AoPS - Problem

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Tags: geometry, circumcircle, parallelogram

Leicich

23.08.2014 00:08

Let $ABCD$ be an inscribed quadrilateral in a circumference with center $O$ such that it lies inside $ABCD$ and $\angle{BAC} = \angle{ODA}$ . Let $E$ be the intersection of $AC$ with $BD$ . Lines $r$ and $s$ are drawn through $E$ such that $r$ is perpendicular to $BC$ , and $s$ is perpendicular to $AD$ . Let $P$ be the intersection of $r$ with $AD$ , and $M$ the intersection of $s$ with $BC$ . Let $N$ be the midpoint of $EO$ . Prove that $M$ , $N$ , and $P$ lie on a line.

djmathman

23.08.2014 02:11

Let

$\omega$ be the circumcircle of

$ABCD$ , and let

$X$ be the second intersection point of

$OD$ with

$\omega$ . Then the condition

$\angle BAC=\angle ODA$ rearranges to

$\angle BDA=\angle XCD\implies \widehat{AB}=\widehat{XC}$ . Therefore

$\widehat{AB}+\widehat{CD}=\widehat{XC}+\widehat{CD}=\pi\implies AC\perp BD.$ Now consider lines

$r$ and

$s$ . Let

$Y$ and

$Z$ be the projections of

$E$ onto

$BC$ and

$AD$ respectively. Trivially

$\triangle EAD\sim\triangle EBC$ , so

$\angle PED=\angle YEB=\angle ZEA$ . Hence

$EZ$ and

$EP$ are isogonal wrt

$\triangle AED$ , but since

$EZ$ is an altitude

$EP$ must pass through the circumcenter of

$\triangle AED$ , which since

$\angle AED=90^\circ$ is the midpoint of

$AD$ . This implies that

$P$ is the midpoint of

$AD$ and similarly

$M$ is the midpoint of

$BC$ . Now since

$OP\perp AD$ and

$OM\perp BC$ ,

$EPOM$ is a parallelogram, implying the result.

Radar

23.08.2014 22:30

Well, if you change $D$ and $E$ and angle-chase a little, than it's http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3109660&sid=64df69e2c903958785a9c858b780a3de#p3109660

anthony_ecu

24.08.2014 01:31

Let be <ODA= a and <ODB=<OBD=b => <DAB= 90-b so <CAD=<DBD=90-a-b so <BEC=90 by brahmagupta teorem, PA=PD and MB=MC, so PO// EM and OM//EP, so POME is a parallelogram so M, N and P are collinear.

hatchguy

26.08.2014 12:19

This is far from being an original problem. A very similar configuration (and the same idea) was used in Iberoamerican 2010 Problem 5 (for part b of course) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2029577&sid=6d200e14ae41b691506c45f0c1a74c44#p2029577 All of the countries that participate in the Cono Sur Olympiad also participate in the Iberoamerican Olympiad. I find it weird that no one noticed the similarity.

CrazyInMath

18.08.2023 09:58

First we have $AC\perp BD$ . Then let $T$ be the foot from $E$ to $AD$ , we want to prove $M,E,T$ are colinear. we have $\measuredangle TED=\measuredangle DAC=\measuredangle DBC=\measuredangle DBM=\measuredangle MEB=\measuredangle MED$ yielding the result. So $M$ is the midpoint of $BC$ , similarly $P$ is the midpoint of $AD$ . We have $POME$ is a parallelogram and we're done.