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### BAYMAX is cyclic!

Posted: **Mon Apr 09, 2018 5:06 pm**

by **Ananya Promi**

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

### Re: BAYMAX is cyclic!

Posted: **Mon Apr 09, 2018 5:21 pm**

by **Ananya Promi**

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