Track / Stage 5 / 256 of 400 #856 of 1964

Problem 856

AIME late

Geometry Difficulty 5.7 Prove it

2. Given the circumcircle $\Gamma$ of $\triangle A B C$ , the midpoints of sides $A B$ and $A C$ are $M$ and $N$ respectively, and the midpoint of the arc $\text{B C}$ of circle $\Gamma$ not containing point $A$ is $T$ . The circumcircles of $\triangle A M T$ and $\triangle A N T$ intersect the perpendicular bisectors of sides $A C$ and $A B$ at points $X$ and $Y$ respectively, and $X$ and $Y$ are inside $\triangle A B C$ . If line $M N$ intersects $X Y$ at point $K$ , prove: $K A=K T$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

2. As shown in Figure 1, let the center of circle $\Gamma$ be $O$ . Then $O$ is the intersection of $M Y$ and $N X$ .

Let the perpendicular bisector of segment $A T$ be $l$ . Then line $l$ passes through point $O$ . Denote the reflection transformation about line $l$ as $r$ .

Since $A T$ is the angle bisector of $\angle B A C$ , the line $r(A B)$ is parallel to $A C$ .

Also, $O M \perp A B, O N \perp A C$ , thus, the line $r(O M)$ is parallel to $O N$ and passes through point $O$ .
Therefore, $r(O M)=O N$ .
Since the circumcircle $\Gamma_{1}$ of $\triangle A M T$ is symmetric about $l$ , then $r\left(\Gamma_{1}\right)=\Gamma_{1}$ .
Thus, the reflection of point $M$ about line $l$ is the intersection of line $O N$ and the arc $\text{A M T}$ of circle $\Gamma_{1}$ , i.e., point $r(M)$ coincides with point $X$ .
Similarly, point $r(N)$ coincides with point $Y$ .
Therefore, $r(M N)=X Y$ .
Hence, the intersection point $K$ of line $M N$ and $X Y$ lies on line $l$ .
Thus, $K A=K T$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.