IMO 1970 SL 8

Origin: POL

Problem

Given a point M on the side AB of the triangle ABC, let r1 and r2 be the radii of the inscribed circles of the triangles ACM and BCM respectively and let \rho1 and \rho2 be the radii of the excircles of the triangles ACM and BCM at the sides AM and BM respectively. Let r and \rho denote the radii of the inscribed circle and the excircle at the side AB of the triangle ABC respectively. Prove that r1 \rho1 r2 \rho2 = r \rho.

Solution

Let AC = b, BC = a, AM = x, BM = y, CM = l. Denote by I1 the incenter and by S1 the center of the excircle of ∆AMC. Suppose that P1 and Q1 are feet of perpendiculars from I1 and S1, respectively, to the line AC. Then \triangleI1CP1 ∼\triangleS1CQ1, hence r1/\rho1 = CP1/CQ1. We have CP1 = (AC + MC −AM)/2 = (b + l −x)/2 and CQ1 = (AC + MC + AM)/2 = (b + l + x)/2. Hence r1 \rho1 = b + l −x b + l + x. We similarly obtain r2 \rho2 = b + l −y b + l + y and r \rho = a + b −x −y a + b + x + y.

What we have to prove is now equivalent to (b + l −x)(a + l −y) (b + l + x)(a + l + y) = a + b −x −y a + b + x + y . (1) Multiplying both sides of (1) by (a + l + y)(b + l + x)(a + b + x + y) we obtain an expression that reduces to l2x + l2y + x2y + xy2 = b2y + a2x. Dividing both sides by c = x + y, we get that (1) is equivalent to l2 = b2y/(x + y) + a2x/(x + y) −xy, which is exactly Stewart’s theorem for l. This finally proves the desired result.