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2016 USAMO Problems/Problem 3

Revision as of 15:06, 2 October 2022 by Vvsss (talk | contribs) (→‎Solution 2)

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

This problem can be proved in the following two steps.

1. Let $I_A$ be the $A$-excenter, then $I_A,O,$ and $P$ 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 implies $\overline{OI_A}\perp\overline{YZ}.$ This can be proved by multiple applications of the Pythagorean Thm.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Solution 2

2016 USAMO 3a.png

We find point $T$ on line $YZ,$ we prove that $TY \perp OI_A$ and state that $P$ is the point $X(24)$ from ENCYCLOPEDIA OF TRIANGLE, therefore $P \in OI_A.$

Let $\omega$ be circumcircle of $\triangle ABC$ centered at $O.$ Let $Y_1,$ and $Z_1$ be crosspoints of $\omega$ and $BY,$ and $CZ,$ respectively. Let $T$ be crosspoint of $YZ$ and $Y_1 Z_1.$ In accordance the Pascal theorem for pentagon $AZ_1BCY_1,$ $AT$ is tangent to $\omega$ at $A.$

2016 USAMO 3b.png

Let $I_A, I_B, I_C$ be $A, B,$ and $C$-excenters of $\triangle ABC.$ Denote \[a = BC, b = AC, c = AB, 2\alpha = \angle CAB, 2\beta = \angle ABC, 2\gamma = \angle ACB,\] \[\psi = 90^\circ – \gamma + \beta, X = AI_A \cap \omega, X_1 = BC \cap AI_A,\] \[I = BI_B \cap CI_C, U= YZ \cap AI_A, W = Y_1Z_1 \cap AI_A,\] $V$ is the foot ot perpendicular from $O$ to $AI_A.$

$I$ is ortocenter of $\triangle I_A I_B I_C$ and incenter of $\triangle ABC.$

$\omega$ is the Nine–point circle of $\triangle I_A I_B I_C.$

$Y_1$ is the midpoint of $II_B, Z_1$ is the midpoint of $II_C$ in accordance with property of Nine–point circle $\implies$ \[Y_1Z_1 || I_B AI_C || VO, IW = AW \implies TW \perp AI.\] \[\angle AXC = 180 ^\circ – 2\gamma – \alpha = 90 ^\circ – \gamma + \beta = \psi.\] \[\angle TAI = \angle VOA = 2\beta + \alpha = 90 ^\circ – \gamma + \beta = \psi.\] \[I_A X_1 = IX_1 = BX_1 = 2R \sin \alpha \implies\] \[\cot \angle OI_A A = \frac {VI_0}{VO} = \frac {R \sin \psi + 2R \sin \alpha}{R \cos \psi} = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}.\] Bisector $AX = \frac {2 b c \cos \alpha}{b + c},$ $\frac {AI}{IX}= \frac {AC}{CX}= b : \frac {ab}{b+c}= frac {b+c}{a} \implies AI = AX \frac {b+c}{a+b+c},$ $AW = \frac {AI}{2}, UW = AU – AW,$ $\frac {AU}{UX} = \frac {m + nk}{k+1},$ where $n = \frac {a}{b}, m = \frac{a}{c}, k=\frac {b}{c},$ $\frac {AU}{UX} = \frac{2a}{b+c} \implies AU = AX \cdot \frac {b+c}{2a +b +c}.$ $\frac {AU – AW}{AW} = \frac {b+c} {2a + b + c}.$

See also

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