Track / Stage 6 / 58 of 400 #1538 of 2444

Problem 1538

National Olympiad, first round

Geometry Difficulty 6.0 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia

Let $ABC$ be a triangle with $A'$ , $B'$ , $C'$ as the midpoints of $BC$ , $CA$ , $AB$ respectively. The circle $(\omega_A)$ of center $A$ with sufficiently large radius cuts $B'C'$ at $X_1$ , $X_2$ . Define circles $(\omega_B)$ , $(\omega_C)$ with $Y_1$ , $Y_2$ , $Z_1$ , $Z_2$ similarly. Suppose that these circles have the same radius. Prove that $X_1$ , $X_2$ , $Y_1$ , $Y_2$ , $Z_1$ , $Z_2$ are concyclic.

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

Let $H$ be the orthocenter of $\triangle ABC$ . Since $AH \perp X_1X_2$ and $AX_1 = AX_2$ , then $HX_1 = HX_2$ . Similarly, $HY_1 = HY_2$ and $HZ_1 = HZ_2$ .

Denote $R$ as the radius of the three circles $(\omega_A)$ , $(\omega_B)$ , $(\omega_C)$ . As $HB \perp Y_1C'$ , $HC \perp Z_1B'$ , we have
$\begin{aligned} & HY_1^2 - R^2 = HY_1^2 - BY_1^2 = HC^2 - BC^2 \\ & HZ_1^2 - R^2 = HZ_1^2 - CZ_1^2 = HB'^2 - CB'^2 \end{aligned}$
In addition, $AH \perp B'C'$ , so
$HC'^2 - BC'^2 = HC'^2 - AC'^2 = HB'^2 - AB'^2 = HB'^2 - CB'^2$
From that $HY_1 = HZ_1$ . Similarly, $HX_1 = HY_1 = HZ_1$ and $HX_2 = HY_2 = HZ_2$ .

From (1) and (2), the orthocenter $H$ is indeed the center of the circle which goes through the six points $X_1$ , $X_2$ , $Y_1$ , $Y_2$ , $Z_1$ , $Z_2$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.