3. From a right prism with a square base of side length $L_{1}$ and height $H$, we extract a frustum of a pyramid, not necessarily a right one, with square bases of side lengths $L_{1}$ (for the lower base) and $L_{2}$ (for the upper base), and height $H$. The two resulting pieces are shown in the following image.
![](https://cdn.mathpix.com/cropped/2024_04_26_a21e0e58f8f87e982718g-6.jpg?height=434&width=550&top_left_y=2053&top_left_x=800)

If the volume of the frustum of the pyramid is 2/3 of the total volume of the prism, what is the value of $L_{1} / L_{2}$?
Solution. If we extend a height $h$ of the frustum of a pyramid to obtain a complete pyramid of height $H+h$, it will have a section as shown in the following figure.

![](https://cdn.mathpix.com/cropped/2024_04_26_a21e0e58f8f87e982718g-7.jpg?height=1102&width=402&top_left_y=800&top_left_x=884)

An argument of similarity of triangles allows us to verify that

$$
\frac{h+H}{L_{1}}=\frac{h}{L_{2}}
$$

and, therefore,

$$
h=\frac{H L_{2}}{L_{1}-L_{2}}
$$

Moreover, we can observe that

Volume of the frustum of the pyramid $=\frac{1}{3}\left(L_{1}^{2}(H+h)-L_{2}^{2} h\right)$

$$
\begin{aligned}
& =\frac{1}{3}\left(\frac{H L_{1}^{3}}{L_{1}-L_{2}}-\frac{H L_{1}^{3}}{L_{1}-L_{2}}\right) \\
& =\frac{H}{3} \frac{L_{1}^{3}-L_{2}^{3}}{L_{1}-L_{2}}=\frac{H}{3}\left(L_{1}^{2}+L_{1} L_{2}+L_{2}^{2}\right)
\end{aligned}
$$

Thus, considering that

$$
\text { Volume of the frustum of the pyramid }=\frac{2}{3} \text { Volume of the prism, }
$$

we have the equation

$$
\frac{H}{3}\left(L_{1}^{2}+L_{1} L_{2}+L_{2}^{2}\right)=\frac{2}{3} H L_{1}^{2}
$$

which transforms into

$$
\left(\frac{L_{1}}{L_{2}}\right)^{2}-\frac{L_{1}}{L_{2}}-1=0
$$

whose only positive solution is $\frac{L_{1}}{L_{2}}=\frac{1+\sqrt{5}}{2}$. That is, the sides must be in the golden ratio.