# 2006 IMO Problems/Problem 1

## Problem

Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA = \angle PBC+\angle PCB$. Show that $AP \geq AI$, and that equality holds if and only if $P=I.$

## Solution

We have $$\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PCB - \angle PCA).$$

and similarly $$\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PBA - \angle PBC).$$ Since $\angle PBA + \angle PCA = \angle PBC + \angle PCB$, we have $\angle PCB - \angle PCA = \angle PBA - \angle PBC.$

It follows that $$\angle IBP = \frac{1}{2} (\angle PCB - \angle PCA) = \frac{1}{2} (\angle PBA - \angle PBC) = \angle ICP.$$ Hence, $B,P,I,$ and $C$ are concyclic.

Let ray $AI$ meet the circumcircle of $\triangle ABC\,$ at point $J$. Then, by the Incenter-Excenter Lemma, $JB=JC=JI=JP$.

Finally, $AP+JP \geq AJ = AI+IJ$ (since triangle APJ can be degenerate, which happens only when $P=I$), but $JI=JP$; hence $AP \geq AI$ and we are done.

By Mengsay LOEM , Cambodia IMO Team 2015

latexed by tluo5458 :)

minor edits by lpieleanu