4. If an acute-angled triangle with sides $a, b, c$ rotates around each of its sides and three bodies are obtained, find the ratio of the volumes of these bodies.
Solution. Let's find the volume $V_{c}$ of the solid when the triangle $ABC$ rotates around the side $AB$ (see the diagram). The resulting solid is composed of two cones with the same radius $h_{c}$ and heights $c_{1}$ and $c_{2}$, respectively. For the volume $V_{c}$, we have:

$$
V_{c}=\frac{1}{3} h_{c}^{2} \pi c_{1}+\frac{1}{3} h_{c}^{2} \pi c_{2}=\frac{1}{3} h_{c}^{2} \pi\left(c_{1}+c_{2}\right)=\frac{1}{3} h_{c}^{2} \pi c
$$

where $c=\overline{AB}=c_{1}+c_{2}$.

Similarly, for the volumes $V_{a}$ and $V_{b}$ of the other two solids, we will get:

$$
V_{a}=\frac{1}{3} h_{a}^{2} \pi a, \quad V_{b}=\frac{1}{3} h_{b}^{2} \pi b
$$

so

$$
V_{a}: V_{b}: V_{c}=h_{a}^{2} a: h_{b}^{2} b: h_{c}^{2} c
$$

However, $a h_{a}=b h_{b}=c h_{c}=2 P$ (where $P$ is the area of triangle $ABC$), so

$$
V_{a}: V_{b}: V_{c}=h_{a}: h_{b}: h_{c}=\frac{1}{a}: \frac{1}{b}: \frac{1}{c}
$$

## Second Year

![](https://cdn.mathpix.com/cropped/2024_06_05_c6c83167c56ccaf27c0cg-2.jpg?height=349&width=541&top_left_y=282&top_left_x=941)