When everyone is busy solving USA(J)MO 2017,I am solving2016

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Thamim Zahin
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When everyone is busy solving USA(J)MO 2017,I am solving2016

Unread post by Thamim Zahin » Thu Apr 20, 2017 11:47 pm

The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.

Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point.

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Atonu Roy Chowdhury
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Re: When everyone is busy solving USA(J)MO 2017,I am solving

Unread post by Atonu Roy Chowdhury » Fri Apr 21, 2017 12:15 pm

Solution with angle chasing only. But seems ugly to me.
Let $M$ be the midpoint of arc $BPC$. We will show that $M$ is our desired point. So, it suffices to show $\angle I_BMI_C = \angle I_BPI_C = \frac {180 - \angle A}{2} = \angle B $ or $ \angle I_BMI = \angle I_CMC $ where $I$ is the incenter of $\triangle ABC$ .
WLOG, $P$ lies on the arc $MC$ not containing $B$ .
It is well known that $\angle AIC = 90 + \angle B $ and $\angle AI_CC = 90 + \angle APC = 90 + \angle B $. So, $AII_CC$ is cyclic and similarly $AII_BB$ is also cyclic.
So, $\angle II_CC = 180 - \angle IAC = 180 - \angle IAB = \angle BI_BI $ and $\angle I_CCI = \angle I_CAI = \frac {\angle A}{2} - \frac {\angle PAC}{2} = \frac {\angle BAP}{2} = \angle BII_B $ along with $BI = CI$ implies $\triangle II_CC $ and $\triangle BI_BI$ are congruent. So, $II_B = CI_C$ .
Again, $\angle MCI_C = \angle PCI_C - \angle MCP = \frac {\angle PCA}{2} - (\angle PCA - \angle MCA) = 90 - \frac {\angle PCA}{2} = \frac {\angle ABP}{2} = \angle ABI_B = \angle MII_B $ along with $II_B = CI_C $ and $MC = MI$ implies $\triangle MII_B $ and $\triangle MCI_C$ are congruent. The conclusion follows.

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