# Difference between revisions of "Carnot's Theorem"

m |
m |
||

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 | + | 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, | ||

## Revision as of 20:19, 5 June 2020

**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! hgfffv

# Problems

### Olympiad

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