Problem 1846

Let $O$ be the center of $\omega$, thus $O = M Y \cap N X$. Let $\ell$ be the perpendicular bisector of $A T$ (it also passes through $O$). Denote by $r$ the operation of reflection about $\ell$. Since $A T$ is the angle bisector of $\angle B A C$, the line $r(A B)$ is parallel to $A C$. Since $O M \perp A B$ and $O N \perp A C$, this means that the line $r(O M)$ is parallel to the line $O N$ and passes through $O$, so $r(O M) = O N$. Finally, the circumcircle $\gamma$ of the triangle $A M T$ is symmetric about $\ell$, so $r(\gamma) = \gamma$. Thus the point $M$ maps to the common point of $O N$ with the arc $A M T$ of $\gamma$ - that is, $r(M) = X$.

Similarly, $r(N) = Y$. Thus, we get $r(M N) = X Y$, and the common point $K$ of $M N$ and $X Y$ lies on $\ell$. This means exactly that $K A = K T$.

Let $X^{\prime}$ be the common point of the line $N X$ with the external bisector of $\angle B A C$; notice that it lies outside the triangle $A B C$. Then we have $\angle T A X^{\prime} = 90^{\circ}$ and $X^{\prime} A = X^{\prime} C$, so we get $\angle X^{\prime} A M = 90^{\circ} + \angle B A C / 2 = 180^{\circ} - \angle X^{\prime} A C = 180^{\circ} - \angle X^{\prime} C A = \angle X^{\prime} C L$. Thus the triangles $X^{\prime} A M$ and $X^{\prime} C L$ are congruent, and therefore
$$\angle M X^{\prime} L = \angle A X^{\prime} C + (\angle C X^{\prime} L - \angle A X^{\prime} M) = \angle A X^{\prime} C = 180^{\circ} - 2 \angle X^{\prime} A C = \angle B A C = \angle M A L .$$
This means that $X^{\prime}$ lies on $\gamma$.
Thus we have $\angle T X N = \angle T X X^{\prime} = \angle T A X^{\prime} = 90^{\circ}$, so $T X \parallel A C$. Then $\angle X T A = \angle T A C = \angle T A M$, so the cyclic quadrilateral $M A T X$ is an isosceles trapezoid. Similarly, $N A T Y$ is an isosceles trapezoid, so again the lines $M N$ and $X Y$ are the reflections of each other about the perpendicular bisector of $A T$. Thus $K$ belongs to this perpendicular bisector.

Let $L$ be the second common point of the line $A C$ with the circumcircle $\gamma$ of the triangle $A M T$. From the cyclic quadrilaterals $A B T C$ and $A M T L$ we get $\angle B T C = 180^{\circ} - \angle B A C = \angle M T L$, which implies $\angle B T M = \angle C T L$. Since $A T$ is an angle bisector in these quadrilaterals, we have $B T = T C$ and $M T = T L$. Thus the triangles $B T M$ and $C T L$ are congruent, so $C L = B M = A M$.

Let $X^{\prime}$ be the common point of the line $N X$ with the external bisector of $\angle B A C$; notice that it lies outside the triangle $A B C$. Then we have $\angle T A X^{\prime} = 90^{\circ}$ and $X^{\prime} A = X^{\prime} C$, so we get $\angle X^{\prime} A M = 90^{\circ} + \angle B A C / 2 = 180^{\circ} - \angle X^{\prime} A C = 180^{\circ} - \angle X^{\prime} C A = \angle X^{\prime} C L$. Thus the triangles $X^{\prime} A M$ and $X^{\prime} C L$ are congruent, and therefore

Similarly, $r(N) = Y$. Thus, we get $r(M N) = X Y$, and the common point $K$ of $M N$ and $X Y$ lies on $\ell$. This means exactly that $K A = K T$.