, 2017

Solution:
Let $M$ be the midpoint of arc $B C$ not containing $L$ and let $D$ be the point where the incircle of triangle $A B C$ touches $B C$. Also let $N$ be the projection from $I$ to $A K$. We claim that $M$ is the desired fixed point.
By Simson's Theorem on triangle $K U V$ and point $I$ we have that points $T, D, N$ are collinear and since quadrilateral $N K D I$ is a rectangle we have that line $D N$ passes through the midpoint of $I K$. Thus it suffices to show that $M$ lies on line $D N$.
Now, let $I_{a}, I_{b}, I_{c}$ be the $A, B, C$-excenters of triangle $A B C$ respectively. Then $I$ is the orthocenter of triangle $I_{a} I_{b} I_{c}$ and $A B C$ is the Cevian triangle of $I$ with respect to triangle $I_{a} I_{b} I_{c}$. It's also well-known that $M$ is the midpoint of $I I_{a}$.
Let $D'$ be the reflection of $I$ over $B C$ and let $N'$ be the reflection of $I$ over $A K$. Clearly $K$ is the midpoint of $D' N'$. If we could prove that $I_{a}, D', K, N'$ were collinear then by taking a homothety centered at $I$ with ratio $\frac{1}{2}$ we would have that points $M, D, N$ were collinear as desired. Thus it suffices to show that points $I_{a}, D', K$ are collinear.
Let lines $B C$ and $I_{b} I_{c}$ intersect at $R$ and let lines $A I$ and $B C$ intersect at $S$. Then it's well-known that ( $I_{b}, I_{c} ; A, R$ ) is harmonic and projecting from $C$ we have that $\left(I_{a}, I ; S, A\right)$ is harmonic. But $K S \perp K A$ which means that $K S$ bisects angle $\angle I K I_{a}$. But it's clear by the definition of $D'$ that $K S$ bisects angle $\angle I K D'$ which implies that points $I_{a}, K, D'$ are collinear as desired. This completes the proof.