Can You Recrack The Case Of The Crystal Key?

Riddler Classic

Earlier this year, Dakota Jones used a crystal key to gain access to a hidden temple, deep in the Riddlerian Jungle. According to an ancient text, the crystal had exactly six edges, five of which were 1 inch long. Also, the key was the largest such polyhedron (by volume) with these edge lengths.

However, after consulting an expert, Jones realized she had the wrong translation. Instead of definitively having five edges that were 1 inch long, the crystal only needed to have four edges that were 1 inch long. In other words, five edges could have been 1 inch (or all six for that matter), but the crystal definitely had at least four edges that were 1 inch long.

The translator confirmed that the key was indeed the largest such polyhedron (by volume) with these edge lengths.

Once again, Jones needs your help. Now what is the volume of the crystal key?

Solution

Similar to last time, I will use the following Heron-type formula.

There are two cases.

Case I. Opposite Sides are Variable.

Let $U$ and $u$ be variable, and $V = W = v = w = 1$. It follows that:


It follows that:


$$Vol = \dfrac{\sqrt{\left(2u \cdot \sqrt{\left(-U^2+4\right)\left(-u^2+4\right)} - 2Uu^2\right) \cdot \left(4Uu\right) \cdot \left(4Uu\right) \cdot \left(2u \cdot \sqrt{\left(-U^2+4\right)\left(-u^2+4\right)} + 2Uu^2\right)}}{192u}$$
$$Vol = \dfrac{\left(4Uu\right) \cdot 2u}{192u}\sqrt{\left(\sqrt{\left(-U^2+4\right)\left(-u^2+4\right)} - Uu\right) \cdot \left(\sqrt{\left(-U^2+4\right)\left(-u^2+4\right)} + Uu\right)}$$
$$Vol = \dfrac{Uu}{24}\sqrt{\left(\sqrt{\left(-U^2+4\right)\left(-u^2+4\right)} - Uu\right) \cdot \left(\sqrt{\left(-U^2+4\right)\left(-u^2+4\right)} + Uu\right)}$$
Since $U,u$ are symmetric with respect to Volume, we can set $U = u$.

$$Vol = \dfrac{U^2}{24}\sqrt{\left(-U^2+4 - U^2\right) \cdot \left(-U^2+4 + U^2\right)}$$


$$= \dfrac{U^2}{12} \cdot \sqrt{-2U^2+4}$$
Solving for $V' = 0$
$$0 = \dfrac{1}{12} \cdot \left(2U \cdot \sqrt{-2U^2+4}\right) + \dfrac{1}{12} \cdot \left(U^2 \cdot \dfrac{1}{2}\dfrac{-4U}{\sqrt{-2U^2+4}}\right)$$
$$0 = 2U \cdot \left(-2U^2+4\right) + U^2 \cdot\left( \dfrac{1}{2} \cdot -4U \right)$$
$$0 = -3U^2 +4$$
Substituting $U = \dfrac{2}{\sqrt{3}}$ into Volume:
$$Vol = \dfrac{4 / 3}{12} \cdot \sqrt{-2 \cdot \dfrac{4}{3} + 4}$$
$$= \dfrac{1}{9} \cdot \sqrt{\dfrac{4}{3}}$$
$$= \dfrac{2}{9\sqrt{3}} \approx 0.128$$

Case II. Adjacent Sides are Variable.

Let $U$ and $V$ be variable, and $W = u = v = w = 1$. It follows that:


It follows that:


Since $p,q,r,s$ are symmetric with respect to Volume, it follows that $U,V$ are symmetric with respect to volume as well. Therefore we can set $U = V$. It follows that:


$$Vol = \dfrac{\sqrt{\left(4\right) \cdot \left(4\right) \cdot \left(2U\sqrt{-3U^2+12}+2U^2-4\right) \cdot \left(2U\sqrt{-3U^2+12}-2U^2+4\right)}}{192}$$
$$= \dfrac{1}{24} \cdot \sqrt{\left(U\sqrt{-3U^2+12}+\left(U^2-2\right)\right)\left(U\sqrt{-3U^2+12}-\left(U^2-2\right)\right)}$$
Applying difference of squares:
$$= \dfrac{1}{24} \cdot \sqrt{\left(-3U^4+12U^2\right) - \left(U^4-4U^2+4\right)}$$
$$= \dfrac{1}{24} \cdot \sqrt{-4U^4+16U^2-4}$$
$$= \dfrac{1}{12} \cdot \sqrt{-U^4+4U^2-1}$$
Solving for $V' = 0$
$$0 = \dfrac{1}{12} \cdot \left(\dfrac{1}{2}\frac{-4U^3+8U}{\sqrt{-U^4+4U^2-1}}\right)$$
$$= -4U^3+8U = 4U(-U^2+2)$$
Substituting $U = \sqrt{2}$ into Volume:
$$Vol = \dfrac{1}{12} \cdot \sqrt{-4+8-1}$$
$$= \dfrac{\sqrt{3}}{12} = \dfrac{1}{4\sqrt{3}} \approx 0.144$$

Answer

Maximum Volume is $\dfrac{1}{4\sqrt{3}}$, which is slightly greater than the previous Crystal Key volume of $\dfrac{1}{8}.$

Rohan Lewis

2021.09.13