# Difference between revisions of "Carnot's Theorem"

(New page: '''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 </math>AB<math>, perpendiculars to the sides...) |
|||

Line 1: | Line 1: | ||

− | '''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 </ | + | '''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>. |

==Proof== | ==Proof== | ||

Line 6: | Line 6: | ||

==Problems== | ==Problems== | ||

===Olympiad=== | ===Olympiad=== | ||

− | <math>\triangle ABC< | + | </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]]) |

==See also== | ==See also== |

## Revision as of 08:03, 27 August 2008

**Carnot's Theorem** states that in a triangle with , , and , perpendiculars to the sides ACABA_1B_1C_1A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$.

==Proof== {{incomplete|proof}}

==Problems== ===Olympiad===$ (Error compiling LaTeX. ! Missing $ inserted.)\triangle ABCD, E, FBC, CA, ABA, B, CEF, FD, DE$ respectively are concurrent. (Source)