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
VectorAB
07.11.2024 18:32
https://mp.weixin.qq.com/s?__biz=MzkzMDE5NDExMQ==&mid=2247501442&idx=1&sn=f45287696f1f5ae0f131d25473525bea&chksm=c27f7152f508f8449682a0e13fdd8b529376c3ef61f1a7702d7b433524e84d4fbe4a3a4bd461#rd
VectorAB
07.11.2024 18:39
HFE similar ABC DK:AK=HK':AK=HH':AH
Ianis
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.
Geolympics
08.11.2024 01:24
My solution with similarity and Thales (in both English and Spanish).
Hint1Let $S$ be the foot of the perpendicular from $H$ to $PQ$ and let $R$ be the intersecction of $AM$ and $HS$. Prove that
$R$ is the orthocenter of $\triangle HPQ$.
Hint2Prove that $\triangle HRP \sim \triangle AHB$. Then, do angle chase.
SolutionLet $S$ be the foot of the perpendicular from $H$ to $PQ$ and let $R$ be the intersecction of $AM$ and $HS$. To show that
$R$ is the orthocenter of $\triangle HPQ$, we will prove that $\triangle HRP \sim \triangle AHB$ using $SAS$.
Since $HR \perp SM$ and $CM \perp SM$, it follows that $HR || CM$, implying that $\angle QHR= \angle HCB$, but we know that
$\angle HCB = \angle BAH$. Hence,
\[
\angle QHR = \angle BAH.
\]
On the other hand, let $D$ be the foot of the perpendicular from $A$ to $BC$. Given that $HR || DM$, it follows from Thale's Theorem that
\[
\frac{HR}{DM} = \frac{AH}{AD}.
\]Therefore,
\[
HR=\frac{AH \cdot DM}{AD}.
\]Similarly, $HD || QM$, which by Thale's Theorem means that
\[
HQ = \frac{DM \cdot CH}{CD}.
\]From here, we see that
\[
\frac{HR}{HQ}=\frac{AH \cdot CD}{AD \cdot CH}.
\]However, it's not difficult to see that $\triangle CDH \sim \triangle ADB$. As a consequence of this,
\[
\frac{CD}{CH} = \frac{AD}{AB}.
\]Combining the last two equations,
\[
\frac{HR}{HQ} = \frac{AH}{AB}.
\]So, by $SAS$, $\triangle HRQ \sim \triangle AHB$, implying that $\angle RQH = \angle HBA$. Some angle chasing leads to
$QR \perp HP$, proving that $R$ is indeed the orthocenter of $\triangle HPQ$, from which the conclusion follows.
Hint1Sea $S$ el pie de la perpendicular de $H$ a $PQ$ y sea $R$ la intersección de $AM$ y $HS$. Demuestra que
$R$ es el ortocentro del $\triangle HPQ$.
Hint2Demuestra que $\triangle HRP \sim \triangle AHB$. Después, haz caza de ángulos.
SoluciónSea $S$ el pie de la perpendicular de $H$ a $PQ$ y sea $R$ la intersección de $AM$ y $HS$. Para probar que
$R$ es el ortocentro del $\triangle HPQ$, procederemos a mostrar que $\triangle HRP \sim \triangle AHB$ usando $LAL$.
Ya que $HR \perp SM$ y $CM \perp SM$, se sigue que $HR || CM$, implicando que $\angle QHR= \angle HCB$, pero se sabe que
$\angle HCB = \angle BAH$. Luego,
\[
\angle QHR = \angle BAH.
\]
Por otro lado, sea $D$ el pie de la perpendicular de $A$ a $BC$. Dado que $HR || DM$, se sigue del Teorema de Thales que
\[
\frac{HR}{DM} = \frac{AH}{AD}.
\]Por tanto,
\[
HR=\frac{AH \cdot DM}{AD}.
\]Similarmente, $HD || QM$, por lo que el Teorema de Thales nos lleva a que
\[
HQ = \frac{DM \cdot CH}{CD}.
\]De aquí que,
\[
\frac{HR}{HQ}=\frac{AH \cdot CD}{AD \cdot CH}.
\]No obstante, no es difícil verificar que $\triangle CDH \sim \triangle ADB$. En consecuencia,
\[
\frac{CD}{CH} = \frac{AD}{AB}.
\]Combinando las dos ecuaciones previas,
\[
\frac{HR}{HQ} = \frac{AH}{AB}.
\]Así, por $LAL$, $\triangle HRQ \sim \triangle AHB$, de lo cual se sigue que $\angle RQH = \angle HBA$. Haciendo caza de ángulos se llega a que
$QR \perp HP$, probando así que $R$ es el ortocentro del $\triangle HPQ$, de lo cual se sigue la conclusión.
Attachments:
gnoka
20.12.2024 16:33
V.A. Yasinsky Geometry Olympiad 2023 IX p4 , Ukraine