A regular pentagon with a unit side, by drawing two diagonals from one vertex, results in the net of a pyramid with a regular triangle base. What is the volume of the pyramid?
Solution. Using the notation in the figure. We know that in the equilateral triangle $ABC$, $FB = \frac{\sqrt{3}}{2}$, and $AD$ and $CD$ were originally the diagonals of a regular pentagon, so their lengths are $\frac{1+\sqrt{5}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_73007d2d15e4063b3663g-1.jpg?height=466&width=487&top_left_y=344&top_left_x=800)

Figure 1

From the right triangle $AFD$:

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

From the triangle $FBD$:

$$
\begin{aligned}
1^{2} & =\frac{5+2 \sqrt{5}}{4}+\frac{3}{4}-2 \cdot \sqrt{\frac{5+2 \sqrt{5}}{4}} \cdot \frac{\sqrt{3}}{2} \cdot \cos \alpha \\
\frac{2+\sqrt{5}}{2} & =2 \sqrt{\frac{5+2 \sqrt{5}}{4}} \cdot \frac{\sqrt{3}}{2} \cdot \cos \alpha \\
2+\sqrt{5} & =\sqrt{15+6 \sqrt{5}} \cdot \cos \alpha \\
\cos \alpha & =\frac{2+\sqrt{5}}{\sqrt{15+6 \sqrt{5}}}
\end{aligned}
$$

Thus, $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{\frac{10-2 \sqrt{5}}{15}}=\frac{m}{\sqrt{\frac{5+2 \sqrt{5}}{4}}}$,

$$
\begin{aligned}
m & =\sqrt{\left(\frac{10-2 \sqrt{5}}{15}\right)\left(\frac{5+2 \sqrt{5}}{4}\right)}=\sqrt{\frac{50+20 \sqrt{5}-10 \sqrt{5}-20}{60}}= \\
& =\sqrt{\frac{30+10 \sqrt{5}}{60}}=\sqrt{\frac{3+\sqrt{5}}{6}}=\sqrt{\frac{6+2 \sqrt{5}}{12}}=\sqrt{\frac{(\sqrt{5}+1)^{2}}{3 \cdot 4}}= \\
& =\frac{\sqrt{5}+1}{2 \sqrt{3}}=\frac{\sqrt{15}+\sqrt{3}}{6} \\
V & =\frac{T_{ABC} \cdot m}{3}=\frac{\frac{\sqrt{3}}{4} \cdot \frac{\sqrt{15}+\sqrt{3}}{6}}{3}=\frac{1+\sqrt{5}}{24}
\end{aligned}
$$

The volume of the pyramid is: $\frac{1+\sqrt{5}}{24}$.