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)=1/2 angle(ABC) - angle(PBC)=1/2 [ angle(PBA) - angle(PBC)] (1) and similarly angle(ICP)=angle(PCB) - angle(ICB) = angle(PCB) - 1/2 angle(ACB)= 1/2[ angle(PCB) - angle(PCA)] (2) since angle(PBA)+angle(PCA)=angle(PBC)+angle(PCB) , then angle(PBA) - angle(PBC) = angle(PCB) - angle(PCA) (3) (1) , (2) and (3) therefore : angle(IBP)=angle(ICP) i.e BIPC is con cyclic . Let J be a point of intersection between (AI) and the circumcise of triangle ABC , then IJB is isosceles traingle because of angle(BIJ)=angle(IBJ) so JB=JC=JI , then JI=JP (because BIPC con cyclic) .

In triangle APJ : AP+JP >= AJ=AI+IJ but JI=JP so AP >= AI .

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2006 IMO Problems/Problem 1

Problem

Solution