusions of algebra (See COMPLETE SCHOOL ALGEBRA, 104). THIRD FORM. x2 - px + rq' = o.Draw any two lines, as OM, OP, meet- P ing at O. Take OA = p, OB = r or q', E A and OC = q' or r. Erect perpendiculars at the middle points of OA, and BC; and from their intersection F as a centre, with a radius FB, strike a circum- o C. ference. Then OE and OD are the values of a. For OE = x, OE x OD = OE x EA = OE (OA- OE) = z(p - ) FIG. 439. = rq', or a2 -a + rqt' = o. For OD =, D x OE = OD (OA - AE) = OD (OA- OD) = (p - x) = rq', orT2 -px + rq' =o. Observe that the former value of x is greater than the latter, but that neither is negative. So also, we may readily see that the roots may become equal, and also, imaginary. Thus if the circle were tangent to OA, the roots would be equal, and if it did not touch OA they would both be imaginary. (See Algebra, as above.) FOURTH FORM. X2 + px + rq' = o.-The construction is the same as the last, only both values of x are negative. Thus, (- a) [p - (- x)] = - ax) (p + x) = rq', - px - xz - rq' = o, or' 2 + px + 9'q' = o. Sce.-Thus we see that we can construct any equation of the second degree containing but one unknown quantity, which has real roots. Hence, if the algebraic solution of a geometrical problem requires only the resolution of such an equation, the algebraic solution will lead to the geometrical construction. 833. We have now given sufficient illustrations of this most interesting and important subject, so that the student should have caught the spirit of this method of using algebra to subserve the purposes of geometrical investigation. We shall simply append a list of problems, upon which the student can put in exercise both his algebraic and geometric knowledge. But we cannot refrain from repeating the advice, that the learner should not rest satisfied with the mere algebraic resolution of the problem. He should be ambitious to tree, as fully as possible, the wonderful relations which exist between the abstract operations of algebra, and the more concrete relations of geometry. 286 EXERCISES IN GEOMETRICAL INVENTION. EXAMPLES. 834. Given the perimeter of a right angled triangle and the radius of the inscribed circle, to determine the triangle. FIG. 440. 835. Given the hypotenuse of a right angled triangle and the side of the inscribed square, to determine the triangle. 836. In a right angled triangle, given the radius of the inscribed circle, and the side of the inscribed square, the right angle of the triangle constituting one angle of the square, to determine the triangle. SuG's. —Letting x and y be the sides, z the hypotenuse, r the radius of the inscribed circle, and s the side of the inscribed square, we have s = xy - r (x + y + z), and + y - z + 2r. Whence z —= 2 etc 83". In any triangle whose sides are a, b, c, to find the radius of the inscribed circle. 838. Show that the area of a regular dodecagon inscribed in a circle whose radius is 1, is 3. - av 839. Find the area of a regular octagon ai'' whose side is a. Result, 2 (2 X2 + 1) a. 840. Find the radii of three equal circles deFIG. 441. scribed in a given, circle, tangent to the given circle and to each other. 841. The space between three equal circles tangent to each other is a; what is the radius? 842. In a triangle, given the ratio of two sides, and the segments of the third side made by a perpendicular let fall from the angle opposite. APPLICATIONS OF ALGEBRA TO GEOMETRY. 287 843. In a triangle, given the base and altitude, and the ratio of the other sides, to determine the triangle. 844. Given the base, the medial line, and the sum of the other sides of a triangle, to determine the triangle. 845. To determine a right angled triangle, knowing the perimeter and area. SUG'S. x2 + y2 = z, x + y + z = 2p, and xy=2s2, give y + x=2p-, x'2 + 2xy + y2 = 4p' - 4pz + z2, z' + 42 = 4sp' -- 4pz + z2; whence p2 s.2 p.2 + s_ z 2-. Now use y +x = 2p-z-= P +-, and xy 2s. 846. To determine a right angled triangle, knowing the perimeter, andc the sum of the hypotenuse, and the perpendicular upon the hypotenuse from the right angle. SUG's. x + y -2 z2 x + y + z = 2p, z + u = a, xy = zu. Then 2 t + 2x y2 - 4p - pz + z2; whence 2xy = 4p' - 4pz, and hence 2z (a - z) 4p - 4pz, etc. 84'. The volume, the altitude, and a side of one of thle baseq of the friustum of a square pyramid being known, to determine a side of the other base. 848. To determine a right angled triangle, knowing the perimeter, and the perpendicular let fall from the right angle upon the hypotenuse. 849. To determine a triangle, knowing the base, the altitude, and the difference of the other sides., 850. To determine a triangle, knowing the base, the altitude, and the rectangle of the other sides. 851. To determine a right angled triangle, knowing the hypotenuse and the difference between the lines drawn from the acute angles to the centre of the inscribed circle. SuG's. —Let fall CD a perpendicular upon AO produced. Now,. since the the angles BAC and ACB are bisected,, and COD = OAC + OCA, and ICD = IAB, they being ( - D complements of the equal angles CID, IAB, we have, COD A B = OCD, and CD = OD = x/ CO. Hence, putting AC FIG. 44'. = h, CO = x, and AO = x + d, we have (x + d + 1/ )2 + (V/ x)2 = h2. From this x is readily found. The student should then be able to complete the solution. 288 INTRODUCTION TO MODERN GEOMETRY. 852. Given two sides of a triangle and the bisector of their included angle, to determine the triangle. 853. Given the three medial lines, to determine a triangle. 854. Given the three sides of a triangle, to determine the radius of the circumscribed circle. 855. Four equal balls whose radius is 9r are placed on a plane so that each is tangent to the other three, thus forming a pyramid; what is its altitude? 856. Given the base of a triangle, the bisector of the opposite angle, and the radius of the circumscribing circle, to determine the triangle. C SuG's.-First to find ED - x. Since EM -- r - -/'2 - b2, it may be considered known and put equal to c. We then have DM = V/c2 + x; and also, DM x a = AD x DB = b2 -22 or DM Whence /Vc2 + x2 a= and z is readily found. Calling ED = s, the student will have no difficulty in FIG. 443. proceeding with the solution. CHAPTER II, INTRODUCTION TO MODENi GEOMIETRY.* SECTION I. OF LOCI. 85T. The term Locus t, as used in geometry, is nearly synonymous with geometrical figzure, yet having a latitude i'n its use which the other does not possess. The locus of a point is the line (geo* With strict propriety only the latter sections of this chapter belong to the Modern GeonLetry, technically so called. But, as the entire chapter is composed of matter which has not hitherto found place in our common text-books, and the relative importance of which is becoming more fully appreciated in modern times, the author has ventured to embrace the whole under this title. t The word Locus is the Latin for place. OF LOCI. 289 metrical figure) generated by the motion of the point according to some given law. In the same manner, a surface is conceived as the locus of a line moving in some determinate manner. ILL'S.-The locus of a point in a plane, which point is always equidistant firom the extremities of a given right line, is a straight line perpendicular to the given line at its middles point. c Thus, suppose AB a fixed line, and the locus of a point equidistant from its extremities is required; that point may be anywhere in a perpendicular to AB at its middle point, and cannot be anywhere else in this plane. A B This perpendicular is the locus (place) of a point subject to the given law. Again, a boy on the green is required to keep at just D 20 feet from a certain stake; where may he be found? FIG. 444. i. e., what is his locus (place)? Evidently, the circumference of a circle whose radius is 20 feet. Thus, the locus of a point in a plane, equidistant from a given point, is the circumference of a circle. This is the place of such a point. What is the locus in space of a point equidistant from a given point? What is the locus of a point in space equidistant from the extremities of a given line? A plane. What is the locus of a line moving so that each point in it traces a right line? In general, a plane; if it move in the direction of its length, a straight line. What is the locus of a right line parallel to and equidistant from a given line? What is the locus of a right line intersecting a given line at a constant angle-? * A conical surface of revolution. What is the locus of a semicircle revolving on its diameter? PROPOSITIONS AND PROBLEMS IN DETERMINING LOCI. [NOTE. -The student should be required to give every demonstration in form, and in detail. Frequent exercise in writing out demonstrations, is almost the only method of securing a good, independent style il demonstration.] 858. Thieo.-The locus of a _point in a plane, equidistant from the extremities of a given line, is a perpendicular to that line at its middle point. SUG.-To prove this we have simply to show that every point in such a perpendicular is equidistant fiom the extremities of the given line, and that no other point has this property (PART II., 129). * That is, an angle which remains of the same size. 19 290 INTRODUCTION TO MODERN GEOMETRY. 859. Prob.-Find the locus of a poinit at any constant distance m from a straight line. Of what proposition in PART II. is this the converse? SvG's.-To prove the proposition which the answer to this question asserts, it will be necessary to show that every point in the affirmed locus is at the same distance fiom the given line and that no other point is at that distance. We affirm that the locus is two iP E c!, D ~right lines parallel to the given line and at a distanzce AP'm B mi therefrom. The formal demonstration is as follows:' 1n Let AB be the given line, and OE, OE', perpendiculars D thereto, each equal to m. Through E and E' draw CD and C'D' parallel to AB; then is CD, C'D', the FrIG. 445. locus required.* For, by Part II. (156), every point in CD, C'D', is at the distance m fi'om AB; and we may readily show that any other point, as P or P', is at a distance greater or less than i friomAB. Htence CD, C'D', is the locus required. 860. Theo.-In a circle, the locus of the, centre of a chord parallel to a givens line is a diameter. L DrM.