X W V \/ A TREATISE ON THE GEOMETRY OF THE CIRCLE. A TREATISE ON THE GEOMETRY OF THE CIRCLE AND SOME EXTENSIONS TO CONIC SECTIONS BY THE METHOD OF KECIPKOCATION WITH NUMEROUS EXAMPLES BY WILLIAM J. MCCLELLAND, M.A. PRINCIPAL OF THE INCORPORATED SOCIETY S SCHOOL, SANTRY, DUBLIN MACMILLAN AND CO. AND NEW YORK 1891 All rights reserved QA ft f 3 PEEFACE. MY object in the publication of a treatise on Modern Geometry is to present to the more advanced students in public schools and to candidates for mathematical honours in the Universities a concise statement of those propositions which I consider to be of funda mental importance, and to supply numerous examples illustrative of them. Results immediately suggested by the propositions, whether as particular cases or generalized statements, are appended to them as Corollaries. The Examples are printed in smaller type, and are classified under the Articles containing the principal theorems required in their solution. The more difficult ones are fully worked out, and in most cases hints are given to the others. The reader who is familiar with the first six books of Euclid with easy deductions and the elementary formulae in Plane Trigonometry will thus experience little difficulty in mastering the following pages. I have dwelt at length in Chap. II. on the Theory of Maximum and Minimum. Chap. III. is devoted to the more recent develop ments of the geometry of the triangle, initiated in 1873 by Lemoine s paper entitled " Sur quelques pro- prietes d un point remarquable du triangle." vi PREFACE. The study of the Brocardian Geometry is appro priate at this stage, as I have shown that the deductions of M. Brocard and of other geometers, both in England and on .the Continent, are simple and direct inferences of the well-known property of Art. 19, which has been called the Point Theorem. Chap. IX. gives an account of the researches of Neuberg and Tarry on Three Similar Figures. A feature of the volume is the application of Reciprocation to many of the best known theorems by which the corresponding properties of the Conic are ascertained. This method and that of Inversion are pursued as far as is admissible within the scope and limits of an elementary treatise on Geometry. In the preparation of the book, I consulted chiefly the writings of Mulcahy, Cremona, Catalan, Salmon, and Townsend, and hereby acknowledge my indebted ness for the valuable stores of information thus placed at my disposal. Many of the Examples are from the Dublin Univer sity Examination Papers, and more especially from those set by Mr. M Cay. I have as far as possible indicated my additional sources of information, and given the reader references to the original memoirs from which extracts have been taken. WILLIAM J. MCCLELLAND. SANTRY SCHOOL, 1st November, 1891. CONTENTS. CHAPTEK I. INTRODUCTION. ARTICLE P AGB 1. Symmetry, Convention of Positive and Negative, ... 1 3. Anharmonio Ratios, ........ 3 Examples, .......... 4 Euler s Theorem, ........ 5 4. Limiting Cases, and oo, ....... 6 Examples, .......... 7 5. 6. Envelopes, .......... 9 Examples. Bobillier s and Mannheim s Theorems, . . 10 7. Casey s Theorem 23. 14 + 3T. 24+12". 34=0, . . . .11 Examples. Feuerbach s Theorem, Hart s Extension, . . 13 CHAPTER II. MAXIMUM AND MINIMUM. SECTION I. INTRODUCTION. 8. Explanation of Terms, . . . . . . . .15 Examples, . . . . . . . . . .17 9. Theorem, 17 Examples, . .......... 18 10. Theorem, 21 Examples, .......... 22 11. Problem, 23 12. Theorem. Corollaries, ........ 23 Examples, .......... 26 13. Problem, 26 Examples, . . . . . . . . . ,27 vii viii CONTENTS. ARTICLE PAGE 14. Theorem. Examples, 27 15. Theorem, 30 16. Problem, 30 Examples, . . . . . . . . . .31 SECTION II. METHOD OF INFINITESIMALS. 17. 18. Explanatory, 32 Examples, .....*.... 35 SECTION III. 19. The Point O Theorem, . . . . . . .40 20. Similar Loci, 42 Examples, .......... 43 21-23. Additional Theorems, 44 Examples, 45 Extension of Ptolemy s Theorem, . . . .48 24. Theorem, 49 Examples, .50 25. Theorem. Centre of Similitude, 51 Examples, 53 SECTION IV. 26. MISCELLANEOUS PROPOSITIONS. 59 CHAPTER III. RECENT GEOMETRY. SECTION I. THE BROCARD POINTS AND CIRCLE OF A TRIANGLE. 27. Brocard Points (13 and O ) 60 28. Brocard Angle (w), 61 Cot u = Cot A + Cot B + Cot <7, 61 Examples 62 Brocard Circle and Brocard s First Triangle, ... 63 Polar Equation of a Circle, 65 CONTENTS. ix SECTION II. THE SYMMEDIANS OF A TRIANGLE. ARTICLE PAGE 29. Fundamental Property of Symmedians, 66 30. Expression for Length of Symmedian. Examples, ... 67 31. Antiparallels, 68 32. The Pedal Triangles of the Brocard Points, .... 69 33. Theorems, . . 70 SECTION III. 34. TUCKER S CIRCLES. 71 35. Construction for Tucker s Circles, . .... 72 36-38. Theorems 73 SECTlOiV IV. TUCKER S CIRCLES. PARTICULAR CASES. 39. Cosine Circle, 74 40. Ham s Property of the Symmedian Point, .... 75 42. Triplicate Ratio ("T.R." ) Circle, 75 45. Taylor s Circle, 76 48. Common Orthogonal Circle of the three ex-Circles of a Triangle, 78 Examples, 79 CHAPTER IV. GENERAL THEORY OF THE MEAN CENTRE OF A SYSTEM OF POINTS. 50. Theorem, 84 51. Theorem, 85 Examples, 86 52. Theorem. Examines, ........ 87 Isogonal and Isotomic Conjugates, ..... 88 53. Method of finding the Mean Centre, 90 Corollaries, .......... 91 Examples, 92 WeilPs Theorem, 96 54. 55. Properties of Mean Centre. Corollaries, .... 98 Examples, 100 x CONTENTS. ARTICLE PAGE 56. Theorem. Corollaries, 103 Examples, .......... 104 RECIPROCAL THEOREMS. 57. Central Axis, 106 58. Method of finding Central Axis, ...... 107 60. Diameter of a Polygon, ........ 108 61. Za . AP 2 a Minimum, 108 Examples, .......... 109 J5S CHAPTER V. COLLINEAR POINTS AND CONCURRENT LINES. 62. Theorem . |I . |f=l ,, 2 63. Theorem |f. ->; 4f= -I, 113 64. Equivalent Relations. Particular Cases, . . . .114 Examples, .......... 117 Centre of Perspective, . . . . . . . .120 65. Criterion of Perspective of Two Triangles, .... 120 66. Axis of Perspective. Homology, 121 Examples, 121 Stall s Theorem, 123 fi7 BX.BX CY. OY AZ.AZ _ } CX. CX AY.AY BZ\~Z ~ Corollaries. Pascal s Theorem, ...... 124 ,, Brianchon s Theorem, ..... 125 ,, Complete Cyclic Quadrilateral, . . . 127 Examples, .128 Tarry s Point, 131 68. Harmonic Properties of the Quadrilateral: Its Diagonal Triangle, 132 Problem, 133 Examples. Maximum Polygon of any Order escribed to a Circle, 134 Note on Pascal and Brianchon Hexagons, .... 136 CONTENTS. x i CHAPTEK VI. INVERSE POINTS WITH KESPECT TO A CIRCLE. ARTICLE PAGE 69. Particular Cases, 139 70, 71. Theorems, 139 Examples, 140 72. Theorem. Corollaries, 140 Examples. Radical Centre and Product, Conjugate Lines of a Quadrilateral, Circle of Apollonius, .... 143 CHAPTER VII. POLES AND POLARS WITH RESPECT TO A CIRCLE. SECTION I. CONJUGATE POINTS. POLAR CIRCLE. 73. Definitions. Elementary Properties of Conjugate Points, . 149 74. Theorem, 149 Examples, 150 75. 76. Theorems. Reciprocal Polar Triangles, .... 151 77. Self-Conjugate Triangles. Polar Circle, . . . .151 78. Expressions for the Radius of Polar Circle. Orthocentric System of Points, 152 Examples. Polar Centre, . . . . . . .153 SECTION II. 79. SALMON S THEOREM. 155 Corollaries, . . . . - 156 Examples, 157 Tangential Equations of In- and Circum-Circles, . . 159 SECTION III. RECIPROCATION. 80. Reciprocal Polar Curves. Their Elementary Properties, . 162 81. Reciprocation of a Circle from any Origin, .... 164 Examples, 167 xii CONTENTS. CHAPTER VIII. COAXAL CIRCLES. SECTION I. COAXAL CIRCLES. ARTICLE PAGE 82. Definitions, Radical Axis, . . . . . 173 83. Conjugate Coaxal Systems, 173 Circular Points at Infinity, 174 84. Modulus of a Coaxal System, . . . . . .174 85. To construct a Circle of given radius of a Coaxal System, . 175 86. Theorem, . . .175 87. Limiting Points. Extreme Cases, . . . . . .175 88. Theorems. Corollaries, .176 Examples, 178 M Cay s proof of Feuerbach s Theorem, . . . .183 89. Theorem, Corollaries, Centres and Circle of Similitude, . 184 Examples, 186 Poncelet s Theorem, 187 91. Theorems. Diagonal Line of a Quadrilateral, . . . 189 Examples, 190 SECTION II. ADDITIONAL CRITERIA OF COAXAL CIRCLES. 92. First Criterion. Corollaries, 191 Examples, 192 93. Second Criterion. Corollaries, 194 Examples, 195 SECTION III. CIRCLE OF SIMILITUDE. 94. 95. Fundamental Properties, 197 Examples, 198 Orthocentroidal Circle, 200 Miscellaneous Examples, ....... 200 Extension of Weill s Theorem, 201 CONTENTS. xiii CHAPTEK IX. THEORY OF SIMILAR FIGURES. SECTION I. TWO SIMILAR FIGURES. ARTICLE PAGE 96. Homothetic Figures, 204 97. Fundamental Properties. Homothetic Centre, Centre of Similitude, .205 98. Double Points of Similar Polygons, 205 99. Further Properties, Instantaneous Centre, .... 206 SECTION II. THREE SIMILAR FIGURES. 100. Triangle of Similitude, 207 Circle of Similitude, Invariable Points. Invariable Triangle, 208 101, 102. Theorems, 209 103. Adjoint Points, Director Point, 209 104. Theorems by Neuberg, . 210 105. Particular Cases, 210 106. M CAY s CIRCLES. 211 107. To find Centres and Radii of M Cay s Circles, . . . 213 Examples, . .216 Feuerbach s Theorem, . . . . . . .217 CHAPTER X. CIRCLES OF SIMILITUDE AND OF ANTISIMILITUDE. SECTION I. 109. CENTRES OF SIMILITUDE. 219 110. Properties of Homologous and Antihomologous Points. . . 220 111. Theorem. Homologous Chords are Parallel, . . . 221 112. Theorem. Antihomologous Chords meet on their Radical Axis 221 113. Products of Antisimilitude. Their Values. Note, . . 222 Examples, 223 Feuerbach s Theorem, ....... 225 xiv CONTENTS. SECTION II. CIRCLES OF ANTISIMILTTUDE. ARTICLE PAGE Definitions, 227 114. Fundamental Properties, 227 Examples, 229 Constructions for the eight circles touching three given ones, . . . 230 CHAPTER XI. INVERSION. SECTION I. INTRODUCTORY. 115. Definitions 234 116. Species of Triangle obtained by Inversion of the Vertices of a given one, 236 118. Relations between the Sides of the two, .... 237 119. Theorem. Corollaries, 237 120. Inversion of a System of Four Points. Particular Cases. Note, 238 121. Relations between the sides of A BCD and A BC D . Corollaries. Inversion of the Vertices of an Harmonic Quadrilateral. 239 Examples, 240 122. Theorem, 241 Examples, .......... 242 SECTION II. ANGLES OF INTERSECTION OF FIGURES AND OF THEIR INVERSES. 123. General Relations between a Circle and its Inverse from any Origin, 243 124. Problem. Corollary, 245 Examples, .......... 246 125. Theorem, 246 126. Angle between two Curves remains unaltered by Inversion, . 247 127- Important Considerations arising from the Theorem of Art. 126, 248 CONTENTS. xv SECTION III. ANHARMONIC RATIOS UNALTERED by INVERSION. ARTICLE PAGE 128. Theorem, 250 129. Inversion of a Regular Cyclic Polygon. Harmonic Polygons, 250 130. Cosymmedian Triangles, 252 Example, 253 Miscellaneous Examples, ....... 254 CHAPTER XII. GENERAL THEORY OF ANHARMONIC SECTION. SECTION I. ANHARMONIC SECTION. 131. Definitions, 261 132. The Six Anharmonic Ratios of Four Points, .... 261 133. Expressions for all the Ratios in terms of any one of them, . 262 Examples, .263 SECTION II. ANHARMONIC SECTION OF AN ANGLE. 134. Anharmonic Equivalents, 265 135. Properties of Transversals to a Pencil, 266 136. Theorem. Corollaries, 267 Homographic Division, ....... 268 Examples, 268 137. Directive Axis, 270 138. Directive Centre, 271 139. Points whose Conjugates are at infinity, .... 272 Examples, 272 Note on Art. 137 (Fig.), 274 140. Theorem. Corollaries, . 275 142. Third Method of describing Homographic Systems, . . 277 Corollaries, 278 143. General Relation connecting Corresponding Points, . . 280 Examples, ....... . 281 xvi CONTENTS. CHAPTEE XIII. INVOLUTION. ARTICLE PAGB 144. Equations of Involution, . . . . . . . 282 145. Property of Directive Axis and Corresponding Property of Directive Centre, 283 146. Particular Case of an Involution, ..... . 283 147. Fundamental Theorem, 284 Examples 285 148. DESARGUES THEOREM. 286 Corollaries, 287 Examples 291 CHAPTER XIV. DOUBLE POINTS. 149, 150. Constructions for the Double Points, . . . .293 Examples, 295 CHAPTER T. INTEODUCTION. Definitions. Right lines passing through a point are called a Concurrent System. The point is the Vertex of the system, and the lines are a Pencil of Rays. Collinear points are those which lie on a right line. Symmetry. Convention of Positive and Negative. 1. The letters A, B, 0, ..., are generally used to denote points and positions of lines, and a. 6, c, lengths, e.g., the vertices of a triangle are A, B, G, and the opposite sides a, 6, c. By AB is meant the distance from A to B measured from A towards B, and by BA the same distance measured in the opposite direction. Thus AB= -BA or AB+BA=Q. Similarly for three collinear points A, B, C: AB+BC=AC= -CA, therefore BC+CA+AB = 0. 2. If four points A, B, C, D, be taken in alphabetical order on a circle, we have by Ptolemy s Theorem 2 INTRODUCTION. the six linear segments being measured from left to right, or we shall say positively, in figure ; hence, by transposing, BC.AD + CA . BD+AB. CD = 0. Again, since each chord is proportional to the sine of the angle it subtends at any fifth point on the circle, this equation reduces to sinBOC ainAOD + sinCOA sinBOD + sinAOB sin(70D = 0, a result which is therefore true for any pencil of four lines, and is deduced directly from Ptolemy s Theorem by describing a circle of any radius through its vertex. In this equation it is implied that AOC denotes the magnitude of the angle measured from A towards (7, and that therefore sin A 00= sin CO A. 3. Let . ABCD denote a system of lines concurrent at ; A, B, C, D, the points in which a line L meets it; and p the distance of the vertex from L. Then 2BOC=BC .p = OB . 00 sin BOG, and 2AOD = AD.p=OA.ODsmAOD-, by multiplication * = OA .OB>OC.OD.mnBOC.AOD;...(l) INTRODUCTION. 3 similarly CA.BD.p*=OA.OB.OC.OD. sin CO A . sin BOD ; (2) dividing (1) by (2) we have BC.AD:CA.BD = $mBOC.$mAOD:smCOA.sii\BOD.(3) The student will observe that three pairs of angles are formed by taking any pair of rays with the remaining or Conjugate pair. Thus BOG and A OD may be conveniently denoted by a and a, CO A and BOD by ft and , and AOB and COD by y and y. With this notation (3) is written BC . AD : CA . BD = sin a sin a : sin /3 sin /3 , and generally we infer from symmetry that COR. 1. If we draw four parallels to the rays of the pencil, we in general obtain a triangle and a transversal to its sides. Moreover, if we denote the angles of the triangle by a, ft, y, those made by the transversal with its sides are the opposites a, ft , y ; hence for any triangle and transversal we have always sin a sin a H- sin ft sin ft -f sin y sin y 0. COR. 2. Let the line ABCD be divided harmonically or such that AB/BC=AD/CD, then BC.AD = AB. CD ; hence by (3) the pencil is divided harmonically, i.e., the angle CO A is divided internally in B and externally in D in the same ratio of sines. Defs. The three ratios and their reciprocals on the left side of (3) are termed the Anharmonic Ratios of the four points on the line ; and those on the right the Anharmonic Ratios of the pencil . ABCD. 4 INTRODUCTION. Their equivalence is expressed thus : A variable line drawn across a pencil is cut in a constant anharmonic ratio ; or any pencil and transversal to it are Equianhar- monic. The foot of the perpendicular from a point on a line is the Projection of the point on the line, and the per pendicular is called its Projector. If A and B be the projections of A and B on a line L, A B is called the Projection of AB, and is equal to AB cos 6, where is the angle between AB and L. EXAMPLES. 1. The sum of the projections of the sides of a polygon on any right line = ; and generally if lines be drawn equally inclined and proportional to the sides of a polygon, the sum of their projections and the sum of the sines of the series of angles is also equal to 0. [For they are proportional to the projections of the sides of a regular polygon on two lines at right angles.] 3. In any quadrilateral whose sides are a, 6, c, c?, to prove that d 2 =a? + b 2 + c 2 - Zbc cos be - 2ca cos ca - 2ab cos aft, where be denotes the angle between the sides b and c. [For completing the parallelogram whose sides are b and c and drawing x we have d? INTRODUCTION. 5 where x is the projection of x on the parallel b ; but by Ex. 1. of = a cos ab + c cos be, substituting for x its value and for or 2 , a 2 + c 2 -2occos ac, the above result is obtained.] 4. Euler s Theorem.* For three collinear points A, B, C and any fourth P to prove the relation BC. AP 2 +CA . BP 2 + AB.CP*= - BC. CA . AB. [By Euc. II. 12, 13, AP 2 = AB 2 + BP 2 - ! 2AB.BPcosB ............ (1) and CP 2 = BC 2 + BP*+2BC.BPcosB ........... (2) multiplying (1) by BC and (2) by AB and adding to eliminate cos By the above follows on reduction.] 4A. Having given the base c of a triangle and Ia? + mb 2 = const., find the locus of the vertex, I and m being given quantities. 5. If APC is a right angle the relation in Ex. 4 is equivalent to BC 2 . AP* + A B 2 . CP 2 = A C 2 . BP 2 . [This follows from Ex. 4 or is obtained directly thus ; let fall perpendiculars BX and BY on CP and AP, then multiplying the equation BP*= BC 2 sm 2 C + AB 2 sm 2 A by AC 2 , therefore, etc.]. 6. If the transversal to a harmonic pencil is parallel to one ray D, the intercept AC is bisected by B the conjugate of D. * "Catalan s The oremes et Problemes de Geom^trie Elementaire," 1879, p. 141. 6 INTRODUCTION. 7. If a line L turn around a fixed point P and meet two fixed lines OA and OB in A and B ; the locus of the harmonic conjugate Q of P with respect to A B is a line passing through ; and .................... (By Ex. 6.) Note. By Euc. VI. 2 if the variable PQ is bisected at Q the locus of Q is a parallel to OQ and Hence for any three lines A , B, C we find in the same manner that 1111 PA + PB + PC ~PQ* where Q describes a right line. 8. For any system of lines A, B, C, D ... the locus of Q such that 111 1 v _l _1 \ ^PA ~P ) is a right line. [See Exs. 6 and 7.] 9. For a regular cyclic polygon, if P coincides with the centre [Through P draw the line parallel to one of the sides, etc.] 10. If parallels be drawn through any point to the four lines in Ex. 4, the relation may be written sin P sin y sin 7 sin / sin a sin /^ _ i sin P sin 7 sin 7 sin a sin a sin fi 1 1. From the formula BC . A D + CA . BD + AB . CD = 0, prove that if A, B, C be three collinear points and P any fourth point BCcotA + CA cotB+ABcot (7=0, the angles being all measured in the same aspect; and hence find the locus of the vertex, having given the base c and I cot A + m cot B= const. 4. Limiting Cases. and o> . Def. The Angle of intersection of two circles is that between the tangents drawn to them at either point of INTRODUCTION. 7 intersection ; it is therefore equal to the angle between the radii drawn to either common point.* (Euc. III. 19.) If the circles touch Internally this angle is 0, if Externally 180. They are said to intersect Orthogonally when the angle is 90. The Angle made by a line and circle is that between the line and the tangent to the circle at its intersection. EXAMPLES. 1. To find the angles between the circum- and ex-circles of a triangle ABC. [Since 6\ 2 = J S 2 + 2/?r 1 , etc., we easily obtain with similar expressions for 2 and $ 3 .] 2. To find the angle of intersection of the in- and circum-circles. [8 2 = R 2 - 2/fr, therefore 2 sin \B = . / where */ - 1 = i.~\ 3. If two concentric circles cut orthogonally one is real and the other imaginary, and their radii are of the forms p, ip. * If Oj, rj ; 2 , r z , be the circles, 5 the distance 0^0^ 6 the angle of intersection, and t the direct common tangent, we have 8 2 = r 1 2 + r 2 8 - 2^2(5080 hence 5 2 - fa - r 2 ) 2 = 4r x r 2 sin 2 ^, .................................... (1) or A * Similarly 5 2 - (r^ + r 2 ) 2 = - 4r x r 2 cos 2 \ 6, hence if t be the transverse common tangent, t>*= - 4^7*2 cos 2 ^0 (2) Multiplying (1) and (2) and reducing we have = 21.^281110 (3) where >/ - 1 = { ; also if 7 denote the length of the common chord, of the circles (real or imaginary) since 2r 1 r 2 sin = yd, t . t = i . y . 5. It is obvious that either the transverse common tangent to the circles or their angle of intersection is imaginary. 8 INTRODUCTION. Let AX be a variable chord passing through a fixed point A at which a tangent is drawn. According as the chord AX and angle TAX diminish in magnitude X approaches the tangent. When X is indefinitely near to A, AX is said to have reached its limiting position and may then be considered to coincide with the tangent. Hence a tangent to a circle is in the direction of the infinitesimal chord at its point of contact, or is the chord joining two indefinitely near points. Again, let the tangent T and its point of contact be fixed and the chord AX given in length. As the radius of the circle increases the curvature diminishes, and the point X obviously approaches the tangent. Hence X may be made to move as near as we please to the tangent by continually increasing the value of the radius of the circle. In the limit, when the latter is indefinitely great, the distance of X from T is so very small that we may consider the point to lie on the line. Hence a finite portion of a circle of indefinitely great radius opens out into a right line, the remainder being, of course, at a distance infinitely great, i.e., at infinity. INTRODUCTION. 9 5. Envelopes, Let a variable line turn around a fixed point and meet any fixed line. According as its angle of inclination to the perpen dicular OM increases, the segments OA, OB, 00 continue to increase and the angles A, B, to diminish. In the limit it reaches a position at right angles to OM. Here the angle between it and the fixed line vanishes, and their point of section is at infinity. In this case the lines are parallel (Euc. I. 28) ; hence Parallel lines may be regarded as having angles of inclination = or lines intersecting at infinity. Thus a system of parallels is a pencil of rays whose vertex is at infinity. 6. Let A and X be any two points on a curve of which A is fixed and X variable, and TA and TX tangents. It appears as before that as X approaches A the chord AX and the base angles A and X of the triangle TAX gradu ally diminish and ultimately vanish. But as the base angles diminish the vertex T approaches the base and a fortiori the element of curve AX. Hence in the limiting position, i.e., when the tangents are con secutive, their point of intersection is on the curve. A curve touched by a variable line is called the Envelope of the line. Thus the envelope of a line which 10 INTRODUCTION. varies according to any law is the locus of the intersec tion of its consecutive positions. EXAMPLES. 1. The envelope of equal chords in a circle is a concentric circle (Euc. III. 14). 2. Bobillier s Theorem. If two sides of a given triangle touch fixed circles the third side also touches, or envelopes, a circle. [Let ABC be the given triangle. Through O l and 2 , the centres of the given circles, draw parallels to the sides meeting the base in A and B and each other in C . Describe a circle O-^O^C ^ and draw C 3 parallel to A B. Since 2 C 3 is a given angle ( = 4), 3 is a fixed point. But A B C is given in all respects save position ; hence the distance p of 3 from A B is a known quantity. The envelope of the base AB is therefore a circle whose centre is 3 and radius = pJ] 3. To find the radius (p) of a circle which touches the sides AC and EG of a triangle and the circum-circle ABC. [Let /"denote the in- and the circum-centre of the triangle ; M the centre of the circle whose radius is required is on the line CI. Then OM=R-p, OP=R*-2Rr, 7<7=r/sin|<7, J/tf=/)/sin|<7, and J//=(p-r)/sh4<7. Also, since C, 7, M are three points in a line and any fourth point, by Euler s Theorem we obtain on reducing (1) INTRODUCTION. 11 Again, if the circle M, p has external contact with the circum -circle, it can be similarly proved that r 3 = pcos^C (2) NOTE. The relation (1) is otherwise expressed : Since r/p = cos 2 <7, (p - r)/p = sin 2 | C. But (p - r)/p = MIJMC and p 2 /MC 2 = sin^C, hence MI.MC=p 2 (3) or the chord of contact PQ of the circle M, p ivith the sides of the triangle passes through the centre of the inscribed circle.] 4. Mannheim s Theorem. Having given the vertical angle and radius of the in- or corresponding ex-circle, the envelope of the circum-circle is a circle. [By Ex. 3.] 7. We shall conclude the present chapter with the following useful property, of the common tangents to four circles which touch a fifth, due to the late Dr. Casey. Denote the circle whose centre is and radius r by 0, r ; and let the four circles 0^, 2 r 2 , 3 r 8 , 4 r 4 , touch a fifth 0, R at the points A, B, C, D. Let the distance 2 3 be 23 , and the direct common tangent to the cor responding circles be 23. Then 23 2 = S 2 * - (r 2 - r 3 ) 2 . 12 INTRODUCTION. In the triangle 00 2 3 we have - 00 3 ) 2 + 400 2 . 00 2 3 2 = 00 2 2 + 00 3 2 -200 2 . 00 3 cos or 23 2 = 400 2 . 00 3 sin 2 J50C7= 00 2 . 00 Similarly hence by multiplication and reduction 3 . 14 = (00! . 00 2 . 00 3 . O and by Ptolemy s Theorem 23. l+3i. 24 + 12. 34 = ........................ (1). The contacts in the figure are similar, or all of the same kind, but it will be observed that if the fifth circle touches any two with contacts of opposite species, their transverse common tangents must be substituted in (1). We let 12 denote the transverse tangent to O v r x and 2 , r 2 ; then T2 * = V-(n + 2) 2 - For example, if the circle O v r x is external and the remaining circles internal to 0, R the relation is written 23 14 + 3T 24 + 12 > S=0, with analogous expressions for all other cases. NOTE. The student must carefully observe that of the three terms of the equation two are positive and one negative ; the latter corresponding to the pairs of circles whose contacts are alternate. Thus in the figure, O lf ^ and 3 , r 3 have alternate contacts with the given circle, therefore the term 31 . 24 is negative, and taking the absolute values only the equation is 23^.14 + 12.34 = 31.24! This is of great importance, and should be borne in mind in the following Examples. INTRODUCTION. 13 EXAMPLES. 1. What does the general property reduce to when the circles become points ? Ptolemy s Theorem. 2. Express a condition that the circum-circle of a given triangle may touch another circle. [If a, b, c be the sides and t lt t 2 , t 3 the tangents from the vertices to the other circle we have a^ + bt 2 + ct 3 = 0.] 3. Feuerbach s Theorem. The nine points circle of a triangle touches the in- and ex-circles. [The middle points of the sides and the in-circle are four circles satisfying the equation of Ex. 2. For 23=a and 14 = (6-c); therefore 223.14 = Sa(6 - c) = 0*.] 4. If a, 6, c be the sides of a triangle inscribed in a circle, and X, /*, v the distances of its vertices from any tangent, show that the equation in Ex. 2 reduces to Wx + W/ + c*Tv = 0. t 5. More generally if X, /*, v denote the distances from any line, give the geometrical interpretation of the equation and hence find a relation connecting the sides of a triangle with the distances of its vertices from a given line. [The roots of the quadratic in x are the distances from the line of the tangents to the circle parallel to it, etc.] 6. Hart s Extension of Feuerbach s Theorem. If the sides of a triangle be replaced by three circles, and four circles correspond ing to the in- and ex-circles of the triangle described to touch them ; the group of four is touched by a circle. [Let the triangle formed by the circles be ABC, and let a<b<c. Then s-a>s-b>s-c. If the in- and ex-circles are numbered * This proof is an application of the converse of Dr. Casey s relation. t This result may be otherwise shown as follows : Let P be the point of contact of the tangent. Then BC . AP + CA . BP + AB . OP=0. But AP 2 =2r\, P 2 =2r/i, and CP 2 =2rv, substituting these values; therefore, etc. 14- INTRODUCTION. 1, 2, 3, 4 respectively, the side a is touched by the four circles and the transverse tangents are drawn to 2 ; also the order of the con tacts is 3, 1, 2, 4 ; hence the equation is -23 .T4 + 31.24 + 12 .34=0 (1) For the side b the transverse tangents are drawn to 3, and the order of the contacts is 2, 1, 3, 4 ; hence -23 .T4 + 3l .24 + l2.3T = (2) For the side c the transverse tangents are drawn to 4, and the order of the contacts is 3, 4, 1, 2 ; hence ^.14 -3l.24 +12.34 = (3) Adding (1) and (3) and subtracting (2) we get 23 . 14 -3T . 2T+I2 .34 = 0, showing that 2, 3, 4 have similar and 1 opposite contacts with a circle which touches all four.] CHAPTER II. MAXIMUM AND MINIMUM INTRODUCTION. 8. When the base and vertical angle of a triangle are given the locus of the vertex is a segment of a circle described on the base, containing an angle equal to the vertical angle. (Euc. III. 21.) Let a number of tri angles be constructed satisfying the given conditions, and it will be observed that as the vertex recedes from either extremity of the base the altitude and area both increase up to a certain point, after which they begin to diminish. This point is obviously the middle point of the seg ment the vertex of the isosceles triangle with the given parts or the point at which the tangent to the arc is parallel to the base. Here the area and altitude are said to have attained their maximum values. Again since the rectangle under the sides AC and BC is equal to the rectangle under the diameter of the circum-circle and altitude (ab = dp) ; ab and p are maxima simultaneously. Also since a 2 + 6 2 = 2( Jc) 2 + 2/3 2 , where /3 is the median to the side c ; when ft is a maxi mum or minimum, a 2 + 6 2 is maximum or minimum. 15 OF THK ^ I TTlNTTXTTP-ooTHrv- i 16 MAXIMUM AND MINIMUM. And if N be the middle point of the arc of the circle below the base, then, since AN = BN ( = x say) by Ptolemy s Theorem, we have ax + bx = c . CN, or x(a + b) = c . ON, from which it appears that a + b and CN are maxima together ; that is when the vertex C is at the middle point M of the arc AB. On the other hand it is manifest that the difference of base angles ( A B) and difference of sides (a b) both diminish as the vertex C approaches M and vanish at that point; and after C passes through this point each difference begins to increase. At C they are said to have their minimum values, though this need not necessarily be nothing. Thus generally : a variable quantity which, under certain conditions, increases up to a definite limit and then begins to diminish, is said to have attained its maximum value at the limit ; and if, after diminishing, it again begins to increase, it attains a minimum value at the stage where it has ceased to diminish. MAXIMUM AND MINIMUM. 17 The foregoing remarks may be thus summed up : Of all triangles having a given base and vertical angle the isosceles has the following maxima area, altitude, rect angle under sides, sum of sides, bisector of base, and sum of squares of sides.* EXAMPLES. 1. The triangle of greatest area and perimeter inscribed in a circle is equilateral. [For each vertex must lie mid-way between the other two, or the area and perimeter would both be increased by removing any vertex to the middle point.] 2. A regular polygon of n sides inscribed in a circle has a greater area and perimeter than any other inscribed polygon of the same order. [By Ex. 1.] 9. Theorem. // two sides AC and AB of a triangle are given in length the area of the triangle AEG is a maximum when they contain a right angle. Let ABC denote the right-angled triangle, and ABC any other triangle formed with the given sides. Draw C X perpendicular to AC. Since AC=AC and ACT > AX ; therefore AOAX, hence (Euc. I. 41) the triangle ABC > ABC , and similarly for any other position ; therefore, etc. * The vertical angle is supposed to be acute. B 18 MAXIMUM AND MINIMUM. EXAMPLES. 1. If the ends of a string of given length are joined, the area of the figure enclosed is a maximum when it takes the form of a semi circle. [Take any point A on the string ABC and join AS and AC. Consider the segments into which the string is divided at A to be rigidly attached to the lines AB and A C. If the angle at A is not right, by rotating AC around A until it is perpendicular to AB, the area of the triangle ABC, and therefore also of the whole figure, is increased. Similarly for any other point A ; hence the area enclosed is a maximum when the joining line BC subtends a right angle at every point on the string.] 2. A closed curve of given perimeter is of greatest area when its form is a circle. [Let A be any point on the curve, and take B such that A MB and ANB are equal in length. Then the areas A MB and ANB are MAXIMUM AND MINIMUM. 19 each a maximum when A B is the diameter of semicircles on opposite sides ; therefore, etc.] 3. Having given the four sides a, 6, c, d of a quadrilateral, its area is a maximum when it is cyclic. [Let A BCD be the cyclic quadrilateral with the given sides, and consider the segments on the sides to be rigidly attached to them.* If then the figure be distorted in any way into a new position *The construction of the cyclic quadrilateral whose four sides are given is as follows : Draw CE making LDGE=tBA C. Since by Euc. iii. , 22, LCDE= LA BC, the triangles ABG and CDE are similar ; therefore DE : c = b : a (Euc. vi. 4) ; hence DE is known and E is a fixed point. Again, AC : CE=a :c; therefore in the triangle ACE we have the base AE and ratio of sides ; the locus of C is therefore a circle (Euc. vi. 3) ; this locus intersects the circle described with D as centre and c as radius at the point C ; therefore, etc. 20 MAXIMUM AND MINIMUM. A BCD , the area of the circle A BCD > A B C D (Ex. 2), but the segments AB=A B , BC=B C , etc.: take away these equal parts and there remains the quadrilateral A BCD greater than A B C D .] * 4. If three sides a, b, c, of a quadrilateral are given in magnitude, the area is a maximum when the fourth side d is the diameter of the circle through the vertices ; and generally, When all the sides but one of a polygon of any order are given in magnitude, the area is a maximum when the circle on the closing side as diameter passes through the remaining vertices. f [Proof as above.] 5. Having given of a quadrilateral the diagonals 8 and 8 and a pair of opposite sides BC and AD, its area is a maximum when BC is parallel to AD. [Take any position of the quadrilateral and through C draw CE parallel and equal to 8. Join BE and AE. The triangles BDE and BCD are equal (Euc. I. 37) ; to each add ABD, therefore ABCD = ABED. * The student should learn the proof of the Trigonometrical expression for the area of any quadrilateral in terms of the four sides and the sum of either pair of opposite angles. (Area) 2 = (s- a)(s - b)(s - c)(s -d)- abed coeP&A + C). (Casey s Plane Trig., art. 152, cors. 3, 4.) f To construct the quadrilateral. Let he the angle between a and b, and A C = x. Then d 2 = c 2 + z 2 = a 2 + 6 2 + c 2 - 2abcos0 ; ...................... (1) but cos = - c/d ; substituting in (1) and simplifying we have the following expression ford: d 3 -d(a 2 + 6 2 + c 2 )-2a&c = 0, an equation which has only one positive root. (Burnside and Paiiton s Theory of Equations, Art. 13.) In the particular case when a = b = c, the equation for d reduces to hence d = 2a, thus showing that the quadrilateral is half the regular inscribed hexagon. MAXIMUM AND MINIMUM. 21 Now, A BDE is a maximum when A D and DE are in the same straight line; hence A BCD is a maximum when EC is parallel to AD.] 6. The diagonals of a quadrilateral are 9 and 10 feet and two opposite sides 5 and 3 feet ; find when its area is a maximum. 10. Theorem. Having given the base AB of a triangle and the locus of the vertex a line L meeting the base pro duced, the sum of the sides AC-}- EG is a minimum when L is the external bisector of the vertical angle. Let fall a perpendicular BL and make B L = BL. Join AB and let G be its intersection with Z. Take any other point P on the line and join AP and B P. 22 MAXIMUM AND MINIMUM. The triangles BCL and B GL are equal in every respect (Euc. I. 4) ; hence BC= BG. Similarly BP = BP. Hence since- (Euc. I. 20) AP + BP > AB it follows that AP+BP>AC+BC. COR. 1. If the line L cuts the base internally the difference of the sides (ACBC) is a maximum when it bisects internally the angle G. EXAMPLES. 1. The triangle of minimum perimeter inscribed in a given one is formed by joining the feet X, I 7 , Z of the perpendiculars * let fall from the vertices on the opposite sides. [For the joining lines are equally inclined to sides on which they intersect (Euc. III. 21).] 2. The polygon of least perimeter that can be inscribed in a given one is that whose angles are bisected externally by its sides. (By Ex. 1.) 3. The base and area of a triangle being given, the perimeter is least when the triangle is isosceles. [For the line L is parallel to the base.] 4. If from 0, the point of intersection of the diagonals of a cyclic quadrilateral, perpendiculars are drawn to the sides and their feet P, Q, R, S joined, the quadrilateral PQRS is of minimum perimeter. 4a. If points P, Q , R , S be taken on the sides of the given quadrilateral, such that PQ , Q R , R S are parallel to PQ, QR, RS, then P S is parallel to PS and the perimeters of the quadrilaterals are equal. [Euc. VI. 2 and I. 5.] 5. The value of the minimum perimeter of the indeterminate inscribed quadrilateral in Ex. 4 is 2SS /D, where D is the diameter of the circum-circle. 6. Given a triangle ABC, find a point such that OA + OB + OC is a minimum. [Where BOC= COA^A OB=120.] * These are generally known as the Perpendiculars of the Triangle, and . X YZ as the Pedal Triangle of ABC. MAXIMUM AND MINIMUM. 23 11. Problem. Given an angle C of a triangle and a point P on ike base, construct the triangle of minimum area. Through P draw APB such that AP = BP. The triangle ABC is less than any other A B C. For draw AX parallel to BB . Then the triangles APX and BPB f are equal in all respects (Euc. I. 4); hence AA P > BB P. To each add APB C, therefore A B C > ABC] hence the triangle of least area is that whose base is bisected at this point. 12. Theorem. Given an angle and any curve concave to its vertex C. The tangent AB which forms ivith the sides of the angle a triangle ABC of minimum area is bisected at its point of contact (P). For this tangent cuts off a less area than any other line A B through P, because it is bisected at P. Now draw any other tangent XY, and let PA E be parallel to it. Since the curve is concave to (7, A B C < XYC \ a fortiori ABC< XYC. COR. 1. In the particular case when the curve is a circle whose centre is at C the triangle is isosceles. This 24 MAXIMUM AND MINIMUM. property may be stated otherwise. When the vertical angle and altitude of a triangle are given, the base and area are both minima when the triangle is isosceles. On account of its importance an independent proof of this pro perty of the isosceles triangle is given. Let ABC be an isosceles triangle and A B C any other, having the same vertical angle and altitude CM. Now BC> B C (Euc. III. 8), but BC=AC< A C, hence A C>B C. Let CD = B C, join AD. The triangles ACD and BBC are equal in every respect (Euc. I. 4), hence AA C>BB C\ therefore A BC> ABC-, therefore, etc. COR. 2. When the curve is a circle touching the sides of the angle the tangent AB and area AEG are each minima when the triangle is isosceles. MAXIMUM AND MINIMUM. 25 COR. 3. If we consider the portion of the circle in Cor. 2, which is convex to (7, the intercept of a variable tangent made by the sides of the angle subtends a con stant angle a at the centre of the circle (2a = 7r C). Hence the variable triangle J.^Ohas a constant vertical angle (a) and altitude (y), and therefore its base and area are minima when A 1 = B 1 0. In this case the point of contact P 1 is the middle point of A^B^ Therefore, having given a circle and two fixed tangents, the portion of a variable tangent intercepted by the fixed tangents becomes a minimum in two positions, viz., when its point of contact bisects the arc XY internally or exter nally. In the latter case the area cut off (ABC) is a minimum but in the former a maximum ; For A&C = CXO Y- 2A 1 B 1 ; therefore, since CXOY is constant, when A^fi is a mini mum, A^B^G is a maximum. 26 MAXIMUM AND MINIMUM. EXAMPLES. 1. The triangle of least area and perimeter escribed to a circle is equilateral. [For the point of contact of each side bisects the arc between the other ; cf. Art. 8, Ex. 1.] 2. The polygon of least area and perimeter escribed to a circle is regular. (By Ex. 1.) 3. Having given the vertical angle C of a triangle in position and magnitude, and the in- or corresponding ex-circle, to prove that the line LM joining the middle points of the sides forms with the centre of the circle a triangle of constant area. [For the ex-circle : if p be the perpendicular of ABC drawn from C to the base, and r 3 the radius, we have WLM = %c(%p + r 3 ) =ABC+AOB=$OCXY=conBt. 1 etc.] 13. Problem. Given an angle of a triangle and a point P on the base, construct it such that AP . BP is a minimum. Through P draw AB so that the triangle ABO is isosceles. Describe a circle touching the sides of the angle at A and B, and draw any other line A PB . It is evident that AP.PB< AT . B P, and is therefore a minimum. MAXIMUM AND MINIMUM. 27 EXAMPLE. 1. Through the point of intersection P of two circles draw a line APE such that PA . PB is a minimum. [This reduces to describe a circle touching the two given ones at A and B such that J, B and P are in a line. It will be afterwards seen that this line passes through a point $, on the line of centres 0^0^ of the circles where QOJQO^ihe ratio of the radii.] 14. Theorem. If a right line be divided into any two parts a and b, their rectangle is a maximum when the line is bisected. For Euc. (II. 5) ab + fc^f = (^T^ = const hence ab is a maximum when a b = or when a = b. COR. The continued product of the segments of a line is a maximum when the parts are equal. EXAMPLES. 1. Through any point P on the base of a triangle parallels PX and Pl r are drawn to the opposite sides ; the area of the parallelogram PXCY is a maximum when the base AB is bisected at P. [For the triangles A PX and BP Y are constant in species, hence PX . PY&AP. BP. But the area of the parallelogram = PX.PY sin C&PX. PY\ therefore, etc.]* 2. The maximum rectangle inscribed in a given segment of a circle is such that if tangents EC and AC be drawn at its vertices X and Y y then BX= CX and CY= A Y. [For NX is the maximum rectangle that can be inscribed in the triangle BCN, and therefore greater than any other X N. Hence from the symmetry of the figure the rectangle on the side XY is greater than that on X Y , and a fortiori greater than that on X"Y". * Hence, the maximum parallelogram inscribed in a triangle is half the area of the triangle. 28 MAXIMUM AND MINIMUM. The construction of the maximum rectangle is as follows : Let EL be drawn perpendicular to OL, the diameter of the circle parallel to AB. Join OX and let it meet EL in P. Since the triangles OCX and BPX are equal in every respect (Euc. I. 26) PX=OX=r. Also OXBL is a cyclic quadrilateral, therefore Euc. (III. 36), PB.PL = PO.PX=2r z , but PL - PB is given ; hence the segment PB is known, and since it is equal to OC, C is determined. In the general case the line AB does not meet the circle, the seg ment is therefore imaginary, and the proposition may be thus stated : given a line AB and a circle; construct the maximum rectangle, having two of its vertices X and Y on the circle and the remaining two on the line.] 3. Draw a chord XY of a circle in a given direction such that the area of the quadrilateral ABXY, where AB is a given diameter, is a maximum. [Draw a diameter YX t and AYBX is a rectangle, hence AX is equal and parallel to BY. Join jBJJTand XX t and draw B C parallel to XX . Then since triangle A XX + triangle BXY= triangle ABX t reject the common part A MX and let BMX be added to each, BXX =ABXY. The quadrilateral is a maximum therefore when BXX 1 is a MAXIMUM AND MINIMUM. 29 maximum. It is easy to see that the latter is half the rectangle inscribed in a given segment BC. For since EG is parallel to XX , AC is perpendicular to XX and therefore parallel to PX, hence BAC=a. The problem is thus reducible to Ex. 2.] 4. If a given finite line be divided into any number of parts , 6, c ... ; to find when a a b@ fl ... is a maximum, where a, /?, y are given quantities. [This expression is a maximum when but a/a is one of the a equal parts into which the segment a may be divided ; hence (a/a) a is the product of the equal subdivisions. Similarly (b//3}P is the product of the /5 equal subdivisions of b, and so on. Therefore (1) attains its greatest value when the subdi visions of a, 6, c ... are all equal ; i.e., when a b _c -, a=/rr 5. Find a point with respect to a triangle such that the product of the areas (BOC)(COA)(AOB) is a maximum. [Since BOG+ CO A + A OB is constant, when BOC= CO A = A OB, by Ex. 4, or when is the centroid of the triangle.] 6. The maximum triangle of given perimeter is equilateral. [From the formula A 2 = *(* - a)(s - b)(s - c) ; since the sum of the factors on the right hand side is constant, A is a maximum when s-a=s-b = s-c; therefore, etc.] 30 MAXIMUM AND MINIMUM. 7. The maximum parallelogram of given perimeter and angles is equilateral. 8. If p 1} p 2 , Ps denote the perpendiculars from any point on the sides of a triangle, the maximum value of p^p^Ps is 8A 3 /27a6c, and is then the centroid of the triangle. (By Ex. 5.) [Otherwise thus : Since 4abp 1 p 2 =(ap l + bp 2 } 2 (ctpi bp 2 } 2 for any point on the base c, p^p 2 is maximum when ap^ - bp 2 vanishes, since ap l + bp 2 equals 2A. Then is the middle point of the base. Now if p s be supposed constant, is on the median through C. Similarly by regarding p l as constant, would be found on the median through A ; and so on. Therefore if the three perpendi culars vary, their product is a maximum for the point of intersection of the medians.] 15. Theorem. If a right line be divided into any two parts a and b the sum of their squares is a minimum when the line is bisected. For (Euc. II. 9, 10) Hence a 2 -f 6 2 is minimum when a b is minimum, because a+b is constant ; that is when a = b. COR. The sum of the squares of the segments of a line is a minimum when the segments are equal. 16. Problem. If a right line be divided into any number of parts a,b,c..., to find when a 2 b 2 c 2 , +-^4- +... ^s mimmum P 7 where a, /3, y are known quantities. Let the segment a be divided into a equal parts ; each part is therefore a/a and the sum of squares of the parts /a\ 2 a 2 a(-) = . W a Similarly if the segment b be divided into p equal parts the sum of squares of the subdivisions = b 2 //3 ; and so on. MAXIMUM AND MINIMUM. 31 Hence the above expression denotes the sum of the squares of the subdivisions of the parts a, 6, c ..., and is therefore a minimum when these are equal ; i.e., when a b c EXAMPLES. 1. Divide a line into two parts a and b such that 3a 2 + 46 2 is a minimum. [When + is minimum, i.e., when - = -, hence 3a = 46.] 2. To find a point P such that the sum of the squares of its distances, x, y, z, from the sides of a triangle is a minimum. [Let A 1} A 2 , A 3 denote twice the areas of the triangles subtended by the sides of the given one at the point. Now since A 1 = a^r, A 2 =6y, and A 3 = c$, and is consequently a minimum when !-&- ................................. < 2 > since AI + A 2 + A 3 = const. From (2) it is obvious that =^?... ...(3) a b c This result may also be seen from the identity (a 2 + b 2 + c 2 )O 2 + j/ 2 + 2 ) - (ax + by + czf = (bz - cyf + (ex - azf + (ay - bx?, with which the student should be familiar.] NOTE. This point is termed the Symmedian Point of the triangle, as it is obvious from (3) that the lines joining it to the vertices of the given triangle make the same angles with the sides as the corresponding medians ; also since x _y _z_ax + by + cz_ 2A a~b~c~~ a? + b 2 + c 2 ~ a* + 6 2 + c 2 2aA 26A .32 MAXIMUM AND MINIMUM. 