Find the height of a regular tetrahedron with edge $a$.

#
Let $A B C D$ be a regular tetrahedron, $D O$ its height, and $M$ the midpoint of $B C$. Since $A M$ is the height of the equilateral triangle $A B C$, and $O$ is the center of this triangle, we have

$$
A M = a^{\frac{\sqrt{3}}{2}}, \quad O A = \frac{2}{3}, \quad A M = \frac{a \sqrt{3}}{3}
$$

From the right triangle $O A D$, we find that

$$
D O = \sqrt{D A^{2} - O A^{2}} = \sqrt{a^{2} - \left(\frac{a \sqrt{3}}{3}\right)^{2}} = a \sqrt{1 - \frac{1}{3}} = a \sqrt{\frac{2}{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-34.jpg?height=105&width=1312&top_left_y=279&top_left_x=1)

How many different pyramids exist, all of whose edges are equal to 1?

## Solution

Consider the $n$-sided angle at the vertex of a regular $n$-sided pyramid, all of whose edges are equal to 1. The lateral faces of such a pyramid are equilateral triangles with side length 1. Therefore, all the plane angles of the considered $n$-sided angle are $60^{\circ}$. Since the sum of the plane angles of a convex polyhedral angle is less than $360^{\circ}$, we have $60^{\circ} \cdot n < 360^{\circ}$. Thus, $n < 6$. Therefore, the condition of the problem can only be satisfied by three pyramids: a triangular pyramid (regular tetrahedron), a quadrilateral pyramid, and a pentagonal pyramid. It is possible to construct a quadrilateral and a pentagonal pyramid, all of whose edges are equal to 1.

## Answer

Three.