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

For discussing Olympiad level Geometry Problems
Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

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

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.

WARNING: DON'T USE GEOGEBRA
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

Atonu Roy Chowdhury
Posts: 40
Joined: Fri Aug 05, 2016 7:57 pm
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.