International Mathematical Olympiad Shortlisted Problems

Let $O$ be the center of $\omega$, thus $O=MY\cap NX$. Let $\ell$ be the perpendicular bisector of $AT$ (it also passes through $O$). Denote by $r$ the operation of reflection about $\ell$. Since $AT$ is the angle bisector of $\angle BAC$, the line $r(AB)$ is parallel to $AC$. Since $OM\perp AB$ and $ON\perp AC$, this means that the line $r(OM)$ is parallel to the line $ON$ and passes through $O$, so $r(OM)=ON$. Finally, the circumcircle $\gamma$ of the triangle $AMT$ is symmetric about $\ell$, so $r(\gamma )=\gamma$. Thus the point $M$ maps to the common point of $ON$ with the arc $AMT$ of $\gamma$ - that is, $r(M)=X$.

Similarly, $r(N)=Y$. Thus, we get $r(MN)=XY$, and the common point $K$ of $MN$ and $XY$ lies on $\ell$. This means exactly that $KA=KT$.

Let $X^{\prime }$ be the common point of the line $NX$ with the external bisector of $\angle BAC$; notice that it lies outside the triangle $ABC$. Then we have $\angle TA{X}^{\prime }=90^{\circ }$ and $X^{\prime }A=X^{\prime }C$, so we get $\angle X^{\prime }AM=90^{\circ }+\angle BAC/2=180^{\circ }-\angle X^{\prime }AC=180^{\circ }-\angle X^{\prime }CA=\angle X^{\prime }CL$. Thus the triangles $X^{\prime }AM$ and $X^{\prime }CL$ are congruent, and therefore $$\angle M{X}^{\prime }L=\angle A{X}^{\prime }C+(\angle C{X}^{\prime }L-\angle A{X}^{\prime }M)=\angle A{X}^{\prime }C=180^{\circ }-2\angle X^{\prime }AC=\angle BAC=\angle MAL.$$ This means that $X^{\prime }$ lies on $\gamma$. Thus we have $\angle TXN=\angle TX{X}^{\prime }=\angle TA{X}^{\prime }=90^{\circ }$, so $TX\parallel AC$. Then $\angle XTA=\angle TAC=\angle TAM$, so the cyclic quadrilateral $MATX$ is an isosceles trapezoid. Similarly, $NATY$ is an isosceles trapezoid, so again the lines $MN$ and $XY$ are the reflections of each other about the perpendicular bisector of $AT$. Thus $K$ belongs to this perpendicular bisector.

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Alternative solution.

Let $L$ be the second common point of the line $AC$ with the circumcircle $\gamma$ of the triangle $AMT$. From the cyclic quadrilaterals $ABTC$ and $AMTL$ we get $\angle BTC=180^{\circ }-\angle BAC=\angle MTL$, which implies $\angle BTM=\angle CTL$. Since $AT$ is an angle bisector in these quadrilaterals, we have $BT=TC$ and $MT=TL$. Thus the triangles $BTM$ and $CTL$ are congruent, so $CL=BM=AM$.

Let $X^{\prime }$ be the common point of the line $NX$ with the external bisector of $\angle BAC$; notice that it lies outside the triangle $ABC$. Then we have $\angle TA{X}^{\prime }=90^{\circ }$ and $X^{\prime }A=X^{\prime }C$, so we get $\angle X^{\prime }AM=90^{\circ }+\angle BAC/2=180^{\circ }-\angle X^{\prime }AC=180^{\circ }-\angle X^{\prime }CA=\angle X^{\prime }CL$. Thus the triangles $X^{\prime }AM$ and $X^{\prime }CL$ are congruent, and therefore

Similarly, $r(N)=Y$. Thus, we get $r(MN)=XY$, and the common point $K$ of $MN$ and $XY$ lies on $\ell$. This means exactly that $KA=KT$.