A body is bounded by 8 equilateral triangular faces, each with a side length of 1 unit, and 2 square faces, the planes of which are parallel. Calculate the volume of the body.

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The body is bounded by 8 equilateral triangular faces, each with a side length of 1 unit, and 2 square faces, the planes of which are parallel. Calculate the volume of the body.
The questionable solid $T$ has $l=10$ faces, the number of its edges is half the total number of sides of all faces, which is $8 \cdot 3 + 2 \cdot 4 = 32$, because this way we counted each edge twice, once for each face it belongs to, so $e = 16$. We only consider the case where $T$ is convex. In this case, Euler's polyhedron theorem applies to it, so the number of its vertices is $c = e + 2 - l = 8$. According to this, $T$ has no other vertices than those of the square faces, $T$ is identical to the solid depicted in the last figure of problem 1320, the upper square face $N_{2}$ is rotated $45^{\circ}$ relative to the lower square face $N_{1}$ about the line connecting their centers, and the distance $m$ between these faces satisfies $m^{4} = 1 / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_da373669b7c33d618a4fg-1.jpg?height=387&width=542&top_left_y=438&top_left_x=781)

Consider the 4 triangular faces that share an edge with $N_{1}$, and extend the 3 adjacent faces of each to their common point. The selected faces, together with these extensions, form a tetrahedron $H$, these are congruent to each other, and they complete $T$ into a regular truncated pyramid $C$ (thinly drawn edges). The area of the base square of $C$ is 2, because its sides are equal to the diagonals of $N_{1}$, so its volume is

$$
V_{cs} = m(2 + \sqrt{2} + 1) / 3
$$

The base area of the $H$-tetrahedra is $1 / 4$, so their total volume, and finally the volume of $T$ is

$$
\begin{aligned}
V_{h} = 4 \cdot (1 / 4) \cdot m / 3, \quad V_{T} = & V_{cs} - V_{h} = (2 + \sqrt{2}) / (3 \cdot \sqrt[4]{2}) = \sqrt[4]{2}(1 + \sqrt{2}) / 3 \approx \\
& \approx 0.956 \text{ volume units }).
\end{aligned}
$$

József Lipták (Esztergom, Temesvári Pelbárt g. IV. o. t.)

Remarks. 1. The volume can also be calculated using the known formula from problem 1240:

$$
V = m(t_{a} + 4t_{k} + t_{f}) / 6
$$

where $t_{a}$, $t_{f}$, and $t_{k}$ are the areas of the lower base, the upper base, and the section cut by the plane perpendicular to the height $m$ at its midpoint, respectively, because according to the second remark in the cited solution, $T$ is a prismatoid. The middle section is a regular octagon with side length $1 / 2$ units.

2. By replacing the squares with any regular $n$-gons ($n \geq 3$) and using $2n$ regular triangular faces, similar convex solids can be formed, which are called Archimedean antiprisms. These, along with Archimedean prisms (faces: $n$ squares and 2 regular $n$-gons, with parallel sides in pairs), belong to the class of semi-regular polyhedra (every face is a regular polygon, and the "environment" of every vertex is congruent). For $n=3$, the antiprism has all faces congruent, resulting in the (Platonic) regular octahedron; the (Archimedean) antiprism for $n=5$ and the (Platonic) regular 20-faced solid (icosahedron) appear in the point system of problem 1472, part I. Part II of this problem also involves a semi-regular solid.[^0]

[^0]:    ${ }^{1}$ K. M. L. 32 (1966) 101-104.

    ${ }^{2}$ K. M. L. 28 (1964) 12.

    ${ }^{3}$ Such a solid appeared in problems 983 and 1046, K. M. L. 20 (1960) 100 and 22 (1961) 159.