IMO 1988 SL 30

Origin: USS

Problem

A point M is chosen on the side AC of the triangle ABC in such a way that the radii of the circles inscribed in the triangles ABM and BMC are equal. Prove that BM 2 = ∆cot B 2 , where ∆is the area of the triangle ABC.

Solution

Let ∆1, s1, r′ denote the area, semiperimeter, and inradius of triangle ABM, ∆2, s2, r′ the same quantities for triangle MBC, and ∆, s, r those for \triangleABC. Also, let P ′ and Q′ be the points of tangency of the incircle of \triangleABM with the side AB and of the incircle of \triangleMBC with the side BC, respectively, and let P, Q be the points of tangency of the incircle of \triangleABC with the sides AB, BC. We have ∆1 = s1r′, ∆2 = s2r′, ∆= sr, so that sr = (s1 + s2)r′. Then s1 + s2 = s + BM ⇒ r′ r = s s + BM . (1) On the other hand, from similarity of triangles it follows that AP ′/AP = CQ′/CQ = r′/r. By a well-known formula we find that AP = s −BC, CQ = s −AB, AP ′ = s1 −BM, CQ′ = s2 −BM, and therefore deduce that r′ r = s1 −BM s −BC = s2 −BM s −AB ⇒r′ r = s1 + s2 −2BM 2s −AB −BC = s −BM AC . (2) It follows from (1) and (2) that (s −BM)/AC = s/(s + BM), giving us s2 −BM 2 = s \cdot AC. Finally,

BM 2 = s(s −AC) = s \cdot BP = s \cdot r cot B 2 = ∆cot B 2 .