Difference between revisions of "1997 USAMO Problems/Problem 2"
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<math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. | <math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. | ||
==Solution== | ==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 < | + | Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By [[Carnot's Thereom]], 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> | ||
| + | |||
| + | But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC. | ||
QED | QED | ||
{{USAMO box|year=1997|num-b=1|num-a=3}} | {{USAMO box|year=1997|num-b=1|num-a=3}} | ||
Revision as of 09:08, 16 August 2008
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
But this is clearly true, 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 | ||
