How to analytically find the length of the gap between five tetrahedra

Question

The tetrahedra are regular with side lengths 1 and are joined face-to-face. How do I determine the length $AB\approx0.111111\dots$ analytically?

Answer 1

The very good answer by @Parcly Taxel can be presented with an analytical geometry varnish. Here is how; have a look at the following figure which represents the scene from above.

Let $\theta=\arccos(1/3)$. Let $L=\frac{\sqrt{3}}{2}$.

We are looking for distance $d=AB$ between $A(L,0)$ and $B(L\cos(5\theta),L\sin(5\theta))$. We have:

$$d^2=AB^2=L^2((\cos(5\theta)-1)^2+\sin(5\theta)^2)$$

Letting $c=\cos(5\theta)$, we have:

$$d^2=L^2(c^2-2c+1+1-c^2)=2L^2(1-c)$$

Knowing that $\cos(5(\theta))=\cos(5(\arccos(1/3)))=T_5(1/3)=\frac{241}{243}$

(where $T_n$ is the $n$-th Chebyshev polynomial: see paragraph "Trigonometric definition" in (https://en.wikipedia.org/wiki/Chebyshev_polynomials)), we finally have:

$$d^2=2\frac{3}{4}\left(1-\frac{241}{243}\right)=\frac{1}{81} \ \implies \ d=\frac{1}{9}$$

Answer 2

The first thing we need is the dihedral angle between two faces of a tetrahedron, which is well-known to be $\cos^{-1}\frac13$ radians. Therefore the gap occupies an angle of $\theta=2\pi-5\cos^{-1}\frac13=0.128\dots$ radians.

The length of $AB$ is then a simple matter, because the six points not on the common edge shared by the five tetrahedra lie on a circle of radius $\frac{\sqrt3}2$. It is the length of the chord subtending an angle of $\theta$ at the circle's centre: $$|AB|=2\cdot\frac{\sqrt3}2\sin\frac{2\pi-5\cos^{-1}\frac13}2$$ $$=\sqrt3\sin\left(\pi-\frac52\cos^{-1}\frac13\right)$$ $$=\sqrt3\sin\left(\frac52\cos^{-1}\frac13\right)$$ $$=\sqrt3\sqrt{\frac{1-\cos\left(5\cos^{-1}1/3\right)}2}$$ $$=\sqrt3\sqrt{\frac{1-241/243}2}=\sqrt{\frac3{243}}=\frac19$$ So the length of $AB$ is just the simple fraction $\frac19$. In fact, the decimal expansion you posted pretty much gives it all away.