Let $O$ be a circumcenter of acute triangle $ABC$ and let $O_1$ and $O_2$ be circumcenters of triangles $OAB$ and $OAC$, respectively. Circumcircles of triangles $OAB$ and $OAC$ intersect side $BC$ in points $D$ ($D \neq B$) and $E$ ($E \neq C$), respectively. Perpendicular bisector of side $BC$ intersects side $AC$ in point $F$($F \neq A$). Prove that circumcenter of triangle $ADE$ lies on $AC$ iff $F$ lies on line $O_1O_2$
Problem
Source: Bosnia and Herzegovina EGMO Team Selection Test 2018
Tags: geometry, circumcircle, perpendicular bisector
paperinikk
25.07.2019 00:30
Unfortunately it's a bit long
Let $G$ be the circumcentre of $\triangle ADE$
As $\triangle AOB$ is isosceles and $\angle AOB=2\gamma$ we have $\angle ABO=\angle ADO=\frac{\pi}{2}-\gamma$
$\angle AFO=\pi-\angle OFC=\frac{\pi}{2}+\gamma$. $\angle AFO+\angle ADO=\pi$ hence $A,F,O,D,B$ are concyclic.
$\angle AEC=\angle AOC=2\beta \Rightarrow \angle AED=\pi-2\beta$. $\angle DO_1A=2\angle DBA=2\beta$ hence $A,D,E,O_1$ are concyclic
$\angle ADB=\angle AOB=2\gamma \Rightarrow \angle ADE=\pi-2\gamma$. $\angle AO_2E=\angle ACE=2\gamma$ hence $A,D,E,O_2$ are concyclic.
Thus $A,D,E,O_1,O_2$ are concyclic.
$\frac{\pi}{2}-\beta=\angle OAC=\angle OED$ so $OE$ is parallel to the height through $C$, thus $OE$ is the perpendicular bisector of $AB$ and $OE$ passes through the midpoint of $AB$. Hence $O_1,O,E$ are collinear ($AB$ is a chord of the circle with center $O_1$)
Similarly $OD$ is the perpendicular bisector of $AC$ and $D,O,O_2$ are collinear.
$O_1,F,O_2$ are collinear iff $\angle FO_1O=\angle O_2O_1E$:
Let $K=OF\cap BC$. $\angle FO_1O=\pi-2\angle O_1OF=\pi-2\angle KOE=\pi-2\beta$
$\angle O_2O_1E=\angle ODE=\frac{\pi}{2}-\gamma$
Hence $O_1,F,O_2$ are collinear iff $\pi-2\beta=\frac{\pi}{2}-\gamma$ $\Rightarrow$ $2\beta-\gamma=\frac{\pi}{2}$
Now we prove that $G,E,C,O_2$ are concyclic iff $2\beta-\gamma=\frac{\pi}{2}$:
$\angle EGO_2=2\angle EDO_2=2\left(\frac{\pi}{2}-\gamma\right)=\pi-2\gamma$
$\angle O_2CE=\gamma+\angle O_2CA=\gamma+\left(\frac{\pi}{2}-\frac{\angle AO_2C}{2}\right)=\gamma+\frac{\pi}{2}-\frac{2\pi-2\angle AOC}{2}=\gamma+\frac{\pi}{2}-\pi+\angle AOC=\gamma-\frac{\pi}{2}+2\beta$
$G,E,C,O_2$ are collinear iff $\angle EGO_2+\angle O_2CE=\pi$ $\Leftrightarrow$ $\pi-2\gamma+\gamma-\frac{\pi}{2}+2\beta=\pi$ $\Leftrightarrow$ $2\beta-\gamma=\frac{\pi}{2}$
To conclude we prove that $G$ lies on $AC$ iff $2\beta-\gamma=\frac{\pi}{2}$:
$G\in AC$ $\Leftrightarrow$ $\angle AGC=\pi$ $\Leftrightarrow$ $\angle AGD+\angle DGE+\angle EGC=\pi$
$\angle AGD=2\angle AED=2\pi-4\beta$
$\angle DGE=2\angle DO_1E=2(\pi-2\angle O_1OD)=2(\pi-2(\pi-\angle O_1OF-\angle DOK))=2(\pi-2(\pi-\beta-\gamma))=2(\pi-2\alpha)$
$\angle EGC=\angle EO_2C=\pi-2\angle O_2EC=\pi-2(\pi-\angle O_2EO_1-\angle OED)=\pi-2(\pi-\alpha-\frac{\pi}{2}+\beta)=2\alpha-2\beta$
$\angle AGD+\angle GDE+\angle EGC=\pi$ $\Leftrightarrow$ $2\pi-4\beta+2\alpha-2\beta+2\pi-4\alpha=\pi$ $\Leftrightarrow$ $4\pi-2\alpha-6\beta=\pi$ $\Leftrightarrow$ $4\pi-2\alpha-6\beta-2\gamma+2\gamma=\pi$ $\Leftrightarrow$ $2\pi-4\beta+2\gamma=\pi$ $\Leftrightarrow$ $2\beta-\gamma=\frac{\pi}{2}$
This completes the proof
Tafi_ak
16.03.2023 10:29
Let $M$ be the midpoint of $BC$ and $X$ be the center of $(ADE)$. Here are some claims: Claim: The point $F$ lies on $(AOB)$ and $D$ lies on the perpendicular bisector of $AC$. Similarly $E, H\in (AOC)$, where $E=OO_1\cap BC$, $H=OM\cap AB$. Proof. Notice that $\angle BFO=\angle OFC=90^\circ-\angle C=\angle BAO$. So $ADBFO$ is cyclic. Let the perpendicular bisector of $AC$ intersects $AB$ at $D'$. So $\angle OD'C=\angle 90^\circ-\angle C$, but we know $\angle ODC=90^\circ-\angle C$. Hence $D'=D$. $\square$ Claim: Points $A, O_1, D, E, O_2$ are cyclic. Proof. Notice $\angle DO_2A=\angle OO_2A=2\angle OEA=\angle CEA=\angle DEA$, so $AO_2ED$ is cyclic. Similarly $AO_1ED$ is cyclic. $\square$ Now come to the conclusion. \begin{align*} \text{$X$ lies on $AC$}&\iff\text{$\triangle ADO_2$ is isosceles} \\ &\iff AO\parallel CH \\ &\iff \text{$AOCH$ is isosceles trapezoid} \\ &\iff \text{$O_1, F, O_2$ are collinear.} \end{align*}As desired.