Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 USAMO Problems/Problem 3"

(→‎Bold text)
Line 21: Line 21:
  
 
QED
 
QED
 +
{{MAA Notice}}

Revision as of 12:32, 4 July 2013

Given a nonisosceles, nonright triangle $ABC,$ let $O$ denote the center of its circumscribed circle, and let $A_1, \, B_1,$ and $C_1$ be the midpoints of sides $BC, \, CA,$ and $AB,$ respectively. Point $A_2$ is located on the ray $OA_1$ so that $\triangle OAA_1$ is similar to $\triangle OA_2A$. Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1,$ respectively, are defined similarly. Prove that lines $AA_2, \, BB_2,$ and $CC_2$ are concurrent, i.e. these three lines intersect at a point.


Solution


LEMMA 1:

In $\triangle ABC$ with circumcenter $O$, $\angle OAC = 90 - \angle B$.

PROOF of Lemma 1:

The arc $AC$ equals $2\angle B$ which equals $\angle AOC$. Since $\triangle AOC$ is isosceles we have that $\angle OAC = \angle OCA = 90 - \angle B$. QED

Define $H \in BC$ s.t. $AH \perp BC$. Since $OA_1 \perp BC$, $AH \parallel OA_1$. Let $\angle AA_2O = \angle A_1AO = x$ and $\angle AA_1O = \angle A_2AO = y$. Since we have $AH \parallel OA_1$, we have that $\angle HAA_2 = x$. Also, we have that $\angle A_2AA_1 = y-x$. Furthermore, $\angle BAH = 90 - \angle B = \angle OAC$, by lemma 1. Therefore, $\angle A_1AC = 90 - \angle B + x = \angle BAA_2$. Since $A_1$ is the midpoint of $BC$, $AA_1$ is the median. However $\angle A_1AC = \angle BAA_2$ tells us that $AA_2$ is just $AA_1$ reflected across the internal angle bisector of $A$. By definition, $AA_2$ is the $A$-symmedian. Likewise, $BB_2$ is the $B$-symmedian and $CC_2$ is the $C$-symmedian. Since the symmedians concur at the symmedian point, we are done.

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_USAMO_Problems/Problem_3&oldid=54726"