IMO 2006 Shortlisted Problems

Let $\angle A=\alpha$, $\angle B=\beta$, $\angle C=\gamma$. Since $\angle PBA+\angle PCA+\angle PBC+\angle PCB=\beta +\gamma$, the condition from the problem statement is equivalent to $\angle PBC+\angle PCB=(\beta +\gamma )/2$, i.e. $\angle BPC=90^{\circ }+\alpha /2$.

On the other hand, $\angle BIC=180^{\circ }-(\beta +\gamma )/2=90^{\circ }+\alpha /2$. Hence $\angle BPC=\angle BIC$, and since $P$ and $I$ are on the same side of $BC$, the points $B$, $C$, $I$ and $P$ are concyclic. In other words, $P$ lies on the circumcircle $\omega$ of triangle $BCI$.



Let $\Omega$ be the circumcircle of triangle $ABC$. It is a well-known fact that the centre of $\omega$ is the midpoint $M$ of the arc $BC$ of $\Omega$. This is also the point where the angle bisector $AI$ intersects $\Omega$.

From triangle $APM$ we have $$AP+PM\ge AM=AI+IM=AI+PM$$ Therefore $AP\ge AI$. Equality holds if and only if $P$ lies on the line segment $AI$, which occurs if and only if $P=I$.