## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_2595713ee0b414c2653fg-15.jpg?height=772&width=897&top_left_y=606&top_left_x=533)

The volume of the larger frustum of a cone is

$$
V_{1}=\frac{v \pi}{3}\left[\left(R_{1}+\frac{a}{2}\right)^{2}+\left(R_{1}+\frac{a}{2}\right) R_{1}+R_{1}^{2}\right]
$$

The volume of the smaller frustum of a cone is

$$
V_{2}=\frac{v \pi}{3}\left[R_{1}^{2}+R_{1}\left(R_{1}-\frac{a}{2}\right)+\left(R_{1}-\frac{a}{2}\right)^{2}\right]
$$

The volume of the solid of revolution is

$$
\begin{aligned}
V & =V_{1}-V_{2}= \\
& =\frac{v \pi}{3}\left[\left(R_{1}+\frac{a}{2}\right)^{2}+\left(R_{1}+\frac{a}{2}\right) R_{1}+R_{1}^{2}\right]-\frac{v \pi}{3}\left[R_{1}^{2}+R_{1}\left(R_{1}-\frac{a}{2}\right)+\left(R_{1}-\frac{a}{2}\right)^{2}\right]= \\
& =a v R_{1} \pi
\end{aligned}
$$

From the right triangle \( B C D \), we calculate the base and height of the given isosceles triangle

$$
a=2 b \sin \frac{\alpha}{2}=20 \sin 15^{\circ} \mathrm{cm}, \quad v=10 \cos 15^{\circ} \mathrm{cm}
$$

The radius of the circumscribed circle is \( R_{1}=\frac{a}{2 \sin 30^{\circ}}=20 \sin 15^{\circ} \) (or \( R_{1}=\frac{a \cdot b \cdot b}{4 P}=20 \sin 15^{\circ} \), where \( P=\frac{1}{2} b^{2} \sin 30^{\circ}=25 \mathrm{~cm}^{2} \)).

$$
\begin{aligned}
& V=a v R_{1} \pi=20 \sin 15^{\circ} \cdot 10 \cos 15^{\circ} \cdot 20 \sin 15^{\circ} \pi=1000 \sin 15^{\circ} \pi \\
& \sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)=\frac{\sqrt{6}-\sqrt{2}}{4} \quad \text { or } \quad \sin 15^{\circ}=\sin \frac{30^{\circ}}{2}=\frac{\sqrt{2-\sqrt{3}}}{2} \\
& V=250(\sqrt{6}-\sqrt{2}) \pi \mathrm{cm}^{3} .
\end{aligned}
$$

## NATIONAL MATHEMATICS COMPETITION

## 4th grade - high school - B variant

Poreč, April 26, 2012.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.
Task B-3.5. An isosceles triangle with the length of the legs $b=10 \mathrm{~cm}$, and the measure of the angle between the legs $\alpha=30^{\circ}$, has a circumscribed circle. Let $t$ be the tangent to this circle that is parallel to the height to the base. Calculate the volume of the solid formed by rotating the given triangle around the tangent $t$.