2016 USAMO Problems/Problem 3

Problem

Let $\triangle ABC$ be an acute triangle, and let $I_B, I_C,$ and $O$ denote its $B$ -excenter, $C$ -excenter, and circumcenter, respectively. Points $E$ and $Y$ are selected on $\overline{AC}$ such that $\angle ABY = \angle CBY$ and $\overline{BE}\perp\overline{AC}.$ Similarly, points $F$ and $Z$ are selected on $\overline{AB}$ such that $\angle ACZ = \angle BCZ$ and $\overline{CF}\perp\overline{AB}.$

Lines $I_B F$ and $I_C E$ meet at $P.$ Prove that $\overline{PO}$ and $\overline{YZ}$ are perpendicular.

Solution

There are two major steps of a proof.

1. Let $I_A$ be the $A$ -excenter, then $I_A,O,P$ are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for $triangle I_AI_BI_C.$

2. Show that $I_AY^2-I_AZ^2=OY^2-OZ^2,$ which shows $\overline{OI_A}\perp\overline{YZ}.$ This can be proved by multiple applications of the Pythagorean Thm.

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

2016 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6
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2016 USAMO Problems/Problem 3

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