Can You Crack The Case Of The Crystal Key?

Riddler Classic

The crystal key is a polyhedron with exactly six edges, five of them $1$ inch long; the sixth edge is whatever makes the volume largest. What is that volume?

Solution

Six edges force a tetrahedron, and five unit edges mean two of its faces are unit equilateral triangles sharing a common edge; only the remaining “hinge” edge $x$ (joining the two apexes) is free. Take one triangle as the base, with area $\tfrac{\sqrt3}{4}$, and swing the other about the shared edge: $$V = \frac13 \cdot \frac{\sqrt3}{4}\cdot h,$$ where $h$ is the height of the apex of the second triangle above the base plane. That apex lies at distance $\tfrac{\sqrt3}{2}$ (the triangle’s own height) from the hinge, so its height $h$ is largest, equal to $\tfrac{\sqrt3}{2}$, exactly when the second face stands perpendicular to the base. Then $$V = \frac13 \cdot \frac{\sqrt3}{4}\cdot\frac{\sqrt3}{2} = \boxed{\tfrac18}.$$ Equivalently, writing the volume in terms of the hinge length, $V = \tfrac{x\sqrt{3 - x^2}}{12}$, which peaks at $x = \sqrt{3/2}$, again giving $\tfrac18$. (Contrast the four-unit-edge key, whose maximum was the smaller $\tfrac{\sqrt3}{12}$.)

The computation

Fix five edges at $1$, leave the sixth free, and maximise the Cayley–Menger volume over that one length.

import numpy as np
from itertools import combinations
from scipy.optimize import minimize_scalar
edges = [frozenset(e) for e in combinations(range(4), 2)]
def vol(x, free):
    d = {**{e: 1.0 for e in edges}, edges[free]: x}
    M = np.ones((5, 5)); M[0, 0] = 0
    for i in range(4):
        for j in range(4):
            M[i + 1, j + 1] = 0 if i == j else d[frozenset((i, j))]**2
    return np.sqrt(max(np.linalg.det(M) / 288, 0))
best = max(-minimize_scalar(lambda x: -vol(x, f),
           bounds=(0.01, 1.73), method='bounded').fun for f in range(6))
print(best)                                    # 0.125 = 1/8