-Let mn be any circle, and AB a given line. A ( wS.,\ Then is the locus of the celntre of a chorcl parallel to AB, \\ ~P H a diameter of the circle. P For, let DH be any cllord pai'allel to AB. Through the centre of thle circle C, and P, the middle point of DH, B draw EL. Now EL is perpendicular to DH (?), and conFIe. 446. sequently to AB (?). Then will EL be perpenlicular to any and every chord parallel to DH (?), and hence will bisect such chord (?). Therefore the locus of the centre of a chord parallel to AB is a diameter. Again, any point in the circle and out of the line EL is not the middle point of chord parallel to AB. Thus, letting P' be such a point, draw a chord through P' parallel to AB. As there can be but one such chord (?), and as EL bisects it (?), P' is without the diameterl(?). 861. Theo. —The locus of the centre of/ a circumference passing through, two given points is a stracight line. SuG. —Consnlt PART II. (159, 163, 197). The student should put the argument in for1im. 862. Theo.-The locus of the centre of a circle which is tan* It is important that the student think of these two lines as one locus, or as parts of one and the same locus, if this will aid the conception. A locus may consist of any number of detached parts: all that is necessary being that the given conditions be fulfilled. In this respect the word locus has a more enlarged meaning than the term geometrical figure. OF LOCI. 291 gent to a given circle at a given point, is a straight line 2assing throug7h the centre of the given circle. DEM.-Let C be the centre of the given circle, T and B the point in the circumference to which the circle * shall be tangent, the locus of whose ( - centreis required. Through B drawTL tangent x i- c to the given circle. Now, a circle passing throughB, and tangent to the given circle, will have TL for its tangent (?), and as a radius is perpendicu- L lar to a tangent at its extremity, and only one FIG. 447. perpendicular can be drawn to TL through B, the centre of a circle tangent to the given circle at B must be in this straight line. Moreover, as the given circle is tangent to hlie right line TL at B, its centre is in the perpendicular AX. Hence AX is the locus required. 863. Theo.-The locus of the centre of a circle of given raditus R, and tangent to a given straight line, is two p2arallels to this line at.a distance R therefronm, on each side. Give proof in form. 864. Prob. —Find the loculs of thie centre of a circle of givenz radius R, whose circunmference passes through a given point. Give proof in form. 865. Theo.-The locus of the centre of a line of constant lengtlh, having its extremities in two fixed lines which cut each other at rgglilt angles, is the circumference of a circle. SUG's.-Let MN be the length of the c given line, and CD, and AB, the two lines intersecting at right angles, in which the extremities of MN are to remain. Now, PN M in whatever position MN may be placed, B N its middle point, P, is at the same distance (4MN) from O (?). To show that any point M not in this circumference, as p, is not the middle point of a line equal to MN passing M through it, and limited by the fixed lines, D firom p as a centre, with a radius jMN cut FIG. 448. CD, as in C; and from C as a centre with the same radius strike the arc pP. If p is without the circle, CB > MN, if within, less. Hence, the required locus is a circumference whose centre is 0, and whose radius is JMN. * Observe the form of expression. We say "the circle," and not "the circles," using the term in a generic sense, as including all which have the required property, i. e., all vwhich are tangent to the given circle at B. 292 INTRODUCTION TO MODERN GEOMETRY. 866. P'rob.-Find the locus of the centre of a chord of constant length, in a given circle. SuG.-We say, once again, always give the proof in, form. 86'i. Prob.-Find the locus of the vertex of the right angle of a riyght angled triangle of a constant hypoteneuse. 868. Prob.-Find the locus of the middle point of the chord intercepted on a line through a given point, by a given circumnference, when the given point is without the circumference, when, it is in, and when it is qwithin the circum2ference. 869. JProb.-Find the locus of a point the sum of whose dislanlces from two fixed intersecting lines is constant, i. e., is equal to a given line. SOLUTION. —Let AB and CE be the fixed'A,-.' ~ lines, and m the constant distance. Draw p,',?p"'.,'V.... P;, MN parallel to AB, and at a distance m firom it. Bisect the angle CPN. Then is -, \;' i~t~al LR (a part of) the locus required. For ict"' D [the student will here show that the sum IV> Doss of the distances from any point in MN to AB and CE, as PD, P'D' + P'd', P"D" FIG. 449. + P"d", P"'D"', PIVDIv - PvdIv, is constant and equal to m], observe that when one of the perpendiculars measuring the distance fiom a point in the locus, changes fiom one side to the other of the line on which it is let fall, its sign changes. Thus P'D', PD, P"D" being considered +, P'd' and P'VD'v are to be considered -. This is a general principle in mathematics. See PART II. (215), and foot note. Finally, LR is only a part of'the locus, since there is another line on the op. posite side of AB, obtained by drawing the auxiliary MN on that side, which fulfills the same condition. The student should show what the result is when we dllaw the auxiliary parallel to CE, and on either side of it, also that any point not in one of these lines cannot fulfill the required condition. The complete locus is four indefinite right lines intersecting each other at right angles, so as to inclose a rectangle. 8'YO. Prob.-Find the locus of a point such that the sum of the squares of its distances from two fixed points shall be equivalent to the square of the distance betwleenz the fixed points. 8''1. Prob.-Find the locus of the intersection of two secants OF LOCI. 293 drawn through, the extremities of a fixed diameter in a given circle, one of the secants being alwoays perpendicqular to a tangent to the circle at the point where the other c cuts it. A SuG's. P being the point, show that PB = AB, for any position of AP and BP. Hence, any point in the circumference having B for its centre, and AB fobr its radius, fulfills the conditions. Show that any point out of the circumference does not fulfill the /p conditions. 450. FIG. 450. 8'$2. Prob.-Find the locuts of the intersection of two lines drawn from the acute angles of a right E antgled triangle, throtgh the points where the perpendicular to the hypotenuse cuts the opyosite sides, or sidles prodtcecl. Sua's.-The locus of P is lrequirecl. Prove that APC is always a right angled triangle, wherever the A C perpendicular EF to the hypotenuse AC is drawn. FIG. 451. 873. Prob.-Find the locus of a point which, divides a line drawn from a fixed point to a fixed line in a fixed ratio. SuG'. —Most problems in finding loci such as are treated in Elementary Plane Geometry, viz., right lines and circles, are readily solved by constructing a few points according to the given conditions, A whence we can determine by inspection whether the required 0c locus is a right line or the circumnfirence of a circle; and, having'discovered this fact by inspection, it will remain to show zwhy it should be so. Thus, in the present problem, 0 o c being the fixed point, alnd AB the fixed line, drawing a few P lines, OC, according to the requirements, and dividing them C in the same ratio (in the figure 3:2), we find a few points P in the locus. We then discover at once that the locus is B a right line parallel to AB, an-d can easily see why it should FI br.. 452. be so. 874. Prob.-Two fixed circumferences intersect: to find the locus of the middle point of the line dravwn through one of the points of intersection and terminated by its other intersection with the circumferences. 294 INTRODUCTION TO IMODERN GEOMETRY. SuG's. —We will first give an example of the course which the mind of the student might take in his efforts to discover the solution. He would naturally draw two unequal * circles, as M and N, M and, through one of the points of intersection, as A, draw BC, and bisect it at P. It is / D' ~p A the locus of P that is desired. Now, supBt D i C pose the line BC to revolve about A, B - ~ i,, passing towards B', and C towards A. It is the path of the middle point that he seeks. VWhen C reaches A, the line becomes tangent B C' to N, and P is the middle point of the chord FIG. 453. AB'. In a similar manner, he sees that the middle point of the chord AC', tangent to M at A, is also a point in the locus. Again, he observes that as B moves towards M, and C towards N, P moves towards A, and when AC = AB, P is at A. It now appears probable that the locus of P is a circumference. Proceeding on this hypothesis, he reasons, that, if this is true, AP' and AP" are chords of the locus, and, bisecting them with perpendiculars, he will have the centre of the locus. Locating O thus, he observes that it appears to be in the line joining the centres 0' and 0", and about midway between them. This leads him to see, whether, by assuming the middle point of O'O" as the centre of a circle, and OA as a radius, he can prove that any suchline as BC drawn through A is bisected by this circumference, as at P. This lhe can readily prove by means of the perpendiculars OD, O'D', and O"D",' which bisect the chords AP, AB, and AC. For, since these perpendiculars are parallel, and 0'0 = 00, D'D = DD"; whence D'P = AD", and, adding AP to each, D'A = PD", or D'B = PD". Adding to BD', D'P, and D"C (= AD" = D'P), there results BP = PC, and the hypothesis is true. But, in giving this problem as a recitation, the student will proceed as follows: Letting M and N be the two fixed circumferences, intersecting at A, join their centres O' and 0", and bisect 0'O" as at O. With O as a centre and OA as a radius, describe a circle. Then is this circumference the locus required. For, let BC be any secant line passing through A, we may show that P is the middle point of BC. [Having done this, as above, and shown that any point not in this circumference is not the middle of the- secant line passing through A, his solution is complete.] 8'5f. Prob. —If the liZhe AB is divided at C, fild the locus of p, so tchat angle APC = ansle BPC. SuG's.