The acute triangle $ABC$ satisfies $\overline {AB}<\overline {BC}<\overline {CA}$. Let $H$ a orthocenter of $ABC$, $D$ a intersection point of $AH$ and $BC$, $E$ a intersection point of $BH$ and $AC$, and $M$ a midpoint of segment $BC$. A circle with center $E$ and radius $AE$ intersects the segment $AC$ at point $F$($\neq A$), and circumcircle of triangle $BFC$ intersects the segment $AM$ at point $S$. Let $P$($\neq D$), $Q$($\neq F$) a intersection point of circumcircle of triangle $ASD$ and $DF$, circumcircle of triangle $ASF$ and $DF$ respectively. Also, define $R$ as a intersection point of circumcircles of triangle $AHQ$ and $AEP$. Prove that $R$ lies on line $DF$.
Problem
Source: 2021 Korea Winter Program Test2 Day1 #3
Tags: geometry
space10
14.02.2021 18:05
(step 1) $\angle{BSC}=\angle{BFC}=180-\angle{BFA}=180-A$, it is a well-known that $S$ is the foot of perpendicular from $H$ to $AM$(Humpty point). $HDMS$ cyclic $\implies AE\cdot AD=AH\cdot AD=AS\cdot AM$ (step 2) Let function $\Phi$: Inversion of a circle with center $A$ and radius $\sqrt{AH\cdot AD}$. Let $\Phi (F)=F'$. $\Phi (E)=C, \Phi (H)=D, \Phi (S)=M, \Phi (\odot (ASD))=MH, \Phi (\odot (ASF))=MF', \Phi (DF)=\odot (AHF')$. $P'=\Phi (P)=\Phi (DF) \cap \Phi (\odot (ASD))=\odot (AHF')\cap MH$, $Q'=\Phi (Q)=\Phi (DF) \cap \Phi (\odot (ASF))=\odot (AHF')\cap MF'$ $\Phi (\odot (AHQ))=DQ', \Phi (\odot (AEP))=CP'\implies R'=\Phi (R)=\Phi (\odot (AHQ))\cap \Phi (\odot (AEP))=DQ'\cap CP'$ (step 3) (ETS) $\Phi (R)\in \Phi (DF)$ Let $R''=CP'\cap \odot (AHF')$. By Pascal's theorem on hexagon $HP'R''Q'F'A$, $M, C, R''Q\cap AH$ lies on a straight line$\implies MC, R''Q, AH$ meet at one point, so we get $D\in R''Q, R'=R''$ $\Phi (R)=R'=R''\in \odot (AHF')=\Phi (DF)$, so Q.E.D.
KPBY0507
15.02.2021 05:05
Inversion at the circle with center $A$ and radius $\sqrt{AH*AD}$ and Pascal's theorem kills this problem.