Let $ABC$ an acute triangle with orthocenter $H$. Let $M$ be a point on segment $BC$. The line through $M$ and perpendicular to $BC$ intersects lines $BH$ and $CH$ in points $P$ and $Q$ respectively. Prove that the orthocenter of triangle $HPQ$ lies on the line $AM$.
Problem
Source: 2024 Mexican Mathematical Olympiad, Problem 4
Tags: geometry, Mexico
07.11.2024 18:32
https://mp.weixin.qq.com/s?__biz=MzkzMDE5NDExMQ==&mid=2247501442&idx=1&sn=f45287696f1f5ae0f131d25473525bea&chksm=c27f7152f508f8449682a0e13fdd8b529376c3ef61f1a7702d7b433524e84d4fbe4a3a4bd461#rd
07.11.2024 18:39
HFE similar ABC DK:AK=HK':AK=HH':AH
07.11.2024 19:28
Let $H'$ be the orthocentre of $HPQ$ and let $D=PH'\cap BC$, $E=QH'\cap BC$, $F=PQ\cap AB$ and $G=PQ\cap CA$. Observe that $PH'\perp HQ\perp AB$ and $QH'\perp HP\parallel CA$. Hence if $\angle BAC=\alpha$, $\angle CBA=\beta$ and $\angle ACB=\gamma$, we get\begin{align*}\frac{MD}{MB} & =\frac{MP}{MF} \\ & =\frac{\sin \angle MBP}{\sin \angle PMB}\frac{\sin \angle BFP}{\sin \angle PBF} \\ & =\frac{\sin (90^\circ -\gamma )}{\sin 90^\circ}\frac{\sin (90^\circ -\beta )}{\sin (90^\circ -\alpha )} \\ & =\frac{\sin \angle QCM}{\sin \angle CMQ}\frac{\sin \angle QGC}{\sin \angle GCQ} \\ & =\frac{MQ}{MG} \\ & =\frac{ME}{MC}. \end{align*}Therefore, $ABC$ and $H'DE$ are homothetic with centre $M$, which gives that $A,H',M$ are collinear, as desired.
08.11.2024 01:24
My solution with similarity and Thales (in both English and Spanish).
Attachments:

20.12.2024 16:33
V.A. Yasinsky Geometry Olympiad 2023 IX p4 , Ukraine