# Difference between revisions of "Carnot's Theorem"

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

'''Only if:''' Assume that the given perpendiculars are concurrent at <math>M</math>. Then, from the Pythagorean Theorem, <math>A_1B^2=BM^2-MA_1^2</math>, <math>C_1A^2=AM^2-MC_1^2</math>, <math>B_1C^2=CM^2-MB_1^2</math>, <math>A_1C^2=MC^2-MA_1^2</math>, <math>C_1B^2=MB^2-MC_1^2</math>, and <math>B_1A^2=AM^2-MB_1^2</math>. Substituting each and every one of these in and simplifying gives the desired result. | '''Only if:''' Assume that the given perpendiculars are concurrent at <math>M</math>. Then, from the Pythagorean Theorem, <math>A_1B^2=BM^2-MA_1^2</math>, <math>C_1A^2=AM^2-MC_1^2</math>, <math>B_1C^2=CM^2-MB_1^2</math>, <math>A_1C^2=MC^2-MA_1^2</math>, <math>C_1B^2=MB^2-MC_1^2</math>, and <math>B_1A^2=AM^2-MB_1^2</math>. Substituting each and every one of these in and simplifying gives the desired result. | ||

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''' If:''' Consider the intersection of the perpendiculars from <math>A_1</math> and <math>B_1</math>. Call this intersection point <math>N</math>, and let <math>C_2</math> be the perpendicular from <math>N</math> to <math>AB</math>. From the other direction of the desired result, we have that <math>A_1B^2+C_2A^2+B_1C^2=A_1C^2+C_2B^2+B_1A^2</math>. We also have that <math>A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2</math>, which implies that <math>C_1A^2-C_1B^2=C_2A^2-C_2B^2</math>. This is a difference of squares, which we can easily factor into <math>(C_1A-C_1B)(C_1A+C_1B)=(C_2A-C_2B)(C_2A+C_2B)</math>. Note that <math>C_1A+C_1=C_2A+C_2B=AB</math>, so we have that <math>C_1A-C_1B=C_2A-C_2B</math>. This implies that <math>C_1=C_2</math>, which gives the desired result. | ''' If:''' Consider the intersection of the perpendiculars from <math>A_1</math> and <math>B_1</math>. Call this intersection point <math>N</math>, and let <math>C_2</math> be the perpendicular from <math>N</math> to <math>AB</math>. From the other direction of the desired result, we have that <math>A_1B^2+C_2A^2+B_1C^2=A_1C^2+C_2B^2+B_1A^2</math>. We also have that <math>A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2</math>, which implies that <math>C_1A^2-C_1B^2=C_2A^2-C_2B^2</math>. This is a difference of squares, which we can easily factor into <math>(C_1A-C_1B)(C_1A+C_1B)=(C_2A-C_2B)(C_2A+C_2B)</math>. Note that <math>C_1A+C_1=C_2A+C_2B=AB</math>, so we have that <math>C_1A-C_1B=C_2A-C_2B</math>. This implies that <math>C_1=C_2</math>, which gives the desired result. |

## Revision as of 08:51, 3 July 2011

**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

**Only if:** Assume that the given perpendiculars are concurrent at . Then, from the Pythagorean Theorem, , , , , , and . Substituting each and every one of these in and simplifying gives the desired result.

** If:** Consider the intersection of the perpendiculars from and . Call this intersection point , and let be the perpendicular from to . From the other direction of the desired result, we have that . We also have that , which implies that . This is a difference of squares, which we can easily factor into . Note that , so we have that . This implies that , which gives the desired result.

## Problems

### Olympiad

is a triangle. Take points on the perpendicular bisectors of respectively. Show that the lines through perpendicular to respectively are concurrent. (Source)