1997 USAMO Problems/Problem 2
Problem
is a triangle. Take points 
on the perpendicular bisectors of 
respectively. Show that the lines through 
perpendicular to 
respectively are concurrent.
Solution
Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Thereom, The three lines are concurrent if ![\[FA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0\]](http://latex.artofproblemsolving.com/8/0/9/809e55b9f19c76e6e6bfb84221217700ce9a2c4f.png)
But this is clearly true, because, since D lies on the perpendicular bisector of BC, BD = DC.
QED
| 1997 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
