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

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.

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.