Difference between revisions of "2016 USAMO Problems/Problem 3"
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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== | ||
− | + | This problem can be proved in the following two steps. | |
− | 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> | + | 1. Let <math>I_A</math> be the <math>A</math>-excenter, then <math>I_A,O,</math> and <math>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. | 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}} | ||
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==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 13:25, 17 February 2017
Problem
Let be an acute triangle, and let and denote its -excenter, -excenter, and circumcenter, respectively. Points and are selected on such that and Similarly, points and are selected on such that and
Lines and meet at Prove that and are perpendicular.
Solution
This problem can be proved in the following two steps.
1. Let be the -excenter, then and are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for
2. Show that which shows 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.
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 |