trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles?

Question

What are the dihedral angles in a disphenoid with four identical triangles, each having one edge of length $2$ and two edges of length $\sqrt{3}$? Tried to look it up, but couldn't find it...

Answer 1

Here's a hedronometric approach ...

Let $a$ and $d$ be the (necessarily opposite) edges of length $2$. These edges are orthogonal to each other in space; moreover, their projections into a mutually-parallel plane form the diagonals of a square of side-length $\sqrt{2}$. This square is a "pseudo-face" of the tetrahedron, and has area $H = 2$.

The other pairs of opposite edges ---say, $b$ & $e$ and $c$ & $f$--- determine the remaining two pseudo-face areas $J$ and $K$. Although the tetrahedron's symmetry dictates that these areas must be equal, the values are a little harder to intuit than $H$. However, the Sum of Squares formula tells us that $$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$ where $W$, $X$, $Y$, $Z$ are the areas of the tetrahedron's (ordinary) faces. In this case, by Heron's Formula, $W = X = Y = Z = \sqrt{2}$. We deduce that $J = K = \sqrt{2}$.

Finally, if we assume that faces $Y$ and $Z$ meet along edge $a$, bounding dihedral angle $A$, and that $Z$ and $X$ meet along edge $b$, bounding dihedral angle $B$, then the tetrahedral Law of Cosines says $$\begin{align} H^2 &= Y^2 + Z^2 - 2 Y Z \cos A \qquad\to\qquad \cos A = \,\,0 \qquad\to\qquad A = \frac{\pi}{2} \\[4pt] J^2 &= Z^2 + X^2 - 2 Z X \cos B \qquad\to\qquad \cos B = \frac{1}{2} \qquad\to\qquad B = \frac{\pi}{3} \end{align}$$

Answer 2

Say the four vertices of the shape are $ABCD$, with $\overline{AB} = \overline{CD} = 2$, and the rest of the segments having length $\sqrt{3}$. Let $O$ denote the midpoint of $AB$, then the triangle $AOC$ is a right triangle with hypotenuse $\sqrt{3}$ and one of the legs $1$. We get that $\overline{OC} = \sqrt{2}$. Similarly, $\overline{OD} = \sqrt{2}$. So the triangle $COD$ is an isosceles triangle with two sides of length $\sqrt{2}$ and the third side of length $2$, which implies that it is a right triangle(!).

So one of the dihedral angles is $90^{\circ}$. Now we need to compute the angle between two faces meeting on a side of length $\sqrt{3}$- for example, between $ABC$ and $DBC$. One way to do this is to assign coordinates to the vertices. From the previous analysis, a particularly nice choice is $A = (-1,0,0)$, $B = (1,0,0)$, $C = (0,\sqrt{2},0)$, $D = (0,0,\sqrt{2})$.

The plane containing $ABC$ is given by the equation $z = 0$, and the plane containing $DBC$ is given by $x\sqrt{2} + y + z = \sqrt{2}$. Then the angle between the planes is equal to the angle between the normal vectors $\mathbf{n}_1 = (0,0,1)$ and $\mathbf{n}_2 = (\sqrt{2},1,1)$:

$$\cos(\theta) = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{1}{1 \cdot 2} = \frac{1}{2}$$

It follows that the other dihedral angle is $60^{\circ}$.

Answer 3

Let $\alpha$ be the vertex angle of a face, and $\beta$ a base angle. Then, by the Law of Cosines: $$\cos\alpha = \frac{-2^2 + \sqrt{3}^2 + \sqrt{3}^2}{2\cdot\sqrt{3}\sqrt{3}} = \frac{1}{3} \qquad \cos\beta = \frac{2^2 - \sqrt{3}^2 + \sqrt{3}^2}{2\cdot 2\sqrt{3}} = \frac{\sqrt{3}}{3}$$ Also, $$\sin\alpha = \frac{2\sqrt{2}}{3} \qquad \sin\beta = \frac{\sqrt{6}}{3}$$

Each vertex of your tetrahedron is surrounded by one $\alpha$ angle and two $\beta$ angles. Let $A$ be the dihedral angle opposite face angle $\alpha$ at a vertex; and let $B$ be opposite a $\beta$. By the Spherical Law of Cosines, we can calculate $$\begin{align} \cos A &= \frac{\cos\alpha-\cos^2\beta}{\sin^2\beta} = \frac{1/3-1/3}{2/3} = 0 \\[4pt] \cos B &= \frac{\cos\beta - \cos\alpha\cos\beta}{\sin\alpha\sin\beta} = \frac{\sqrt{3}/3\,(1-1/3)}{2\sqrt{12}/9} = \frac{1}{2} \end{align}$$

Therefore,

$$A = \frac{\pi}{2} \qquad\qquad B = \frac{\pi}{3}$$

Answer 4

Here's a filthy coordinate-based approach — but one that can be done by just plugging through the coordinate geometry, without needing to remember any formulas substantially more complicated than Pythagoras. As with other answers, note that (by easy symmetry arguments, if nothing else) the edges of length 2 are orthogonal to each other; this means that projecting along the axis mutually orthogonal to both of them, we can use them as axes for the orthogonal plane — in other words (choosing the $Z$ axis as our projection axis and putting one of these edges on the $XY$ plane), the vertices of the tetrahedron are $(\pm1, 0, 0)$ and $(0, \pm1, C)$ for some $C$. Now, Pythagoras gives us $C$: $(\pm1)^2+(\pm1)^2+C^2=(\sqrt3)^2$, or in other words $C=1$. This means the four vertices of the disphenoid can be placed at $(1,0,0)$, $(-1,0,0)$, $(0,1,1)$ and $(0,-1,1)$.

Now, for instance, we can look at one of the faces — e.g., the one with verts $A,B = (\pm1,0,0)$ and $C=(0,1,1)$. Two vectors within this face are $(2,0,0)$ (representing the edge $AB$) and $(1,1,1)$ (representing the edge $BC$), so an orthogonal vector to this face is $(2,0,0)\times(1,1,1)=(0,-2,2)$ and a normal is $\hat{n}_{ABC}=(0, -\frac12\sqrt2,\frac12\sqrt2)$. Similarly, a normal to the face $ABD$ (where $D=(0,-1,1)$) is $\hat{n}_{ABD}=(0, \frac12\sqrt2,\frac12\sqrt2)$, and so the dihedral angle across the edge $AB$ is $\arccos(\hat{n}_{ABC}\cdot\hat{n}_{ABD})=\arccos(0)=\frac\pi2$. Obviously the dihedral across the edge $CD$ is the same, and the dihedrals across any of the 'vertical' edges can be gotten in similar fashion, after finding e.g. a normal to the face $ACD$.