Let $A, B, C$ be the vertices of an equilateral triangle. Determine the geometric locus of points $P$ in the plane of the triangle for which
a) $P A^{2}+P B^{2}=P C^{2}$
b) $P A^{2}+P B^{2}=2 P C^{2}$.
Solution. Let's set up a Cartesian coordinate system such that the $A$ vertex of the triangle is at $(-1 ; 0)$, the $B$ vertex is at $(1 ; 0)$, and the $C$ vertex lies on the positive $y$-axis. The height of an equilateral triangle with side length $a$ is $\frac{a \sqrt{3}}{2}$, so the coordinates of $C$ are $(0 ; \sqrt{3})$.

The square of the distance from any point $P(x ; y)$ to points $A, B, C$ is, according to the known formula:

$$
P A^{2}=(x+1)^{2}+y^{2}, \quad P B^{2}=(x-1)^{2}+y^{2}, \quad \text { and } \quad P C^{2}=x^{2}+(y-\sqrt{3})^{2}
$$

In case a), $P$ belongs to the geometric locus if and only if

$$
\left((x+1)^{2}+y^{2}\right)+\left((x-1)^{2}+y^{2}\right)=x^{2}+(y-\sqrt{3})^{2}
$$

Performing the squaring and then rearranging:

$$
\begin{gathered}
2 x^{2}+2 y^{2}+2=x^{2}+y^{2}-2 \sqrt{3} y+3 \\
x^{2}+(y+\sqrt{3})^{2}=4
\end{gathered}
$$

We performed equivalent transformations, so the sought geometric locus is the figure described by equation (1), a circle whose center is the reflection of the $C$ vertex of the $\triangle ABC$ over the $AB$ side, and whose radius is equal to the side length of the triangle (Figure 1).

In case b), $P$ belongs to the geometric locus if and only if

$$
\begin{gathered}
\left((x+1)^{2}+y^{2}\right)+\left((x-1)^{2}+y^{2}\right)= \\
=2\left(x^{2}+(y-\sqrt{3})^{2}\right)
\end{gathered}
$$

which means if

$$
2 x^{2}+2 y^{2}+2=2 x^{2}+2 y^{2}-4 \sqrt{3} y+6,
$$

or if

$$
y=\frac{\sqrt{3}}{3}
$$

We performed equivalent transformations, so the sought geometric locus is now the figure described by equation (2), a line parallel to the $AB$ side of the $\triangle ABC$ and passing through the centroid of the triangle (Figure 2).