When a triangle with sides $a, b$, and $c$ is rotated around these sides, double cones are formed; let the volumes of these cones be denoted by $K_{a}, K_{b}, K_{c}$, and their surface areas by $F_{a}, F_{b}, F_{c}$.

$K_{a}$ is the sum of the volumes of two cones, $K_{b}$ and $K_{c}$ are the differences of the volumes of two cones, and $F_{a}, F_{b}$, and $F_{c}$ are the sums of the lateral surface areas of two cones each (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_76f36209e2a31e341d69g-1.jpg?height=697&width=916&top_left_y=351&top_left_x=580)

Thus,

$$
\begin{aligned}
K_{a} & =\frac{m_{a}^{2} \pi\left(a-a_{1}\right)}{3}+\frac{m_{a}^{2} \pi a_{1}}{3}=\frac{m_{a}^{2} \pi a}{3} \\
K_{b} & =\frac{m_{b}^{2} \pi\left(b+b_{1}\right)}{3}-\frac{m_{b}^{2} \pi b_{1}}{3}=\frac{m_{b}^{2} \pi b}{3} \\
K_{c} & =\frac{m_{c}^{2} \pi\left(c+c_{1}\right)}{3}-\frac{m_{c}^{2} \pi c_{1}}{3}=\frac{m_{c}^{2} \pi c}{3} \\
F_{a} & =m_{a} \pi b+m_{a} \pi c=m_{a} \pi(b+c) \\
F_{b} & =m_{b} \pi a+m_{b} \pi c=m_{b} \pi(a+c) \\
F_{c} & =m_{c} \pi a+m_{c} \pi b=m_{c} \pi(a+b)
\end{aligned}
$$

Using the fact that $m_{a}: m_{b}: m_{c}=\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$ (which can be proven from the two expressions for the double area: $a m_{a}=b m_{b}=c m_{c}$; see the solution to exercise 299 in the previous issue on page 117), we obtain that

$$
\begin{gathered}
K_{1}: K_{2}: K_{3}=\frac{1}{a^{2}} \cdot a: \frac{1}{b^{2}} \cdot b: \frac{1}{c^{2}} \cdot c=\frac{1}{a}: \frac{1}{b}: \frac{1}{c} \\
F_{a}: F_{b}: F_{c}=\frac{b+c}{a}: \frac{a+c}{b}: \frac{a+b}{c}
\end{gathered}
$$

István Trembeczki (Sárospatak, Rákóczi g. IV. o. t.)
Given a triangle with sides $a, b, c$. The angle opposite to $a$ is an obtuse angle. Determine the ratio of the surface areas and volumes of the solids of revolution formed by rotating this triangle about its sides $a, b$, and $c$.

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