Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Omega$. The tangent line of the circumcircle of triangle $BHC$ at $H$ meets $AB$ and $AC$ at $E$ and $F$ respectively. If $O$ is the circumcenter of triangle $AEF$, prove that the circumcircle of triangle $EOF$ is tangent to $\Omega$.
Problem
Source: Thailand 2023 TSTST1/1
Tags: geometry
trinhquockhanh
22.08.2023 16:39
You can look for a harder property of this configuration at 2023 Kazakhstan JMO G9 P6
X.Allaberdiyev
04.09.2023 19:59
Let $(E H B)\cap (F H C)=P$, it is trivial that $P$ lies on $ABC$. Then we have $\angle EPF=\angle EPH+\angle HPF=\angle HCA+\angle HBA=180-2\angle A$ and since $\angle EOF=2\angle A$ we have $(E O F P)$ cyclic. Then let $KP$ be the tangent to $(A B C )$ (such that $K$ lies on the half plane with $C$, according to line $AP$), then $\angle KPF=\angle CPF+\angle CPK=\angle CHF+\angle CAP=\angle CBP+\angle HBC=\angle HBP=\angle HEP=\angle FEP$, then $KP$ is also tangent to $(E O F P)$, which completes the solution.
CT17
05.09.2023 01:41
The inversion at $A$ swapping $(E,B)$ and $(F,C)$ maps $O$ to the reflection of $A$ over $BC$, $(EOC)$ to $(BHC)$, and $(ABC)$ to line $EF$.