Let's create a body that is bounded by $8$ unit-sided regular triangular plates and $2$ square plates. The square plates share one edge. Compare the volume of this body with the volume of the body examined in problem 1464.
I. The solid bounded by the given faces can be constructed in several ways; let us first consider the following. Let the $A B C D$ and $A B E F$ square faces be the side faces of a regular triangular prism $H$ (1. a and b figures). We construct the open net of a regular quadrilateral pyramid $N$ with apex $G$ from four triangular faces, and attach it to the $C D F E$ square face of $H$, and from three more triangles, we construct the net of a regular tetrahedron $T$ with apex $K$, and attach it to the $A D F$ cover face of $H$, finally, as the eighth triangular face, we take the $B C E$ base face of $H$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2e370e125a942bbcdbd1g-1.jpg?height=602&width=339&top_left_y=393&top_left_x=714)
a)

![](https://cdn.mathpix.com/cropped/2024_05_02_2e370e125a942bbcdbd1g-1.jpg?height=569&width=315&top_left_y=431&top_left_x=1054)

1. figure

The volumes of $H, N, T$ can be easily calculated in a known manner, so the volume of the solid is

$$
V=\frac{\sqrt{3}}{4}+\frac{\sqrt{2}}{6}+\frac{\sqrt{2}}{12}=\frac{\sqrt{3}+\sqrt{2}}{4} \approx 0.786 \ldots
$$

volume units; it is smaller than the volume of the solid bounded by the same faces, which was examined in problem 1464, which is $0.956 \ldots$ units.

II. The combination of solids $H N$ is symmetric with respect to the perpendicular bisector plane of the edge $A B$ - and also with respect to the plane $A B G$ - therefore, we get a congruent solid if we attach $T$ to the base face (outward). We cannot attach $T$ outward to the side faces of $N$, because then two faces of $N$ and $T$, adjacent to the attachment face, would lie in the same plane, forming a rhombus face. Indeed, by drawing a segment $G K^{*}$ from $G$ parallel and equal to $C D$ and $E F$, the quadrilaterals $C D K^{*} G$ and $E F K^{*} G$ are rhombuses, so $K^{*} D$ and $K^{*} F$ are also equal to the common edge length, thus $D F G K^{*}$ is a regular tetrahedron, and a tetrahedron can only be attached outward to the face $D F G$ in one way.

III. We can also bound a solid with a smaller volume than $V$ using the given faces, as we can attach either $T$ or $N$ inward to the appropriate face of $H$ - as if carving out the pyramid from $H$. However, we cannot turn both inward at the same time, because then certain faces of theirs would intersect each other, resulting in a spatial figure reminiscent of looped polygons, for which volume cannot be defined. It is easy to see that - due to the mentioned symmetries - if we attach $T$ inward to three faces of the $H N$ combination, we get different solids, for example, to the faces $A D F, F D G$, and $C D G$. If we turn $N$ inward, we can attach $T$ outward to the same three faces. In the latter two cases, what is "outward" relative to the carved-out $H N$ is "inward" relative to $N$ itself.

![](https://cdn.mathpix.com/cropped/2024_05_02_2e370e125a942bbcdbd1g-1.jpg?height=382&width=334&top_left_y=2216&top_left_x=733)
a)

![](https://cdn.mathpix.com/cropped/2024_05_02_2e370e125a942bbcdbd1g-1.jpg?height=515&width=287&top_left_y=2063&top_left_x=1060)
b)

The 2nd figure shows the case where $T$ is attached inward to the $C D G$ face of $N$; both b) figures show the solid of the a) figure in two projections, omitting the coordinate axes and the dash marks.

In the case of such solids, the volume of $T$, or $N$, must be subtracted from the sum of the volumes of the other two elements.

István Vályi (Budapest, Fazekas M. Gyak. G. III. o. t.)

Remarks. 1. The problem did not intend to present all possible solids. However, some solutions pointed out the multiple possibilities and then selected the one in the 1st figure, saying they were focusing only on the convex case. For this reason, it should be noted that - as can be clearly seen from the front view of the 1st b) figure - this solid is not convex either.

2. Some solutions thought they solved the problem by cutting a regular octahedron paper model into two quadrilateral pyramid nets (i.e., along an edge and the perpendicular edges running into its endpoints), and then closing these pyramids with the square faces. Although the two pyramids are still connected along their common edge, this is a masterful solution. Others attached a regular tetrahedron between the two pyramids. However, in this case, the base squares, like the pyramid faces, lie in the same plane; we did not accept this as a solution.