Given $6$ thin sticks of different lengths, their lengths are $a, b, c, d, e, f$; any three of them can form a triangle. How many different tetrahedron frameworks can be assembled from the sticks, which cannot be transformed into each other by rotation or reflection?
I. solution. First, we determine the number of ways to pair up the 3 opposite (non-adjacent) edge pairs (rod pairs), and then how many ways these 3 pairs can be assembled into a complete frame.

In the pairing, every rod is used. The rod of length $a$ can be paired with 5 others. Once we have chosen a pair, consider the next rod in the list from the remaining 4. For this, we can choose 3 different pairs from the remaining, and the two unchosen rods form the third pair. Thus, each of the 5 initial choices can be continued in 3 ways, giving us $5 \cdot 3 = 15$ pairings. These are all essentially different, and we have covered all possible pairings.

In the following, it is enough to consider, for example, the pairing $a b, c d, e f$, since we can design the same number of complete frames from each pairing. The 4 different endpoints of the $a b$ pair are the 4 vertices of the tetrahedron, so $c$ must connect to one endpoint of $a$ and the other to $b$, and $d$ will connect the remaining free endpoints of $a$ and $b$. The resulting skew quadrilateral has 2 pairs of opposite vertices, and we can choose the endpoints of the $e$ rod in 2 ways: either connecting the $a c$ and $b d$ connections, or the $a d$ and $b c$ connections, thus the endpoints of the $f$ edge are also determined. Although each such choice can be realized in 2 ways in space, for example, the first choice as shown in the first figure - the top views of the two are identical, but in one $e$ is above, in the other $f$ is above - these two frames can be transformed into each other by reflecting in the plane of the drawing, so they should not be considered different.

Thus, the number of possible frames cannot be more than $15 \cdot 2 = 30$.

Árpád Szalontai (Budapest, Berzsenyi D. Gymnasium, 2nd year) Miklós Horváth (Veszprém, Lovassy L. Gymnasium, 1st year)

Note. The obtained number 30 is only an upper limit for the number of frames that can be formed. It is possible, for example, that the two triangles connected to the $a$ edge can be placed in the same plane - on the same and opposite sides of $a$ - and the distance between the two vertices not connected to $a$ is smaller, or larger, than the sixth edge to be placed opposite $a$.
![](https://cdn.mathpix.com/cropped/2024_05_02_d180a99745c552f671f2g-1.jpg?height=602&width=530&top_left_y=1112&top_left_x=770)

Figure 1

For example, the tetrahedron as shown in Figure 1 does not form with the dimensions $a=44, b=24, c=46, d=23, e=26, f=42$ (units), because the vertex opposite $a$ in the $a, e, d$ triangle is farther from $b$ than the vertex opposite $a$ in the $a, c, f$ triangle (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_d180a99745c552f671f2g-1.jpg?height=547&width=526&top_left_y=1952&top_left_x=802)

Figure 2

It can also be recognized that we do not always get a tetrahedron, because the angles between the 3 edges meeting at one vertex would add up to more than $360^{\circ}$, which is impossible. In the previous example, each of the angles $\gamma, \alpha, \varphi$ between the edges $b, e, d$ is greater than $120^{\circ}$ (the $b, c, e$ and $f, b, d$ faces have also been folded into the plane of the $a, c, f$ face).

II. solution. The three sides of the base triangle can be chosen in $6 \cdot 5 \cdot 4 = 120$ ways, but these are identical in groups of 6, for example, $a b c, a c b, b a c, b c a, c a b, c b a$, so we can form 20 different base triangles. Fixing one such base, the two additional edges of the side triangle that fits on the first base edge can be chosen in $3 \cdot 2 = 6$ ways from the remaining rods - because the two endpoints of the base edge are already distinguished by the other two base edges leading to them - and this also determines the position of the last remaining rod. The $20 \cdot 6 = 120$ frames thus conceived are identical in groups of 4, only 30 are different, because the tetrahedron can stand on any of its 4 faces.

Gábor Petravich (Budapest, Eötvös J. Gymnasium, 3rd year) László Horváth (Hódmezővásárhely, Bethlen G. Gymnasium, 1st year)

Note. We arrive at a completely analogous solution if we group the frames according to the possible selection of edges leading to one vertex, instead of the edges forming one triangle.

Júlia Simon