# Volume of a tetrahedron whose 4 faces are congruent.

Suppose that I have a tetrahedron such that all four faces consist of congruent triangles, says with the lengths $a,b$ and $c$ for each side. Is there a beautiful method to compute its volume?

PS. The reason for me tagging calculus and linear algebra is that I figured that the technique used to calculate such a problem may come from this areas.

• It's not beautiful, but Heron's formula generalizes to allow one to find the volume of a tetrahedron given the lengths of its sides. Sep 7 '18 at 8:36
• Thank you, that was very informative. Still, I'd want to see an answer that utilize the symmetry we see here. Sep 7 '18 at 8:54
• See Disphenoid - Wikipedia, referring to John Leech, Some Properties of the Isosceles Tetrahedron, The Mathematical Gazette Vol. 34, No. 310 (Dec., 1950), pp. 269-271. Sep 7 '18 at 13:23
• See also Isosceles Tetrahedron -- from Wolfram MathWorld, referring to Klee, V. and Wagon, S. Old and New Unsolved Problems in Plane Geometry and Number Theory, rev. ed. Washington, DC: Math. Assoc. Amer., 1991. Sep 7 '18 at 14:15
• See also the following recent paper: Konrad Voss, Tetraeder mit inhaltsgleichen Seitenflächen, Elemente der Mathematik, 73 (2018), 45–55. Among other things Voss proves that tetrahedra with faces of equal area are automatically tetrahedra with congruent faces. Sep 9 '18 at 14:28

The volume of a tetrahedron satisfies

$$36V^2 = a^2 b^2 c^2\left(\;1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma\;\right) \tag{1}$$

where $a$, $b$, $c$ are lengths of edges coinciding at a vertex, and $\alpha$, $\beta$, $\gamma$ are the angles between those edges ($\alpha$ between $b$ and $c$, etc).

For the tetrahedron in question, we see that $a$, $b$, $c$ and $\alpha$, $\beta$, $\gamma$ are also elements of a triangle. Since $\alpha+\beta+\gamma = 180^\circ$, the trig factor of $(1)$ reduces, and we have $$36 V^2 = a^2 b^2 c^2 \cdot 4 \cos \alpha \cos \beta \cos \gamma \quad\to\quad 9V^2 = a^2b^2c^2\cos\alpha\cos\beta\cos\gamma \tag{2}$$ which, by the Law of Cosines, we can write as

$$72V^2 = (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2) \tag{3}$$

Another way to get at this result is with the Pseudo-Heron Formula for volume. (See my note, "Heron-like Hedronometric Results for Tetrahedral Volume" (PDF).) If $W$, $X$, $Y$, $Z$ are the face-areas of a tetrahedron, and $H$, $J$, $K$ are the pseudoface-areas (see below) then

$$81V^4 = \left|\;\begin{array}{ccc} H^2 & X Y - W Z & Z X - W Y \\ X Y - W Z & J^2 & Y Z - W X \\ Z X - W Y & Y Z - W X & K^2 \end{array}\;\right| \tag{4}$$

In an equihedral tetrahedron ($W=X=Y=Z$), this reduces to

$$81V^4 = \left|\;\begin{array}{ccc} H^2 & 0 & 0 \\ 0 & J^2 & 0 \\ 0 & 0 & K^2 \end{array}\;\right| = H^2 J^2 K^2 \quad\to\quad 9 V^2 = H J K \tag{5}$$

A pseudoface of a tetrahedron is the quadrilateral shadow of the figure in a plane parallel to two opposite edges. If edges $a$, $b$, $c$ meet at a vertex, and have respective opposite edges $d$, $e$, $f$, then our $H$, $J$, $K$ are related to respective edge-pairs $(a,d)$, $(b,e)$, $(c,f)$, and we have, for instance, $$16 H^2 = 4 a^2 d^2 - \left(\; b^2 - c^2 + e^2 - f^2 \;\right)^2 \tag{6}$$ In the tetrahedron in question, $a=d$, $b=e$, $c=f$, so that \begin{align} 4 H^2 = a^4 - \left( b^2 - c^2 \right)^2 &= \left(\phantom{-}a^2 - b^2 + c^2 \right)\left(\phantom{-}a^2 + b^2 - c^2 \right)\\ 4 J^2 \;\;\phantom{= a^4 - \left( b^2 - c^2 \right)^2} &= \left(\phantom{-}a^2+b^2-c^2\right)\left(-a^2+b^2+c^2\right) \\ 4 K^2 \;\phantom{= a^4 - \left( b^2 - c^2 \right)^2} &= \left( -a^2+b^2+c^2\right)\left(\phantom{-}a^2-b^2+c^2\right) \end{align}\tag{7} and then $(3)$ follows from $(5)$.

Edited to add some hedronometric context to @Calum's answer ...

• The rectangular faces of the tetrahedron's bounding cuboid are exactly the figure's pseudofaces. Indeed, we can deduce that the pseudofaces must be rectangles: each corresponding pseudofacial projection of a tetrahedron with three pairs of opposite congruent edges is necessarily a quadrilateral with two pairs of opposite congruent edges (hence, a parallelogram) and a pair of congruent diagonals (hence, a rectangle).

