# 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.

• By compass and straightedge construction this is quite easy, so keep perservering (there are not too many unknowns). Nov 18, 2012 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.
– user856
Nov 18, 2012 at 22:17
• Thanks to everyone for all the solutions!
– Mark
Nov 19, 2012 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.

• 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, 2012 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, 2012 at 0:48 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 $(x_c,y_c)$, $(x_3,y_3)$