Difference between revisions of "Carnot's Theorem"
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| − | '''Carnot's Theorem''' states that in a [[triangle]] <math>ABC</math> with <math>A_1\in BC</math>, <math>B_1\in AC</math>, and <math>C_1\in AB</math>, [[perpendicular]]s to the sides <math>BC, < | + | '''Carnot's Theorem''' states that in a [[triangle]] <math>ABC</math> with <math>A_1\in BC</math>, <math>B_1\in AC</math>, and <math>C_1\in AB</math>, [[perpendicular]]s to the sides <math>BC</math>, <math>AC</math>, and <math>AB</math> at <math>A_1</math>, <math>B_1</math>, and <math>C_1</math> are [[concurrent]] [[if and only if]] <math>A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2</math>. |
==Proof== | ==Proof== | ||
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==Problems== | ==Problems== | ||
===Olympiad=== | ===Olympiad=== | ||
| − | < | + | <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. ([[1997 USAMO Problems/Problem 2|Source]]) |
==See also== | ==See also== | ||
Revision as of 07:03, 27 August 2008
Carnot's Theorem states that in a triangle 
with 
, 
, and 
, perpendiculars to the sides 
, 
, and 
at 
, 
, and 
are concurrent if and only if 
.
Contents
Proof
Problems
Olympiad
is a triangle. Take points 
on the perpendicular bisectors of 
respectively. Show that the lines through 
perpendicular to 
respectively are concurrent. (Source)
