Barycentric Coordinates of the circumcenter of an arbitrary triangle

Question

Given points $A(1, 0, 0), B(0, 1, 0), C(0, 0, 1)$ in barycentric coordinates, and points $P(x_P, y_P, z_P), Q(x_Q, y_Q, z_Q), R(x_R, y_R, z_R)$, what would the barycentric coordinates of the circumcenter of $\triangle PQR$ be? The circumcircle can be found in a straightforward manner by using the equation for a circle in barycentric coordinates.

Answer 1

You want the barycentric coordinates for any triangle: that is, you want $A$, $B$, $C$ to have coordinates $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ respectively, so that $P$, $Q$, $R$ will have "general" barycentric coordinates.

But you can simplify: saying that $M$ has barycentric coordinates $(\alpha,\beta,\gamma)$ w.r.t. $A,B,C$ means that

$$\overrightarrow{OM}=\frac{\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}+\gamma\overrightarrow{OC}}{\alpha+\beta+\gamma}$$

So, you can get the barycentric coordinates of the circumcenter of $PQR$ w.r.t. $P, Q, R$, which is the formula I linked to above. You get coordinates $(\sin2\hat P,\sin2\hat Q,\sin2\hat R)$, where I denote the angles of triangle $PQR$ by $\hat P,\hat Q,\hat R$.

This means you have, for circumcenter $K$ (not the usual notation for circumcenter, but anyway).

$$\overrightarrow{OK}=\frac{\sin(2\hat P)\ \overrightarrow{OP}+\sin(2\hat Q)\ \overrightarrow{OQ}+\sin(2\hat R)\ \overrightarrow{OR}}{\sin(2\hat P)+\sin(2\hat Q)+\sin(2\hat R)}$$

Or equivalently

$$\overrightarrow{OK}=\alpha\overrightarrow{OP}+\beta\overrightarrow{OQ}+\gamma\overrightarrow{OR}$$

With $\alpha+\beta+\gamma=1$ and

$$\alpha=\frac{\sin(2\hat P)}{\sin(2\hat P)+\sin(2\hat Q)+\sin(2\hat R)}$$ $$\beta=\frac{\sin(2\hat Q)}{\sin(2\hat P)+\sin(2\hat Q)+\sin(2\hat R)}$$ $$\gamma=\frac{\sin(2\hat R)}{\sin(2\hat P)+\sin(2\hat Q)+\sin(2\hat R)}$$

It will be simpler in the following to have sum $1$.

You can further expand this using the barycentric coordinates of $P,Q,R$ w.r.t. $A,B,C$, that is the equations:

$$\overrightarrow{OP}=\frac{x_P\overrightarrow{OA}+y_P\overrightarrow{OB}+z_P\overrightarrow{OC}}{x_P+y_P+z_P}$$

$$\overrightarrow{OQ}=\frac{x_Q\overrightarrow{OA}+y_Q\overrightarrow{OB}+z_Q\overrightarrow{OC}}{x_Q+y_Q+z_Q}$$

$$\overrightarrow{OR}=\frac{x_R\overrightarrow{OA}+y_R\overrightarrow{OB}+z_R\overrightarrow{OC}}{x_R+y_R+z_R}$$

Again, you can admit without loss of generality that $x_P+y_P+z_P=x_Q+y_Q+z_Q=x_R+y_R+z_R=1$, to simplify.

When expanding, you will get a relation of the form

$$\overrightarrow{OK}=\alpha'\overrightarrow{OA}+\beta'\overrightarrow{OB}+\gamma'\overrightarrow{OC}$$

With

$$\alpha'=\alpha x_P+\beta x_Q+\gamma x_R$$ $$\beta'=\alpha y_P+\beta y_Q+\gamma y_R$$ $$\gamma'=\alpha z_P+\beta z_Q+\gamma z_R$$

And it's pretty straightforward to check that $\alpha'+\beta'+\gamma'=1$, so $(\alpha',\beta',\gamma')$ are barycentric coordinates of $K$ w.r.t. $A,B,C$.

Answer 2

This answer explains how to get the barycentric coordinates of the circumcenter of an arbitrary $n$-simplex. For triangles, it boils down to the trilinear coordinates of the circumcenter being $$ \begin{split} \lambda_0 &= \frac{\|v_0 - \frac{1}{2}(v_1 + v_2)\|^2 - \frac{1}{4}\|v_1 - v_2\|^2}{2h_0},\\ \lambda_1 &= \frac{\|v_1 - \frac{1}{2}(v_2 + v_0)\|^2 - \frac{1}{4}\|v_2 - v_0\|^2}{2h_1},\\ \lambda_2 &= \frac{\|v_2 - \frac{1}{2}(v_0 + v_1)\|^2 - \frac{1}{4}\|v_0 - v_1\|^2}{2h_2}, \end{split} $$ where the $h$ are the respective altitudes. Multiply those by the edge length to get the barycentric coordinates.

(For the general simplex, the $ \frac{1}{2}(v_1 + v_2)$ are the circumcenters of the faces and the $\frac{1}{4}\|v_1 - v_2\|^2$ are the squared circumradii.)