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Belarus competition problems

Geometry Difficulty 5.4 AIME, harder Prove it Belarus

The incircle of the right triangle $ABC$ is tangent to the hypotenuse $AB$ at point $P$ and is tangent to the legs $AC$ and $BC$ at points $Q$ and $R$ respectively. Points $C_1$ and $C_2$ are symmetric to $C$ with respect to the lines $PQ$ and $PR$ .
Find the angle $C_1IC_2$ where $I$ is the incenter of the triangle $ABC$ . (Mikhail Karpuk)

Solution

Denote $PR \cap CC_2 = X$ . Let's do some angle-chasing.
$\angle RCC_2 = 90^\circ - \angle CRX = 90^\circ - \angle PRB = \frac{\angle B}{2}.$
Therefore $RI = RC = RC_2$ , i.e. $R$ is the center of the circumcircle of the triangle $ICC_2$ . Hence

$\angle RIC_2 = 90^\circ - \frac{\angle IRC_2}{2} = 90^\circ - \angle ICC_2 = 90^\circ - (45^\circ + \angle RCC_2) = 45^\circ - \frac{\angle B}{2}.$

Similarly, $\angle QIC_1 = 45^\circ - \frac{\angle A}{2}$ . Then

$\angle$ C_1IC_2 = $90^\circ + \angle$ QIC_1 + $\angle$ RIC_2 = $\text{180 -}$ A + $\angle$ B}{2} = $135^\circ.$

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