Let $ABC$ be a scalene triangle. Let $\ell$ be a line passing through point $B$ that lies outside of the triangle $ABC$ and creates different angles with sides $AB$ and $BC$ . The point $M$ is the midpoint of side $AC$ and the ponts $H_a$ and $H_c$ are the bases of the perpendicular lines on the line $\ell$ drawn from points $A$ and $C$ respectively. The circle circumscribing triangle $MBH_a$ intersects AB at the point $A_1$ and the circumscribed circle of triangle $MBH_c$ intersects $BC$ at point $C_1$. The point $A_2$ is symmetric to the point $A$ relative to the point $A_1$ and the point $C_2$ is symmetric to the point $C_1$ relative to the point $C_1$. Prove that the lines $\ell$, $AC_2$ and $CA_2$ intersect at one point. (Yana Kolodach)
Problem
Source: V.A. Yasinsky Geometry Olympiad 2023 X-XI p6 , Ukraine
Tags: concurrency, concurrent, geometry
14.12.2023 02:34
First, let the $B$-antipodes wrt $(MBH_A), (MBH_C)$ be $X, Y$ respectively; then $A_1, C_1$ are the projections of $X, Y$ onto $BA, BC$ respectively, and furthermore $AX \parallel CY \implies \triangle{MAX} \cong \triangle{MCY} \implies XA = XA_2 = YC = YC_2$. Now let $\ell \cap AC = Z$; then by Menelaus it suffices to show that $\frac{BA_2}{AA_2} \cdot \frac{CC_2}{BC_2} = \frac{CZ}{AZ}$. If $\angle{ABH_A} = \angle{AXA_1} = \alpha$ and $\angle{CBH_C} = \angle{CYC_1} = \gamma$ then by the Ratio Lemma on $\triangle{ABC}$, $\frac{CZ}{AZ} = \frac{BC}{BA} \cdot \frac{\sin \gamma}{\sin \alpha}$. Furthermore, $\frac{CC_2}{AA_2} = \frac{2 \cdot XA \sin \gamma}{2 \cdot XA \sin \alpha} = \frac{\sin \gamma}{\sin \alpha}$, thus it suffices to show that $\frac{BA_2}{BC_2} = \frac{BC}{BA}$, which is equivalent to $A_2AC_2C$ being cyclic. Claim 1: $A_2AC_2C$ is cyclic. Proof: This is equivalent to the perpendicular bisectors of $AA_2, CC_2, AC$ concurring. Let $X', Y'$ be on the perpendicular bisector of $AC$ such that $XX' \perp AA_2, YY' \perp CC_2$; then by LoS, $MX' = MX \cdot \frac{\sin \angle{MBA}}{\sin \angle{A}}$, which by Ratio Lemma on $\triangle{ABC}$ equals $MX \cdot \frac{\sin \angle{MBC}}{\sin \angle{C}} = MY'$, thus $X' = Y' \implies A_2AC_2C$ is cyclic, and we are done. $\square$