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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.
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.
If the center is $O(x_0, y_0)$ and the vertex given has coordinates $(x_1,y_1)$ then the radius of the circumcircle is $r=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$. Therefore the three vertices are on the circle of equation $$(x-x_0)^2+(y-y_0)^2=(x_1-x_0)^2+(y_1-y_0)^2$$ To find the other two vertices solve the system of equations
$$\left\{ \begin{array}{ll}(x_2-x_0)^2+(y_2-y_0)^2=(x_3-x_0)^2+(y_3-y_0)^2=(x_1-x_0)^2+(y_1-y_0)^2\\ (x_2-x_1)^2+(y_2-y_1)^2=(x_3-x_1)^2+(y_3-y_1)^2=(x_3-x_2)^2+(y_3-y_2)^2\end{array}\right.$$
Don't forget that $x_1, \ y_1,\ x_0,\ y_0$ are known, so we have just $4$ unknowns with $6$ equations sufficient to solve for $x_2,\ y_2,\ x_3,\ y_3 $.
The first equations implies that the three vertices of the triangle are on the circle while the second equations implies that the three sides of the triangle are equal.
Let the given vertex of the equilateral triangle be $(x_1,y_1)$ and the centroid be $(x_c,y_c)$. Let $(x_2,y_2)$ and $(x_3,y_3)$ be the other two vertices. Then we have that $$\dfrac{x_1 + x_2 + x_3}{3} = x_c \,\,\,\,\,\,\,\,\,\, (1)$$ $$\dfrac{y_1 + y_2 + y_3}{3} = y_c \,\,\,\,\,\,\,\,\,\, (2)$$ Also, the line joining vertices $2$ and $3$ is perpendicular to the line joining $(x_1,y_1)$ and $(x_c,y_c)$. Hence, the equation of the line is given by $$y = - \left( \dfrac{x_1 - x_c}{y_1 - y_c}\right) x + k \,\,\,\,\,\,\,\,\,\, (*)$$ The foot of the altitude from $1$ to the base $23$ be $(x_h,y_h)$. Then we have that $$\dfrac{2x_1 + x_h}{3} = x_c$$ and $$\dfrac{2y_1 + y_h}{3} = y_c$$ Hence, $$x_h = 3 x_c - 2x_1 \,\,\,\,\,\, \text{ and } y_h = 3 y_c - 2y_1$$ lies on the equation $(*)$. Hence, $$k = \left( \dfrac{(3 y_c - 2y_1)(y_1 - y_c) + (3 x_c - 2x_1)(x_1 - x_c)}{y_1 - y_c}\right)$$ Also, $(x_2,y_2)$ and $(x_3,y_3)$ lie on equation $(*)$. This gives us $$y_2 = - \left( \dfrac{x_1 - x_c}{y_1 - y_c}\right) x_2 + k \,\,\,\,\,\,\,\,\,\, (3)$$ and $$y_3 = - \left( \dfrac{x_1 - x_c}{y_1 - y_c}\right) x_3 + k \,\,\,\,\,\,\,\,\,\, (4)$$ Now make use of equations $(1)$,$(2)$,$(3)$ and $(4)$ to solve for $(x_2,y_2)$ and $(x_3,y_3)$. However, these $4$ are linearly dependent. The other equation can be acquired by equating the distance between $(x_2,y_2)$, $(x_c,y_c)$ and $(x_c,y_c)$, $(x_3,y_3)$
