Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 AIME, harder Prove it

$\mathrm{P}$ is a fixed point in the plane. Show that amongst triangles $\mathrm{ABC}$ such that $\mathrm{PA}=3, \mathrm{~PB}=5, \mathrm{PC}$ $=7$ , those with the largest perimeter have $\mathrm{P}$ as incenter.

Solution

Given points $\mathrm{P}, \mathrm{B}, \mathrm{C}$ and a fixed circle center $\mathrm{P}$ , we show that the point $\mathrm{A}$ on the circle which maximizes $\mathrm{AB}+\mathrm{AC}$ is such that $\mathrm{PA}$ bisects angle $\mathrm{BAC}$ . Consider a point $\mathrm{A}^{\prime}$ close to $\mathrm{A}$ . Then the change in $A B+A C$ as we move $A$ to $A^{\prime}$ is $A A^{\prime}(\sin P A C-\sin P A B)+O\left(A A^{\prime 2}\right)$ . So for a maximal configuration we must have $\sin \mathrm{PAC}=\sin \mathrm{PAB}$ , otherwise we could get a larger sum by taking A' on one side or the other. This applies to each vertex of the triangle, so $\mathrm{P}$ must be the incenter.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.