In the paper Space Vectors Forming Rational Angles a special set of tetrahedra is mentioned.

"The remaining three are in the R-orbit of the tetrahedron with dihedral angles ($π/7, 3π/7, π/3, π/3, 4π/7, 4π/7$)."

What is a set of edge lengths or vertices for this tetrahedron? I've written a function that converts edges to angles, but I need the reverse.

I solved it in a very messy way: https://community.wolfram.com/groups/-/m/t/2169279. An elegant solution would still be nice.

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    $\begingroup$ Does this even specify a unique tetrahedron? It's not obvious to me that one couldn't arrange these dihedral angles in multiple ways to produce non-congruent solids. $\endgroup$ Jan 21, 2021 at 6:48
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    $\begingroup$ Once you associate five of the angles to five of the edges of a tetrahedron, the sixth dihedral angle is determined. This is unlike six edge lengths, where any six values within certain bounds can form a valid tetrahedron. This lost degree of freedom is because the angles do not specify the scale of the solid. I have no idea what the effect is of different arrangements of the first five angles on the sixth angle, and agree with @RavenclawPrefect that the arrangement might not always be unique. After all, six edge lengths can also produce different non-congruent tetrahedra. $\endgroup$ Jan 21, 2021 at 12:36
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    $\begingroup$ I solved it in a very messy way. community.wolfram.com/groups/-/m/t/2169279 An elegant solution would still be nice. $\endgroup$
    – Ed Pegg
    Jan 21, 2021 at 16:01

2 Answers 2


The original paper already gives the relative configuration of dihedral angles: if $\alpha_{ij}$ is the dihedral angle at the edge between points $i$ and $j$ $$(\alpha_{12},\alpha_{34},\alpha_{13},\alpha_{24},\alpha_{14},\alpha_{23})=\left(\frac\pi7,\frac{3\pi}7,\frac\pi3,\frac\pi3,\frac{4\pi}7,\frac{4\pi}7\right)$$ so the tetrahedron is rotationally symmetric about the line between the midpoints of edges $12$ and $34$.

Now consider four non-collinear points $p,a,b,c$. Let $\angle bpc=A$ and the dihedral angle between planes $apb$ and $cpa$ be $A'$; define $B,B',C,C'$ cyclically. Then Wikipedia gives the following formula: $$\cos A'=\frac{\cos A-\cos B\cos C}{\sin B\sin C}$$ Inverting the three formulas obtained from this by cyclic permutation of $A,B,C$ gives, again with cyclic permutations, $$\boxed{\cos A=\frac{\cos A'+\cos B'\cos C'}{\sin B'\sin C'}}$$ which allows determining the minimal polynomials of the cosines of all face angles: $$\angle412,\angle123:x^3-3x^2-x+\frac{13}7\ (0.708065\dots)$$ $$\angle421,\angle213:x^3-\frac{\sqrt{21}}3x^2-x+\frac{13\sqrt{21}}{63}\ (0.606682\dots)$$ $$\angle142,\angle231:x^3+\frac{\sqrt{21}}3x^2+\frac x3-\frac{\sqrt{21}}{63}\ (0.131776\dots)$$ $$\angle423,\angle314:x^3-\sqrt{21}x^2+\frac{17}3x-\frac{29\sqrt{21}}{63}\ (0.935327\dots)$$ $$\angle243,\angle431:x^3-\frac{\sqrt{21}}3x^2+\frac x3+\frac{\sqrt{21}}{63}\ (-0.131776\dots)$$ $$\angle234,\angle341:x^3+x^2-x+\frac17\ (0.473952\dots)$$ If edge $12$ has length $1$, the laws of sines and cosines finally give minimal polynomials for edge lengths: $$24,31:x^3-\sqrt{21}x^2+\frac{3\sqrt{21}}7\ (0.712358\dots)$$ $$14,32:x^3+2x^2-x-1\ (0.801937\dots)$$ $$34:x^3-9x^2-x+1\ (0.286208\dots)$$


Let a tetrahedron $OABC$ have edges $$a := |OA| \qquad b := |OB| \qquad c := |OC| \qquad d := |BC| \qquad e := |CA| \qquad f := |AB|$$ and dihedral angles $A$, $B$, $C$, $D$, $E$, $F$ along respective edges. Let faces $W$, $X$, $Y$, $Z$ be opposite vertices $O$, $A$, $B$, $C$.

The reader is invited to verify these straightforward expressions for edge-lengths in terms of face-areas and volume:

$$(a, b, c, d, e, f) = \frac{2}{3V}\left( YZ\overline{A}, ZX\overline{B}, XY\overline{C}, WX\overline{D}, WY\overline{E}, WZ\overline{F} \right) \tag{$\star$}$$

where $\overline{\theta} := \sin\theta$. So, "all we have to do" is determine the face-areas and volume. This is not difficult.

Viewing $X$ as the combined shadows of $W$, $Y$, $Z$ (and then viewing $Y$ and $Z$ similarly) gives rise to these analogues of the $a= b\cos C+c\cos B$ relations for a triangle: $$\begin{align} X &= W \ddot D + Y \ddot C + Z \ddot B \\ Y &= X \ddot C + W \ddot E + Z \ddot A \\ Z &= X \ddot B + Y \ddot A + W \ddot F \end{align} \tag1$$ where $\ddot{\theta}:=\cos\theta$. This comprises a simple linear system in $X$, $Y$, $Z$. Solving gives $$\begin{align} X \Delta &= W \left(\; \ddot A \left(-\ddot A \ddot D + \ddot B \ddot E + \ddot C \ddot F\right) + \ddot D + \ddot C \ddot E + \ddot B \ddot F \;\right) \\ Y \Delta &= W \left(\; \ddot B \left(\phantom{-}\ddot A \ddot D - \ddot B \ddot E + \ddot C \ddot F\right) + \ddot E + \ddot A \ddot F + \ddot C \ddot D \right) \\ Z \Delta &= W \left(\; \ddot C \left(\phantom{-}\ddot A \ddot D + \ddot B \ddot E - \ddot C \ddot F \right) + \ddot F + \ddot B \ddot D + \ddot A \ddot E \right)\end{align} \tag2$$ where $\Delta := 1 -2\ddot A\ddot B \ddot C - \ddot{A}^2 - \ddot{B}^2 - \ddot{C}^2$. From these, we can calculate volume: $$V^2 = \frac{2}{9} X Y Z \sqrt{\Delta} \tag3$$ which in turn allows us to find the edge-lengths via $(\star)$. Done! $\square$

For the problem at hand, we take $W=1$ and $$(A,B,C,D,E,F)=\left(\frac\pi7,\frac\pi3,\frac{4\pi}7,\frac{3\pi}7,\frac\pi3,\frac{4\pi}7\right)$$ then throw everything into Mathematica. The exact trig expressions are a mess, but here are the numerical values: $$(a,b,c,d,e,f) = (3.1457\ldots, 2.2409\ldots, 2.5227\ldots, 0.9003\ldots, 2.2409\ldots, 2.5227\ldots)$$ Re-scaling so that $a=1$ gives $$(a,b,c,d,e,f) = (1, 0.7123\ldots, 0.8019\ldots, 0.2862\ldots, 0.7123\ldots, 0.8019\ldots)$$ confirming @Parcly's values.


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