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 Theorem]], The three lines are [[concurrent]] if | Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By [[Carnot's Theorem]], The three lines are [[concurrent]] if |
Revision as of 16:13, 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
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 |