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

(creation)
 
Line 2: Line 2:
 
<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==
{{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 <cmath>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</cmath> But this is clearly true, because, since D lies on the perpendicular bisector of BC, BD = DC.
 +
 
 +
QED
  
 
{{USAMO box|year=1997|num-b=1|num-a=3}}
 
{{USAMO box|year=1997|num-b=1|num-a=3}}

Revision as of 14:40, 15 August 2008

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 Thereom, The three lines are concurrent if \[FA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0\] But this is clearly true, because, since D lies on the perpendicular bisector of BC, BD = DC.

QED

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