3. Find a point P such that the sum of squares of its distances from the vertices of a triangle may be a minimum. [If CP be supposed constant while AP and BP vary, the point P describes a circle around C as centre, and if M be the middle point of the base A P 2 + BP 2 = 2 A M 2 + 2 J/P 2 . Hence AP* + BP* + CP* is minimum when 2PM 2 +CP 2 is minimum, since AM is constant. Therefore P is a point on the median CM such that CPjPM= 2, i.e., the centroid. Similarly by supposing AP or BP to remain constant we find the same point. Hence the centroid is the required point when AP, BP and CP all vary.] SECTION II. METHOD OF INFINITESIMALS. 17. It has probably been observed in the preceding section that the positions of maximum and minimum of a quantity, varying according to a given law, are sym metrical with respect to the fixed parts of the figure. Thus when the base and vertical angle of a triangle are given, the altitude, rectangle under sides, area, etc., etc., are maxima when the triangle is isosceles. MAXIMUM AND MTNTMUM. 33 In Art. 9 the triangle of maximum area is found by placing the two given sides at right angles: Again, a figure of given perimeter and of maximum area is circular. As the variable line AB in Art. 11 rotates in a positive direction around P, according as PB recedes from the perpendicular from P on BC, the segments AP and BP approach an equality, and the triangle ABG is a minimum when AP = BP. 18. The several parts, of a geometrical figure which varies according to a definite law, can always be expressed in terms of the fixed parts of the figure and those quantities which are sufficient to define its position. Take for example the figure of Art. 8. In any posi tion of the vertex (7, by assuming the triangle to be of given altitude; the variable parts, a, 6, area, and other functions of the sides or angles can be found in terms of the base c, vertical angle (7, and altitude. Thus the variables may be regarded as functions of the given parts and the co-ordinates of their position. It follows, then, that if the latter vary continuously those functions must do likewise.* Hence a very small change in position will cause a very slight change or increment in the magnitude of the function. Suppose in Art. 8 the circle to be divided into an indefinitely great number of equal parts, and let the vertex C occupy each point of section from A towards B. As the altitude thus receives indefinitely small increments so does the area. Let AB\)Q the base of a triangle and any curve C f (7 1 (7 2 the locus of its vertex. * See Burnside and Panton s Theory of Equations, Art. 7. c 34 MAXIMUM AND MINIMUM. In the figure as the vertex approaches G on the curve from left to right the intercept AX made by the per pendicular may be taken as the co-ordinate of its position, since if AX is known the position of is also known. Thus while AX continues to receive positive incre ments, the area, altitude, and other functions of it are sometimes decreasing, as from G to C v and sometimes increasing, as from G l to (7 2 . At the points G, C v G 2 the increments in the altitude alter in sign and therefore consecutive values are equal. Here also the tangents to the curve are parallel to the base AB, and at any other point G n the increment of the variable divided by the corresponding increment in the function = cot a, where a is the angle made by the tangent at G n with AB. We have seen that if AX denote the value of a variable in any position, and CX any function of AX, when "the function passes through a maximum or minimum its two consecutive values are in each case equal to one another. Suppose, for example, that a variable chord XY of a circle moves parallel to a certain direction; it gradually increases in length as it approaches the centre and if XY MAXIMUM AND MINIMUM. 35 ( be a diameter and X Y a consecutive chord; since XX and YY are tangents to the circle, and therefore parallel, XYX Y is a parallelogram and XY = X Y (Euc. I. 34). Hence the diameter is the maximum chord in a circle (cf. Euc. III. 15). EXAMPLES. 1. Having given the base and locus of vertex of a triangle ; find when the area is a maximum or minimum. [Let the locus be a curve of any order and it is readily seen (Euc. I. 39) that the tangents at the required points are parallel to the base.] 2. In Ex. 1 when is the sum of the sides a minimum or maximum ? [Let C and C be two points indefinitely near to each other on the locus MN. Draw CX and C Y perpendiculars to AC and EC respectively. Then since in the triangle ACX, X is a right angle and A inde finitely small, ACX is approximately a right angle and A C is nearly equal to AX. Hence in the limit C X=AC -AX=AC -AC. Similarly CY is the increment (negative) of BC. Therefore C X=CY and the right-angled triangles CC X and CC Y are equal in every respect, and L.AC C=LBCC . But AC C ACM when A is indefinitely small; hence the required points C on the locus are such that A C and B C are equally inclined to the curve, i.e., to the tangent at their point of intersection. 36 MAXIMUM AND MINIMUM. It similarly follows that if A and B were upon opposite sides of the curve this relation holds when ACBC is maximum or minimum.*] 3. Given the vertex A of a triangle fixed, the angle A in magni tude and the base angles moving on fixed lines intersecting in ; to construct the triangle ABC of minimum area. [By taking two consecutive positions as in figure, we have AB . AC^AB . AC and LBAB =L.CAC . Hence AB : AB = AC : AC, and the triangles BAB and CA C are similar (Euc. VI. 6). 1 Therefore LABO = LA C = AGO in the limit In the required position the sides AB and AC are equally inclined to the given lines. Here again we have an illustration of the symmetry of the figure when the triangle is minimum. If the angle A is 180 the property (Art. 13) follows at once.] 4. Given two sides of a triangle fixed in position and a point P on the base ; when is AB a minimum? [Taking two consecutive positions of. AB and drawing perpen diculars A X and BY ; as before A X is the increment of AP and B Yof BP ; hence A X=B Y. Again A X = AX cotA =AP sin P. cot A . Sim ilarly B Y=BY cot B = BP sin P cot B. Therefore in the limit AP cotA=BP cotR *It follows if the curve is of such a nature that AC+BC is constant then for every point on it A C and BC are equally inclined. MAXIMUM AND MINIMUM. 37 But if Q denote the foot of the perpendicular on the base we have BQcotA=AQcot B, hence AP = BQ, or the minimum chord is such that the given point P and the foot of the perpendicular are equidistant from the extremities of the base. This is known as Philo s Line. 5. Through a given point in the diameter produced of a semi circle to draw a secant OBC such that the quadrilateral A BCD may be a maximum. [Take two consecutive positions of the secant O.CC and OB C such that ABCD = AB C D, and join AB, AB , DC, DC , and B C. Now since ABCD^AB C D it follows that BB CC =ABB + DCC or BB C+ CB C = BB A + CC D. 38 MAXIMUM AND MINIMUM. Transposing we have BB C- BB A = CC D - CB C , or since twice the area of a triangle is the product of two sides x the sine of the included angle ; in the limit this relation becomes _ CC (CD* diameter diameter but from similar triangles BB /CC = OB/OC. Hence if AB = a, BC=b, CD = c, AD = d, and the angles subtended at the centre of the circle by the sides a, b, c be denoted by 2a, 2 ft, 2y, this relation may , ... 6 2 -a 2 OC be written -5-^ = ^ which is easily reducible to cos 2a + cos 2y = l, or the projection XY of the intercept is equal to the radius of the circle. The construction of the chord BCwill be afterwards given.] 6. Having given two opposite sides A B and CD of a quadrilateral and the diagonals CA and BD, to construct it so that the area may be a maximum. [Let AB be fixed and draw C and D consecutive positions of C and I). Let be the intersection of AC and BD. Then since CC is small compared with OC and OCC a right angle ; OCC may be considered an isosceles triangle, and OC=OC . Similarly OD = OD ; and since CD=C D the triangles COD and C OD are equal in every respect. From the equal areas A BCD and ABC D take the equals COD and C OD and the common part A OB, and there remains BOC+AOD=BOC + AOD , MAXIMUM AND MINIMUM. 39 or BOG -BOC=AOD-A OD , hence BO. 00= AO. DO, from which it is manifest that CD and AB are parallel. Cf. Art. 9, Ex. 5. A similar proof may be applied to show that when the four sides of a quadrilateral are given the area is a maximum when CO.AO=BO.DO, i.e., when the figure is cyclic. See Milne s Companion to the Weekly Problem Papers, 1888, p. 27.] 7. To draw a parallel to a given line meeting a semicircle in C and D such that A BCD is a quadrilateral of maximum area. [As before, when A BCD is a maximum it is equal to the consecutive area ABC D . Hence CC DD = A CC + BDD , therefore CC D - CC A = DD B - DD C , which in the limit reduces to &2_ a 2 =c 2_ & 2 or 2 & 2 = a 2 + c 2 (1) Again if X and F are the projections of C and D on the diameter d of AB we have AX=a 2 jd, BV=c 2 /d and X T=b cos a. Making these substitutions in (1) we have on reducing 26 2 + dcos a.6-d 2 = (2) NOTE. If a = the quadrilateral is found to be one half of the inscribed hexagon. If a =90 the maximum quadrilateral is the inscribed square. 40 MAXIMUM AND MINIMUM. SECTION III. THE POINT O THEOREM. 19. Theorem.- If points P, Q, and R be taken on the sides of a triangle the circles AQR, BRP, and CPQ pass through a common point 0. For let the circles AQR and BRP meet in 0. Then since (Euc. III. 22) QOR = - 7r -A and ROP = 7 r-B > we have QOP= 27r-(7r-A}-(7r-B)=A+B = Tr-C , there fore the quadrilateral POQC is cyclic. The angles BOG, CO A, A OB, subtended by the sides of the given triangle at 0, are respectively A+P, B+Q, C+R, when is ivithin the triangle ABC. For, applying Euc. I. 32 to the triangles BOG and CO A, it follows that L AOB=C+CAO + CBO. MAXIMUM AND MINIMUM. 41 But CA = QRO since A QRO is cyclic, also CBO = PRO since BPRO is cyclic ; therefore AOB = C+R (a) where R denotes an angle of the triangle PQR. Similarly for the angles BOG and COA. If falls outside the triangle ABC these angular rela tions become somewhat modified. Take for example within the angle C. Then from the cyclic quadrilaterals QRAO and RPBO we have (Euc. III. 20) adding these equations R = OAQ+OBP = or AOB = R-C. Again, since Euc. I. 32, A+ACO = BOC+ABO, by transposing A-BOC=ABO-ACO (1) But ABO = RPO since PRBO is cyclic, and AGO = QPO since PQCO is cyclic. Substituting these values in (1) we have A - BOC= RPO - QPO = P ; therefore OC=A-P. Similarly COA = B-Q (8) It may be shown in the same manner that if the points P, Q, R are such that two of the angles P, Q of the triangle formed by them are greater than A and B respectively BOC=P-A, COA = Q-B, (y) and AOB = C-R. Hence if a triangle PQR of given species be inscribed in a given one ABC, the circles AQR, BRP, and CPQ 42 MAXIMUM AND MINIMUM. pass through either of two fixed points, one of which subtends at the sides of ABC, angles A+P, B+Q, C+R, and the other A-P, B-Q, R-C, or P-A, Q-B, C-R, according as two of the angles of the given triangle are greater or less than the corresponding angles of the inscribed triangle. 20. Let PQR be a triangle of given species inscribed in ABC. We have seen that the point is fixed, and therefore the lines AO, BO divide the angles of ABC into known segments. But the segments of A are equal to the base angles of the triangle QOR ; similarly of B to the base angles of ROP, and of C to the base angles of POQ. Hence each of the triangles POQ, QOR, ROP are given in species. Therefore as the inscribed triangle PQR varies in position OQR, ORP, OPQ remain constant in species, and OP : OQ : OR are constant ratios. Again, since the triangle OPQ is fixed in species and one vertex a fixed point ; if P describes a line BC it follows that the locus of Q is also a line (OA). And generally, when one vertex of a figure of given species is fixed and any other vertex P or point invariably con nected with it describe a locus, the remaining points Q ... describe loci, u hich may be derived from P by revolving it through a knoivn angle POQ and increasing or diminishing OP in the ratio of OQ : OP. The loci thus described are similar, the ratio OP : OQ is termed their Ratio of Similitude and the point the Centre of Similitude. Thus since is a point invariably connected with a variable inscribed triangle PQR of given species, the MAXIMUM AND MINIMUM. 43 ortho-centre, circum-centre, ex-centres, median point, etc., and all other points invariably connected with the triangle, describe right lines which can at once be con structed by the above method. Moreover, we know that if is fixed and P describes a circle, and the variable line or Radius Vector OP be divided in Q, in a given ratio, the locus of Q is a circle. Now if Q be turned around through any given angle the locus is the same circle displaced through the same angle. Therefore if one vertex of a triangle of given species is fixed, and another vertex describe a circle, the remaining vertex and all other points invariably con nected with it likewise, describe circles. EXAMPLES. 1. Having given the diagonals and angles of a quadrilateral ABC2J, construct it. [On one diagonal A C describe segments of circles containing angles respectively equal to B and D. Let ABCD be the required quadri lateral Produce CD to Y and EC to X. Join B Y and A Y. Then since the chord BYoi a given circle subtends a given angle C it is of known length. The triangle ADYis also given in species ; hence the following construction : On BY describe a segment of a circle containing an angle C. The triangle ADY, of given species, 44 MAXIMUM AND MINIMUM. has one vertex T fixed, another A describing the circle AYC, therefore the remaining vertex D describes a circle. Take B as centre and BD as radius, and cut this locus in the point D ; therefore, etc."*] 2. Eequired to place a parallelogram of given sides with its vertices on four concurrent lines (M Vicker). [Let A BCD be the parallelogram situated on the pencil . ABCD. Through C and D draw parallels CP and DP to BO and AO respec tively. Join OP. By Ex. 1 the diagonals and angles of the quadrilateral CDPO are given ; therefore, etc.] 21. When the triangle PQR is given in every respect, the triangles OPQ, OQR, ORP are completely determined ; for in addition to their species we are given the sides PQ> QR, and RP, hence the sides OP, OQ, OR are easily determined. We have therefore four solutions, real or imaginary, to the problem : Having given two triangles ABC and PQR to place either ivith its vertices on the corresponding sides of the other ; for having determined the point 0, the position of which depends altogether on the species of the triangles, we get the position of the vertex P by taking as centre and OP as radius and describing a circle cutting BC. 22. When the line OP is perpendicular to BC, OQ and OR are therefore perpendiculars to CA and AB respec tively, and the circle with as centre and OP as radius touches BC. In this case the two solutions coincide, and PQR is the minimum triangle of given species that can be inscribed in ABC. 23. It is manifest that a given triangle ABC may be escribed to another PQR. For having determined the point 0, the triangles BOO, CO A, and AOB are given in species, and are therefore completely determined, since * For other solutions see " Mathematics from the Educational Times," Vol. XLIV., p. 29, by D. Biddle and Rev. T. C. Simmons. MAXIMUM AND MINIMUM. 45 EC, CA, and AB are given lines. Hence any vertex (C) is found by describing a segment of a circle upon PQ containing an angle equal to C, and with as centre and OC as radius describing circle. Where these circles meet is the required position of C. Again in the triangle BOG when BC is a maximum OC is a maximum, and is therefore a diameter of the circle OPQO. Then OPC is a right angle. Hence the maximum triangle of given species escribed to a given one is that whose sides are perpendicular to OP, OQ, OB. COR. If the sides of the given escribed triangle be X, I*, and v, and a, /5, y the distances of from P, Q, It, \a + jui/3 + vy = a, minimum. Hence required to find a point, given multiples of whose distances from three fixed points is a minimum when any two of the multiples are together greater than the third. EXAMPLES. 1. If d denote the distance of the point from the circumcentre H of the triangle A BC ; prove that twice the area of the minimum triangle PQR is ( 2 - d 2 ) sin A sin E sin C. [For join AO and produce it to meet the circum-circle again in C ; join BC . 4ft MAXIMUM AND MINIMUM. Now since LR = A OB - C= A OB - C = OBC (Euc. 1.32), we have 2PQR=RP.RQ sin R=RP.RQsin OBC (1) but RP = OB sin B and RQ = OA sin A. Substituting these values in (1) and putting OB sin OBC = OC sin C", 2PQR=AO. BO sin A sin B sin OBC = AO . OC sin A sin B sin (7 = (/e 2 - d-) sin ^ sin B sin (7.] NOTE. If the point is on the circum-circle R = d and the area of the triangle vanishes, hence if from any point on the circum-circle of a triangle perpendiculars be let fall upon the sides their feet lie in a line. This is termed a Simson Line of the triangle, and the col- linearity of the points admits of an easy direct proof. 2. If the pedal triangle PQR of a point is constant in area the locus of the point is a circle. [Concentric with the circum-circle by the equation of Ex. 1.] 2a. The theorem holds generally for a polygon. 3. Having given of a triangle the base c, and ab sin (C-a) where a is a given angle, find the locus of the vertex. [In Ex. 1 we have 2PQR = AO . BO sin A sin B sin (AOB- C) co A O.BO sin (AOB-C), and the locus of is in that case a circle. Hence in the triangle A OB we have the data in question ; therefore the locus of the vertex is a circle concentric with //.] 4. To inscribe a quadrilateral of given species PQRS in a given quadrilateral A BCD. Find the point 0^ of the triangle PQR of given species inscribed in a given one, viz., that formed by three of the sides, AB, BC, CD of MAXIMUM AND MINIMUM. 47 the quadrilateral. Similarly find 2 of the triangle PQS inscribed in a given one. Now by Art. 19, since the species of each of the triangles O^PQ and 2 PQ is given, we have z.0 1 / > 2 = Z PQ ~ 1 PQ= a known quantity ; therefore the point P is determined. 5. To escribe a quadrilateral ABCD of given species to a given one PQRS. [Take any quadrilateral abed of the same species as ABCD. Inscribe in it by Ex. 4 a quadrilateral pqrs of the species PQRS. It is obvious that L.SPA =spa, since the figures are similar, hence the problem reduces to drawing lines in known directions through P, Q, 21, S. Otherwise thus : Upon a pair of opposite sides PQ and RS describe segments of circles containing angles equal to B and D respectively. Find a point M such that the arcs PM and QM subtend angles equal to A BD and CBD respectively. Similarly find N such that CDN and ADN may be equal to the known segments of the angle C. Join MN ; where it meets the circles in B and D are two of the required vertices of the quadrilateral ABCD.] 6. To escribe a square ABCD to a quadrilateral PQRS. [By Ex. 5 or simply thus : Join PR and let fall a perpendicular from Q upon it. Make QS = PR. SS is a side of the required square. This construction depends upon the property that any two rectangu lar lines terminated by the opposite sides of a square are equal to one another (Mathesis). 48 MAXIMUM AND MINIMUM. 7. From any point P on the base of a triangle perpendiculars PX and PY are drawn to the sides, find the locus of the middle point N of XY. [Bisect CP in M, join MX, JfFand MN. It is easy to see that MXY is an isosceles triangle of given species, each of its base angles being the complement of C ; and since its vertices Z, M t Y move on fixed lines, any point ^invariably connected with it describes a line. By taking P to coincide alternately with A and B the locus is seen to be the line joining the middle points of the perpendiculars from the extremities of the base of the triangle A BCJ\ 8. The sides of the pedal triangle PQR are in the ratios a.AO-.b.JBO : c.CO. [For QR = AOsmAxa.AO, etc.] 9. Extension of Ptolemy s Theorem. If the three pairs of opposite connectors of four points be denoted by a, c ; &, d ; 6, 8 to prove the relation &8 z =a*c z + b*d*-2abcd cos (0+00, where 6+& is the sum of a pair of opposite angles of the quadrilateral. [Let A, B, C, be the four points. From any one of them let fall perpendiculars OP, OQ, OR on the sides of the triangle ABC formed by the remaining three ; then since PQ2 = QR2 + RP* - 2QR . RP cos R, substituting for PQ, QR, RP the values in Ex. 8, and reducing, the above equation follows at once (M Cay).] 9a. What does this theorem reduce to for the quadrilateral ABCP in the figure of Ex. 7 ? Deduce the relation of Art. 3, Ex. 5, as a further particular case. MAXIMUM AND MINIMUM. 49 10. A variable circle passes through the vertex of an angle and a second fixed point ; find the locus of the intersection of tangents at the extremities of its chord of intersection. 11. If a, (3) y denote the distances of any point from the sides of a triangle ; to prove that +r& where S and S are the rectangles under the segments of a variable chord through 0* of the circum-circles of ABC and of the pedal triangle of the point (M Vicker). [In Ex. 1 let K be the point where RO meets the circum-circle of PQR; iheny=S IOK=S smP/psmOQK. BntsinOQK=sin(A + P)=sinBOC; .-. {3y = S sinP/smBOC. Also a = OB.OCsmBOC/a, therefore aj3y = S . OB. OCsmP/a. Again OB = RP/smB, etc. ... therefore by substitution s RP asinJ5sin<7 A/ 2/2 2^ 12. In the particular cases when coincides with the in- or ex- centres of the triangle ABO, the formula in Ex. 11 reduces to & = R*-ZRr or S^tf + ZHr etc. 24. Theorem. When three points P, Q, R are taken collinearly on the sides of a triangle, the circles circum scribing the four triangles QRA, RPB, PQO, ABC meet in a point. This theorem may be easily proved directly, but it is obviously a particular case of Art. 19, for the circles QRA, RPB, PQG meet in a point (Art. 19) such that CO A = Q B, which in this case is 180 J5; therefore, etc. Euc. III. 22. The transversal PQR to the sides of A BC is the limiting case of a triangle inscribed in ABC, the angles at P and * The constant rectangle under the segments of a variable chord of a circle passing through a fixed point has been termed by Steiner the Power of the Point with respect to the circle. D 50 MAXIMUM AND MINIMUM. R being each and Q = 180. The species of the limiting triangle is determined by the ratios QR : RP : PQ, or their equivalents a . A : b . BO : c . CO. (Art. 23, Ex. 8.) Hence if a transversal is drawn to a triangle such that the ratios of its segments made by the sides is constant ; the ratios AO : BO : GO are known and with them the point 0. As in the general case, the triangles QOR, ROP, POQ are constant in species. It follows then that if P, Q, R be the feet of the perpen diculars from on the sides cf ABC, and the lines OP, OQ, OR rotated through any angle in the same direction, P, Q, R will ahvays remain collinear and the ratios PQ : QR : RP are constant* COR. Ptolemy s Theorem. Since QR:RP:PQ = a. AO: therefore a.AO + c.CO = b. BO. EXAMPLES. 1. Place a given line PQ divided in any point R such that the points P, Q, It may lie in an assigned order on the sides of a given triangle. *Chasles Geom^trie supdrieure, p. 281. MAXIMUM AND MINIMUM. 51 2. Draw a line across a quadrilateral, meeting the sides in PQRS such that the ratios PQ \QR\ RS may be given. 3. The line joining to the orthocentre of ABC is bisected by the Simson line PQR, and intersects it on the nine points circle. 4. The angle subtended by any two points Oi and 2 on the circle is equal to the angle between their Simson lines. 5. The Simson lines of two points diametrically opposite intersect at right angles on the nine points circle. (By Ex. 4.) 25. Theorem. For three positions, PQR, P^R^ P 2 Q 2 R 2 , of the triangle of given species inscribed in a given one ABC ; to prove that PP l : PP 2 = QQ l : QQ 2 = RR, : RR 2 . Since the triangles OPQ, OP^, OP 2 Q 2 are similar, we have OP : OP X = OQ : OQ V also LPOP l = QOQ l since L-POQ^P^OQ^ therefore the triangles POP 1 and QOQ l are similar. Hence Similarly (^ : R^ = OQ:OR; therefore PP l : QQ l : RR l = OP:OQ: OR. Similarly PP 2 : QQ 2 : RR 2 = OP:OQ:OR- > therefore, etc. Now if PiQ^! and P 2 Q 2 R 2 denote two fixed positions of the variable inscribed triangle PQR of constant species, 52 MAXIMUM AND MINIMUM. and PQR any arbitrary position, it follows that a variable line PQ, dividing similarly two linear segments P^ and P 2 Q 2 , subtends a constant angle POQ at a fixed point 0. The point is determined by the intersection of the loci of the vertices of the triangles P-^Qfl and P 2 Q 2 0, whose bases PjQj and P 2 Q 2 are given and ratio of sides ( = P : P 2 : QQ-t), or the intersection of the circles CP^ and <7P 2 Q 2 . Since P-f z and Q^ 2 form similar triangles with 0, this point is termed the Centre of Similitude of the segments. Thus the centre of similitude of two segments AB and CD is the intersection of the circles passing through the two pairs of non-corresponding extremities and the inter section of the given lines. Or it may be regarded as the common vertex of two similar triangles described on the sides. If the points B and D coincide, coincides with them, and the circle ADO meeting CD in coincident points D and therefore touches CD. In the same case the circle BCO touches AB. Con. The centres of similitude of the sides of a triangle taken in pairs are therefore found by describing circles MAXIMUM AND MINIMUM. 53 on BC and AC touching the sides AC and BO respectively. The second point of intersection of these circles is a centre of similitude of AC and BC } similarly for each of the remaining pairs of sides. EXAMPLES. 1. Draw a line L dividing three linear segments A 1 A 2 ^ B-^B.^, and CC 2 in the same ratio. (Dublin Univ. Exam. Papers.} [Let the required line intersect the segments in P, Q and R respectively, O l and 2 the centres of similitude of the pairs of lines A ^2, B^BZ and B^ C&. Then in the triangle 0^0 2 we know the base 1 (9 2 and vertical angle, since it is equal to ISQ-O.^QP- O^QR ; therefore, etc.] 2. The centres of similitude of the sides of a triangle taken in pairs are the middle points of the symmedian chords of the circum- circle. [Let X, Y, Z denote the middle points of the sides of the triangle ABC ; CD and CE the median and symmedian chords of the circle respectively ; J/the middle point of CE. Join ZE, AM and BM. Then since L.ACD=BCE and L.CAZ=CEB, the triangles A CZ and ECB are similar, and Y and M being the middle points of a pair of corresponding sides, CYZ and CMB are therefore similar. Hence L.CBM = CZr=BCZ=ACM. Similarly L.CA M= BCM ; there fore the triangles BCM and CAM are similar.] 54 MAXIMUM AND MINIMUM. 3. Prove the following results from Ex. 2 : 1. CEZ= difference of base angles (B-A). 2. The triangles A DZ and BEZ equal in every respect. 3. CZ. CE=ab. 4. CM = ab/ \/a 2 + 6 2 + 2ab cos C. 5. BMC=CMA = TT-C. 6. The circum-circle of ABM passes through the centre of the circle ABC. 4. Having given the base (c) bisector of base (CZ) and difference of base angles (B A) ; construct the triangle. [The triangle CEZ is readily constructed ; therefore, etc.] 5. Having given the bisector of base (CZ) rectangle under sides (ab) and difference of base angles (B-A) ; construct the triangle. [As in Ex. 4.] 6. Having given the base, median, and symmedian of a triangle ; construct it. SECTION IV. MISCELLANEOUS PROPOSITIONS. 26. Prop. I. Through a point P to draw a line across an angle such that the intercepted segment MN may subtend at a fixed point Q a triangle of maximum area. MAXIMUM AND MINIMUM. 55 The transversal PMN such that the parallels OH and ON to the sides of the angle intersect on PQ is the required line. For draw any other line PM N . Join M N. Then the triangles MON and M ON are equal (Euc. I. 37), but M ON > M ON ; therefore MON > M ON . MON MQN , MON W tudes = PO/PQ. Similarly = PO/PQ; therefore MQN > M QN . To find the point 0. Evidently by similar triangles PA/PO = PM/PN = PO/PB ; therefore PA . PB = PO 2 . Prop. II. On the sides BO and GA of a triangle, to find points M and N suck that if the lines AM and BN meet in the triangle MON may be a maximum. Regarding A as a point on the base produced of BCN and AOM a transversal to the sides, MON is maximum when ON and MN parallels to these sides respectively meet on AC. Similarly since B is on the base produced of ACM and BON a transversal to the sides, OM and NM parallels to the sides meet on the base. 56 MAXIMUM AND MINIMUM. Then we have ANM O and CN OM equal parallelo grams (Euc. I. 36), therefore AN=CN t also BM=CM . But by Prop. I. AN . AC=AN \ therefore AN.AC = CN* ; similarly BM . BC = CM 2 , or the sides of the triangle ABC are divided in extreme and mean ratio, the greater segments being measured from the vertex. Prop. III. Through one extremity A of the diameter APB of a semicircle draw a chord AMN to meet a perpendicular through P to the diameter AB in M and the circle in N, such that the triangle MEN may be a maximum. Suppose a tangent is drawn at the required point N. Let it meet PM in S. Join AS. From the centre C let fall CX perpendicular on AS. Join CN. By Prop. I. the parallels MQ to the tangent and NQ to PS meet on AB, for then with respect to the angle PSN the triangle MEN is maximum ; therefore a fortiori it is the maximum triangle whose vertex N lies on the circle ANB. MAXIMUM AND MINIMUM. 57 = ABN=ANQ. Hence since MN, the diagonal of a parallelogram MSNQ, bisects the angle N, the figure is a rhombus, and NQ = NS. Then the triangles ANS and ANQ are equal in every respect (Euc. I. 4), therefore ASN is a right angle ; hence CNSX is a rectangle, and SX is equal to the radius of the circle. Also CPSX is a cyclic quadrilateral, therefore AS.AX = AC.AP which is known. Therefore we have the rectangle and difference of A 8 and AX, from which data these lines are at once determined. Then we can construct the right- angled triangle ACX, which fixes the point X ; there fore, etc. COR. In the particular case when PM8 is a vertical radius, if 8N meet the tangent AT in T and AB in T t we have AS. AX = r 2 , therefore by parallels AT. AC=CT 2 . 58 MAXIMUM AND MINIMUM. Similarly TT . TS=T S\ but TT . TS = AT 2 = TN 2 ; therefore TN=T S, and TS=T N. But when a line TT is divided in extreme and mean ratio in 8 and from the greater segment a part T N is taken equal to the less TS, T S is divided into extreme and mean ratio. Ex. Draw the transversal AMN such that the quadrilateral MNBP may be a maximum. Prop. IV.* Through a given point in the tangent at C to a circle draw a secant AB such that the triangle ABC may be of maximum area. Draw tangents at A and B to meet in T. The required triangle is such that the parallels through A and B to the tangents at these points meet on OC in P. For since and T, and C, are pairs of conjugate points with respect to the circle, CT is the polar of 0. * This Proposition may be omitted on the first reading. MAXIMUM AND MINIMUM. 59 Let 0(7 , the second tangent from 0, meet PT in (7, Since PET A is a rhombus, AB is at right angles to PT ; also since TC MP is a harmonic row, we have TC f /C f M=TP/PM=2 , therefore Tif or PM = 3M7. Then a given angle (70(7 is divided by the required line AB, such that the ratio of the tangents of its segments is known ; therefore, etc. Ex. If a, 6, c denote the sides of the maximum triangle ABC, prove that /,\ OA c 2 a 2 < 2 > - CHAPTER III. EECENT DEVELOPMENTS OF POINT O THEOEEM. SECTION I. THE BROCARD POINTS AND CIRCLE OF A TRIANGLE. 27. Brocard Points Q, 2 . In Art. 20 if the inscribed triangle PQR is similar to ABC and P = A, Q = B, R = C, then BOG = A+P = 2A, similarly CO A = 25 and 4 05 = 2 C; therefore is the centre of the circum-circle. Secondly, let P = B, Q = C and R = A. Then similarly and Thirdly, let P=C, Q = A and R = R It follows as in the last case that BOC=7r-B, COA = 7r-C and Thus we see that a triangle PQR similar to a given one may be inscribed in the latter in three different ways; and that the point in each case may be found as in the general method by describing segments of circles on two of the sides containing given angles. In the second and third positions the points of inter section of the circles are usually denoted by the letters Q arid Q . They are termed the Brocard Points of the 60 RECENT GEOMETRY. 61 triangle ABC, and are distinguished as Positive (fi) and Negative (Q ). 28. Brocard Angle (o>). Since BIC is the supplement of G y ttBC+ttCB=C or QBC=QCA. For a similar reason QOA = ttAB, hence 18(7 = Q(M = &AB = o> (say). The angle eo is called the Brocard Angle of the triangle ABC. We may remark that the angle subtended at 2 by the base c is the supplement of B, the angle at the right extremity of AB, and at 1 equal to the supplement of A, the angle at the other extremity of AB. The same relations hold for the sides a and b ; hence the names Positive and Negative Brocard points. The value of w as a function of the sides or angles is thus found. Let x, y, z denote the lengths of ^Q, Q and (X2 respectively. Then in the triangle BtiC sin a> 2ay sin CD Similarly in the triangles CQA and AQB 62 RECENT GEOMETRY. rot ^ 4OU It is proved in like manner for fi that and that the value of these angles is also given by (1). Again cot A = , ."""^ 26csmJ. for cot j5 and cot G. Hence Again cot A = , ."""^ = + c ~" a with similar values 4A ) 4A 4A or cot co = cot A + cot .B + cot G ......... (2) EXAMPLES. 1. Prove that (1) cosec 2 w (2) rin^- 2. The distances of 12 from the sides of ABC are D 2#sm 2 a>-,2flsin 2 (u-; and of 12 , 2#sm 2 o>^, 2/2sin 2 w c , 2R sin 2 o>?. c a c a b [For let the distances of 12 be denoted a, /?, y. Then a=y smu = therefore, etc. sin z> The ratios of the distances* are evidently as follows : a : /3 . y = c 2 a : a 2 6 : 6 2 c, and a :/3 :/ = a& 2 : tc 2 : ca 2 , and also <xa = ^ = yy = 4/^ 2 sin 4 w.] * Or Trilinear Co-ordinates of the points with respect to the triangle, which is also called the Triangle of Reference. RECENT GEOMETRY. 63 3. AD is the bisector of the angle A of a triangle ABC, and o^, w 2 the Brocard angles of the triangles AED and ACD respectively ; prove that cot ta x 4- cot o> 2 = 2 cosec .4 + cot ^4 -f cot to, with similar expressions for the triangles formed by the bisectors of the angles B and C. 4. If ci^ and <o 2 denote the Brocard angles of the triangles CAD and BAD, where AD is the median to the side BC, b 2 ~ c 2 COt 0^ - COt 0> 2 = -^ , with similar expressions for the medians BE and CF. 5. Hence prove that cot o> 1 + cot co 3 + cot o> 5 = cot a> 2 + cot to 4 + cot w 6 , and 2 cot ^W+g + V. 6. If -4jB(7 is divided as in the previous exercises by the symmedians, prove that 2(6 2 + c 2 ) (coto^ cot wo) = 0. 7. 12 and 12 are Brocard points of their pedal triangles PQR and F Q R . (Euc. III. 21.) 8. The triangles PQR and P Q R are equal in area. [For SIPQ and &BC are similar ; hence (Euc. VI. 19) 12P# : a5<7= OP 2 : 12.B 2 = sin 2 w ; similarly !2<?/j! : &CA = $IRP : 12J=sin 2 (o ; therefore PQR = P qR =ABC . sinV] 9. The Brocard points are equidistant from the circum-centre. [By Ex. 8 and Art 23, Ex. 1.] 10. If A , B , C be the points of intersection of the pairs of lines y, z : z, x : x, y , prove that the six points A , B , C , 0, 12, 12 lie on a circle. [For the triangles BOA , CAB and ABC are isosceles and similar, their base angles each being equal to to, hence OA , OB , OC are the bisectors of their vertical angles. In the quadrilateral 01212 A we have 012 = 012 and OA the bisector of the angle 12^1 12 ; therefore is a point on the circum-circle of !2^4 12 , and the quadrilateral is therefore cyclic. Similarly B and C are on the circum-circle of the triangle 01212 .] DEF. This is called the Brocard Circle, and A B C the First Brocard Triangle of ABC. 64 RECENT GEOMETRY. 11. To find the distance of the Brocard points from the circum- centre(012 = 012 = o-). [By Art. 23, Ex. I, but (Ex. 8) hence R* - 8 2 = 4 tf 2 sin 2 a> or 12. The angle subtended at the circum-centre by 12 12=2o>. (By Ex. 10 and Euc. III. 22.) 13. To find the distance 1212 between the Brocard points. [Since 01212 is an isosceles triangle, 1212 = 2012 sin <o = 2R sin w \/l-4shrw, by Ex. 1 1 .] 14. The diameter of the Brocard circle is equal to R sec w v 1 4 sin 2 a>. [For it equals 8 1 sin 2o> ; therefore, etc.] 15. The altitudes of the similar isosceles triangles BCA , CAB , ABC are equal to the distances of the symmedian point (K) from the sides. [For C Z*= \c tan to = -J^ c ^^ ; therefore, etc., by Art. 28, (1).] 16. The circle on OK as diameter is the Brocard circle. [For KA is parallel and OA perpendicular to BC, hence OA subtends a right angle at A; similarly for the points B and C ; therefore, etc.] 17. Brocard s first triangle is Inversely Similar to ABC; i.e., by rotation in the plane of the paper their sides cannot be brought into a position of parallelism with each other. [For B C subtends equal angles at A and A", but KB and KG are respectively parallel to CA and AB, and therefore contain an angle A ; similarly the angles B and C are equal to B and C.] 18. Having given the base c and Brocard angle o> of a triangle ABC, find the locus of the vertex (Neuberg). [Let p be the median CZ and the angle between it and PZ. Since cotw = (a 2 + 6 2 + c 2 )/2c. CR and a 2 + 6 2 = %c 2 + 2p 2 , we have RECENT GEOMETRY. 65 72 = 2c cot W. p cos#, or p 2 -c cot to. /o NOTE. Comparing this result with the standard form of the equation in the footnote we have by equating coefficients c cot w = 2rf and d 2 - r 2 = f c 2 , or d = \c cot w and r 2 = |c 2 cot 2 w - |c 2 . It is evident that the locus is a curve symmetrical with respect to the perpendicular bisector of the base, as to each position of the vertex C there is a corresponding one, C" of the inversely similar triangle ABC" described on the base. The distance of C , a vertex of Brocard s first triangle, from c = %c tan w ; therefore ZC . Z0=(%c) 2 where is the centre of the required locus. This example is a particular case of : Having given the base c and (Ia 2 + mb 2 +nc 2 )l&. to find the locus of the vertex; a solution of which is similarly obtained. ISa. Six similar triangles are constructed on a given base and on the same side of it. Prove that their vertices C^ C 2 , ... C 6 are con- cyclic. (Mathesis, t. 2, p. 94.) * This is known by Analytical Geometry to be the Polar Equation of a Circle. If we take any point Z and draw a variable line (Radius Vector) to a given circle (0, r) and let d = ZO, the equation connecting p and 6 is for all points on the circle p 2 - 2pd cos d + d 2 - r 2 = ; and p and are called the Polar Co-ordinates of the point P. E 66 RECENT GEOMETRY. 19. Having given the base c, and Brocard Angle to, find the locus of the centroid of ABC. [A circle whose equation is formed from that in Ex. 18 by changing p into 3/> ; hence 12p 2 - 4c cot w . p cos + <? = 0. It has many important properties, which will be found in the Transactions of the Royal Irish Academy, vol. xxvni. xx, where M Cay names it the " (7" circle of the triangle ABC.} 20. The lengths of the tangents drawn from A, B, C to the Brocard Circle are inversely proportional to a, b, c, and the sum of their squares = 2A cosec 2w. SECTION II. THE SYMMEDIANS OF A TKIANGLE. 29. Let K be the symmedian point of ABC, a and /3 the distances of Z from BG and CA respectively. Then a If? = alb = BZ sinB/AZ sin A, hence [11 BZ a 2 " or the symmedians divide each side in the duplicate ratio of the remaining two. RECENT GEOMETRY. 67 Again from (1) A&/C- 6 2 /(ct 2 +6 2 ) or AZ = 6 2 c/(a 2 +6 2 ) ; similarly BZ = a 2 c/(a 2 + 6 2 ) (2) Also CZ jCK = a! I a = (a 2 + 6 2 + c 2 )/(a 2 + 6 2 ), hence COR. If 0- 90 then Off = J22" (Euc. I. 47) and K is the middle point of the perpendicular on the hypotenuse. 30. The length of the symmedian CZ is found as follows : In the formula b*BZ + a*AZ = cAZ . BZ +cCZ 2 sub stitute the values in (2) and reduce. We easily obtain V^+& 2 +2a6cos(7 a/b + b/a with similar expressions for the lines through A and B. EXAMPLES. 1. The symmedian is divided harmonically at A", and Q its point of intersection with the perpendicular to the base of the triangle at its middle point Z. therefore CQ/QZ - CK\KZ = (a 2 + 6 2 )/c 2 . 1 2. Since Z. CKZ Q is an harmonic pencil any line through K is cut harmonically by its rays, hence if KG is parallel to one ray, it is bisected at by the conjugate ray CZ. Also the parallel through K to P^L is bisected at K. 3. The vertices of Brocard s first triangle and the symmedian point are equidistant from the extremities of the parallels through K to the sides of ABC. [Let be the middle point of MN. Since OJ/=O^Vand (Ex 2) OKOC y subtracting these results ; therefore, etc.] 68 RECENT GEOMETRY. 4. The lines joining the middle points of the sides of ABC to the middle points of the perpendiculars on them meet in a point. [By Ex. 2 the point of concurrence is the symmedian point. The ratios of the segments into which the joining lines are divided at K are easily seen to be be cos A/a?, etc., etc.] 5. Prove that cot KBC+ cot KG A + cot KAB=2 cot w. 6. The sides of the pedal triangle of K are at right angles to the medians of ABC. ANTIPARALLELS. Def. A straight line meeting the sides a and 6 of a triangle at angles A and B is parallel to the base. If a line meet these sides at angles A and B respectively it is said to be Antiparallel to c. 31. The following are the fundamental and obvious properties of antiparallels to the sides of any triangle : (1) Antiparallels to the sides a and b meet c at equal angles (C). (2) They are parallels to the sides of the pedal triangle. (3) Or to the tangents at A, B, to the circum-circle. RECENT GEOMETRY. 69 (4) The locus of the middle point of a variable anti- parallel to a side, c, is the corresponding symmedian chord CK. (5) Antiparallels through K to each side are bisected at the point, and are equal to one another. The latter part follows from (1). (6) The median and symmedian to c of the triangle ABC are respectively the symmedian and median of the triangle A B C cut off by any antiparallel A B . (7) The extremities of a parallel and antiparallel to any side of a triangle are concyclic. THE PEDAL TRIANGLES OF THE BROCARD POINTS. 32. From 2 let fall perpendiculars on the sides and denote their feet as in figure by A B C . It follows conversely since AQB is the supplement of B (Art, 28), and is equal to C+A (Art. 19) that A = A] similarly B = B and (7 = 0. Also A", B", C" are respec tively equal to A, B an r! C. 70 RECENT GEOMETRY. 33. Theorems I. 2 is the common positive Brucard point of ABO and A B C . Since AC A tl is a cyclic quadrilateral &AB = SlC A = o> (Euc. III. 21); similarly tiB C and QA B are each equal to . It follows also that Q is the common negative Brocard point of ABC and A"B"C". II. The sides of A B C and A"B?C" are equally inclined to the corresponding sides of ABC. For by (1) CB C = AC A = BA B = 90-*>, and BO B" = AB"A" = CA C" = 90 - o>. III. The six points A , B , C, A", B", G" are coney die. For the angles AC A = AB"A", therefore A A"B"C is cyclic (Euc. III. 22). Similarly B B"C"A and C C"A"B are cyclic. But generally if three pairs of points on the sides of a triangle are such that every two pairs are cyclic, the six points lie on a circle.* For if they do not the tangents to the three circles from A, B and C are easily seen to be equal, which is impossible. IV. The lines B"C , C"A , A"B are parallel to the sides a, b, c respectively. % We know that each pair of sides of ABC with Q and 1 form similar triangles, i.e., BM and Aft C, CClA and BQ, A t AQB and 00, B are similar ; hence the perpen diculars (or other corresponding lines) through Q and Q divide the opposite sides similarly. In the triangles C&A * For example, if A B G be the middle points of the sides and A"B G the feet of the perpendiculars, it follows immediately that A B C A"B"C is a cyclic hexagon since each pair of points A A and BE form a cyclic quadrilateral. ("Nine Points" Circle.) RECENT GEOMETRY. and B&A we have therefore AC /AC = AB"/AB, or is parallel to a. V. Hence also A A", B B", C C" are antiparallels to the sides a, b, c. (Euc. III. 22.) SECTION III. TUCKER S CIRCLES. 34. By Art. 24 if the inscribed triangle A B C is given in species only it may be conceived to vary its position by rotating around the point Q which is fixed. Let it revolve in a positive direction through any angle and also let A"B"C" revolve in the opposite direction through an equal angle. Then each of the equal angles of inclination of the sides of A B C and A^C" are diminished by 0, therefore for all values of the sides are equally inclined and the vertices of the two triangles are always coney clic. The circles thus described are called the Tucker Circles of the triangle. Thus the lines B"C f and A A" t etc., are always parallel and antiparallel respectively to the. opposite side a, and therefore remain constant in direction. Now since the point Q is fixed and the triangle A B C of constant species ; since the vertices move on given lines all points fixed relatively to the figure describe lines. The locus of the centre of the system of Tucker s circles is therefore a line. (Art. 20.) By taking particular positions of the triangle we find points on the line of centres. In the case where $ = the 72 RECENT GEOMETRY. vertices of ABC and ABC coincide, and the circum- circle is thus seen to be one of Tucker s circles. The line of centres thus passes through the circurn-centre of ABC. Similarly the loci of the other Brocard points of the triangle A B C and A B C" are lines. 35. Let the vertices of the triangle formed by the parallels B"C , C"A, A"B to the sides of ABC be denoted by Z, F, Z. Then AAA X is a parallelogram, as are also BB B Y, CC C"Z ; and since the diagonals bisect each other A X bisects the antiparallel AA". AX, BY, CZ are the symmedians of ABC. Hence the following construction for Tucker s circles : Let K be the symmedian point of ABC. Join AK, BK, CK. Take any point X on AK and draw parallels through it to the sides 6 and c. Let them meet BK and CK in F and Z respectively. YZ is parallel to a, and the hexad of points in which the sides of ABC are cut by these parallels lie on one of the required circles. RECENT GEOMETRY. 73 36. The antiparallels A A", B B", C C" are equal For since A"E f is parallel to c, and A A and B B" are equally inclined to c (at an angle (7), A A" = B?B ; therefore, etc. ; or they are the chords of a Tucker circle intercepted by parallel lines. 37. Theorem. The line OK is the locus of the centre of Tucker s system of circles. For let L be the middle point of the chord A A" of one of the system. Draw L0 1 at right angles to it meeting OK in O r Join AO. Since the tangent at A to the circum- circle is anti- parallel to a, AO and L0 1 are parallel lines. But AK/AX = BKIBY=CK/CZ (Euc. VI 2); there fore AK/AL = BKIBM=CK/CN=OK/00 V or O l is the centre of the Tucker circle. 38. Since Q is the positive Brocard point of the triangles ABC and ABC , and &AB and IA B a pair of similar triangles ; if be the inclination of the sides to those of ABC, we have sin a) ,. ...... This ratio is the Ratio of Similitude of the triangles, and is the constant relation between all corresponding lines of A B C and ABC. For example, if p be the radius of Tucker s circle for any value of 0, *> -^?-T ......... (2) -R sin (0+ ) In (2) we have the following particular cases: when = p = R ............... (circum-circle) ; 6 = o) p = J^Rseco) .......... (T. R. circle); = 90 p = j?tanft) ........... (cosine circle). Also area ABC : ABC=sin*to : sin 2 (0+o>) (Euc. VI. 19). 74 RECENT GEOMETRY. SECTION IV. TUCKER S CIRCLES, PARTICULAR CASES. 39. I. Cosine Circle. As a particular case of the general theorem (Art. 33 v.) we shall consider the anti- parallels A A", B B", C C" to pass through K. The points L, M, N will therefore coincide with K, which is also the centre of the corresponding Tucker s circle. It is otherwise evident that the six segments KA, KA \ etc., of antiparallels through K to the sides are equal, (Art. 31 (5)). Also B C B O", C AC" A", A B A B" are rectangles since their diagonals are equal. in because A B B" is a right-angled triangle A B" = B W cosA B"B = B B" costf, RECENT GEOMETRY. 