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
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if
But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.
QED
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |