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 16:11, 15 April 2012

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

Template:Alternate 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

See Also

1997 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions