A convex body is bounded by two congruent regular octagonal faces, 16 congruent triangular faces, and 8 congruent rhombic faces. The sides of the two octagons are pairwise parallel, the sides of the triangles form the edges of the body with these, and the other sides of the triangles are with one rhombic side each. What is the radius of the smallest sphere that can enclose the body? What is the volume of the body?
I. Let's construct the model of the solid in our mind by placing the faces in the prescribed order side by side. All edges of the solid are equal because each face is an equilateral triangle, let the common length be \( c \). Let the consecutive vertices of the octagonal face \( N' \) be \( A_i (i=1,2, \ldots, 8) \) in increasing order of indices, and the third vertex \( B_i \) of the triangular face \( H_i \) adjacent to the side \( A_i A_{i+1} \) (indices greater than 8 are understood to be reduced by 8), and the rhombus \( B_i A_{i+1} B_{i+1} C_{i+1} = R_{i+1} \) that fits between \( H_i \) and \( H_{i+1} \). Thus, the second triangular face \( B_i C_i C_{i+1} \) that fits between \( R_i \) and \( R_{i+1} \), and the vertices \( C_i \) are the consecutive vertices of the octagonal face \( N'' \) in increasing order of indices.
![](https://cdn.mathpix.com/cropped/2024_05_02_c2ab79a4ecfe14c803a0g-1.jpg?height=912&width=1328&top_left_y=430&top_left_x=386)

We will show that the solid fits into itself by rotations of \( k \cdot 45^\circ \) around the axis connecting the centers of \( N' \) and \( N'' \), and the midpoint of this axis is the center of the solid. \( B_i \) is on the perpendicular bisector plane \( S_i \) of the edge \( A_i A_{i+1} \), and so is \( B_{i+4} \), because the regularity of \( N' \) implies that \( S_i \) also perpendicularly bisects the edge \( A_{i+4} A_{i+5} \). \( B_i \) and \( B_{i+4} \) are also points on the common perpendicular bisector plane of the edges \( C_i C_{i+1} \) and \( C_{i+4} C_{i+5} \), so these two planes are identical, because both are perpendicular to the planes of \( N' \) and \( N'' \), and only one perpendicular plane can be erected on their connecting line \( B_i B_{i+4} \). We will call the planes of \( N' \) and \( N'' \) the base planes. (Here we used that \( B_i \) and \( B_{i+4} \) are different points, and that their connecting line is not perpendicular to the base plane; indeed,

![](https://cdn.mathpix.com/cropped/2024_05_02_c2ab79a4ecfe14c803a0g-1.jpg?height=43&width=1795&top_left_y=1648&top_left_x=154)
the distance of the side is \( c(1+\sqrt{2}) \), and half of this is greater than the height of an equilateral triangle with side \( c \), \( c \sqrt{3} / 2 \).) From the above, it also follows that \( C_i C_{i+4} \| A_i A_{i+4} \), and that \( C_i \) is on the same side of \( S_i \) as \( A_i \).

Now consider the intersection line \( t \) of the plane \( S_{i+1} \) formed by the points with indices one greater, with \( S_i \),

![](https://cdn.mathpix.com/cropped/2024_05_02_c2ab79a4ecfe14c803a0g-1.jpg?height=43&width=1780&top_left_y=1819&top_left_x=167)
the intersection point of \( t \) is the center of the respective face. Indeed, for example, \( O' \) is the common point of the two lines that \( S_i \) and \( S_{i+1} \) cut from the base plane, each of which is a symmetry axis of \( N' \), and two symmetry axes intersect at the center of the regular polygon.

Let the midpoint of \( A_i A_{i+1} \) be \( D_i \), and the midpoint of \( C_i C_{i+1} \) be \( E_i \), then \( D_i O' O'' E_i \) is a rectangle, because there are right angles at \( O' \) and \( O'' \), and the congruence of the octagons implies \( D_i O' = E_i O'' \), so \( D_i E_i \| t \), and \( A_i D_i E_i C_i \) is also a rectangle, so \( A_i C_i \| t \).

According to these, the diagonal \( B_{i-1} B_i \) of \( R_i \) is parallel to the base planes, because it perpendicularly bisects the diagonal \( A_i C_i \), and thus every \( B_i \) is in the perpendicular bisector plane of \( O' O'' \). The centers \( K_i \) of the rhombuses (in this plane) form an octagon \( N \) congruent to \( N' \), with sides pairwise parallel; let the center of \( N \) be \( O \). Then \( O K_i \) bisects the angle \( B_{i-1} O B_i = 45^\circ \), and \( K_i B_{i-1} = K_i B_i \), so \( B_i B_{i-1} \) is perpendicular to \( O K_i \), and \( O B_{i-1} = O B_i \), the points \( B_i \) are the vertices of a regular octagon \( N''' \), \( O \) is indeed the center of the solid, and it fits into itself by any rotation and reflection that maps \( N' \) and \( N'' \) onto themselves.

From this it follows that the points \( A_i \) and \( C_i \), and the points \( B_i \), are on the surfaces of two spheres centered at \( O \), and the larger of their radii gives the radius of the smallest sphere enclosing the solid.

II. Denoting the midpoint of the segment \( K_1 K_2 \) by \( L \), from the right triangles \( O K_1 L \), \( O K_1 B_1 \), \( B_1 D_1 L \), and \( O A_1 K_1 \),

\[
\begin{gathered}
O K_1 = \frac{K_1 L}{\sin 22.5^\circ} = \frac{c}{\sqrt{2-\sqrt{2}}}, \quad O B_1 = \frac{O K_1}{\cos 22.5^\circ} = \frac{c}{\sin 45^\circ} = c \sqrt{2} \\
O L = \frac{c}{2} \operatorname{ctg} 22.5^\circ = \frac{c}{2}(1+\sqrt{2}), \quad \text{and thus} \quad B_1 L = \frac{c}{2}(\sqrt{2}-1) \\
A_1 K_1 = D_1 L = \sqrt{B_1 D_1^2 - B_1 L^2} = \sqrt{\frac{3 c^2}{4} - \frac{c^2}{4}(3-2 \sqrt{2})} = \frac{c}{\sqrt[4]{2}} \\
O A_1 = \sqrt{O K_1^2 + A_1 K_1^2} = c \sqrt{1+\sqrt{2}}
\end{gathered}
\]

The latter is greater than \( c \sqrt{1+1} = O B_1 \), so the radius of the smallest sphere enclosing the solid is \( O A_1 \approx c \cdot 1.55 \).

III. The solid can be divided into 8 congruent pyramids and a prism with base \( N' \) by planes passing through the corresponding sides of \( N' \) and \( N'' \). The volume of the pyramid \( B_1 A_1 A_2 C_2 C_1 \) is \( A_1 A_2 \cdot A_1 C_1 \cdot B_1 L / 3 \), where \( A_1 C_1 = 2 A_1 K_1 \), which is also the height of the prism, so the volume of the solid, based on the known area formula of \( N' \), is

\[
V = A_1 A_2^2 \cdot A_1 C_1 \left( \frac{4(\sqrt{2}-1)}{3} + 2(1+\sqrt{2}) \right) = c^3 \frac{2 \sqrt[4]{8}}{3}(1+5 \sqrt{2})
\]

Work by Judit Szemke (Budapest, Ságvári E. Gyak. G. III. o. t.), with an added proof.