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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.

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By compass and straightedge construction this is quite easy, so keep perservering (there are not too many unknowns). – hardmath Nov 18 '12 at 22:02
If you want to do it via coordinate geometry, the easiest way I can think of is to take a reference equilateral triangle with center at $(0,0)$ and vertices at $(\cos 2n\pi/3, \sin 2n\pi/3)$ and transform it to match your triangle. – Rahul Nov 18 '12 at 22:17
Thanks to everyone for all the solutions! – Mark Nov 19 '12 at 0:42

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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Sorry to ask what I'm sure is a dumb question, but once I have a 2 by 2 matrix, how do I know which are the correct x and y coordinates? – Mark Nov 19 '12 at 0:46
The $2\times 2$ matrix doesn't represent coordinates. The latter $2\times 1$ vectors are the coordinates you want. The first row is $x$ and the second row is $y$. – EuYu Nov 19 '12 at 0:48

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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.

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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 $(x_c,y_c)$, $(x_3,y_3)$

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