75 or A B" = 2/o costf, with similar expressions for FG" and C A". Hence The segments intercepted by the circle on the sides of ABC are proportional to the cosines of the opposite angles* It is from this property the circle derives its name. 40. The middle point M of A"B is on the median through G to the opposite side c ; hence the perpendicular through K to this side passes through M, or as has been shown otherwise (Art. 30, Ex. 4). If a perpendicular be drawn through K to the base meeting it in N and the median in M, MK = NK, from which it follows that the lines joining the middle points of the sides to the middle points of the corresponding perpendiculars meet at the symmedian point (Hain). 41. The sides of the triangles A B G and A"B"C" are perpendicular to the corresponding sides of ABC. The cosine circle may therefore be obtained by rotating the two inscribed triangles in opposite directions until 6 = 90. (Art. 39.) The ratio of similitude of A B G and ABC = tsMa). 42. II. Triplicate Ratio Circle. Let the parallel in figure of Art. 35 pass through K. Then L, M, N are the middle points of AK, BK, and CK, since AA A K, etc., etc., are parallelograms; and the centre of the corresponding Tucker circle bisects OK. The sides of A B G are inclined to those of ABC at an angle = a>. For consider the angles in the equal segments A A", B B", C C", and it is obvious (Euc. III. 21) that A B A" = A C A" = B G B" = B A B" = G A G" = G B C". * See Mathesis, t. i., p. 185 : " Sur le centre des Me*dianes Antiparalleles," Neuberg (1881). 76 RECENT GEOMETRY. Hence K is the negative Brocard point of A Similarly it is the positive Brocard point of A"B"C". It follows generally that the locus of the negative Brocard point of A EG is a line passing through K. 43. The ratio of similitude of A B C and ABC is sin o>/sin 2o> since = oo ; hence p = %R sec w (1) 44. The intercepts B C", C A", A E" made by the circle on the sides are thus determined: The triangles A KB" and ABC are similar, therefore A B"/c = ratio of altitudes a 2 +6 2 +c 2 c with similar expressions for .ZW and (7 J.". The general property of the circle may be thus stated -.Parallels through the symmedian point meet the non-correspond ing sides in six points which lie on a circle; and the intercepts made on each side are in the ratios a 3 : b 3 : c 3 . From the latter property the circle takes its name. For the sake of brevity it is often written " T.R. " Circle.* 45. III. Taylor s Circle. Let the antiparallels A A", B B", C C", which, it will be remembered, are always parallel to the sides of the pedal triangle (PQE) of ABC, pass through the middle points a, /3, y of the sides of PQR Consider the segments into which A A" is divided by ft and y. We have {Sy = %QR, yA" = JPQ (Euc. I. 5), and * An account of the circle will be found in Mathesis in the article by Neuberg already referred to (Art. 39). See also Nouvelles Anncdes, 1873, p. 264. RECENT GEOMETRY. 77 for the same reason /3A = %RP; therefore A A" is equal to the semiperimeter of PQR = J(a cos A -f b cosB+c cos (7) = 2JK sin J. sin B sin (7. Hence generally ^ / ^ / = / J B // = tf / <7 // = 2Esm4 sin^sina...... (1) Again, since B"aC f is an isosceles triangle, the perpen dicular to the chord B"C of Tucker s circle at the middle point bisects the vertical angle a and passes through the in-centre of a/3y. Similarly for the chords C"A and A ff. Hence The centre of the circle coincides with the in-centre of the median triangle (a/3y) of PQR. Many properties of this circle are proved in Neuberg s article in Mathesis, t. 1, p. 185, but it was described independently in England by Mr. H. M. Taylor, and now bears his name. (Proc. Lond. Math. Society, vol. xv. p. 122.) 46. Since aQ = aR = aB" = aC , the circle on QR as diameter passes through B" and (7 and RB"Q = RCQ = 90; or B" and C are the projections of Q and R on the sides AB and AC-, hence 78 RECENT GEOMETRY. The six projections of the vertices of the pedal triangle on the sides of ABC lie on Taylor s circle. 47. The triangle B aC" is isosceles, therefore C^a the bisector of its vertical angle a is at right angles to BC; hence generally The lines C^a, C^/3, 0y are perpendiculars to the sides of ABO. Let H z denote the orthocentre of CPQ ; then QH Z and OjGt are parallel ; similarly PH 3 and Oft are parallel ; hence the triangles PQH Z and a^0 1 are similar, their ratio of similitude being = J, or H Z R is bisected at O r Similarly PH^ and QH 2 are each bisected at O l ; and therefore the triangles H^H.fl^ and PQR are equal in all respects. 48. Theorem. Taylor s circle of the triangle ABC is the common orthogonal circle of the ex-circles of PQR. In the triangle A A A" we have by rule of sines AA" = A A" siiiC/sin-d. = 2J? sin sin 2 (7 (Art. 45 (1)), also AC = ARcosA = 6cos 2 ^l ; multiplying these results and reducing AA .AV = 4R 2 sin 2 ^ sm 2 a cosl, but AQ = ccosA; substituting we obtain A A" ,AC = AQ*sm*B* or the square of the perpendicular from A on QR. Hence the tangent from A an ex-centre of PQR to Taylor s circle * Otherwise from the right-angled triangle AA"P and ACP we have AA* = b*WL*C ; and from the triangles ACR and AC R y ; therefore A A " . A C - 6 2 sin 2 *? cos 2 .4 . RECENT GEOMETRY. 79 is equal to the radius of the ex-circle ; similarly for the ex-centres B and C; therefore, etc.* EXAMPLES. 1. To find the value of the radius p of a circle cutting the ex- circles of a triangle PQR orthogonally. [In figure of Art. 45 p 2 =O l A 2 . But if a perpendicular be drawn from 1 to fty it is equal to the radius of the in-circle of the triangle a(3y or half the radius (Jr) of PQR ; and the distance of its foot from A is equal to the semiperimeter of a(3y i.e., %s of PQR. Hence (Euc. I. 47) p a = l(*+ 8 > Similarly for the radii p ly p 2 , p 3 of the circles cutting two escribed and the inscribed of PQR orthogonally we obtain and by adding these results we have, on reducing, or, the sum of the squares of the radii of the four circles cutting orthogon ally the inscribed and escribed circles of any triangle taken in threes is equal to the square of the diameter of the circum-circle. * In the triangle PQR since perpendiculars PA and QB" are let fall from the extremities of the base PQ on the external bisector A B of the vertical angle R, by a well-known property yA = yB = J sum of sides. But the distance of the middle point of any side from the points of con tact of the ex-circles which touch it externally = \ sum of sides. Hence if a circle be described with y as centre and yA = yB" as radius, it cuts the ex-circles of PQR whose centres are at A and B orthogonally. It follows that the locus of the centre of a circle cutting these two ortho gonally is the line yO v since it is perpendicular to the line of centres ; similarly a0 1 and $0 l are the loci for the centres of circles orthogonal to the remaining pairs of ex-circles, whose centres are at B and <7, C and A respectively. Therefore 1 is the centre and 1 ^ =O 1 5" = etc., the radius of the common orthogonal circle, i.e., Taylor s circle. 80 RECENT GEOMETRY. 2. To find the radius p of Taylor s circle of a triangle ABC. [Taylor s circle for the triangle ABC is the circle in Ex. 1 for PQR; hence we have to express r and s of the latter triangle * in terms of the parts of ABC. We easily obtain p 2 = 4S 2 (sin 2 J. sinlB sin 2 tf+ coa*A cos 2 cos 2 <7) also Pl 2 = 4 J ft 2 (sin 2 ^ cos 2 cos 2 ^ cos 2 ^ sin 2 sin 2 (7) with similar values for /o 2 2 and p 3 2 . From these expressions we have the result given in Ex. 1 : 3. The lines B"C , C"A , A"B , parallels to the sides of ABC, are the chords of contact of the ex-circles of PQR with its sides, t [Let A"B meet PR in the point Q . Then B B"RQ is a parallelo gram, therefore R Q = semiperimeter of PQR, etc.] 4. Employing the notation of Art. 35, prove that the lines joining the corresponding vertices of the two triangles PQR and XYZ are concurrent at the circum-centre of the latter. [Let p and q be the perpendiculars from R on the sides YZ and ZX of the triangle XYZ. Then plq = RB smB/RA amA. But RB IRA = QR/RP= a cos A/b cos B. Substituting and reducing we have pjq cos A/cos B. But if Z be joined to the circum-centre of XYZ, the joining line is the locus of a point such that perpendiculars from it on the sides are in this ratio; hence ZR passes through the circum-centre of XYZ.% And similarly for the lines PX and QY.] * The sides of the pedal triangle are equal to a cos A, bcosB, ccosC, or P sin2A , R sin2B, R sm2(7 ; hence its perimeter = 4/? sin A sin B sin C ; its s-a = 2jF?sm.4cos2?cos<7, its s-b = 2RcosA sin BcosC, etc.; its r = 2R cos A cos BcosC; its r t =27? cos A sin B sin C, etc. fThe polars of the vertices of a triangle with respect to the ex-circles meet the sides in six points which lie on the same circle. Mathesis, t. 1, p. 190. $ PX, Q Y, and RZ are perpendiculars to antiparallels to the sides of XYZ and therefore meet the sides of PQR at right angles. Hence the circum-centre of XYZ is the orthocentre of the triangle PQR. RECENT GEOMETRY. 81 5. The Simson lines of the median triangle LMN of a given one ABC with respect to the vertices P, Q, It of the pedal triangle pass through the centre of Taylor s circle.* [The circum-centre of ABC is the orthocentre of LMN. Hence RO is bisected by the Simson line XYZ of R. Also CZ^RZ ; therefore the line XYZ\* parallel to OC. But the centre of Taylor s circle 1 is (Art. 47) the middle point of RH 3 ; therefore, etc.] 6. The Simson lines of PQR, whose poles are L, M, N, pass through 0,. * The point on the circum-circle from which perpendiculars or other isoclinals are let fall on the sides of an inscribed triangle is called the Pole of the Simson line. F. Mathesis, t. 2, p. 106, " Sur la Droite de Simson," par M. Barbarin. F OF THB 82 RECENT GEOMETRY. [For the perpendicular NZfrom N on PQ bisects it (Euc. III. 3); and the perpendiculars NX and NY are equally inclined to AB (Euc. I. 26), hence the line XYZ is a perpendicular to AB through the middle point of PQ; therefore, etc, (Art. 47.)] 7. Prove that the common inclination (0) of the sides of the triangles A B C and A B C" to those of ABC is given by the equation tan = - tan A tan B tan C. (Taylor) 8. The intercepts made by Taylor s circle on the sides are a cos A cos (B- C\ b cos B cos (C - A\ ccos Ccos (A -B). [A B" = A R + RB"=(a. coaA + b cos B) cos C= etc.] RECENT GEOMETRY, 83 9. The circum-centre of a triangle, its symmedian point, and the orthocentre of its pedal triangle are collinear. (Tucker.) [The orthocentre of the pedal triangle has been shown to be (Ex. 4) the circum-eentre of XYZ t and K is the centre of similitude of ABC and XYZ\ therefore, etc.] 10. The circum-centre and the orthocentre of its pedal triangle are equidistant from, and collinear with, the centre of Taylor s circle. (Neuberg.) [For Cff 3 and ZR are parallel, since both are at right angles to PQ; also RH 3 is bisected at O l (Art. 47), therefore, etc., by Art. 37.] CHAPTER IV. GENERAL THEORY OF THE MEAN CENTRE OF A SYSTEM OF POINTS. 49. We now proceed to the discussion of the general linear relation connecting the distances of a system of points from a given line. Let A, B, C, D ... be the system of points, AL, BL CL ... their distances from any line L, and 2(a . AL) the algebraic sum where a, b, c ... are given quantities. By 2(a . AL) is therefore meant the sum of given multiples of the distances of the system of points from the line ; perpendiculars from points on opposite sides of L being taken with opposite signs. 50. Theorem. For any two lines M and N and systems of points A, B, C ... and multiples a, b, c ... having given 2(a . AM) = and 2( to prove that where L is any line passing through the intersection of MandN. 84 THEOREMS. 85 Join AO and let this line be denoted by R. Then since LMNR is a concurrent system of lines we have sin JO . smLR + siuNL . sinMR + sinLM . sinNR = 0, but, by Art. 2, smLR : sinMR : sinNR = AL: AM : AN ; therefore sinMN. AL + siuNL . AM+siuLM.AN= 0. Similarly for the points B, C ... we have siuMN . BL + siuNL . BM+aiuLM . BN= sinMN . CL + sinNL . CM+ai&LM . CN=0. Multiplying these equations respectively by a, b, c . . and adding sin MN2(a . AL) + &mNLI,(a . AM) + sinJf 2(a . AN) = 0, hence if 2(a . u4 Jf ) = and 2(a . -4jV) = 0, it follows that Def. The point which satisfies the relation 1,(a.AL)=Q for every line L passing through it is termed the Mean Centre of the system of points A, B, C... for the system of multiples a, b, c 51. Theorem. The position of the mean centre for a given system of multiples is either unique or indeter minate. For let O l and 2 be two of its positions, and any point whatever. Join 0^0 and 2 0, and denote these lines by M and N. Since 2(a . AM) = and 2(a . AN) = 0, it follows by Art, 50 that any line L through 0, i.e. any line whatever, satisfies the equation It is obvious, in the general case, that when all the points of the system, and all save one of the multiples are 86 MEAN CENTRE. given ; by assigning a definite value to the last multiple, the position of the mean centre is determinate ; and con versely any point whatever is the mean centre of a given system for multiples, all of which save two may be arbitrarily chosen. EXAMPLES. 1. The middle point of a right line is the mean centre of its extremities (Euc. I. 26). 2. The mean centre of two points A and B for the multiples a and b divides the line AB inversely as the multiples, i.e., AO:BO=b:a. The mean centre of the same points for the multiples a, - 6, divides the line externally such that AO: B0 = b: a. 3. The mean centre of a linear system of points A, B, C ... for multiples each = l satisfies the equation 2L40 = 0. 4. The bisectors L, J/, N of the sides of a triangle ABC are con current. [For 2 AL = 0, 24 M= and ^AJV= 0, hence each line passes through the mean centre (centroid or centre of gravity) of the vertices.] 5. The lines joining the middle points of the three pairs of opposite connectors BC and AD, CA and BD, AB and CD of four points A, B, C, D are concurrent, and each is bisected at the point of concurrence.* * In the particular case when the fourth point D coincides with the orthocentre of the triangle ABC we infer at once the well-known property : The lines joining the middle points of the sides of a triangle with those of the segments towards the angles of the corresponding perpendiculars meet in a point and bisect each other. From this it follows immediately (Euc. I. 4) that the six segments are equal, and that the circle passing through the middle points of the sides passes through the feet of the perpendiculars and hisects the segments of the latter towards the angles. This is the fundamental property of the Nine- Points- Circle. THEOREM. 87 6. The geometrical centre of a regular polygon is the mean centre of the vertices A, B, C .... [Join AO and BO. If the polygon be of an even order these lines (L and J/) will pass through the opposite vertices, and the perpen diculars from the remaining vertices are equal in pairs and opposite in sign ; and if the polygon be of an odd order L and M bisect the opposite sides at right angles ; therefore, etc.] 7. ABCD ... is a regular cyclic polygon and L any line passing through its centre ; prove that AL + RL + CL+ ... =0. 52. Theorem. Any point is the mean centre of the vertices of a triangle ABC for multiples proportional to the areas BOG, CO A, AOB. For letting L coincide with AOX and applying the relation IZaAL we have or disregarding signs BL/CL = c/b. Also since the triangles CO A and AOB are upon the same base A 0, BL/GL = A OB /CO A ; equating these values, b COA therefore Similarly c AOB c AOB a BOG Hence a : b : c = BOG : COA : AOB. If the point is outside the triangle, and within the angle A, the multiples are proportional to -BOG, COA and A OB, with similar results when is within the angles B or G. EXAMPLES. 1. The in-centre of a triangle is the mean centre of the vertices for multiples proportional to the sides. 2. The ex-centres are the mean centres for systems of multiples - a, 6, c ; a, - b, c ; a, b, -c; or quantities proportional to them. 88 MEAN CENTRE, 3. If 0, O lt Oft 2 denote the in- and ex -centres of a triangle, each is the mean centre of the remaining three for multiples, 5 a, s b, s c ; s b,s c, s, etc. [For the areas in the first case are 2 3 0, O^O, 0^0, and these are obviously proportional to s - a, s b, s - c. Similarly for each of the ex-centres. Thus generally since -s : s- a: sb :s c - l/r : l/r x : l/r a : 1/V 3 ; for the points 0, 1} 2 , 3 each is the mean centre of the remaining three for the corresponding multiples of the system - l/r, l/r l5 l/r 2 , l/r 3 .] 4. Prove the following points are the mean centres of the vertices for the system of multiples written opposite to them. ~. ( acos^i, bcosB. ccosC. Circum-centre }. on- /v ( sm2^, sm2 J 5, sm2<7. Orthocentre tan A, tan 2?, tan C. Symmedian Point a 2 , 6 2 , c 2 . Brocard Points * I, \ ; 1, * I. b" C* a" c* or cr " Nine-Points " Centre a cos( - C\ b cos(<7- A\ c cos(^l - B)* 5. The lines drawn from the vertices of a triangle to the points of contact of the in-circle are concurrent at the mean centre of the vertices for multiples r 1} ?* 2 , r 3 . 6. The lines drawn to the internal points of contact of the three ex-circles meet at the mean centre of the vertices for multiples l/r l5 l/r. 2 , l/r 3 . 7. If a point be the mean centre of the vertices for multiples I, m, n, its Isotomic Conjugate t is the mean centre for multiples the reciprocals of ?, m, n. 7a. The Isogonal Conjugate f of is the mean centre for multiples or/7, 6 2 /m, c~/n. * From this it is evident that the sides of the triangle ABG meet the Nine- Points-Circle at angles B-C, C-A,A-B. + Two points X and X equidistant from the extremities of a line BC are called Isotomic Conjugates with respect to the line. It is easy to see, and it will be afterwards proved, that if the sides of a triangle A BC be divided isotomically in the pairs of points X, X ; Y, Y ; Z, Z ; such that AX, BY and CZ are concurrent at a point O ; then J X , EXAMPLES. 89 8. Any point on the segment AB of the circum-circle of an equilateral triangle ABC is the mean centre of the vertices for multiples IJOA, l/OB, - HOC. 9. The mean centre of 0, O lt 2) 3 is in Ex. 3 the circum-centre of the triangle. 10. The centre of Taylor s circle is the mean centre of the vertices of the pedal triangle of ABC for multiples a cosCS-C ), bcos(C-A), ccos(A-B). 11. The mean centre of the vertices of ABC for multiples /, w, n is the mean centre of the vertices of the pedal triangle PQR of for multiples 2 /, 6 2 /m, c^/n. [From the figure of Art. 23, Ex. 1, we have QORiROP: POQ = OQ.OR sinA :OR.OP sinB .OP.OQsinC = a/OP : b/OQ : c/OR (1) But OP : OQ : OR=BOC/a : COA/b : AOB/c = I/ a : rnjb : n/c. Substituting these values in (1) j therefore, etc.] 12. The symmedian point of any triangle is the centroid of the pedal triangle of 0. [For BOC : CO A : A OB = a 2 : 6 2 : c 2 by Art. 16, Ex. 2 (2).] 13. The lines joining A, B y C to the corresponding vertices of Brocard s first triangle are concurrent, and the point of concurrence is the mean centre of the vertices of ABC for multiples the recipro cals of a 2 , 6 2 , c 2 . [For it has been shown that it is the isotomic conjugate of the symmedian point, Art. 30, Ex. 3.] 14. If perpendiculars be let fall from any point P on the sides of a regular polygon ; the mean centre of their feet lies on the line joining P to the circum-centre. BY , CZ are also concurrent at . The points and are termed Isotomic Conjugates with respect to the triangle ABC. If the pairs of lines AX, AX t etc., are equally inclined to the sides 6 and c, etc., they are Isogonal Conjugates with respect to the angles ; and if AX, BY, CZ &re concurrent, AX , BY , CZ are also concurrent. The points of concurrence are Isogonal Conjugates with respect to the triangle. 90 MEAN CENTRE. [Through draw OAA parallel and PA perpendicular to pi. The projection of p l on OP= projection of A A ; but A, B, C, ...and A y B , C , ... are the vertices of regular polygons, whose mean centres are both on OP. Therefore the sum of the projections of p t ... on 0P-0.] 53. Theorem. For any line L to prove that 2a.AL = 2(a Draw M through parallel to L. Then AL = AM+OL, Cl=CM+OL,etc. Multiplying these equations respectively by a, b, c, ... and adding, we have 2(a .AL) = 2(a . AM) + S(a) THEOREM. 91 but *Z(a.AM) = () since M passes through the mean centre; therefore, etc. This property enables us to find the mean centre. For by taking a line L in an arbitrary position and calculating 2(a. AL)fS,(a) we have for the locus of a line parallel to L at this distance from it. Again, take a line in another position and construct the locus of as before. The intersection of these loci is the point required. COR. 1. If 2a . AL is a constant, the line L touches, or envelopes, a circle concentric with 0. COR. 2. If the multiples are all equal 2AL = n . OL, where n denotes the number of points in the system. COR. 3. For systems of points and multiples and their mean centres A n B n C n . . ., a n b n c n ..., O n , the mean centre of all the points and their correspond ing multiples is the mean centre of O v 2 , ... O n for the multiples 2(0^), 2(a 2 ), ... 2(a). [For since 2a l -4 1 i = 2(a 1 )0 1 Z, 2ct 2 ^ 2 Z = 2(a 2 )0 2 Z, etc., on adding these equations Hence the mean centre of a system of points can be found as follows : Find the mean centre O l of two of the points A and B ; next find the mean centre of O l and C for multiples a + b, c. Denote this by 2 , and find the mean centre of 2 and D for multiples a+b+ c, d, and so on. When the entire system has thus been exhausted the last mean centre found is that of the system. 92 MEAN CENTRE. EXAMPLES. 1. The sum of the distances of the vertices of a triangle from any line is equal to three times the distance of its centroid from the line. 2. Draw a tangent to a circle such that ILa . AL may be a maximum, minimum, or have any given value. [The extremities of the diameter passing through the mean centre are obviously the points of contact in the extreme cases. The general case reduces to draw a common tangent to two circles.] 3. If L touches the in-circle 2a . AL = 2A when the multiples are equal to the sides of the triangle. 3a. For the ex-circle to the side c the equation becomes 4. The projection of the mean centre on any line is the mean centre of the projections of the system of points on the line. [Let the projections be denoted by , A , /?, C ... and L the line 00 . Then A = AL, B =BL, etc. Hence 2a . A O = ^a.AL = 0- > therefore, etc.] 5. If 0, 1} 2 , 3 denote the in- and ex-centres of a triangle, * This relation may be otherwise written : 0,L . 0,L . 0,L OL EXAMPLES. 93 [For is the mean centre of the remaining points for multiples s _ a> s _ 5 ? s - c (Art. 52, Ex. 3), and since 2(s- a) =s ; therefore, etc.] C. Let three similar triangles BCA, CAB and ABC be described on the sides of ABC in the same aspect ; to prove that the mean centres of the triangles ABC and A B C coincide (Brocard). [Let X be the middle point of BC and Z of A B . Complete the parallelogram BAG P. Join AX, C Z y Z X and PB . The triangles BPCznd B CA are similar, therefore CP/Cfi=B C/AC (Euc. VI. 4), or by alternation B C\CP=AC\BC ; also the angles B CP and ,4 5 are equal, therefore the triangles BPC and ABC &re similar (Euc. VI. 6) ; hence CS f (S f P=CA/AB- t alternately CB /CA = PB /AB- but CB /CA = C A/AB (hyp.) ; therefore PB /AB=C A/AB from which Again LPB C=LBAC, to these add the equals ^1(7^ and respectively ; therefore PB and .46" are parallel. But Z X is parallel and equal to half of PB ; therefore it is parallel and equal to half of AC . Hence the medians yUTand C Z trisect each other.* Otherwise thus : f Let another triangle ABC" be described below the base AB symmetrically equal to ABC . It is easy to see that * For another proof see Milne s Companion to the Weekly Problem Papers, Art. 123. t Educational Times. Reprint. Vol. liv., p. 102. 94 MEAN CENTRE. the triangles ABA and CBC" are equal in area ; similarly ABB and CAC" are equal. By addition we have ABA + ABB = ABC + ABC" or ABA + ABB -ABC = ABC, i.e. the algebraic sum of the perpendiculars on AB from A , B , <7 = the perpen dicular from C on A B. Similar results are obtained for the sides BC and CA ; therefore, etc. Syamadas Mukhopadhyay.] 7. If two points A and B be displaced to new positions .4 and B , their mean centre M for any multiples is displaced to M found by the following construction : Through M draw lines MP and MQ equal and parallel to AA and BB respectively. Join PQ and divide it in M such that [For since AA PM&nd BB QM are parallelograms, A P*=*AAf&nd Q = BM; therefore by similar triangles PA M and QB Af, 8. If three points A, B and C be displaced to new positions A , B and C", their mean centre M is displaced to M found by the following construction : Through M draw lines MP, MQ and MR equal and parallel to the displacements AA, BB and CC respectively ; M is the mean centre of P, Q, R. [For let X denote the mean centre of A and B, X which is found by Ex. 7 of A and B. Draw MO equal and parallel to XX 1 . Join OX , RC and X C . It is evident by parallels that is the mean centre of P and Q ; also MX OX and MCRC ; therefore in the similar triangles (1) EXAMPLES. 95 hence M is the mean centre of X and C , that is of A , B and C\ for the same multiples that M is of A, B and C. But each of the ratios in (1) is equal to M O/M R therefore M is the mean centre of and R, that is of P, Q and R for the same set of multiples. NOTE. The construction for the displaced mean centre may in the same manner be extended to the quadrilateral and generally to a polygon of any number of sides. Hence for two systems of points A, B, (7, ... and A , B , C t ... and their mean centres J/and M for the same set of multiples a, b, c ... if we draw through M parallels MP, MQ, MR, ... equal to AA , BB , CC y . . . respectively, the mean centre of the third system P, Q, R, . . . for the same multiples coincides with M . 9. If through any point M are drawn MP, MQ and MR parallel and proportional to the sides of a triangle ABC, the mean centre of P, Q and R for multiples each equal to unity coincides with M. [By Ex. 8, or thus : Complete the parallelograms PMQR and draw MR. ffince-^-*? and the angles at P and C equal, the tri- PR ty-"l b angles PMR and ABC are similar, hence MR=MR =2MO, and 96 MEAN CENTRE. is the mean centre of P and Q, and therefore M is the mean centre 10. Prove the similar property for the quadrilateral ; and generally : If through any point M lines are drawn parallel and proportional to the sides of a polygon ; the mean centre of their extremities for multiples each = l coincides with M. 11. If a system of points A, B, (7, ... be displaced to A , B, C t ... such that A A } BB , (7(7 , ... are parallel and proportional to the sides of a polygon, the mean centre of the system remains a fixed point. [By aid of Exs. 8 and 10.] 12. Weill s Theorem. A variable polygon is inscribed to one circle and escribed to another ; to prove that the mean centre of the points of contact of its sides with the latter circle is a fixed point. [Let ABC... denote the polygon, A B C ... a consecutive position, T and T the points of contact of AB and A B with the circle of radius r ; B the small angle between AB and A B , and X their intersection. The triangles AA X and BB X are similar, hence BB jAA = BXjA X and **.-** ** inthelimit-Jg (1) AA + BB BX+A X BX+AX AB WfflLL S THEOREM. 97 Also, since AB and A B are indefinitely near to one another, X is indefinitely near to the point of contact T, and BX and BY are therefore equal because they are tangents from the same point to a circle. Dividing (2) by (1) AB = AA+JBB , 3) BC BB + CC " A aiU TZ-Vr*!? <K e of Sines), hence ^-^n" smA AB but AB (diameter of ABC] x siiL4 and TT = 2rO; therefore TT a A A + BB a AB (by 3). Thus as the polygon ABC... varies, its points of contact are displaced for each consecutive position in the direction of its sides, and proportional to them ; therefore the mean centre is a fixed point. NOTE. If the - side BC is a variable tangent to a third circle of radius r, the result of dividing (2) by (1) is AB_AA + BB BX^ . BC~ BB +CC 7 BY therefore if the three circles are so related that BXjBYis a constant ratio Jc t AB =J . AA + BB BC " BB + CC and TT j 7\ T 7 / = r/kr . ABJBC. ] 13. The mean centres of the vertices of any polygon and of similar triangles similarly described on its sides coincide (M Cay). [Let the vertices of the triangles on the sides AB, BC, CD ... be A, B, C ... respectively. Since A A : BB : CC ... =AB : BC : CD ... and are inclined to the sides of the polygon at the same angle ; we may regard the vertices of the given polygon displaced to A B C ... distances proportional and parallel to its sides turned through that angle (cf. Ex. 6).]* * The proofs of Examples 11-13 were communicated to the Author by Mr. Charles M Vicker. G 98 MEAN CENTRE. 14. Through the centre of a regular polygon any line is drawn meeting the sides in A , B\ C , ... to prove that 2- = 0. (JA [Let M be the middle point of one side, then MA O is a right- angled triangle, and if a perpendicular MM be let fall on the hypotenuse we have OA . OM = r* or 2^L=isOJf = 0. Art. 50. See Art. 3, Ex. 9.] 54. Theorem. For any system of points A, B, C,... their mean centre 0, and any line L ; to prove that Za . AL 2 = 2a . AL 2 + 2(a)OZ 2 , where L is the line through parallel to L. . . AL 2 = AL 2 +OL 2 +2AL .OL-, .-. BL 2 = BL 2 + OL 2 + 2BL . OL; Multiplying these equations by a, b, c, ... respectively and adding results, 2a . AL 2 = 2a . AL 2 + 2(a)OZ 2 + 20LI(a . AL \ but 2a.AL = (Art. 50); therefore, etc. COR. 1. When the multiples are equal also since Z-4 L = n. OL ; OL is the arithmetical mean of the several lines AL, BL, CL ..., and AL , BL ... the several differences between each and their mean. Hence, the sum of the squares of n quantities = n times the square of their mean value-}- the sum of squares of the n differences ; or if the quantities are the segments of a line this property may be stated : the sum of the squares of the unequal parts = the sum of the squares of the equal parts + the sum of the squares of the n differences. This property is obviously an extension of Euc. II. 9, 10. COR. 2. For any two parallel lines L and M, - OM 2 ). THEOREMS. 99 55. Theorem. For any point P to prove that Project the system of points on the line OP and denote their projections by A , B , (7, .... Then (Euc. III. 12-13) Similarly BP 2 = BO 2 + OP 2 + 20P . OR etc. Multiplying these equations by a, 6, c ... and adding the results, but is the mean centre of the system A , B , C ... (Art. 53, Ex. 4) -therefore 2a . OA = Q. COR. 1. If the n multiples are equal COR. 2. For a regular cyclic polygon the sum of the squares of the distances of any point on the circle from the n vertices is constant and = 2nR 2 . COR. 3. If 2& . A P 2 is constant, the locus of P is a circle concentric with the square of whose radius is equal to ~ ^af ^- COR. 4. 2a. AP* is a minimum when P coincides with 0. See Art. 16, Ex. 3. 100 MEAN CENTRE. EXAMPLES. 1. ABCD ... is a regular cyclic polygon, the centre, R the radius, and P any point on the circle to prove that the sum of the squares of the perpendiculars from P on the radii OA, OB, OC ... [Denote the feet of the perpendiculars by A , B , C ... The circle on OP as diameter passes through these points (Euc. III. 31) ; also since A B , B C , ... subtend equal angles (27r/n) at 0, a point on the circle, A B C 1 ... is a regular cyclic polygon. Hence (Cor. 2) Similarly 20A" 2 = %nR\] 2. For any line L passing through 0, ^,AL* [Let L coincide with OP. By similar triangles AL = PA, BL = PB , etc. therefore 2P.1 /2 = 2^ 2 =|^ 2 by Ex. L] 3. The sum of the squares of the perpendiculars p^ jt? 2 , p 3 ... p n EXAMPLES. 101 from any point P upon the sides of the polygon is equal to n(r z + ^S 2 ), where r is the radius of the in-circle and 8 = OP. [Through draw parallels OA t OB , OC ... to the sides of the polygon meeting the corresponding perpendiculars from P in A , I?) C", ... As before A B C ...is a regular cyclic polygon in scribed in the circle on OP as diameter. Since the sum of the perpendiculars is constant and=?w 2pi 2 = w 2 + 2P.4 2 (Art. 54) (1) but 2P^ 2 =!?iS 2 (Ex. 1), substituting this value in (1); therefore, etc.] 4. In Ex. 3 if P is on the in-circle SjOj 2 = f w 2 . 5. If ir lt 7T 2 , 7T 3 , ... denote the distances of the vertices from any line L and 8 = OL, [Through draw L parallel to L and let A t B , C be its inter sections with AL, BL, CL ... respectively. Since ^AL = nOL (Art. 53), 2AL 2 = 7i . OL 2 + ^AA 2 (Art. 54), but ?,AA z = %nR* (Ex. 2) ; therefore by substitution 2 or 27r 1 2 =w(8 5a. If L is a tangent to the circum-circle Zir^-liUR 6. If P be a point on the circum-circle of a regular polygon ABC..., ^PA* = QnR\ [Draw OP and produce it to meet the circle again in Q, and let A\ B , C be the projections of the vertices on this line. Since PAQ is a right-angled triangle, PQ.PA =PA\ 102 MEAN CENTRE. Squaring, we have therefore But and Substituting (Art. 54, Cor. 1.) (Ex. 2.) 7. If a, 6, c denote the sides of a triangle ABC and P any point on the in-circle, 2a . AP 2 = 2a . ^ O 2 + 2rA. 8. If J.SC be an equilateral triangle, and L a tangent to the [For ^Z; + 5Z, + CX = 3r, therefore on squaring ZALP + SZBL. CL 9r 2 . Also ?AL 2 = 3r* + jR?, or since R=2r, 2^Z 2 = 9r 2 ; hence 7^ = 0, therefore, etc.] 9. Perpendiculars are let fall from P on the sides of any polygon ABO... and their feet joined; prove that if the area of the in scribed figure A BC ... is constant, the locus of P is a circle concentric with the mean centre of J, B, C, ... for the multiples in 2 A, sin 25, sin 2(7, .... THEOREMS. 103 [Let be the middle point of AP. Then hence V2A OB =S2PA B - or J . P^ 2 sin 2 A = 2^ 5 C" . . . - A BC . . . . Therefore 2 sin 2A . AP is constant .... For a triangle the mean centre of A, JB, C for multiples sin 2A, sin 2.5, sin 2(7 is the circum-centre, showing Art. 23, Ex. 2, to be a particular case of this theorem. M Vicker.] 56. Theorem. If 2AB denote the sum of the mutual distances of a system of points A, B, C ... from each other; to prove that 2(o* . AB) 2 = 2(a) . 2(a . A O 2 ). In Art. 55 if we suppose P to coincide with each point of the system successively we have the following relations : Multiplying these results by a, b, c ... respectively and adding 2Sa6 . AB* = 2(a) . 2a . A0 2 +2(a) . 2a . A0\ therefore 2a6 . A B 2 = 2(a) . 2a . A O 2 . COR. 1. If the multiples are each equal to unity, COR. 2. The sum of the squares of all the lines joining the vertices of a regular polygon = n 2 R 2 ; where R is the radius of the circum-circle. COR. 3. For three points A,B,C, the sum of the squares of the sides of a triangle = three times the sum of the squares of the lines joining the vertices to the centroid ; or three times the sum of the squares of the sides =four times the sum of the squares of the medians. 104 MEAN CENTRE. COK. 4. If be the in-centre and a, b, c the sides of a triangle ABC 2(ab . AB 2 ) = 2(a) . 2a . A O 2 reduces to (Art. 52, Ex. 1) with analogous results for O lt 2 , and 3 . COK. 5. The sum of the squares of the six lines joining the centres of the in- and ex-circles = 48J2 2 Since the centre of the circum- circle is (Art. 52, Ex. 9.) the mean centre of O lt 2 , 3 , 4 , but rj+rj+rg r=s4B; therefore, etc.* EXAMPLES. 1. If S denote the symmedian point of a triangle, (Art. 52, Ex. 4.) 2. For the Brocard points ft, 12 , (Art. 52, Ex. 4.) 3. The distance OP of any point P from the in-centre of a triangle is given by the equation [Eliminating 2a. AO 2 between the equations, 2a . AP 2 = 2a . A O 2 + 2(a) . OP 2 , and 2(6 . ^J5 2 ) = 2(a) . 2a . ^O 2 , the above result follows.] * Otherwise thus : Since O x is the orthocentre of 2 3 4 , if per pendiculars OX, Y, OZ be drawn to the sides from the circum-centre of Oo0 3 4> OjOosSOX, 1 3 =20r, ... ; also 00 2 =2/?; hence therefore S0 2 3 2 =48 J R 2 . EXAMPLES. 105 4. If P coincides with the circum- centre, prove the following where Z), D lt D 2 , D 3 are the distances of the circum-centre from the in- and ex-centres : D*=R 2 -2Rr ; D^ff + ZRr^ etc., etc. 5. Prove that the distance 8 of the symmedian point S from the circum-centre of a triangle ABC is given by the equation 2 3aW [For any point P (Art. 52, Ex. 4) 2a letting P coincide with ; therefore therefore, etc.] 6. The distances of fl and from the circum-centre are given by the equations Oft = 00! = R \/l - 4 sin 2 o7. 7. For the in-centre O l and the ex-centres 2 , 3 , 4 prove the relations r 8. For any point P (S - CL)PO? + (S- b)P0 2 2 + (S- C)P0 3 2 - SPO 2 = 9. Find the following expression for the square of the distance 8 between the circum- and ortho-centre of a triangle ABC. 8 2 = /? 2 (1 - 8 cos A cos B cos <7) = 2a 2 (a 2 - 6 2 )(a 2 - c 2 )/16A 2 . [By the previous method, or more simply by finding the area of the pedal triangle of ABC, (2 area = # 2 sin 2A sin 2B sin 2(7), and using Art. 23, Ex. 1 , and reducing.] * This expression is equivalent to where w is the Brocard angle. 106 MEAN CENTRE. RECIPROCAL THEOREMS. 57. Theorem. For any two points M and N, and systems of lines A, B, C, ... and multiples a, b, c, ... having given 2a . M. A = and Za . NA = to prove that 2a.LA = 0, where L is any point on the line connecting M and N. For MN.LA+Nl.MA+LM.NA**0. Similarly for the lines B and (7, MN.LB+NL.MB+LM.NB=0, MN . LC+NL . MC+LM. NC=0. Multiplying these equations respectively by a,b,c, ... and adding, we get MN2a.LA+NL2a.MA+LM2a.NA = (1) hence if I,a.MA and ^a.NA each = 0, I.a.LA^Q, for any other point L on the line MN. More generally: If 2a.MA and 2a.NA are equal 2a . L A has the same value. For, let 2a . M. A = 2a . NA = k ; substituting in (1) MNZa . LA + (NL + LM)k = ; dividing by MN( = LN+ML) and transposing CENTRAL AXIS. 107 Hence, the locus of a point L such that the sum of given multiples of the perpendiculars from it upon a system of lines A, B, C, ... is constant (1 l a.LA = k) is a right line. Del When the constant vanishes 2a.LA = Q, the locus is termed the Central Axis of the system of lines for the given system of multiples. It is evident that the central axis is one of a system of parallel lines obtained by taking different values, 1 2 3 " * ^ For if L in (1) lies on then Q .................. (2) Za.MA ML MO ^ VT .. = = (EUC.VL4.) hence the values of the summation corresponding to any point is proportional to the distance of that point from the central line or axis. Otherwise thus : If M l and J\\ are the loci of M and N such that 2a. MA=k 1 and 2a. NA =& 2 and P if possible their point of intersection ; then since P is on both lines 2a . PA k v and 2a. PA = k 2 , which is absurd ; therefore, etc. 58. Problem. To find the Central Axis of a given system of lines A,B,C,. ..for a given system of multiples a,b,c,... Take any three points P, Q, R, and calculate 2a . PA, On QR find a point L such that 2a . Q A _ QL I,a.RA~RL L is by (2) on the required line ; similarly obtaining points M and jftf on the other sides of the triangle P, Q, R, their line of connection is that required. 108 MEAN CENTRE. 59. Let the multiples a, b, c ... denote segments of the given lines A,B,C... respectively; a . LA, b . LB, c . LG . . . are each twice the area of the triangle subtended by the corresponding segment at the point L ; hence, the locus of a point such that the sum of the areas subtended at it by any number of finite lines is constant, (k) is a right line ; and if different values be assumed for k the locus varies in position by moving parallel to itself. 60. Theorem. The locus of the mean centre of the points of intersection A lt B v C v Z) p of a variable line L, moving parallel to itself, with the sides of a given poly gon is a right line. Let a, b, c, etc., be the given multiples and a, 0, y ... the angles at A v B v C l ... made by the variable line with the sides A, B, G ... of the given polygon. By hyp. 2a.^L 1 = 0, but Afl^OA /sin a; Bfl^OB/smft-, C^Otf/sin^etc., substituting these values, a/sin a . OA + 6/sin /3.0B+ (7/sin y . OC+ etc. = 0, hence describes a line, viz., the central axis of the system for the multiples a cosec a, b cosec ft, c cosec y .... Def. This locus of the mean centre for the system of parallels, is termed a Diameter of the Polygon when the multiples a = & = c = ... = l; a name suggested by the property to which the theorem is reducible when the polygon becomes a circle. 61. Problem.- To find a point P such that for any systems of lines A, B, C ... and multiples a,b,c... Sa . PA 2 is a minimum. Let any line L through P meet the sides of the poly gon in A t J5 , G f . . . at angles a, ft } y . . . . Then Sa . PA* PROBLEM. 109 is a minimum when 2a sin 2 a . PA 2 is a minimum, that is when P is the mean centre of A , B } C ... for the multiples a sin 2 , b sin 2 /3 .... As L varies parallel to itself the locus of P is a diameter. Let it meet the sides of the polygon in A lt B v C l ...\ the mean centre of these points for the multiples a sin 2 a, b sin 2 /? ... is obviously the point required. EXAMPLES. 1. If a line is drawn through the centre of an escribed circle to meet the sides in X and Y such that CX=CY ; prove that AY.BX=($XYy-, and conversely, if AY.BX=QXYy, AB is a tangent to the circle. [The angles of the triangles JJOJTand AOYsuce as follows : 90-$B, therefore BOX = W-\A ; 0-%A, therefore AOY=90-$B. Hence they are similar ; therefore, etc.] 2. The diameters of an equilateral triangle envelope the in-circle. [Suppose the multiples to be equal to unity, through B and C draw any two parallel lines terminated by the opposite sides of the triangle and trisect them in X and Y towards the vertices. Since X and Y are the mean centres of their intersections with the sides, 110 MEAN CENTRE. the line XY is a diameter. Draw parallels XX , YY" to the sides AB and AC respectively. Then the triangles XX X" and YY Y" are similar, therefore X X".Y Y"=XX".YY". Again, the triangles CXX" and BYY" are similar, since the sides are parallel, therefore XX" . YY"= CX" . BY"=(\X" Y J, therefore X X" . Y Y"=(%X"Y")* ; therefore, etc., by Ex. 1. M Vicker.] Otherwise thus : Draw any system of parallels AA , BB , CC terminated by the opposite sides and let A t B , C denote the mean EXAMPLES. Hi centres of their points of intersection with the sides of ABC. Let the diameter A B C meet the sides in JT, T t Z\ the parallels through X and Fare bisected at these points, hence AX and BY each bisect CO and therefore meet at its middle point. Then from the com plete quadrilateral ABCXYZ the row A C"BZ is harmonic, therefore AA f C C" and BB are in harmonic progression, or 1121 AA f + BB ~C C"~CC" but 2-j ,=0 is the criterion for the tangent to the in-circle. See AA Art. 55, Ex. 8 ; therefore, etc.] 3. If a system of n points J, B, C, ..., N be situated at equal distances on an arc of a circle 0, r ; required to find the position of their mean centre. [Through draw a parallel L to the chord of the arc A N ; let the angle AOL = a and AON=n(3. Then, if d be the distance of the mean centre from 0, we have (Art. 53) but a + |ft/?=|7r, therefore the above expression becomes, on re duction, r cot|/3 sin|w/2.] NOTE. If the number of points on the arc is infinitely great, it follows, since ft is indefinitely small, that 7 _ chord x radius length of arc CHAPTER V. COLL1NEAE POINTS AND CONCURRENT LINES. 62. Theorem. // a straight line be draiun cutting the sides of a triangle ABC in points X, Y, Z, to prove the relation BX CY AZ CX AY BZ = 1; and conversely, having given this relation to prove the points are collinear. (Menelaus.) For denoting the perpendiculars from the vertices on the transversal by I, m, n ; we have by similar pairs of tri angles, BX_m, CY _n. AZ_l_ n AY~ I B~Z~m 112 CEVA S THEOREM. 113 Multiplying these equations* and reducing, the above result follows at once. Conversely, if the line joining X and F meet the base in Z by the first part of the Proposition, BX CY AZ * ~ , , , , BX CY AZ but by hyp. ^. _._ , AZ AZ hence M^W therefore Z and Z coincide. 63. Theorem.// three lines AO, BO, CO be drawn from the vertices of a triangle ABC through any point to meet the opposite sides in X, F, Z\ to prove the 7 ,. BX CY AZ T . rdation _.__._= -l,t and conversely, if this relation be given the lines AX, BY, CZ are concurrent. (Ceva.) For the triangles BOG and CO A on a common base are proportional to their altitudes, which are in the ratio BZ/AZ. * The proof here given applies equally to the general proposition : Any right line meeting the sides of a polygon AB CDEF... in points X, Y y Z, U, V, W... gives the relation AX BY CZ DU EV FW , BX CY ~DZ ~EU FV GW" t A line drawn across the sides of a triangle meets them either all externally, or two internally and one externally, i.e. the number of sides cut externally is always odd, and therefore the product of the ratios z> V* /^V A 7 ?i iL, is positive. On the other hand, if three points on G X. AY .o^ the sides connect concurrently with the opposite vertices, an odd number is internal and the product of the ratios is therefore negative. H 114 COLLINEAR POINTS. Hence the following equations : BX_AOB CY BOG AZ_COA, CX~AOC y AY~BOA BZ~~ COB on multiplying* and reducing, the above result is obtained. Conversely, let AX and BY meet in 0. Join CO and let it meet AB in Z . Then by what has been proved BX CY AZ CX AY BZ by hyp. = -1, therefore the points Z and Z coincide. 64. The relations of the previous Articles are equivalent to the two following : sin BAX sin CBY sinACZ sin CAX sin ABY sin For by the rule of sines CX sin with similar values for the remaining ratios, compounding and reducing, the above results are obtained. *More generally, if the vertices of a polygon ABCD... of any odd number of sides be joined to any point and the lines produced to meet the opposite sides in X, Y, Z, U, T, W, it follows by similar . ., . AX BY CZ DU , reasoning that BX CY DZ WU - = ~^ PARTICULAR CASES. 115 These formulae may be regarded as criteria of points on the sides of a triangle lying on a line and connecting concurrently with the opposite vertices. We shall now apply them to the following remarkable particular cases : I. Let the points X, Y, Z be at infinity on the sides, thus BX = CX, CY= A Y, &ndAZ=BZ , hence the criterion of Art. 62 is satisfied and it follows that every three and therefore all points at infinity in the same plane may be regarded as lying on a line* II. Let AX, BY and CZ be any three parallel lines. Si BX CY AZ__ CX AY BZ~ every three, and therefore all, parallel lines are con current. Of these properties Townsend says : " Paradoxical as these conclusions appear when first stated, all doubt of their legitimacy has been long set at rest by the number and variety of the con siderations tending to verify and confirm them." Modern Geometry, Vol. I., Art. 136. III. When AC=BC, and is a point on the circle touching the equal sides at A and B. By Euc. III. 32, LBAO = L.CBO; LABO=L.CAO. Substituting in the above equation, and sin A CO _ sin 2 ABO _AO* sin BCO~sm*BAO~WJ* * This conception of elements situated at an infinite distance is due to Desargues. About the year 1640 he showed that parallel straight lines meet at an infinitely distant point ; and that parallel planes may be regarded as intersecting in the line at infinity. More recently the celebrated Poncelet proved that all points at infinity may be considered to lie in a plane. 116 COLLINEAR POINTS. Similarly, if CO meet the circle again in , _ smBCO~~BO 2 Hence : A variable chord 00 of a circle passing through a faced point C divides harmonically the arc AB, intercepted by the tangents CA and CB. Also, since AB is divided harmonically at and , 00 is divided harmonically by AB ; hence the variable pairs of tangents at and intersect on the fixed line AB. IV. Describe a circle about AOB, and let it meet the lines AC, BC, CO again in A B O . Then, for the point 0, smBAO s = sin ABO sinCAO~smACO but CBO = CO B and CAO = CO A . (Euc. III. 22.) Substituting these values and reducing by rule of sines, OB OB _ sin BOO ( OA OA ~sinACO .............. Similarly, for , VB Q B smBCO . _ O A A ~smACO PARTICULAR CASES. 