IMO 1986 SL 18

Origin: TUR

Problem

Let AX, BY, CZ be three cevians concurrent at an inte- rior point D of a triangle ABC. Prove that if two of the quadrangles DY AZ, DZBX, DXCY are circumscribable, so is the third.

Solution

We shall use the following criterion for a quadrangle to be circumscribable. Lemma. The quadrangle AY DZ is circumscribable if and only if DB − DC = AB −AC. Proof. Suppose that AY DZ is circumscribable and that the incircle is tangent to AZ, ZD, DY , Y A at M, N, P, Q respectively. Then DB −DC = PB −NC = MB −QC = AB −AC. Conversely, assume

that DB−DC = AB−AC and let a tangent from D to the incircle of the triangle ACZ meet CZ and CA at D′ ̸= Z and Y ′ ̸= A respec- tively. According to the first part we have D′B −D′C = AB −AC. It follows that |D′B −DB| = |D′C−DC| = DD′, implying that D′ \equivD. A B C D M N P Q X Y Z Let us assume that DZBX and DXCY are circumscribable. Using the lemma we obtain DC −DA = BC −BA and DA −DB = CA −CB. Adding these two inequalities yields DC −DB = AC −AB, and the statement follows from the lemma.