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, r1 of 4 with respect to the same circle of inversion. The relation Σ23.14=0 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 rr1, 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~ r1; 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 on 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 THEORY OF ANHARMONIC SECTION. SECTION I. ANHARMONIC SECTION. 131. Definitions.-Let a line AB be divided by two variable points C and D such that AC/BC÷AD/BD is a constant ratio (=x). The value of x is thus - CA. BD/BC. AD, and is termed the Anharmonic Ratio in which the segment AB is divided by the points C and D. Similarly the anharmonic ratio of CD divided at A and B is CA/DA÷CB/DB or -CA. BD/BC. AD. The points C and D are Conjugate or Corresponding Points in the Row A, B, C, D, and AB and CD are Conjugate Segments. It is obvious that conjugate segments 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 BC divided in A and D=BA/CA÷BD/CD =λ, (1) CA divided in B and D=CB/AB÷CD/AD=μ, (2) and AB divided in C and D=AC/BC÷AD/BD=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 A, μ, 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 AX/BX=k. AX'|BX'. As X approaches B, AX/BX increases; therefore the conjugate point X' approaches B simultaneously. For let AX'=a and BX'=b and we have but a>b, thus it follows that as ' moves towards B the ratio AX'/BX' 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 X and X'. Similarly A is a double point. Again, as X' recedes from B on the line produced, X approaches M the middle point of AB. In the limit when X' is at infinity and AX'/BX' therefore equal to unity, its conjugate X(=P) divides the line in the simple_ratio_AP|PB=k. Similarly when I moves to infinity, its conjugate X'(=Q) gives the relation AQ/BQ=1/k ; and the two points whose conjugates are at infinity are isotomic conjugates 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 ABCD in terms of any one of them (X). Since BC. AD+CA. BD+AB. CD=0; dividing by AB. CD, we have + +1=0, AB. CD AB.CD whence on substituting from Art. 132 Thus generally it follows, by dividing the above equation by each of its terms, that μ+1/λ=1; v+1/μ=1; λ+1/v=1. The six ratios are therefore λ, 1/^, (^—1)/^, ^/(x−1), 1–λ, 1/(1—λ). These may be expressed as trigonometrical functions of an angle. For let Then the ratios taken in the above order reduce sec20. to the following :— sec20, cos20, sin20, cosec20,- tan20, — cot20. If two of the ratios are equal, e.g. λ= (λ − 1)/λ, then λ2−λ+1=0 and λ=w or w2, the imaginary cube roots of unity. In this case the three pairs of ratios have the values w and w2. If λ= -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 notation [ABCD] to denote the ratio into which AB is divided by C and D. [ABCD] is equivalent to AC/BC÷AD/BD, and [ABCD]. [ABDC]=1. EXAMPLES. 1. To prove that [ABCD] = [BADC] = [DCBA] = [CDAB]; and hence when any two constituents of four points are interchanged, 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 κ is equal to its reciprocal, and is therefore unity. The four points form in this case an harmonic system.] 3. To prove for any collinear system of points A, B, C, D, E ..... that [ABCE]/[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'D'] and [ABCE]=[A'B'C'E'], to prove that [BCDE]=[B'C'D'E'], [CADE]=[C'A'D'E'], [ABDE]=[A'B'D'E']. [By Ex. 3.] 5. If [ABCD]=[ABC' D'], prove that [ABCC']=[A BDD']. [Expanding the ratios the required result follows by alternation.] 6. If in Ex. 4 [ABCD]=[A'B'C'D'], [ABCE]=[A'B'C'E'], [ABCF]=[A'B'C'F'], etc., etc.; prove that [ADEF]=[A'D'E'F"], [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']. 2°. [ABCD...]=[A'B'C'D' ...]. [Since [MNAA']=[MNBB']=[MNCC']= ... etc., by Ex. 5. [MNAB]=[MNA'B']; [MNAC]=[MNA'C'], etc. Hence by division we have [MABC]=[MA'B'C'], etc. ... To prove 2°. We have by 1° [MABC]=[MA'B'C'] and [MABD]=[MA'B' D'], 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 interchangeable, therefore [MABC']=[MA'B'C]; dividing these equations, 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. |