Problem 1201

2. Let $K$ and $L$ be the midpoints of $A C$ and $A B$ respectively, $O$ the circumcenter of $\triangle A B C$, and $M \in A C, N \in A B$ such that $P M \| E K$ and $P N \| E L$. Then $\angle P M C=\angle E K C=\angle L K C-\angle L K E=$ $180^{\circ}-\gamma-\angle C B O=2 \alpha+\beta-90^{\circ}$.

Consider the reflection $C^{\prime}$ of point $C$ in the perpendicular bisector of $M N$ $\left(C^{\prime} \not \equiv B\right.$ since $\left.A B \neq A C\right)$. Points $B$, $P, N$ and $C^{\prime}$ lie on a circle for $\angle N C^{\prime} P$

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$=\angle N B P=x$. Thus $\beta-x=\angle C B P=\angle P N C^{\prime}=\angle P M C=2 \alpha+\beta-90^{\circ}$, and therefore the statement.

Second solution. The Ceva theorem in trigonometric form for point $P$ in triangle $A B C$ gives us $\frac{\sin \angle C A E}{\sin \angle E A B}=\frac{\sin \angle P B C}{\sin \angle P C B}=\frac{\sin (\beta-x)}{\sin (\gamma-x)}$. On the other hand, the same theorem for $E$ in $\triangle A K L$ gives $\frac{\sin \angle C A E}{\sin \angle E A B}=\frac{\sin \angle A K E}{\sin \angle E L A}=\frac{\cos (\gamma-\alpha)}{\cos (\beta-\alpha)}$. These two equalities together imply

$$\begin{aligned} 0 & =\sin (\beta-x) \cos (\beta-\alpha)-\sin (\gamma-x) \cos (\gamma-\alpha) \\ & =\frac{1}{2}(\sin (2 \beta-\alpha-x)+\sin (\alpha-x)-\sin (2 \gamma-\alpha-x)-\sin (\alpha-x)) \\ & =\sin (\beta-\gamma) \cos \left(180^{\circ}-2 \alpha-x\right) \end{aligned}$$

so $x=90^{\circ}-2 \alpha$.