-In, seeking for the solution, the follom-ing would be a natural process. Drawing any line, as AB, in the lower part of the figure, taking C, any point in it, and conceiving BP and AP drawn so as to make equal angles with PC, we would naturally discover that, if a circle were circumscribed about BPA, PC produced would bisect the arc below AB. Thus we discover a ready method of locating P; i. e., in the main figure, bisect AB by a perpendicular, as ED, and * Equal circles would probably have special relations. OF LOCI. 295 with any point on ED as a centre, pass a circumference through A and B. Througll D and C draw a line, and P is a point in the locus (?). Any number of points can be found in this way; and, having found a few, as P, P', P", P"', etc., the situation of these will sucggest that, probably, the p' — locus is the circumference of a, circle whose centre is in AB pro- / /... duced. If this should be the fact, I. - A CP' is a cllord of that circle, and, ~ ~' A erecting a perpendicular at the.,, middle point of CP', its inter- \ / section witlI AB produced, as. p... 0, will be the centre of the locus (?). We will now endeavor p to prove that any point in this circumfere'nce, as P', is so situated that BP'C = CP'A, and that no D point out of this circumfer- FIG. 454. ence has this property. We can readily show that the angle OP'B = OCP'- BP'C = OCP' - CP'A (?). But P'AC = OCP' - CP'A (?)... Triangle OP'B is similar to OP'A (?), and OA: OP'::OP' OB, or OA; OC;;OC; OB. Now, for any point in this circumfererce, as P, we shall have OA: OP::OP: OC, since OP = OC, and OA, OC, and OB are constant.- Hence, wherever P is taken (in this circumferenceS, the triangle OPB is similar to OPA, angle OPB = PAB, and BPC = CPA. Finally, that no point out of this circumference possesses this property is evident, since the distance of such a point from O would not equal OC, and the angle OP1B (Pi being such a point) would not equal P1AB. 8'6. Prob.-In a fixed circle, any two chords intersect at right angles in a fixed point; find the locus of the centre of the chord joining their extremzities. Give the proof. 81'T. Prob. —Find the locus of the point in space equidistant from three given points. Give the proof. 878. Prob.-Find the locus of thle point in space equidistant from two given points. Give the proof. 8T9. Prob.-Find the locus of the point in a plane sutcc that the differen2ce of the squares of tIhe distances from it to two fixed points without the pjlane shall be constant. Su&s.-Conceive the two points without the plane joined by a right line, and a perpendicular to this line drawn from either extremity of it; the point where this perpendicular pierces the plane is a point in the locus (?). The required locus is two parallel straight lines. 296 INTRODUCTION TO MODERN GEOMETRY. 880. Prob. —Find the locus of the mniddle point qf a straight line of constant length, whose extremities remain in two lines at right angles to each other, but which are not in the samee lane. Give the proof. 881. Prob.-Find the locus of the point equidistant from two fixed planes. Give pioof. SUG's.-Consider, 1st, When the fixed planes are parallel; and 2d, when they intersect. 882. Prob.- What locus is the intersection of a plane and the surface of a slp)here? Give proof. 883. Prob.- Wltat locus is the intersection of the surfaces of two given spheres? 884. Prob. —Find the locus oqf the point in space such that the ratio of its distance fronm a given right line to its distance fron a fixed point in that line is constant. SECTION II. OF SYMMETRY. 885. DEF.-Two points are said to be symmetrical wcith respect to a third, when the right line joining the two points is bisected by the point of reference, called the Centre of Symmniietry. 886. DEF.-Two loci, or two parts of the same locus, are syruP? m?v2eetrical with respect to a point, when'/ W X every point in one has its symmetrical point / A' / /s/ a — P in the other. P/ /'(a) D o ILL'S.-In (a) P' is symmetrical with P in rem' spect to S, it S is the middle point of PP'. In (b) (b;) we observe that the semi-circumference Am'B is A B symmetrical with the semi-circumference AmB, pi..s p. in respect to the centre S; for any point P in the p-. latter has a symmetrical point P' in the former. B (e) A In (c) the triangle A'SB' is symmetrical with ASB ia. 45. in respect to S (?). OF SYMMETRY. 297 88T'. Jheo.-The symmetrical of a A straight line, with respect to a point, is an. \ p equal straight line.'s SUGo's.-We commence by assuming A'S = AS, and B'S BS, and drawing B'A'. We then P have to show that any point in AB, as P, has its A symmetrical point P' in B'A'. FIG. 456. 888. Theo. —The symmetrical of an angle, wzitl respect to a point, is an equal angle. o SUG's.-To show. the symmetrical of AOB, with respect to P B S, take A'S = AS, B'S = BS, O'S = OS. and draw O'B', O'A'. A Then show that any point in OA has its symmetrical in O'A', A and any point in OB has its symnietrical in O'B'. Hence, A'O'B' is the symmetrical of AOB, with respect to S. Then apply AOB to A'O'B' and show that these symnmetricals o' are equal. FIG. 457. 889. Prob.-Having given a polygon, to A draw its syinmmetrical with respect to a yiven.... p2oint.. 890. Theo.-Any polygon is equal to its o symmetrical with respect to a given point. FIG. 458. Su's. — Proof by revolving the figure about S, keeping it in its own plane, each line, as AS, ES, etc., passing through 180~. 891. DEF. —The lines AS, ES, BS, etc., are radii of symmetry. 892. DEF.-Two points are syzmmtetrical iwith respect to a straight line in the same plane, when the straight line which joins them is bisected at right angles by the line of reference, called the Axis of Symmetry. 893. DEF. —Two loci, or two parts of the same locus, are syrm9metrical with respect to a straight line when every point in the one has its symmetrical point in the other. ILL'S.-Thus in (a) P and P p P' are symmetrical pointsP with respect to the right line X'X, P's -- Ps and X'X X s s is perpendicular to PP'. So \ X the part of (b) above X'X is symmetrical with the part (a)'P' p below, i. e., the curve is sym- () metrical with respect toX'X. FIG. 459. 298 INTRODUCTION TO MODERN GEOMETRY. 894. Theo. —The symmaetrical of a straight line, with respect to a rectilinear axis of symmetry, is an equal righyt line. DEM.-Let AB be any straight line, and X'X the p P PoJ B axis. Let fall the perpendiculars As and Bs, and ^ ~ —~, {, produce them till A's = As, and B's = Bs. Then is 1 i, x AB -- A'B', the symmetrical of AB. For, taking P, X, s, s, any point in AB, letting fall Ps, and producing it to A! *! P', the point P' is symmetrical with P; since re^ p'.' - volving sA'B's upon X'X, A'B' -will coincide with FIG. 4C60. AB, and P' will fiall at P. Hence A'B' = AB, and every point in AB'has its symmetrical point in, A'B' (893). 895. COR. 1. —If two straight linues intersect, their symmznetricals intersect, and the points of intersection are symmetrical. The student should show how this follows from the proposition. 896. COR. 2.-Two rectilinear symmetricals vmeet the axis in the samze:point, and make equal angles therewith. Student give proof. 89". DErF.-A Tracpezoid like ABB'A', having its non-parallel sides equal, is called Isos'celes. O p. 898. Theo. —The symmetrical of ag, angle, A B s vwith respect to a rectilinear axis of symmetry,',,A X is anll equal angle. X {B, SuG's.-The student should be able to give the dem-,pr oniostration from the figure, in a Ilanner altogetlher similar to the preceding; or, drawing AB and A'B', he FIG. 461. can base it upon the preceding. 899. Prob.-Having given. a polyAC yon, to draw its symmtetrical with respect A, to a gqiven axis. } I 1, E,* *900. Theo.-Any plane figure is x', I X equal to its symmetrical, with reference to a rectilinear axis. Proof by applying one to the other by revolution. FIG. 462. OF SYMMETRY. 299 901. DEF.-Two loci in space, or two parts of the same locus (planes or solids), are symmetrical with respect to a point, when every point in one has a corresponding point in the other, such that the line joining them is bisected by the point called the centre of symmetry. ILL's.-Symmetrical triedrals (PART II., 432) afford an illustration of solids symmetrical with respect to a point-the vertex. Tle two hemispheres into which a great circle divides a sphere are symmetrical parts of the solid (sphere) with reference to the centre. 902. DEF.-Two points in space are symmetrical with respect to a plane called the Plane of ASymmetlry, whell the line joining the points, is perpendicular to the plane and bisected by it. 903. DEF.-TWO loci in space (planes or solids), or two parts of the same locus, are symmetrical withll respect to a plane when every point in one has its symmetrical point in the other. 904. DEr. —The corresponding (symmetrical) parts of symmetrical figures are called Homologous parts. 905. Theo. —The sy1mmetrical of a right line, with respect to a plane, is an equal right line. DEM.-Let AB be any right line, and MN the plane of symmetry. Let fall the perpendiculars Bb, Aa, upon A the plane, produce them, inaking B'b = Bb, A'a = Aa, and join A' and B'. Then A'B' = AB, and is its sym- 8 metrical. For ABB'A' being a plane figure (?) and ab, M the intersection of this plane with MN, being a right line bisecting AA' and BB' at right angles (?), we may revolve abB'A' upon ab and bring A'B' into coincidence N with AB. Hence A'B' = AB. Again, P being any point in AB, draw PP' perpendicular to ab, and upon i p revolution P' will fall in P, and P's = Ps. Hence, every A point in AB has its symmetrical in A'B', and the latter FIG. 463. line is the symmetrical of the former. 906. CoR. 1.-A right line and its symmnetrical, with respect to a plane, pierce the plane at the samle point. Student give proof. 907. Theo. -The symmetrical of a plane angle, with respect to a plane, is an equal plane anyle. 300 INTRODUCTION TO MODERN GEOMETRY. I)EM.-Let AOB be any plane angle, and MN the o plane of symmetry. Let P be any point in AO, and Pp/,..p P' any point in OB. Let O' be the symmetrical of O, 4 $'- B NP' of P, and Pi' of Pi; then is A'O' the symmetrical of PA -'" -N AO, O'B' of OB, and angle A'O'B' of AOB. Now by / k / the preceding proposition the two triangles POP, and M,B' P'O'P', are mutually equilateral, whence AOB = its symmetrical A'O'B'. o' QunERY.-When will the triangle pop' exist, and FIG. 464. when not? 908. Theo. —_Any plane pol0/f0on has for its symmetrical, with reference to a plane, an equal plane polyyon. SuG's.