Twenty-four unit-radius spheres were packed into a "pyramid": nine in the bottom layer, four in the middle layer, and one on top. What fraction of the volume of the regular square pyramid that encloses these spheres is filled by the spheres?
The base of the pyramid $ABCDE$ enveloping the spheres is the square $ABCD$, and its apex is $E$. To calculate its volume, we need to know the height of the pyramid and the length of the side of the square.

![](https://cdn.mathpix.com/cropped/2024_05_02_f264c50edaeb9354e2c7g-1.jpg?height=786&width=730&top_left_y=244&top_left_x=684)

The centers of 9 unit-radius spheres placed on the base plane are 1 cm above the base plane,

![](https://cdn.mathpix.com/cropped/2024_05_02_f264c50edaeb9354e2c7g-1.jpg?height=46&width=1791&top_left_y=1077&top_left_x=156)
From the diagram, we can read that $PS = QS = 4 \text{ cm}$, and $PSQ$ is an isosceles right triangle, so $PQ = 4\sqrt{2}$.

Let's draw a plane through the points $A, E, C$. This plane intersects the pyramid in an isosceles triangle and the spheres in circles, as shown in the second diagram (front view), from which

thus

$$
\begin{gathered}
PR = RQ = 4, \quad PT = \frac{1}{2} PQ = 2\sqrt{2}, \quad RT \perp PQ \\
RT = \sqrt{4^2 - (2\sqrt{2})^2} = 2\sqrt{2}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f264c50edaeb9354e2c7g-1.jpg?height=629&width=1084&top_left_y=1510&top_left_x=65)

2nd diagram

![](https://cdn.mathpix.com/cropped/2024_05_02_f264c50edaeb9354e2c7g-2.jpg?height=602&width=697&top_left_y=141&top_left_x=1126)

Let the midpoint of edge $AD$ be $F_1$, and the midpoint of edge $BC$ be $F_2$. Draw a plane through the points $F_1, E, F_2$. The plane section (from the front view) is shown in the 3rd diagram.

$$
\begin{aligned}
ST = 2, & \quad TW = 1, \quad TR = 2\sqrt{2} \quad (\text{from the previous}) \\
RS & = \sqrt{2^2 + (2\sqrt{2})^2} = 2\sqrt{3}
\end{aligned}
$$

Triangle $PER$ is similar to triangle $TRS$; the similarity ratio is $\frac{1}{2}$, so $ER = \sqrt{3}$. Therefore, the height of the pyramid is $m = 1 + 2\sqrt{2} + \sqrt{3}$. The side length of the base, from the similarity of triangles $F_1EW$ and $SRT$:

$$
\begin{gathered}
\frac{a}{2} : 2 = m : 2\sqrt{2} \quad \rightarrow \quad a = \frac{2m}{\sqrt{2}} \\
V_{\text{pyramid}} = \frac{\left(\frac{2m}{\sqrt{2}}\right)^2 \cdot m}{3} = \frac{2m^3}{3}, \quad V_{\text{spheres}} = \frac{56\pi}{3}, \\
\frac{V_{\text{spheres}}}{V_{\text{pyramid}}} = \frac{56\pi}{2(1 + 2\sqrt{2} + \sqrt{3})^3} \approx 0.5116.
\end{gathered}
$$

Thus, the total volume of the spheres is approximately half the volume of the pyramid.

Based on the work of András Fördhécz (Székesfehérvár, Teleki Blanka Hatosztályos Gimn., 1st year)