A tetrahedron's faces are congruent triangles, one of whose angles is $60^{\circ}$. What are the lengths of its edges, if they are integers, and the diameter of the sphere circumscribed around the tetrahedron is $23 \mathrm{~cm} ?$
Let the sides of one face of the tetrahedron be denoted by $a, b$, and $c$. Let the angle opposite to $c$ be $60^{\circ}$. By visualizing the net of the tetrahedron, we can see that the congruence of the faces is only possible if from each vertex, one edge of length $a$, one of length $b$, and one of length $c$ extends. This implies that opposite edges are equal. It is known that a parallelepiped (two) can be circumscribed around any tetrahedron such that the edges of the tetrahedron are the face diagonals of the parallelepiped. In this case, the parallelepiped will be a rectangular cuboid, since the diagonals of opposite faces are equal. Let the edges of the cuboid be $x, y, z$ as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_f96fd13348bad3e75a7dg-1.jpg?height=407&width=506&top_left_y=390&top_left_x=796)

The sphere circumscribed around the cuboid is also the sphere circumscribed around the tetrahedron, as it passes through each vertex of the tetrahedron. Therefore, the space diagonal of the cuboid is $23 \mathrm{~cm}$, and thus

$$
x^{2}+y^{2}+z^{2}=23^{2}
$$

Applying the Pythagorean theorem to the tetrahedron's edges as the face diagonals of the cuboid:

$$
x^{2}+y^{2}=a^{2} ; \quad y^{2}+z^{2}=b^{2} ; \quad z^{2}+x^{2}=c^{2}
$$

Adding these equalities and using (1), we get:

$$
a^{2}+b^{2}+c^{2}=2 \cdot 23^{2}=1058
$$

Applying the cosine rule to one face of the tetrahedron:

$$
c^{2}=a^{2}+b^{2}-a b
$$

This can also be written as:

$$
(a-b)^{2}=c^{2}-a b
$$

from which it follows that $c^{2} \geq a b$. From (2) and (3):

$$
a b+2 c^{2}=1058
$$

Therefore, considering $c^{2} \geq a b$, we have $3 c^{2} \geq 1058$, and thus $c \geq 19$. But $c$ is less than the diameter of the circumscribed sphere, so $19 \leq c<23$. We also see that $c$ cannot be an even number. If $c$ were even, then from (3) it would follow that $a$ and $b$ are also even. But if all three edges are even, then in (2) the left side would be divisible by 4, while the right side would only be divisible by 2. Therefore, $c$ is not even, and thus $c=19$ or $c=21$. Multiply the (4) equation by 3 and add it to (3):

$$
c^{2}+3 \cdot 1058=a^{2}+b^{2}-a b+3\left(a b+2 c^{2}\right), \text { or } 3174-5 c=(a+b)^{2}
$$

This shows that the number $3174-5 c^{2}$ is a perfect square. This is only true for $c=19$, so $c=19$. From the last equation, $(a+b)^{2}=37^{2}$, so $a+b=37$, and from (4) $a \cdot b=336$. From these two equations, $a=16 ; b=21$. The only solution to the problem is therefore

$$
a=16 ; \quad b=21 ; \quad c=19
$$

Based on the work of Péter Erben (Budapest, Berzsenyi D. Gymnasium, 4th grade)

Note. A tetrahedron with congruent faces is often referred to as an equilateral tetrahedron, based on certain similarities.