In the 1980 MAA Annual High School Mathematics Examination (Problem #16), students solved this spatial geometry challenge:

"Four of the eight vertices of a cube are vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron: (A) $\sqrt{2}$ (B) $\sqrt{3}$ (C) $\sqrt{6}$ (D) $3/2$ (E) 2."

Let the cube have edge length $s$. A cube has 6 square faces, so its total surface area is: $$A_{\text{cube}} = 6s^2$$ We can form a regular tetrahedron inside the cube by selecting 4 alternating vertices that do not share any edges of the cube (for example, in 3D coordinates: $(0,0,0)$, $(s,s,0)$, $(s,0,s)$, and $(0,s,s)$).

Notice the length of each edge of this tetrahedron: Every edge connects opposite corners across a square face of the cube. Therefore, each edge of the tetrahedron is a face diagonal of the cube: $$e = \sqrt{s^2 + s^2} = s\sqrt{2}$$ A regular tetrahedron has 4 identical equilateral triangle faces. The area of one equilateral triangle with side $e$ is: $$\text{Face Area} = \frac{\sqrt{3}}{4} e^2 = \frac{\sqrt{3}}{4} (s\sqrt{2})^2 = \frac{\sqrt{3}}{4} (2s^2) = \frac{\sqrt{3}}{2} s^2$$ Multiplying by all 4 faces gives the total surface area of the tetrahedron: $$A_{\text{tetrahedron}} = 4 \times \left(\frac{\sqrt{3}}{2} s^2\right) = 2\sqrt{3} s^2$$ Now, let's take the ratio of the two surface areas: $$\text{Ratio} = \frac{A_{\text{cube}}}{A_{\text{tetrahedron}}} = \frac{6s^2}{2\sqrt{3} s^2} = \frac{3}{\sqrt{3}} = \mathbf{\sqrt{3}} \approx 1.73205$$ Notice that the edge length $s$ cancels out completely!

In this lesson, we write code in Lua to compute the surface areas, verify the cancellation of $s$, and check the ratio against $\sqrt{3}$.