A triangle has two fixed vertices, and the third one moves in such a way that the sum of the squares of the sides is always equal to eight times the area of the triangle. What is the geometric locus of the third vertex?
I. solution. Let us choose the coordinate axes to be the line of segment $AB$ and its perpendicular bisector. Then the coordinates of vertices $A$ and $B$ of triangle $ABC$ are $(-c/2, 0), (c/2, 0)$ where $c = AB > 0$, and the coordinates of vertex $C$ are $(x, y)$. The area of the triangle is $|y| \cdot c / 2$, so the coordinates satisfy

\[
\begin{gathered}
AB^2 + BC^2 + CA^2 = c^2 + \left(x - \frac{c}{2}\right)^2 + y^2 + \left(x + \frac{c}{2}\right)^2 + y^2 = 4c \cdot |y| \\
x^2 + (|y| - c)^2 - \frac{c^2}{4} = 0
\end{gathered}
\]

If $C$ is a point above or below the $x$-axis, i.e., $y \geq 0$ or $y < 0$, then (1) becomes:
(2)
$x^2 + (y - c)^2 - c^2 / 4 = 0$,
or (3)
$x^2 + (y + c)^2 - c^2 / 4 = 0$,

which means $C$ lies on one of the circles with center $(0, c)$ and $(0, -c)$ and radius $c / 2$.

Conversely, if $(x, y)$ is a point on one of these circles, i.e., (2) or (3) holds, then the left side can be transformed into twice the difference of a term that is the sum of the squares of the sides and another term that is eight times the absolute value of the area of the triangle. Therefore, every point on the circumference of the two circles belongs to the geometric locus.

Zaymus Vince (Pannonhalma, Bencés Gimn., I. o. t.)

II. solution. Let the projection of $C$ onto line $AB$ be $D$ and $AD = d$ (positive if $D$ is on the ray $AB$, negative if it is on the extension beyond $A$), and $DC = m_c$. Then, with the usual notation, in every case

\[
\begin{gathered}
a^2 + b^2 + c^2 = m_c^2 + (c - d)^2 + m_c^2 + d^2 + c^2 = 4cm_c \\
(m_c - c)^2 = d(c - d), \quad m_c = c \pm \sqrt{d(c - d)}
\end{gathered}
\]

The expression under the square root, being the geometric mean of $AD$ and $BD$, represents the distances from $AB$ to the two intersection points, $C^*$ and $C^{**}$, of the circle with diameter $AB$ and the line $DC$ - i.e., if we also introduce a direction on the line $DC$. Therefore, the geometric locus of $C$ is given by the translation of the said circle perpendicular to $AB$ by a distance of $AB = c$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a981ca1e762bfd55c7e1g-1.jpg?height=469&width=925&top_left_y=1300&top_left_x=600)

The translation is of course possible in both directions.

Kérchy László (Baja, III. Béla Gimn., II. o. t.)