By Barmak J.A., Minian E.G.

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**Example text**

11). In the former view the points of the plane are arranged in triangles with edges on the points fi. One or three of the edges must be produced. Exercise 2 - If we take two such triangles x, y. z and x.. Y.. z. then x - x.. y - Y•• z - z. are parallel. The corresponding edges y - z and Y. - z. meet at f •. Consider the next case, 8 18 2 8 s8 4 =1. Here 8 1 8 2 and 8 a8 4 are reciprocal stretches with the same fixed point f, the intersection of fl - f2 and fa - f4. Taking any point x, the 28 THE EUCLIDEAN GROUP points x, xSl> XS l S 2, XS l S 2S S form a closed cycle.

If the circles are external, the ring includes the point 00. We consider all the arcs orthogonal to both. These terminate at the image pair II' 12' Any arc meets Cl say at Xl and C 2 say at X 2• Consider the cross-ratio (Xl (Xl - Il)(x 2 - 12) I2)(x 2 -II) First, it is real because the points are on a circle. Second, it is the same for all arcs. For inversion in any of the arcs will not alter it, but will send an arc I l x l x 2I2 into another, say, II Yl yd2' Third, it is positive, for it is manifestly positive for the segment II -12' Of this positive number, which is the same for all arcs, we take the logarithm, A.

The stretch 8 1 sends a circle C with centre y into a circle C8 1 with centre z. The stretch 8 2 sends this latter into a circle with centre x. Thus this circle is C81 8 2 • The stretch 8 s sends this into the original circle, since 8 18 28 s is to be I. The product of ratios PIP2PS is then 1, (1) But there is a further relation. We obtain it by placing y and therefore z at A. Then x8 s is 11 and 1182 is x. We have then 11 - and Is= Ps(x -Is) x - 12=P2(X - 12) whence, eliminating x, (2) This is one of three equivalent forms.