A triangle has two fixed vertices, while the third one moves in such a way that the sum of the squares of the sides is equal to eight times the area of the triangle. On what curve does the third vertex move?
Solution. Let the two fixed vertices of the triangle be $A$ and $B$. Place the triangle in the coordinate system such that $A$ and $B$ are on the $x$-axis, and the perpendicular bisector of the segment defined by $A$ and $B$ is the $y$-axis. The coordinates of point $A$ are $A\left(-\frac{c}{2} ; 0\right)$, the coordinates of point $B$ are $B\left(\frac{c}{2} ; 0\right)$, and the coordinates of the third vertex $C$ are $C(x ; y)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_73b4daa55fd53ada1006g-1.jpg?height=924&width=460&top_left_y=300&top_left_x=819)

According to the condition,

$$
\left(x+\frac{c}{2}\right)^{2}+y^{2}+c^{2}+\left(\left|x-\frac{c}{2}\right|\right)^{2}+y^{2}= \pm 4 c y
$$

Perform the indicated operations:

$$
x^{2}+c x+\frac{c^{2}}{4}+y^{2}+c^{2}+x^{2}-c x+\frac{c^{2}}{4}+y^{2}= \pm 4 c y
$$

From this, we get $x^{2}+(y \pm c)^{2}=\left(\frac{c}{2}\right)^{2}$. This is the equation of two circles. The two circles are reflections of each other across the $x$-axis, with their centers at $(0 ; c)$ and $(0 ;-c)$, respectively, and both have a radius of $\frac{c}{2}$.

From the definition of a circle, it follows that every point on its circumference is an element of the geometric locus.