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, </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>.
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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</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$ respectively are concurrent. ([[1997 USAMO Problems/Problem 2|Source]])
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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. ([[1997 USAMO Problems/Problem 2|Source]])
  
 
==See also==
 
==See also==

Revision as of 08:03, 27 August 2008

Carnot's Theorem states that in a triangle $ABC$ with $A_1\in BC$, $B_1\in AC$, and $C_1\in AB$, perpendiculars to the sides $BC$, $AC$, and $AB$ at $A_1$, $B_1$, and $C_1$ are concurrent if and only if $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$.

Proof

Template:Incomplete

Problems

Olympiad

$\triangle ABC$ 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. (Source)

See also

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