117 Equating these values, AO AO _AO A O* BO BO ~BO BO Hence: If two arcs of a circle AB and AB are divided in and so as to fulfil the relation (3), AA , BB and 00 are concurrent. EXAMPLES. 1. The internal bisectors of the angles of a triangle are con current. 2. Any two external and the internal bisector of the remaining angle are concurrent. 3. The lines joining the vertices (a) to the points of contact of the in-circle (/3) to the internal points of contact of the ex-circles, are concurrent. [The centres of perspective are named respectively f point de Gergonne and point de Nagel of the triangle.] * The function ~ -f ^ _ is termed the Anharmonic Ratio of the BO BO points A, B, 0, ; and (3) may be expressed thus : "If the arcs AB and A B are divided equi-anharmonically in and , the lines AA t BB and 00 are concurrent ; and conversely." t Educational Times, July, 1890. 118 COLLINEAR POINTS. 4. The perpendiculars of a triangle are concurrent. 5. The tangents to the circum-circle at A, B, C meet the opposite sides collinearly. 6. If a circle meet the sides of a triangle in X, X , Y, Y t Z t Z such that either triad JT, T, Z is collinear or connects concurrently with the opposite vertices ; a similar relation exists amongst the remaining points X , I 77 , Z . 7. If three points are collinear, their isotomic conjugates with respect to the sides are collinear. 7a. If they connect concurrently with the vertices, their isogonal conjugates with respect to the angles also connect concurrently. 8. For any triangle ABC and transversal X YZ ; if any point is joined to the six points sin BOX sin CO Y sin CO X siii AOY sin BOZ P* * -"- -N- therefore, etc. ...] 9. If the sides of a triangle and any three concurrent lines * Examples 8 and 9 will be afterwards enunciated as follows : 8. The lines joining any point to the six vertices of a quadrilateral form a pencil of rays in Involution. 9. Any line drawn across the sides and diagonals of a quadrilateral is cut in Involution. EXAMPLES. through its vertices are cut by a transversal in six points X and X , Fand F, Z and Z ; (BC in X, AO in X ...) ZT .XZ , and conversely. [For sm ^^ = *j*L with similar values for sm ^ff and sin 640 YX sin ABO sin 4(70 ,1 . , n - ^ ; therefore, etc.] 10. If AX, BY, CZ are concurrent, the intersections of FZ" and BC(X \ ZX and (74 (F ), XFand AB (Z ) are collinear. [For . . __=l. Compounding this with two similar \*> 2L AY j equations involving F and Z and reducing, we have EX CY 4Z _, -, ~CX AY BZ~ 11. Given two points 4 and 5 on a circle MNP, on the same side of the diameter MN ; find a point P on the other side such that the intersections X and F of AP and .Z?P respectively with MN may be equidistant from the centre. [Let AB and MN meet in Z , then it is easily proved that PX 2 /PY 2 = BZ/AZ , hence the species of the triangle PXY is known ; therefore, etc.] 12. Draw two circles in contact each touching a given line at a given point and having their radii in a given ratio. * Will be afterwards seen to be an Equation of Involution of the pencil. 120 COL LI NEAR POINTS. 13. If lines be drawn from the vertices of ABC to a point 12 such that QBC=-$ICA = IAB = 0, prove that 9 is given by the equation cot = cot.4 + cot J5+cot C, [For sm 3 <9 = sin(,4 - 0} sm(B- 0)sin(C7- 0) ; etc. Of. Art. 28.] 14. In the general case if the lines in Ex. 13 making equal angles (a) with the sides are not concurrent, they form a triangle A B C similar to ABC and the ratio of similitude is equal to cos a - sin a(cot A + cot B+ cot C] : 1. Defs. The Centres of Perspective of two lines AB and A B are the points of intersection of the pairs of lines AB , AB and A A , BB joining their extremities. Two triangles are said to be in perspective when the lines joining corresponding vertices meet in a point. This point is called the Centre of Perspective of the triangles. 65. Criterion of Perspective of Triangles. Theorem. // the perpendiculars from the vertices of a triangle A B C on the sides of another ABC be denoted by p v p z , p 3 ; TRIANGLES IN PERSPECTIVE. 121 q v q 2 , q 3 ; r lt r 2 , r s (i.e. from A on BGp lt A on CAp^ and so on), the two are in perspective if 2i . !!? .Ps = i . an d conversely. ^i P 2 & For let AA meet BG in X . Then sin B AX /sin GAX f =pJp^ with similar values for r z /r lt and qjq^ multiplying these equations together, therefore, etc., by Art. 64, which also proves the converse* proposition. 66. Theorem. // the vertices of two triangles connect concurrently, their pairs of corresponding sides intersect collinearly (BG and KG in X, etc. ...). For, by similar triangles, B X r G Y , AZ Multiplying, we have .2,. !i.a.l f therefore, etc. Def. The line of collinearity is termed the Axis of Perspective or Homology^ of the triangles. EXAMPLES. 1. Any triangle escribed to a circle is in perspective with that formed by joining the points of contact of its sides. [The centre of perspective is the symmedian point of the in scribed triangle.] *0r thus : Let be the centre of perspective of the triangles and a, /3, y the perpendiculars from it on the sides of ABC; since /3/7=j9 2 /p 3 , 7/a = qsfqv and a//3 = r x /r 2 ; multiply and reduce ; therefore, etc. t The term Homology is due to Poncelet who first studied the pro perties of homological figures in space, v. Traitt des proprittes projectives des figures (1822). 122 COLLINEAR POINTS. 2. If three triangles ABC, A^C^ A^B 2 C Z have a common axis of perspective XYZ, their centres of perspective when taken two and two are collinear. [For the triangles (fig. of Ex. 3) BB^ and CC^o are in perspec tive, their centre being at X ; similarly T is the centre of perspec tive of CC-fito AA^Ai and Z of AA^A 2 and BB^. Hence the corresponding sides of these pairs of triangles intersect in collinear points. But these points (e.g. AA^ BB^ are the centres of per spective of the given triangles in pairs ; therefore, etc.] 3. If three triangles ABC, A^B^C^ A 2 B 2 C% have a common centre of perspective, their axes are concurrent. [Consider the three triangles whose sides are respectively the directions BC, B^C^ B^ ; CA, C V A^ C^A- 2 ; AB, A^B^ A 2 B 2 . It is manifest they are in pairs in perspective, the axis of the first pair being CC-^ ; and JT^is a line joining corresponding vertices. Thus the axis of perspective XY of any two and therefore of every two of the given triangles passes through the centre of per spective of the conjugate triad ] NOTE. It will be noticed that the common centre of the three given triangles is the point of concurrence of the axes AA^ BB ly CO) of the conjugate triad, and the common centre of the conjugate triad taken in pairs is the point of concurrence of the axes of the given triangles. 4. Brocard s first triangle is in perspective in three ways with ABO. STOLL S THEOREM. 123 [The Brocard points are evidently two centres of perspective (Art. 28) ; also the lines A A , BB , CC are concurrent, for p z /p 3 found by aid of the property of Art. 28, Ex. 2, to be c 3 /^ 3 ; therefore, etc. The three centres of perspective are the mean centres of the vertices ABC for multiples proportional to (Art. 52) 1,1,1.1,1,1. LI,II a 2 6 2 c 2 6 2 c 2 a 2 c 2 a 2 & 2 J 5. If ft, ft , ft" denote the three centres of perspective of A BC and its first Brocavd triangle A B C , to prove that the corresponding vertices of their three median triangles lie on three right lines. (Stoll.) [For A B C and ABC have a common centroid G (Art. 53, Ex. 6). But ftft ft" has the same centroid ; for its vertices are the mean centres of A. B, C for multiples proportional to =-=> -= s ; _ ; 6 2 c 2 a 2 c 2 a 2 6 2 _?_> _L> 1. ; therefore (Art. 53, Cor. 3) the mean centre of ft, ft , ft" is a b~ c" that for A, B, C for multiples each = I/a 2 + l/6 2 + 1/c 2 . Now let Z, L j L" be the middle points of the corresponding sides of the three triangles such that GA = 2GL, GA = 2GL , and 6 r ft"=2(7Z": since J, A t ft" are collinear ; L, L\ L" are also collinear, and the two lines of collinearity parallel] 67. Theorem. Two triangles ABC and A B C are in perspective when BX^BX CY. CY AZ.AZ _ GX . CX AY.AY BZ .BZ ~ 124 COLLINEAR POINTS. where X and X are the points of intersection of BC with C A and A B y etc. ; and conversely. Using the previous notation, we have by similar Hence BX.BX _q & . wr-6x } -f t therefore the left side of the above equation becomes which is equal to, on reduction, Ti . P* . 3 . therefore, etc. (Art. 65.) ?i T 2 Ps COR. l. Pascal s Theorem. If XX YY ZZ be any cyclic hexagon, then (Euc. III. 36) AY.AY = AZ.AZ ; BZ . BZ = BX . BX t etc. Hence : The two triangles formed by the two triads of alternate sides of any cyclic hexagon are in perspective ; or, the opposite sides of a cyclic hexagon meet in three collinear points. The centre and axis of perspective of any two triangles in perspective are called the Pascal * Point and Line of the hexagon XX YY ZZ , which is termed a Pascal Hexagon. COR. 2. If X, X ; F, F ; Z, Z f coincide in pairs on the circle, the sides of the hexagon become the tangents to the circle at X, Y, Z } and the chords of contact YZ, ZX and XY; the Pascal point is therefore the symmedian point of the triangle XYZ. (Art. 66, Ex. 1.) * When only sixteen years old, Pascal discovered this property of the mystic hexagram. Essai sur les Coniques, Pascal, 1640. PASCAL S THEOREM. 125 COE. 3. smBA XsmCA X smCB YsmAB Y smAC ZsmBC Z X sinOfl Fsi = 1. TFor _ 2 . _ 2 , , sinJWZ -V sin(Li Z -7 8 hence the above expression is equivalent to COR. 4. Brianchon s Theorem. Let AC BA CR be an escribed hexagon and #, 2/, z the intercepts made by the circle on the sides of the triangle A B C ; since smBA XsmCA X _ y* * sin .B^JTsin CA X ~ ~z* with two other similar equations, Cor. 3 in this particular *The property on which this depends is as follows: If from the point of intersection C of two tangents CA , CB to a circle a secant oj length x is drawn dividing the angle A CB into segments a and /3 ; then sin a sin /3 oc x 1 . For if be the centre of the circle and OX a perpendicular to the secant, we have sin a sin j8 = siu 2 |(a + j3) - sin 2 (a - 18) = r 2 /0a 2 - OX*/OC 2 = therefore, etc. 126 COLLINEAR POINTS. case reduces to : The lines connecting the opposite ver tices of an escribed hexagon are concurrent; or, the two triangles formed by joining the alternate vertices of an escribed hexagon are in perspective. The centre and axis of perspective of the triangles are termed the Brianchon * Point and Line of the hexagon AC BA CB , which for the same reason is called a Brianchon Hexagon. COR. 5. If two of the sides AF and EF of an escribed hexagon coincide, the vertex F is the point of contact of the tangent AE (Art. 6) ; hence for an escribed pentagon ABODE, if the lines AD and BE meet in 0, the points C, 0, F are collinear (cf. Art. 63, foot-note). COR. 6. If two pairs of sides BC, CD and AF, EF coincide, the hexagon reduces to a quadrilateral ABDE , hence the diagonals AD and BE meet on CF , similarly they meet on G F \ therefore the internal diagonals of an escribed quadrilateral and of the corresponding in scribed meet in a point. * Published by Brianchon in the year 1806, and derived by him from Pascal s Theorem by the process of reciprocation with respect to the circle. (See Art. 80, 2.) COMPLETE QUADRILATERALS. 127 COE. 7. Consider the cyclic hexagon FFC GGF. Its Pascal line is the line of collinearity of the three points (1) FF, CO i (2) Fff, CF-, (3) FF> CC : but the line joining (2) and (3) is the third diagonal of the inscribed quadrilateral CFC F and (1) is the intersection of the tangents at and F, and therefore one extremity of the third diagonal of the escribed quadrilateral ; hence : the third diagonals of any inscribed and corresponding escribed quadrilaterals coincide. COR. 8. Let PQRS be any cyclic quadrilateral ; and let XX YY ZZ , the corresponding escribed quadrilateral, 128 COLLINS AR POINTS. be regarded as a Brianchon hexagon ZPX Z RX whose two pairs of coincident sides are the tangents from F Then the lines ZZ , PR, XX are concurrent at the Brianchon point B ; similarly, if the pairs of coincident sides are the tangents from F , we have ZZ , QS, XX* concurrent, i.e. the pairs of opposite connectors PR and QS of the inscribed quadrilateral and ZZ and XX of the corresponding escribed cointersect. We see there fore, from Cors. 7 and 8 that any pair of opposite con nectors of an inscribed quadrilateral and the correspond ing pair for the quadrilateral escribed at its vertices are concurrent. The three points of concurrence on the figure are J., B, 0. The points 7, V, W, U , V, W lie in triads on four lines. EXAMPLES. 1. Three pairs of tangents are drawn from the vertices of a triangle to any circle to meet the opposite sides in points XX , YY , ZZ \ show that if X, Y, Z are collinear, X , T, Z are also collinear. [Apply Cor. 4.] 2. ABC is a triangle inscribed in and in perspective with A B C ; the tangents from ABC to the in-circle of A B C meet the opposite sides in three collinear points X, Y, Z (BC in X, etc.). [Let the axis of perspective of the two triangles be X Y Z , therefore by Cor. 4 we have (^J ^)(...X-)=1 > therefore, \CX . CX J etc., by Ex. 1.] 3. If points XX , YY , ZZ 1 be taken on the sides of a triangle infcfV^ BX BX CY CY> AZ AZ 1 CX CX> AY AY-m Z =l > they are the vertices of a Pascal hexagon. 4. The lines joining each pair of points to the opposite vertex (AX and AX , etc.) of the triangle determine a Brianchon hexagon. EXAMPLES. 129 5. (a) Any two transversals XYZ t X Y Z 1 determine on the sides the vertices of a Pascal hexagon. (/3) Two triads of points on the sides which connect concurrently with the opposite vertices determine a Pascal hexagon. (y) A transversal XYZ and three points X , Y , Z which con nect concurrently with the opposite vertices determine a Brianchon hexagon. 6. A hexagon is inscribed in a circle ; prove that the continued products of the perpendiculars from any point on the Pascal line on the alternate sides are equal (xyz x y z 1 ). [Let AB CA BC be the hexagon whose pairs of opposite sides EG , B C-, CA, C A ; AB, A B meet in points X, Y, Z respectively and the Pascal line L (XYZ} at angles a, a, /?, /? , y, y ; then BL . C LBX . C X sin 2 a _sin 2 a ,^ m 36 >, Similarly, J CL.AL sm 2 /3 A L.BL Multiplying these equations and reducing, / sinY; therefore, etc.] 7. From the middle points Z, M, N of the sides of a triangle tangents are drawn to the in-circle ; show that these tangents form a triangle (A B C ) in perspective with that (PQR) obtained by joining the points of contact of the in- or ex-circles with the sides, and the centre of perspective is the median point of ABC. [For since the sides of ABC with any two of the tangents form i 130 COLLINEAR POINTS. an escribed pentagon, e.g , BCMNA , by Cor. 5, the lines BM, CN, A P are concurrent ; that is, A P passes through the centroid (BM, CN). Similarly for B Q, C R ; therefore, etc.] NOTE. If LMN is any inscribed triangle in perspective with ABC, the above reasoning applies to prove that A B C and PQR have the same centre of perspective. 8. If two triangles ABC and A B C are in perspective, A BC, ABC ; ABC, A BC ; ABC , A B C are also in perspective. 9. If AA , BB, CC denote the lengths of three lines whose directions are concurrent, their six centres of perspective (of BB and CC , X and X , etc.) taken in pairs lie in triads on four lines. [For they are the axes of perspective of the triangles in Ex. 8.] 10. If X, Y, Z are on the sides of a triangle and fulfil the relation the perpendiculars through them to the sides are concurrent ; and conversely. 11. If two triangles are such that the perpendiculars from the vertices of either upon the sides of the other are concurrent, then conversely the perpendiculars from the vertices of the latter upon the sides of the former are concurrent. [By Ex. 10.] TARRY S POINT. 131 12. State the particular cases of the Theorem of Ex. 11 for a given triangle taken with the (a) pedal, (/3) median, (y) triangles formed by joining the points of contact with the sides of the in- or ex-circles. 13. If XYZ be a transversal to a triangle ABC, X , Y , Z the harmonic conjugates of X, F, Z, with respect to the sides ; prove that 1. The triads of points Y Z X, Z X Y, X Y Z&re collinear. 2. X Y Z , X YZ, Y ZX, Z XY connect concurrently with the opposite vertices. 14. The middle points of the segments XX , YY , ZZ are collinear. [For they are the middle points of the diagonals of a. complete quadrilateral by Ex. 3. For another proof v. Art. 91.] 15. The perpendiculars from the vertices of a triangle ABC on the sides of A B C , its first Brocard triangle, are concurrent on the circum-circle. (Tarry s Point.) [By the theorem of Ex. 11.] 16. The perpendiculars from the middle points of the sides of A B C on the sides of ABC are concurrent, (Cf. Ex. 15.) 17. The Sim son line of Tarry s point is perpendicular to OK, the line joining the circum-centre to the symmedian point. 18. In the figure of Art. 28 show that OA : OB : OC = cos(A + w) : cos(+o>) : cos(C+a>) ; and deduce the formula for the Brocard angle, sin A cos(A + o>) + sin B cos(Z? + o>) + sin C cos( C+ w) = 0. Note on Tarry s Point. It will appear obvious that the diameter of the circum-circle containing Tarry s point is related to the triangle ABC in the same manner as OK is to A B C ; and that the circum- and Brocard circles are divided similarly by these corresponding diameters. Also, if a, /3, y denote the perpendiculars from Tarry s point on the sides of ABC, a. : ft : y = sec(A + o>) : sec(B + o>) : sec(C+ to). A point of interest may be here noticed. From Art. 28, Ex. 18 (note) it is evident that the centres O lt 2 , 3 of Neu berg s circles with respect to the sides of ABC are the vertices of similar isosceles 132 COLL IN EAR POINT*. triangles described on a, 6, c respectively, whose equal base angles are TT - to. Therefore, if T denote Tarry s point, it easily follows that AT, A0 } -, BT, B0 2 ; CT, C0 3 divide the angles of ABC isogonally. But the isogonal conjugate of a point on the circum- circle is at infinity ; hence the lines A 0^ B0 2 , O0 5 are parallel. HARMONIC PROPERTIES OF THE QUADRILATERAL. 68. Theorem. In any complete quadrilateral each of the diagonals XX , YY, ZZ is divided harmonically by the other two. Consider the triangle ZZ Y and transversal BXX , Z X YX_=KB m Y X ZX ZB " And since YY , YZ, YZ are three concurrent lines through its vertices, we have Z X Y X Z A ~ Equating these results, we have ZA\Z A = -ZB/Z B. Hence the row of points ZZ AB is harmonic. Similarly, BGXX and CAYY are harmonic. COR. 1. The angles of the triangle ABC, formed by the diagonals (the diagonal triangle) are divided harmoni cally by the pairs of lines AX, AX ; BY, BY ] CZ, CZ . COR. 2. If two lines be given in magnitude and posi tion (ZZ and XX ) their two centres of perspective (Y THREE QUADRILATERALS. 133 and F ) joined to their point of intersection (B) form a harmonic pencil. They also divide the line joining their centres of perspective (in A and C) harmonically. Problem.^ determine the number of polygons which can be formed from n points. Each point joined to the remaining n- I points gives n- 1 lines. Taking any one of these lines as the first side of the polygon we have similarly T? 2 selections for the second side, n - 3 for the third side, and so on. Therefore we have (?i- !)(?&- 2) selections for the first two sides, (n \}(n - 2)(w. - 3) for the first three sides, etc. ; hence we have finally \n - 1 equal to twice the number of polygons, since any sequence of sides when reversed gives the same polygon. Thus four points may be joined in three ways as in figure. Fig. 1 is called a Convex, Fig. 2 an Intersecting, and Fig. 3 a Re-entrant Polygon. By application of the general formula to the hexagon we find that six points in general determine a system of sixty hexagons. 134 COLLINEAR POINTS. EXAMPLES. 1. The conditions that the quadrilaterals in the figures are escribed are : 1. BC+AD=AB + CD. 2. BC^Al) = AB^CD. 3. BC<*AD = AB*CD. [Since tangents from any point to a circle are equal] .2. To prove that the quadrilateral whose angles and perimeter are given is of maximum area when it is escribed to a circle. (Hermite.) [Let two of the sides AB and AD be fixed in position and the remaining two vary. It is easy to see that the locus of C is a line. Suppose C l and C 2 to be the positions of C on the fixed lines and C^j C 2 D 2 parallels to the fixed directions CD and CB. The perimeters of the triangles AG 1 D 1 and AC 2 D 2 are each equal to the perimeter of the quadrilateral A BCD ; the ex-circle of A C l D l is the ex-circle of AD 2 C 2 and D 2 C 2 - D 2 C^ = J\C\- D^. Now, for any point P and parallels PP l and PP 2 by similar triangles PC 1 /C 1 C 2 = (PP 1 -P 1 C 1 )/(D 2 C 2 - D 2 C,} and PCi}C&=(PPt - P 2 <7 2 )/(A<?i - A<? 2 ) 5 adding these equations, we get PP l + PP 2 -P&- P 2 C 2 m D 2 C 2 - D 2 C, ; to each side add AC 1 + AC 2 , and A P l + P l P+PP 2 + P 2 A = AD 2 + D 2 C 2 + C 2 A =given perimeter. HERMITE S THEOREM. 135 Eegarding P and G as consecutive points on the locus, the area of the quadrilateral is a maximum when BCPP l = CDPP 2 , i.e. ED is parallel to C^C* Hence the parallels BO and DO to CD and BC respectively form with AB and AD a re-entrant escribed quadri lateral, and therefore AB+ BO=AD + DO^pr (Euc. I. 34) AB + CD ^AD+JBO- therefore, etc.] It may at once be inferred that the maximum polygon of any order ^ of given angles and perimeter , is escribed to a circle, 3. If three common tangents Z>, E> F to three circles A, B, C taken two and two are concurrent ; prove that the conjugate triad Z) , E , F are also concurrent.* 4. Let the lines joining the middle points of the three pairs of opposite connectors, BC and AD, etc., of four points A, /?, (7, D be A, /A, v ; prove by means of the following evident formulae, cosoc, (1) cosoc, (2) (3) the relations, 1. 2( /A 2 + v2) 2. 4(X2 + /jt 2 3. 4A 2 = 6 2 + with similar expressions for //, and v : 4. ^2 _ V 2 = ^ cos c v 2 - A 2 = - 6d cos & ; A 2 * Catalan s Theoremes et Problemes de Geometric Elemeiitaire, pp. 53, 54 (1879). 136 COLLINEAR POINTS. 5. 4(area of quadrilateral) = (6 2 + d 2 - c 2 - 2 )tan 88 . 5a. Hence, or otherwise construct a quadrilateral, having given its four sides and area. 6. To find the cosine of the angle between any pair of opposite connectors. [Equate the values of A 2 , /* 2 , v 2 in 3 with those of (1), (2), (3). 7. If any point D be joined to the vertices of a triangle ABC; the area of the triangle formed by joining the orthocentres of BCD, CD A, DAB is equal to ABC. [Let O lt 2 , 3 denote the orthocentres. DO^ D0 2 , D0 3 are equal to a cot .4, b cot B, ccotC respectively, and are mutually in clined at angles A, B, C ; therefore, etc.] 8. If the vertices of a quadrilateral ABCD be joined to the orthocentres 0, O l} 2 , 3 of the four triangles formed by their vertices taken in triads ; to prove that . ABCD = O l . ABCD = 2 . ABCD = 3 . ABCD. [Let the angle AOD = 6. Taking any of the anharmonic ratios of the pencil . ABCD and reducing, we obtain sin 6 sin B _b sin _bdsin. OAD_bdcosbd_v 2 - A 2 sin(B+B)siuA asm(B + 6) acsinOCD ac CQS ^ v 2 -/x 2 (Ex. 4, 4). It follows generally that the six anharmonic ratios of A 2 - u 2 u 2 - v 2 v 2 - A 2 the pencil . ABCD are ^ ",, ^ ^ 9 , -^ , and their re- A 2 - v 2 /x 2 A 2 v 2 - /x 2 ciprocals. Similarly for the remaining pencils O t . ABCD, etc. Russell.] NOTE ON PASCAL AND BRIANCHON S HEXAGONS. When two triangles ABC and A B C are in perspective, the lines A A, BB , CC are concurrent ; therefore A and A , B and B , C and C may be regarded as the opposite vertices of a Brianchon hexagon, and the centre of perspective of the two triangles is the Brianchon point of the hexagon. But in this case we have three other pairs of triangles in perspective, viz., BCA and B C A, CAB and C A B, ABC and A B C. Hence with the vertices of two triangles in perspective we can form four Brianchon hexagons having the same Brianchon NOTES. 137 point, the opposite vertices of the hexagons being in each case corresponding vertices of the two triangles. Again, if the non-corresponding sides of the triangles intersect as in figure in points X and X , Y and F , Z and Z , and the corresponding sides in UVW, UVWis the axis of perspective. But in this case we have three other pairs of triangles in perspective to the same axis, viz., those obtained by interchanging a pair of corresponding sides, e.g., if Z, J/, N and L y M\ N denote the sides of the given triangles, it is obvious that the triangles LMN and L M N, MNL and M N L, NLM and N L M have the same axis of perspective ; hence with the sides of two triangles in perspective we may form four Pascal hexagons having a common Pascal line, i.e., the axis of perspective of the triangles, the corre sponding sides of the triangles being in each case opposite sides of the hexagons. In the accompanying figure the legs of the angles whose vertices are at 7, Fand W intersect again in twelve points, viz., X, X , F, F , Z, Z , A, A , B, B, C, C", and these we have seen may be grouped in four different ways into two groups of six (XX YY ZZ \ and AA BB CC determining Pascal and Brianchon hexagons respectively ; also that the alternate sides 138 COLLISEAR POINTS. XX and YY } of the Pascal hexagon intersect (in C) in six points, which form a Brianchon hexagon. Again, since sixty Pascal hexagons may be formed from the points XX YY ZZ , and YY and ZZ meet in A, and YX and Z X in A t by taking these lines as pairs of opposite sides of one of the hexagons ( YY XZ ZX \ A A is its Pascal line; similarly BB and CC are Pascal lines of the hexagons XX YZ ZY and XX ZY YZ respectively ; but A A , BB and CC are concurrent, hence the sixty Pascal lines pass in threes through twenty points. Similarly it may be shewn that of the sixty Brianchon hexagons formed by the conjugate hexad of points AA BB CC , their Brian chon points lie in triads on twenty lines. And either property involves the other as will be seen by reciprocation with respect to a circle. CHAPTER VI. INVERSE POINTS WITH EESPECT TO A CIRCLE. Def. Two points P and Q are inverse with respect to a circle when the line PQ passes through the centre and OP . OQ = the square of the radius of the circle. For the circle of unit radius OP . OQ = 1 or OP is the inverse, or reciprocal, of OQ. 69. It appears from the definition (1) That inverse points are in the same direction from the centre when the circle is real and in opposite directions when the radius is imaginary, that is when it is of the form jRv 1. (2) They coincide on the circle ; and when the radius is not real the inverse Q of a point P at a distance OP from the centre is given by the equation OP . OQ = R 2 . (3) When either coincides with the centre the other is at infinity. 70. Theorem. If a line ABbe divided internally and externally in P and Q in the same ratio, P and Q are inverse points with respect to the circle on AB as diameter ; also A and B are inverse points with respect to the circle on PQ. For if M be the middle point of AB, by hyp., AP = AQ AM+MP_AM+MQ BP BQ hence BM-MP QM-MB 139 140 INVERSE POINTS. by taking the sum to difference on each side we have A similar proof applies to show that NA NB = NP 2 = NA Z , where N is the middle point of PQ 71 . Since MP . MQ = MN 2 - PN 2 , (Euc. II. 6) therefore (Art. 70) AM 2 = MN 2 - PN 2 , or transposing, MN* = AM*+PN* Hence for any two segments AB and PQ placed to divide each other harmonically, the square of the distance (MN) between their middle points = the sum of the squares of half the segments. EXAMPLES, 1. The distances of the points of contact of the in- and ex-circles of a triangle with the sides measured from any vertex on either of the sides passing through it are s, s - a, s - b, s - c. 2. If M denote the middle point of the base (c) of a triangle, Q the intersection with the base of the fourth common tangent to the ex-circles 1 and 2) P the foot of the perpendicular from the vertex on the base, MP . MQ = ( - [For Oflz is divided harmonically in C and Q, project 19 0% and C on base and apply Art. 70]. 3. Show also that the rectangle under the distances of the middle point of the base from the feet of the perpendicular and internal bisector of vertical angle = square on half the difference of sides. 72. Theorem. The distances of any point X on a circle from a pair of inverse points have a constant ratio. Since OQ : OX= OX : OP ; the two triangles OQX and OXP are similar (Euc. VI. 6), THEOREMS. 141 OX* i /i7 ITT A^ and (Euc. VI. 4) ,, - PX* OP therefore QX*= OQ> or /te squares of the distances of a variable point (X) on a circle from a pair of inverse points (P, Q) are as the distances of these points from the centre. COR. 1. Let X coincide with each extremity of the diameter AB containing the points, then ___ QX* QA* QB* OQ COR. 2. Given a triangle (PQX), the base (PQ), and ratio of sides, the locus of the vertex is a circle (ABX) with respect to which the extremities of the base are inverse points. COR. 3. If the ratio of sides in Cor. 2 = 1, the locus is a line bisecting the base at right angles, therefore the reflexion of a point is its inverse with respect to the line. COR. 4. From Cor. 1. AX and BX are the bisectors of the angle PXQ. COR. 5. If PX be produced to meet the circle again in X , A and B are the centres of the in- and ex-circles of the triangle QXX . (By Cor. 4.) 142 INVERSE POINTS. COR. 6. The line PQ containing a pair of inverse points bisects the angle (XQX } which any chord through either (P) subtends at the other. COR. 7. The quadrilateral OQXX is cyclic. [For OXX = PQX, but X X = X X therefore, etc, Euc. III. 21.] COR. 8. For any other pair of inverse points P , Q f on the diameter AB ; the angles PXP f and QXQ are equal or supplemental according as the pairs of points are taken in the same or opposite directions from the centre. [The angles PXQ and P XQ have in either case the same bisectors AX and EX.] RADICAL C EXT RE. 143 EXAMPLES. 1. Any circle passing through a pair of inverse points P and Q with respect to a given one cuts the latter orthogonally. [From the definition of inverse points and Euc. III. 37.] 2. To find two points P and Q which shall be inverse with respect to two given circles. [The circle passing through any point and its inverses with respect to each of the given circles meets their line of centres in the points required.] 3. The line L bisecting PQ at right angles is such that the tangents from any point on it to either of the circles in Ex. 2 are equal to OP or OQ. [For the circle with as centre and OP OQ as radius meets the given circles orthogonally : * therefore, etc.] 4. Any two pairs of inverse points are con cyclic. 5. Any chord XY of a circle passing through P is divided har monically by P and the perpendicular to PQ through Q. [For the angle XQYis bisected internally and externally by the lines at right angles.] 6. The radical axes Z, M, N of three circles taken in pairs are concurrent. [For the point (Z/, J/) of intersection of any two is the centre of the circle cutting the three given ones orthogonally.] Def. This point of concurrence is the Radical Centre of the circles, and is such that for any three secants XX , YY , ZZ drawn through it to the circles respectively OX. OX =OY. OY = OZ. OZ . The common value of these rectangles is called the Radical Product of the circles, and is equal to the square of the tangents to them when is outside the circles. (See Art. 23, Ex. 11, footnote.) * Hence the locus of a point from which tangents to two circles are equal is a right line, viz., the axis of reflexion of their common pair of inverse points. It is termed the Radical Axin of the circles, and is their chord of intersection, real or imaginary. 1 44 IN VERSE POIN TS. 7. The radical axis of two intersecting circles is their chord of intersection ; hence show that the common chords of three circles taken in pairs are concurrent. 8. Describe a circle meeting three given circles at right angles. 9. For any triangle ABC find a point such that OA : OB : OC= given ratios. 10. For any four collinear points A, B, C, D find the loci of points (1) such that the angles AOB and COD are equal, (2) BOC is supplement of AOD. 11. For any six collinear points taken in the order ABCC B A find such that the angles BOC, COA } AOB are respectively equal to B OC , C OA, A OB . [By Ex. 10.] 12. The four sides of an escribed quadrilateral ABCD being given in magnitude and AB in position ; find the locus of the centre of the in-circle. [Make AD = AD and BC = BC. Since OA, OB, OC, OD are the bisectors of the angles of the quadrilateral, it is easy to see that AOB+COD=7r. Again the triangles A0l> and AOD are equal in every respect (Euc. I. 4) ; hence LA DO A D O ; similarly LBCO = BC ; therefore by addition it follows that LC OD = COD or AOB+C OD = ir, and the required locus is a circle having A, B and C , D pairs of inverse points. Dilworth.] 13. The centres of perspective Pand Q of any two parallel chords A A and BB of a circle are inverse points with respect to the circle, and the circle touching the chords at their middle points. EXAMPLES. 145 [For we have PA=PA , PB=PB t QA = QA and QB=QB ; hence PA(QA =P/QB=etc. ; therefore, etc. The second part follows, since MN is divided harmonically by P and Q. Art. 70.] 13a. To what does the theorem reduce when A A and BB coincide ? 14. For any two pairs of inverse points P, Q and P, Q prove that PP .Pq_OP (_PA 2 _P&\ QP.QQ OQ* \ QA* BQ?r [PPQQ is a cyclic quadrilateral (Ex. 4) ; hence the triangles OPP and OQQ are similar ; so also are OPQ and OPQ ; therefore, etc. (Euc. VI. 4). Otherwise if p and q denote the perpendiculars from P and Q on OPQ , we have PP . PQ =p . D, and QP . QQ =q . D ; 15. If P, Q, R be any three collinear points on the diagonal triangle of a quadrilateral ; their harmonic conjugates PQ R with respect to the diagonals XX , YT, ZZ are also collinear. [For XX is divided harmonically in B and C (Art. 68) and P and P ; hence, by Ex. 14, BP.BP BX*^BL CL Similarly LX ). = 31^ == ^ (whereJ/]r=J/r): 6tC> 146 INVERSE POINTS. Multiplying, we have BP CQ AR BP^ CQ AR _BL CM AN # CP AQ BR CF AQ BR CL AM W but P, Q, R are in a line ; f therefore, etc.] 16. To what does Ex. 15 reduce when the line PQR is at infinity ? 17. The angles subtended by the diagonals of a complete quadri lateral at any point have a common angle of harmonic section, real or imaginary. [0 is the point of intersection of the lines PQR and P Q R in Ex. 16 ; therefore, etc.] 18. The circles on the diagonals of a complete quadrilateral pass through two points, real or imaginary. [In Ex. 17, if two of the angles XOX , TOY are right; ZOZ must also be right,^ since it is divided harmonically by PQR and P Q R .] 19. Any transversal to the pencil in Ex. 17 is cut in six points which, taken in pairs, have a common segment of harmonic section. 20. To what does Ex. 17 reduce when is at infinity ? 21. If the sides of a triangle ABC are divided harmonically in XX , YY f , ZZ ; if X, Y, Z are collinear, the middle points L, M, N of these segments are collinear. 22. If perpendiculars be let fall on the sides of a triangle from a pair of inverse points and and their feet joined ; the triangles PQR and PQR thus formed are similar and their areas are as the distances of and from the circum-centre. [For QR = AO sin A , and Q R = A O sin A, therefore QRfQ R = A OJA ; similarly EP/RT = BO I BO , etc. Art. 72.] * Hence also the middle points L t M, N oj the diagonals of a complete quadrilateral are collinear. t PQR and P Q R are termed Conjugate Lines of the quadrilateral. I Generally, For a number of angles at a common vertex having a common angle of harmonic f ection if any two are right, all the others are also right. EXAMPLES. 147 23. Through a point P in the diameter of a semi-circle draw a chord AB such that the area of the quadrilateral ASA S , where A B is the projection of AB on the diameter, may be a maximum. [Let Q be the inverse of P with respect to the circle ; draw QQ at right angles to A B . Project M the middle point of AB on A B and let X be the intersection of MM with the semi-circle on Q O. Then area S of quadrilateral ASA S = A B . MM , hence 2 =4JO/ 2 . A M *=40M . PM . M P.M Q , by Art 70, = 4PJ/ 2 . OM . M Q = 4PM 2 . M X* ; or S is equal to the area of the maximum rectangle that can be inscribed in a given circle, one of whose sides is parallel to a given line. Art. 14, Ex. 2.] 24. Six perpendiculars are drawn from the inverse of the inter section of the diagonals of a cyclic quadrilateral to the sides and diagonals. Show 1. The feet of those to the sides are collinear. 2. The line of collinearity bisects at right angles the line joining the feet of perpendiculars on the diagonals. [By method of Ex. 22.] 25. If XX ; TT ; ZZ denote the feet of the bisectors of the angles of a triangle ABC, show that the pedal triangles of two points and inverse to any of the circles on these segments as diameters, with respect to ABC, are inversely similar. (Neuberg.) [Let and be inverse with respect to ZZ C* PQR and Q PR their pedal triangles respectively. M the middle point of ZZ . 148 INVERSE POINTS. Then M0 ,_ H R P AO siu A BO sin B ,, , A . RP m = BO^B = AO^TA (by EX U) = RQ ; also the angles R and R are equal ; therefore etc. NOTE. If is on the circle ZZ C the pedal triangle is isosceles, similarly if it is the point of intersection of the circles ZZ C and YY B it is isosceles in a double aspect, i.e. equilateral. Hence we may infer that the circles AXX , BYY\ atid CZZ pass through tivo points and which are inverse (Ex. 22) with respect to the circum-circle of ABC and whose pedal triangles with respect to ABC are equilateral.] * Le cerde d Apollonius du triangle ABC par rapport &AB. V. Educ. Times, Dec., 1890. CHAPTER VII. POLES AND POLAES WITH RESPECT TO A CIRCLE. SECTION I. CONJUGATE POINTS, POLAR CIRCLE. 73. Def. The perpendicular to the line joining a pair of inverse points passing through either is the Polar of the other with respect to the circle. In the figure of Art. 74 G and Z are inverse points ; and C and the line AB are termed Pole and Polar with respect to the circle. Any point A or B on the polar is the Conjugate of (7, hence the polar of a point is the locus of its conjugates. Again, since the circle on EG as diameter passes through Z and therefore cuts the given one orthogonally : 1. The circle described on the line joining two conjugate points cuts the given circle orthogonally. 2. The distance between two conjugate points is equal to twice the length of the tangent to the circle from the middle point of the line connecting them. 74. Theorem. For any two conjugate points B and (?, to prove that each lies on the polar of the other with respect to the circle. Suppose the polar of G to be AB, we require to prove that the polar B passes through C. Join AO, draw a 149 150 POLES AND POLARS. perpendicular to it CX. Then evidently (Euc. III. 36) OA.OX = OC.OZ=r*\ hence CX is the polar of A. Thus as the point B moves along the line AB its polar turns around, or envelopes, C. At Z therefore the polar is the chord of contact of tangents through that point to the circle. EXAMPLES. 1. The extremities of any diameter of a circle which cuts a given one orthogonally are conjugate points with respect to the latter. 2. If a variable chord AB of a circle pass through a fixed point P ; the locus of the intersection of tangents to the circle at A and B is a line. [The polar of P with respect to the circle.] 3. The diameter ^17? of a circle is the polar of a point at infinity in a direction perpendicular to AB. 4. The locus of a point which has a common conjugate with respect to three circles is their common orthogonal circle. POLAR CIRCLE. 151 75. Theorem. // A and B be any two points and L and M their polars ivith respect to a circle, the point LM is the pole of the line AB. For LM is conjugate to both A and B, hence the line joining A and B is its polar (Art. 73), or " the line of connexion of any tiuo points is the polar of the point of intersection of the polars of the points." Townsend. 76. More generally for three points A, B, C and their polars L, M, N, denoting the points MN, NL, LM by A, B y C respectively ; we see as above that A , B , C are the poles of BC, CA, AB ; hence, for any two triangles if the vertices of either are the poles of the corresponding sides of the other ; then, reciprocally, the vertices of the latter are the poles of the corresponding sides of the former. Def. Such triangles are said to be Reciprocal Polars with respect to the circle. 77. In the particular case when ABC and AB C coin cide, the triangle is Self -Reciprocal with respect to the circle. It is manifest, since each vertex is the pole of the opposite side, every two of its vertices are conjugate points ; and the triangle is therefore termed Self -Conju gate with respect to the circle. Its centre coincides with the orthocentre of ABC and the square of its radius (p) is given by P 2 = OA . OX=OB.OY=OC.OZ, where X, Y, Z are the feet of the perpendiculars of the triangle. This circle is called the Polar Circle of the triangle. NOTE. In order that the polar circle may be real, the pairs of points A and Jf, B and Y, C and Z, which are inverse with respect to it, must lie in the same direction from its centre 0. It is therefore 152 POLES AND POLARS. real when the triangle is obtuse angled, and imaginary for acute angled triangles. 78. Expressions for the Radius ( p ) of the Polar Circle. 08) Let be the ortho-centre of ABC, then it appears that A, B, C are the ortho-centres of BOG, CO A, and AOB respectively. For this reason the four points A, B, C, are said to form an Orthocentric System. Also the circum-circles of the four triangles BOG, GO A, AOB, and ABC are equal. Hence since a and AO, chords of equal circles, subtend complementary angles at the circumferences, a?+A0 2 =tf+B0 2 = c 2 +C0 2 =d; 2 , ............ (1) also (fig. 0) a 2 = 2 + C0 2 +2C0 . OZ, (Euc. II 13) therefore by substitution from (1) or -(70.0^=d 2 -i(a 2 + & 2 +c 2 ) = p 2 ............ (2) This formula is equivalent to /o 2 = c 2 cos A cos B cos C, by reduction or independently, as follows : -p* = OC. OZ=OG. OA OB =d%osA cos B cos Q... (3) d since a chord is equal to the diameter of the circle into the sine of the angle it subtends. EXAMPLES. 153 EXAMPLES. 1. The four polar circles of the triangles BOG, CO A, AOB and ABC are mutually orthogonal.* [Let their radii be p a , p b , p c , p. Since their centres are at A, B, C t 0, by Euc. II. 2, AB*=AB . AZ+AB . BZ=p a ?+p l> 2 ; therefore, etc.] 2. jB and C, C and J, A and .5 are pairs of conjugate points with respect to the polar circles of BOC, CO A and AOB respectively. 3. The square of the distance BC betw r een any two conjugate points is equal to the sum of the squares of the tangents drawn from them to the circle. [By Ex. 1 the tangents from B and C to the circle p a are the radii of p b and p c , but BC* = p b 2 + p c * , therefore, etc.] 4. Prove that AZ. EZ=t\ where t is the tangent to the polar circle from Z, the Polar Centre t of A B ; and conversely. [By similar triangles ACZ&nd OBZ, AC : CZ=OZ:BZ, etc.] 5. Conjugate points A and B with respect to any chord MN are conjugate with respect to the circle. [For the polar centre Z of AB is the middle point of MN ; but (hyp.) ZA . ZB=ZM*= -ZM.ZN or the square of the imaginary tangent from Z to the circle ; therefore, etc., by Ex. 4.] 6. If a number of circles have a common orthogonal circle, the extremities of any diameter of the latter are conjugates with respect to the entire system. 7. On a given line find two points which shall be conjugates to each of two given circles. [The middle point of the required segment is such that the tangents from it to the circles are equal ; therefore, etc., by Art. 72, Ex. 3.] * Hence : If four circles are mutually orthogonal, their centres form an orthocentric system and one of the circles is imaginary. + Z the foot of the perpendicular from the centre on A B is also called the Middle Point of the line. (Cf. Euc. III. 3.) OF TKK TT "NT T"VT TT-D O 154 POLES AND POLARS. 8. On a given circle find two points A and B which shall be conjugates to each of the circles (7, ?\ ; Z), r. [The middle point M of the required chord is on the radical axis L of the given circles (Art. 72, Ex. 3). Let 2 be the length of AB ; then CM****+i$rt+;AM*r-+rt-OM* ; hence CM 2 + OAT 3 is known, and the triangle COM is completely determined ; therefore, etc.] 9. Place a chord of given length in a circle so that its extremities may be conjugates with respect to another. [See Ex. 8.] 10. If a right line AB meet either ((7, r) of two circles in con jugate points (A t B} with respect to the other ; then reciprocally it meets the latter ((7, /) in conjugate points (A and B ) with respect to the former. [For by Ex. 5 AB divides A B harmonically, hence A B divides AB harmonically ; therefore, etc.] 11. Find the locus of the middle points M and N of the chords A Band A B in Ex. 10. [CM 2 + C M 2 = (Art. 71.) = const. ; hence the required locus is a circle whose centre is at the middle point of CC and the square of whose radius is equal to SALMON S THEOREM. 155 where 2& = CC . It evidently passes through the intersections of the given circles.] 12. Show that CM.C N= const. [Draw CX at right angles to C N. Join OX. Since OC X is an isosceles triangle and N a point in the base produced, CM. C N=--C N.NX=ON*-OX*=ON*-OC 12 where 6 is the angle between the given circles ; therefore, etc.] 13. Any circle described around the polar centre of a triangle ABC meets the corresponding sides of the median triangle in A , 2?, C such that AA = BB = CC . 14. A tangent is drawn from the polar centre to the circum- circle, and from the point of contact a tangent is drawn to the polar circle, show that the angle between these lines is 45. 15. Draw through P a line cutting each of two given circles in conjugate points with respect to the other. [By Exs. 10 and 11.] 16. Draw a line cutting each of two circles X and Fin conjugate points with respect to a third (Z). [Let the required line meet Z in the points A and B. The middle point M of AB is the intersection of two known circles passing through the intersections of Z find JTand Zand F(Ex. 11), and is thus determined ; therefore, etc.] SECTION II. 79. Salmon s Theorem. The distances of any two points A and B from the centre of a circle are pro- portional to the distances AM and BL of each from the polar of the other. Draw AB and BA perpendiculars to OB and OA respectively. 156 POLES AND POLARS. Then OA . OL=OB.OM=r, and since AA BB is a cyclic quadrilateral, OA . OA =OB . OB ; therefore OA _ OB _ OM_ OM- OB _ B M_ AM . OB~OA ~OL~OL-OA ~AL~BL therefore, etc. By alternation OA/AM=OB/BL. COR. 1. If if is a fixed line and OA/AM a constant ratio, B is a fixed point and the envelope of L is a circle ; or, the pole of a variable tangent to a circle ivith respect to another given circle is such that its distance from the centre of the latter bears a fixed ratio to the distance from a fixed line. COR. 2. If A and B are both on the circle (0, r) ; OA = OB } and therefore AM=BL\ or, the points of con tact of tangents to a circle are equidistant from the tangents as is otherwise evident (Euc. I. 26). COR. 3. Let B and its polar M vary and the different positions be denoted by B v B 2 , B y ..., M v M 2> M s , ...; then since OAOB OAOB OA OB 2 ^ SALMON S THEOREM. 157 by multiplication of ratios, we have OA n OB. AM.AM l . AM 2 ...