IMO 2004 SL G7

Origin: RUS | Category: Geometry

Problem

For a given triangle ABC, let X be a variable point on the line BC such that C lies between B and X and the incircles of the triangles ABX and ACX intersect at two distinct points P and Q. Prove that the line PQ passes through a point independent of X.

Solution

The proof uses the following observation:

Lemma. In a triangle ABC, let K, L be the midpoints of the sides AC, AB, respectively, and let the incircle of the triangle touch BC, CA at D, E, respectively. Then the lines KL and DE intersect on the bisector of the angle ABC. Proof. Let the bisector ℓb of \angleABC meet DE at T . One can assume that AB ̸= BC, or else T \equivK \inKL. Note that the incenter I of \triangleABC is between B and T , and also T ̸= E. From the triangles BDT and DEC we obtain \angleITD = \alpha/2 = \angleIAE, which implies that A, I, T, E are concyclic. Then \angleATB = \angleAEI = 90◦. Thus L is the circumcenter of \triangleATB from which \angleLTB = \angleLBT = \angleTBC ⇒LT \parallelBC ⇒T \in KL, which is what we were supposed to prove. Let the incircles of \triangleABX and \triangleACX touch BX at D and F, respec- tively, and let them touch AX at E and G, respectively. Clearly, DE and FG are parallel. If the line PQ intersects BX and AX at M and N, respectively, then MD2 = MP \cdot MQ = MF 2, i.e., MD = MF and analogously NE = NG. It follows that PQ is parallel to DE and FG and equidistant from them. The midpoints of AB, AC, and AX lie on the same line m, parallel to BC. Applying the lemma to \triangleABX, we conclude that DE passes through the common point U of m and the bisector of \angleABX. Analogously, FG passes through the common point V of m and the bisector of \angleACX. Therefore PQ passes through the midpoint W of the line segment UV . Since U, V do not depend on X, neither does W.