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 $\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PCB - \angle PCA)$ (1) and similarly $\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PBA - \angle PBC)$ (2). Since $\angle PBA + \angle PCA = \angle PBC + \angle PCB$ , we have $\angle PBA - \angle PBC = \angle PCB - \angle PCA$ (3).

By (1), (2), and (3), we get $\angle IBP = \angle ICP$ ; hence $B,I,P,C$ 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 :)

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2006 IMO Problems/Problem 1

Problem

Solution