# Carnot's Theorem

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

$OO_A+OO_B+OO_C=R+r$

$[asy] pair a,b,c,O,i,d,f,g; a=(0,0); b=(4,0); c=(1,3); O=circumcenter(a,b,c); i=incenter(a,b,c); draw(a--b--c--cycle); draw(circumcircle(a,b,c)); draw(incircle(a,b,c)); dot(i); dot(O); label("A",a,W); label("B",b,E); label("C",c,N); label("I",i,N); label("O",O,N); d=foot(O,b,c); dot(d); draw(O--d); label("O_A",d,N); draw(rightanglemark(O,d,b)); f=foot(O,a,b); dot(f); draw(O--f); draw(rightanglemark(O,f,a)); label("O_C",f,S); g=foot(O,c,a); dot(g); draw(O--g); draw(rightanglemark(O,g,a)); label("O_B",g,W); [/asy]$

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. Explicitly,

$OO_A+OO_B+OO_C=\frac{abc(|\cos{A}|+|\cos{B}|+|\cos{C}|)}{4|\Delta|}$

where $\Delta$ is the area of triangle $\Delta ABC$.

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

# Below is not Carnot's Theorem

Carnot's Theorem states that in a triangle $ABC$ with $A_1\in BC$, $B_1\in AC$, and $C_1\in AB$, perpendiculars to the sides $BC$, $AC$, and $AB$ at $A_1$, $B_1$, and $C_1$ are concurrent if and only if $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$.

#### Proof

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

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

#### Problems

$\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ respectively are concurrent. (Source)