A regular quadrilateral pyramid with base edge $a$ and height $m$ has a cube inscribed in it such that one face of the cube lies on the base of the pyramid, and the other 4 vertices of the cube are a) on each edge, b) on each slant height. Calculate the volume of these cubes as a function of $a$ and $m$. How do these volumes compare when $m=a$?
a) Let's first consider the four vertices of the cube on each edge. If we draw a plane through two opposite edges of the pyramid, it cuts a triangle from the pyramid, whose side in the base of the pyramid is $a \sqrt{2}$, and from the cube, it cuts a rectangle whose shorter side is the edge $x$ of the cube, and the longer side is the diagonal of a cube face: $x \sqrt{2}$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_ab831cb92c7530240faeg-1.jpg?height=480&width=571&top_left_y=318&top_left_x=758)

The ratio of the heights and sides of the similar triangles is the same:

$$
\frac{m-x}{\sqrt{2} x}=\frac{m}{\sqrt{2} a}
$$

From this,

$$
x=\frac{a m}{a+m}
$$

The volume of the cube is:

$$
V_{1}=\left(\frac{a m}{a+m}\right)^{3}
$$

b) If the four vertices of the cube are on each face height, then the cube is essentially placed in a pyramid with a base of a square with side $\frac{a}{\sqrt{2}}$ and height $m$, in the manner described in part a) (the side of the square formed by connecting the midpoints of the sides of a square with side $a$ can be calculated using the Pythagorean theorem as $\left.\frac{a}{\sqrt{2}}\right)$. Thus, applying the previous result, the volume of the cube is:

$$
V^{2}=\left(\frac{\frac{a}{\sqrt{2}} m}{\frac{a}{\sqrt{2}}+m}\right)^{3}=\left(\frac{a m}{a+\sqrt{2} m}\right)^{3}
$$

The ratio of the volumes when $m=a$ is

$$
V_{1}: V_{2}=\frac{a^{6}}{8 a^{3}}: \frac{a^{6}}{(1+\sqrt{2})^{3} a^{3}}-\left(\frac{1+\sqrt{2}}{2}\right)^{3}=\frac{7+5 \sqrt{2}}{8}=\frac{7+\sqrt{50}}{8} \approx 1.759
$$

Judit Németh (Kecskemét, Közg. t. III. o. t.)