Task B-4.6. The lengths of the sides of a triangle are consecutive natural numbers (in centimeters). The largest angle of the triangle is twice the smallest. Determine the volume of the solid formed by rotating the triangle about its longest side.
## Solution.

Let the sides of the triangle be $a=n, b=n+1, c=n+2$.

Opposite the side $a$ is the smallest angle $\alpha$, and opposite the side $c$ is the largest angle $\gamma$. (1 point) It holds that $\gamma=2 \alpha$.

From the Law of Sines, we have

$$
\frac{n}{\sin \alpha}=\frac{n+2}{\sin \gamma}=\frac{n+2}{2 \sin \alpha \cos \alpha}
$$

so

$$
\cos \alpha=\frac{n+2}{2 n}
$$

From the Law of Cosines, we have

$$
\begin{aligned}
n^{2} & =(n+1)^{2}+(n+2)^{2}-2(n+1)(n+2) \cos \alpha= \\
& =(n+1)^{2}+(n+2)^{2}-2(n+1)(n+2) \cdot \frac{n+2}{2 n}
\end{aligned}
$$

and after multiplying by $n$, we get

$$
\begin{aligned}
n^{3} & =n(n+1)^{2}+n(n+2)^{2}-(n+1)(n+2)^{2}= \\
& =n^{3}+2 n^{2}+n+n^{3}+4 n^{2}+4 n-(n+1)\left(n^{2}+4 n+4\right)= \\
& =2 n^{3}+6 n^{2}+5 n-n^{3}-4 n^{2}-4 n-n^{2}-4 n-4
\end{aligned}
$$

The result of simplification is the quadratic equation

$$
n^{2}-3 n-4=0
$$

The only possible solution is $n=4$.

Then the sides of the triangle are $a=4 \text{ cm}, b=5 \text{ cm}, c=6 \text{ cm}$.

By rotating this triangle around the side $c$, two cones with a common base are formed.

![](https://cdn.mathpix.com/cropped/2024_05_30_8c954f607b7f1570b01bg-20.jpg?height=586&width=572&top_left_y=233&top_left_x=733)

The radius of the base is the height of the given triangle. Let's first calculate $\sin \alpha$,

$$
\begin{aligned}
& \cos \alpha=\frac{n+2}{2 n}=\frac{6}{8}=\frac{3}{4} \\
& \sin \alpha=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}
\end{aligned}
$$

Since $\sin \alpha=\frac{r}{5}$, it follows that $r=5 \frac{\sqrt{7}}{4} \text{ cm}$.

The volume of the resulting body is

$$
V=\frac{r^{2} \pi \cdot x}{3}+\frac{r^{2} \pi \cdot(6-x)}{3}=\frac{r^{2} \pi \cdot 6}{3}=5^{2} \cdot \frac{7}{16} \cdot \pi \cdot 2=\frac{175 \pi}{8} \text{ cm}^{3}
$$