BALKAN 2013 Problem 1

We denote the angles of the triangle by $\alpha, \beta$ and $\gamma$ as usual. Since $\angle K M P=90^{\circ}-\frac{\beta}{2}$, it suffices to prove that $\angle K L P=90^{\circ}+\frac{\beta}{2}$, which is equivalent to $\angle K L C=\frac{\beta}{2}$. Let $I$ be the incenter of triangle $A B C$ and let $D$ be the tangency point of the incircle with $A B$. Since $C K \| I B$ and $C L \| I A$, it holds that $\angle K C L=\angle A I B$. Moreover, from $C N=A D=\frac{b+c-a}{2}$ and $\angle K C N=\frac{\beta}{2}$ we obtain $C K=C N \cos \frac{\beta}{2}=$ $A D \cos \frac{\beta}{2}=A I \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$ and analogously $C L=B I \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$, which imply $\frac{C K}{C L}=\frac{A I}{B I}$. Hence the triangles $K C L$ and $A I B$ are similar, and thus $\angle K L C=$ $\angle A B I=\frac{\beta}{2}$.