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2006 IMO Problems/Problem 1

Revision as of 18:49, 20 May 2025 by Spacephysics (talk | contribs) (→‎Solution)

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

$\textbf{Second solution:}$

Firstly, call $x=\angle PBC + \angle PCB =\angle PBA + \angle PCA$ Then, by the triangle sum of $\triangle ABC$ and $\triangle BPC$, we have: $\angle BAC +2x=180°$ and $\angle BPC +x=180°$ $\Rightarrow \angle BPC = 90°+\frac{\angle BAC}{2}$

Therefore, since $\angle BAC$ is fixed and looks at fixed segment $BC$, $P$ is contained within a circle $\pi$ that passes through $B$, $C$ and $I$ (since we can quickly access that it satisfies $\angle BIC = 90°+\frac{\angle BAC}{2}$). Hence, to prove $AP \geq AI$ it suffices to show that $AI$ meets the center $O$ of circle $\pi$, since that would directly imply that $I$ is the closest point to $A$ on the circle. \[\angle BOC=2(180°-\angle BPC)=180°-\angle BAC\] It follows that $O$ is contained within the circumcircle of $\triangle ABC$.But since $O$ is the center of circle $\pi$, $OB=OC=Radius$, meaning $O$ sits at the middle point of arc $BC$ of the circumcircle, therefore proving that it is contained in the angle bissector of $A$ $\boxed{}$.

By Pietro Leão Baruffato(a.k.a Spacephysics) :P

See Also

2006 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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