**Problem 2 of the IberoAmerican Mathematical Olympiad 2001**

The inscrite circumference of the triangle ABC has center at O and it is tangent to the sides BC, AC and AB at the points X, Y and Z, respectively. The lines BO and CO intersect the line Y Z at the points P and Q, respectively. Show that if the segments XP and XQ have the same length, then the triangle ABC is isosceles.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

We have ∠BOQ = ∠BCO + ∠OBC = ½ (180° - ∠A ) = ∠YZA = ∠ZOA and ∠OZQ = ∠ZAO (2 sides perpendicular to each other) or ∠BOQ + ∠BZQ = ∠YZA + 90° + ∠ZAO = 180°

Therefore, BZQO is concyclic and ∠BQO = ∠BZO = 90°. Similarly, ∠CPO = 90° and since ∠BXO = ∠CXO = 90°, BZQOX and CYPOX are both concyclic. We also note that BQPC is also concyclic.

Therefore, ∠QXO = ∠QBO = ∠PCO = ∠PXO and triangles QXO and PXO are congruent which leads to OQ = OP and ∠QOX = ∠POX or

∠QBX = 180° - ∠QOX = 180° - ∠POX = ∠PCX (1)

Since ∠OQP = ∠OPQ (OQ = OP) and ∠OZY = ∠OYZ (OZ = OY = radius of the circle), triangles OZQ = triangle OYP or ZQ = YP. Furthermore, ∠ZQX = 180° - ∠XQP = 180° - ∠XPQ = ∠YPX and triangle ZQX = triangle YPX which leads to ∠ZXQ = ∠YXP

Adding ∠ZBQ to both sides of (1) ∠QBX = ∠PCX, we have

∠ZBX =∠ZBQ +∠QBX =∠ZXQ +∠QBX = ∠YXP +∠PCX = ∠YCP +∠PCX = ∠YCX or AB = AC and the triangle ABC is isosceles.