-ABCDE being any plane polygon, and / c MN the plane of symmetry, by constructing A r D A',B',C',D',E' synmmetrical with A, B, C, D, E, we have by the preceding propositions A'B'C'D'E'. equilateral and equiangular with ABCDE; whence ___i it only remains to show that A'B'C'D'E' is a plane / (not a alrped) surface. Let F be any point in the /;"-t- -.../ angle AED, draw HI, and let H' and I' be the symN mletricals of H and I (895). Draw H'l'. Then is Iz-l'' every point in HF within the angle BAE has its ApD...i F "D synmletrical in H'F' (894). Thus, by taking three' c' npoints, not in a straight line, in the angle BAE, we FIG. 465. can show that their symmetricals are in the plane B'A'E', and also in A'E'D'. In like manner, all the angles of A'B'C'D'E' can be shown to be in the same plane. 909. CoR.-If tw1o planes intersect, their symmetricals intersect, and the two intersections are symmetrical rig/ht lines. The student should show how this grows out of the proposition. C / 910. Theo.-The synmmetrical of a diedral is an equal diedral. 0 B A DEM. AOB being the measure of the diedral A-OC-B, and A'-O'C'-B' thle sylmmetrical diedral, and O' the symnn metrical of 0, the symmetrical of AO being A'O', the angle A' A'O'C' is right, and in like manner B'O' being the symmet6o rical of E30O, B'O'C' is right. But BOA = B'O'A' (?), whence the diedrals are equat. FIG. 466. OF SYMMETRY. 301 911. Theo. —Two polyedrons, symmetrical H G with respect to a plane, have their faces equal, E each to each, and their homologous solid angles symnletrical. M A SUG's. -This is an immediate consequence of preceding propositions. Thus E' being the symmetrical solid i N homologous with E, the homologous plane faces including them are equal (908). Again, the facial angles B being equal, but not similarly disposed, the solid angles G I G are symmetrical. E FIG. 467. 9 12. CoR. —Two symmetrical polyedrons can be decomposed into the samne number of tetraedlrons, symmzetrical each to each. For we can decompose one of the polyedrons into tetraedrons having for their common vertex one of the vertices of this polyedron, and each of these tetraedrons will have its symmetrical in the other. 913. Theo. —Two symmetrical poyedrons are equivalent. DEM.-From the last corollary it will appear that it is sufficient for the demonstration of this proposition to show that two symmetrical tetraedrons are equivalent (?). Let S-ABC, and S'-ABC be two tetraecdrons symmetrical with respect to their common base. They have a common base and equal altitudes (?), hence they are equivalent..914. GENERAL ScROLIUM. —We may speak of two loci, or two parts of the same locus, as symmetrical with-respect to a line or plane, whenever all the points in one have symmetrical points in the other, even though the line joining the symmet- p p p rical points be not perpendicular to the axis, or / / P the plane, of symmetry; observing, however,,, / // / that this line is always bisected by the axis or plane. Thus, the ellipse in the figure is symmet- P<' rically divided by the line X'X, since every point p, /,/ in one portion has a symmetrical point in the p othei, as Ps = P's, for every point in the curve. FIG 46. In such a case the parts cannot be brought into coincidence by simple revolution: one part must be reversed. 302 INTRODUCTION TO MODERN GEOMETRY. SECTIO N NI. OF MAXIMA AND MINIMA. 915. DEF. —A liaximmcn value of a magnitude conceived to vary continuously in some specified way, is a value, which is greater than the preceding alnd succeeding values of the magnitude. ILL'S.-Thus, suppose in a given circle, a chord passing 4/ \ tlllhrough a fixed point, P, revolves so as to take successively 32 4A~ 2" the positions la, 2b, Ac, 3d, 4e, etc. It is at a maximum when FIG. 469. it passes through the centre, as Ac. The chord is the magnitude which is conceived to vary in the way specified, and Ac is a value greater than the preceding and the succeeding:/............values. Again, conceive a circle to be compressed or ex/ tended, as in the direction mn, so as to take the forms in0, 1, O, / dicated by the dotted lines, its area will be diminished, the perimeter remaining the same. That is, of all figures of a given perimeter, the circle has the maxiFIG. 470. nmum area. 916. DEF.-A Minimum value of a magnitude conceived to vary continuously in some specified way, is a value which is less thcan the preceding and succeeding values of the magnitude. ILL's.-Thus, conceive the varying magnitude to be a X 1 A 4 X straight line from the fixed point P to the fixed line X'X; FiG. 471. that is, suppose such a line to start from some position P 1, and move through the successive positions P2, PA, P3, P4. PA is a minimum, since it is less than the preceding and succeeding values. PROPOSITIONS CONCERNING MAXIMA AND MINIMA. 91'. Axiom.- The mtinimum distance between two points is a straight line. 918. Theo.-The minimum distance.from a point to a line is a straight line perpendicular to the given line. Student give proof. 919. Theo.-Thbe maximum line which can be inscribed in a given circle is a diameter. OF MAXIMA AND MINIMA. 303 Proof based on the fact that the hypotenuse of a right angled triangfle is the greatest side. 920. Theo.-The sum of the distances froml A two points on the same side of a line, to a point in the line, all being in the same plane, is a minim71um when the lines measuring the distances mnake equal angles with, the given line. Student prove AP + BP < AP' + BP'. FIG. 472. 921. Theo. —If a triangle have a constant base and altitude, its vertical angle is a mzaximum wihen the triangle is isosceles. SUG.-By what is the vertical angle measured? FIG. 473. 922. Theo.-The base and area of a triangle being constant, its per-imeter is a mi-nimum when the triangle is'isosceles.;' SuG's.-The area and base being constant, the vertex remains in a line parallel to the base, for all values of the other sides. The figture will suggest the demonstration, which is based on the fact that any side of a triangle is less than the sum of the other two. FIG. 4T4. 923. Theo. —The difference between the distances from two points on. opposite sides of a fixed line to a p]oint in that line, is a mnaximum, when the lines measuring these distances make equal angles with the fixed line...... SU's. P'O = AP - AP'; but P'O > A'O (= A'P) - A'P'. o FIG. 475. QUERY.-Having the points P, P', and the fixed line given, how is the point A found by geometrical construction? 924. Theo. —-The lengths of two sides of a triangyle being constanlt, the area is a maximum when the included angle is right. FrI. 476. 304 INTRODUCTION TO MODERN GEOMETRY. P p 925. Theo.-Tlhe sum of two adjacent sides of a rectangle being constant (AB), the area is a maximum when thle sides are equal. A' D B FIG. 477. ISOPERIMIETRY. 926. Isoperimetric Figures are such as have equal perimeters, i. e., bounding lines of equal length. Problems in isoperimetry are a species of problems in Maxima and Ainima. Thus, of all figures whose perimeters are mn (say 10 inches), to find that which has the greatest area, is a problem in isoperimetry. Again, what must be the form of a pentagon whose perimeter is?m2, in order that its area may be a maximum? 92T. Theo. —Of isoperimetric triangyles with a constant base, the isosceles is a 9maximum. SuG's.-By means of the figure to Theorem (921), we can readily show that any triangle having the same base as the isosceles triangle, and its vertex either in or beyond the line through the vertex of the isosceles triangle and parallel to its base, has a greater perimeter than the isosceles triangle. Hence, the isoperimetric triangle on the given base has its vertex below this parallel, except when isosceles; and consequently the isosceles is the maximum. 928. CoR. —Of isoperimetric trianyles, the equilateral has the qmnaximum area (?) 92.9. Prob. —Given any triangle with, a constant base, to construct the maxinunz isoperizetric trianyle. 930. Prob.-Given, any triangle, to construct the maximum isoperimetric triangle. D C 931. Theo.-Of isoperbimetric quadrilaterals, /'. %.,~o' the square has the maxizmtm area. DEM.-Let ABCD be any quadrilateral. If AD is not A. equal to DC, the triangle ADC can be replaced' by the A:1SUJ B isoperimetric triangle AD'C, and the area of the quadrilateral increased. So ABC can be replaced by AB'C. Therec " a fore AB'C'D > ABCD. In like manner if AD' is not equal to AB', D'AB' can be replaced by the maximum isoperimetric triangle D'A'B'. So also D'CB' can be replaced by D'C'B'. Therefore A'B'C'D' > AB'CD' > ABCD. Now, A' B A'B'C'D' is a rhombus (?), and the student can show that FIG. 4178. the square on A'B' is greater than any rhombus with the same side. ISOPERIMETRY. 305 932. Prob. —Having given a quadrilateral, to construct the maximugm isoperinetlric quadrilateral. 933. Theo. —Of isoperimetric quadrilaterals with a constant base, the maximum has its three remaincing sides equal each to each, and the angles which they include equtal. DEM.-Let ABCD be the maximum isoperimetric quadrilateral on the base AD, then AB = BC = CD, and angle ABC = BCD. For, if AB is not equal to BC, draw AC, and replacing the triangle ABC with its isoperimetric isosceles triangle, we shall have a quadrilateral isoperimetric with ABCD, and greater than ABCD, i. e., greater than the maximum, which. is absurd. Again, if angle ABC is not equal to'BCD, let ABC < BCD, whence BCE < EBC, and BE < EC. Take EF = EC, and EC = EB, whence the triangles FEC A D and BEC are equal, and FC = BC. Also, since AB FIG. 479. + BC + CO- = AE + ED - (EB + EC) + BC, and AF + FC + CD = AE + ED - (FE + EC) + FC, it follows that AFCD and ABSD are isoperimetrical, and, since ABCD = AED - BEC, and AFCD = AED - FEC, that AFCD and ABCD are equal. Therefore, AFCD is a maximum, and by the preceding part of the demonstration AF = FC = BC = AB, which is absurd; and there can be no inequality between angles ABC and BCD. 934. Theo.-Of isoperimetric polyqgons of a given number of sides, the regular polygon has the maximum area. DEM.-First, the polygon must be equilateral; for, c if any two adjacent sides, as AB, BC, are unequal, the triangle ABC can be replaced by its isoperimetric A- --- D isosceles triangle, and thus the area of the polygon be increased. Second, the polygon must be equiangular; for, if any two adjacent angles, as B and C, are unequal, the FIG. 480. quadrilateral ABCD can be replaced by its isoperimetric quadrilateral with B = C, and thus the area of the polygon be increased. 935. Theo.