Given are regular pentagons with sides $a$ and $b$. If the pentagon with side $a$ is rotated around one of its sides, the volume of the resulting solid is equal to the volume of the solid obtained by rotating the pentagon with side $b$ around one of its diagonals. Determine the ratio $a: b$.
Solution. If we completely rotate a regular pentagon with side $b$ around its diagonal, the resulting solid of revolution is as if we were rotating a trapezoid consisting of three sides and one diagonal.

![](https://cdn.mathpix.com/cropped/2024_05_02_851d4846cba0e9191df3g-1.jpg?height=385&width=382&top_left_y=263&top_left_x=861)

The volume of the solid of revolution is equal to the product of the area of the rotated polygon and the arc length traveled by its centroid (Pappus-Guldin theorem): $V=T \cdot R_{S} \cdot \alpha, \alpha=2 \pi$, since we rotate it completely. $R_{S}$ is the distance of the centroid from the axis.

The area of a regular pentagon with side $a$, by breaking it into triangles:

$$
T_{a}=\frac{5 a^{2} \operatorname{tg} 54^{\circ}}{4}
$$

The distance of the centroid from the axis of rotation, which is one of the sides here: $R_{S a}=\frac{a}{2} \operatorname{tg} 54^{\circ}$. Thus, the volume of the solid of revolution obtained by rotating a regular pentagon with side $a$ around one of its sides is:

$$
V_{a}=T_{a} R_{S a} 2 \pi=\frac{5 a^{2} \operatorname{tg} 54^{\circ}}{4} \cdot \frac{a}{2} \operatorname{tg} 54^{\circ} \cdot 2 \pi
$$

The diagonal of a regular pentagon with side $b$, which is also the base of the resulting trapezoid: $c=2 b \sin 54^{\circ}$, the height of the trapezoid, $h=b \sin 72^{\circ}$, and its area is

$$
T_{b}=\frac{b+c}{2} \cdot h=\frac{b+2 b \sin 54^{\circ}}{2} \cdot b \sin 72^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_851d4846cba0e9191df3g-1.jpg?height=393&width=398&top_left_y=1484&top_left_x=858)

The centroid of a trapezoid with bases $b$ and $c$ and height $h$ is at a distance $R_{S b}=\frac{h}{3} \cdot \frac{2 b+c}{b+c}$ from the longer base $(c)$. Substituting the previously obtained values:

$$
R_{S b}=\frac{b \sin 72^{\circ}}{3} \cdot \frac{2 b+2 b \sin 54^{\circ}}{b+2 b \sin 54^{\circ}}
$$

Thus, the volume of the solid of revolution obtained by rotating a regular pentagon with side $b$ around its diagonal is:

$$
V_{b}=T_{b} R_{S b} 2 \pi=\frac{b+2 b \sin 54^{\circ}}{2} \cdot b \sin 72^{\circ} \cdot \frac{b \sin 72^{\circ}}{3} \cdot \frac{2 b+2 b \sin 54^{\circ}}{b+2 b \sin 54^{\circ}} \cdot 2 \pi
$$

According to the problem, $V_{a}=V_{b}$, from which, after simplifications and reductions:

$$
\frac{2}{3} b^{3} \sin ^{2} 72^{\circ} \cdot\left(1+\sin 54^{\circ}\right)=\frac{5}{4} a^{3} \operatorname{tg}^{2} 54^{\circ}
$$

The exact values of the trigonometric functions:

$$
\sin 72^{\circ}=\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}}, \quad \sin 54^{\circ}=\frac{1}{4}(1+\sqrt{5}), \quad \operatorname{tg} 54^{\circ}=\sqrt{1+\frac{2}{\sqrt{5}}}
$$

Substituting these and performing the simplifications:

$$
b^{3}(3 \sqrt{5}+5)=6 a^{3}(\sqrt{5}+2), \quad \text { from which } \quad \frac{a^{3}}{b^{3}}=\frac{3 \sqrt{5}+5}{6 \sqrt{5}+12}
$$

The desired ratio:

$$
\frac{a}{b}=\sqrt[3]{\frac{3 \sqrt{5}+5}{6 \sqrt{5}+12}} \approx 0.772
$$

Based on the work of Gergő Schwarz