Let $ABC$ be an acute triangle inscribed in a circle of center $O$. If the altitudes $BD,CE$ intersect at $H$ and the circumcenter of $\triangle BHC$ is $O_1$, prove that $AHO_1O$ is a parallelogram.
Problem
Source: Greece JBMO TST 2015 p2
Tags: geometry, circumcircle, parallelogram, orthocenter
AlastorMoody
29.04.2019 13:13
Let $H'$ be the reflection of $H$ over $BC$, and since, center of $\odot (BHC)$ is the reflection of $O$ over $BC$ $\implies$ $HO'$ $\cap$ $H'O$ $=$ $G$ $\in$ $BC$ $\implies$ $\angle H'HO'$ $=$ $\angle AH'O$ $=$ $\angle HAO$ $\implies$ $AHOO'$ is parallelogram
Com10atorics
29.08.2020 22:51
$O_1$ is obviously the reflection of $O$ in $BC$ so we have to parts to the solution, 1)$AH$ is parallel to $OO_1$ Yes since $AH\perp BC$ and $OO_1\perp BC$ 2)$AO$ is parallel $HO_1$ Let $F= AO\cap CE$ $\angle CHO_1 =\angle HCO_1=\angle ECB+\angle BCO=\angle C$ $\angle CFO= \angle AFE= 90-\angle BAO=\angle C$
Jishnu4414l
26.10.2023 16:52
Reflect the orthocentre $H$ to $H_A$ over $BC$. It is a well known fact that $H_A$ lies on $(ABC)$.
Now notice that $O$ and $O_1$ are reflections over $BC$. Now $HO_1$ and $OH_A$ must meet at $BC$ at say, $F$. We have $\angle FHH_A=\angle FH_AH=\angle OH_AH=\angle OAH_A$ which gives $AO \parallel HO_1$. Also we trivially have $AH=OO_1$. Our proof is thus complete.