How Far Down Does The Little Cube Reach?

Three unit cubes lie flat on the ground, their corners touching so as to enclose a small equilateral-triangle gap. A fourth unit cube rests on top, its lowest corner dropping snugly into the gap. How far above the ground is that bottommost corner?

The Riddler, FiveThirtyEight(original post)

Solution

The top cube sits balanced on a corner, its long diagonal vertical and its three top faces symmetric about that axis. Put its bottommost vertex at the origin and slice horizontally at height $z$ above it. The three vertices next to the bottom one lie at height $\tfrac{1}{\sqrt3}$ (the vertical part of a unit edge when the long diagonal is upright) and form an equilateral triangle of side $\sqrt2$, so each horizontal slice is an equilateral triangle whose side grows linearly, $$\text{side}(z) = \sqrt2 \cdot \frac{z}{1/\sqrt3} = \sqrt6\,z.$$ The gap between the three floor cubes is an equilateral triangle of side $1$, sitting at the height of their tops, namely $1$. The cube drops until its slice there fits the gap exactly, $\sqrt6\,z = 1$, so the bottom vertex is $z = \tfrac{1}{\sqrt6}$ below that level. Its height above the ground is $$1 - \frac{1}{\sqrt6} \approx \boxed{0.592}.$$

The computation

Confirm the slice relation: the height $h$ above the bottom vertex at which the triangular slice has side $1$, equivalently the right triangle linking that vertex, the slice’s center and a side midpoint.

import sympy as sp
h = sp.symbols('h', positive=True)
# right triangle: h^2 + (inradius of unit triangle)^2 = (half-face slant)^2
sol = sp.solve(h**2 + (1/(2*sp.sqrt(3)))**2 - sp.Rational(1, 2)**2, h)[0]
print(sol, float(1 - sol))                     # 1/sqrt(6); 0.5918