<s>[INST] How to find the other vertices of an equilateral triangle given one vertex and centroid If I know the coordinates of the center and one vertex of an equilateral triangle, how do I find the coordinates of the other vertices?
I'm thinking I need to find (x,y) such that the distance to the known vertex is the square root of 3 times the distance to the center, so just need to solve those two distance equations simultaneously, but I keep coming up with too many unknowns.
 [/INST] Suppose we have centroid $M = (x_0,\ y_0)$ and vertex $A=(x_1,\ y_1)$. 
First let us center the triangle at the origin with shifted vertex $$A' = (x_1',\ y_1') = (x_1 - x_0,\ y_1 - y_0)$$
The other vertices will be reached from this one by a rotation about the origin $120^\circ$ clockwise and counter-clockwise. The counter-clockwise rotation matrix is
$$R_{120^\circ} = \begin{pmatrix}\cos120^\circ & -\sin120^\circ \\ \sin120^\circ & \cos120^\circ\end{pmatrix} = \begin{pmatrix}-\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}$$
with the clockwise rotation matrix as
$$R_{-120^\circ} = \begin{pmatrix}\cos120^\circ & \sin120^\circ \\ -\sin120^\circ & \cos120^\circ\end{pmatrix} = \begin{pmatrix}-\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}$$
Your vertices are then
$$B' = \begin{pmatrix}-\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}x_1' \\ y_1'\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}x_1' - \frac{\sqrt{3}}{2}y_1' \\ \frac{\sqrt{3}}{2}x_1' - \frac{1}{2}y_1'\end{pmatrix}$$
$$C' = \begin{pmatrix}-\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}x_1' \\ y_1'\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}x_1' + \frac{\sqrt{3}}{2}y_1' \\ -\frac{\sqrt{3}}{2}x_1' - \frac{1}{2}y_1'\end{pmatrix}$$
Adding $M$ to each coordinate shifts back the triangle to the original spot.
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