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== | ||
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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 | ||
Revision as of 09:17, 19 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
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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/d/6/b/d6b20981adfbb3753b05f37c72c3239f774fc7c2.png)
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 |