• Specifically, the rectangular pseudoface $H$ (the projection into a plane parallel to the $a$ edges) has congruent diagonals $a$, and edges $b^\prime := \sqrt{b^2-h^2}$ and $c^\prime := \sqrt{c^2-h^2}$, where $h$ is distance between planes containing the $a$ edges (that is, $h$ is the corresponding "height" of the cuboid, what I call a pseudoaltitude of the tetrahedron). Since $a^2 = (b^\prime)^2 + (c^\prime)^2$ in the rectangle, we have that $h^2=\left(-a^2+b^2+c^2\right)/2$, whence $b^\prime = \sqrt{\left(a^2-b^2+c^2\right)/2\;}$ and $c^\prime=\sqrt{\left(a^2+b^2-c^2\right)/2\;}$. As $H = b^\prime c^\prime$, we reconfirm equation $(7)$.

• As shown in @Calum's cuboid figure, each face of the given tetrahedron is the "hypotenuse-face" of a right-corner tetrahedron whose "leg-faces" are the three half-pseudofaces. (I hadn't really noticed this before!) By de Gua's Theorem , $$\left(\frac12H\right)^2+\left(\frac12J\right)^2+\left(\frac12K\right)^2 = W^2 \quad\to\quad H^2 + J^2 + K^2 = 4 W^2$$ This is consistent with the Sum of Squares identity that holds for any tetrahedron: $$H^2 + J^2 + K^2 = W^2 + X^2 + Y^2 + Z^2$$

• @Calum expresses the tetrahedron's volume in terms of pseudoaltitudes, which I tend to label $h$, $j$, $k$, instead of $x$, $y$, $z$; my $(5)$ expresses the volume in terms of pseudoface areas $H$, $J$, $K$. As it happens, any tetrahedron's volume is given by each pseudoface-pseudoaltitude pair: $$3 V = Hh = Jj = Kk$$ This gives us a bridging relation between the two formulas: $$27 V^3 = H J K \cdot h j k$$

• You might be interested in the paper quoted in my comment to the question. Sep 9 '18 at 14:30
• @ChristianBlatter: Interesting, but I don't have access to the journal. :/ BTW: Proving that "four equal face-areas" implies "four congruent faces" (that is, "three pairs of congruent edges") is easy. If faces $Y$ and $Z$ meet along $a$, which is associated with pseudoface $H$, then it happens that $9a^2 V^2 = [H,Y,Z]$, where $[p,q,r] := (p+q+r)(-p+q+r)(p-q+r)(p+q-r)$. Likewise, the opposite edge $d$ satisfies $9 d^2 V^2 = [H,W,X]$. Consequently, if $W=X=Y=Z$, then necessarily $a=d$ (and, similarly, $b=e$ and $c=f$).
– Blue
Sep 9 '18 at 17:26

If the given triangle is acute-angled, define positive numbers $x, y, z$ by: \begin{align*} x^2 & = \tfrac{1}{2}(b^2 + c^2 - a^2), \\ y^2 & = \tfrac{1}{2}(c^2 + a^2 - b^2), \\ z^2 & = \tfrac{1}{2}(a^2 + b^2 - c^2). \end{align*} Let $\mathbf{u}, \mathbf{v}, \mathbf{w}$ be mutually orthogonal vectors of unit length, and define: $$\mathbf{t}_{ijk} = ix\mathbf{u} + jy\mathbf{v} + kz\mathbf{w} \quad (i, j, k = 0, 1).$$ The cuboid with these 8 vertices has volume $xyz$. It contains a tetrahedron $T$ with vertices $\mathbf{t}_{100}\mathbf{t}_{010}\mathbf{t}_{001}\mathbf{t}_{111}$, and faces which are triangles with the same sides: \begin{align*} a & = \sqrt{y^2 + z^2}, \\ b & = \sqrt{z^2 + x^2}, \\ c & = \sqrt{x^2 + y^2}. \end{align*} The remainder of the cuboid consists of 4 tetrahedra, each with sides $x, y, z$ meeting at right angles at one of the vertices $\mathbf{t}_{000}, \mathbf{t}_{011}, \mathbf{t}_{101}, \mathbf{t}_{110}$, and each therefore having volume $\tfrac{1}{6}xyz$, implying that $T$ has volume $\tfrac{1}{3}xyz$.

Wikipedia illustration (with different notation):

Another Wikipedia illustration, to accompany the comment:

• I can't visualise this (but see Tetrahedral-octahedral honeycomb), but if you extend the dissection to infinity by stacking the cuboids in the obvious way, space becomes honeycombed by polyhedra with all faces congruent to the given triangle. Letting $i,j,k$ take on all integer values, a point with $i+j+k$ even is the centre of an equifacial octahedron of volume $\tfrac{4}{3}xyz$, and one with $i+j+k$ odd is a vertex incident on 12 edges given by selecting two terms from $\pm x\mathbf{u}\pm y\mathbf{v}\pm z\mathbf{w}$ (I think). Sep 8 '18 at 18:23