## Riddler Classic

Dakota Jones is back in action! To gain access to a hidden temple deep in the Riddlerian Jungle, she needs a crystal key.

She already knows the crystal is a polyhedron. And according to an ancient text, it has exactly six edges, five of which are $1$ inch long. Cryptically, the text does not specify the length of the sixth edge. Instead, it says that the key is the largest such polyhedron (i.e., with six edges, five of which have length $1$) by volume.

Once again, Dakota Jones needs your help. What is the volume of the crystal key?

## Solution 1

Given the distances between the vertices of a tetrahedron the volume can be computed using the Cayley–Menger determinant:

$$288\cdot V^{2}= \begin{vmatrix} 0&1&1&1&1\\ 1&0&d_{12}^{2}&d_{13}^{2}&d_{14}^{2}\\ 1&d_{12}^{2}&0&d_{23}^{2}&d_{24}^{2}\\ 1&d_{13}^{2}&d_{23}^{2}&0&d_{34}^{2}\\ 1&d_{14}^{2}&d_{24}^{2}&d_{34}^{2}&0 \end{vmatrix}$$

where the subscripts $i, j \in \{1, 2, 3, 4\}$ represent the vertices and $d_{ij}$ is the pairwise distance between them – i.e., the length of the edge connecting the two vertices.

If $a$, $b$, $c$ be the three edges that meet at a point, and $x,y,z$ the opposite edges. The volume $V$ of the tetrahedron is given by

$$V = \frac{\sqrt{4a^2b^2c^2-a^2X^2-b^2Y^2-c^2Z^2+XYZ}}{12}$$

where

\begin{aligned} X = b^2 + c^2 - x^2 \\ Y = a^2 + c^2 - y^2 \\ Z = a^2 + b^2 - z^2 \end{aligned}

In our case, let $a=b=c=y=z=1$, we have

\begin{aligned} X = 2 - x^2 \\ Y = 1 \\ Z = 1 \end{aligned}

We need to find the value of $x$ that maximizes

$$V = \frac{\sqrt{4- (2-x^2)^2-1-1+(2-x^2)}}{12} = \frac{x\sqrt{-x^2 + 3}}{12}$$

$V$ attains the maximum value of $\frac{1}{8}$ when $x = \sqrt{\frac{3}{2}}$.

## Solution 2

A polyhedron with $6$ edges has to be a tetrahedron. In this particular case, we have a tetrahedron where two faces are equilateral triangles of side length $1$ which share a common edge. The volume of tetrahedron (which is a triangular pyramid where the base is an equilateral triangle) is given by

\begin{aligned} Vol &= \frac{1}{3} \times base \times height \\ &= \frac{1}{3} \times \frac{\sqrt{3}}{4} \times height \end{aligned}

The volume is maximized when the height is maximized i.e. when the second equilateral triangular face which shares an edge with the base is perpendicular to the base. The height of an equilateral triangle of side $1$ is $\frac{\sqrt{3}}{2}$.

Therefore the volume of the crystal key is

$$Vol_{max} = \frac{1}{3}\times \frac{\sqrt{3}}{4} \times \frac{\sqrt{3}}{2} = \frac{1}{8}$$

## References

https://en.wikipedia.org/wiki/Tetrahedron