~BL .B^.B^L... or, the product of the distances of a point (A) from any number of lines (M) is to the product of the distances of their poles (B) from the polar (L) of the point as the n ih power of the distance of the point from the centre is to the product of the distances of the poles from the centre. COR. 4. If M, M v M 2 in Cor. 3 form an inscribed polygon, B, B v B 2 , ... are the vertices of the correspond ing escribed one ; hence the product of the distances of any point from the sides of an in-polygon is to the product of the distances of the vertices of the correspond ing ex-polygon from the polar of the point as the n ih power of the distance of the latter from the centre is to the product of the distances of the vertices of the ex- polygon from the centre. COR. 5. The rectangle under the distances of the extremities of any chord from a tangent is equal to the square of the distance of its point of contact from the chord. EXAMPLES. 1. The opposite vertices of an escribed quadrilateral are AA , BB , CO ; to prove that OA . OA : OB. OB : 00. OC = AX. A X : EX. B X : CX . C X, where X is a tangent to the circle at any point P. [Let the corresponding pairs of sides of the in-quadrilateral be Z, L ; M, M ; N, N ; then since OAOP , OA OP multiplying these equations, but PL . PL = PM . PM = PN . PN ; therefore, etc.] 158 POLES AND POLARS. 2. If a, (3, y denote the perpendiculars from any point on the circum-circle on the sides of an in-triangle, fty sin A + y a sin B + a/3 sin C= a ++- c =o. a (3 y 3. If A, fi,, v be the perpendiculars from the vertices of any triangle upon a variable tangent to the in-circle, cot^J. cot^ff cot| C_ Q A jj, v [Let A , ff, C , P be the points of contact with the sides and any tangent, then -r = , where a is the perpendicular from P on B C , Hence 2-r2..0; (Ex.2) A a but 0^ . W = 2r 2 cotJ^l ; substituting, we have A particular case of this has been noticed in Art. 55, Ex. 8.] 4. If the perpendiculars from the vertices on any tangent to the circum-circle of a triangle be A, /A, v ; to prove that [If P be the point of contact of the tangent to the circle, by Ptolemy s Theorem, a.AP+b.P+c.CP=0, but AP 2 =2r\, etc., hence 5. For any point P on the in-circle whose distances from the sides are a, /?, y ; to prove that cos | AJa + cos J/V/3 + cos \C*Jy = 0. [Let A , //, v be the distances of the points of contact A t B t C of the sides of ABC from the tangent at P; a , /? , y the distances of P from the sides of A B C . By Ex. 4, 2aV A =0 or 2-^ = 0, but V/?i7~=a =\//ty ; (Art. 79, Cor. 5.) *The angles of ^4 ,5 C" are respectively 90-^4, 90 -2?, 90- W ; therefore a : b : c = cos^A : cos^B : cos^C. EXAMPLES. 159 hence, on substituting, since a = 2rcos^J[, 2aVX 0=ooflJ^a, therefore, etc.] NOTE. The equations in Exs. 2 and 5 are known in Analytical Geometry to be those of the circum- and in-circles respectively, the given triangle ABC being taken as the triangle of reference. The expressions in Exs. 3 and 4 are the Tangential Equations of the In- and Circum-Circles. 6. If two triangles ABC, A B C are reciprocal polars, they are in perspective. [Let the perpendiculars from A B C on the sides of ABC be Pit PI, Pt 5 <?i> q* 3 ; n, r* r 3 respectively ; then, by Salmon s Theorem, 2^ = ?? ^=Tl OA -P i OC V OA p s OB qi multiplying these equations we have ? . 2? . ?^ = 1 . therefore, etc. (Art. 65.)] p 3 qi r, 7. A triangle inscribed in a circle is in perspective with the corresponding escribed one. [By Ex. 6.] 8. Any two triangles may be so placed that the vertices of either are the poles of the sides of the other with respect to a circle. [At the centre of the required circle the sides of each triangle subtend angles similar to those of the other triangle. Find points 160 POLES AND POLARS. satisfying these conditions with respect to each triangle and place the latter with the points coincident and AO at right angles to B C ; then OB and OC will be at right angles to C A and A B . Again, since the perpendiculars from ABC on the sides of ABC are con current, those from A B C on the sides of ABC are also concurrent ; it follows obviously that OA } OB , OC are perpendicular to the sides of ABC ; and OA. OX=OB. OY=... = OA . OX = ...=p\ 9. To find the radius p of the circle in Ex. 8. area ffO(7_ Off. 0(7 _ p 4 _ p 4 4COA . AOB AB C _ v__J _ = P 4 AEG 4 BOC . COA 4 BOG . COA . AOB -\ J Similarly, - - etc. Adding these results, we have v__J 4 BOC . CO 4 BOC. CO A . AOB. A B C (ABC? 10. The area of the reciprocal polar A B C of a given triangle with respect to a circle is given by the equation of Ex 9. 11. The minimum value of A B O is obtained when the centre coincides with the centroid of ABC ; and = P^. [In this case BOC=COA = AOB. Art. 14, Ex. 5.] 12. The reciprocal polar of the median triangle with respect to the in-circle or ex-circles of the given one is equal to ABC. 13. The reciprocal polar triangle may be of any species. [Species depends on the position of the centre 0.] 14. In Ex. 8 is one or other of two fixed points. [One of them is obviously within both triangles and the sides of each subtend at it angles equal to the supplements of the angles of the other. The other is the common intersection of the circles described externally on the sides of ABC containing angles equal to TT- A , TT-B ^ir- G . On making the figure it will be observed that these circles, intersecting in pairs at the vertices of the triangle, can only EXAMPLES. 161 meet again in one point ; hence, if a point be reflected with respect to the three sides of a triangle, the circles BC0 1} CA0 2 , AB0 3 meet in a point* 15. If the triangles ABC and A B C are similar the second centre is any point on the circum-circle of ABC ; also if P be joined to A, B, and and X, Y, Z be the middle points of these lines and Z the middle point of AB ; XYZZ is a cyclic quadrilateral for LXZT=AOB and XZ Y=APB = Tr-AOB ; hence XZ7+ XZ F= TT ; therefore Z the middle point of OP is on the nine-points-circle of ABP. Similarly it is on the nine-points-circles of the triangles with BC and AC as bases and P as vertex. Hence for any four points A t B, C, P, the nine-points-circles of three of the triangles formed by them are concurrent. It is therefore obvious that all four nine-points-circles of the four triangles BCP, CAP, ABP, ABC are concurrent.^ 16. A triangle reciprocates into a similar one from either of the Brocard points as origin. (Art. 27.) * The points and P are reciprocally related to the triangle ABC. For it will be seen that, if P be reflected with respect to the sides, the circles BCP V CAP 2 , ABP 3 will meet in 0. It follows thence that the nine points circles of the triangles BCO, CAO and A BO also pass through this point of concurrence. t Van de Berg, Mathesis, t. 2, p. 141. L 162 POLES AND POLARS. SECTION III. RECIPROCATION. 80. If ABC ... be any polygon and A B C ... another derived from it by taking the poles A , B , C", ... of the sides BC, CA, AB, etc., with respect to any circle, then we have seen (Art. 76) that the vertices A, B, C, etc., of the former are the poles of the sides of the latter, and the two polygons are said to be Reciprocal Polars with respect to the circle. The process of deriving A B C ... is termed Reciprocation, and the circle, radius, and centre are the Circle, Radius, and Centre, or Origin of Reciprocation. More generally, if A BC ... be any curve to which tangents T v T 2 , T B , ... are drawn at the points A, B,C, ..., the locus of their poles is the Reciprocal Polar Curve of ABC . . . with respect to the circle. If the tangents at A and B are indefinitely near, their poles A , B are also indefinitely near on the reciprocal curve ; but the point T^T^ is (Art. 76) the pole of the line A B ; hence in the limit the point A is the pole of the tangent at A f . The point A and tangent at A are said to correspond. Thus, of two polar reciprocal curves any tangent to either corresponds to a point on the other, and each point of contact and the corresponding tangent are pole and polar with respect to the circle. The following fundamental properties of two Recipro cal figures will appear obvious : 1. The line joining any two points of either is the RECIPROCATION. 163 polar of the intersection of the corresponding lines of the other. 2. Concurrent lines reciprocate into collinear points. 3. The angle subtended by any two points of one at the origin is equal to the angle between the correspond ing lines of the other. 4. For any two figures X and Y and their reciprocals X and Y t the points of intersection of X and Y correspond to the common tangents to X and Y ; in other words, a common tangent to two curves corresponds to a point of intersection of their reciprocals. 5. If X and Y touch, their reciprocals X and Y also touch, and each point of contact is the pole of the common tangent at the other. 6. Since two circles have four common tangents, real or imaginary, they reciprocate into curves which inter sect in four points. (By 4.) 7. Any point connected with X and the tangents through it to the curve corre spond to a line and its points of intersection with the recip rocal curve X . 8. The reciprocal of a circle is a curve of the second degree, i.e. one which meets every line in two points, real or imagin ary. (By7.) 9. The pencils determined by any four collinear points A, B, C, D at the origin 8 and the corresponding lines A . B , C , D f are similar. [For the corresponding rays of pencils are at right angles.] 164 POLES AND POLARS. 10. Harmonic rows of points reciprocate into har monic pencils of rays ; and in the particular case when one point D of the row A, B, C, D coincides with the origin 8 ; SA, SB , SC f are in arithmetical progression. 11. Parallel lines reciprocate into points collinear with the origin. 12. A point and its polar reciprocate into a line and its pole with respect to the reciprocal curve. (Cf. 7.) EECIPROCATION OF THE CIRCLE. 81. Let the origin S be outside the circle (0, ?*) ; OS =(5; L the polar of with respect to the Circle of Reciprocation, and P the pole of any tangent to the circle at Z. For the two points and P we have, by Salmon s Theorem, ^ = ^= * = const. = e (say). rL UZ r The locus of P given by the equation SPjPL = e is a Conic Section, of which S is termed a Focus, L a Directrix, and e the Eccentricity. (See Art. 79, Cor. 1.) When e > 1, the conic is called a Hyperbola, ,,6=1, Parabola, e<l, Ellipse. Thus the reciprocal polar of a circle is a hyperbola, parabola, or ellipse, according as the origin is outside, upon, or within the circle. In the particular case when the origin coincides with the centre of the given circle, the reciprocal curve is a concentric circle. Since the tangents to a circle are real and distinct from any points outside it, and reciprocate from S as RECIPROCATION. 165 origin to two points at infinity; their points of contact X and F reciprocate into two real tangents to the conic, meeting in C the correspondent of XY, whose points of contact are at infinity. These lines are termed the Asymptotes of the hyper bola. They are imaginary for the ellipse, though they intersect in a real point, and coincident with the line at infinity for the parabola. The tangents A and B at the extremities of the diameter OS correspond to points A and B called the Vertices of the conic ; also since the distances of S from A , XY, B, are in H.P., SA, SC, S.B their reciprocals are in A.P. ; hence G is the middle point of the segment AB, and it is obviously the point at which the asymptotes intersect.* * When the origin is outside the circle its polar divides the circum ference into two parts which are respectively concave and convex to it. These portions reciprocate into two distinct curves convex and con cave to the origin as shown in the figure, and both branches reach to 166 POLES AND POLARS. Also since SA , SO and SB are in A.P., their reciprocals 8 A y SL, SB respectively are in H.P. The tangents from any point K, on XY, to the circle, with JTFand KS form an harmonic pencil (Art. 78, Ex. 5) ; hence by reciprocation any line through C meets the conic in an harmonic row of points, one of which, corre sponding to the ray KS, is at infinity. Thus every chord of the conic through C is bisected. On account of this property C is termed the Centre of the curve. Again, the tangents to the circle from any point on the perpendicular through 8 to RS and the lines joining that point to R and 8 form an harmonic pencil ; hence by reciprocation any line parallel to OS meets the conic in an harmonic row of points, one of which, correspond ing to the ray through 8, is at infinity ; another, that corresponding to the ray through R, is on M the per pendicular through C to OS. It is therefore manifest that the conic is symmetrically situated with respect to this line. It is moreover symmetrical with respect to ON. These rectangular lines OM, ON through the centre C are termed the Axes of the curve. infinity. If, however, we assume in general that consecutive tangents to the circle reciprocate into consecutive points on the conic, by taking two tangents indefinitely near, one on the convex and the other on the concave part of the circle, we are led to the conclusions that the points at infinity on the opposite branches of the curves are indefinitely near, that the asymptotes are tangents at the points of coincidence, and that the hyperbola is a continuous curve. RECIPROCATION. 167 EXAMPLES. 1. A circle, any point and its polar with respect to the circle, e.g. Circle, centre and line at in finity. Circle, origin and polar of origin. Circle and inscribed polygon. Circle (or conic) and self con jugate triangle.* 2. The opposite sides of a cyclic hexagon meet in three coliinear points. (Pascal.) A conic, a line and its pole with respect to the conic. Conic, directrix and focus. Conic, line at infinity and centre of conic. Conic and escribed polygon. Conic and self conjugate tri angle. The opposite vertices of an escribed polygon connect by three concurrent lines. (Brianchon.) This result follows when the circle described about the hexagon is taken as the circle of reciprocation. In general, from any origin, the theorem of Pascal with respect to a circle reciprocates into Brianchon s property for a conic. 3. Four points on a circle sub tend at a variable point on it equianharmonic pencils. Four fixed tangents to a circle meet a variable tangent to it in equianharmonic rows ; hence, generally from any origin, the property of Euc. III. 21 becomes : A variable tangent to a conic meets four fixed tangents in rows of points which are equianharmonic ; and reciprocally four fixed points on a conic subtend equianharmonic rows at a variable fifth point on it. And again it follows conversely that, if two points connect equi- anharmonically with four others, all six lie on a conic ; hence : Any two of the hexad of points connect equianharmonically with the remaining four. This system is sometimes called an Equi anharmonic Hexagon. (Townsend, Mod. Geom. vol. II. p. 168.) 4. Concentric Circles. Conies having same focus (origin) and directrix. * If the origin is taken at one of the vertices of the triangle the cor responding side of the reciprocal triangle is therefore at infinity, and its other two sides are diameters (conjugate) of the conic. See Exs. 8, 9. 168 POLES AND POLARS. 5. Circles having a common Conies having a common focus pair of inverse points (from and centre, either point as origin). From the symmetry of the conic we infer that such a system has a second common focus ; hence : Coaxal Circles reciprocate from either of their common pair of inverse points into a system of Confocal Conies. Euc. III. 35, 36. The rectangle under the dis tances of either focus from a pair of parallel tangents is constant ; hence from symmetry \ve infer that the rectangle under the distances of the foci from any tangent is constant ; and conversely, the envelope of a variable line, the product of whose distances from two fixed points is constant, is a conic having the fixed points for foci. The locus of the intersection of rectangular tangents to a conic is a circle. (Director Circle.) The variable chord of contact 7. A chord of a circle which subtends a right angle at the origin envelopes a conic. 8. A variable chord of a circle passing through a fixed origin is divided harmonically by the point and its polar. of two parallel tangents passes through and is bisected at the centre of the conic. Def. The diameter of a conic parallel to a tangent is said to be Conjugate to that which passes through its point of contact. 9. Conjugate points with re spect to a circle (from the pole of line joining them as origin). 10. If a variable point P moves on a line through the origin, S its polar passes through Q the pole of the line with re spect to the circle ; and the tangents from P and the lines PQ and PS form an harmonic pencil. Conjugate diameters of a conic. If a variable chord of a conic moves parallel to a fixed direc tion, the harmonic conjugates of the points on it at infinity (i.e. the middle points) are collinear ; RECIPROCATION. 169 hence the locus of the middle points of any is a line. tern of parallel chords 11. Conjugate points coincide on the circle. 12. The rectangle under their distances from the middle of the line joining them is constant. 13. Euc. III. 21, 22. 14. The locus of intersection of tangents containing a given angle is a concentric circle. Their chord of contact en velopes a concentric circle. 15. If the vertex of an angle of given magnitude is on a circle, its variable chord of intersection envelopes a concentric circle. 16. If the angle is right, the chord envelopes the centre (from vertex as origin). 17. The perpendiculars of a triangle are concurrent. Each asymptote is its own conjugate. The product of the tangents of the angles made by a pair of conjugate diameters with either axis of the conic is constant. The angles subtended at a focus by either pair of opposite sides of an escribed quadrilateral are equal or supplemental. The envelope of a chord which subtends a constant angle at the focus is a conic having the same focus and directrix. The locus of the point of inter section of the tangents at the extremities is another conic having same focus and direc trix. If two points are taken on a fixed tangent so as to subtend a constant angle at the focus, the locus of the intersection of the tangents through them is a conic having same focus and directrix. The locus of intersection of rectangular tangents to a para bola is the directrix. The diagonals of a complete quadrilateral each subtend a right angle at a certain point ; or the circles on the diagonals are concurrent. 170 POLES AND POLARS. It follows, because their centres lie on a line, that they pass through a second point, the reflexion of the first with respect to the line, i.e., they are coaxal. The line joining the centre of a conic to the foot of the per pendicular from focus on any tangent is constant. 18. Having given the base and ratio of sides of a triangle, the locus of the vertex is a circle to which the extremities of the base are inverse points (origin at either). The locus of the foot of the perpendicular is called the Auxiliary Circle of the conic. The circle and conic evidently touch at the extremities of the major axis. Since the centre of a parabola is at infinity, its auxiliary circle degenerates into the tangent at the vertex. 19. Common tangents to two circles subtend right angles at either common inverse point. 20. The feet of the perpen diculars from any point on a circle on the sides of an inscribed triangle are collinear. Confocal conies cut at right angles. The perpendiculars through the vertices of a triangle, escribed to a parabola, to the lines joining them to the focus are concurrent ; in other words, the circum-circle of a triangle described about a parabola passes through the focus (cf. Ex. 18). We infer that the circum-circles of the four triangles formed by four tangents (that is any four lines whatever) meet in a point. RECIPROCATION. 171 It follows also, since any point (origin) on the circum -circle and the orthocentre are equidistant from the Simson line of the point, that the locus of the orthocentre of a variable triangle escribed to a parabola is the directrix. 21. Having given base and If the extremities of a variable vertical angle, the locus of the line, which subtends a constant vertex of the triangle is a circle. angle at a fixed point, move on (Euc. III. 21.) two fixed lines, it envelopes a conic to which these lines are tangents. It therefore cuts them equianharmonically. 22. Since inverse segments subtend similar angles at any point on the circle, the segments of a line drawn across two circles subtend similar angles at either common inverse point. 23. All circles meet in two imaginary points on the line at infinity. 24. The polars of a point with respect to a system of coaxal circles are concurrent. 25. The two points in Ex. 24 are in perpendicular directions from either common inverse point. 26. The sum of the squares of the segments of two rectangular chords of a circle is constant. The pairs of tangents to con- focal conies from any point are equally inclined. Confocal conies have pairs of imaginary common tangents passing through the foci. The poles of a line with re spect to a system of confocal conies are collinear. The locus of the poles is a line perpendicular to the given one. The sum of the squares of the reciprocals of the distances of the foci from two rectangular tangents is constant ; hence if p lt p 2 , TT^ 7r 2 denote the distances of the foci from the tangents 2 l/pi 2 = constant. 27. In Ex. 26, if the square of the radius of reciprocation is the power of the point with respect to the circle. PI" + p-2 Z + TTi 2 + 7T-2 2 = constant ; or the locus of the intersection of rectangular tangents is a con centric circle (Director Circle). 172 POLES AND POLARS. 28. From the properties of the conic, rectangular tangents, director circle, centre and line at infinity. A variable chord of a conic which subtends a right angle at any point envelopes a conic ; and the focus and directrix of the envelope are pole and polar with respect to the given conic. If the point is on the given conic the envelope reduces to a point * on the perpendicular to the tangent passing through its point of contact. (The Normal.} 29. The base EG of a triangle ABC inscribed in a circle is fixed and the origin taken at its pole. Applying the formula of Art. 79, Ex. 10, we have the area of the reciprocal triangle constant, hence : the area cut off ,by any tangent with the asymptotes is constant. And conversely, given the vertical angle in position and area of a triangle, the envelope of the base is a conic ; and the sides are divided equianharmonically by the extremities of the base. 30. Show by reciprocating from a vertex of a self conjugate triangle with respect to a circle that a. The sum of the squares of any two conjugate diameters of an ellipse is constant. /3. The difference of the squares of any two conjugate diameters of a hyperbola is constant. 31. Find by the methods of Art. 79, Exs. 3 and 4, the tan gential equations of a conic circumscribed or inscribed to the triangle of reference. * This is proved independently as follows : If two right lines are drawn at right angles through a fixed point and intercept a variable segment AB on a fixed tangent to a circle ; the locus of the intersection of tangents through A and B is a line. For it is a locus that can only meet the given tangent in one point ; therefore, etc. , by reciprocation. CHAPTER VIII. SECTION I. COAXAL CIRCLES. 82. Definitions. The Radical Axis L of two circles A, TJ and B, r 2 is the line perpendicular to AB and dividing it so that AL 2 ~ BL 2 = rf ~ r 2 2 . Cf. Art. 72, Ex. 3. It follows from the definition that L is the common chord of the circles when they intersect, and we may generalize this statement by regarding the radical axis as their chord of intersection real or imaginary. Thus all circles having a common radical axis pass through two real or two imaginary points. Such a group is termed a Coaxal System. 83. It has been seen, Art. 72, Ex. 3, that a variable circle cutting two given ones orthogonally passes through two fixed points, viz., their common pair of inverse points ; this orthogonal system is therefore coaxal ; and from their mutual relations the two groups are said to be Conjugate Coaxal Systems. It is obvious that if either set possesses real points of intersection, the other does not; also the common points of one set are the common pair of inverse points with respect to the other Art 72, Ex. 1. Since the line of centres AB bisects the common chord 173 174 COAXAL CIRCLES. MNit is the axis of reflexion of each common point with respect to the other. NOTE. If two circles are concentric their radical axis is the line at infinity ; therefore a system of concentric circles passes through two imaginary points at infinity. These are called the Circular Points. If the circles touch, their radical axis is the common tangent at the point of contact. If the circles reduce to points, the radical axis of two points is their axis of reflexion. 84. Let A, i\ ; B, r 2 ; G, r 3 ... denote the circles of a coaxal system. Then, since RADICAL CENTRE. 175 r*-r*, AL*-CI? = rf-r* t etc, we have by transposing AI?-r l * = BL*-rf = Cl*-r *=...= W ...(1) The common value of these quantities (& 2 ) is the Modulus of the system. It is positive for a non-inter secting system and negative for the intersecting or common point species. 85. It follows from Art. 84 (1) that the position of the centre C of any circle of given radius of a coaxal system is determined, and conversely. In the former case CL 2 = AL 2 r l 2 + r s z = Si known quantity. Two values of CL equal in magnitude but of opposite signs are thus found. Hence the reflexion of every circle of the system with respect to the radical axis is also a circle of the system. The radical axis is therefore the line around which the entire group is symmetrically disposed. 86. The radical axes of three circles taken in pairs are concurrent (Art. 72, Ex. 6). In the particular case when their centres are collinear the axes are parallel, and the point of concurrence (Radical Centre) is at infinity. If the circles are coaxal the radical axes coincide and the tangents from any point on this line to the three circles are therefore equal. Conversely, if three circles whose centres are collinear have a radical centre not at infinity they form a coaxal system. 87. Limiting Values of the Radius given by the equa tion AL 2 r x 2 = const. Since AL z r^ is constant, AL and r x increase and diminish in value together ; or according as the centre 176 COAXAL CIRCLES. approaches to or recedes from the radical axis, the radius diminishes or increases. It follows in the limit when is at infinity that the circle loses its curvature, and a portion of it coincides with the radical axis. The remainder being at infinity is the line at infinity; hence we regard the line at infinity, and the radical axis, together as forming the circle oj the system whose radius is infinitely great* Again, since AL* - r^ = CL 2 - r* if r 3 = 0, CL 2 = AL 2 -r* ........................ (1) The two values of CL in this equation determine there fore the positions of the centres of the circles of infinitely small radii. These are the Points or evanescent Circles of the group, and are termed the Limiting Points. By(l) r 1 * = AL*-CL* = (AL- where C is the reflexion of C with respect to the radical axis ; therefore the limiting points are the common pair of inverse points of the coaxal system. (Cf. Art. 72, Ex. 1.) Hence the radical axis of a circle and point is the axis of reflexion of the point and its inverse with respect to the circle. 88. Theorems. I. The radical axis of a coaxal system is the locus of a point the tangents from which to the circles are equal. Let the tangents from P be t : and f 2 . * Since two circles meet on their radical axis, we infer that any two circles pass through two imaginary points on the line at infinity. Also, because every two circles intersect on this line, therefore all circles pass through the same two imaginary points, i.e. the Circular Points at Infinity. THEOREMS. 177 Then t* hence, by subtraction, <i 2 ~ ** = PA 2 - P & - Oi* ~ V) = 5 ( Art - 82 ) therefore, etc. II. More generally, The difference of the squares of the tangents (t^ ~ t 2 2 ) from any point P to two circles = twice the rectangle under the distance between their centres and the distance of P from their radical axis ; or For, draw PP perpendicular to AB and take M the middle point of AB. Then t* = AP 2 - rf, and 2 2 = BP* - r 2 z ; hence tf -t* = AP*- BP* - (r^ - r) = AP 2 - J5P /2 - (A L 2 - BL Z ) (Euc. I. 47) = 2AB.P M + 2AB.ML-, (Euc. H. 5 or 6) therefore t* - 1* = 2 A B . PL. COR. 1. If P be any point on one of the circles (B, r 2 ), t 2 = 0, and t* = 2AB . PL, or t^ oc PL ; or, if the square of the tangent from a variable point to a given circle varies as its distance-from a fixed line, 178 COAXAL CIRCLES. the locus of the point is a circle coaxal with the given circle and line. COR, 2. More generally, if C be the centre of a circle coaxal with A and B passing through P, ^ and t 2 the tangents from P, we have, by Cor. 1, t* = 2AC.PL (1) and t* = ZBC.PL (2); dividing (1) by (2), we have hence the locus of point such that the ratio of the tangents drawn from it to two circles is constant is a coaxal circle ivhose centre is determined by (3). COR. 3. The common tangents to tivo circles each sub tend right angles at the limiting points. For, if M\)Q a limiting point, XY one of the common tangents, and L its intersection with the radical axis, LX = LY = LM\ therefore, etc. COR. 4. If a variable chord XY of a circle be divided at P such that PZ.PFocP^/ 2 , where M is a fixed point ; the locus of P is a circle coaxal with the given circle and point. The line PM is the tangent from P to the limiting point M ; therefore, etc. EXAMPLES. 1. If a variable chord (AB) of a circle (0, r) subtend a right angle at a fixed point (M), the loci a. of its middle point N ; /3. of N the foot of the perpendicular on it from M ; y. of the pole P of A B are circles each coaxal with the given point and circle. EXAMPLES. [To prove a and /3 ; we have 179 hence N and N lie on the same circle coaxal with M and 0, r, whose centre bisects internally the interval OM, by Cors. 2 and 4. To prove y. Since JV describes a circle, its inverse P describes a circle coaxal with 0, r and the locus of N. For the locus of P is a circle ; and it is coaxal with the other two, because the three circles have a common pair of points real or imaginary.] 2. The orthocentre of a triangle is the radical centre of the circles described on the sides as diameters ; and the common value (Art. 77) of the rectangles under the segments of the perpendiculars is the radical product of the point with respect to the circles. 3. The middle points of the four common tangents to two circles the collinear. [Each point of bisection is on the radical axis.] 180 COAXAL CIRCLES. 4. Find the radical centre and product of the ex-circles of a given triangle. [The middle point of the base is the middle point of the common tangent to the two circles which touch the base externally ; there fore the line through it parallel to the internal bisector of the vertical angle, i.e. at right angles to their line of centres, is their radical axis. Similarly for each of the remaining pairs. Hence the radical centre is the in-centre of the median triangle ; and, generally, the ex-centres of the median triangle are the radical centres of the three triads of circles formed by taking the in-circle and two ex-circles of the original triangle. For the values of the radical products, see Art. 48, Ex. 1.] 5. The circum-centre of a triangle is the radical centre of any three coaxal systems which have B and C, C and A, A and B for limiting points. 6. The extremities of any two secants to two given circles which intersect on their radical axis are concyclic. 7. Any circle P, R cutting two circles A, r ; B, r% at angles a and ft meets the radical axis at an angle 9 given by the equation cos = r i cosa - r - cos ff [Denote the secants by PXX and PYY . Applying the formula ?,2 _ t? = 2.1 B . PL, we have = R( XX - YY } = 27?(ncos a - r 2 cos (3) ; hence M^coBa-r^g But PZ//? = the cosine of the angle in the segment of P, R made by the intercept on the radical axis ; therefore, etc.] 8. The axis of perspective of ABC and its pedal triangle is the radical axis of the circum- and nine-points-circles. [By Art. 88, I. and Euc. III. 36.] 8a. The line joining the orthocentre and circum-centre is at right angles to the axis of perspective of ABC smd the pedal triangle. [It is the line of centres of the circum- and nine-points-circles.] EXAMPLES. 181 9. Two circles touch at M and a chord AB of either touches the other at F ; prove that PMis a bisector of the angle A MB. [By Art. 88, Cor. 2, AP/AM=BPfBM.] 10. For any cyclic quadrilateral whose diagonals intersect in J/ ; prove that, if the bisectors of the angles between the diagonals meet the four sides in X, Y, X , 7 , AL.BL.CL. DL = XL . YL . X L . Y L, where L is the radical axis of the circle and point. 11. If //, J/, N denote the radical axes of three pairs of circles X and A, X and B, X and C, and L , M , Af the radical axes of Y and A t Y and B, Y and C ; to prove that the two triangles LMN and L M N are in perspective ; and that the centre of perspective is the radical centre of A , B and C ; and their axis of perspective the radical axis of X and Y. [For MN is a point on the radical axis of B and C (Art. 72, Ex. 6) ; similarly M N a vertex of the triangle L M N is on the same line ; therefore, etc.] 12. If three lines A X, BY, CZ be drawn from the vertices of a triangle to the opposite sides ; the radical centre of the circles on these lines as diameters is the orthocentre and their common orthogonal circle the polar circle of the triangle. [The perpendiculars of the triangle are respectively chords of these circles ; therefore, etc. Art. 77.] 13. For any three circles A, B, C and three others taken with them such that 5, (7, X \ C, A, Y; A, B, Z form three coaxal systems ; to prove that, 1, the system of six circles have the same radical 182 COAXAL CIRCLES. centre and product ; and, 2, if the centres of Jf, 7", Z are collinear, these circles are coaxal. [In 1 the radical centre and product is obviously that of the circles A, JB, C ; 2 follows at once since, if the circles be not coaxal, their radical centre is at infinity. Art. 86.] 14. Two coaxal systems have a common circle ; find the locus of the points of contact of the circles which touch. [Let L and Z , the radical axes of the systems, meet at 1\ and I 7 be one point of contact. The common tangent at T passes through P, and PT is the radius of the common orthogonal circle of the two systems, which is therefore the required locus.] 15. The radical axis of any two circles bisects the distance be tween the polars of the centre of each circle with respect to the other. *16. Three circles are described each touching two sides of a triangle and the circum-circle internally in points L, J/, and N ; to prove that the triangles ABC and LMN are in perspective. [Let one of the circles touch the sides a and b in the points P and Q and the circum-circle in ^V. Then N being a limiting point of the two circles AQ 2 /AN* = BP*IBN 2 = (R-p)/R, where p is the radius of the inner circle; but AQ=b- CQ = b - ab/s, Art. 6, Ex. 3; similarly, BP a- ab/s; substituting these values and reducing we get =lZ^/ll&. Also, Ai\ 7 jBjV=the ratio of the perpen diculars from N on the sides b and a respectively. (Euc. III. 22.) Similarly, the ratios of the perpendiculars from L and M on the corresponding pairs of sides of ABC are il-2/lZJf? and ln^/lll?; b I c c I a therefore, etc., by Art. 65.] *17. If circles are described as in Ex. 16 touching the circum- circle externally in points L ,M , N\ the triangles ABC and L M N are in perspective. *18. The centres of perspective in Exs. 16 and 17 are respectively the isogonal conjugates of the centres of perspective of ABC and * Professor de Longchamps, Educ. Times, July, 1800. FEUERBACH S THEOREM. 183 the triangle formed by joining the internal points of contact of the escribed circles with the sides (point de Nagel} ; and of ABC and the triangle formed by joining the points of contact of the in- circle with the sides (point de Gorgonne). [Make use of the property given in Art. 64, Ex. 3.] 19. The nine-points-circle of a triangle touches the in- and three ex-circles. [Let ABC be the triangle, and / the centres of the circum- and iii-circles, PP the common diameter, ATiZ aiid X Y Z the Simson lines of P and P t R their point of intersection, L, M, j\ the middle points of the sides, L , M , N the points of contact of the in-circle. Since OP= OP , NZ= NZ . But the Simson lines of two points diametrically opposite meet at R at right angles on the nine points circle ; therefore NZ = NZ = NR. Again, OP 1 01 = NZJNN ^NRINN i therefore NRJNN = MR/MM = LRJLL ; hence it follows that R is a limiting point of the in-circle and the circuni- circle of the triangle LMN. See Art. 83 Note. This elegant proof of the well-known property is due to M Cay.] 20. A variable circle 0, p touches two circles A, r\ ; B, r-> ; prove that the polar M of its centre with respect to either (A, r) envelopes a fixed circle. [Since it touches the two circles, it cuts their radical axis L at a constant angle (Art. 88, Ex. 7), or pjOL = const. Draw a parallel 184 COAXAL CIRCLES. L to L such that p/OL=r 1 /LL ) then each of these ratios =AO/OL . Let P be the pole of L with respect to A, TI ; by Salmon s Theorem, we have AOjOL =AP/PM, therefore PMis constant, and the envelope of M is the circle described with P as centre and PM as radius.] NOTE. If four positions O v 2 . 3 , 4 of the centre and their corresponding polars M v J/ 2 , J/ 3 , M are taken ; since the anharmonic ratios made by the four tangents on any variable one M is constant, therefore (Art. 80, 9), the envelope circle reciprocates into a curve of such a nature that the anharmonic ratios of the pencils joining four fixed points on it to a variable fifth are equal. This we have seen Art. 81, Ex. 3, to be a conic section ; and the ratio AOjOL is the eccentricity, A the focus, and L the directrix of the conic. 89. Theorem. A straight line is drawn to meet two circles A, r x ; B y r% in points X, X and Y, Y respectively, to prove that the tangents at these points intersect in four points P, Q, R, S which lie on a circle coaxal with the given ones. Let a and 6 be the angles of intersection of the line with the circles. Then sin a/sin /3 = PY IPX = QY/QX = RY/RX therefore, since the ratios of the tangents each of the points P, Q, R, S to the given circles are THEOREM. 185 equal, they lie on a coaxal circle, whose centre C is given A C* cnn2/3 / 2 by the relation = ^f = X (Art. 88, Cor. 2) COR. 1. Since sin a = XX /2r 1 and sin ft = FF /2r 2 , we have by division tjt. 2 = sin 0/sin a = TY /XX +rJ^ ; ........... (1) therefore, ^/ ^e intercepts made by two fixed circles on a variable line are in a constant ratio (XX IYY ), the tangents at the points of intersection meet on a fixed circle coaxal with the given ones. COR. 2. If the intercepts in Cor. 1 have the ratio of the radii t l = t 2 , a = /3, C is at infinity, and the locus of the intersection of the tangents is the radical axis. COR. 3. If the intercepts are in the sub-duplicate ratio of the radii XX" 2 /YY 2 = r,/r 2 , then * The two points Ci and C 2 satisfying this relation are easily seen to be the points of intersection of the direct and transverse common tangents to the two circles and are called their Centres of Similitude. The corresponding coaxal circles are the External and Internal Circles of Anti- similitude of the two given ones. 186 COAXAL CIRCLES. hence the circle coaxal with two given ones whose centre divides the distance between their centres in the ratio of the radii is the locus of a point, the tangents from which to the given circles are in the sub-duplicate ratio of the radii. COR. 4. If the intercepts are equal, XX =YY, the tangents are in the ratio of the radii and the locus of their intersection is called the Circle of Similitude of the given ones ; its centre is given by the equation rf/r* .................. (Cor. 1.) (1) COR. 5. Since AB is divided internally and externally in <7 2 and G 1 such that ^^~ = -^f = and again in C, **V| -^^2 ^*2 by Cor. 4, such that -0^ = -% it follows (Art. 70) that ti\J Tn C is the middle point of the segment C-f/% and that the circle of similitude is the circle on it as diameter. COR. 6. If the line XX YY passes through the inter sections (QS, PR and PS, QR) of opposite connectors of the quadrilateral; when PQ and RS are parallel; the circles A and B reduce to points and are therefore the limiting points of the system ; i.e. the common pair of inverse points of the circum-circle of the trapezium PQRS and that touching the parallel sides at Z and Z . (Art. 72, Ex. 13.) EXAMPLES. 1. Any line meeting a pair of opposite sides of a cyclic quadri lateral at equal angles makes equal angles with each of the remaining pairs (Euc. III. 21, 22) ; intersects them in points XX , YY } ZZ such that the circles touching the pairs of opposite con- PONCELETS THEOREM. 187 nectors at these points are coaxal with the given one ; and one of them lies on the side of the radical axis opposite to the other two. 2. A variable quadrilateral inscribed in a circle moves so that a pair of opposites envelope a circle, then each of the remaining pairs of opposites always touch circles coaxal with the given ones. 3. A variable triangle ABC is inscribed in a circle of a coaxal system, and two of its sides each envelope a circle of the system ; to prove that the third side AC envelopes another. [Let A B C be any other position of the given triangle. Then ABA B is a cyclic quadrilateral, and one pair of opposites AB and A B touch a given circle, therefore AA and BB touch one circle of the system. Similarly BB and CC touch one circle of the system. But BB can touch only one circle of the group on either side of the radical axis, Art. 92, Ex. 6 ; hence AA\ BB , CC touch the same circle. Now consider the quadrilateral AA CC ; it is obvious by Ex. 2 that AC and AC touch one circle ; therefore the envelope of AC is a coaxal circle.*] 4. Poncelet s Theorem. If a variable polygon inscribed in a circle of a coaxal system moves so that all the sides but one touch fixed circles of the system, the last side also touches in every position a fixed circle of the system. [By Ex. 3.] * Dr. Hart, Quarterly Journal, Vol. II. p. 148. 188 COAXAL CIRCLES. 5. The problem " to describe a polygon having all its vertices on a given circle and all its sides touching another " is either impossible or indeterminate. [Let all the circles in Ex. 4 touching all the sides but one of the polygon coincide ; it follows therefore that if the last side touches this circle in one position it touches it in every position.] 6. To find the relation connecting the radii 1\ and r^ of two circles and the distance 8 between their centres so that a quadrilateral may be inscribed to one and circumscribed to the other. (Art. 88, Ex. 1.) [By Ex. 5, when this is possible the position of the quadrilateral is indeterminate. Assuming it to have the position of symmetry, i.e., with a pair of opposite vertices at the extremities of the common diameter, and 6 the angle between any side and this dia meter. By right-angled triangles we have the relations -^\ = sin and - = cos r. 2 -8 squaring and adding these results 7. If A, 7*1, B, r- 2 , (7, r s be three coaxal circles such that a variable quadrilateral whose pairs of opposite sides envelope A and B is inscribed in (7, prove that (?*3- where 6\ and 8- 2 denote the distances AC and BC. [By the method of Ex. 6.] 8. If a variable line L meet two circles Ar^ Br so that the chords intercepted, 2c and 2c are in a constant ratio K ; to show that two points A , B may be found on the line AB to satisfy the relation A L. B L = const. [For hence or (AL + KBL)(AL-KBL) = const, but THEOREMS. 189 arid A L - *BL = ( 1 - *) B L, where A and J5 divide the line AB internally and externally in the ratio K : 1.] NOTE. The variable line in the present article is thus seen to envelope a conic of which the points A and B are the foci. 90. We have seen, Art. 86, that in general three circles have bat one common orthogonal circle, and in the particular case when more than one can be drawn the three form a coaxal system. This property is sometimes of use in determining whether circles are coaxal, and may be regarded as a criterion of coaxality. The following illustrations are due to Walker. 91. Let ABC be a triangle and XYZ any transversal to its sides. Join AX, BY, CZ. These lines are drawn from the vertices of each of the four triangles AYZ, BZX, CXY, ABC, and terminated by the opposite sides ; therefore, Art. 88, Ex. 12, the orthocentres of the four triangles are each the radical centre of the circles de scribed on AX, BY, CZ as diameters. Hence we have the following theorems : 1. The orthocentres of the four triangles formed by any four lines are collinear. 190 COAXAL CIRCLES. 2. The middle points of the diagonals AX, BY, CZ of a complete quadrilateral are collinear. 3. The line of collinearity of the orthocentres is at right angles to the line in 2, called the Diagonal Line of the Quadrilateral. 4. The circles on the three diagonals as diameters are coaxal. 5. The polar circles of the four triangles belong to the conjugate coaxal system. EXAMPLES. 1. A, B, 0) D are the vertices of a convex quadrilateral taken in order; A e , B e , C ey D e and A h B t , d, D* the external and internal bisectors of the angles ; prove that a. The sixteen centres of the circles touching the sides of the four triangles formed by taking the sides of the quadrilateral in triads, lie in fours on these bisectors. /2. The following groups of quadrilaterals are cyclic : A e B> C t D e( } A t B t C e D e \ ( . y. Groups (a) and (c) are coaxal, and groups (b) and (d) con- jugately coaxal. [These properties are proved by employing Euc. III. 32 to show that any circle of either group is cut orthogonally by any circle of the other group. Eussell.] 2*. A, B, C, D are four points on a circle. Omitting each point in turn we have four triangles ; prove that the sixteen centres of the circles touching the sides of these triangles lie in fours on four parallel lines, and also in fours on four lines each perpendicular * Educational Times, Reprint Vol. LI. p. 65. CRITERION. to the former set ; and that the two sets of lines are parallel to the bisectors of the angle between AC and BD. (M Cay.) 3. ABC is a triangle, A A a, diameter of the circnm-circle and If the orthocentre ; show that A and H are equidistant from the base EC\ and hence deduce the theorem "the Simson line of any point is equidistant from the point and orthocentre of the triangle." SECTION II. ADDITIONAL CRITERIA OF COAXAL CIRCLES. 92. I. Relation connecting ike distances between the centres and the radii of three circles of a coaxal system. Let the circles be denoted by A, r^ ; B, r 2 ; (7, r 3 . Then for any point P on the radical axis, we have hence if t be the length of the tangent from P to the circles, since AP z = rf + t 2 , etc., by substituting in this equation and reducing, BC.r* + CA.r*+AB.r*=-BC.CA.AB, ..... (1) a result from which the radius r 3 of any circle of the system may be found when the position of its centre is known ; and conversely. COR. 1. If r s = 0, G is a limiting point (Art. 87), by letting AGx in (1) we obtain a quadratic in x, the last term of which is r*. Hence the product of the distances of the limiting points from the centre of any circle of the system = the square of its radius. Cf. Art. 87. COR. 2. If r. = r = 0, the criterion reduces to 192 COAXAL CIRCLES. EXAMPLES. 1. If tj t t-2, t 3 denote the tangents from any point P to three circles of a coaxal system ; to prove that SC.tJ+CA . t.? + AB.t = Q. [For J3C.AP 2 +CA.BP 2 + AB.CP 2 =-BC. CA.AB, (1) and BC.rJ+CA.r + AB,r=-BC.CA.AB (2) Subtracting (2) from (1) ; therefore, etc.] 2. Deduce as a particular case of Ex. 1 the theorem : The locus of a point, the tangents from which to two given circles are in a constant ratio, is a coaxal circle. [Let * 3 = 0.] 3. Explain the formula of Ex. 1 when ^ 2 = ^ 3 = 0. 