In an acute triangle $ABC$, the incircle $\omega$ touches $BC$ at $D$. Let $I_a$ be the excenter of $ABC$ opposite to $A$, and let $M$ be the midpoint of $DI_a$. Prove that the circumcircle of triangle $BMC$ is tangent to $\omega$. Patrik Bak (Slovakia)
Problem
Source: 2021 Czech-Polish-Slovak Match, P2
Tags: geometry, Computer problems
Steve12345
03.08.2021 12:38
https://artofproblemsolving.com/community/c6h17323p118682 Post 15
hakN
03.08.2021 15:11
By ISL 2002 G7, we know that if $X = DI_a \cap \omega$, then $(BXC)$ is tangent to $\omega$. Let $E,F$ be touch points of incircle with $AB$ and $AC$ respectively. Let $D'$ be the antipode of $D$ in $\omega$.
It is well known that $EF , D'X , BC$ concur at say $P$. And so since $(P,D;B,C) = -1$ and $\angle PXD = 90$, we have $XD$ bisects $\angle BXC$.
Let $N = AI \cap (ABC)$. Since $N$ is the midpoint of $II_a$, we have $MN \parallel ID \perp BC \implies MN \perp BC$. So $M$ lies on the perpendicular bisector of $BC$, and combining this with the fact that $\angle BXM = \angle BXD = \angle CXD = \angle CXM$, we have $BXCM$ are concyclic, as desired.
rafaello
03.08.2021 22:06
@above, much easier solution here: Let $T$ be the intersection of $\omega$ and $DI_a$. By IMO SL 2002 G7, we know that $BTC$ is tangent to $\omega$. By homothety sending $\omega$ to $(BTC)$ centered at $T$, we get that if $S=DI_a\cap (BTC)$, then $BC\perp ID\parallel OS$, where $O$ is the centre of $(BTC)$. Also if $N$ is the midpoint of $BC$, we know that $ON$ is the perpendicular bisector of $BC$, hence we get that $NS$ is the midline of $\triangle DI_aG$, where $D$ is touch point of incircle with $BC$ and $G$ is touch point of $A$-excircle with $BC$. We conclude that $S$ is the midpoint of $DI_A$.