Let $P$ be a point on the circumcircle of triangle $ABC$. Let $A_1$ be the reflection of the orthocenter of triangle $PBC$ about the reflection of the perpendicular bisector of $BC$. Points $B_1$ and $C_1$ are defined similarly. Prove that $A_1,B_1,C_1$ are collinear.
Problem
Source: Sharygin 2019 Finals Day 1 Grade 9 P2
Tags: geometry, Sharygin Geometry Olympiad, perpendicular bisector
30.07.2019 16:23
Straightforward complex bash Set $\odot{ABC}$ as unit circle. First we present a lemma Lemma wrote: Let $AB\parallel CD$ be points on unit circle. Then $ab=cd$ Proof Straightforward Now back to the original problem. Clearly if $P'$ is reflection of $P$ over $\text{perp bisector of BC}$ then $A_1$ is orthocenter of $P'BC$.Hence $A_1=\dfrac{bc}{p}+b+c$ and similar variants. Now it's straightword. $\blacksquare$
30.07.2019 16:39
Cute and easy. MathPassionForever wrote: Let $P$ be a point on the circumcircle of triangle $ABC$. Let $A_1$ be the reflection of the orthocenter of triangle $PBC$ about the reflection of the perpendicular bisector of $BC$. Points $B_1$ and $C_1$ are defined similarly. Prove that $A_1,B_1,C_1$ are collinear. Let $X,Y$ be points on $\odot(ABC)$ such that $PX||BC$ and $PY||AC$. Let $H$ be the orthocenter of $\triangle ABC$. Now, $A_1,B_1$ are orthocenters of $\triangle XBC$ and $\triangle YAC$ respectively. Therefore $AXA_1H$ and $BYB_1H$ are parallelograms. Lastly $AY = CP = BX$, so $AX||BY$, so $H,A_1,B_1$ are collinear and we are done. $~\square$
06.08.2019 12:01
Let $P'$ be the $P-$antipode WRT $\odot (ABC)$, then due to reflection, it follows that $\overline{A_1B_1C_1}$ is the Steiner Line WRT $P'$
23.07.2022 12:52
The first official solution uses a very nice argument. We move $P$ with constant velocity on $\odot(ABC)$, and so on...