1997 USAMO Problems/Problem 2: Difference between revisions
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Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By [[Carnot's | Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By [[Carnot's Theorem]], The three lines are [[concurrent]] if | ||
<center><math>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</math></center> | <center><math>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</math></center> | ||
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But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC. | But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC. | ||
QED | QED | ||
==See Also== | ==See Also== | ||
{{USAMO newbox|year=1997|num-b=1|num-a=3}} | {{USAMO newbox|year=1997|num-b=1|num-a=3}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 15:10, 15 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 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 | ||
