Difference between revisions of "1997 USAMO Problems/Problem 2"

Revision as of 10:47, 21 February 2015

Problem

$\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ respectively are concurrent.

Solution

Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if

$FA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0$

But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.

QED

@@ Line 10: / Line 10: @@
 QED
-==Solution 2==
-These three lines concur at the radical center of the three circles centered at <math>D, E, F</math> and passing through <math>C, A, B</math> respectively. (Indeed, each line passes through one intersection point of a pair of circles, and because it is perpendicular to the line of centers it must the the radical axis of these circles.)
-{{alternate solutions}}
 ==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

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Difference between revisions of "1997 USAMO Problems/Problem 2"

Revision as of 10:47, 21 February 2015

Problem

Solution

See Also