3. An equilateral triangle with side $a$ is inscribed in a circle. A point $M$ is chosen on the circle. Determine the sum of the squares of the distances from point $M$ to the vertices of the triangle.
Solution. Let $M$ lie on the arc $BC$ and let $D$ be a point on the segment $AM$ such that $\overline{MB}=\overline{MD}$. We have $\measuredangle ACB=\measuredangle AMB=60^{\circ}$, as peripheral angles over the same arc, so $\triangle DBM$ is equilateral. From $\measuredangle BAD=\measuredangle BCM$ (peripheral angles over the same arc), $\overline{AB}=\overline{CB}$ and $\overline{BD}=\overline{BM}$ it follows that $\triangle ABD \cong \triangle CBM$, so

![](https://cdn.mathpix.com/cropped/2024_06_05_3d165451ddef8224b771g-2.jpg?height=301&width=339&top_left_y=364&top_left_x=1162)
therefore $\overline{AD}=\overline{CM}$. Thus,

$$
\overline{AM}=\overline{AD}+\overline{DM}=\overline{CM}+\overline{MB}
$$

Let $E$ be the foot of the altitude in the equilateral triangle $\triangle DBM$ and let $\overline{AD}=x, \overline{AD}=\overline{DE}=y$. Then $h=\overline{BE}=y \sqrt{3}$ and since triangle $ABE$ is right-angled, we get

$$
a^{2}=(x+y)^{2}+(y \sqrt{3})^{2}=x^{2}+2xy+4y^{2}.
$$

According to this,

$$
\overline{AM}^{2}+\overline{BM}^{2}+\overline{CM}^{2}=(x+2y)^{2}+(2y)^{2}+x^{2}=2\left(x^{2}+2xy+4y^{2}\right)=2a^{2}
$$