How does the volume of a regular tetrahedron relate to that of a regular octahedron with the same edge length?
Let's label the sides of the regular tetrahedron $ABCD$ and mark the points with the numbers $1, 2, 3, 4, 5, 6$. These points divide the tetrahedron into 5 smaller solids.

![](https://cdn.mathpix.com/cropped/2024_05_02_5d33dd90e3273798f32fg-1.jpg?height=439&width=458&top_left_y=239&top_left_x=823)

Let the edge length of the tetrahedron $ABCD$ be $2a$. Then the tetrahedra $A146$, $B251$, $C326$, and $D453$ will be regular tetrahedra, each with an edge length of $a$, which is half the edge length of the original tetrahedron. The solid $123456$ will be a regular octahedron, as each of its faces is a congruent equilateral triangle with an edge length of $a$. Since the small tetrahedra are similar to the large one and their edges are half the length of the large one, their volume will be one-eighth of the large one. (The volumes of two similar solids are in the same ratio as the cubes of corresponding distances, such as the edges.) Denoting the volume of the small tetrahedra by $K_t$ and the volume of the octahedron by $K_0$, we have:

$$
4 K_{t} + K_{0} = 8 K_{t} \quad \text{which means} \quad K_{0} = 4 K_{t}
$$

Thus, the volume of the regular octahedron is four times the volume of a regular tetrahedron with the same edge length.