# Is it possible to find the vertices of an equilateral triangle given its center point?

I was wondering how to find the vertices of an equilateral triangle given its center point?

Such as:

        A
/\
/  \
/    \
/   M  \
B /________\ C


Provided that AB, AC, BC = x and M = (50,50) and M is the middle of the triangle, I want to find A, B and C.

Thanks.

• This is unsolvable since any rotation of a possible solution gives another. Are you enforcing that one of the points lies directly above the center? Then it's solvable Jun 30, 2015 at 15:31
• Is $x$ a known value? And do you want line segment $BC$ horizontal, as in the diagram? Without further restrictions such as these, there are infinitely many answers. Jun 30, 2015 at 15:31
• Not without further information regarding the orientation of the triangle. Without that information, imagine rotating the triangle arbitrarily. Jun 30, 2015 at 15:32
• @AlexR. Yeah, I actually want that point A will be above the center (same Y value). @RoryDaulton Yeah x will be a known positive number, and BC should be horizontal. Jun 30, 2015 at 15:33

Let's assume that $$x$$, the side of the equilateral triangle, is a known positive quantity and that side $$BC$$ is horizontal (or that point $$A$$ is directly above point $$M$$). Let's also assume you are using Cartesian coordinates (where increasing the first coordinate means moving right and increasing the second coordinate means moving up) and that $$M$$ is the point $$(50,50)$$.

Then $$x$$ is the side of the equilateral triangle $$ABC$$. By simple geometry we know that the altitude is $$\frac{\sqrt 3}2x$$. Point $$M$$ is the centroid of the triangle and is on the altitude which is also the median. This means the distance $$AM$$ is two-thirds the altitude, namely $$\frac{\sqrt 3}3x$$.

Therefore point $$A$$ is $$\left( 50,50+\frac{\sqrt 3}3x \right)$$.

Point $$B$$ is $$\frac{\sqrt 3}2x$$ down from $$A$$ and $$\frac x2$$ to the left, so point $$B$$ is $$\left( 50-\frac x2,50-\frac{\sqrt 3}6x \right)$$, and point $$C$$ is $$\left( 50+\frac x2,50-\frac{\sqrt 3}6x \right)$$.

I tested this answer with Geogebra, and it checks.

• Indeed works.. Thanks! I accepted this answer because it was the easiest to implement programmatically. Jun 30, 2015 at 15:51
• Yoou should probably define point $M$ as $(50, 50)$ somewhere. Apr 2, 2020 at 9:51
• @creativecreatorormaybenot: Yes, you are right. I have added that to my answer. Thanks! Apr 3, 2020 at 10:27

Draw the circle of radius $\frac {\sqrt 3}3x$ around $M$. Pick an arbitrary point $A$ on this circle. Then intersect the circle of radius $X$ around $A$ with the first circle to determine $B,C$ as intersection points.

Note that $A$ could be picked anywhere on the circle, hence the result is not unique.

Note that we are assuming that $BC$ is parallel to the coordinate axis. Now, we can get all the horizontal and vertical distances we need to solve this problem.

Let $P$ be the midpoint of $BC$. Note that we have $PC = BP = x/2$ and $MP = (x/2)/\sqrt{3}$. Finally, $MA = x/\sqrt{3}$. This should be enough to get all the coordinates.

EDIT: A better method, especially if the triangle can be in any orientation, is to first shift everything so that the center is at $(0,0)$. Now, we have that the distance from the center to any point is $R = x/\sqrt{3}$. Now, in complex numbers, we can express the three points as $R \cdot e^{i\theta}, R\cdot e^{i\theta + 2\pi/3}, R\cdot e^{i(\theta - 2\pi/3)}$. This is because each point is $120^\circ$ apart.

Now simplify shift everything back so that the center is $(50,50)$.