Let $ABC$ be a triangle with $\angle A=60^\circ$; $AD$, $BE$, and $CF$ be its bisectors; $P, Q$ be the projections of $A$ to $EF$ and $BC$ respectively; and $R$ be the second common point of the circle $DEF$ with $AD$. Prove that $P, Q, R$ are collinear.
Problem
Source: 10.7 of XX Geometrical Olympiad in honour of I.F.Sharygin
Tags: geo, geometry
NO_SQUARES
07.08.2024 09:56
WLOG $AB<AC$. It is known that Feuerbach point $K$ of $\Delta ABC$ lies on $(DEF)$. In other side, if $\angle A=60^\circ$, then $K$ lies on $AD$, because in this case $AH=AO$ and so $AI$ is perpendicular bisector to $OH$ and $O_9 \in AI$; but $K \in IO_9$ by Feuerbach theorem.
By this reason we get that $R$ is Feuerbach point and so it is lies on $\omega _9$. Note that if $X=EF \cap BC$, then $AX \perp AD$ and $APQX$ is cyclic. So $\angle DQP \angle XAP=\angle ALF$, there $L = AD \cap EF$.
If $I$ is incenter of triangle $ABC$, then $AEIF$ is cyclic, so we can easily find $\angle ALF$. Also we can easily find $\angle DQR$ and get that $\angle DQR = 30^\circ +\frac{1}{2}\angle C=\angle DQP$, so $R\in PQ$.
Parsia--
18.08.2024 21:19
Here is my solution from the exam: Let $AD \cap EF = M$, $EF \cap BC = D'$ and $PQ \cap AD = X$ Claim 1: $AD$ is tangent to $(APQD')$ Proof: We have $\angle APD' = 90 = \angle XAD'$ and so $AD$ is tangent to $(APQD')$ $\newline$ Claim 2: $P,Q,M,D$ are concyclic Proof: Since $\angle D'PA = \angle D'AM = 90$ we have $\triangle D'PA \sim \triangle D'AM$ and so $\angle PQD = \angle PAD' = \angle AMP$ which proves the claim. $\newline$ Claim 3: $X$ is the midpoint of $AI$ Proof: By combining claims 1 and 2 we have $XA^2 = XP \cdot XQ = XM \cdot XD$ and because $(AI,MD) = -1$ we get that $X$ is the midpoint of $AI$ $\newline$ To finish notice that $\angle EIF = 120 = 180 - \angle BAC$ so $A,E,I,F$ are cyclic and so we have $ME\cdot MF = MI \cdot MA = MX \cdot MD$ which gives that $X$ is on $(DEF)$ and so $X\equiv R$ which proves the problem.