Let's fix the vertices $A$ and $B$ of $\triangle ABC$. What is the geometric locus of points $C$ such that the sum of the squares of the variable sides is

$$
a^{2}+b^{2}=d^{2}
$$

where $d$ is a given segment.
Solution: We will show that the medians from $C$ in the triangles that satisfy the condition are equal. Let the length of side $AB$ be $c$, and the distance of the perpendicular projection of point $C$ onto $AB$ from the midpoint of $AB$ be $x$ (Figure 4).
![](https://cdn.mathpix.com/cropped/2024_05_02_77002de91b80ec5e9441g-1.jpg?height=408&width=882&top_left_y=262&top_left_x=602)

Figure 4

Then, by the Pythagorean theorem,

\[
\begin{aligned}
& a^{2}=\left(\frac{c}{2}-x\right)^{2}+m^{2}=\left(\frac{c}{2}\right)^{2}-c x+x^{2}+m^{2} \\
& b^{2}=\left(\frac{c}{2}+x\right)^{2}+m^{2}=\left(\frac{c}{2}\right)^{2}+c x+x^{2}+m^{2}
\end{aligned}
\]

and

\[
m^{2}+x^{2}=s_{c}^{2}
\]

Thus,

\[
d^{2}=a^{2}+b^{2}=2\left(\frac{c}{2}\right)^{2}+2\left(x^{2}+m^{2}\right)=2\left(\frac{c}{2}\right)^{2}+2 s_{c}^{2}
\]

which implies

\[
s_{c}^{2}=\frac{d^{2}}{2}-\left(\frac{c}{2}\right)^{2}
\]

This value must be positive, otherwise, it is not possible to have a triangle that satisfies the conditions of the problem. Since $d$ and $c$ are given, the length of $s_{c}$ is independent of the position of point $C$. Therefore, all points that satisfy the conditions lie on the circle centered at the midpoint of segment $AB$ with radius

\[
r=\sqrt{\frac{d^{2}}{2}-\left(\frac{c}{2}\right)^{2}}
\]

Let $C^{\prime}$ be any point on this circle, and let its distances from $A$ and $B$ be $b^{\prime}$ and $a^{\prime}$, respectively. Then, the median from $C^{\prime}$ in triangle $ABC^{\prime}$ is $r$, so by the above formula,

![](https://cdn.mathpix.com/cropped/2024_05_02_77002de91b80ec5e9441g-1.jpg?height=110&width=576&top_left_y=1815&top_left_x=769)

which implies

\[
{a^{\prime 2}}^{2}+{b^{\prime 2}}^{2}=d^{2}
\]

Thus, point $C^{\prime}$ satisfies the conditions of the geometric locus.

The sought geometric locus is therefore the circle centered at the midpoint of segment $AB$ with radius $r$.