Cor. 4. Since the three circles of similitude are of the c.f.species, two points may be found from which any three circles Subtend equal angles. These are the common points to the three circles of similitude. (288°, Cor. 2) Cor. 5. The groups of circles on the following triads of segments as diameters are severally co-axal, 295°. Any two circles Z and Z', which touch three circles S1, S2, Sa similarly, cut their circles of antisimilitude orthogonally (292”, Cor. I), and therefore have their centres at the radical centre of the three circles of antisimilitude. (276°, Cor. 4) But Z and Z' have not necessarily the same centre. ‘.. the three circles of antisimilitude of the circles S1, S2, and S3 are co-axal, and their common radical axis passes through the centres of Z and Z'. 296°. Theorem.––If two circles touch three circles similarly, the radical axis of the two is an axis of similitude of the three ; and the radical centre of the three is a centre of similitude of the two. Proof–The circles S and S' touch the three circles A, B, and C similarly, I. Since S and S' touch A and B sinnilarly, the radical axis of S and S' passes through a Centre of similitude of A and B. (292”, Cor. 3) Also, the radical axis of S and S' passes through a centre of similitude of B and C, and through a centre of similitude of C and A. , the radical axis of S and S is an axis of similitude of the three circles A, B, and C. 2. Again, since A and B touch S and S', the radical axis of A and B passes through a centre of similitude of S and S'. For similar reasons, and because A, B, and C touch S and S' similarly, the radical axes of B and C, and of C and A, pass through the same centre of similitude of S and S. But these three radical axes meet at the radical centre of A, B, and C. ... the radical centre of A, B, and C is a centre of simili tude of S and S'. Q.e.d. 297°. Problem.--To construct a circle which shall touch three given circles. In the figure of 296”, let A, B, and C be the three given circles, and let S and S' be two circles which are solutions of the problem. Let I, denote one of the axes of similitude of A, B, and C, and let O be their radical centre. These are given when the circles A, B, and C are given. Now L is the radical axis of S and S (296', 1), and O is one of their centres of similitude. But as A touches S and S' the chord of contact of A passes through the pole of L with respect to A (29.3°). Similarly the chords of contact of B and C pass through the poles of L with respect to B and C respectively. And these chords are concurrent at O. (292’) Hence the following construction — Find O the radical centre and L an axis of similitude of A, B, and C. Take the poles of L with respect to each of these circles, and let them be the points /, q, r respectively. Then Off, Oy, Or are the chords of contact for the three given circles, and three points being thus found for each of two touching circles, S and S', these circles are determined. (This elegant solution of a famous problem is due to M. Gergonne.) Cor. As each axis of similitude gives different poles with respect to A, B, and C, while there is but one radical centre O, in general each axis of similitude determines two touching circles; and as there are four axes of similitude there are eight circles, in pairs of twos, which touch three given circles, Putting i and e for internal and external contact with the touching circle, we may classify the eight circles as follows: PART V. ON HARMONIC AND AN HARMONIC RATIOS— HOMO GRAPHY, INVOLUTION, ETC. SECTION I. GENERAL CONSIDERATIONS IN REGARD TO HARMONIC AN ID AN HARMONIC DIVISION, 298°. Let C be a point dividing a segment AB. The position of C in relation to A and B is determined by the ratio H.--— # AC : BC. For, if we know this ratio, we A C D * know completely the position of C with respect to A and B. If this ratio is negative, C lies between A and B ; if positive, C does not lie between A and B. If AC : BC = – I, C is the internal bisector of AB ; and if AC : BC = + 1, C is the external bisector of AB, i.e., a point at oo in the direction AB or B.A. Let D be a second point dividing AB. The position of D is known when the ratio AD : B D is known. ZDef.--If we denote the ratio AC : BC by 7/7, and the ratio AD : BD by 21, the two ratios //; : 7; and 7: ; //z, which are reciprocals of one another, are called the two a/m/harmonic rafios of the division of the segment A B by the points C and D, or the harmonoids of the range A, B, C, D. 252 Either of the two anharmonic ratios expresses a relation between the parts into which the segment AB is divided by the points C and D. Evidently the two anharmonic ratios have the same sign, and when one of them is zero the other is infinite, and vice Z/€7°S (?. These ratios may be written — AC AD or AC. BD AC. BD 2. –––– : — O1 --> - Ol' ------- & BD BC BI) . AC AC. B.D The last form is to be preferred, other things being convenient, on account of its symmetry with respect to A and B, the end-points of the divided segment. 299°. The following results readily follow. I. Let o be +. Then . and ; have like signs, and therefore C and D both divide AB internally or both externally. (298") In this case the order of the points must be some one of the following set, where AB is the segment divided, and the letters C and D are considered as being interchangeable : CDAB, ACI) B, CABD, ABCI). 2. ILet o be — . Then . and ; have opposite signs, and one point divides AB internally and the other externally. The order of the points is then one of the set CADB, ACBD. 3. When either of the two anharmonic ratios is + I, these ratios are equal. . 4. Let *- + I. Then ...-...} and C and D are both internal or both external. |