# Ib AMO 2001 Problem 2

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.