-Qf isoperimetric regular polygons, the one of the greater ntumber of sides is thle greater. DE-M.-Let ABC be an equilateral (regular) triangle. A Join any vertex, as A, with any point, as D, in the opposite side. Replace the triangle ACD with the isosceles triangle \ E AED. Then is the quadrilateral ABDE > the triangle ABC..But, of isoperimetric quadrilaterals, the regular (the B D C square) is the greater. Hence, the regular quadrilateral (the FIG. 481. square) isoperimetric with the triangle ABC, is greater than 20 306 INTRODUCTION TO MODERN GEOMETRY. the triangle. In the same manner the regular pentagon isoperimetric with the square can be shown greater than the square; and thus on, ad libitum. 936. CoR.-Of plane isop~erimetric figures, the circle has the maximum area, since it is the limiting form of the regular polygon, as the number of its sides is indefinitely increased. SECTION IV. OF TRANSVERSALS. 93'. DEr.-A Transversal is a line cutting a system of lines. A transversal of a triangle is a line cutting its sides;* it either cuts two sides and the third side produced, or the three c sides produced. In speaking of the transversal of a triangle (or polygon), the distances on any A/ X- e side (or side produced) from the intersection of AU O R the transversal with that side to the angles, are c Segments. Of these there are six. A djacent segments are such as have an extremity of each at the same point. NonA7/ IB ~~C~.R adjacent segments are such as have no extremity common. IT-/ l ILL'S. TR is a transversal of the triangle FiG. 482. ABC; aA, aC, bC, bB, cA, cB are adjacent segments two and two; aC, bB, cA, and aA, bC, cB are the two groups of non-adjacent segments. 938. THE TWO FUNDAMENTAL PROPOSITIONS OF THE THEORY OF TRANSVERSALS. 939. Theo.-The product of three non-adjacent segments of the sides of a triangle cut by a transversal, is equal to the product of the other three. DEM.-ABC being cut by the transversal TR, aA x bC x cB = aC x bB x cA. Draw BD parallel to AC, and from the similar triangles we have aA cA DB aC C c and — bD whence, multiplying, aAR B C x cA' A; g. B 8 c _R VB = 77jx CB, or aA x bC x cB/7 -I caC x bB x cA. FiG. 483. * Or, sides produced-this expression being usually omitted in higher Geometry as all lines are to be considered indefinite unless limited in the problem. OF TRANSTERSALS. 307 940. CoR.-Conversely, If three points be taken in the sides of a triangle (as a, b, c) such that the product of three non-adjacent segments equals the product of the other three, the points are in the sanme straight line. For, passing a line through a and b, let it cut the third side in c'. Then, by the proposition, aA x bC x c'B = aC x bB x c'A. But, by hypothesis eB cA aA x bC x cB = aC x bB x cA. Whence B= A and c and c' must coincide. Scm. —This theorem is known among mathematicians as The Ptolemaic Theorem, and is usually attributed to Claudius Ptolemy, an Egyptian mathematician and philosopher who flourished in Alexandria during the first half of the second century. But it is thought to be more properly due to Menetaus, who lived a century before Ptolemy. 941. Theo.-The three angle-transversals* of a triangle, passing through a conmmon, point, divide the sides into segments such that the product of three non-adjacent C segments equals the product of the other three. b DEM.-From the triangle ACe cut by the transversal aB, A B we have aA x CO x cB = AB x aC x Oe; and from CBe cut by bA, Oc x bC x AB = CO x bB x cA. Multiplying, c we obtain aA x bC xc B = aC x bB x cA. 942. COR. 1.-Conversely, If the three angle- A —transversals of a triangle divide the sides into segmzents such that the product of three non-adjacent FIG. 484. segments equals the product of the other three, the transversals pass through a conmmon point. For, the sides being divided at a, b, and c, so that aA x bC x cB = aC x bB x cA, draw Cc, and Ab, and let O be their intersection. Now, let a' be the point in which BO cuts AC. Then, by the proposition, a'A x bC x cB = a'C x bB x cA. aA aC Whence - -A = -', and a and a' coincide. 943. COR. 2. —If any one of the sides is bisected, the line joining the other points of division is parallel to this side. For, let bC = bB. Then aA x bC x cB = aC x bB x cA, becomes aA x cB = C x cA; or Aa: aC::Ac: cB. QUERY.-How does this apply to the second figure? 944. COR. 3. —If the line joining two points of division is parallel to the third side, the latter side is bisected. * The transversals passing through the angles. 308 INTRODUCTION TO MODERN GEOMETRY. For, if ab is parallel to AB, aC: bC::aA: bB, whence aC x bB = bC x aA. And, since aA x bC x cB -= aC xbB x cA, cA = cB. 945. We will now give a few problems to illustrate the use of the theory of transversals. 946. Prob. —To 7show that te nmedial lines of a trianzgle pass c through, a commnzon point. a/ SOLUTION.-Since aA = aC, bC = bB, and cB = cA, by multiplying, we have aA x bC x cB = aC x bB x cA; whence A a.~~ - -- B by the last corollary these transversals pass through a comrnFIr. 485. mon point. 94T. Prob. — To show that the bisectors of the angles of a triangyle pass through a convmmon point. SOLUTION.-In the last figure let aB, bA, cC be the bisectors. aA AB bC AC cB CB aA x bC x< cB Then a multiplying, - 1, or Then C CB' bB AB' cA AC' ltplygaC x bB x cA aA x bC x cB = aC x bB x cA. Therefore these transversals pass through a common point. 948. Prob.-To show that the altitudes of a triangle pass througeh a cogmmon point. Svu's.-In the last figure, if aB, bA, cC, were the perpendiculars, there would aA AO bC CO cB OB be three pairs of similar triangles giving -B -BO' - AO' - CO whence, as in the last. 949. Prob.-To show that thle angle-transversals terminating in the points of tangency of the sides of the triangle with its inscribed circle, pass through a conzmoon point. SuG's.-In the last figure, if a, b, e were the points of tangency we should have aA = cA, bC = aC, cB = bB; whence aA x bC x cB = aC x bB x cA. Which shows that the transversals pass through a common point. 950. Theo. —If two sides of a triangle are divided proportionC ally, startingfrom the vertex, the anlle-transversals from the extremities of the other side / to the correspoding points of division, iz- tersect in the rmedial line to this third side..-.'.. / —t DEM. —Since AC and CB are divided proportion=-' —'= — -= i~=====n \ ally at a and a', aA x a'C = aC x a'B; and as A D B DB = -DA, aA x a'C x DB = aC x a'B x DA, FIG. 486. the angle-transversals Aa', Ba intersect in CD. The same may be shown of any other angle-transversals from A and B, dividing CB and CA proportionally. OF TRANSVERSALS. 309 951. COR.-In any trapezoid the tranzsversal passing through the intersection of the diagoncals, and the intersection of the nonparallel sides, bisects the partallel sides. SuG. —Joining aa' in the last figure, CD is such a transversal. The student will readily see the connection with the proposition. 952. Prob.-Throvygh a given point to draw a line which shall nleet two given lines at their intersection in an invisible, inaccessible point. SoLUTION.-Let Mm, Nn be the two given lines which meet in the invisible,' inaccessible point S, and P the given I point through which a line is to be lo- - - cated which will meet Mm, Nn in S. B Through P draw any convenient trans-. c versal, as BF, and any other meeting this, H as AF. Now, considering MS as a trans-................. versal of the triangle CDF, we have M AFx BC x SD -AD x BF x SC;whence N SD AD x BF FiG. 487. SD AD x BFC But, HD being drawn FIG 4 HD SD AD x BF AD'x BF x PC parallel to BF, we have = SC=AF BC or HD AF BF PC ePC __ SC __ AF -x E~BC' 0 AF x bC whence HD is known, as AD, BF, PC, AF, BC can be measured. The points P and H determine the required line. 953. DEF.-The Complete Quadrilateral is the figure formed by four lines meeting in six c points. The complete quadrilateral has three diagonals. ILL.-ABCDEF is a complete quadrilat- era], and its diagonals are CF, BD, and AE, L the latter being spoken of as the third or A —------- -------- ---------— E exterior diagonal. FIG. 488. 954. Theo.-The middclep oints of the three diagonals of a comnp7te quadrilateral are in the same straight line. DEM.-m, n, o being the centres of the diagonals of the complete quadrilateral, in the preceding figurle, are in the same straight line. Bisect the sides of the triangle FDE, as at I, N, L, and draw IN, IL, LN. Since IN is parallel.to BE, and bisects DF and DE, it also bisects DB (9) and hence passes through n. For like reasons IL passes through mn, and LN through o. Now, AC being a transversal of the triangle FDE gives CD x BE x AF = CE x BF x AD. Therefore, 310 INTRODUCTION TO MODERN GEOMETRY. noticing that CD -- l, BE = nN, 1AF = oL, JC9 = mL, ~BF = n!, and 4AD = oN, we have ml x nN x oL = mL x nl x oN. Hence these three points m, n, o lie in a transversal to the triangle ILN. 955. Theo. —In the complete quadrilateral, any diagonal is divided harmonically by the other c ~ ~ ~ two. DEM.-Thus, AFH is divided harmoni/ 0.........cally at C and H. For, considering BH as a transversal of the triancgle ACF, we have A;""V_-. H HF x DC x BA = HA x DF x BC. And CC, AD, FB being angle-transversals of the FIG. 489. same triangle, we have CF x BA x DC = CA x BC x DF. Whence, dividing, H HA, i.e., AH is divided harmonically. I -C -A, Again, if CH is drawn, CA, CC, CF, CH constitute a harmonic pencil, and BH a, transversal of it, is cut harmonically at B, I, D, H. Finally, if F and I be joined, FH (or FA), FB, Fl, FD constitute a harmonically, and hence CC is cut harmonically at C, I, E, C. 956. COR.-An angle-transversal of a triangle, and a line passing through the feet of the other angle-transversals, divide the third side harmnonically. SECTIONV VY HARMONIC PROPORTION AND HARMONIC PENCILS. 95'. IDEF.-Three quantities are in Harmonic Proportion when the difference between the first and second is to the difference between the second and third, as the first is to the third. ILL.-6, 4, 3 are in harmonic proportion, since 6 - 4 4 - 3:: 6: 3. In general, a, b, e are in harmonic proportion, if a - b: b - c:: a: c. 958. Theo.-If a given line be divided internzally and externally in the same geometric ratio, the distance between thepoints of division is a h]armonic mean between the distances of these points from the extremity of the given line, not included between the points. DEM. - Let AB be the given line; and let O and O' be so taken that AO:BO::AO':BO'; then is 00' a harA O B 0, monic mean between AO' and BO'. For AO = AO'-OO', and BO = 00' - BO'; FIG. 490. whence AO', 00', and BO' are such that AO - OC': 00OO' -BO:: AO,: BO', that is, they are in harmonic proportion. HARMONIC PROPORTION AND HARMONIC PENCILS. 