IMO 1976 SL 1

Origin: BUL

Problem

Let ABC be a triangle with bisectors AA1, BB1, CC1 (A1 \in BC, etc.) and M their common point. Consider the triangles MB1A, MC1A, MC1B, MA1B, MA1C, MB1C, and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then AB = BC = CA.

Solution

Let r denote the common inradius. Some two of the four triangles with the inradii \rho have cross angles at M: Suppose these are \triangleAMB1 and \triangleBMA1. We shall show that \triangleAMB1 ∼= \triangleBMA1. Indeed, the altitudes of these two triangles are both equal to r, the inradius of \triangleABC, and their interior angles at M are equal to some angle ϕ. If P is the point of tangency of the incircle of \triangleA1MB with MB, then r \rho = A1M+BM+A1B A1B , which also implies r−2\rho \rho

A1M+BM−A1B A1B

2MP A1B = 2r cot(ϕ/2) A1B . Since similarly r−2\rho \rho

2r cot(ϕ/2) B1A , we obtain A1B = B1A and consequently \triangleAMB1 ∼= \triangleBMA1. Thus \angleBAC = \angleABC and CC1 \perpAB. There are two alternatives for the other two incircles: (i) If the inradii of AMC1 and AMB1 are equal to r, it is easy to obtain that \triangleAMC1 ∼= \triangleAMB1. Hence \angleAB1M = \angleAC1M = 90◦, and \triangleABC is equilateral. (ii) The inradii of AMB1 and CMB1 are equal to r. Put x = \angleMAC1 = \angleMBC1. In this case ϕ = 2x and \angleB1MC = 90◦−x. Now we have AB1 CB1 = SAMB1 SCMB1 = AM+MB1+AB1 CM+MB1+CB1 = AM+MB1−AB1 CM+MB1−CB1 = cot x cot(45◦−x/2). On the other hand, AB1 CB1 = AB BC = 2 cos2x. Thus we have an equation for x: tan (45◦−x/2) = 2 cos 2x tan x, or equivalently 2 tan  45◦−x

sin  45◦−x

cos  45◦−x

= 2 cos2x sin x. Hence sin 3x−sin x = 2 sin2
45◦−x  = 1−sin x, implying sin 3x = 1, i.e., x = 30◦. Therefore \triangleABC is equilateral.