Vtyjkyuk

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.

The volume of a tetrahedron satisfies

$$36V^2 = a^2 b^2 c^2left(;1+2cosalphacosbetacosgamma-cos^2alpha-cos^2beta-cos^2gamma;right) tag1$$

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
$$36 V^2 = a^2 b^2 c^2 cdot 4 cos alpha cos beta cos gamma quadtoquad 9V^2 = a^2b^2c^2cosalphacosbetacosgamma tag2$$
which, by the Law of Cosines, we can write as

$$72V^2 = (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2) tag3$$

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

$$81V^4 = left|;beginarrayccc
H^2 & X Y - W Z & Z X - W Y \
X Y - W Z & J^2 & Y Z - W X \
Z X - W Y & Y Z - W X & K^2
endarray;right| tag4$$

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

$$81V^4 = left|;beginarrayccc
H^2 & 0 & 0 \
0 & J^2 & 0 \
0 & 0 & K^2
endarray;right| = H^2 J^2 K^2 quadtoquad 9 V^2 = H J K tag5$$

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,
$$16 H^2 = 4 a^2 d^2 - left(; b^2 - c^2 + e^2 - f^2 ;right)^2 tag6$$
In the tetrahedron in question, $a=d$, $b=e$, $c=f$, so that
$$beginalign
4 H^2 = a^4 - left( b^2 - c^2 right)^2 &= left(phantom-a^2 - b^2 + c^2 right)left(phantom-a^2 + b^2 - c^2 right)\
4 J^2 ;;phantom= a^4 - left( b^2 - c^2 right)^2 &= left(phantom-a^2+b^2-c^2right)left(-a^2+b^2+c^2right) \
4 K^2 ;phantom= a^4 - left( b^2 - c^2 right)^2 &= left(
-a^2+b^2+c^2right)left(phantom-a^2-b^2+c^2right)
endaligntag7$$
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 $b^prime := sqrtb^2-h^2$ and $c^prime := sqrtc^2-h^2$, 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 $h^2=left(-a^2+b^2+c^2right)/2$, whence $b^prime = sqrtleft(a^2-b^2+c^2right)/2;$ and $c^prime=sqrtleft(a^2+b^2-c^2right)/2;$. 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 , $$left(frac12Hright)^2+left(frac12Jright)^2+left(frac12Kright)^2 = W^2 quadtoquad H^2 + J^2 + K^2 = 4 W^2$$ 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: $$27 V^3 = H J K cdot h j k$$

If the given triangle is acute-angled, define positive numbers $x, y, z$ by:
beginalign*
x^2 & = tfrac12(b^2 + c^2 - a^2), \
y^2 & = tfrac12(c^2 + a^2 - b^2), \
z^2 & = tfrac12(a^2 + b^2 - c^2).
endalign*
Let $mathbfu, mathbfv, mathbfw$ be mutually orthogonal vectors of unit length, and define:
$$
mathbft_ijk = ixmathbfu + jymathbfv + kzmathbfw quad (i, j, k = 0, 1).
$$
The cuboid with these 8 vertices has volume $xyz$. It contains a tetrahedron $T$ with vertices $mathbft_100mathbft_010mathbft_001mathbft_111$, and faces which are triangles with the same sides:
beginalign*
a & = sqrty^2 + z^2, \
b & = sqrtz^2 + x^2, \
c & = sqrtx^2 + y^2.
endalign*
The remainder of the cuboid consists of 4 tetrahedra, each with sides $x, y, z$ meeting at right angles at one of the vertices $mathbft_000, mathbft_011, mathbft_101, mathbft_110$, and each therefore having volume $tfrac16xyz$, implying that $T$ has volume $tfrac13xyz$.

Wikipedia illustration (with different notation):

Another Wikipedia illustration, to accompany the comment: