EGMO 2023 | Problem 6 | Solution - 17 April 2023

Problem 6.
Let $ABC$ be a triangle with circumcircle $\Omega$ . Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs $AC$ and $AB$ that do not contain the third vertex. Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$ ). Let $I$ be the incenter of $ABC$ . Let $\omega_b$ be the circle that is tangent to $AB$ and internally tangent to $\Omega$ at $S_b$ , and let $\omega_c$ be the circle that is tangent to $AC$ and internally tangent to $\Omega$ at $S_c$ . Show that the line $IN_a$ , and the lines through the intersections of $\omega_b$ and $\omega_c$ , meet on $\Omega$ .

Solution
Suppose that $\omega_b$ is tangent to $\overline{AB}$ at $D$ and $\omega_c$ is tangent to $\overline{AC}$ at $E$ . The homothety at $S_b$ sending $\omega_b$ to $\Omega$ sends $D$ to $S_c$ , so $D$ lies on $\overline{S_bS_c}$ . Similarly, $E$ lies on $\overline{S_bS_c}$ . Since $S_bA=S_bI$ and $S_cA=S_cI$ by the incenter-excenter lemma, $\overline{S_bS_c}$ is perpendicular to $\overline{AI}$ , so $AD=AE$ . Thus, $A$ lies on the radical axis of $\omega_b$ and $\omega_c$ .

Let $\overline{IN_a}$ intersect $\Omega$ again at $X$ , and let $S_a$ be the antipode of $N_a$ in $\Omega$ . Let the tangents to $\Omega$ at $S_b$ and $S_c$ intersect at $T$ . We have $TS_b=TS_c$ , so $T$ lies on the radical axis of $\omega_b$ and $\omega_c$ . Also, notice that $-1=(B,C;S_a,N_a) \overset{I}{=} (S_b,S_c;A,X),$ so $T$ lies on $\overline{AX}$ . Since $A$ and $T$ both lie on the radical axis of $\omega_b$ and $\omega_c$ , we know that $X$ lies on the radical axis, as desired.