Barycentric coordinate of the excenter proof?

Question

I'm trying to show that the barycentric coordinate of excenter of triangle ABC, where BC=a, AC=b, and AB=c, and excenter opposite vertex A is Ia, is Ia=(-a:b:c). I've gotten to the point where after a lot of ratio bashing I have that it's (ab/(b+c)):CP:BP, where P is the incenter, but I have no idea where to go from there. What would be the proof?

Answer 1

One of the best ways to "catch" barycentric coordinates is by reasoning on oriented areas. In fact, it can be easily established that the barycentric coordinates of any point $Q$ with respect to $\triangle ABC$ are:

$$\tag{1}\left(\dfrac{[QBC]}{[ABC]}, \dfrac{[AQC]}{[ABC]}, \dfrac{[ABQ]}{[ABC]}\right) \ \ \text{i.e., proportional to} \ \ ([QBC], [AQC], [ABQ]),$$

where $[MNP]$ denotes the oriented area of $\triangle MNP,$

i.e., $\boxed{+}$ ordinary area of $\triangle MNP$ if this triangle has a positive orientation, $\boxed{-}$ ordinary area otherwise.

Mnemonic remark for (1): substitute, in $[ABC]$, $Q$ in the first place, then in the second place, then in the third place, precisely because there is a linear system behind all that (see explanation at the bottom of this page).

Thus, letting $r_A$ be the radius of this excircle, the barycentric coordinates of $I_A$ are proportional to:

$$[I_ABC]=\boxed{-}\frac12r_A \times a, \ \ \ [AI_AC]=\boxed{+}\frac12r_A \times b, \ \ \ [ABI_A]=\boxed{+}\frac12r_A \times c$$

(triangle area = half of basis times height) whence the result by suppressing the common factor $\frac12r_A.$ (only $I_ABC$ has negative orientation).

Explanation for formulas (1):

Consider $\triangle MNP$, $M(x_M,y_M),N(x_N,y_N),P(x_P,y_P)$ with respect to a certain coordinate system. We have (see for example (http://aleph0.clarku.edu/~ma130/determinants1.pdf.)):

$$\tag{2}\text{area of} \ \triangle MNP \ := \ [MNP] \ = \ \frac12 \begin{vmatrix}x_M&x_N&x_P\\y_M&y_N&y_P\\1&1&1\end{vmatrix}$$

Now, let us write that a certain point $Q$ is the barycenter of $M,N,P$ with resp. weights $m,n,p$:

$$mM+nN+pP=Q,$$

which is equivalent to the two first equations of the following system:

$$\tag{3}\begin{cases}mx_M&+&nx_N&+&px_P&=&x_Q\\my_M&+&ny_N&+&py_P&=&y_Q\\m&+&n&+&p&=&1\end{cases},$$

the third one being the normalizing condition.

One can already observe that the determinant of system (3) is the determinant that has been given in (2).

Thus, it suffices to solve this system by applying Cramer formulas to obtain.

$$\left(m=\dfrac{[QNP]}{[MNP]}, n=\dfrac{[MQP]}{[MNP]}, p=\dfrac{[MNQ]}{[MNP]}\right).$$

See the nice document (https://cp4space.files.wordpress.com/2012/10/moda-ch10.pdf) where, instead of "barycentric coordinates" they use the older name "areal coordinates".) See also (http://www.cut-the-knot.org/triangle/barycenter.shtml).

Answer 2

The trilinear coordinates of the $A$-excenter are trivially $[-1:1:1]$ by the definition of excenter as the center of a peculiar circle. The conversion between trilinear and baricentric coordinates is straightforward (have a look after equation $(9)$) and you get that the barycentric coordinates of the $A$-excenter are $[-a:b:c]$.