2016 USAMO Problems/Problem 3
Contents
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 implies This can be proved by multiple applications of the Pythagorean Thm.
Solution 2
We find point on line we prove that and state that is the point from ENCYCLOPEDIA OF TRIANGLE, therefore
Let be circumcircle of centered at Let and be crosspoints of and and respectively. Let be crosspoint of and In accordance the Pascal theorem for pentagon is tangent to at
Let be and -excenters of Denote is the foot ot perpendicular from to
is ortocenter of and incenter of
is the Nine–point circle of
is the midpoint of is the midpoint of in accordance with property of Nine–point circle In segment cross segment where
Let be the base triangle with orthocenter center of Nine-points circle be the Euler line of
is orthic triangle of
is orthic-of-orthic triangle.
is perspector of base triangle and orthic-of-orthic triangle.
Therefore is point of ENCYCLOPEDIA OF TRIANGLE CENTERS which lies on Euler line of the base triangle. [[1]]
Claim Proof
vladimir.shelomovskii@gmail.com, vvsss
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.