BAYMAX is cyclic!

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Ananya Promi
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BAYMAX is cyclic!

Unread post by Ananya Promi » Mon Apr 09, 2018 5:06 pm

Let $M$ be the midpoint of the side $AC$ of triangle $ABC$. Let $P$ on $AM$ and $Q$ on $CM$ be such that $PQ=AC/2$. Let $(ABQ)$ intersect with $BC$ at $A_X$ other than $B$. $(BCP)$ intersects with $BA$ at $A_Y$ other than $B$. Prove that $BA_YMA_X$ is cyclic

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Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm
Location: Naogaon, Bangladesh

Re: BAYMAX is cyclic!

Unread post by Ananya Promi » Mon Apr 09, 2018 5:21 pm

Let $N$ and $R$ be the midpoints of $AB$ and $BC$ resp.
Then we get $NRQP$ is a parallelogram.
Again, $AA_Y.AB=AP.AC$ means, $AA_Y.AN=AP.AQ$
It gives $NA_YPQ$ cyclic.
Similarly, $QMRA_X$ is cyclic.
We are working with directed angle
$\angle{APN}=-\angle{AA_YM}$
Again, $\angle{APN}=\angle{AQR}=\angle{MQR}=\angle{MA_XR}$
So, $\angle{MA_XR}=-\angle{AA_YM}$
$\angle{MA_XB}=\angle{MA_YB}$
So, $BA_YMA_X$ is cyclic :D

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