Volume of regular tetrahedron in a cube

Question

Given a cube with side length $a$, a regular tetrahedron is constructed such that two vertices of the tetrahedron lie on the cube’s body diagonal and the other two vertices lie on the diagonal of one of the faces of the cube. Determine the volume of the tetrahedron.

All I can do is to visualise such a tetrahedron, but unable to see how to get the side of it since it's vertices could be anywhere on the diagonals though at equal distances from the vertices of the cube. If somehow, I can get it's side $l$ , then it's volume would be $\dfrac{l^3}{6\sqrt{2}}$.

Another fact that I know is that volume of a tetrahedron is $1/6$ of the volume of the parallelopiped formed by it's three side vectors, but don't know if it's useful here.

Any help is appreciated!

Thanks.

Answer 1

The key insight is to first determine the minimum distance between the body diagonal and the face diagonal. Without loss of generality let the cube have unit side length and take the body diagonal to be the segment joining $(0,0,0)$ to $(1,1,1)$; then there are six face diagonals that do not intersect the body diagonal. These can be partitioned into two groups of three such that each group lies in a plane, and form the sides of equilateral triangle of side length $\sqrt{2}$ perpendicular to the body diagonal, which passes through the centers of these triangles.

Consequently, the minimum distance between the body diagonal and any such face diagonal is simply $$\frac{\sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{6}}.$$

Since the tetrahedron is regular, with two vertices on the body diagonal and two on the face diagonal, this means distance between two non-adjacent edges of the tetrahedron is $1/\sqrt{6}$. If the side length of the tetrahedron is $2s$, then this implies the distance between $(s, -\frac{1}{2\sqrt{6}}, 0)$ and $(0, \frac{1}{2 \sqrt{6}}, s)$ is $2s$; i.e., we require $$s^2 + \frac{1}{6} + s^2 = 4s^2,$$ or $$s = \frac{1}{2\sqrt{3}}.$$

Therefore, the tetrahedron's volume is $$V = \frac{(2s)^3}{6\sqrt{2}} = \frac{1}{18\sqrt{6}},$$ and the volume for the original cube of side length $a$ is $$V(a) = \frac{a^3}{18 \sqrt{6}}.$$

We could also have found this by noting that the circumscribed cube to the tetrahedron has edge length $1/\sqrt{6}$, thus the edge of the tetrahedron, being also the face diagonal of the circumscribed cube, has length $\sqrt{2}$ times this. And the circumscribed cube's volume is just $\frac{1}{6 \sqrt{6}}$, which is $3$ times the volume of the inscribed tetrahedron, since the four congruent tetrahedra inside the cube but outside the regular tetrahedron each has volume $1/6$ that of the cube. So the desired tetrahedron's volume is again $1/(18 \sqrt{6})$.

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For your understanding and enjoyment, please see the animation below, which illustrates the six tetrahedra that can be thus formed with a single body diagonal. A representative set of coordinates for a single tetrahedron, up to symmetry, is $$\left\{ \left(\frac{6 - \sqrt{6}}{12}, \frac{6 + \sqrt{6}}{12}, 0 \right), \left(\frac{6 + \sqrt{6}}{12}, \frac{6 - \sqrt{6}}{12}, 0 \right), \left(\frac{1}{6}, \frac{1}{6}, \frac{1}{6}\right), \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) \right\}.$$

Answer 2

Following the suggestion in the above comment by "Semiclassical", I considered the body diagonal extending from $(0,0,0)$ to $(1,1,1)$ and the face diagonal from $(1, 0, 1)$ to $(0, 1, 1)$, and I picked points $A$ and $B$ on the body diagonal as follows:

$A = (t_1, t_1, t_1) $

$B = (t_2 , t_2, t_2) $

where $t_1, t_2 \in [0, 1]$ and $t_2 \gt t_1$

and I picked points $C$ and $D$ on the face diagonal symmetrically as follows

$C = (1 - t_3, t_3, 1)$

$D = (t_3, 1- t_3, 1)$

where $0 \le t_3 \le 0.5 $

Next, I set up the equations

$\overline{AB}^2 = \overline{AC}^2 = \overline{AD}^2 = \overline{BC}^2 = \overline{BD}^2 = \overline{CD}^2 $

These lead to three independent equations:

$ 3 (t_1 - t_2)^2 = (t_1 + t_3 - 1)^2 + (t_1 - t_3)^2 + (t_1 - 1)^2$

$3 (t_1 - t_2)^2 = (t_2 + t_3 - 1)^2 + (t_2 - t_3)^2 + (t_2 - 1)^2$

$\sqrt{3} (t_2 - t_1) = \sqrt{2} (1 - 2 t_3)$

Solving with WolframAlpha.com, gives one valid solution:

$t_1 = \dfrac{1}{2}$

$t_2 = \dfrac{5}{6} $

$t_3 = 0.295875854768069$

From this the length of the edge of the tetrahedron is $\sqrt{3} (t_2 - t_1) = \dfrac{\sqrt{3}}{3} $

For a cube of edge length $a$ , the edge length of the tetrahedron will be (by direct scaling)

$e = a \dfrac{ \sqrt{3}}{3}$

Hence, the tetrahedron volume is

$V= \dfrac{a^3 /(3\sqrt{3})}{6 \sqrt{2}} = \dfrac{a^3}{18 \sqrt{6}}$

Answer 3

The regular tetrahedron inscribed in the cube is the cube from which $4$ disjoint tetrahedrons of $\frac{1}{6}$ the cube volume are shaved off. Hence the volume is $a^3( 1 - \frac{4}{6})= \frac{a^3}{3}$.

$\bf{Correction:}$ We just showed above that a regular tetrahedron has volume $\frac{d^3}{3}$, where $d$ is the distance between the opposite edges ( the supporting lines). Now we just follow the solution of @heropup.