4. Find the locus of a point P if the product of the tangents from it to two circles bears a constant ratio to the square of the tangent to any circle coaxal with them (ktit<2 = t^}. [In Ex. 1, substituting the given condition, the equation reduces to the form (ti-mt- 2 )(ti-nt- 2 } = ; hence P describes two coaxal circles, since the ratio of the tangents t\ and t-2 m, or ??.] 5. If the common tangent ZZ to two circles meet a coaxal circle in the points A and B ; to prove that J/Z and MZ? are the bisectors of the angles subtended by the chord AB at either limiting point. [For AZ, AM and BZ t BM being pairs of tangents drawn from two points A and B on the same circle to two circles of the system, it follows that AZJAM=BZIBM, by alternation AMfBM=AZ/BZ, and for a similar reason =AZ jBZ ; therefore, etc.] EXAMPLES. 193 6. To describe two circles of a coaxal system touching a given line. [In Ex. 5 divide the line AB internally and externally in Zand Z in the given ratio AMfBM; therefore Z and Z are the required points of contact. It will be noticed that the circles lie one on each side of the radical axis.] 7. A triangle ABC is inscribed in a circle of a coaxal system ; prove that the points of contact X, X , Y, Y , Z, Z of the three pairs of circles of the system which touch the sides BC, CA, and AB respectively, a. Lie three and three on four lines, j8. Connect with the opposite vertices by six lines, passing three and three through four points. [Apply the relations in Ex. 5 to the three sides ; therefore, etc. Arts. 62 and 63.] 8. Apply the criterion of the Article to show that the nine-points-, circum- and polar circles are coaxal. 9. If points B and D are taken on any two circles whose centres are and and joined to the limiting point M such that BMD is a right angle, the locus of the intersection of tangents at B and D to the circles is a coaxal circle. [Let the line BD meet the circles again in A and C ; then MO MC 2 MB.MC AB.BD 00 AC.CD (AB .AC. BD. MA 2 _MO _ MD 2 _ MA.MD AB.AC 00 BD.CD (AB . AC. BD.CDf 194 COAXAL CIRCLES. MB.MCMO wh MA.MDMO " But since BMD=90, AMC=9Q (Art. 72, Cor. 8), and therefore B MC+ A MD = 1 80 ; hence BC _MB . MC _MO ~AD ~ by (1), a constant quantity ; therefore, etc. (cf. Art. 89, Ex. 8).] 10. A quadrilateral PQRS is inscribed to one circle and escribed to another at the points A, B, C, D ; prove that its position is indeterminate, and the diagonals PR and QS, BC and AD of the two cyclic quadrilaterals intersect (the latter at right angles) at the limiting point M. [By Art. 89, Ex. 6. See also Art. 88, Ex. 1, and Art. 67, Cor. 6.] 11. Construct a quadrilateral in a given circle symmetrical with respect to a given diameter and circumscribed to a circle having its centre at a fixed point on the diameter. [Find the radius of the second circle by Art. 89, Ex. 6.] 93. II. A variable circle cuts three others of a coaxal system at angles a, /3, y, to prove the relation BC . rjcos a + CA . r 2 cos /3-1- AB . r 3 cos y = 0. Let P, p be the variable circle meeting the given ones at the points R, S, T respectively ; join PR, PS, PT, and produce the lines to meet the circles again in R f , S , T. By Art. 92, Ex. 1, BC . t* + CA . t* + AB . t* = 0, but tf = PR . PR = p(p + RR) = p(p + S^cos a), with similar values for f 2 and 3 . Substituting these values in the equation and reducing, we obtain the required result. COR. 1. If two of the circles are cut orthogonally, every circle of the system is cut orthogonally. For if a = ft = 90, two terms of the equation vanish, therefore AB.r 3 eosy = or y = 90. COR. 2. If the variable circle touch two of the given ones, it cuts the circle C, r s coaxal with them at an angle CRITERION. 195 determined by the equation A B. r 3 cosy = BC.rCA.r 2 ; like signs being taken when the contacts are similar and unlike signs when the contacts are dissimilar. The four possible values arising from the selections of sign on the right side of the equation give the values of y correspond ing to each assigned species of contact. COR. 3. In Cor. 2, if cos y = 0, the centres C of the par ticular circles of the system which are cut at right angles are given by the relation or AC/BC=r l /r 2 . Hence, the variable circle having similar contacts with two given circles cuts at right angles the coaxal circle whose centre is their external centre of similitude ; and, if the contacts are dissimilar, the coaxal circle whose centre is the internal centre of similitude. COR. 4. If a = j8 and y = 90, the equation reduces to AC/BC= r 1 /r 2 , as in Cor. 3. Hence, the variable circle cutting two others at equal or supplemental angles cuts at right angles their external or internal circle of anti- similitude respectively. COR. 5. Let the radius of the variable circle be infinite ; hence (Cor. 3) all lines cutting two circles at equal or supplemental angles are diameters of their external or internal circles of antisimilitude. EXAMPLES. 1. To describe a circle cutting any three circles J, TI; B, r 2 ; (7, r s at given angles a, ft, y. [The required circle cutting B, r 2 ; C, r s at given angles, therefore touches a known circle coaxal with them by Cor. 2 ; similarly for each of the remaining pairs of the given circles ; hence the problem 196 COAXAL CIRCLES. reduces to " describe a circle touching three given circles with assigned contacts" There .are in consequence eight solutions. These are given in a subsequent chapter.] 2. Show that Ex. 1 cannot be reduced to describing a circle cutting three given circles orthogonally. [For let X be the circle coaxal with B and C which is cut ortho gonally by the required circle, and constructed by putting y = 90 in the relation of the present Article ; similarly let T coaxal with C and A, and ^coaxal with A and B, be circles cut orthogonally by it. Their centres, being found by the relations BX _r&o& y CF_riCOsa AZ_r. 2 cosft ~CX~ r. 2 cos ft* AY~r 3 cos y BZ~r l cosa y are collinear, Art. 62, and their common orthogonal circle therefore indeterminate.] 3. A variable circle P, p touches two others A, ri ; B, r 2 ; show that the square of the common tangent t, to it and any third circle C, 7*3 coaxal with them, varies as its radius (t 2 oc p). [By Cor. 2 it cuts C, r 3 at a constant angle y. But (Art. 4 (1)) 4 sin 2 ^y = t z lp . 7*3 ; therefore, etc. In the particular case when C, r 3 is a limiting point we have the theorem : " if a variable circle touch two fixed circles, its radius is in a constant ratio to the square of the tangent to it from either of the limiting points." Also, " the ratio of the tangents from the limiting points is constant"] 4. A variable circle cuts two fixed circles at angles a and ft, tan gents are drawn from its centre to the circles, and tangents ti and t 2 from the points of contact to the variable circle ; prove that ti 2 /t-2 2 =r l coB a/rocos ft, and deduce the properties of Ex. 3 as particular cases (Preston). See Spherical Trigonometry, Art. 159, Ex. 15. 5. Find the locus of the centre of a circle cutting any three circles at equal or supplemental angles. [By Cor. 4.] 6. The vertex and base of a triangle are fixed in position and the vertical angle given in magnitude ; find the envelope of the circum- circle. CIRCLE OF SIMILITUDE. 197 SECTION III. CIRCLE OF SIMILITUDE. 94. Let A, r x ; By r 2 be any two circles, Z and Z the points of section of AB such that AZ_AZ _r lt BZ BZ r 2 then the segments AB and ZZ divide each other harmoni cally, and the circle C, r 3 on ZZ as diameter is termed their Circle of Similitude. The points Z and Z are the Internal and External Centres of Similitude. 95. The circle of similitude has the following funda mental properties : 1. Its centre C and radius r 3 are connected by the relation CA . CB = r* (Art. 70), or the centres of the given circles are inverse points with respect to their circle of similitude. 198 COAXAL CIRCLES. 2. The points Z and Z are the intersections of the transverse and direct common tangents. 3. It is coaxal with the given circles. [For Z and Z are on the same circle coaxal with A and B, since the ratios of the tangents from them are each equal to the ratio of the radii, and only one circle coaxal with A, T! and B, r 2 can contain these points, viz. that on the line ZZ as diameter.] 4. From Cor. 3 it is the locus of a point such that the tangents drawn from it to the circle have the constant ratio of the radii. [Of. Art. 88, Cor. 2.] This follows independently, since PZ and PZ are the bisectors of the angle A PB, hence PA/PB = AZIBZ=AR/BS ; therefore, etc., by Euc. VI. 7. 5. The circles subtend equal angles at any point on it. (By 4.) 6. In the particular case when the circle B, r 2 becomes a right line the centre B is at infinity, its inverse A (Cor. 1) coincides with (7, therefore the centres of similitude of a line and circle are the extremities of the diameter of the circle perpendicular to the line. EXAMPLES. 1. The circles of similitude of any three circles taken in pairs are coaxal [Their centres are collinear, Art. 72, Ex. 21 ; therefore, etc., Art. 88, Ex. 13, 2.] 2. A circle cuts two at angles a and (3 ; find the angle it makes with their circle of similitude. EXAMPLES. 199 3. The tangents from any point P on the circle of similitude to the circles A, r and B, r 2 meet them at R and S; prove (a) the chords which the circles intercept on the line JRS&re equal to one another ; (/3) The tangents from R and S to the circles B and A are equal. [Compare Art. 89, Cor. 4.] 4. The circle on the third diagonal of a complete cyclic quadri lateral is the circle of similitude of those described on the remaining two. [Let ABCD be the quadrilateral, LMN its diagonal line, PP the third diagonal, BD= 2r l5 CA - 2r 2 , PP = 2r 3 . Join PM t PN. The triangles PAC and PBD are similar, Euc. III. 21 ; hence, since PN and PM are homologous lines, PBM&iul PCNare similar ; therefore PM\PN=r^. Similarly, P MIP N=r^ ; therefore P and P lie on a circle to which M and N are inverse points. Also the circles on the three diagonals are coaxal ; therefore, etc. It follows also by 1 that LM . LN=r 3 *.] 5. Having given the three diagonals of a cyclic quadrilateral ; to construct it. [Let be the centre of the circle and r 1} r 2 , r s the diagonals. By Ex. 4 LM. LN=r 3 z , and is therefore known. Also LM/LN^r^/rJ ; hence the lines LM&nd LN are determined. LM= rir s /r 2 , LN= rir. 2 /r 3 , and MN = ?*_( r i*- r * 2 \ But OM and ON are known (Euc. I. 47), fy*\ rir 2 / consequently the triangle OMN is completely determined.] 200 COAXAL CIRCLES. 6. Six circles pass through two points P and Q on the circum- circle of a triangle ABC and touch the sides ; prove that the points of contact JT, X ; F, Y ; Z, Z lie in threes on four lines. [Let the line joining the points P and Q cut the sides of the tri angle in L, Mj and N respectively, and we have obviously LX=LX and LB . LC=LX 2 =LX" 2 , with similar relations on the remaining sides of the triangle ; therefore, etc.] 7. From any point on a given line tangents are drawn to a circle ; a circle is described touching the fixed circle and variable pair of tangents to it ; prove that the envelope of the polar of its centre is a circle. 8. The circle of similitude of the circum- and nine-points-circle of a triangle is that described on the interval between the centroid and orthocentre as diameter. [Let be the circum-centre, H orthocentre, N the nine-points centre, and E the centroid. By a well-known property of these four collinear points OEIA 7 E=Off/NH= 2 = ica,tio of radii of circum- and nine-points-circles ; therefore, etc. [It is called the Orthocentroidal Circle of the triangle.] MISCELLANEOUS EXAMPLES. 1. Prove that the equation of the two circles touching three given ones with contacts of similar species are 23V ^+ 3lV^+ 12 v^= 0, where S lt S 2 , S 3 denote the powers of any point on either of the tangential circles with respect to the given ones. 2. If a variable chord A Z? of a circle is such that the sum of the tangents from A and B to another given circle is proportional to the length of AB, it envelopes a circle coaxal with the two. 3. If a variable circle touches two fixed circles and cuts a circle concentric with either in the points A and B : required to find the envelope of AB. (Dublin Univ. Exam. Papers, 1891.) [Applying Casey s relation between the common tangents to four MISCELLANEOUS EXAMPLES. 201 circles to the points A and B and the two given circles, it follows by Ex. 2 that the envelope of AB is a coaxal circle.] 4. Prove that the circles cutting three given ones orthogonally passing through their circles, and bisecting the circumferences are coaxal. 5. Eeciprocate the following theorem from a limiting point : The square of the distance of any point on a circle from a limiting point varies as its distance from the radical axis. [The rectangle under the distances of the foci from any tangent to a conic is constant.] 6. Prove that the limiting points of any two circles lie on a pair of opposite connections of their common escribed quadrilateral. 7. If 8 denote the distance between the limiting points and y the length of their imaginary common chord, prove that 8 iy. 8. Tf two circles whose radii are TI and r 2 are so related that a hexagon can be inscribed to one and circumscribed to the other, then + _ __ _ (ri 2 - 8 M ) 2 + 4fV? (rj - 8 2 ) 2 - 9. If an octagon can be inscribed to one and circumscribed to the other, 10. The mean centre of the vertices of a cyclic quadrilateral lies on the circumference of the nine-points-circle of the harmonic triangle of the quadrilateral. (Russell.) 11. If a variable polygon is inscribed to one circle and escribed to another, the locus of the mean centre of any number (r) of consecutive points of contact is a circle. (Weill). Cf. Art. 53, Ex. 12. 12. Prove the following extension of WeilPs theorems : If a variable polygon of any order be inscribed in a circle of a coaxal 202 COAXAL CIRCLES. system having all its sides touching respectively fixed circles of the system ; there exists a set of multiples for which the mean centre of the points of contact of the sides with the circles is a fixed point. [Let any circle of the system be denoted by (0, r, 8) where 8 is the distance of its centre from the circumcentre of the polygon, and let a, /?, y, and c be the displacements of the points of contact of the sides AB, BC, CD, etc. for consecutive positions. Then, by Art. 53, Ex. 12, we have I __ __L__ AB ~ EC ~ CD ~~ hence the mean centre of the points of contact remains fixed for the system of multiples */O\/PI, \^oV/*2, \^oVr 3 , etc.] 12a. The locus of the mean centre of r consecutive points of contact for their respective multiples is a circle. [For, join the extremities of the r sides thus forming a polygon of r + 1 sides, and let the last side touch a fixed circle (O r+ i, r r+1 , 6V+i) of the system. (Art. 89, Ex. 4.) By Ex. 12, the mean centre of ther+1 points of contact for the corresponding multiples is a fixed point (X). Let Fbe the mean centre for the r points and ,2" the point of contact of the last side. Then Y divides the line XZ in a constant ratio, and since ^describes a circle, therefore, etc.] * * The following is an independent proof of the generalization of Weill s theorem. Let A BCD .. and A B C D ... be any two positions of the variable polygon; T lt T 2 , 7 3 , 7\ , T 2 , T^ ... points of contact of the sides AB, BG, ... ; A B , B C , ... with the corresponding circles O lt r lt 6 1 ; O 2 , r 2 , 5.,, ... of the system ; R the point of intersection of AB and A B and 9 the angle between them ; S the intersection of A A and BB , and the angle between them. Then A A , BE , CC ... touch a circle (ii, />, X) coaxal with the given system. Let L, M, N ... be its points of contact with A A , BB , CC , etc. ... and we have ?J?l - r -l si ?-M = r j. R - ^2 ^ ~LM~ psinl$~ ~p ~M\~~p therefore . T^ / LM=*> . T Z T 2 / Jf#=etc. r i r 2 MISCELLANEOUS EXAMPLES. 203 13. If the diagonals of a cyclic quadrilateral are conjugate lines and a homothetic quadrilateral be described with their intersection as homothetic centre ; prove that the consecutive pairs of sides of the one quadrilateral intersect the corresponding pairs of the other in eight points which lie on a circle coaxal with the circum -circles" of the quadrilaterals. See Art. 96. [Use the theorem of Art. 92, Ex. 2.] i.e., multiples \ / 5 1 /r 1 , \ 8^Jr^ v^/r-g of the displacements T^T-^, T 2 7 2 2 . . . are proportional to the sides of the polygon ; therefore, etc. Bowesman. ] CHAPTER IX. SECTION I. Two SIMILAR FIGURES. 96. Two figures similar and similarly placed are said to be Homothetic, and their homologous parts are called Corresponding Points, Lines, etc. It is plain, if a line of either figure is displaced through an angle 9, that every line of it is displaced through the same angle. For let AB be displaced to A R. It follows (Euc. III. 21, 22), since B = B , that the angle between EC and B C is equal to 0. Also, since corresponding lines meet at equal angles, a variable pair of corresponding lines passing through a pair of corresponding points A and A intersect on the circumference of a circle described on AA containing an angle ; and conversely. 204 TWO SIMILAR FIGURES. 205 Corresponding lines are made up of corresponding points ; and the point of intersection of any two lines of either figure is the correspondent of the points of inter section of the corresponding lines of the other. 97. We have seen how to find a point S which, with the extremities of two linear segments AB and A B , forms similar triangles (Art. 25), and that it possesses the properties. a. A variable line XX dividing the segments similarly AX : BX = A X :B f X subtends a constant angle at it; and /3. Its distances from the lines are proportional to their lengths (Euc. VI. 19). Now, if similar polygons be similarly described on AB and A &, it follows, as in Euc. VI. 20, that 1. The distances of 8 from each pair of corresponding lines are proportional to these lines. 2. All pairs of corresponding points P and P of the polygons subtend the same angle at it, and with it form a triangle of constant species. 3. The polygons can be made homothetic by the revolution of either around it (2). For this reason it is called the Homothetic Centre of the Polygons, or their Centre of Similitude. The ratio of SP to SP f is the Ratio of Similitude of the figures. 98. Since to each point P of one figure corresponds a point P of the other such that PSP is a triangle of con stant species, if P coincides with 8, P also coincides with 206 SIMILAR FIGURES. it ; and therefore S taken as a point of either figure is its own correspondent in the other. Hence it is a Double Point of the polygons. 99. From these considerations we make the following inferences : I. If upon the lines joining a fixed point S to the ver tices of any polygon F l similar and similarly situated triangles are constructed, their vertices form a polygon F 9 similar to the given one, and S is their double point. II. If the lines joining corresponding points of two directly similar figures are divided in the same ratio, the points of section form a polygon similar to the given ones (H. Van Aubel). III. If the vertices of a polygon, constant in species, move on curves of any nature, to each position of it there is a corresponding centre of similitude. This is called the Instantaneous Centre for the position, and is such that the lines drawn from it to all points A, B, C ... X of the figure make equal angles with the tan gents at these points to their respective loci. [This is seen by taking two indefinitely near positions of the polygon.] IV. Reciprocally : If the lines L, M, N of the figure, moving as in the previous case, envelope curves, the lines joining the contacts of any position to S make equal angles with L, M, N. [For the points of contact are the intersections of two consecutive positions of the moveable figure and are therefore corresponding points.] THREE SIMILAR FIGURES. 207 SECTION II. THREE SIMILAR FIGURES. 100. Let F v F 2 , F 3 be any three directly similar figures ; 8 l the double point of F 2 and F z ; $ 2 and S 3 the double points of the remaining pairs F y F l and F v F 2 ; a lt a 2 , a 3 the lengths of corresponding lines d v d^ d 3 , a lt a 2 , a 3 the angles of the triangle D^Dy whose sides are d lt d# d y Then, by Art. 96, 1. The variable triangle D^D^Dy formed by any three corresponding lines, is constant in species. 2. The distances of S l from d 2 and d 3 are proportional to a 2 and a y and similarly for $ 2 and $ 3 (Art. 97 (/3)) ; therefore, the lines joining S v S# S 3 to the corresponding vertices of DJ)^}^ divide the angles D v D 2 , D 3 each into constant parts, and are concurrent (Art. 65). Hence the triangle SfijSp whose vertices are the centres of similitude of F v F y F 3 taken in pairs {Triangle of 208 SIMILAR FIGURES. Similitude), is in perspective with all homologous triangles DflyDy etc. ; and the centre of perspective K is a point such that its distances from any triad of homologous lines are in the ratios a l : a z : a y 3. Since the base angles of each of the triangles DJD 3 K, D 3 DJK:, D^D^K are constant (Art. 100, 2) as D r D 2 , D 3 vary, the angles subtended by the sides of /SySf 2 $ 3 at K are each constant, and the locus of K is therefore the circum-circle ; hence, Any triangle formed by three homologous lines is in perspective with SjSJS^ at a point on the circum-circle of the latter ; or the locus of the centre of perspective of S 1 S 2 S 3 and any triangle formed by three homologous lines is the circum-circle of the former. This is called the Circle of Similitude of F lt F z , F 3 . 4. The chords KP V KP y KP 3 drawn parallel to d lt d d 3 are homologous lines, for they intersect at angles a, /3, y, and their distances from d lt cZ 2 , d 3 are in the ratios a l :a 2 : a y * Moreover, they meet the circle in fixed points, since the angle SJKP^ is constant and $ 2 a h xed point ; therefore P^ is fixed, and similarly P 2 and P 3 are fixed points. They are termed the Invariable Points, and Pff^ the Invariable Triangle, of F v F y F y 4. May be enunciated as follows : A II concurrent triads of homologous lines pass through the invariable points and intersect on the circle of simili tude; and reciprocally: the lines joining P Jt P 2 , P 3 to any three homologous points B v E v B 3 meet in a point on the * These lines are therefore the sides of an evanescent triangle D^^D^ of constant species. THEOREMS. 209 circle of similitude ; and all triangles whose vertices are three homologous points are in perspective with P 1 P 2 P 3 and the locus of their centre of perspective is the circle of similitude. 101. Theorem. The triangle of similitude and the invariable triangle are in perspective ; and the distances of the centre of perspective E from the sides of the latter are inversely, as the ratios a : : a 2 : a 3 . Since S l is its own correspondent with respect to P 2 and F y P 2 $j and P 3 $j are homologous lines and lengths of these figures, therefore Sfs8f, = asa, (1) but (Euc. III. 22) ^P 2 : ^P 3 as the distances of , from P^ and PJP^a^-.a^ by (1), with similar relations for the points $ 2 and $ 3 ; therefore, etc., Art. 65. 102. Theorem. The invariable triangle is inversely similar to D^Df)^ Follows by Euc. TIL 22. 103. Adjoint Points.* Let / be the point of F l which corresponds to S 1 of the figures F z and F 3 . Then S^S^ is a particular case of a triangle formed by three homologous points, and is therefore (Art. 100, 4) in perspective with PjP 2 P 3 at a point on the circle of similitude ; hence the lines PJSIf, P&, P 3 S l are concurrent. Their common point is therefore 8 l ; that is to say, P^ passes through E and $ x (Art. 101) ; hence, The lines Sfi^, $ 2 $ 2 , Sj8 3 meet each other in E and the circle of similitude at the invariable points. * The theorems contained in Arts. 100-103 are due to Tarry. Matkesis, 1882, p. 72. 210 SIMILAR FIGURES. Defs. The point E is called the Director Point, and /, #/, 3 the Adjoint Points of F v F v F v 104. Theorems.* In any three similar figures there exists an infinite number of triads of homologous points C v Cg, C s ivhich are collinear. 2. The loci of these points are circles passing through E. 3. The variable line GjC 2 G 3 turns around E. Neuberg. The triangles ^(7 2 3 , SjO t C v S&C, are constant in species (Art. 97, 2) ; hence the angles 8&S V 8,C 2 S V Sfift are given, and therefore the loci of the points are circles passing through each pair of double points. Again, since SjCI^ is a constant angle, the variable line Gfit meets the locus of C l in a fixed point, and similarly it meets the loci of (7 2 and C 3 in fixed points. Therefore the fixed points are coincident ; that is to say, the circular loci have a point in common. In the particular case of the collinear triads S^S^, S 2 S 2 S 2 , S 3 S 3 S 3 it has been proved (Art. 103) that their lines of collinearity pass through E ; therefore, etc. The points $/, S 2 , $ 3 are on the corresponding circles. 105. Particular Cases. Let the three similar figures F v F y F 3 be described on the sides of a triangle ABC. It has been shown that the middle points of the symmedian chords of the circurn-circle ( are the common vertices of directly similar triangles described on the sides, taken in pairs (Art. 25, Ex. 2), and they are therefore the three double points. Hence, 1. Brocard s second triangle is the triangle of simili- * Mathexw, 1882, pp. 76-8. t The middle points of the symmedian chords of the circum-circle are the vertices of the triangle known as Brocard s Second Trianyle. M l CArS CIRCLES. 211 tude, and the Brocard circle the circle of similitude, of three directly similar figures described on the sides of a triangle. 2. Brocard s first triangle is their invariable triangle, Art. 29, Ex. 3. 3. Brocard s second triangle and the given one are in perspective at a point on the circum-circle of the former whose distances from the sides of ABC are in the ratios of their lengths (Art. 100, 2). See also Art. 16, Ex. 2. 4. The centre of perspective is the symmedian point of ABC. 5. The locus of the intersection of concurrent triads of homologous lines is the Brocard Circle, Art. 100, 4. 6. Brocard s first and second triangles are in perspec tive (Art. 101), and their centre of perspective E, or director points, is the centroid of ABC. (Art. 53, Ex. 6.) 7. All collinear triads of homologous points lie on a variable line passing through E, and each point describes a circle passing through two vertices of Brocard s second triangle and the centroid of ABG. M/CAY S CIRCLES. 106. The loci in 7 of the previous Article are fully described by M Cay in his memoir " On Three Circles related to a Triangle."* Amongst many other properties they possess those given in this and the following Article. The notation employed is as follows: ABG is the given triangle ; A^Bf/ v A 2 B 2 C 2 Brocard s first and second triangles ; E centroid ; A , B , C f three homologous col- linear points ; M middle point of AB ; H circum-centre ; * Transactions of the Royal Irish Academy, vol. xxviii. Science. 212 SIMILAR FIGURES. A y By G 3 the homologues of A^ B z , C 2 respectively as double points of F v F# F y P^ the c correspondent of P regarded as an a point, and L M and L ab the c and b correspondents of any line L regarded as an a line ; the circular loci the "A," "B," and "C" circles of the triangle. 1. The mean centre of any three collinear homologous points is at E (Art. 53, Ex. 6). 2. If one of them C coincides with E, A B is a tangent to the "0" circle and EA = EB f or EE ca = EE cb ; simi larly we have EE^EE^ and EE bc = EE ba . 3. If one of them coincides with a double point A 2 , the line of collinearity is A^EA^ (Art. 103) and Similarly the lines B 1 B. 2 B 3 and C}JB Z each pass through E t which is the common point of trisection of the segments C,C r 4. The circles cut each other at angles A, B, and C. 5. Their centres are on the perpendicular bisectors of the sides. PROBLEM. 213 This is proved for the " (7" circle as follows : On the sides of ABC construct three directly similar triangles BC A , CAB , ABC t each inversely similar to ABC. Their centres of gravity are therefore correspond ing points. But they lie on a parallel through E to AB; hence E , the centroid of ABC , is on the "0" circle and E and E are reflexions with respect to the perpendicular bisector of AB. 107. Problem. To find the Centres and Radii of M Cay s Circles. This is done by finding where the circles again cut the corresponding medians. We take, for example, the "C" circle and require to find C . Let L denote the median CM, and take it an a line. Since it makes an angle BCM with the side a, we draw the corresponding b and c lines by making angles CAB and ABC equal to BCM. From similar triangles MBC and MCB we have MC . MC = MB 2 = MC . MI ; hence M C = ML This also follows, since the triangles ABI and BAG are similar. Again, the triangle CBC Z is inversely similar to ABC , but it is (hyp.) directly similar to BAC 3 . Hence BAC^ and ABC are inversely similar; therefore C is the 214 SIMILAR FIGURES. reflexion of <7 3 with respect to the perpendicular bisector of the base. The connection between three collinear points A , B , C on the median to the side c of the given triangle and (7 2 , (7 2 , (7 3 has thus been established. The triangles EGA , CAB , ABC are similar to one another, and to GEG y ACC Z , and BAC 3 , and therefore A, (7 2 ; B , (7 2 ; G\ C 3 are reflexions of one another with respect to the corresponding perpendicular bisectors of the sides of ABC. It follows that if the median and symmedian cut the circum-circle in /and J, and these points be joined to M, the lines Jf/and MJ produced through M pass through C and (7 3 respectively; MJ=MC 3 and MI=MG, or G f and C 3 are the reflexions of I and J with respect to the base AB. Let d be the distance of the centre of the " C " circle from AB, m the median, and 9 the angle it makes with the base, t the tangent from M to the circle. Then (D M CAY S CIRCLES. 215 Again, 2c?, S in0 by (1); whence cZ = ccota>, and the radius of the "0" circle is given by the equation p = JW^tf = icx/cot 2 o> - 3 (cf. Art. 28, Ex. 19). Also, since the highest and lowest points of the circle are distant from the base p + d and p 8, these quantities are the roots of the quadratic equation 12/t 2 -4ccotft)./i + c 2 = 0; ............... (3) or, putting h = ^c tan <, 3 tan 2 2 coto>. tan + 1=0, ............... (4) an equation which reduces by an easy transformation to sin (ft> + 20) = 2 sin &> ......... (5). The forms (4) and (5) are remarkable inasmuch as they express as a symmetric function of the angles ; hence, Three similar isosceles triangles may be constructed on the sides of ABC, whose vertices are a triad of collinear homologous points. Let P, Q, R be the vertices of these triangles. Since COS0 with similar values for HP and HQ; also, from the collinearity of P, Q, R we have 2 -7Tp- = 0. By substitution, we obtain sin A siuB sin C _ n^nf ~U i _t\ ^^^/ /^ I _t\ \ / an equation which is therefore identical with the forms (4) and (5). Let 7ij and h z be the roots of (3), then 1 , 1 4coto) 2 2 7 1 J~ : 216 SIMILAR FIGURES. where (7 is the vertex of Brocard s first triangle ; there fore The vertices of Brocard s first triangle and the cor responding sides of ABC are pole and polar with respect to the "A" "B," and " C" circles. Many other beautiful properties of these circles are given in the memoir from which the preceding are extracts. 108. If A t B , C be the feet of the perpendiculars of ABC, the triangles AB V, ABC\ and A HC are similar, and may therefore be taken as portions of three directly similar figures F v F^ F s whose double points are A , B , C , homologous lines in the ratios cos A : cos B : cos C, the middle points of the segments of the perpendiculars towards the angles A , B", C" , the invariable points A" , B ", C ", points of concurrence of homologous lines middle points of sides, and the nine-points-circle the circle of similitude (Neuberg). EXAMPLES. 1. If similar figures FI, F 2) F 3 be described on the perpendiculars A A , BB , CC of a triangle, their circle of similitude is the ortho- centroidal circle. [For the orthocentre being the point of concurrence of three corresponding lines is on the circle of similitude (Art. 100, 4). Also the parallels through the centroid E to the sides of the triangle trisect the perpendiculars at right angles, and are therefore also corresponding lines ; therefore, etc. We note that the parallels meet the corresponding perpendiculars in P, Q, /?, the invariable points of FI, F* F 3 .] 2. The lines joining the in- and circum-centres of the copedal triangles B C A, C A B, A B Cuieet at the point of contact of the nine-points and in-circle of ABC. EXAMPLES. 217 [By Art. 108, the three triangles being parts of similar figures have the nine-points-circle of ABC for circle of similitude, and the middle points of the segments of the perpendiculars for invariable points ; hence (Art. 100, 4), if 7i, 7 2 , / 3 and O l5 2 , O s denote the in- and circum-centres of the triangles, the lines IiOi, / 2 2 , 7 3 3 correspond, and are concurrent on the circle of similitude. Dr. Casey * proves the remainder of the property, which includes Feuerbactis Theorem, as follows : Let .y be the nine-points-centre ; then J\ 7 3 = ^R. Draw IP parallel to j.\"0 3 . Now, if PI is proved to be equal to the radius of the in-circle, the line I 3 S is the join of parallel radii, and therefore passes through a centre of similitude of the circles ; similarly for 7i0! and 1 2 2 . Since C01 and C0 3 7 S are corresponding parts of similar figures, they are similar; therefore the angle D10 = 7/ 3 P, and ODI = OCI = CIP, since N0 3 is parallel to OC. Hence the triangles ODJ and z are similar, and R ID 2Rr 2/e since C//(7/ 3 =l/cosC , the ratio of similitude of ABC and A B C (Art. 108).] 3. If A and A , corresponding points of two similar figures, are conjugate points with respect to a fixed circle, required to find their loci. * Casey s Sequel to Euclid, fifth edition, p. 202. 218 SIMILAR FIGURES. [Take S the double point, M the middle point of A A . Then SAA is a triangle of constant species ; therefore SMI MA is a constant ratio. But MA=t, the tangent from M to the given circle (Art. 73, 2). Hence SMjt is constant and M describes a circle (Art. 72, Ex. 3) ; therefore also A and A describe circles.] 4. If XiX. 2 X s be a triangle formed by joining a triad of corre sponding points of three similar figures such that X\X 2 : XiX s = const., the locus of each vertex is a circle. [The triangle 83X1X3 is constant in species, Art. 97 ; similarly for 8^X1X3 ; hence S^XilXiX^ and 8- 2 Xi/XiX 3 are constant ratios. Dividing one by the other, we have the base S a S 3 and ratio of sides of the triangle 8-283X1 ; therefore, etc. It is to be noted that as the ratio 8-2X1/83X1 varies in magnitude the vertex Xi describes a coaxal system of which >S1> and S 3 are the limiting points.] 5. If the area of XiX. 2 X 3 is given, each vertex describes a circle. [For XiX% . XiXs sin Xi varies as 8% Xi . 83X1 sin(Xi - 0) ; there fore, etc. (Art. 23, Ex. 3). X 2 and X s similarly describe circles.] 6. If a side or an angle of XiX 2 X s is given, its vertices describe circles. 7. If the area of a variable triangle formed by three correspond ing lines be given, its sides envelope circles whose centres are the invariable points of F it P\ F 3 . These and many other excellent illustrations of the theory of three directly similar figures are to be found in Casey s Sequel to Euclid, to which the student is referred. See fifth edition, Miscellaneous Examples, pp. 231-248. CHAPTER X. SECTION I. CENTRES OF SIMILITUDE. 109. If A,T^\ B, r 2 be any two non-intersecting circles, P and Q the points of intersection of the direct and transverse common tangents, it is easily proved that A t B, P, Q are collinear, and that AP/BP = A Q/BQ = rjr^ ; hence the centres of similitude of two circles arc the points of intersection of the direct and transverse common tangents* In the case of intersecting circles, if C be a point of intersection, we infer from these equations that the bisectors of the angle between the circles meet the line of centres in P and Q (Euc. VI. 3). For the in- and ex-circles of a triangle taken in pairs the twelve centres of similitude are the vertices and the points where the bisectors of the angles meet the opposite sides. The centres of similitude of a line L and circle A are the extremities of the diameter perpendicular to L. For the common tangents to the circle and line are * Therefore the common tangents, real or imaginary, to any two circles always intersect in real points. 219 220 CENTRES OF SIMILITUDE. parallel to the latter, and the line of centres is the diameter at right angles to L ; therefore, etc. 110. It has been seen as a particular case of a general property of coaxal circles (Art. 93) that any line A^A^B^B^ through C, a, cuts the circles at equal angles and, /3, that the intercepted chords A : A 2 and B^B^ are ^ n ^ ne ra tio f the radii. These are obvious by the following method : Join AA 1 and BB r Since CA/CB^rJr^^ AAJBB l the triangles CAA l arid CBB 1 are. similar (Euc. VI. 7) ; there fore AA 1 is parallel to BB V and similarly AA 2 to BB 2 . Hence the isosceles triangles AA^A^ and BB^B^ are similar, whence, , the angles A^AA Z and B 1 BB 2 are equal, and, /3, A l A 2 /B l B 2 = r 1 /r 2 . Definitions. A l and B l are termed Homologous Points; and since the radii AA l and BB l through them are parallel, the tangents at homologous points on the circles are THEOREMS. 221 parallel. Thus the tangents at A 2 and B 2 are parallel. More generally any two points A n and B n which connect through C such that CA n /CB^^rJr t are homologous. A l and B 2 are termed Antihomologous Points, and since the radii A A 1 and BB 2 through them make equal angles with their line of connexion, the tangents at antihomo- logous points meet on the radical axis. Let a second transversal through G meet the circles in A^A^B^B^. The chords A^A^ and BJS^ joining pairs of homologous points are termed Homologous Lines, and those joining pairs of antihomologous points Antihomo logous Lines. Thus A 2 A, B^B^ and A^A^ B 2 B 4 are pairs of antihomologous lines, 111. Theorem. Homologous chords (A^, B^B^) of any two circle^ are parallel. For it has been shown that AA 1 and BB V AA 2 and BB 2 are pairs of parallel lines; hence the two isosceles triangles AA^ S and BB^ have equal vertical angles, and are therefore similar (Euc. VI. 6). NOTE. Since any line through G meets homologous lines A^AZ and B : B B in homologous points A n and B n , therefore A n , B n are in general the corresponding inter sections of pairs of homologous lines. The two points A^Ap A 2 A and B-^B^ B 2 B are homologous. 112. Theorem. Antihomologous chords (A 2 A, B^B^) of any two circles meet on their radical axis. By Art. Ill, we have CAJCA^CBJCB^ but (Euc. 111.36) CAJCA 9 = CAJCA Z ; hence CBJCB 3 = OAJCA 2 or CA 2 . CB l = CA 4 . CB 3 ; thus : any two points are coney die with the corresponding pair of antihomologous points; therefore, etc. (Art. 88, Ex. 6). 222 CENTRES OF SIMILITUDE. PRODUCTS OF ANTISIMILITUDE. 113. By the previous Article, we have from the cyclic quadrilateral A^A.JBJS^ CA 2 . CB l = CA, . CB y We may therefore infer that the rectangle under the distances of either centre of similitude from a pair of antihomologous points is constant. If the circles A, i\ ; B, r 2 be regarded as portions of two geometrical figures, any point A n of one is antihomo logous to B n of the other when the line A n B n passes through a centre of similitude C, and CA n . CB n is equal to the above constant, which is termed the Product of Antisimilitude (External or Internal). To find the values of the products, we take the extreme positions of the variable line CA^B^ which for real intersections are the common tangents. We have therefore CA 2 . CB^CT, . CT 2 (1) Again, since T^T 2 subtends a right angle at each of the limiting points M and N (Art. 88, Cor. 3), CT^.CT^CM.CN ....(2) These constant values which may be expressed in terms of the distance (<$) between the centres of the given circles and their radii (r l and r 2 ) are of importance in the theory of coaxal circles, and will frequently be made use of in the next chapter. Join A T! and BT 2 . Let A GT^ = 0. Then CTj . CT 2 = r^ 2 eot 2 = r,r 2 EXAMPLES. 223 Similarly the internal product of antisimilitude is found to be equal to NOTE. It should be noticed when the two circles lie wholly out side each other 8 > 1\ + r 2 , if they intersect 8 < r + r 3 and > 1\ ~ r- 2 (Euc. I. 20), and when one lies completely within the other S<TI - r. 2 (Euc. III. 12) ; hence it follows from (3) that the external product of antisimilitude is negative only when one circle lies wholly within the other. Also from (4) the internal product is negative when the circles are external to one another and positive in every other case. In the case where both products are positive 8 > r\ - r 2 and < ?*i -f r- 2 ; therefore 8, r^ r- 2 form a triangle (Euc. I. 20), or the circles intersect in a pair of real points. EXAMPLES. 1. If a variable circle touch two circles with contacts of similar species, its points of contact are antihomologous points. [By Art. 112, if AA 2 and BBi be produced to meet in X, XBi = XA 2 . In the case of internal contact the points of contact are AI, -5 2 .] 2. Describe a circle passing through a given point (P) and touching two fixed circles (A, 7*1) (B, r}. [By Art. 110, the required circle passes through an antihomo logous point P , and the problem thus reduces to " describe a circle passing through two fixed points and touching a given circle"] 3. The polars of the external centre of similitude with respect to two circles are equidistant from the radical axis, and therefore also from the limiting points. 4. The line at infinity is an axis of perspective of two circles. [Regard the circles as similar polygons of an infinite number of sides, and join their corresponding vertices (i.e. the homologous points). Thus the ex-centre of similitude is a Centre of Perspective of the circles. Again, the corresponding sides (i.e. homologous lines) intersect on the axis of perspective. In this case they are parallel. 224 CENTRES OF SIMILITUDE. Hence the line at infinity is the axis of perspective of every two circles. (Cf. Art. 87).] 5. The radical axis is also an axis of perspective of two circles. [For since antihomologous points B\, A 2 connect through a centre of similitude (7, the circles may be regarded as polygons of an infinite number of sides whose corresponding vertices are antihomologous points and whose corresponding sides are therefore antihomologous lines ; but these latter intersect on the radical axis (Art. 112), which is therefore the axis of perspective.* 6. The poles A n , B n of the chords AiA 2 and B\B* are homologous points. [For they are the intersections of pairs of homologous lines, viz. the tangents at A\, A 2 and B\, B z respectively.] 7. In Ex. 6 the lines A\B^ and A n B n are conjugate with respect to both circles. 8. If C, C denote the centres of similitude of two circles which cut orthogonally at X ; the inverse (C") of the point C with respect to the circle A is the inverse of C with respect to the circle B. [Since C and C" are inverse points, AC"X=AXC = 45 ; hence AC"X=BXC, therefore CBIBX=BX\BC", therefore etc. 9. A variable circle touches two equal circles with contacts of opposite species : show that the product of the intercepts on their transverse common tangents made by the perpendiculars from the centre and measured from their point of intersection is constant. 10. The centres of similitude, the centre of the circle of similitude, and the centre of either circle B are pairs of inverse points with respect to a circle concentric with A . * Two circles are thus shown to be doubly in perspective to each centre of similitude ; the two axes of perspective forming the coaxal circle whose radius is infinitely great, viz., the radical axis and the line at infinity. It follows that " for every two circles in the same plane, however circumstanced as to magnitude and position, the radical axis and the line at infinity, being both axes of perspective, are both chords of intersection ; the corresponding points of intersection, real or imaginary, according to circumstances in the case of the former, being of course from the nature of the figures always imaginary in the case of the latter." (Townsend.) FEUERBACH S THEOREM. 225 11. The poles AI, BI of the radical axis of two circles (A, j ; B, r 2 ) are inverse points with respect to their circle of similitude. [For since AA 1 . AL = r^ angle A PL = A A also since BB l . EL = r 2 2 , angle BPL = BB^P. By addition APB = PA l B l + PB 1 A l = ir- A^PB^. Thus AI, BI and A, B, since they subtend similar angles at P, are pairs of inverse points with respect to the circle of similitude (Art. 72, Cor. 8).] 12. If a variable circle V cut two circles A and B at constant angles, show that the centre of similitude of any two positions Fi and F 2 is on L the radical axis of A and B. [For FI and F 2 meet the line L at equal angles (Art. 88, Ex. 7) ; therefore it passes through their ex-centre of similitude.] 12a. Hence show that if the circles A and B each cut three fixed circles FI, F 2 , F 3 at the same angles a, (3, y, an axis of similitude of the three is the radical axis of the two. 13. Construct a quadrilateral, having given the four sides, and that two adjacent angles are equal. (Mathesis, 1881.) 14. Feuerbach s Theorem. To prove by an elementary method that the nine-points-circle touches the in-circle. Draw C X the fourth common tangent to the in- and ex-circles to the side c of the triangle ABC. We shall prove that the line joining M, the middle point of the base, to the point of contact X passes through the point of contact Y of the in- and nine-points- circles . Let T be the point of contact of the in-circle, P the foot of the perpendicular, and C the foot of the internal bisector of .7. p X^ OP .TWH 226 CENTRES OF SIMILITUDE. By Art. 71, Ex. 3, MP . Mi = J( ~ b?= MT* = MX. MY. Hence XYPC is a cyclic quadrilateral and angle MC X= MYP ; but MC X=MC C-XC C=A ~B- hence MYP=A~B, and therefore Y is on the nine-points-circle, since the latter cuts the base AB at this angle. Therefore the circles cut or touch at Y. But the tangents at M and X to the circles are parallel, since they both meet the base at the same angle A ~ B. M and X are thus homo logous points. 15. The straight lines joining the points of contact of the fourth common tangents to the in- and three ex-circles to the middle points of the corresponding sides are concurrent. (Dublin Univ. Exam. Papers.} [By Ex. 14, the point of concurrence is where the nine-points- touches the in-circle.] 16. A right line A BCD is drawn across two circles cutting them at angles a and ft respectively ; show that if a variable circle cuts the given ones at the same angles in the points A , B , C , 2) , AA , BB , CC , DD are concurrent ; and find the locus of their point of concurrence. [The given circles meet the line ABCD and circle A B C D at equal angles ; hence A A etc. are antihomologous points with respect to the external centre of similitude of the latter. Therefore A A etc. meet on the circle A B C D at a point (P) the tangent at which is parallel to ABCD. The locus of P is the radical axis of the fixed circles by Ex. 1 2.] CIRCLES OF ANTISIMILITUDE. 227 SECTION II. CIRCLES OF ANTISIMILITUDE. Definitions. The circle described with either centre of similitude of two given circles as centre, the square of whose radius is equal to the corresponding product (Art. 113) of antisimilitude, is known as a Circle of Antisimili tude. Thus there are two circles of antisimilitude, External and Internal, according as the centre coincides with the external or internal centre of similitude of the given circles. From the definition it is evident that all pairs of antihomologous points are inverse points with respect to the circle of antisimilitude, or, more generally, that each of the two given circles is the inverse of the other with respect to either circle of antisimilitude. In the next chapter this latter circle, from this funda mental property, will be otherwise known as the Circle of Inversion of the two given ones. 114. The following theorems are of importance in the geometry of these circles. 