Let $\omega$ be the circumcircle of triangle $ABC$, $O$ be its center, $A'$ be the point of $\omega$ opposite to $A$, and $D$ be a point on a minor arc $BC$ of $\omega$. A point $D'$ is the reflection of $D$ about $BC$. The line $A'D'$ meets for the second time at point $E$. The perpendicular bisector to $D'E$ meets $AB$ and $AC$ at points $F$ and $G$ respectively. Prove that $\angle FOG = 180^\circ - 2\angle BAC$.
Problem
Source: Sharygin Finals 2023 10.3
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023
keglesnit
02.08.2023 14:28
Let $H$ be the orthocenter, $AH$ intersects $(ABC)$ again in $H'$ and $DH'$ intersects $D'H$ on $BC$ at $X$. We will prove that $H$ is the isogonal conjugate of $O$ in quad $FGBC$, if so, then $\angle FOG=180^{\circ}-\angle BOC=180^{\circ}-2\angle BAC$. To do that, we'll show that $X$ lies on $FG$, and then prove that $XH$ and $XO$ are isogonal in $\angle(FG,BC)$. This suffices since $BH$, $BO$ are isogonal in $\angle B$ and $CH$, $CO$ are isogonal in $\angle C$. Claim. $X$ lies on $FG$. Proof. We will prove something even stronger: $X$ is the center of $(ED'D)$. Simply note that $$\angle DED'=\angle DEA'=\angle DAA'=90^{\circ}-\angle AA'D=90^{\circ}-\angle AH'X=\angle BXH'=0.5\angle DXD'$$which finishes the proof since $BC$ is the perpendicular bisector of $DD'$. Claim. $XH$ and $XO$ are isogonal in $\angle(FG,BC)$. Proof. Let $U=XO\cap DE$ and $V=DD'\cap BC$. Note that $XUVD$ is cyclic (right angles). Now we have $$\angle(FG,XH)=0.5\angle EXD'=\angle EDD'=\angle UDV=\angle UXV=\angle (XO,BC)$$which finishes the claim. Remark. Very similar to geometry marathon P140 https://artofproblemsolving.com/community/u969437h2665078p27093690
i3435
03.08.2023 17:20
$\angle AED'=\angle AEA'=90^{\circ}$, so the midpoint of $AD'$ is the foot from $O$ onto the perpendicular bisector of $D'E$, or the foot from $O$ to $\overline{FG}$ is the midpoint of $AD'$. Since $D'$ lies on the reflection of $D$ across $\overline{BC}$, the midpoint of $AD'$ is on the nine-point circle of $\triangle ABC$. The feet from $O$ to $\overline{BC},\overline{CG},\overline{BF}$ also are on the nine-point circle of $\triangle ABC$, so the feet from $O$ to the sides of $BCGF$ are cyclic. Angle chase means $\angle FOG+\angle BOC=180^{\circ}$, as desired.
VicKmath7
03.08.2023 19:41
Delete $D$ and rename $D'$ to $D$ being a point on $(BHC)$. We will prove that $O$ and $H$ are isogonal conjugates in the quadrilateral $BFGC$, which will imply $\angle FOG+\angle BOC=180^{o}$. Thus, it is sufficient to show that the foot of the perpendicular $P$ from $O$ to $FG$ lies on the nine-point circle of $ABC$. As $FG \parallel AE$, $P$ lies on the perpendicular bisectors of the sides $EA$ and $ED$ of the right-angled triangle $AED$, so $P$ is the midpoint of $AD$. It is known that the reflections of $A$ with respect to the vertices of the orthic triangle lie on $(BHC)$, so the homothety at $A$ with coefficient $\frac{1}{2}$ sends $(BHC)$ to the NPC, meaning that $P$ lies on the NPC, as desired.
alinazarboland
03.08.2023 22:08
Just note that $D'$ lies on $(BHC)$ where $H$ is defined as usual.let $H'$ be the reflection of $H$ across $FG$.So $HD',EH'$ are symmetric wrt $FG$ and the circles $(BHC),(BAC)$ are congruent which means $H'$ lies on $(BAC)$ by symmetry arguement.So the midpoint $M$ of $HH'$ which is also on $FG$ , lies on npc of $ABC$.So since npc is a circle formed by the perpendiculars feom $H$ to the sides of $BCFG$ ,$H $has isogonal conjugate in $BCFG$ and clearly it's $O$ which implies $\angle FOG + \angle BOC =0$ and we're done.