A regular quadrilateral pyramid's height is twice the length of the base edge. What fraction of the pyramid's volume is the volume of the inscribed cube? (The inscribed cube has four vertices on the pyramid's side edges, and four vertices on the base plane.)
Solution. Let the base edge of the pyramid be $x$, its height $m=2x$, and the edge of the cube $a$. Let us place a plane on the axis of the pyramid that is parallel to one of the cube's face planes. This plane cuts an isosceles triangle out of the pyramid, the base of which is equal to the base edge of the pyramid, i.e., $x$, and its height is $2x$. The plane intersects the cube in a square with side $a$.

Using the notation from the diagram, $\triangle BDC \sim \triangle BFA$, and since $FA = \frac{1}{4} BF$, it follows that

$$
\frac{\frac{a}{2}}{2x - a} = \frac{1}{4}
$$

From this, $a = \frac{2}{3} x$
![](https://cdn.mathpix.com/cropped/2024_05_02_208d10fa163240477f15g-1.jpg?height=702&width=746&top_left_y=674&top_left_x=686)

Let's write down the volumes.

The volume of the pyramid:

$$
\frac{2x \cdot x^2}{3} = \frac{2x^3}{3}
$$

The volume of the cube:

$$
\left(\frac{2}{3} x\right)^3 = \frac{8}{27} x^3
$$

The ratio of the volumes of the cube and the pyramid:

$$
\frac{\frac{8}{27} x^3}{\frac{2}{3} x^3} = \frac{8}{27} \cdot \frac{3}{2} = \frac{4}{9}
$$