1. Any two circles A and B and their circles of anti- similitude are coaxal. For the constant product GA Z . CB l (Art. 113) has been proved equal to CM . CN; hence M and N are a common pair of inverse points to the four circles. 2. The squares of the tangents ^ and t 2 from any point of either circle of antisimilitude to A and B are in the ratio of the radii ; or t^ : t = r^ \ r%. 228 CENTRES OF SIMILITUDE. Since the circles are coaxal, t* : t/ = CA:CB^ i\ : r 2 . (Art. 88, Cor. 2.) 3. The external circle of antisimilitude cuts ortho gonally all circles cutting A and B at equal angles. Since AA 2 and BB l are equally inclined to the line A. 2 B V if they are produced to meet in X, then XB 1 A 2 is an isosceles triangle, and X is therefore the centre of a circle cutting A and B at equal angles. Thus any circle cutting A and B at equal angles passes through a pair of inverse points A 2 and B l with respect to the ex-circle of antisimilitude ; therefore, etc. See also the method of Art. 93, Cors. 3, 4. . 4. Any circle intersecting A and B at supplemental angles is orthogonal to the internal circle of antisimili tude. {Proof similar to 3.] 5. Any circle intersecting A and B orthogonally is orthogonal to both their circles of antisimilitude. For in this particular case A and B are cut at angles which are at once both equal and supplemental ; there fore, etc. by 3 and 4 combined. EXAMPLES. 229 EXAMPLES. 1. A variable circle passing through a fixed point and cutting two given ones at equal angles passes through a second fixed point. [In every position it passes through the inverse of the fixed point with respect to the ex-circle of antisimilitude.] 2. A variable circle passing through a fixed point and cutting two fixed circles at supplemental angles passes through a second fixed point. [The inverse of the given one with respect to the ill-circle of antisimilitude.] 3. Two circles X, Y intersecting two others A and B at equal angles have for radical axis a line passing through the centre C of the ex-circle of antisimilitude of A and B. [For if X and T intersect in a point P, each must pass through the inverse of P with respect to CJ\ 3a. If the angles are supplemental, the radical axis of A" and Y passes through the in-centre of antisimilitude. 4. If three circles A", 7, Z meet two others A and B at equal or supplemental angles, the radical centre of the three coincides with the external or internal centre of similitude C or C of the two. [For by Ex. 3 the radical axes of J~, Z ; Z, X ; A, Y each pass through C or C according as the angles of section are equal or supplemental ; therefore, etc.] NOTE. In this example it may be noticed that in the first case the circles A and B each cut A", Y, and Z at equal angles ; therefore they cut the ex-circles of antisimilitude of Y, Z ; Z^ A"; A", Y&t right angles (Art. 114). But the ex-circles of antisimilitude are coaxal ; hence a variable circle A cutting three others X, Y, Z at equal angles describes a coaxal system, the conjugate of that formed by the circles of antisimilitude o/A", Y, Z taken two and two. More generally, a variable circle cutting three others X, Y, Z at similar angles describes four coaxal systems whose radical axes are the four axes of similitude of X, Y, Z. Also, since the common ortho- 230 CENTRES OF SIMILITUDE. gonal circle of the three cuts them at once at equal and supple mental angles, it belongs to each of the four coaxal systems. 5. If two circles A and B touch with similar contacts three others X, Y, Z, the radical axis of A and B is the line joining the ex-centres of similitude of Jf, I 7 ", Z taken in pairs. [A particular case of the foregoing.] 6. The eight circles that can be described to touch three given ones arrange themselves in pairs coaxal with the four axes of similitude of the given ones. 7. In Ex. 5 the chords of the three circles joining the points of contact with the two meet at the in-centre of similitude of A and B, and therefore at the radical centre of A", Y, Z. 8. The chords of contact pass through the poles of the radical axis of A and B with respect to each of the circles Jf, Y, Z. [For the tangents at the extremities of the chord of contact of X being equal intersect on the radical axis of A and B.~\ NOTE. Gergonne deduces by means of the foregoing properties a simple geometrical construction for the eight circles of contact of EXAMPLES. -231 three given ones A , F, Z. The circles having similar contacts are found as follows : Find the ex-centres of similitude of X^ F, Z taken in pairs ; the line L joining them is the radical axis of the required circles A and B. Next find C the centre of the common orthogonal circle of the given ones. C is the in-centre of simili tude of A and B. Now obtain the inverses A r/ , F , Z of L with respect to AT, F, Z respectively. Join C"A", C Y t and C Z ; these lines meet the given circles at the required points of contact ; therefore, etc. The remaining circles may be similarly found. Otherwise, thus : By Casey s relation in Art. 7, if we number the given circles 1 , 2, 3 and let 4 be the required point of contact with 1, we have the ratio of the tangents from 4 to 2 and 3, a given quantity k. Similarly for the second circle which has the similar contacts with the three given ones, the ratio of the tangents from its point of contact (5) to 2 and 3 the same ratio Jc ; therefore, etc. (Art. 88, Cor. 1). 9. Let AiAz, BiB. 2 be the extremities of the common diameter of two circles ; J/, N their limiting points ; prove that the circles on AiBi, AzB-2, MN as diameters are coaxal. [For their centres are collinear, and they each cut the internal circle of antisimilitude orthogonally (Art. 114, 4) ; therefore, etc.] 10. A variable circle cutting three given ones at equal angles passes through two fixed points, real or imaginary. [For it cuts the external circles of antisimilitude of the given ones taken two and two orthogonally, and these (Art. 88, Ex. 13. 2) are coaxal ; therefore the variable circle passes through their limit ing points, real or imaginary.] 11. Two variable circles X and Y touch externally two fixed circles A, i\ and Z>, r. 2 at four points BI, A 2 and A- 2 , BI in a right line ; prove that o. The line joining their centres passes through a fixed point. (3. The sum of their radii is constant. y a . The foot of their radical axis describes a circle. 232 CENTRES OF SIMILITUDE. [a. Since the diagonals of a parallelogram bisect each other, AT bisects and is bisected at the middle point Z of A B. /3. Let L be the radical axis of A , r 1} and B, r., ; then XLIp = const. (Art. 88, Ex. 7), and therefore + = const., but the numerator is constant by a ( = 2ZL) ; therefore, etc. y. The circle on CZ is evidently the locus.] 12. Circles are described touching two fixed circles (as in Fig. of Ex. 8) ; find the locus of the limiting points of these circles taken in pairs. [The internal circle of antisimilitude of the two given circles (Art. 114, 3).] 12a. Circles are described touching one another, and each touch ing two given circles ; find the locus of their points of contact. [The points of contact are the coincident limiting points of the touching circles ; hence the required locus is the internal circle of antisimilitude of the two given ones.] 13. If n points be taken on a circle, prove that (1) the mean centres of the n systems of n-l points formed by omitting each EXAMPLES. 233 point in succession, lie on a circle S n ; (2) if another point be taken on the original circle, the centres of the n + 1 circles () obtained by omitting each point in succession lie on an equal circle ; and so on ad infinitum. (St. Clctir.)* [Let G be the mean centre of the system of n points. Produce AG to a, making AG : Ga = n- 1:1; then a is the mean centre of the w- 1 points formed by excluding A. In the same manner we get BG : Gb = n-l : 1, etc. ; hence the points a, b, ... lie on a circle ; and G is a centre of similitude of the locus circle and the given one. * Educational Times, February, 1891. CHAPTER XL INVERSION. SECTION I. INTRODUCTORY. 115. It has been seen (Art. 74) that the inverse of every point on a line with respect to a circle lies on a circle described on the line joining the centre of the given circle with the pole of the line. This circle is said to be the inverse of the line with respect to the given circle ; and it may be generally inferred that the inverse of a line is a circle passing through the centre of the given circle; and conversely. This latter is named the Circle of Inversion, and its centre the Origin or Centre of Inversion. We shall now proceed to discuss the inversion of a system of points which are not collinear. Take the simplest case the vertices of a triangle ABC. Let their inverses with respect to a circle of inversion 0, r be respectively A , B y C". It is obvious that the three quadrilaterals BCB C , CAC A , ABA B are cyclic ; hence we have the angular relations : A C O = OAC, B C O = OBC, etc. (Euc. III. 22), 234 THEOREM. 235 and thence by addition, AOB=C+C (1) Similarly, BUC = A+A (2) and COA=B + B (3) If the base AB and origin are fixed, and C given in magnitude, C is also given in magnitude by (1) ; hence : // a variable point (C) describes a circle (circuni- circle of ABC), the locus of its inverse (C") is a circle (A B C )* Two circles or, more generally, any two curves so related that every point of one has a Corresponding Point on the other inverse to it with respect to a given circle, are Inverse Figures with respect to the circle of inversion. It has thus been proved that in general a line or circle * Tins statement is equivalent to the following : If a variable line OPP is drawn from a fixed point to a given circle and divided at X such that OP . OX const. ; the locus of X is a circle, which may be thus proved independently. Since OP . OP and OP . OX are both constant, OX : OP = const. Through X draw XC parallel to CP . From similar triangles OX : OP = OC : OG C X : CP = const. Hence C is a fixed point, and C X is of constant length. The locus of X is therefore a known circle ; and the circle of inversion is obviously a circle of antisimilitude of the given one and its inverse. 236 INVERSION. inverts into a circle ; and in the particular case when the origin is on the circle, its inverse is a line. 116. Species of A B C . Let the points A, B, C be fixed. Since AOB= C+G , C may have any value de pending on the position of the point 0. The following particular cases are worthy of notice, and may be readily inferred : 1. If is the circum-centre of ABC, A=A ,B = B , C=C . 2. If is the right (or positive) Brocard point of ABC, AOB = C+C =TT-B- hence C = A. Similarly A = B and R = C. 3. If is the left (or negative) Brocard point, ABC and A B C are again similar. 4. If is one of the vertices (C 2 ) of Brocard s second triangle, AOB = 2(7= C+C , therefore C=C -, and also Hence the triangles are similar when the centre of inversion coincides with any of the six points 0, Q, Q , A 2 , B 2 , C 2 , or their inverses. (Art. 72, Ex. 22.) 5. If is on the circum-circle, C = 0, and the points A , B , C are collinear. 6. Let BOC, CO A and AOB be equal respectively to 60 + ^, 60 + , 60 + Then A = H = C f =60; there fore the vertices of any triangle may be inverted into tliose of an equilateral ; or one of any given species. 117. In the preceding figure the point has been taken inside the triangle. It is easy to verify the analogous angular relations when the centre of inversion is outside ABC. INVERSE FIGURES. 237 It will be observed from tbe relations of Art. 116 that, if a variable triangle of the species of A B C be inscribed in the given one, the fixed point in connexion with the figure determined by the method of Art. 19 coincides with the centre of inversion. 118. Relations between the sides of ABC and A B C . From similar triangles AOB and A OB , AB^AB * = OA . OB/0 A . OR ; but OA - r*/OA and OB - r*/OB, therefore by substitution ABjAB f =OA.OBjr\ or c/c =OA.OB/r\ By dividing the similar relations a/a = OB . OC/r 2 and &/& = OC. OA /r 2 , we have a la OB T / T = 7TT = COnst. b/ b OA Hence : If the base and ratio of sides of a triangle are given, the base and ratio of sides after inversion are also known. In each case the locus of the vertex is a circle having the extremities of the base for a pair of in verse points (Art. 70) ; and since the loci are inverse figures, we have the following important theorem : Every circle and a pair of inverse points invert into a circle and a pair of inverse points ; and more generally, A circle and a pair of figures each the inverse of the other with respect to it, retain this relation after inversion from any origin. 119. Theorem. Any circle X, its inverse X and the circle of inversion are coaxal, i.e. have a pair of common points, real or imaginary. 238 INVERSION. Let P and Q be the common pair of inverse points of the circles and X. It is manifest that they are inverse points to X . For X, P, Q invert respectively into X , Q, P, which by the last Article are a circle and pair of inverse points ; therefore, etc, The theorem requires no proof when the intersections of the circles are real, as the coaxal system is of the common point species. COR. 1. The circle of antisimilitude is the circle of inversion of either of two given ones with respect to the other ; hence, Two circles and their circles of antisimili tude are coaxal. COR. 2. The inverses of the vertices of any triangle with respect to the polar circle, real or imaginary, are the vertices of the pedal triangle; hence, The circum- and nine-points-circles are inverse figures with respect to the polar circle of the triangle; and the three circles are coaxal. 120. Inversion of a System of Four Points. Let A, B, C, D and A t B , (7, D be any four points and their inverses with respect to a given circle of inversion 0, r. The quadrilaterals BCB C , CDC D ,... are cyclic. Hence the angular relations : OC D =ODC, IN VERSE FIG URES. 239 from which we obtain TT .......................... (1) Also AQC=B+B ; therefore by substituting in (1), B + B +D + iy = 2Tr , ......................... (2) similarly, A +A + C+ C = 2?r, or the sums of corresponding pairs of opposite angles of the two quadrilaterals are together equal to four right angles. The following particular cases are noticed : 1. If B + D = TT, then also B +D = Tr , i.e., a cyclic system of points inverts into a cyclic system. Cf. Art. 11 5. 2. I{B f = D f and A = C simultaneously, A KG IT is a parallelogram, and its angles are given by the equations and NOTE. The centres of inversion in this case are easily found; for AOC=B+B = B + 7r- %(B+D\ and BOD similarly equals A + TT J(J5+D) ; hence there are two centres of inversion from which the vertices of any quadrilateral invert into the vertices of a parallelogram in an assigned order, viz., the intersections of the known circles COA and BOD. Four other points might be similarly found from the intersections of pairs, of circles BOC t AOD,*ndAOB t COJ). 3. A cyclic, system of four points may be inverted into the vertices of a rectangle. 121. Relations between the sides of ABCD and A ffG L . By Art. 118, BC/B C ^OB . OC/r* and AD/A IT = OA .OD/r*. Multiplying these relations we have OA OB OC OD 240 INVERSION. CA .BD OA.OB.OC. OD similarly, f , = - - .............. (2) etc. etc. ; hence BC.AD:CA.BD:AB. CD = RC . A D : C A . RD : A R . C D ......... (3) COR. 1. If A, B, C, D be a harmonic system of points on a circle ; A y B , C , D are also a harmonic cyclic system. For if the ratios on the left side of (3) are equal, those on the right are also equal. COR. 2. Combining 3 of the last Article with the previous corollary, it follows that a harmonic system of cyclic points may be inverted into the vertices of a square. EXAMPLES. 1. Any two triangles may be placed such that the vertices of the one may be inverses of those of the other taken in any assigned order. 2. Any four points may be inverted into an orthocentric system. [For the latter quadrilateral has the following angles : A , 90- A y 180 + .4 , 90 -A ; hence since BO D = A + A , COA = 5 + 90- A , and A + C+A + 7r + A = l80 ; the centres of inver sion are the intersections of two known circles BOD and COAJ\ 3. Each side of a triangle divided by the perpendicular on it from any origin remains unchanged by inversion. 3a. If the origin is the symmedian point of the one triangle, it is also the symmedian point of the other. 4. If a, /3, y denote the perpendiculars from any point on a circle, on the sides of an inscribed triangle, then J3y sin A + y a sin B + afi sin C= 0. [For let A B C be any three points on a line Z-, and the origin ; since EXAMPLES. 241 after inversion is on the inverse circle L and ^+^ + ^=0, or +* + <-<>; o ft y a (3 7 therefore, etc.] 5. Prove generally for any cyclic polygon that 2(a/a) = 0. . (Casey.) 6. The inverse of a figure with respect to a line is its reflexion with respect to the line, and is equal in every respect to the given one. 7. The inverses A , B , C ... of the points of intersection A, B, C, D of any two figures are the corresponding points of inter section of the inverse figures ; and the lines A A , BB , CC ... are concurrent at the centre of inversion. 7 a. If two curves touch at A, their inverses touch at A the inverse of A. 8. A circle coincides with its inverse when the circle of inversion is orthogonal to it. 9. A variable chord AB of a circle, the inverse C of a fixed point C on it and the centre are concyclic. [Since the points A, B, C, co are collinear ; their inverses with respect to the given circle are concyclic ; i.e., ABC O is a cyclic quadrilateral.] 10. From any point P on the circum-circle a line is drawn through the symmedian point A", cutting the sides of the triangle ABC in A , B , C , prove the relation 2l/PA = 3/PK. [Employ the properties of Ex. 4 and Art. 15, Ex. 1 (3).] 122. Theorem. The inverse of the circum-circle of a triangle ABC with respect to the in-circle is the nine- points-circle of the triangle PQR formed by joining the points of contact. Let X, T, Z be the middle points of the sides of PQR. From similar triangles we get OA .OX=OB . OY=OC . OZ=r*-, therefore, etc. Q 242 INVERSION. Mr. Piers C. Ward has applied this property in the following elegant proof of Mannheim s Theorem : Inverting with respect to the in-circle, the circum- circle inverts into XYZ, that is, a circle passing through a fixed point Z and of constant radius ( = |-r). It there fore envelopes a circle concentric with Z whose radius is equal to the diameter of XYZ ; therefore, etc., by Art. 121, Ex. 7a. EXAMPLES. 1. A variable triangle ABC is inscribed to one and escribed to another circle ; prove that the mean centre of the points of contact P t Q, R is a fixed point. [This particular case of WeiWs Theorem (Art. 53, Ex. 12) is easily seen. For the mean centre of P, Q, R is the point of trisection of the line joining its circmn- and nine-points-centres, both of which are fixed ; therefore, etc.] 2. If a quadrilateral A BCD be inscribed to one circle and circum scribed to another ; prove that the mean centre of its points of contact P, Q, R, S with the inner circle is a fixed point. [Let Tf, X, Y, Z\>e the middle points of the sides of the cyclic quadrilateral JP, Q, R, S. Then W, X, Y, Z is a cyclic parallelo gram, and is therefore a rectangle. The mean centre of P, Q, R, S is evidently that of the system IF, A", Y, Z, or the centre of the circle inverse to ABCD with respect to the other given circle.] ANGLES OF INTERSECTION. 243 3. The four nine-points-circles of the four triangles formed by taking the vertices of a cyclic quadrilateral in threes pass through a point. [For the nine-points-circles invert into the circum-circles of the triangles formed by drawing tangents to the circle at the vertices of the quadrilateral ; therefore, etc. The more general property for any quadrilateral has been independently demonstrated. Art. 79, Ex. 15.] SECTION II. ANGLES OF INTERSECTION OF FIGURES AND OF THEIR INVERSES. 123. The general relations existing between the centres and radii of a circle, its inverse, and the circle of inver sion are as follows : Let (7, C", be the centres of the three circles ; AB, A B , MN the extremities of their common diameter ; SS and TT the direct common tangents intersecting in 0. Join ST and S T . 244 INVERSION. Since AB and A B are inverse segments with respect to the circle of inversion, the three circles are coaxal. (Art. 114, Ex. 9.) Let I and / denote the points of intersection of ST and S T with the line of centres ; by comparing equal triangles OIS and OIT, etc., it follows that ST and 8 T are both perpendicular to AB. The quadrilateral CSST is therefore cyclic; hence the inverse of C is / ; and similarly the inverse of C is / with respect to the circle of inversion, and therefore : The centre C of any circle inverts into the inverse I of the centre of inversion with respect to the inverse circle C ; and The inverse I of the centre of inversion ivith respect to any circle C inverts into the centre C of the circle inverse to the circle C* In the particular case when the inverse circle is a line, the inverse of the centre of a given circle is the reflexion of the origin with respect to the line. The inverse of ST is the circle on 0(7 as diameter. Again, by similar triangles 00/00 = OS/OS f = CS/C 8 . or, say d/d = t/t = r/r ..... , .................. (1) To find d t t , and r, we have where R is the radius of inversion. Hence d = - a relation which gives the position of the centre C of the inverse circle. *Townsend, Modern Geometry of the Point, Line, and Circle, 1863, p. 373. PROBLEM. 245 From (1) we have therefore generally d r f R 2 1 79 9> \ d r d?~ r 2 from which the position of the centre and magnitude of the radius of the inverse circle may be determined. COR. If the centre of inversion is on the circle ; d = r and r oo , thus verifying that the inverse of a circle from any origin on its circumference is a right line. 124. Problem. To invert two circles such that the ratio of the radii of their inverses may be a given quantity K. Let r v r 2 be the radii of the given circles ; d lt d 2 the distances of their centres from the origin 0; R the radius of inversion ; t v t. 2 the tangents, real or imaginary, from to the given circles. Then if p v p. 2 denote the radii of the inverse circles, we have, by Art. 123, Pi = 12V A 2 and p 2 = R 2 r 2 /t/. Dividing these equations, Pl = Ti. t ^ = K p 2 r 2 t* The centre of inversion is therefore on a locus such that tangents drawn from any point on it to the given circles have a constant ratio; i.e. a circle coaxal with them. COR. Any two circles may be inverted into equal circles; and the locus of the centre of inversion is either circle of antisimilitude. For when p 1 = p. 2 ; t^jt.f = rjr, 2 ; therefore, etc. (Art. 114, 2.) Otherwise thus : Since a circle and two inverse figures invert into a circle and two inverse figures ; if the origin 246 INVERSION. be taken on either circle of antisimilitude this circle inverts into a line. Therefore any two figures the inverse of each other ivith respect to a circle invert into reflexions of each other with respect to a line. (Art. 121, Ex. 6.) EXAMPLES. 1. Show how to invert any three circles into equal circles. [The centres of inversion are the points of section of the circles of antisimilitude of the given ones taken in pairs.] 2. How many centres of inversion are there in the solution of Ex. 1 ? [The three external circles of antisimilitude are coaxal (Art. 88, Ex. 13), and therefore meet in two real or imaginary points. Also since every two internal and one external circles of anti- similitude are coaxal, there are in all eight centres of inversion real or imaginary. J 3. Any three circles are unaltered by inversion with respect to their common orthogonal circle. For this reason the latter lias been named the Circle of Self -Inversion of the given ones. 4. To invert the sides of a triangle into a. Three equal circles. /5. Three circles whose radii have any given ratios p : q : r. [o. The centres of the in- and ex-circles are the four origins. /3. The distances of the origin from the sides are in the inverse ratios p : q : r.] 125. Theorem. The tangents at corresponding points A and A of two inverse figures make equal angles with their line of connexion A A . For take the corresponding points B and B on the curves which are consecutive to A and A. Join AA and BB ; they each pass through 0. The lines AB and A B joining consecutive points may be regarded as tangents to the respective curves ; also THEOREM. 247 since ABA E is a cyclic quadrilateral and the angle at indefinitely small, we have (Euc. III. 22) therefore TAA is an isosceles triangle. 126. Theorem. The angle of intersection of two curves is similar* to that of their inverses -at the corresponding point. For the angle between any two curves is the angle between the tangents at their points of intersection. But the tangents determine two isosceles triangles (Art. 125) on the line A A ; therefore, etc. If the centre of inversion is external or internal to both circles the angle remains unaltered ; if on the other hand it is external to either and internal to the other, the angles of intersection before and after inversion are supplemental. * The angle of intersection of two circles undergoes as a figure no change of form under the process of inversion, but often does as a magnitude, change into its supplement, under that process. "In the application of the theory of inversion to the geometry of the circle, this circumstance must always be attended to. The two cases of contact, external and internal, come of course under it as particular cases ; and in but one case alone, that of orthogonal intersection, which presents no ambiguity, can the pre caution ever be entirely dispensed with. " Townsend s Modern Geometry of the Point, Line, and Circle, Art. 407. 248 INVERSION. 127. Amongst the various results which follow from the preceding Articles, we note 1. Any two circles meeting at an angle a invert from either point of intersection into two lines inclined at the same angle, e.g. two orthogonal circles into two lines at right angles. 2. Three mutually orthogonal circles, e.g. the three real polar circles of the triangles formed from an orthocentric system of points, invert from any of their points of intersection into a circle and two perpendicular diameters. 3. Any three circles invert from any centre on their common orthogonal circle into three others whose centres are collinear ; the line of collinearity being the inverse of the common orthogonal circle. 4. A system of circles having more than one ortho gonal circle inverts into a system having more than one orthogonal line. 5. In 4 the intersections of the common orthogonal circles are evidently the limiting points of the given system which is coaxal. (Art. 86.) Hence for any centre of inversion : a. A coaxal system inverts into a coaxal system ; or b. A circle and a pair of inverse points invert into a circle and a pair of inverse points ; and for a centre of inversion at either of the limiting points : c. A coaxal system inverts into a concentric system, the common centre being the inverse of the second limiting point with respect to the circle of inver sion. PARTICULAR CASES. 249 6. A system of concurrent lines inverts into a coaxal system of the common point species, the common points being the centre of inversion and the in verse of the point of concurrence. 7. An angle and its bisectors invert into two circles and their circles of antisimilitude. (Art. 109.) 8. If two circles, concentric with the extremities of the third diagonal of a cyclic quadrilateral, are described cutting the given one orthogonally; they are mutually orthogonal, and their points of inter section O l and 2 are therefore inverse points with respect to the given circle. Hence if we take O l and 2 as centres of inversion we arrive at the following results : The three circles invert into a circle and two rectangular diameters ; the vertices of the quadrilateral, which are inverse points with respect to the circles, invert into inverse points in the same order with respect to the lines, i.e. form the vertices of a rectangle. Thus the vertices of any cyclic quadrilateral may be inverted into those of a rectangle, and the centres of inversion are inverse points with respect to the circle. 9. A circle may invert into a circle having its centre at a given point A. For let A the inverse of A be the centre, and AA the radius of inversion. Then the given circle and pair of points A and A inverse to it, invert into a circle and a pair of inverse points ; but the inverse of the centre of inversion A is at infinity; therefore A is the centre of the inverse .circle. 250 INVERSION. 10. Two parallel lines invert into two circles touching externally if the origin is between the lines ; and internally if the lines are on the same side of the origin. 11. If a quadrilateral A BCD inverts into a parallelo gram from an origin ; the pairs of circles BOC, AOL and CO A, BOD touch at 0.* SECTION III. ANHARMONIC RATIOS UNALTERED BY INVERSION. 128. Theorem. // A, B, C, D be any four concydic points and A , B , C , D their inverses ivith respect to any circle of inversion, then This property has been shown to hold for any four points and their inverses, and is therefore true in the particular case when they lie on a circle ; hence the an- harmonic ratios of four coney clic points are equal to the anharmonic ratios of their inverses with respect to any circle of inversion. Particular cases have been noticed in Art. 121, Cors. 1, 2. 129. Problem. To invert a regular cyclic polygon ABC. . . from any origin P. The circumcircle ABC... inverts into a circle a/3y. ..; the diameters AA y BB , CC . . . into circles passing through the origin P and cutting ajBy. . . orthogonally in act , /3/3 , 77"" : " Hence a construction for the required centres of inversion. HARMONIC POLYGON. 251 They therefore pass through Q the inverse of P with respect to the inverse circle and thus form a coaxal system of the common point species. (Art. 127, 6.) Also the chords aa, /3/3 , yy ... meet in a point K on PQ (Art. 72, Ex. 6). On the primitive figure any side BC of the polygon and any diameter A A meet the circle in a harmonic row of points ; therefore (Art. 1 28) on the inverse figure fiyaa is an harmonic row ; hence (3a/ya = /8a /ya , or, by Euc. III. 22, the diagonal aa of the quadrilateral is the locus of a point such that its distances from either pairs of sides which meet at its extremities are proportional to the lengths of the sides; similarly for the quadrilaterals y(5/3/3 , etc. Therefore the distances of the point K from the sides of the polygon a/5y... are proportional to the sides. For an Harmonic Quadrilateral, K is evidently at the intersection of the diagonals ; and the inverse of the regular polygon possessing, as has been shown, a corre- 252 INVERSION. spending and more general property has been termed by Casey an Harmonic Polygon. Definitions. The point K is called the Symmedian Point of the Polygon ; and if the ratio of any perpen dicular from K to half the side on which it falls is tan w, then o> is the Brocard Angle of the Polygon. For the properties of harmonic polygons the reader is referred to Casey s Sequel to Euclid, Supplementary Chapter, Section VI. 130. Cosymmedian Triangles. Let ABC be a tri angle K, its symmedian point, and let the lines AK, BK, CK meet the circum-circle again in A , B , . If the circle of inversion be K, p where the vertices of ABC invert into A, B\ C . Also since BCAA is a harmonic quadrilateral, therefore B C A A is harmonic, or A A is a symmedian of the triangle A B C ; similarly the other symmedians are B B and C C. It appears thus that the two triangles have the same symmedian lines, symmedian point, Brocard Circle, Brocard Angle, Brocard Points, etc. On account of these relations they have been termed Cosymmedian Tri angles* * Their properties were first stated by Casey before the Royal Irish Academy in December, 1885. A further account of them will be found in Milne s Companion. EXAMPLE. 253 EXAMPLE. 1. If ABC be a triangle and G its centroid ; AA , BB , CC chords of the circum-circle passing through O ; the symmedian point of A B C is on the diameter which contains Tarry s point. (Vigarie.) [Let the circle be self -in verted from G as origin and the points J, B, C invert into A , B , C respectively. Let A A", BB", CC be the symmedian chords meeting in K. If a circle CGC" meet GK in the point L then KQ.KL~KC.KCTi and similar relations hold for the circles A GA " and BGB" : there fore these three circles meet in a second common point Z, which is the inverse of K , the symmedian point of A B C . Let J be the inverse of K with respect to the circum-circle ABC, and it follows that KO.KJ=KG . KL = t\\e power of K with respect to the circum-circle. Hence OGJL is a cyclic figure, and the angle GOK=L. It has been shown (Art. 67, Ex. 18) that Tarry s point on the circum-circle corresponds to the circum-centre on the Brocard Circle with respect to ABC and Brocard s first triangle, and that G is their common centroid; hence angle GNO=GOK and GRO=GKO = GF 0. Therefore OGKF is a cyclic quadrilateral, and (Euc. III. 21) the points F, K, F are collinear. There fore KO.KJ = KG .KL = KF. KF or F, J, L are collinear, the line being the inverse of the circle OGKF with respect to K as origin. 254 INVERSION. Now the circum-circles of ABC and GFL cut each other orthogonally since the angle OFG = L ; hence the inverse of the latter from G is the diameter NR, and therefore L inverts into a point K on it ; therefore, etc. This solution is due to M Cay.*] MISCELLANEOUS EXAMPLES. 1. The six circles that can be described to touch three given ones A, B, C, two externally and one internally and two internally and one externally, are in pairs the inverses of one another with respect to the common orthogonal circle of A, B, C. [Invert with respect to the common orthogonal circle of A, B, C, and since A , B, C remain unaltered after inversion, three of the circles of contact invert into the remaining three ; therefore, etc.] 2. The eight circles of contact with A, B, C have a common circle of antisimilitude. [As in Ex. 1 they are in pairs the inverses of each other with respect to the common orthogonal circle of A, B, and C.] 3. Three circles are described touching the ex-circles of a triangle, two externally and one internally ; prove that they each pass through the centre of Taylor s Circle. [Invert with respect to Taylor s Circle and the circles in question invert into, the remaining circles of contact, which in this case are the sides of the triangle ; and since the circles invert into lines they each pass through the centre of inversion.] 4. If ABC be a triangle ; C, p a circle of inversion, A and B the inverses of A and B ; to prove that 2.s = p 2 sin C/r where / is the radius of the in-circle of A B C. [\Ve have AC=p*/A C, BC=p 2 /B C and AB A B ^pt/A C . B C, hence by addition * "Mathematical Questions with their Solutions," from the Educational Times, vol. lii., p. 73. EXAMPLES. 255 5. Mannheim s Theorem.* Having given the vertical angle C and radius / of the in-circle of a triangle A B C : the envelope of the circum-circle is a fixed circle. [From Ex. 4 by inverting from the vertex with respect to a circle of inversion (7, p, the inverse of the circum-circle is the base AB of a triangle of known perimeter ; and since the inverse envelopes a circle, viz., the ex-circle of the triangle ABC ; therefore, etc.] 6. A variable circle touches the base of an isosceles triangle at its middle point ; prove that the chords of intersection with the sides that meet within the circle envelope a fixed circle. (M Vicker.) [See the property of Art. 61, Ex. 1.] Qa. By inverting from the vertex derive Mannheim s Theorem. 7. Two circles meet at an angle w, and are such that 2 cos u=Jr/R ; prove that a triangle may be inscribed to one and circumscribed to the other. Hence find the locus of a point from which two circles may be inverted into two others, so that a triangle may be inscribed to one and circumscribed to the other. 8. A variable chord XX of a circle 0, r passes through a fixed point Q ; to prove that the circum-circles of the triangles QOX and QOX envelope coaxal systems. [Let P be the inverse of Q with respect to the given circle. The circles in question invert into the right lines PX and PA , which by Art. 72, Cor; 5, touch each of two concentric systems, viz., the in- and ex-circles of the triangle PXX .} 9. Prove that the vertices of a triangle and the reflexions ^u ^2> ^5 f anv point with respect to the sides may be inverted into the vertices of a triangle and three collinear points on the sides. (Russell.) [The circle BCO^ CA0. 2 , AB0 3 meet in a point P (Art. 79, Ex. 15), which is seen from Euc. III. 22 to be on the circum-circle of 0!0 2 03- Inverting from P ; therefore, etc.] * This well-known property is thus seen to be the inverse of : Having fjiven the vertical angle C and either of the quantities s or s-c ; the envelope of the base in a circle. 256 INVERSION. 10. Any triangle ABC and a Sinison line ATZmay be inverted from the pole of the line into a triangle X Y Z and Sinison line A B C . 11. If four circles be mutually orthogonal, and if any figure be inverted with respect to each in succession ; the fourth inversion will coincide with the original figure. [The following proof has been given by M Cay : Invert the four orthogonal circles from a point of intersection of any two of them. The latter invert into rectangular lines ; a third circle becomes one p, cutting these lines at right angles ; and the fourth after inver sion (p), since it cuts the third at right angles and is concentric with it, satisfies the relation p 2 + /> 2 =0 or p 2 = - p" 2 . Let PI, P 2 , PS denote the successive inversions of the point P on the inverse figure ; since OP 2 . OP s =p 2 and OP- 2 = -OP, therefore OP. 0/3= -p 2 , or the inverse of P 3 with respect to the imaginary circle of radius ip, whose centre is at 0, coincides with P ; there fore, etc.] 12. "The centres of the four circles circumscribed about the four triangles formed by four right lines are concyclic." Prove this theorem by inversion from the point P common to the four circum- circles, and show that the circle passes through P. [It is evident that, 1, the four lines invert into four circles passing through P ; 2, the four circles into lines joining the remaining pairs of intersections of the circles in 1 ; 3, the centres of the four circles into the reflexions of P with respect to the four EXAMPLES. 257 lines on the inverse figure by Art. 123 ; but these are collinear ; therefore, etc.] 13. Let T be a common tangent to two circles, t and t the tangents to them from any point ; if the circles are inverted from as origin prove that T 2 /tt is unaltered. 14. The vertex C of a given angle ACE is fixed ; required to find the envelope of the circle ACE where A and E are points on a given line. 15. A chord AE of a circle passes through a fixed point P ; find the locus of the point of intersection of the circles passing through P and touching the given one at A and B. 16. If two circles be inverted into any two others ; for each pair the square of the common tangent divided by the product of the diameters are equal. [Compare Art. 126 and Art. 4, footnote.] 17. Prove Casey s relation among the common tangents to four circles all of which are touched by a fifth (Art. 7) by the inversion of a system of four circles touching a line. 18. Draw two parallel lines and describe a number of circles touching the lines and each other in succession. Invert this system from a point on a diameter of any circle perpendicular to the lines and deduce the following theorem : A, B, C are three collinear points, and circles 2T, JF, Z are described on the segments BC, OA, AE respectively. A system of circles is drawn as in figure to touch each other and the given ones, if C n) p denote the nth circle to prove that the distance of its centre from AB = 2np. (Pappus.) R 258 INVERSION. 19. If three circles Ar lt J5r 2 , O 3 touch one another in pairs ; prove by inversion that the radii of the circles which touch them with contacts of similar species are *Wa Sr^ 2A where 2A is the area of the triangle ABC. [Invert from the point of contact of Br^ Cr 3 with a radius equal to the tangent to Ar^ ; etc.] 20. The rectangle under the distances of the ex-centre of simili tude of two circles from their radical axis and in-centre of simili tude is equal to the constant product of antisimilitucle. [The circle of similitude inverts from either centre of similitude into the radical axis of the given circles. 20a. Prove that the poles of the radical axis of two circles with respect to the circles are harmonic conjugates with respect to the centres of similitude. [This is the inverse of the theorem : The polars of either centre of similitude with respect to two circles are equidistant from their radical axis ; the circle of antisimilitude being taken as circle of inversion.] 21. A variable circle ABCD touching two fixed circles externally meets their radical axis in L and and the pair of transverse common tangents in A, C and B, D respectively ; prove the follow ing properties of the figure : 1. The limiting points M and N of the circles are the middle points of the parallel sides of the quadrilateral PQRS. 2. The lines AB and CD move parallel to the direct common tangents PQ and RS respectively. 3. The vertices of ABCD lie on the lines joining and L to the limiting points. 4. BC and AD envelope circles concentric with M and N respec tively. To prove 1. Since the four common tangents to the two given circles form a common escribed quadrilateral, the diagonals of which are concurrent with the diagonals of the corresponding inscribed quadrilaterals ; therefore, etc. See Art. 67, Cor. 6. EXAMPLES. 259 2. Let the points A and B and the given circles be numbered 1, 2, 3, 4. Apply Casey s relation connecting the common tangents to four circles all touched by a fifth and reduce, it follows that AZ+BZ QC AB. Hence AB is constant in direction and PQ is a particular position of it, therefore AB and PQ are parallel ; similarly CD and RS are parallel. 3. To prove that the points 2), Z, N are collinear. Invert the figure from D as origin. The circles, their radical axis and pair of inverse points invert into three coaxal circles, one of which passes through the origin, and their limiting points ; also the circle ABCD inverts into the direct common tangent of the latter system. It follows easily (Art. 92, Ex. 5) that the inverses of ^V and L pass through D : therefore, etc. 4. BM bisects externally the base angle B of the triangle ZBC\ since LO bisects internally the vertical angle of the isosceles triangle LMN \ similarly CM bisects externally the other base angle, therefore M is the ex-centre of BCZ. NOTE. This property, communicated by Mr. Charles M Vicker, is a manifest extension of Mannheim s Theorem. For if either of the circles is reduced to a point Z, we have of the triangle BCZ the vertical angle Z fixed in magnitude and position and the ex-circle ; since the variable circum-circle BCZ (i.e. ABCD) envelopes a circle to which the vertex and centre of the ex-circle are a pair of inverse points ; therefore, etc. 260 INVERSION. 22. Prove the converse of Casey s Theorem (Art. 7), showing the relation which holds between the common tangents to four circles, all of which are touched by a fifth. [Invert the circles 1, 2, 3 into equal circles (Art. 124) A, r; B, r\ C, r; and find the inverse D, r of 4 with respect to the same circle of inversion. The relation 223. 14 = holds for the four circles after inversion (Art. 126) ; also the tangents 23, 31, 12 are equal to the sides of the triangle ABC formed by joining the centres of the equal circles. Now describe a circle concentric with D and a radius equal to r <*> r^ and the tangents from A, B, C to it are respectively equal to 14, 24, 34. Hence the general relation has been reduced to the corresponding one for three points and a circle. It is easy to see that the circum-circle of ABC touches D, r *> r : ; for by the converse of Ptolemy s Theorem the limiting points of the two circles are on ABC ; therefore, etc. Fry.] NOTE.- The method of inversion so useful in Modern Geometry was discovered by the Rev. Dr. Stubbs of Trinity College, Dublin, in the year 1843. His valuable memoir 011 the subject is to be found in the Philosophical Magazine, Nov., 1843, p. 338. About the same time, Dr. Ingram published his researches in the Transactions of the Dublin Philosophical Society. See vol. i., p. 145. CHAPTER XII GENERAL THEOEY OF ANHAEMONIC SECTION. SECTION I. ANHAEMONIC SECTION. 131. Definitions. Let a line AB be divided by two variable points C and D such that AC/BC-t-AD/BD is a constant ratio ( = AC). The value of AC is thus -CA.BD/BC.AD, and is termed the Anharmonic Ratio in which the seg ment AB is divided by the points C and D. Similarly the anharmonic ratio of CD divided at A and B is CAIDA + CB/DB or -CA . BD/BC . AD. The points C and D are Conjugate or Corresponding Points in the Roiv A, B, C, D, and AB and CD are Conjugate Segments. It is obvious that conjugate seg ments divide each other Equianharmonically, i.e. the anharmonic ratio of AB divided at C and D is equal to that of CD divided at A and B. 132. Let the four points A, B, C, D be divided into three pairs of opposite segments BC, AD ; CA, BD- } AB, CD ; then the anharmonic ratios of 261 262 ANHARMONIC SECTION. BG divided in A and D = BA/CA +BD/CD=\, (1) CA divided in B and D = OB/45 -5- CJ9/4D = /*, (2) and AB divided in (7 and D = AC/BC+-ADjBD = v, (3) or their reciprocals ; since a segment divided in A and D is divided in the reciprocal anharmonic ratio by D and A. These three fractions X, /m, v and their reciprocals are the six anharmonic ratios of the four points A, B, C, D. NOTE. Let a line AB be divided internally in a variable point X and externally in X such that AXfBX=k . AX jBX . As X approaches B, AXJBX increases ; therefore the conjugate point X approaches B simultaneously. For let AX a and BX = b and we have ax > or < " according as a> or < b. b-x b but a>b, thus it follows that as X moves towards B the ratio A X jBX continually increases, and becomes infinitely great when the variable point coincides with B. Here also it coincides with its conjugate X, and the point B is thus a Double Point of the systems described by the variables A" and X . Similarly A is a double point. Again, as X recedes from B on the line produced, A" approaches Jf the middle point of AB In the limit when A" is at infinity and AX JBX therefore equal to unity, its conjugate X(=-l f ) divides the line in the simple ratio AP]PJ3 = k. Similarly when A" moves to infinity, its conjugate X { Q) gives the relation AQ/BQ = I/& ; and the two points whose conjugates are at infinity are isotomic conju gates with respect to AB. We may note here, and we shall see presently, that when the corresponding points of the two systems move in the same direction the double points are imaginary. 133. Problem. To express all the Anharmonic Ratios of A BCD in terms of any one of them (X). Since BC . AD + CA . BD + AB . (7D = 0; dividing by AB . CD, we have PROBLEM. 263 BC.AD CA.BD , 1 _ AB.CD^AB.CD whence on substituting from Art. 132 Thus generally it follows, by dividing the above equa tion by each of its terms, that The six ratios are therefore X, 1/X, (X-1)/X, These may be expressed as trigonometrical functions of an angle. For let A = sec 2 $. Then the ratios taken in the above order reduce to the following : .sec 2 0, cos 2 <9, sin 2 (9, cosec 2 0, - tan 2 0, - cot 2 0. If two of the ratios are equal, e.g. A = (A - I)/ A, then A. 2 - A + 1 = and A,=w or <o 2 , the imaginary cube roots of unity. In this case the three pairs of ratios have the values o> and <o 2 . If A = 1 the points form an harmonic row, and the remaining ratios are - 1, - 2, - 1/2, 2, 1/2. In speaking of the anharmonic ratio of four points on a line the order in which the points are taken is to be understood. Dr. Salmon introduced the convenient nota tion [ABCD] to denote the ratio into which AB is divided by C and D. [ABCD] is equivalent to AC/BC+AD/BD, wid [ABCD]. [ABDC] = l. EXAMPLES. 