Difference between revisions of "1997 USAMO Problems/Problem 2"
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 17:13, 12 April 2012
Problem
is a triangle. Take points 
on the perpendicular bisectors of 
respectively. Show that the lines through 
perpendicular to 
respectively are concurrent.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Thereom, 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 | ||