311 959. COR. 1. —AO, AB, and AO' are in harmonic proportion,, i.e., AB is a harmonic mean between AO and AO'. For AB - AO (= BO): AO' - AB (= BO'):: AO: AO'. 960. COR. 2. — Whe AO, 00', BO' are in hargmzonic proportion, AO x BO'- BO x AO'. 961. Con. 3.-Conversely, 7When a line is divided into three parts such7 that the rectangle of the extreme parts equals the rectangle of the gmean part into the whole line, the line is divided harmonically. Thus, letAO' he the line, and AO x BO' = BO x AO'; then AO: 80:: AO': BO', whence, by the proposition, 00' is a harmonic mean between AO' and BO'. DEF.-The points 0 and o' are called Harmonic Conjugates. 962. Theo. — f two lines be drawn, one bisecting the interior and the other the adjacent exterior angle of a triangle, and mneeting the op- c posite side,* they divide this line liarsnonically. Sn. —By means of (358, 359, PART A 0 B II.) the student will be enabled to establish FIG. 491. the relation AO: BO:: AO': BO', whence, by the last proposition, AO, 00', AO' are in harmonic proportion. 963. Prob. —Given a right line to locate two harmonic conjugate points. SOLUTION.-Let AB be the line. O may be taken at pleasure between A and B. We are then to find 0', so that AO: BO:: AO': BO', Taking this by division, we have AO - BO: BO:: AO' - BO' (= AB): BO'. The first three terms being known, the other can be constructed. Or, we may first locate O' at pleasure, and then find 0. 964. Theo.-If from the given point C in a line the distances CO, Cs, CO' be taken in thle samie di- A C O B O' rection, so that CO x CO'= b.2; and if FIG. 492. CA = CB be taken in the opposite direction, AO' will be divided harmonically at 0 and B. DEM. -From CO x CO' C, we readily write CO: CB:: CB: CO', CB + CO (= AO): CB - CO (= BO):: CO' + CB (= AO'): CO - CB (= 80'). * The bisector of the exterior angle meets the side produced; but in higher geometry, as it is always understood that lines are indefinite unless limited by hypothesis, such specifications are deemed unnecessary. 312 INTRODUCTION TO MODERN GEOMETRY. 965. Cob. 1.-Conversely, If a line AO' be cut harmqonically at 0 and B, and either of the harmonic means be bisected, as AB at C, the three segments CO, csB, co' will be in geometric proportion. For, since AO': BO':: AO: BO, AO' + BO': AO' - BO':: AO + BO: AO - BO, or 2CO': 2CB::2CB:2CO, and CO': CB:: CB: CO. 966. Con. 2. —In a given line, as AB, as O approaches the centre C, O' recedes, and when 0 is at C, o' is at ifinity, since co' = - 96'. Theo.-The geometric mzean between two lines is also the geometric gmean between their arithT mzetic and harmizonic mOeans. DEM.-Let AO' and BO' be the two lines. On their difference, AB, draw a semicircle, A C o Bs' draw the tangent O'T and let fall the perpenFIG. 493. dicular TO. Then O ind O' are harmonic conjugates, since CO x CO' - CB'2 (?), CO' is the arithmetic mean (that is, ~ the sum) of AO' and BO' (?) and TO' is the geometric mean (?).. AO': TO':: TO': BO' (?). QIERIEs.-Which is the greatest, in general, the arithmetic, geometric, or harmonic mean between two quantities? Are they ever equal? 968. ScH. —Tis proposition affords a ready method of finding either of the'harmonic conjugates 0 or 0', when the other is given. The student will show how. 969. CoR. I.-Thle rectangle of the harmonic means and the sugm of the extreines, is equivalent to twice the rectangle of the extremes. For, CO' x 00' = TO'2 - AO' x BO', whence 2CO' x 00' = 2AO' x BO'; and, since 2CO' = AO' + BO', (AO' + BO') x 00' = 2AO' x BO'. 9W0. CoR. 2. —The rectangle of the harmonic mnean and the difference of the di ferences of the 1st and 2Cnd, and the 2nd and 3rd, is equivalent to twice the rectan)gle of these differences. That is, 00' x [(AO' - 00')- (00'- BO')] = 2 (00' - AO') (00' - BO'), or 00' x (AO - BO) = 2AO x BO. Let the student give the proof.. 9T1. CoR. 3.-If three quantities are in harmo2nic proportionr their reciprocals are in arithmetic proportion (i.e., the difference between the 1st and 2nd equals the difference between the 2nd and 3rd). For, from AO', 00', BO, we have the reciprocals AO' O"o' BO' Now 1 1 AO' - 00 AO 1 1 00 - BO' 00' - AO' OO X' O' xAO'; and -BO,O O' x BO" DO' 00'00' xB' HARMONIC PENCILS. 313 BC AC BO AO BO 1 00' x BO" But00' AO' - 00' x s = O' BO' since 00' 1 1 1 AO'- BO' 00".9T2. Prob. Given the harmonic mean and the difference betwieen the extremes, to find the extremes. SUG's.-We have 00' and AB, (Fig. Art. ) given. Then CO x CO' -2' 2 = C-T = iAB, and CO'- CO = 00', whence CO'2 - 00' x CO' = AB2. From this equation CO' can be constructed ( ), and the problem solved. 973. Theo.- When two circles cut each other orthogonally (i. e., so that the tangents at the common point are at right angles), any line passinz through the centre of one, and cuttiing the other, is divided harmonically by the circumferences. T 1)EM.-The tangents being perpendicular to each other pass through the centres, CO B hence CO x CO' = CT. But CB = CT. Therefore AO is cut harmonically. FIG. 494. 974. Prob.-To find the altitude of a triangle in terms of the radii of the escribed circles touching the adjacent sides. SOLUTION. -Let r and r' be the radii of the escribed circles, and p the altitude. Now RT, AT, and QT are in harmonic proportion; since, considering the triangle ACT, CQ bisects its interior and RC its exterior angle (?), we have r QT: QA: RT: RA. But r, p, r', sustain the same relation to each other as RT, AT, QT; B hence r, p, r' are in harmonical proportion. 27'r' FIG. 495. Therefore, by( )p (r + r') = 2rr'; or p= 49 HARMONIC PENCILS. 9T5. DEF.-A Pencil of lines is a series of lines diverging from a common point. DEF.-A Iliarntonic Pencil is a pencil of four lines cutting another line harmonically. ILL.-In the following figure OA, OB, OC, OD constitute a H2armnonic Pencil, since they divide the line mn harmonically at A,-B, C, D. 314 INTRODUCTION TO MODERN GEOMETRY. 9W6. Theo.-A harmonic pencil divides harmonically every line which cuts it. DEM.-OA, OB, OC, OD being a harmonic pencil, that is, AB, BD, AD, being in- harmonic proportion, A'D', any other line cutting the pencil, is divided harmonically, so that A'B', B'D', A'D', are in harmonic propor/ / \ Cd tion. Through C and C' draw parallels to OA. B D' as LK and L'K'. Now, from similar triangles, A' c L AB: BC:: AO: CK, and AO:CL:: AD: CD. Am / n But AD: CD:: AB: BC, since AD is harmoni~9 B/.I D4 cally divided. Hence AO:CK:: AO: CL, and CK = CL. Hence from similar triangles C'K' = C'L'. Again A'B': B'C':: A'O': C'K'(?), FIG. 496. and A'D': C'D':: A'O: C'L' (= C'K') (?), whence A'B': B'C':: A'D': C'D', or A'B', B' D', C'D' are in harmonic proportion. ScH.-If the line through C' cut the prolongation of AO beyond O, it is still harmonically divided; and, in fact, it is scarcely necessary to make this statement, since in all general discussions lines are to be considered indefinite, unless limited by hypothesis. 9TT. DEF.-The alternate legs of a harmonic pencil are called cozjugate, as OA and OC, OB and OD. 9t8. Theo.-If tqvo conjugate legs of a harmonic pencil be at right angles, one of thenm bisects the anglye'included by the otherpair, and the other the supplement of this angle. SUG.-This is the converse of ( ), remembering that the bisectors of two adjacent supplemental angles are at right angles. SECTIO N VZ ANHARMONIC RATIO. 9'9. DEF. —The Anharmonic Ratio -of four points in a right line is the ratio of the rectangle of the distance between the first and fourth into the distance between the second and third to the rectangle of the distance between the first and second into the distance between the second and fourth. A- B C D FIn. 497. ILL.-The anharmonic ratio of the four points A, B, C, D isAD x BC:AB x CD. AN HARMONIC RATIO. 315 980. The relation AD x BC: AB x CD is expressed for brevity [ABCD]. ILL. —-ThUS [ABCD] means AD x BC: AB x CD; [ADBC] means AC x DB: AD x BC; [BACD] means BD x AC: BA x CD; etc. The ratio [ABCD], or AB AD AD x BC: AB x CD is evidently the same as B-C: CD 981. ScH.-The appropriateness of the term aniharmonic (not-harmonic) AB AD will be seen when we observe that, if AD is harmonically divided, BC equals C AB AD If; therefore, BC is not equal to - which is the general case of division, irre-'BC CD' spective of the position of the points B and C, we may consider the ratio of AB AD BC t CD This general ratio, or, what is the same thing, AD x BC: AB x CD, is called the anharmonic ratio. 982. Theo.-T-71e anharmonic ratio offour points is not changed by interchanging two of the letters, provided the other two be interchanged at the same time. DExIL [ABCD] = [DCBA] = [BADC] = [CDAB], i. e., AD x BC: AB x CD = DA x CB: DC x BA = BC x AD:BA x DC = CB x DA:CD x AB, which are evidently identical. [The student should notice the different segments of the line indicated by the different forms.] 983. ScH.-But [ACBD] is a different anharmonic ratio from [ABCD]; since AD x CB: AC x BD is not necessarily equal to AD x CB: AB x CD. Now, as there can be twenty-four permutations of four letters, there may be formed six different anharmonic ratios from four given points in a line. 984. Theo. —I.f a pencil of four lines is cut by any transversal, the anharmonic ratio of the four poinlts of interseition is constant. DEM. SL, SM, SN, SO, or, as we may read it, S-L,M,N,O, p being such a pencil, and AD any transversal, draw through A C NP parallel to SO. Then, ADCO AB CN: AS AD x BC: AB x CD:: AD C C { C CN CP. D But CN: CP is constant for all positions of C on SM. There-. 498 fore AD. x BC: AB x CD is constant for any transversal. 985. SCo. —Other constant ratios may be written from, the preceding proposition and scholium. The anharmonic ratio [ABCD] is called the anharmonic ratio of the pencil. The angles of the pencil are the six angles included by the rays. 316 INTRODUCTION TO MODERN GEOMETRY. 986. —Co. 1.1.-f two pencils are tmutually equiangular th7eir anharvmonic ratios are equal. S QUERY.-Is the converse of this corollary true? M-.2 - - - -— N 987. CoR. 2.-If two pe1ncils have their interN sections in the sae right line, their acnharnonic ratios are equal. FIG. 499. 988. DEF.-The anharmonic ratio of four points on the circumference of a circle is the anharmonic ratio of the pencil formed by joining these points with any point in the circumference. ILL.-Thus, the anharmonic ratio of the points A, B, C, D is the anharmonic ratio of the pencil O-A,B,C,D, it being immaterial where in the circumference the point O is taken, A B C D since by COR. 1, preceding, the ratio is the same for any posiFIG. 500. tion of O (?). 989. Theo.-If four fixed tangents to a circle are cut by afifth, \th e anharmlonic ratio of the four ploints ---—: B -.- -v of infersection, calledl tuhe anharmonic ratio of the tangents, is constant. I DEMr. A, B, C, D being the fixed points of D tangency, any transversal, as TV, cuttingr the tangents, has the anharmonic ratio [LMNP] FIG. 501. constant. For the pencil O-L,M,N,P has its angles constant. - Thus LOM is measured by ~ are (AX - IX) = JAB, which is constant. And in like manner MON is measured by I are BC, and NOP is measured by ~ arc CD. Hence, by the first of the preceding corollaries, the anharmonic ratio [LMNP] is constant. 990. The theory of anharmonic ratio is applied with great facility to the demonstration of theorems showing that several points are in a right line, and that several lines intersect in a common point. We' give three specimens of each class. 991. Theo. —If two pencils have the samre anlzarmonic ratio and a homoloyous ray cozmmon, the intersection of the other homoloM S - gous rays are inv the same right line. K E F H A IGB B C2 D DEM. —Let S-A,B,C,D and S'-A',B',C',D' FIG. 502. be two pencils having the same antharmonic AN HARMONIC RPATIO. 317 ratio,- and the rays be SA, S'A coincident; then the intersections E, F, H are in the same right line. Let the line passing through E and F intersect SA in K, and suppose it intersect SD in H', and S'D' in H". Then, since the anharmonic ratios of the two pencils are equal LKEFH'] = [KEFH"]; whence H' and H" are the same point, and must be the intersection of the two lines SD, S'D', that is, H. 992. Theo.-If in two right lines four points in the one have the same anh7armonic ratio as four points in the other-,.1andz one hom2ologyous point in common, the three lines passing thlrough the othler pairs of homologous points meet in a commonO point. s DE. —Let A be common, and [ABCD] = [A'B'C'D']. Draw SA and SD'. Call the point in which SD' D L cuts AL D" (for the time being). Then [AB'C'D'] = [ABCD"]. But by hypothesis [AB'C'D'] = [ABCD]. A Therefore [ABCD] = [ABCD"], and D and D" are one and the same point. Hence the three lines which FIG. 503. pass through B and B', C and C', D and D' meet in a common point S. 993. Theo. — If the lines passing throeugh the corresponding vertices of twvo triangles mleet in a common point, the intersections of their homologous sides lie in the samle right line. DEM. -Let ABC and A'B'C' be two triangles so situated that the lines AA', BB', CC' meet in the com- /./ D mon point S; then L, M, N, the intersections of the homologous sides, are in a right line. For the pencil ----. S-L, B, A, C being cut by the-' "- N__ two transversals LD, LD', gives B [LBAD] = [LB'A'D'] (984). But C-L,B,A,D, and C'-L,B',A',D', have these harmonic ratios, hence C-L,'O,M,N, and C'-L,O,M,N, their equivalents, and having a common ray CC', have equal anharmonic ratios,./ and consequently L, M, N are in the same right line, by the last proposi- FIG. 504. tion. 994. Theo.-If the intersection qf the corresponding sides of t1wo triangles are in the samne right line, the lines passing through their corriesponding angles meet in a cogmmzon point. DEM.-In the last figure, if AB and A'B', AC and A'C', BC and B'C' have their 318 INTRODUCTION TO MODERN GEOMETRY. intersections in the same right line, as LN, the lines passing through B and B', A and A', C and C' meet in a common point, as S. By (987) C-L,O,M,N has the same anharmonic ratio as C'-L,O,M,N, whence [LBAD] = [LB'A'D'], and the truth of the theorem follows from (992). 995. Theo. —The opposite sides of an inscribed hexagon have their intersections in the same straight line.,1lL DEM.-The anharmonic ratios of the pencils B-A,E,D,C,* and F-A,E,D,C being equal ( ), LCDE, which intersects the first, is divided in the same anharmonic ratio as NHDC, which cuts the secc ond, or [LGDE] = [NCDH]. But these Al/ ~ G\t M lines have a common homologous point D, hence the lines joining the other pairs of 9 \ XH homologous points, as LN, CG, EH, meet in a common point, as M. Therefore L, M, N are in the same right line. 996. Sce-. This theorem is due to PASCAL, whose wonderful achievements in his brief life of thirty-nine years (1623N 1662) have been the admiration of all succeeding generations. F[G. 505. 99'. Theo. —The diagonals joining the opposite vertices of a cirp cncumscribed hexagon intersect in a common point. \A Q - / /M DEM.-Consider AB, BC, CD, EF four E - — B fixed tangents cut by ED and FA. Then.a'-' "' [PNDE]=[AQMF](989). Hencetheanharmonic ratios of B-P,N,D,E,* and \>Y7C C-A,Q,M,F are equal ( ); and since they have a common ray (CQ, BN) the intersections A, O, D, of their homologous rays, are in the same right line. Therefore the diagonals pass through a common FIG. 506. point. * The student can conceive the rays BE, BD. etc,, as drawn, without encumbering the figure with them. POLE AND POLAR IN RESPECT TO A CIRCLE. 319 SEC TI ON VII POLE AND POLAR IN RESPECT TO A CIRCLE. 998. DEF.-If a secant to a circle be revolved about a fixed point in the plane of the circle, the locus of the harmonic conjugate of the fixed point, in reference to the in- T tersections, is the Polar of the fixed c point. The fixed point is the Pole of the Polar Lisne. The terms pole A and polar as here used are correlative, A and neither has any significance A B without the other. ILL.-Let AP be a secant revolving about A\' the fixed point P, and let C be so taken that (in every position) AC: CB:: AP BP, then is the locus of C the Polar of P, and P is FIG. 507. the Pole of the locus of C. 999. Theo. —The Polar of a given point in respect to a circle is a right line. DEM.-Let P be the pole, AP any secant passing through P, and a point in the polar. The locus of C is required. Draw PL through N A the centre, and let fall the perpendicular /B CC'. Draw AL, AH, AC'F, and C'B. Since / "' - AC: CB:: AP: BP, C'P bisects the angle / " ". BC'F, the exterior angle of the triangle L O C \ H P AC'B (?); hence, as LAH is a right angle, AL bisects NAC', the exterior angle of the triangle C'AP (?). Therefore, PL is harmonically divided at C', and H; and, C' being a fixed point, and C any point in the locus, the locus FaG. 508. is the perpendicular TCC'V. 1000. COR. 1.-Since, as the secant revolves, the points A and B will vanish in C", C" is the point at which a tangent from the pole P touches the circle. 1001. COR. 2.-Drawing OC", C"P, we see that O6c" (or OH' =OC' x OP. 3 0 INTRODUCTION TO. MODERN GEOMETRY. 1002. COR. 3. —The polacr of a point in? the circujlference is a tacngent at that point. For, as OC' x OP is constant and equal to OH2,OP diminishes as OC'increases, and when OP = OH, OC' = OH also. 1003. Prob. To draw the polar to a givenlpole in respect to a given circle. CoR. 1 effects the solution. 1004. Prob.-To find the pole of a polar to a given circle. Through the centre draw a perpendicular to the polar. [The student should be able to complete the solution.] 1005. DEF.-The point C' where the polar cuts the line passing through the pole and the centre of the circle is called the Polar Point. T1006. Theo. —Te pole andpolar point are interchancgeable. / C.[B'. DEM.-TV being the polar to P, we are nL_...... to show that T'V', parallel to TV and passo c P R ing through P, is polar to C'; i.e., that any \B secant, as AC'C", passing through C', is divided harmonically in the intersections with c the circumference, C', and the intersection,v~ v' with T'V'. Drawing AP, since P is the FIG. 09. pole of TV, we have, as in the last demonstration, angle APB bisected by PC'; and consequently RPB bisected by PC". Therefore AC': C'B:: AC": BC". Q. E. D. 100T. Theo. —T/le polars of all the points in a right line pass through the pole of that line; and, conT versely, The poles of all straight lines which pass through a given point are in the polar of that point. DEM.-lst. TV being a given line and P its pole, we are to show that the polar to any point, as N, N passes through P. Draw through P a perpendicuAv' lar to ON; then P' is the polar point to N. For, OP:ON::OP':OB (?): whence ON x OP' FIG. 510. Y = OP x OB = OA2 Thlerefore, T'V' is the polar of N (?) 2ud. P beilg any point and TV RECIPROCAL POLARS. 321 its polar, the pole of any line, as T'V' passing through P, is in TV, as at N.'Draw ON perpendicular to T'V'. Then, as before, ON x OP' =OP x OB= OA, and N is the pole of T T'V'. T 1008. CoR.-The pole of a straight line P is the intersection of the polars of any two yN of its points; and, conversely, The polar of 0 B any point is the straight line joining the poles of any two straight lines passing V through that point. FrI. 511. RECIPROCAL POLARS. 1009. DEF. —If two polygons be constructed, one within, or inscribed in, a circle, and the other without, or circumscribed- about the same circle, such that the vertices of the one are the poles of the sides of the other, P the two polygons are called V Beciprocal Polars; and the circle is called the Auxiliary Circle. The possibility of constructing such polygons is apparent from the last theorem. When the points P, P', P", P"' FI, 51. are in the circumference, TV, TT', T'V', V'V become tangents, as appears from (1002). 1010. Prob. —Havtng given one of two reciprocal polars, to construct the other. The student should De alie to maxe tne construction. 1011. By means of the relation between reciprocal polars a large class of propositions relating to the relative positions of lines and points, become, as it were, double; i.e., one proposition being proved, another can be inferred. The process by which the inference is made is called reciprocation. We will give an example. 322 INTRODUCTION TO MODERN GEOMIETRY. 1012. Prob.