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?
Using the same setup of as the earlier post, without loss of generality, Dakota will assume that the equilateral triangle is situated along the $xy$-plane with vertices as $(0,0,0),$ $(1,0,0)$ and $(1/2, \sqrt{3}/2,0).$ Let $v = (x,y,h)$ be the remaining vertex. Again, without loss of generality, let's assume that the distance from $(0,0,0)$ to $v$ is $1,$ that is, $$x^2 + y^2 + h^2 = 1.$$ Furthermore, the volume of such a triangular pyramid would be $$V(x,y,h) = \frac{1}{3} B h = \frac{\sqrt{3}}{12} h.$$
Therefore, Dakota now needs to find \begin{align*} \max \,\, & \frac{\sqrt{3}}{12} h \\ \text{s.t.} \,\, & x^2 + y^2 + h^2 = 1,\end{align*} which is solved by the optimal vertex of $\hat{v} = (0,0,1),$ for a maximal volume of $$V_\max = V(1) = \frac{\sqrt{3}}{12}.$$
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