A regular quadrilateral pyramid has all its edges of equal length. There are two spheres that touch every edge of the pyramid. What is the ratio of their volumes?
I. Solution. Use the notations of Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_206f1ac4dcac1204004cg-1.jpg?height=655&width=746&top_left_y=220&top_left_x=668)

Figure 1

One of the spheres - the so-called internal tangent sphere - touches the pyramid's edges at internal points, and by symmetry, it touches the base edges $AB, BC, CD, DA$ at their midpoints. This sphere touches the edge $EC$ at point $F$. The other sphere - the external tangent sphere - also touches the base edges at their midpoints, and the side edges at their extensions, for example, $EC$ at point $H$. According to the equality of tangent segments from a point outside a sphere, $CF = CG = \frac{a}{2}$ and $CG = CH = \frac{a}{2}$ (where $G$ is the point of tangency of the spheres on edge $BC$, and $a$ is the edge length of the pyramid). Thus, $EF = EC - FC = a - \frac{a}{2} = \frac{a}{2}$, and $EH = \frac{3a}{2}$. In the central similarity centered at $E$, $EF : EH = 1 : 3$. Therefore, the ratio of the volumes of the spheres is $1 : 27$.

Remarks: 1. The apex $E$ of the pyramid is the center of a central similarity. The two edge-tangent spheres can also be derived by taking a "small" sphere that touches the edges emanating from $E$, and whose center is inside the pyramid. We enlarge this sphere from $E$ until it touches the base edges, and then continue enlarging until this happens again.

2. Let the center of the square $ABCD$ be $O_1$. It is easy to see that the triangle $EO_1C$ is isosceles and right-angled. Therefore, the perpendicular bisector plane of segment $EC$ intersects the pyramid's height $EO_1$ at $O_1$. Thus, the radius of the internal tangent sphere is $O_1F = \frac{a}{2}$, and the radius of the other sphere is $\frac{3a}{2}$.
3. Reflecting our pyramid over its base, we obtain a regular octahedron. From this, it is evident that the center of the smaller sphere lies within the base square.

Galambos István (Zalaegerszeg, Zrínyi M. Gym., III. grade)

Kálmán Tamás (Fazekas M. Fóv. Gyak. Gym., III. grade)

II. Solution. We use the central similarity mentioned in the first solution.

The plane of the pyramid's side face $EAB$ intersects both spheres in circles. Since the pyramid's edges are tangents to the spheres, the edges will also be tangents to these circles.

![](https://cdn.mathpix.com/cropped/2024_05_02_206f1ac4dcac1204004cg-1.jpg?height=607&width=548&top_left_y=1960&top_left_x=775)

On the second figure, $S$ and $T$ are the common points of the line $EA$ and these circles. It is known that the lengths of the tangent segments $ES$ and $ET$ are $s-a$ and $s$, respectively. Since the central similarity centered at $E$ maps these circles onto each other, the ratio of the radii of these circles, and thus the radii of the spheres, is:

$$
\frac{s-a}{s} = \frac{\frac{3}{2}a - a}{\frac{3}{2}a} = \frac{1}{3}
$$

Thus, the ratio of the volumes is $1: 27$.

Remark: Some solvers noted that the ratio of the radii of the two spheres is 1:3 for any regular pyramid with equal edge lengths. We note that such a pyramid can have at most five sides. Indeed, in a regular hexagonal (or more-sided) pyramid, the side edge is always longer than the base edge.

Csörnyei Marianna (Fazekas M. Fóv. Gyak. Gym., II. grade) Kálmán Tamás (Fazekas M. Fóv. Gyak. Gym., III. grade)