-To deduce the reciprocal of Pascal's theorem (995). N SOLTrION. —Draw tangents at the six ver. tices of the inscribed hexagon. Thus, a cirp,/ \ \\ cumscribed hexagon is formed whose sides E M/ V are the poles of the vertices of the inscribed Q \ / hexagon, through the verltices of which they F z _ \ / respectively pass (1002). Now, drawing the diagonals PM, NQ, OL, they are the polars \\ VA of the intersections of the opposite sides of the inscribed hexagon, as PM, polar to the intersection of DE and CB (?); and hence they pass through a colmmon point, as V. Thus we have Brianchon's theorem, viz. L The lines joining the op posite angles of a cirFIG. 513. cumscribed he.xagyon pass through a common point. 1013.-The three following theorems are of frequent use in applying the theory of reciprocation. 1014. Theo. - The angle included by two straight lines is equal to the angle included by the lines joiningy their poles to the ceztre of the aucxiiary A circle. ) DEM.-The pole of a line being in the perpendicular fiom the centre of the auxiliary circle upon the line (1000), O' is the supplement of O; hence o = 0. FIG. 514. 1015. Theo.-The distances of any two points from the centre of the auxiliary circle are to each other as the dr istances of each point friom the polar of the other. C E D C[ DEM. P, P' being the points, and TV, T'V their D polars respectively, we are to show that ~8D CP: C: P':: PD": P'D"'. By( - ) R CP x CD = CP' x CD', R being T' D' D" the radius of the auxiliary circle. Whence CP:CP'::CD':CD. But CP:CP':: CF:CE (?),and FIG. 515. there follows CF: CE.:: CD': CD, CD' - CP (= P'D"'): CD - GP (= PD"):: CF: CE:: CP: CP'. SCEn.-This is known as Sacmon's Theorem. RADICAL AXIS AND CENTRES OF SIM ILITUDE OF CIRCLES. 323 1016. Theo. — The anharvmlonic ratio of four points in a straight line is equal to that qf the pencilformed by the foar polars of these points. DEMr.-1st. The polars pass through a common point and thus form a pencil (?). 2d. The angles included by the lines joining the four points with the centre, and those included by the polars are equal (?), hence the two pencils have the same anharmonic ratio. —/ FIG. 516. SECTION VIII. RADICAL AXES AND CENTRES OF SIMILITUDE OF CIRCLES. i01T. DEF.-The Powier of a Point in the plane of a circle is the rectangle of the distances from the point to the intersections of the circumference by a line passing through the point. A LI,. —Thus, the powoer of a poitnt P, without the circle, B is PA x PB;-the power of a point within, as P', is' P'A' x P'B'; the power of a point in the circumference is zero, since one of the distances is then 0;-the power B of the centre is the square of the radius. FIG. 517. 1018. CoR. —Te power of a given point w ith respect to a given circle is a constant quantity. Thus PA x PB = the square of the tangent from P to the circle, in whatever position PB lies, so long as it passes through P. So also P'A' x P'B' = the rectangle of the segments of any other chord passing through P'. 10 19. DEF. -The Radical -Axis of Twzo Circles is the locus of the point whose powers with respect to the two circles are equal. 324 INTRODUCTION TO MODERN GEOMETRY. 1020. Prop. — The Radical Axis of two circles is a right line. P P IT 0 C~rue 0 FIG. 518. DEi. —Join the centres of the two circles O, 0', and take a point R on this linesuch that OR2 - O'R = OT - O'T —, or OR2 _ -T = ORR2 _ OT2, and erect PR perpendicular to 00'. Then P being any point in this perpendicular, OP2 - OR O2 - OR2. Adding this to the preceding equation, we have Op _ T = 2 T —, or PT... PT = P'T', PT and PT' being tangents to the circles from any point in PR. Hence PV is the radical axis of the two circles. 1021. CoR.- When the circles do not intersect, the Radical Axis lies betweemz them, touchiny neither; when they are tangent, either externzally or internally, the radical axis is the commzon tangent; when they cut each other, the axis of the common cord produced. 1022. ScH. —When the circles intersect it might seem that the above demonstration fails fobr points within, as in the common chord. But, the powers of any point in this chord are still equal. Thus, at the intersections the powers are zero; and at any other point in the chord, as a, ab x ac = ad x ae, since each is equal to ao x os. 1023. CoR.-There is an infinite number of circles having their ceztres in the same right line, which have the samze radical axis as anzy two given circles. Thus, in the first figure, PV being the radical axis of the circles 0, 0', letting circle 0 remain fixed, O' may vary indefinitely so that O'R2 - O'T-' remains constant, and equal to OR-2 - OT2. 1024. Prob.-Given two circles, to draw their radical axis. SOLUTION.-In any case, draw a common tangent, bisect it, and through the point of bisection draw a line perpendicular to the line joining the centres. mThen the circles are tangent to each other, the distance between the points of tangency is 0; hence the perpendicular is erected at this point. When they intersect, produce the common chord, or use the first method. CENTRES OF SIMILITUDE. 325 1025. Prop.- When two circles cut each other ortlhogonaZlly, that is, at right angles, the square of the radius of either is equal to the power of its cen-tre with respect to the other. DEM.-The power of 0 with respect to circle O' is P Oa x On = OP2, and of O' with respect to circle O, O'b x O'n - o-P2-; since, as the circles cut each other orthogonally, their tangents are at right angles, and the tangent to either passes through the centre of the FIG. 519. other. 1026. Prop). — The radical axes of a system of three circles whose centres are nzot in the same straight line, intersect at a common po int. V' DEM.-Since O, 0', 0" are not in a straight line, tlie radical axes of O, 0', and 0, 0", as PV" and o —... PV intersect. Let P be their P v S > common point. Now the power a:7 of P with respect to O' is equal, 6 \. to its power with respect to 0",, since each is equal to its power with respect to O. Hence P is a FIG. 520. point in the radical axis of O', 0" 1027. CoR.-If the centres are in the samce straight line, their radical axes are parallel, and the commonpoint is at infinity. 1028. DEF.-The intersection of the radical axes of three circles is called their Radical Centre. CENTRES OF SIMILITUDE. 1029. DEF.-If the line joining the centres of two circles be divided externally, as at C, and internally, as at C', in the ratio of the radii, E.. these points are respectively -- ---- ---—. - thle External and the In- 0\ _cp_ ternal Centres of Similitude of the two circles. FIG. 521. ILL.-If CO: CO':: EO: E'O', C is the external centre of similitude; and, if C'O: C'O':: EO: E'O', C' is the internal centre of similitude. 326 INTRODUCTION TO MODERN GEOMETRY. The student should construct the figure when the circles are tangent externa.lly, —when they are tangent internally,-and when one is wholly within the other. QUERY. —How are the centres of similitude situated in the three different relative positions of the circles? 1030. Prop. —In two circles the line passing through, the extrealities of two parallel radii on the same side of the line passing through# the cenetres, intersects this line in the externzal centre of sim-,lit cude, aned if the radii are on ol2posite sides of this line'the intersetion is the iltersnal centre of siilitihde. The proof consists in showing that the line passing through the centres is divided as above. Let the student show it for the three different positions of the circles. 1031. COR. 1.-Conversely, If any transversal be drawn from either centre of similitude, the radii drawn to the ilntersections are parallel. Thus in the last figure, since CO: CO': EO: E'O', and the triangles have the angle C common, EO and E'O' are parallel. 1032. COR. 2.-Tangenzts drawn at the alternate intersections of a ranzsversal through the exterzal centre of.similitude are parallel; a(1so, those at the mzean intersections, and those at the extreme intersections, if the transversal be drawvn through, the internal centre of simiititde. This follows as a consequence of the parallelism of the corresponding radii, to which the tangents are perpendicular. Thus, tangents at E and E' are parallel, as are those at F and F'. So, also, tangents D and D', and at E' and E" are parallel. 1033. DEF.-The extremities of two parallel radii on the same side of the line joining the centres are called Homoloyous Points, anld those of non-parallel radii where the transversal cuts the circumferences, as E, F', are called Anti-Homologous Points. 1034. Con. 2. —The distances of a centre of similitude from two hcmlologous poiints are to each other as the radii. 1035. Con. 3. —The centtres of similitude and the centres of the circles are four har,)Zmonic pioi7nts. 1036. Prop).-]f a circle touch twvo others, the line joining their poin[ts of' coitact passes through, the exter!nacl centre of similitude of CENTRES OF SIMILITUDE. 327 the latter if the contact is external; and through thlie internal centre of similitude if the contact is internal. E E/oo N E E Fin. 522. DEM.-In either case let 0" be the circle tangent to 0, and 0'; and through the points of tangency draw E'C'. The angle O"E'F = O"FE' = EFO = FEO; whence OE and O'E' are parallel, and the similar triangles CEO, CE'O' give CO:CO':: OE: O'E'. Q. E. D. 103T. Prob.-To draw a li2e parallel to a given lize so that the distance betiween the extremze intersections with two yivez circles shall be a ma2Cximum. SOLUTION.-Draw a line through the internal centre of' similitude and parallel to the given line. Now, at the extreme intersections draw tangents, and it will become evident that the line first drawn is a maximum. [The student should make the figure and fill out the proof.] If the circles are wholly exterior to each other, the distance between the mean intersections is a minimum. 1038. CONCLUDING ~fOTE.-Our limits preclude our pursuing these topics farther. We have given enough to make the language of the Modern Geometry intelligible, and to afford some insight into its character. One of the best elementary resources for the English student who wishes to pursue the subject at greater length, is.MULCHAY'S Principles of Mfodern Geomnetry, Dublin, 1862. It is, however, much to be regretted, that there is no English treatise which presents the elements of this sulblect with the philosophic elegance of the French. The best of the latter is RoUCmU and COMBEROUSSxTi'S Treatise on Elementary Geometry. For a more extended view of the subject, SALMON or WHITWORTR will furnish the English student; but he who would be proficient must read the works of CHASLES and PONCELET, who are the great authorities.