Weird Line Passes Through Pole of Side

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Weird Line Passes Through Pole of Side

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: LiOG Epsilon 12.3

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

63 posts

#1 h Apr 29, 2025, 9:45 AM

Y by

Let $\triangle ABC$ be a triangle with orthic triangle $\triangle DEF$ and orthocenter $H$ and midpoint of $\overline{BC}$ as $M$ . If $P=\overline{MH} \cap \overline{EF}$ then prove that $\overline{PD}$ passes through pole of $\overline{BC}$ wrt $(ABC)$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

245 posts

#2 h Apr 29, 2025, 9:52 AM

Y by

What a nice problem.

All pole-polars are wrt $(ABC)$ . We define some new points.

$\bullet$ Let $X_A$ and $Q_A$ be $A$ -Ex and Queue points respectively.
$\bullet$ Let $Y$ , $Z=\overline{EF} \cap (ABC)$ .
$\bullet$ Let $T$ be pole of $\overline{BC}$ .

Now $\overline{BC}$ is polar of $T$ and by La Hire we have $T$ lies on Polar of $X_A$ . Also $(X_AD;BC)=-1$ and so $D$ lies on Polar of $X_A$ .

So all it is left to prove is $P$ lies on Polar of $X_A$ or that $(X_AP;YZ)=-1$ . By PoP we have $X_AZ \cdot X_AY=X_AB \cdot X_AC=X_AD \cdot X_AM=X_AQ_A \cdot X_AA=\lambda$ Hence invert at $X_A$ with radius $\sqrt{\lambda}$ and since inversion preserves cross ratios we have $(X_A,P;Y,Z)=(\infty,P^*;Y,Z)$ So we need to prove that $P^*=(X_AAD) \cap \overline{EF}$ is midpoint of $\overline{YZ}$ but see that $\overline{AP^*} \perp \overline{EF}$ and it is well known that $\overline{AO} \cap \overline{EF}$ so $A$ , $P^*$ , $O$ collinear and obviously $\overline{AO}$ bisects $\overline{YZ}$ and so we are done.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a