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.
 A: 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.
A: 
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.
A: 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 $(x_c,y_c)$, $(x_3,y_3)$
