Competition problem — Geometry — Maths Olympiad Prep

Library / /259 of 520

Geometry Difficulty 6.5 National olympiad Prove it

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

Solution

Let $O$ be the center of the circle $\omega$ . We have $O=(M Y) \cap(N X)$ . Let $\ell$ be the perpendicular bisector of $[A T]$ . It passes through $O$ .
Let $s$ be the reflection with respect to $\ell$ . Since $(A T)$ is the angle bisector of $\widehat{B A C}$ , the line $s(A B)$ is parallel to $(A C)$ . Since $(O M) \perp(A B)$ and $(O N) \perp(A C)$ , the line $s(O M)$ is parallel to $(O N)$ . Moreover, it passes through $O$ , so $s(O M)=(O N)$ .
Furthermore, the circumcircle $\gamma$ of $A M T$ is symmetric with respect to $\ell$ , i.e., $s(\gamma)=\gamma$ , so $s(M)$ belongs to both $\gamma$ and $s(O M)=(O N)$ . Necessarily, $s(M)=X$ . Similarly, $s(N)=Y$ . We deduce that $s(M N)=(X Y)$ . The intersection point $K$ of $(M N)$ and $(X Y)$ thus lies on the line $\ell$ , which implies that $K A=K T$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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