In how many ways can the network of the solid examined in Exercise 1120[² be constructed so that the total length of the edges to be joined is seven times the length of the original cube edge? We do not consider two networks different if they only differ in the pairwise exchange of letters.[^0]


[^0]:    ${ }^{2}$ K. M. L. $36(1968) 64$. o.
The solid in question was formed from a cube defined by the base $A B C D$ and the side edges $A E, B F, C G, D H$, by cutting out the triangular pyramids $A D G H$ and $B C E F$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_4356800a3f5f4dca0700g-1.jpg?height=526&width=542&top_left_y=274&top_left_x=778)

Figure 1

We observed that the total length of the edges to be glued in the net is shortest - exactly 7 cube edges, if only the original cube edges are left as glueable, in other words, if the edges resulting from the 4 face diagonals and 2 space diagonals of the cube are formed by folding the paper in the net.

Accordingly, the combined set of the $A B E, B E C, E C F$, and $C F G$ triangular faces remains connected, and the same applies to the combined set of the mirror images of these faces with respect to the center $O$ of the cube (the common midpoint of the diagonals $C E, A G$, and the edges). Let's call these combined sets of faces complexes. When these complexes are unfolded, their boundary lines form a centrally symmetric hexagon with a side length of one cube edge, $A B C G F E=\mathrm{I}$, and $C D A E H G=\mathrm{II}$, which are congruent to each other. The first three angles (in the order of the vertices) are $90^{\circ}, 135^{\circ}, 135^{\circ}$. Thus, we only need to examine how the two complexes and the $A B C D=I I I, E F G H=I V$ square faces can be connected to form a net, or in other words, how the 3 cube edges (out of the 10 remaining on the solid) that are formed by folding the paper can be chosen.

We note that the line passing through $O$ and parallel to $A E$ is a rotational symmetry axis of the solid, a $180^{\circ}$ rotation about this axis maps the solid onto itself, swapping the vertex pairs $(A, C),(B, D),(E, G),(F, H)$, and also swapping I with II. Similarly, the line passing through the midpoints of the edges $A E$ and $C G$ is also a symmetry axis, a $180^{\circ}$ rotation about this line swaps the two complexes and also the two square faces.

First, we consider nets where I and II are adjacent. Their common edges, $A E$ and $C G$, can be mapped onto each other by the first symmetry, so it is sufficient to examine the case where the connecting edge is $A E$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_4356800a3f5f4dca0700g-1.jpg?height=441&width=640&top_left_y=1601&top_left_x=710)

Figure 2

Their combined set is centrally symmetric about the midpoint $K$ of the segment $A E$. The III square face connects along one of the edges $3_{1}, 3_{2}, 3_{3}, 3_{4}$, and the IV square face similarly along one of the edges $4_{1}, 4_{2}, 4_{3}, 4_{4}$. Choosing the connecting edges independently, $4 \cdot 4=16$ pairings can be thought of.

However, among these, the pairings $3_{i}, 4_{j}$ and $3_{j}, 4_{i}$ (where $i$ and $j$ are different indices from 1, 2, 3, 4) are not distinct, as they can be rotated into each other about $K$. The choice $3_{i} 4_{i}$, on the other hand, gives a different net for each value of $i$. The number of such pairings is 4, and of the remaining 12 pairings, only half, 6, are distinct.

Moving on, if I and II are not adjacent, then one of the square faces connects to both, and by the second symmetry, it is sufficient to choose III. I and II connect to III either along two opposite sides or along two adjacent sides. There are two distinct possibilities for the former, for example, the edge $A B$ is not equivalent to the edge $B C$, as different shaped faces connect to them on the solid (Figures 3 and 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_4356800a3f5f4dca0700g-2.jpg?height=608&width=418&top_left_y=127&top_left_x=840)

Figure 3

![](https://cdn.mathpix.com/cropped/2024_05_02_4356800a3f5f4dca0700g-2.jpg?height=455&width=621&top_left_y=865&top_left_x=730)

Figure 4

The combined set of the I and III faces is centrally symmetric about the center of the III face in both cases, and the 4 edges (drawn thickly) suitable for connecting the IV face are pairwise images of each other. Thus, the connecting edge can be chosen in 2 ways on both figures.

If I and II connect to III along adjacent sides, the common vertex of these two sides can only be $A$ and $C$ - for example, the edges $B A$ and $B C$ both connect I to III. By the first symmetry, it is sufficient to take $A$ as the common vertex (Figure 5).

![](https://cdn.mathpix.com/cropped/2024_05_02_4356800a3f5f4dca0700g-2.jpg?height=369&width=525&top_left_y=1713&top_left_x=781)

Figure 5

The combined set of the first three faces is thus not symmetric, and connecting the IV face along any of the possible edges results in a different net.

According to all this, the number of essentially different nets is $(4+6)+2+2+4=18$. We list the 3 cube edges formed by folding in these nets:

| $A E, C B, G H$, | $A E, C B, F G ;$ | $A B, C D, E F$, |
| :--- | :--- | :--- |
| $A E, B A, H E$, | $A E, B A, E F$, | $A B, C D, F G ;$ |
| $A E, A D, E F$, | $A E, B A, F G ;$ | $A B, A D, E F$, |
| $A E, D C, F G ;$ | $A E, A D, F G ;$ | $A B, A D, F G$, |
| $A E, C B, H E$, | $B C, A D, E F$, | $A B, A D, G H$, |
| $A E, C B, E F$, | $B C, A D, F G ;$ | $A B, A D, H E$. |