### 2013 IMO SL #G2

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

Let $O$ be the circumcenter of $ABC$. We claim that $Y$ is the reflection of $N$ over the perpendicular bisector of $AT$. To show this, denote $Y'$ as the reflection of $N$ over the perpendicular bisector of $AT$. Note $ANY'T$ is an isosceles trapezoid and therefore cyclic. Thus, to prove that $Y=Y'$, it suffices to show that $Y'$ is on the perpendicular bisector of $AB$, or that $YM\perp AB$. Because $OM\perp AB$, it suffices to show that $Y,O,M$ are collinear. Now, note that $AMON$ is cyclic and thus $\angle MON = 180-\angle A$. Because $AT \parallel NY'$, we know that $\angle ONY'$ is equal to the angle between lines $NO$ and $AT$. Since $NO\perp AC$, this is equal to $90-\frac{\angle A}{2}$. Next, because $N$ and $Y'$ are reflections across a line containing $O$, $ON=OY'$, we have $\angle OY'N=90-\frac{\angle A}{2}$. Therefore $\angle NOY'= \angle A$. Lastly, since $\angle MON = 180-\angle A$, we see that $Y',O,M$ are collinear. Hence $Y=Y'$, as desired by the claim.

Similarly, we can show that $X$ is the reflection of $M$ over the perpendicular bisector of $AT$. So, $MN$ and $XY$ are reflections over the perpendicular bisector of $AT$. Thus, they meet on the perpendicular bisector of $AT$, which shows $KA=KT$. $\square$

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