# Barycentric coefficients of an orthocenter

#### Dodecahedron75

Hi,
The Barycentric coefficients of an orthocenter of any triangle are as follows :

(A, tan(A)), (B, tan(B)), (C, tan(C))

Coefficient a = tan(A), coefficient b = tan(B), coefficient c = tan(C)

$$\displaystyle a.\vec{MA} + b.\vec{MB} + c.\vec{MC} = \vec{0}$$

My question is : what is the explanation of that? Why tan and not sin or cos?

#### mrtwhs

Let $D$ be the foot of the perpendicular from $A$ to side $BC$. Define $E$ on $AC$ and $F$ on $AB$ in a similar way. Using the right triangles in the diagram you can get barycentric coordinates for these three points:

$D=(0,b \cos C,c \cos B)$, $E=(a \cos C,0,c \cos A)$, and $F=(a \cos B,b \cos A,0)$

Since the coordinates are homogeneous we can divide the first by $\cos B \cos C$, the second by $\cos A \cos C$, and the third by $\cos A \cos B$ to get:

$D = \left(0, \dfrac{b}{\cos B},\dfrac{c}{\cos C} \right)$, $E = \left(\dfrac{a}{\cos A},0,\dfrac{c}{\cos C} \right)$, $F = \left(\dfrac{a}{\cos A},\dfrac{b}{\cos B},0 \right)$.

With these we now get the coordinates of the orthocenter as:

$\left( \dfrac{a}{\cos A}, \dfrac{b}{\cos B}, \dfrac{c}{\cos C} \right)$.

But since the sides are proportional to the sines of the angles (by the law of sines) we can write:

$\left( \dfrac{\sin A}{\cos A}, \dfrac{\sin B}{\cos B}, \dfrac{\sin C}{\cos C} \right) = (\tan A, \tan B, \tan C)$.

Amazing.