Problem 59: ELMO Shortlist 2012, G3

Posts: 312
Joined: Mar 10, 2015

by henderson, Jan 21, 2017, 2:33 PM

$\color{red}\bf{Problem \ 59}$ $ABC$ is a triangle with incenter $I.$ The foot of the perpendicular from $I$ to $BC$ is $D,$ and the foot of the perpendicular from $I$ to $AD$ is $P.$ Prove that $\angle BPD = \angle DPC.$

Proposed by Alex Zhu
$\color{red}\bf{Solution}$ Let the incircle touch the sides $AB,AC$ at $E,F,$ respectively and let $X=EF \cap BC.$ Then $X$ lies on the polar of $A$ with respect to the incircle, so $A$ lies on the polar of $X.$ Thus $AD$ is the polar of $X.$ In particular, $IX \perp AD,$ so $\angle XPD=90^{\circ}.$ Since $AD,BE,CF$ are concurrent, $(B;C;D;X)=-1,$ and it is a well-known fact that the last two equalities imply $\angle BPD=\angle DPC.$

This post has been edited 4 times. Last edited by henderson, Jan 21, 2017, 3:20 PM

Math Path

Problem 59: ELMO Shortlist 2012, G3

by henderson, Jan 21, 2017, 2:33 PM

0 Comments