## Riddler Express

I recently found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. Here’s a photo I took of this arrangement:

Now, if each of the four cubes has side length 1, then how far above the ground is the bottommost corner of the cube on top?

## Solution

The part of the fourth cube on the top enclosed by the three cubes on the floor is an inverted tetrahedron whose base is an equilateral triangle. Using symmetry it is easy to see that the bottommost corner of the tetrahedron is directly above the center of the base. The triangle formed by the following line segments is a right angled triangle:

1. The line segment from the bottommost corner to the center of the base. Let this be $h$ units.

2. The line segment from the center of the base to the midpoint of a side of the base. The length of this line segment is $1/2\sqrt{3}$ units.

3. The line segment from the midpoint of the base to the bottommost corner along the face of the tetrahedron. This is the hypotenuse of the triangle. The length of this line segment is $1/2$ units. This follows from the fact that the face of the tetrahedron is an right angled isoceles triangle with base $1$ unit.

Therefore, we have $h^2 + (\frac{1}{2\sqrt{3}})^2 = (\frac{1}{2})^2 \implies h = \frac{1}{\sqrt{6}}$.

The center of the base of the inverted tetrahedron is at a height of $1$ unit (the length of the side of a cube) from the ground. The bottommost corner is therefore at a height of $\bf{1-\frac{1}{\sqrt{6}}}$ from the ground.