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Dodecagon star puzzle

June 8, 2025 at 3:19 PM by Dr. Drang

Looking through the June edition of Scientific American this morning, I saw this puzzle (that’s an Apple News link; here’s a regular one):

The regular dodecagon and the blue star inside it both have a side length of 1 unit. What is the area of the star?

Dodecagon with inset star

There’s a clever way to solve this, which is how SciAm does it. I like clever solutions as much as the next person, but being an engineer, I go for brute force when a clever solution doesn’t come to me. In this case, that’s getting the area of the dodecagon and subtracting off the twelve light blue equilateral triangles.

There’s a relatively simple formula for the area of any n-sided regular polygon:

A=ns24cot⁡(180°n)=ns24tan⁡(180°/n) where s is the side length.

We can prove to ourselves that this works by checking some areas we already know. For an equilateral triangle, the area is

3s24tan⁡60°=3s243=34s2

which may not be immediately familiar, but you’d get the same answer by doubling the area of a 30-60-90 triangle with a hypotenuse of s.

For a square, it’s

4s24tan⁡45°=4s24⋅1=s2

And for a regular hexagon, it’s

6s24tan⁡30°=6s24(1/3)=332s2

This is the area of six equilateral triangles, which is how you can form a regular hexagon.

So the area of the star is

Astar=Adodec−12Atri=12⋅124tan⁡15°−1234⋅12

which would be nice if I had the tangent of 15° on the tip of my tongue. Luckily, I do know the trig function values for 30° and the half-angle formulas for tangent are easy to find.

tan⁡θ2=sin⁡θ1+cos⁡θ=1−cos⁡θsin⁡θ

So

tan⁡15°=sin⁡30°1+cos⁡30°=12+3

Because tangent is in the denominator of the area formula, I’m choosing the half-angle formula that puts the square root in its denominator. That’ll put the square root in the numerator of the area formula.

Therefore,

Astar=3(2+3)−33=6

which is pretty darned simple and is obviously the reason the puzzle exists. That this is equal to the area of six unit squares is a decent clue to the clever solution.

You could argue that solving the problem by brute force makes it not a puzzle at all—the whole point of puzzles is to be clever. But doing it this way reminded me of some formulas I’d forgotten1 and, as I said, gave me a clue to the clever solution.

  1. If you woke me from a dead sleep and asked me for the double angle formulas, I’d be able to tell you—at least for sine and cosine. But I’ve never had a good grasp on the half-angle formulas. ↩

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