Given the equilateral triangle $A B C$. Where are those points $P$ in the plane of the triangle for which $P A^{2}=$ $P B^{2}+P C^{2} ?$
Place the triangle $ABC$ with side $a$ in the coordinate system as shown in the figure. The coordinates of its vertices are:

$$
A\left(0 ; \frac{a \sqrt{3}}{2}\right) ; \quad B\left(\frac{a}{2}, 0\right) ; \quad C\left(-\frac{a}{2}, 0\right)
$$

The coordinates of point $P$ are: $P(x ; y)$. Using the known distance formula, we can write the squares of the distances $PA$, $PB$, and $PC$ as:

$$
P A^{2}=x^{2}+\left(y-\frac{a \sqrt{3}}{2}\right)^{2}, \quad P B^{2}=\left(x-\frac{a}{2}\right)^{2}+y^{2}, \quad P C^{2}=\left(x+\frac{a}{2}\right)^{2}+y^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e23663d7a58c788d660ag-1.jpg?height=547&width=417&top_left_y=534&top_left_x=843)

Writing the equality for the squares, we perform the indicated operations. This leads us to the following equation:

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

Rearrange the equation and complete the square for the terms involving $x$ and $y$. This gives us:

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

This is the equation of a circle. The center of the circle is $O\left(0 ; \frac{-a \sqrt{3}}{2}\right)$, and its radius is $a$.

The points $P$ lie on this circle. The steps are reversible, so every point on the circle belongs to the set of points we are looking for.