Let $ABC$ be an acute triangle and let $B'$ and $C'$ be the feet of the heights $B$ and $C$ of triangle $ABC$ respectively. Let $B_A'$ and $B_C'$ be reflections of $B'$ with respect to the lines $BC$ and $AB$, respectively. The circle $BB_A'B_C'$, centered in $O_B$, intersects the line $AB$ in $X_B$ for the second time. The points $C_A', C_B', O_C, X_C$ are defined analogously, by replacing the pair $(B, B')$ with the pair $(C, C')$. Show that $O_BX_B$ and $O_CX_C$ are parallel.
Problem
Source: Romania TST 2022
Tags: geometry, romania, Romanian TST
VicKmath7
15.05.2022 11:41
Prove that the angle between $X_BO_B$ and $BC$ is $90+\beta-\gamma$ with angle chasing. The angle between $X_CO_C$ and $BC$ is the same by symmetry and we are done.
motannoir
02.10.2022 13:30
Let's consider the circle (ABC) and let t be the tangent at B to $(ABC)$. By angle chasing we can prove that t is also tangent to $(BB'_aB'_c)$ so the two circles are tangent. Let O be the circumcenter of $(ABC)$ so $B,O,O_b$ are collinear. But $BOA$ and $BO_bX_b$ are both isosceles so we get easily that $OA \parallel X_bO_b$ . Similary we get for C so we are done.