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

2006 IMO (Problems) • Resources
Preceded by [[2006 IMO Problems/Problem {{{num-b}}}\|Problem {{{num-b}}}]]	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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2006 IMO Problems/Problem 1

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