Problem 4.
Let triangle ABC satisfy AB < AC < BC. Let $I$ and $\omega$ be the incenter and the incircle of triangle $ABC$, respectively. Let $X$ be a point on line $BC$, different from $C$, such that the line through $X$ and parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be a point on line $BC$, different from $B$, such that the line through $Y$ and parallel to $AB$ is tangent to $\omega$. Line $AI$ intersects the circumcircle of triangle ABC at P ≠ A. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively.
Prove that ∠KIL + ∠YPX = 180°.
Proposed by Dominik Burek, Poland

Solution

Notice that such a point $X$ and $Y$ are unique (For $X$, let $J$ be the intersection of the tangent with $(AB)$. Then since $\omega$ is inscribed inside $AJXC$ we have $AC+JX = AJ+XC \Longrightarrow AC+ k AC = (1-k) AC + (1-k) BC$, by Thales where $k = \frac{JX}{AC}$. Solving this equation, we have $k$ unique hence $X$ is unique. ).

A construction of such points $X$ and $Y$ is easy : take $S$ the symmetric of $A$ about $I$. Draw the tangents of $S$ to $\omega$ and their intersections $X$ and $Y$ with $(BC)$. The parallelism is obvious since we have a rhombus.

Notice that $A, S, P$ (the south pole) belong to the bisector of $A$.

Now the following lemma : $B,X,S,P$ are concyclic (as well as $P,S,Y,C$). Sketch of the proof : Angle chase with $\angle BXS =_{\text{parallelism}} = \pi - \angle BCA$ and $\angle BPS = \angle BPA =_{\text{A,B,P,C concyclic}} \angle BCA$.

Remark with Thales theorem we have $\angle LIK = \angle BSC$. Finally $\angle BSC + \angle YPX = \angle BSC + (\angle BXP + \angle PYC) - \pi = \angle BSC + \angle BSP + \angle PSC = 2 \pi - \pi = \pi$, as desired.