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Difference between revisions of "2016 USAMO Problems/Problem 3"

(Created page with "==Problem== Let <math>\triangle ABC</math> be an acute triangle, and let <math>I_B, I_C,</math> and <math>O</math> denote its <math>B</math>-excenter, <math>C</math>-excenter,...")
 
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Lines <math>I_B F</math> and <math>I_C E</math> meet at <math>P.</math> Prove that <math>\overline{PO}</math> and <math>\overline{YZ}</math> are perpendicular.
 
Lines <math>I_B F</math> and <math>I_C E</math> meet at <math>P.</math> Prove that <math>\overline{PO}</math> and <math>\overline{YZ}</math> are perpendicular.
 
==Solution==
 
==Solution==
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{{solution}}
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There are two major steps of a proof.
 +
 
 +
1. Let <math>I_A</math> be the <math>A</math>-excenter, then <math>I_A,O,P</math> are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for <math>triangle I_AI_BI_C.</math>
 +
 
 +
2. Show that <math>I_AY^2-I_AZ^2=OY^2-OZ^2,</math> which shows <math>\overline{OI_A}\perp\overline{YZ}.</math> This can be proved by multiple applications of the Pythagorean Thm.
  
 
{{MAA Notice}}
 
{{MAA Notice}}
 
==See also==
 
==See also==
 
{{USAMO newbox|year=2016|num-b=2|num-a=4}}
 
{{USAMO newbox|year=2016|num-b=2|num-a=4}}

Revision as of 14:24, 17 February 2017

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. AMC Logo.png

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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