1. To prove that [ABCD] = [BADC] = [DCBA] = [CDAB]; and hence when any two constituents of four points are inter changed, the anharmonic ratio of the system remains unaltered, provided the remaining pair be likewise interchanged. 2. If [ABCD] = [ABDC] = K find the value of K. [It is plain that K is equal to its reciprocal, and is therefore unity. The four points form in this case an harmonic system.] 264 A3 HARMONIC SECTION. 3. To prove for any collinear system of points A, B, C, J), E ... that [ABCE]j[ABCD] = [ABDE]. [Expanding the ratios on the left side and reducing ; therefore, etc.] 4. For any two collinear systems of points A, B, C, D, E ... A , B, C , D , E ... having given [ABCD] = [A B C U] and = [A BC E f ], to prove that [B C D E \ \CADE} = [C A D E \, [ABDE]=[A B D E ]. [By Ex. 3.] 5. If [ABCD] = [ABC D f ], prove that [ABCC ] = [ABDI>]. [Expanding the ratios the required result follows by alternation.] 6. If in Ex. 4 [ABCD] = [A B C D }, [ABCE] = [A B C JE l [ABCF] = [A B C F ], etc., etc. ; prove that [ADEF] = [A D E F l [BDEF} = [B D E F \, etc (1) and [DEFG ... } = [DE F G ... ]. 7. If a segment MN is divided equianharmonically by pairs of points A, A , B, B , C, C , etc. ; to prove that 1. [MABC ... ] = [MA B C ... ] and [NABC ... ] = [NA B C 1 ]. 2. [ABCD...-\ = [A B C D ...]. [Since [MNAA ] = [MNBB ] = [MNCC ]= ... etc., by Ex. 5. [MNAB] = [MNA JB ] ; [MA T AC] = [MNA C l etc. Hence by division we have [MABC] = [MA JB C ], etc. ... To prove 2. We have by 1 [MABC] = [MA B C f ] and [MABD} = \_MA B D l therefore by division [ABCD] = [A B C D ]. 8. If a segment MN is divided harmonically by points A and A , B and B , C and C ; to prove that the anharmonic ratio of four of the six points taken in any order is equal to that of their four conjugates, [ABCC ] = [A B C C]. [By Ex. 7. [MABC} = [MA B C ] ; but (hyp.) C and C are inter changeable, therefore [MABC ] = [MA B C] ; dividing these equa tions, therefore, etc., as in Ex. 4.] 9. To prove the converse of Ex. 8, i.e., for any six collinear points A, B, C, A , B , C , if the anharmonic ratio of any four is equal to that of their four conjugates [CABA ] = [C A B A] then 1. The anharmonic ratio of every four is equal to that of their four conjugates. EXAMPLES. 265 2. The segments A A , BB , CO have a common segment of har monic section. [To prove 1. By hyp. since [CABA ] = [C A JB A] ; on rearranging, by Ex. 1, we get [AA BC] = [A AJ3 C ] = [AA C B ]. Therefore by alternation (Ex. 5) [AA BC ] = [AA CJ3 ] = [A AC ] -, similarly for all other combinations. To prove 2. Let MN divide the segments AA and BB harmonically, it divides CC also harmonically. For [MABA ] = [MA B A] (by Ex. 7) and [NABA } = \NA B A] ; also by 1 [CABA } = [C A B A] and [C ABA ] = [C A B A\ hence (Ex. 6) [MiVCC ] = [MNC C] ; therefore, etc. (Ex. 2)]. 10. Show generally for two equianharmonic systems if any two conjugates A and A are interchangeable, e.g., if [ABCD] = [ A B C D ] and [A BCD] = [AB C D ] that 1. Every four are equianharmonic with their four opposites ; 2. The segments AA , BB , CC , DD have a common segment of harmonic section. [By the method of Ex. 9.] SECTION II. ANHARMONIC SECTION OF AN ANGLE. 134. It has been explained in Art. 3 that the anhar- monic ratio of four points A, B, G, D is equal to that of the pencil . A BCD formed by joining them to any point 0. It follows then that all the properties of four collinear points stated in the previous section involve correlative properties of a pencil of rays, and that the latter are immediately derived from the former by aid of the equation BG.AD-.GA.BD-.AB.GD = sin BG . sin AD : sin UA . sin BD : sin AB . sin CD. Also by describing a circle through the vertex of the 266 ANHARMONIC SECTION. pencil 0. A BCD, and denoting by A, B, C, D the points where it meets the legs of the pencil again; since the sines of the angles at are in the ratios of the chords opposite to them we may further obtain from the anhar- monic properties of collinear points corresponding relations amongst points which lie on a circle. 135. The following properties will appear evident : 1. All transversals to a pencil of rays are cut equian- harmonically. 2. A transversal to a pencil drawn parallel to one of its rays D is divided by the remaining three in the simple ratio AC/BC; which is the anharmonic ratio of the pencil. 3. In 2, if the pencil is harmonic, any transversal A B C parallel to D is such that A B ^B C . 4. For any two equianharmonic rows of points A, B, C,D,... and A , B\ , D f , ..., if the lines A A , BB , and CC are concurrent at ; DD and all other lines joining corresponding points of the given systems pass through 0. [This important property is the converse of 1 and follows easily by an indirect proof.] PENCILS IN PERSPECTIVE. 267 136. Theorem. If two lines be divided equianhar- monically such that a pair of corresponding points coincide at their intersection [OABC...] = [OA B C ...] the systems are in perspective ; and reciprocally if two equianharmonic pencils are such that a pair of corre sponding rays coincide on the lines joining their vertices they are in perspective. Let A A and BB meet in P. Join PC, and if possible let PC cut the other axis in C". Then [OABC] = [OA RC"l since the rows are in perspective. But [OABC] = [OA RV] (hyp.) ; therefore {OA RC^^OA RC"], i.e. C and C" coincide. Reciprocally for any two pencils P . ABC, . . . and P. A B G , ... if the rays A, A and B, B f intersect respec tively in X and F, it follows that C and C meet on the line XY. Otherwise thus : The rows [XYZW] and [XYZ W] are equi- anharmonic ; therefore Z and Z coincide. COR. 1. If two pencils are equianharmonic, any two rows passing through the intersection of a pair of corre sponding rays are in perspective. 268 ANHARMONIC SECTION, COR. 2. Through a given point P a line may be drawn across a triangle ABC, cutting its sides in the points Q, R, S, such that [PQRS] = a given anharmonic ratio. [For the pencil (A . PQRS) formed with the row at any vertex A of the triangle is given, and since three of its rays are given the fourth is known.] Def. Lines divided equianharmonically are also said to be divided Homographically. The term homographic is applied in general to the equianharmonic division of figures of the same kind, e.g. lines, circles, etc., etc. EXAMPLES. 1. Every tangent to a circle is cut harmonically by the sides of the escribed square. [In the limiting position when the variable tangent coincides with a side of the square the row of points determined on it are harmonic ; therefore, etc., Art. 81, Ex. 3.] 2. To express the anharmonic ratios in which a variable tangent is divided by four fixed tangents, in terms of the chords of contact of the tangents. [Let P, Q, R, S denote the points of contact of the sides of the escribed quadrilateral, which meet the variable tangent at in A, B, C, D ; the centre of the circle. Then ABCD = . ABCD = O.PQRS, since O A, OP ; O B, OQ ... are four pairs of perpendi cular lines ; therefore the required expressions are QR. P>S:RP.QS:PQ.RS.] 3. For any quadrilateral escribed to a circle at the points P, Q, R, S, each pair of diagonals and a corresponding pair of opposite connectors of the inscribed quadrilateral PQRS are con current. (See Art. 67, Cor. 8.) [To prove that the sets of lines QR, PS, YY 1 , ZZ 1 HP, QS, ZZ\ XX PQ, RS, XX , YY 1 are each concurrent. EXAMPLES. 269 Consider each of the four tangents at the points P, Q, R, S a transversal to the quadrilateral XX YY ZZ . Since consecutive tangents meet on the circle, the tangents at P and Q are cut in the same order at the points P, Z, Y, X and Z, Q, X, Y ; therefore [ PZYX ] = [ZQX Y 1 ] = [QZY X]. Hence PQ, YY , XX are concur rent. Similarly RS, YY and XX are concurrent ; therefore, etc.] NOTE. As the above properties are more generally true for the Conic, we consider an interesting case which arises in the parabola when the fourth tangent is at infinity (Art. 81). Let tangents AC and BC be drawn to a parabola at the points A and B, and a third tangent XY meeting BG and CA in X and Y respectively. Then the equianharmoriic relations easily reduce to BA /CX= CY/A Y ; or a variable tangent divides two fixed tangents in the same ratio. It also subtends a constant angle at the focus. Therefore the foci of the three parabolas described to touch each pair of sides (b, c, etc.) of a triangle ABC at the extremities of the third side (BC) are the vertices of Brocard s second triangle. 4. If a circle touch four others the anharmonic ratios of the points of contact are equal to 23 . 14 : 3l . -24 : 1:2 . 34. [By Art. 7.] 5. The anharmonic ratios of the points of contact of the nine- points-circle with the in- and three ex-circles of the triangle ABC are [As in Ex. 4.] 6. If the anharmonic ratios of four points A , B, C, D on a circle (or conic) be denoted by A, p, v, etc., to prove that the anharmonic ratios of the pencil P. ABCD are X 2 , /x 2 , i> 2 , etc., where P is the pole of the line AB. [Let PC, PD meet the conic again in C", D , and AB in E, O ; then CD , DC , and AB are concurrent at F ; and since C . ABCD=iy. ABCD, [ABCD] = [ABEF] = [ABFG] = X (say); 270 ANHARMONIC SECTION. whence or But [ABEG} = P. ABEG=P. ABCD ; therefore, etc.] 137. Directive Axis. For any two homographic rows of points ABC . . ., AB C ... on different axes L and L , if any pair of corresponding points A and A be each joined to all the points on the other axis, the two pencils A . A B C ...,A. ABC ... are in perspective (Art. 136), i.e. the intersections of the pairs of lines AB^AB^")-, AC , A C (B"); ^D ,4 D,etc., are collinear. We are thus enabled to find a point P on the line L corresponding to a given point P on L. For having obtained the line B"C", join A P and let it meet B"C" in P"; then AP" meets the axis L in the required point. An important point arises out of the consideration of the correspondents to the intersections 0, P, and P of the axes L, L , L" taken in pairs. By means of the general method given above we find that P on the axis L corresponds to on the axis L\ and that P on the axis L corresponds to on the axis L. This shows that the axis L" of perspective of the pencils A. AB C ..., A.ABC... y DIRECTIVE AXIS AND CENTRE. 271 whose vertices A and A were arbitrarily chosen as any pair of correspondents of the given homographic systems, is a fixed line, since it meets each axis in a point cor responding to their intersection regarded as a point on the other. Hence : all pairs of corresponding connec tors (XT , X Y) of pairs of non-corresponding points lie on a line. This line is called the Directive Axis of the given homographic systems. Otherwise thus : Take the two homographic pencils at A" and L and L as transversals to them respectively, then [BCPO] = [C B P 0]-, similarly for the vertex B" it follows that [CAPO] = [A CT Ol therefore by division (Art. 133, Ex. 3) [ABPO] = [B A P O], i.e. the lines AB , A B, PP are concurrent. The same proof applies to the more general case of two systems of points on a conic. 138. Directive Centre. The following property of two homographic pencils is derived from Art. 187 by reciprocation : For any two homographic pencils of rays 0. ABC . . . and . A B C . . . the lines joining pairs of cor- 272 ANHARMONIC SECTION. responding intersections (AB , A B) of non-corresponding rays (A, B and A , B) are concurrent. The point of concurrence is termed the Directive Centre of the systems, and its property just stated may be proved by methods analogous to either of those given in Art. 137 for the directive axis. These are left as useful exercises for the student. 139. Problem. To find a point X on either axis L ivhose correspondent on the other is at infinity (oo ). Since the lines joining A, oo and A , X meet on the directive axis, we have the following construction: through A draw a parallel to L , join A to its point of intersection with the directive axis ; this line meets L in the required point. EXAMPLES. 1. Having given two homographic pencils of rays at different vertices ; to find a ray of either corresponding to a given one of the other. [By means of their directive centre.] 2. If two homographic rows of points are such that the points QO, GO at infinity on the axis correspond, the lines are divided similarly. [For [ABC <x>} = [A BC <*> }, hence A B : BC=A B \ B C \ there- fore, etc.] 3. Having given the vertical angle in magnitude and position of a triangle of constant species, the extremities of the base divide the sides homographically. 4. If the lines A A , BB , CC connecting the corresponding ver tices of two triangles ABC and A B C are concurrent at a point 0, the intersections X, Yj Z of the pairs of sides EC, B C , etc., are collinear (cf. Art. 66). [Join XT and let it meet the lines AA , BB , CC in X , Y f , Z respectively. Then X. OBY B =X. OCZ C =Y. OCZ C = Y. OAX A ; EXAMPLES. 273 therefore [OBY B ] = [OAX A ~\, and since the point is common to both rows the pairs of connectors AS, X Y , A B are concurrent. Therefore also the centre and axis of perspective L of the two triangles divide the corresponding segments AA , BB , CC equian- harmonically.] 5. A variable triangle moves with its vertices on three concurrent lines such that twp of its sides pass through fixed points X and T ; then the third side passes through a fixed point on the line XY. [By Ex. 4.] 6. The lines joining pairs of corresponding points of any two figures in perspective are cut homographically by the centre and axis of perspective. 7. Any line passing through either centre of perspective of two circles is cut in a constant anharmonic ratio by their radical axis. 8. Every four of the six points A", F, Z, X , Y , Z in Ex. 4 are equianharmonic with their four opposites. 274 ANHARMONIC SECTION. 9. In the figure of Art. 137 prove the relations 1. [BCPO] = [B C OP] = [B C"PPl [CAPO]=[C A OP] = [C"A"PP-}. 2. [ABCP]=[A B C 0] = [A"B C"Pl [ABCO] =\A ffC P] = [A"B C"P l NOTE. It will be seen that the triangle AB C" is inscribed to A BC and escribed to B C A", and more generally that of this system of three triangles each is inscribed to one and escribed to the other of the remaining two. The vertex A and opposite side B"C"of the triangle A B"C"form with the extremities B and C of the corresponding side of A BG to which it is inscribed a row of points B, C, A, P. Similarly the vertex A and opposite side BC of A BC form with the correspond ing side B C of the triangle A B C to which it is inscribed a row B, C , A , 0. But these rows are equianharmonic (Ex. 8, 2) ; hence for such a system of triangles the vertex and the opposite side of each divide homographically the corresponding side of the triangle to which it is inscribed. Again, B C"PP is the row of points formed by the extremities of the base B"C" and its intersections with the corresponding sides BC and B C of the remaining triangles. But hence the sides of each are cut homographically by the corresponding sides of the other two. Let the point C vary along the axis L . Then the lines A C and BC turn around the fixed points A and B A" and B" move on the lines A C and BC , and the directive axis passes through the fixed point C". In this case A B C is a variable triangle inscribed to A B C and escribed to ABC", both of which are fixed. Hence for a variable triangle A B C inscribed to a given one A B C, if two of its sides pass through the vertices A and B of a triangle escribed to the latter, its third side passes through the third vertex C". Let us now consider two positions of the variable triangle A B C . Since its sides pass through the fixed points A, B, C" respectively, ABC" is a common inscribed triangle. Hence when two triangles are each inscribed to a third A B C, if the sides A"B", etc., and opposite ver tices C , etc., divide the corresponding side A B of A B C in a constant CORRESPONDING POINTS. 275 anharmonic ratio [A B C P ], the intersections of their corresponding sides determine a common inscribed triangle ABC" which is escribed to AB C. And the vertex C" and opposite side AB cut the corresponding sides B"C", etc., in the above constant anharmonic ratio. 140. Theorem. For any tiuo homographic rows of points ABC...X and A B W ... X\ if X and X be the points whose correspondents oo and oo are at infinity ; to prove the relations AX .A X = BX. B X = CX . G X = etc. Since A, A -, B, B \ X, oo ; oo, Jf are four pairs of cor responding points [ABXoo] = [A B w X ]. Expanding and reducing, this relation becomes AX/BX=l~A X /B X -, therefore AX. A X = BX. B X , etc., etc.; or:// vari able points A and A be taken on fixed lines L and L respectively such that the rectangle under the distances from two fixed points X and X on the lines is constant, they describe homographic systems. COR. 1. When the vertical angle of a triangle of con stant area is given in magnitude and position, the extremities of the base divide the sides homographically. In this case the points X and X , whose correspondents are GO and oo, are supposed to coincide at the intersection of the axes. By Art. 81, Ex. 3, we see that the envelope of the base is a conic ; and by Ex. 29 of the same article the curve is a hyperbola whose asymptotes are the given axes. COR. 2. Any two homographic rows of points may be so placed that the corresponding segments A A , BB , etc., may have a common segment of harmonic section. Place the systems so that the axes L and L and the 276 AN HARMON 1C SECTION. points X and X are coincident. The equations of the article are then written XA . XA = XB . XB = XC . XC = P \ Describe a circle with X as centre having A, A \ B, J9 ; etc., pairs of inverse points, and let it cut the axis in M and N. MN is the common segment of harmonic section by Art. 70, but it is imaginary when A and A lie in opposite directions from X. Def. Two homographic systems of points on any axis which have a common segment of harmonic section are said to be in Involution, and the corresponding points A,A \ B, B ; etc., are Conjugate Points of the Involution. We have seen in Cor. 2 that there always exists a pair of points, real or imaginary, each of which regarded as belonging to either system is coincident with its corre spondent of the other. These are the Double Points (M t N) of the involution, and are connected with the systems by the equations [MNBC\ = [MNB C l [MNCD] = [MNOD l etc., etc., [MABC...] = [MA B C ...] and [NABC...] = [NA B C ...~]. See Art. 133, Ex. 7. COE. 3. In any two homographic rows of points on a common axis the double points M and N are found from the equations * XA .X A = XB. X B . ..=XM. X M= XN . X N ; they are therefore equidistant from X and X . * If the distances OA, OA from any point on the axis be x, x , it follows that (x - OX)(x - OX ) = const., a result of the form = (cf. Art, 143). PROBLEM. 277 141. For any two homographic rows of points we have seen how to find the correspondent P of any point P, , by means of the directive axis, Art. 137, and /3 by the formula XP . X P = const. It will now be proved that two given homographic rows can be generated by the revolution of either of two determinate angles around fixed vertices, the positions of the latter and the mag nitude of the angles depending on the equal values [ABCD. . .] and [AB C D . . .] and the positions of the axes. 142. Problem. If ABC ... and A E C ... be any two homographic rows of points; to find two points such that the angles subtended at them by the segments A A, BB, etc., joining pairs of corresponding points are equal. Let E and F be the required points ; X, X the corre spondents of oo and <x> (Art. 139). Since AEA is a constant angle, if any point P on L coincides with X, EP is parallel to the axis L . Similarly if Q and X coincide, EQ is parallel to L. Hence the lines EX and EX are equally inclined to L and L , or the angles AXE and AX E are equal. 278 ANHARMONIC SECTION. Again, the angles subtended at E by any two points A and X and their correspondents A and oo are equal (hyp.) ; therefore in the two triangles AEX and EA X we also have the angles AEX and EA X equal, and the triangles are similar. Hence (Euc. VI. 4) AXJXE=EX IX A and EX.EX = AX.A X = coust. (Art. 140). Now in the triangle XEX we are given the base XX fixed, the difference of base angles and rectangle under the sides ; therefore the vertex E is one or other of two fixed points E or F, which are obviously the opposite vertices of a parallelogram with XX as diagonal. COR. 1. The angles AEA, AXF, and A X Faxz equal. For if A and X coincide, EA is parallel to L ; there fore AEA is equal to the angle between EX and L or between FX and L, since EX and FX are parallel. COR. 2. The triangles AEA , AXF, and EX A are similar. [For by similar triangles AEX and EA X we have AX/AE=EX /EA , but EX = FX, hence AX/AE = FX/EA , or by alternation AX/XF=AE/EA ; therefore, etc. (Euc. VI. 6).] COR. 3. If denote the point of intersection of the axes L and L , the points E and F are isogonal conjugates with respect to the variable triangle OAA. [By Cor. 2, FAX = EAA and FA X = EA A ; there fore, etc.] COROLLARIES. 279 COR. 4.* The product of the perpendiculars p and p from E and F on the variable line A A is constant (pp = k 2 ). [By Cor. 3.] COR. 5.* The locus of the intersection of every two rectangular positions of A A is a circle the square of whose radius (/o) is given by the equation p 2 = 2k z + S 2 , where 2S = EF. COR. 6. A variable line cutting two fixed lines homo- graphically cuts all positions of itself in a system of points A"B"C"... such that [ABCD . . .] = [A ffViy. . .] = [A"B"G"D". . .]. Draw the directive axis XYZ ... of the system as in figure. Then OX and OA", divide the angle LOL of the quadrilateral PXP O harmonically (Art. 68). Similarly for OFandO".... Hence we have [O.XY...] = [O.A"B"...] Art. 133, Ex. 7. But Therefore [A B C . . .] = [A"B"C". . .]. * These properties respectively may be otherwise stated : A variable line A A cutting two fixed axes homographically envelopes a conic of which E and F are the foci. The locus of intersection of rectangular tangents is a circle (the Director Circle). 280 ANHARMONIC SECTION. Con. 7. If a variable line meet two fixed circles in a harmonic row of points, it intersects all positions of itself homographically. [For the rectangle under its distances from the centres of the circles is constant, Art. 78, Ex. 12; therefore, etc., Cor. 4] COK. 8. A variable line meeting two fixed circles such that the chords intercepted by them are in a fixed ratio cuts all positions of itself homographically. [By Art. 90, Ex. 8.] 143. If the distances of any point from four points A, B, C, D on a line L passing through it be denoted by a, /3, y, x, and the distances of any point measured along another line L to A , B , C , D be similarly a, f?, y , x, the two systems of points are homographic if which when multiplied out is of the form Axx +Bx + Cx +D = 0, .................. (1) an equation which enables us to determine the position of any point of either system corresponding to a given one in the other. (See Art. 140, Cor. 3.) We have seen that the lines joining corresponding- points envelopes a conic touching L and L . In the par ticular case when #=oo in (1) the simultaneous value of x is also oo, and the corresponding conic is therefore touched by the line at infinity. It follows obviously that when A = in the above equation the conic is a parabola. Thus if a variable line be drawn cutting the sides a PARTICULAR CASES. 281 and b of a triangle ABC in X and Y such that lAY+mBX = const, it envelopes a parabola to ivhich the tiuo sides of the triangle are tangents. If the axes L and L are coincident and B = C in (1), x and x are interchangeable in the equation and, as will be more fully explained in the next chapter, the two systems are in Involution. The double points of two systems on a common axis are found from (1) by putting x = x , in which case the equation reduces to the form Ax z +(B + C)x+D = 0. EXAMPLES. . 1. If the distances of two pairs of collinear points A, B and A t B from an origin on the line be denoted by the roots of the equa tions ax 2 + Zbx + c Q and a x 2 + 2b x + c = Q, they form a harmonic row if ac + a c-266 = 0. 2. Having given two of the auharmonic ratios of four collinear points equal, prove that CHAPTER XIII. INVOLUTION. 144. When of two systems of points A, B, C, ...; A , B , C , ... on any line or circle any three pairs A, A \ B, B ; C, C which correspond are connected by a relation of the form [BCAA] = [B C f A A\ it has been proved in Art. 133, Ex. 9, 1. that every four and their four oppo- sites are equianharmonic ; 2. that A A , BB , CC , ... have a common segment of harmonic section. By Art. 140, Def., we may therefore regard either of these properties as a criterion of points in Involution. Now since [BCA K] = \EC AE\ t by expanding and reducing we get \ BA CIT_ AC GA AH BC" a result previously arrived at in Art. 64, where it was shown by the application of Ceva s Theorem that a straight line drawn across a quadrilateral is cut in involu tion; the conjugate points A, A , etc., being the intersections of the line with the pairs of opposite connectors of the figure. Again, if a pencil of six rays be taken and a circle described through the vertex cutting the rays in points 282 EQ UA TIONS OF IN VOL UTION. 283 A, A ; B, B ; C, C , they form a system in involution if sin BOA sin COB s sin CO A sin A OB sin BOC The criteria (1) and (2) are called Equations of Involu tion. 145. It has been noticed in Art. 1S$, Ex. 10, that when any two conjugates A and A of two homographic systems are interchangeable, every two are interchangeable, and A A , BB , CC ... have a common segment or angle of harmonic section. It follows that " when any one point on an axis, or ray through a vertex, has the same correspondent to which ever system it be regarded as belonging, then every point on the axis or ray through the vertex possesses the same property." * In illustration of this theorem, let the correspondents be joined in pairs to any point (A") on the directive axis of the systems (Art. 137). Then the corresponding rays A"B, A E are interchange able, their productions through A" being A"C , A"C \ therefore The locus of a point at which two homographic rows subtend a pencil in involution is their directive axis; and similarly, or by reciprocation, a variable line meeting tivo homographic pencils at a system of points in involu tion passes through their directive centre. 146. A system of points in involution on a line is com pletely determined when two pairs of its conjugates A, A \ B, B are given ; and the conjugate G f of any point * Townsend, Modern Geometry, vol. ii. p. 276. 284 INVOLUTION. G is its inverse with respect to the circle described with AB and A E as a pair of inverse segments. If the radius of the circle is indefinitely great, one of the double points (N) is at infinity, and therefore (Art. 72, Cor. 3) MA=MA f , MB = MB , etc., etc. ; that is, if one of the double points of a system in involution is at infinity, the segments A A , BB , CC f ... have a common centre, viz., the other double point. Also a variable segment AA of constant length moving along a given axis determines two systems of points in involution the double points of which are imaginary. 147. Theorem. If two chords AA\ BE of a circle meet in (7, any line through C which meets the circle in and determines a system of points A, A , B, B ; 0, in involution. Let AB and 00 meet in Z (Art. 64, iv. fig.). Then the pencil B. AB OO is equianharmonic with the row of points ZCOO it determines on the transversal to it through C. For a similar reason [ZCOO ] = A . BA f OO = [A BO 0], from which relation it follows that every four of the six cyclic points and their four opposites are equianharmonic. The concurrency of the chords AA\ BB , 00 , being involved in this relation, furnishes a geometrical explana tion of the theorem of Art. 133, Ex. 9 (1). The following generalized statement is a direct inference of the preceding : If through any point P, inside or outside a circle (or conic) a number of chords be drawn to cut the curve in A, A ; B, B -, (7, C , ..., the two systems ABC..., A B O ... EXAMPLES. 285 are in involution, and (Art. 64, III.) the polar of P meets the circle in the double points, real or imaginary.* EXAMPLES. 1. A variable line passing through either centre of similitude of two circles cuts them in four equianharmonie systems of points. 2. A variable circle cutting two given ones at equal or supple mental angles divides them equianharmonically. 3. If two circles Fj, F 2 cut two others at the same angles a and ft in the points A, B, C, 1) and A\ B , C , D , prove that [AA 1 , BB, CC , Dlt are concurrent at the external centre of similitude of F 1} F 2 . Cf. Art. 113, Ex. 12.] 4. More generally for any number of circles FI, F 2 , ... V n , prove that \AA A\ . .] = [BB B". . .] = [CC C". . .] = [DD D". . .]. 5. In Ex. 3, if the angles a and /? are right, the anharmonic ratio of the four points of intersection of the variable circle is equal to that of the four points on their common diameter. 6. If two triangles ABC, A B C inscribed in the same circle are in perspective at 0, and from any point P on the circle lines PA , PB , PC" are drawn meeting the sides of ABC in X, Y, Z, the points X, Y, Z, are collinear. [The Pascal hexagons PB BACC , PC CBAA , PA ACBB have YOZ, ZOX, A"0F as Pascal lines ; therefore, etc.] 7. If P denote the point on the circle corresponding to P in the perspective, and the lines FA, P B, PC meet the sides of A B C in X , Y , Z , 1. X , Y , Z are collinear with X, F, Z and the six points are in involution ; 2. [XYZO] = [X Y Z 0]. (Townsend, vol. ii. p. 208.) * When the point is outside its polar cuts the circle in real points M and N which divide A A , BB , CC ... harmonically, and are therefore the double points of the involution ABC ..., A B C .... t It follows directly that the anharmonic ratio of four points on a circle is unaltered by inversion; the circle of inversion in this case being either circle of antisimilitude of F] and F 2 . 286 INVOLUTION. 8. A variable circle cutting three fixed circles at equal or similar angles determines six home-graphic systems of points on the circles. [Take two positions of the variable circles cutting the given ones at equal angles a and ft respectively ; then each of the given ones cuts a coaxal system (Art. 114, Ex. 10) at the same angles a and (3 ; therefore, etc. It is evident that the three pairs of double points of the homographic systems on each circle are the points of contact of the corresponding circles of contact.] 9. Describe a circle touching three given ones with contacts of assigned species. [By Ex. 7.] 10. Describe a circle passing through a fixed point and cutting two given arcs on each of two circles equianharmonically. 11. Describe a circle cutting three pairs of arcs on three given circles equianharmonically. 12. The line joining the centres of perspective of any two chords of a circle is divided harmonically both by the circle and the chords. 13. Equal arcs of a circle are divided equianharmonically by the two circular points at infinity. DESARGUES THEOREM. 148. Any transversal to a cyclic quadrilateral ABCD meets the three pairs of opposite connectors BO and A D, etc,, etc., in X t X \ Y, Y \ Z, Z and the circle in W and W in eight points in involution. DES ARGUES THEOREM. 287 For the pencils B . ADWW and C. ADWW are equal, and therefore [ZY WW ^YZ WW ^Z YW Wl or the two triads F, Z, W\ F, Z , W are in involution. Again, because C.BLWW = A.BDWW it follows similarly that Z, X, W and Z\ X , W are in involution ; and since A . CDW W = B. CDW W, X, Y, W and X t F, W are in involution ; therefore, etc., Art. 144. COR. 1. By reciprocation with respect to the given circle we obtain the correlative theorem : For any escribed quadrilateral the lines joining any point P to the three pairs of opposite intersections X, X \ F, F ; Z, Z and the pair of tangents P W, P W are in involution. COR. 2. By reciprocation from any origin it follows that the theorem and Cor. 1 are more generally true for a quadrilateral inscribed or escribed to conic. COR. 3. In the particular case when a pair of opposite sides of a cyclic quadrilateral, or one inscribed in a conic,- coincide, the remaining pair become tangents, and the transversal (Z) meets their chord of contact in a double point. 288 INVOLUTION. Also the line (M ) passing through their point of inter section, which is therefore a double point, is divided harmonically ; i.e. A variable chord of a conic passing through a fixed point is divided harmonically by the point and its polar. COR. 4. When the transversal (N) is a tangent to the conic, the points of contact (WW) and (FT 7 ) are the double points. COR. 5. As a particular case of Cor. 4, let the transversal be parallel to the chord of contact. Then one of the double points (FF ) is at infinity, and the other is there fore the middle point ofXX\ hence we have the following property : The chord of contact of two parallel tangents (i.e. a diameter) bisects every parallel chord of the conic, or the locus of the middle points of parallel chords of a conic is a right line. COR. 6. Since a parabola touches the line at infinity (Art. 81) and the chord of contact of any tangent and the line at infinity is a diameter, any chord (TPTF ) of a para bola meets a tangent at a point X, which is the centric, DES ARGUES THEOREM. 289 and the diameter through its point of contact at a double point (YY f ) of the involution. Hence also or by drawing the ordinates WP, W P , OP. OP = OF 2 . COR. 7. Since the asymptotes of a hyperbola and the line at infinity are a particular case of a quadrilateral inscribed in a conic, any transversal WW is divided similarly at X and X y because one of the double points ( YY ) is at infinity. The other double point is therefore the middle point of WW , and the intercepts WX and W X between the curve and the asymptotes are equal. Also, the portion of any tangent to a hyperbola inter cepted by the asymptotes is bisected at the point of contact. 290 INVOLUTION. COR. 8. If the point P in Cor. 1 is such that two pairs of opposite connectors PX, PX \ PY } PY r are at right angles, the tangents from P to the circle are likewise at right angles. But the circle reciprocates from Pas origin into an equilateral hyperbola; therefore if an equilateral hyperbola be circumscribed to a triangle, it passes through the orthocentre. More generally, if an equilateral hyperbola be described about a quadrilateral, it passes through the orthocentre of the four triangles formed by taking the vertices in triads. The property of Art. 68. Ex. 8, will now appear obvious. It follows also that the locus of the centres of equilateral hyperbolas described about a triangle is its nine-points- circle. COR. 9. If the sides of the quadrilateral be numbered 1, 2, .3, 4, and the perpendiculars from TFand W on them be denoted by p lt p 2 , p s , p ; q v q 2 , q s , g 4 , since [ WW XX*] = [ TFTT FF] = [WW ZZ l , ,, WX WZ WZ WX and therefore - - -7 = etc etc " we have hence ^Ps/Z^iP* ^ s ^ constant value for all points on the conic, or the locus of a point such that the products of the perpendiculars from it to the three pairs of opposite sides of a quadrilateral have constant ratios is a conic passing through its vertices ; and by reciprocation we derive the correlative theorem: If a quadrilateral is circumscribed to a conic, the rectangles under the distances of the pairs DES ARGUES THEOREM. 291 of opposite vertices from a variable tangent have are to each other in constant ratios.* COR. 10. If either asymptote of a hyperbola be taken as a transversal to an inscribed quadrilateral, the double points of the involution are both at infinity, and the seg ments XX , FF , ZZ have a common middle point; therefore the lines joining a variable point on a hyper bola to a pair of fixed points on it intercept segments of constant length on each of the asymptotes. This property is thus stated in Townsend s Modern Geometry, Art. 340 : " For every two homographic pencils of rays through different vertices there exist two lines, real or imaginary, on each of which the several pairs of corresponding rays intercept equal segments." EXAMPLES. 1. A pencil whose rays are parallel to the three pairs of opposite connectors of a quadrilateral determines a system in involution. [Since the line at infinity is a transversal cut in involution by the sides of the quadrilateral ; therefore, etc.] 2. The three pairs of parallels drawn through the vertices and the extremities of the third diagonal of a quadrilateral cut any transversal in a system of points in involution. 3. If the fourth vertex D of the quadrilateral ABCD is the ortho- centre of ABC, prove the following particular case of the general theorem of Art. 148 : For any pencil of rays in involution, if tivo pairs of conjugates are at right angles, then all pairs of conjugates are at right angles. 4. Hence deduce "The circles on the diagonals of a complete quadrilateral are coaxal." * Chasles, Sections coniques, Art. 26. 292 INVOLUTION. 5. Any line or circle intersects a coaxal system at points in involution. 6. The parallels through any point to the sides of a triangle and the lines connecting that point to the vertices form an involution. 7. Every two circles and their two centres of perspective subtend at any point a pencil in involution. 8. For every two self-reciprocal triangles with respect to the same circle any two vertices connect equianharmonically with the remaining four. CHAPTER XIV. DOUBLE POINTS. 149. The solutions of a large number of problems of every variety in Geometry are frequently made to depend on the finding of the double points of two homographic systems. On account of the great importance of these points various constructions have been given for them. Thus in the last corollary they are easily found when we have obtained the points whose conjugates are at infinity on the axis by the equations XA . X A = XM . X M= XN. X N. We give in the following article two additional construc tions for homographic rows on an axis and append a sufficient number of examples, some of which have apparently no connexion with our present subject, to enable the student to form an idea of their extensive applications. 150. For any two systems of points on a circle (Art. 67, Ex. 6) the pairs of lines B(7, B C; GA\ C A ; A&, A E intersect respectively in points X, Y, Z, which are col- linear; and the line of collinearity meets the circle in points M and N y real or imaginary, given by the equations [ABCM] = [A B C M] and 293 294 DOUBLE POINTS. But since the anharmonic ratios are unaltered by inversion, if the origin be taken on the circle, the cyclic system inverts into points lying on a line and the double points of the former invert into the double points of the latter system. Hence the following construction for the double points of two homographic systems ABC ... and A B C ... on a line. Take any arbitrary point and describe the circles BOO , HOC meeting again in JT; COA, C OA in F; and AOB\ A OB in Z. Then 0, X, Y, Z lie on a circle which meets the axis in the required points M and N, real or imaginary. (Chasles.) Otherwise thus: Since [BCAM]=[RC A M], we have . BA IBM B A IB M CAj CM~V A \ C M which gives on reduction the ratios MB . MC jMB . MC, a known quantity. But the numerator and denominator are respectively the squares of the tangents from M to the circles described on the segments BC and B C as diameters ; therefore, etc., by Art. 88, Cor. 2. It should be noticed that two homographic systems ivhose double points are imaginary may be generated by the revolution of a constant angle about either of two fixed vertices which are reflexions of one another with respect to the axis. For if A A, BH, and CC f subtend equal angles at a point P (Art 72, Cor. 8), then since [ABCD ...]= [A B C D . . .]. EXAMPLES. 295 EXAMPLES. 1. Through a given point P draw a line meeting two given lines L and 11 divided homographically in corresponding points X, X . [Join PA , PB, PC, and let these lines meet the axis L in A", B", C", then ABC...=A"B C"... since the systems are in perspective at P, therefore A"B"C"...=A B C ..., and if any point of either coincides with its correspondents of the other, what is required is done ; hence lines joining P to the double points of these systems give the two solutions of the problem.] 2. Draw a line through a point P cutting four lines L\, _Z/ 2 , I/ 3) L in a row of points A, B, C } D having a given anharmonic ratio k. [Take points AI, A 2 , A 3 , ... on the axis L^ and draw lines cutting the remaining axes in systems of points such that AiBidfii ...=A 2 B. 2 C,D 2 ...=A 3 B 3 C 3 D 3 .. . The angle L^L^ is thus divided homographically by the pairs of rays through Ci t A ; <7 2 , A ; <7 3 , A ..-, etc., and the systems Ci<7 2 <7 3 ..., DiD z D 3 ... are therefore equianharmonic.* Join PC\, PC- 2 , PC 3 , ..., and let the joining lines meet Z 4 in D\, A > A > It follows, as in Ex. 1, that Di/) 2 D 3 ... = A A A ---) and the lines joining their double points to P are those required.] 3. Draw a line intersecting five lines such that the anharmonic ratio of any four of the points of intersection is equal to that of any other four. 4. Given two homograph ic pencils, find the pairs of corresponding rays which intersect on a given line L. [Let the line meet the pencils in points ABC, A B C ; the required rays therefore pass through the double points of the homographic rows so determined.] 5. In Ex. 4 find the pair of corresponding rays which intersect at a given angle. [Join the vertices and of the pencils, and on 00 describe a segment of a circle containing the given angle ; let this circle cut the pencils in the points ABC..., A B C ..., and find the double points of these homographic systems ; therefore, etc.] * This is otherwise evident as all the lines touch the same conic. 296 DOUBLE POINTS. 6. Find the direction of the parallel rays ; and hence draw a transversal to two homographic pencils which shall be divided similarly by them. 7. Find two points on a given line which shall be isogonal conju gates with respect to a given triangle. 8. Construct a triangle with its sides passing through given points and its vertices on given lines, or on a circle. 9. Let the line L joining the vertices of two homographic pencils regarded as a ray of each system have for conjugates LI and L- 2 ; prove that any transversal through the point L\L- 2 is cut in involu tion (cf. Art. 145). 10. Through a given point Pdraw a line intersecting five lines in the points A, A ; B,B \ P in any assigned order forming with P an involution. [Let the lines containing A, B meet in ; those containing A t B in . Since (hyp.) . ABPP = O. A B P P= . B A PP and the pairs of rays which correspond OB, OB \ OB, O A are fixed; therefore the variable rays OP and O P divide the fifth line L homographically and the double points give the required solutions.] 11. Find a point on a given line such that if joined to five given points any two pairs of connectors shall be in involution with the line and fifth. 12. Describe a circle touching three circles with contacts of assigned species. INDEX. ANGLE, Brocard (of a Tri angle), - - 61, 105 Brocard (of a Poly gon), - - -252 of Intersection of Figures, - - 6 Antisirailitude, Products of, 222 Antiparallels, 68 Axis, Central, - - - 107 Directive, - - 270 of Perspective, - 121 Radical, - - - 143 Barbarin, - - - - 81 Biddle, Mr. D., - - - 44 Bowesman, Mr. George W., 203 Burnside, Professor, - - 33 Brianchon, - - - - 167 Brocard, M., - - - 93 Casey, - - 20, 217, 218, 241 Catalan, .... 4 Chasles, .... 50 CENTRE Directive, - 271 Instantaneous, - 206 Perspective, - - 120 Radical, - - - 143 of Similitude, 42, 52, 185 Chords, Homologous, and Antihomologous, - - 221 PAGE CIRCLE of Apollonius, - 148 Auxiliary, - - 170 Brocard s, - - 131 Coaxal, - - 168 Concentric, - - 167 Cosine, - - 74 Director, - 171, 279 M Cay s ("A," "B,""C," circles), 211 Neuberg s, - - 131 Nine Points, 70, 86 Orthocentroidal, - 200 Polar, - - - 149 of Self-Inversion, - 246 of Similitude, - 197 Taylor s, - - 76 Triplicate Ratio ("T.R." circle), 75 Tucker s, - 71 Conjugate Segments, - - 261 Desargues, - - - 115, 286 Diameter (of a Polygon), - 108 Dilworth, Mr. William J. (T. C. D.), --- 144 Envelopes, .... 9 Equianharmonicism, - - 4 Fry, Mr. Matthew W. J. (F.T.C.D.), - - - 260 297 298 INDEX. PAGE Hain, 75 Hart, Sir Andrew S. f - 187 HEXAGONS, Brianchon, - 126 Equianharmonic, 167 Pascal, - - 128 Homology, Axis of, - - 121 Homothetic Figures, - - 204 Homographic Division, - 268 Ingrain, Dr. J. K. (SF.T.C.D.), 260 Involution, - - 118, 276 Equations of, 119,273 Isogonal Conjugates, - 132 Isotomic ,, - - 118 LINES, Antihomologous, - 221 Brianchon, - 126 Conjugate, - - 146 Diagonal, - - 190 Homologous, - - 221 Pascal, - - - 124 Philo s, - - 37 Simson, - - 46, 50 ,, (pole of), 81,256 Longchamps, Professor de, - 182 Mathesis, 47, 75, 80, 81, 94, 161, 209, 260 Milne, Rev. J. J., 39,92,252 Mukhopadhyay, Syamadas, - 94 M Cay, Mr. W. S. (F.T.C.D.), 48, 66, 97, 183, 191, 211, 254, 256 Neuberg, M., - 64, 75, 76, 216 Nouvelles Annales, - - 76 Parallel Lines, ... 9 Philo s Line, - - - 37 Projection of a Point, - - 4 Pascal, .... 167 Poncelet, - - - 115, 121 Preston, Mr Thomas, - - 196 POINTS, Adjoint, - - - 209 Antihomologous, - 221 PAGE POINTS, Brianchon, - - 126 Brocard, - - 60, 104 two Circular, - - 174 Conjugate, - - 149 Director, - 210 Double, - - 206, 276 de Gergonne, - - 117 Homologous, - - 220 Invariable, - - 208 Limiting, - -176 Middle of a Line, - 153 deNagel, - - 117 Pascal, - - - 124 Symmedian, - - 31 Tarry s - - 136 Polar Equations, - 65 POLYGONS, Convex, - - 133 Intersecting, - 133 Re-entrant, - 133 Harmonic, - 252 QUADRILATERALS, Harmonic, properties of, - 132 Diagonal, Line of, 192 Harmonic, 252 Radius Vector, - - 43, 65 Ratios, Anharmonic, 3,117,261 of Similitude, - - 42 Reciprocation, - - - 162 Russell, Mr. Robert (F.T.C.D.), - 190,201,255 Salmon, Rev. Dr., - - 263 Simmons, Rev. T. C., - - 44 Stubbs, Rev. Dr., - - 260 Steiner, .... 49 Symmetry, 1 Systems of Circles (Coaxal), 173 Conjugate Coaxal, 173 INDEX. 299 PAOE PAGE Tangential Equations, - 159 THEOREMS by Tarry, - - 209 THEOREMS by Bobillier, 10 Weill, - 96, 201 Brian chon, 125 Townsend, 151, 224, 244, 247, Casey, 11 283, 285 Ceva, 113 TRIANGLES, Brocard s First, 122 Desargues, 286 ,, Second, Euler, 4 210, 269 Feuerbach, 13, 217 Cosymmedian, - 252 Hart, - 13 Diagonal, - - 132 Hermite, 134 Invariable, - 208 Mannheim, 11, 255 Median, - - 160 ,, (ex Pedal, - - 22 tension of), - 259 of Reference, - 63 Menelaus, 112 Self -Reciprocal, 151 Neuberg, 210 Similitude, - 207 Pascal, - 124 Tucker, .... 83 Poncelet, 187 Van de Berg, - - 161 Ptolemy, 48 Vigari^, - - - 253 Russell, - 201 Walker, - - 1S9 Salmon, - 155 Ward, Mr. Piers C. (T.C.D.), 242 Stoll, 123 Weill, - - - - 201, 242 PRINTED BY ROBKKT MACLJiUUbK AT THE UNIVERSITY 1 KKSS, GLASGOW. Messrs. Macmillan and Co. s Books for Students. ANALYTICAL GEOMETRY. 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CLARK RUSSELL. Volume for 1891. Containing goo pages, and about 500 Illustrations. Price Bs. Among the chief Contents of the Volume are the following Complete Stories and Serials : The Witch of Prague. By F. MARION CRAWFORD. The Wisdom Tooth. By D. CHRISTIE MURRAY and HENRY HERMAN. Wooden Tony. By Mrs. W. K. CLIFFORD. Two Jealousies. By ALAN ADAIR. Gentleman Jim. By MARY GAUNT. Harrow School. Winchester College. Fawsley Park. Ham House. Westminster Abbey. Norwich. The New Trade-Union Movement. Russo- Jewish Immigrant. Queen s Private Garden at Osborne. MACMILLAN AND CO., LONDON. vi.50.9.91. :; 4 u If 13-*.) 5u- \/ 9 J