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Team Selection Test for the 54th IMO

Geometry Difficulty 8.0 National Olympiad, round 2 Prove it Bulgaria

The incircle of $\triangle ABC$ touches the sides $BC$ and $AC$ at points $A_1$ and $B_1$ , respectively. The lines $B_1A_1$ and $AB$ are concurrent at $X$ such that $A$ lies between $X$ and $B$ . If $\angle CXB = 90^\circ$ and $BC^2 = AB^2 + BC \cdot AC$ find the angles of $\triangle ABC$ .

Solution

Let $Y \in BC$ be such that $AY \parallel XA_1$ . Thus $BY = a-b$ and since $\frac{a}{c} = \frac{c}{a-b}$ we have that $\triangle ABC \sim \triangle YBA$ . Therefore $\angle AXA_1 = \angle BAY = \gamma$ which implies that quadrilateral $XAA_1C$ is cyclic. It follows from $\angle XAC = \angle XA_1C = 90^\circ - \frac{\gamma}{2}$ that $\alpha = 90^\circ + \frac{\gamma}{2}$ . Moreover $\angle AA_1C = 90^\circ$ , i.e. $AA_1$ is simultaneously altitude and angular bisector. Hence $\gamma = \beta$ and $90^\circ + \frac{\gamma}{2} + 2\gamma = 180^\circ$ , i.e. $\gamma = 36^\circ$ . The angles of $\triangle ABC$ are $36^\circ, 36^\circ$ and $108^\circ$ .

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