Let $AH_A, BH_B, CH_C$ be the altitudes of the triangle $ABC$. Points $A_1$ and $C_1$ are the projections of the point $H_B$ onto the sides $AB$ and $BC$, respectively. $B_1$ is the projection of $B$ onto $H_AH_C$. Prove that the diameter of the circumscribed circle of $\triangle A_1B_1C_1$ is equal to $BH_B$. Proposed by Anton Trygub
Problem
Source: Kyiv City MO 2024 Round 2, Problem 10.3
Tags: circumcircle, geometry
Iveela
05.02.2024 06:38
First, we relabel $A, B$ such that $B$ and $A$ switch places. Note that $\triangle AH_CH_B \sim \triangle ACB$ and hence $\frac{H_BB_1}{B_1C} = \frac{BH_A}{H_AC} = \frac{A_1H_B}{A_1H_C}$. It follows that $A_1B_1 \parallel CH_C$. Similarly, $A_1C_1 \parallel BH_B$. Therefore, $\angle B_1A_1C_1 = 180 - \angle C_1AB_1$. Also note that the quadrilateral $AB_1H_AC_1$ is cyclic since $\angle H_AC_1A + \angle H_AB_1A = 180$. It follows from the law of sines that $AH_A = \frac{AH_A}{\sin 90} = \frac{B_1C_1}{\sin \angle C_1AB_1} = \frac{B_1C_1}{\sin \angle C_1A_1B_1}$ which is equal to the diameter of $\odot A_1B_1C_1$. $\blacksquare$
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IAmTheHazard
05.02.2024 22:11
The main claim is that $A_1B_1C_1H_B$ is a parallelogram. It suffices to prove (by symmetry) that $\overline{A_1H_B} \parallel \overline{B_1C_1}$, or $\overline{CH_C} \parallel \overline{B_1C_1} \iff \frac{H_AC_1}{CC_1}=\frac{H_AB_1}{H_CB_1}=\frac{AH_B}{CH_B} \iff \overline{AH_A} \parallel \overline{H_BC_1}$, which is true. Now $(A_1B_1C_1)$ is congruent to $(BA_1H_BC_1)$, which clearly has diameter $\overline{BH_B}$.