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

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

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.

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.

User avatar
Atonu Roy Chowdhury
Posts: 40
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

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.

Post Reply