# Difference between revisions of "Carnot's Theorem"

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− | 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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====Proof==== | ====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! | 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! | ||

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=Problems= | =Problems= |

## 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)