# Difference between revisions of "Carnot's Theorem"

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− | + | '''Carnot's Theorem''' states that in a [[triangle]] <math>ABC</math>, the signed sum of [[perpendicular]] distances from the [[circumcenter]] <math>O</math> to the sides (i.e., signed lengths of the pedal lines from <math>O</math>) is: | |

<math>OO_A+OO_B+OO_C=R+r</math> | <math>OO_A+OO_B+OO_C=R+r</math> | ||

Line 37: | Line 37: | ||

</asy> | </asy> | ||

− | where r is the [[inradius]] and R is the [[circumradius]]. The sign of the distance is chosen to be negative iff the entire segment OO_i lies outside the triangle. | + | where r is the [[inradius]] and R is the [[circumradius]]. The sign of the distance is chosen to be negative iff the entire segment <math>OO_i</math> lies outside the triangle. |

Explicitly, | Explicitly, | ||

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− | = | + | =Carnot's Theorem= |

'''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]] [[iff|if and only if]] <math>A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2</math>. | '''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]] [[iff|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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+ | ''' 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. | ||

+ | |||

+ | =Carnot Extended= | ||

+ | Let <math>P,Q,R</math> be points in the plane of triangle <math>ABC</math>. Then the perpendiculars from <math>P,Q,R</math> to <math>BC,CA,AB</math> respectively are concurrent if and only if <cmath>PB^2-PC^2+QC^2-QA^2+RA^2-RB^2=0</cmath> | ||

+ | |||

+ | ====Proof==== | ||

+ | Let <math>X,Y,Z</math> be the feet of perpendiculars from <math>P,Q,R</math> to <math>BC,CA,AB</math> respectively. Note that <math>PB^2-PC^2=XB^2-XC^2</math> from the application of pythogorean theorem to triangles <math>PXB,PXC</math>. Now with similar relations for <math>Y</math> and <math>Z</math>, Carnot's theorem finishes the job! | ||

− | |||

− | + | =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]]) | <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]]) |

## Latest revision as of 01:45, 18 February 2021

**Carnot's Theorem** states that in a triangle , the signed sum of perpendicular distances from the circumcenter to the sides (i.e., signed lengths of the pedal lines from ) is:

where r is the inradius and R is the circumradius. The sign of the distance is chosen to be negative iff the entire segment lies outside the triangle. Explicitly,

where is the area of triangle .

Weisstein, Eric W. "Carnot's Theorem." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html

## Contents

# Carnot's Theorem

**Carnot's Theorem** states that in a triangle with , , and , perpendiculars to the sides , , and at , , and are concurrent if and only if .

#### 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.

# Carnot Extended

Let be points in the plane of triangle . Then the perpendiculars from to respectively are concurrent if and only if

#### Proof

Let be the feet of perpendiculars from to respectively. Note that from the application of pythogorean theorem to triangles . Now with similar relations for and , Carnot's theorem finishes the job!

# Problems

### Olympiad

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