We rotate a triangle around each of its three sides. Let the volumes of the three resulting solids be $V_{a}, V_{b}, V_{c}$. What is the largest angle of the triangle if

$$
\frac{1}{V_{a}^{2}}=\frac{1}{V_{b}^{2}}+\frac{1}{V_{c}^{2}} ?
$$
Let's denote the sides and altitudes of the triangle in the usual way as $a, b, c, m_{a}, m_{b}, m_{c}$. If we rotate the triangle around one of its sides, the resulting solid is either a cone, or the sum or difference of two cones, depending on whether there is no obtuse angle or there is one among the angles on the axis of rotation. The volume of the resulting solid in each case (using the notation in the diagram):
![](https://cdn.mathpix.com/cropped/2024_05_02_7d66de0a624e4f23811dg-1.jpg?height=1328&width=370&top_left_y=347&top_left_x=866)

\[
\begin{gathered}
V_{1}=\frac{\pi}{3} \cdot A C^{2} \cdot A B=\frac{\pi}{3} \cdot m_{c}^{2} \cdot c \\
V_{2}=\frac{\pi}{3} \cdot C T^{2} \cdot A T+\frac{\pi}{3} \cdot C T^{2} \cdot B T=\frac{\pi}{3} \cdot m_{c}^{2} \cdot c \\
V_{3}=\frac{\pi}{3} \cdot C T^{2} \cdot B T-\frac{\pi}{3} \cdot C T^{2} \cdot A T=\frac{\pi}{3} \cdot m_{c}^{2} \cdot c
\end{gathered}
\]

Thus, if we rotate the triangle around the side $c$, the volume of the resulting solid is $\frac{\pi}{3} \cdot m_{c}^{2} \cdot c$ in all cases. Similarly, $V_{a}=\frac{\pi}{3} \cdot m_{a}^{2} \cdot a$ and $V_{b}=\frac{\pi}{3} \cdot m_{b}^{2} \cdot b$. Substituting these into our original equation and rearranging, we get:

\[
\frac{1}{m_{a}^{4} \cdot a^{2}}=\frac{1}{m_{b}^{4} \cdot b^{2}}+\frac{1}{m_{c}^{4} \cdot c^{2}}
\]

If the area of the triangle is $t$, then $16 t^{4}=m_{a}^{4} \cdot a^{4}=m_{b}^{4} \cdot b^{4}=m_{c}^{4} \cdot c^{4}$. Using this, we can rewrite (1) as:

\[
\frac{1}{\frac{t^{4}}{a^{2}}}=\frac{1}{\frac{t^{4}}{b^{2}}}+\frac{1}{\frac{t^{4}}{c^{2}}}
\]

or:

\[
a^{2}=b^{2}+c^{2}
\]

From this, by the converse of the Pythagorean theorem, it follows that the angle opposite the side $a$ is a right angle; thus, the largest angle of the triangle is $90^{\circ}$.