6. Find all tetrahedra that have a net in the shape of a deltoid and exactly four edges of a given length a. (A deltoid is understood to be a convex quadrilateral symmetric with respect to only one of its diagonals; thus, neither a square nor a rhombus is considered a deltoid.)
SOLUTION. In the first (more significant) part of the solution, we will find all tetrahedra that have a net in the shape of a kite; then it will be relatively easy to determine which of the found tetrahedra have exactly four equal edges.

Consider, therefore, an arbitrary tetrahedron $ABCD$ and describe the lengths of its edges with the letters $x, y, z, u, v, w$ as shown in Fig. 3. We will divide all nets of the tetrahedron $ABCD$ into two groups. The first group will include those nets in which one face of the tetrahedron is adjacent to the other three faces;
the second group will include the remaining nets, in which each face is adjacent to at most two other faces. Since we have not specified the labeling of the tetrahedron's vertices in advance, we will further consider only one net from each of the two groups, namely the nets depicted in Figs. 4 and 5. We will deal with each of them separately.

The net in Fig. 4 is (in general) a hexagon $AD_3BD_1CD_2$, it will be a quadrilateral only if two of its angles at the vertices $A, B, C$ are right (i.e., have a size of $180^\circ$). It is clear that a right angle cannot be at any of the vertices $D_1, D_2, D_3$. Considering the already mentioned arbitrariness of labeling, let us assume that the angles $D_2AD_3$ and $D_3BD_1$ (marked in Fig. 4) are right. Our net is then a quadrilateral $D_2D_3D_1C$, whose sides (in the order in which they cyclically follow each other) have lengths $2u, 2v, w$, and $w$. If this quadrilateral is a kite (and not a rhombus), it must certainly be true that $u=v$ and $2u \neq w$ (Fig. 6a). From the axial symmetry with respect to the line $D_3C$, we then find that $y=x$; the tetrahedron with the "kite-shaped" net from Fig. 6a is shown in Fig. 6b. It is a tetrahedron symmetric with respect to the plane of symmetry of the edge $AB$. Let us add that

![](https://cdn.mathpix.com/cropped/2024_04_17_e538feb3203e564d0a7dg-13.jpg?height=488&width=458&top_left_y=350&top_left_x=1381)

Fig. 3 in addition to the inequality $2u \neq w$, the inequality $z \neq x$ must also hold; in the net from Fig. 7a, it must be true that $2u = x + v$ and $x = v$, hence $x = u = v$. In the triangle $D_2D_1C$, the segment $AB$ is the midline (Fig. 7a), so $w = |D_2C| = 2|AB| = 2z$. The corresponding tetrahedron is shown in Fig. 7b.

![](https://cdn.mathpix.com/cropped/2024_04_17_e538feb3203e564d0a7dg-14.jpg?height=490&width=607&top_left_y=590&top_left_x=412)

Fig. 6a

![](https://cdn.mathpix.com/cropped/2024_04_17_e538feb3203e564d0a7dg-14.jpg?height=479&width=613&top_left_y=1300&top_left_x=412)

Fig. 7a

![](https://cdn.mathpix.com/cropped/2024_04_17_e538feb3203e564d0a7dg-14.jpg?height=341&width=483&top_left_y=740&top_left_x=1149)

Fig. 6b

![](https://cdn.mathpix.com/cropped/2024_04_17_e538feb3203e564d0a7dg-14.jpg?height=336&width=483&top_left_y=1442&top_left_x=1149)

Fig. 7b

Let's summarize the results of our previous considerations: Only two types of tetrahedra (Figs. 6b and 7b) have a net in the shape of a kite. Our task is now to determine when these tetrahedra have exactly four equal edges (of a given length $a$). Let's first consider the tetrahedron from Fig. 6b, whose edges have lengths $x, x, z, u, u, w$. Suppose, therefore, that exactly four of them are equal to $a$, which ones are they? First, it must be true that $x = a$, otherwise it would have to be true that $a = z = u = w$, which is in contradiction with the inequality $z < 2u$, hence $x > z$. In the same net, the triangle $ABC$ has an obtuse interior angle at vertex $B$, because its exterior angle $ABD_1$ is the interior angle at the base $AB$ of the isosceles triangle $ABD_1$, so it is necessarily acute. Therefore, the longest side of the triangle $ABC$ is the side $AC$, which we can write as: $y > \max \{x, z\}$. Together, we get $y > x > z$, with respect to the equality $x = a$, it follows that $2z = a$. The tetrahedron $ABCD$ is thus uniquely (up to congruence) determined by the found conditions. The length of the last edge $AC$ we calculate as the median to the side $D_1D_2$ of the triangle $CD_1D_2$ with sides $2a, 2a, a$. We get $y = \frac{1}{2}a\sqrt{6}$. The solution to our problem is the unique tetrahedron from Fig. 8a, and its net in the shape of a kite is in Fig. 8b.

![](https://cdn.mathpix.com/cropped/2024_04_17_e538feb3203e564d0a7dg-15.jpg?height=374&width=504&top_left_y=908&top_left_x=366)

Fig. 8a

![](https://cdn.mathpix.com/cropped/2024_04_17_e538feb3203e564d0a7dg-15.jpg?height=336&width=794&top_left_y=949&top_left_x=920)

Fig. 8b

Answer. The sought tetrahedron is unique: its three edges of length $a$ emanate from one vertex, the edges of the opposite face have lengths $a, \frac{1}{2}a, \frac{1}{2}a\sqrt{6}$. One of the nets of this tetrahedron has the shape of a kite with sides $a, a, 2a, 2a. ^{6}$[^0]


[^0]:    5 In the case $w = z$, the "kite-shaped" net from Fig. 6a has a right angle at vertex $C$, so it is not a kite but a triangle.

    6 We recommend that solvers cut out such a kite from paper and then fold the model of the tetrahedron.