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.
Problem
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Tags: geometry, circumcircle, parallelogram
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.