Proof without words: Twelve corners

This is again a proof without words and the circle number \pi. What is the area of a regular dodecagon with a radius r=1? The puzzle you have in front of you provides an answer in an amazingly simple way: Put the pieces together in such a way that they form three equal squares of edge length one. Since equality of content follows from equality of decomposition, the area of the given dodecagon must be exactly A=3. This is also close to the circle number \pi, which is the area of the unit circle (and 3\lt\pi).


And now … the mathematics of it:

We can easily calculate the area of the regular dodecagon. However, it is not immediately obvious why this results in such a nice integer. Nevertheless, we want to do this briefly here. To do this, we divide the dodecagon into twelve equal sectors — as in Figure 1 below. Each of these sectors is an isosceles triangle whose angle at the apex is equal to \alpha=2\pi/12=\pi/6. This is composed of two equal right triangles whose hypotenuse has length r=1 and each has an acute angle of \alpha/2=\pi/12 (they join along the bisector of \alpha). The area of one of these triangles A_\Delta is thus given by

    \[A_\Delta=\frac{a\cdot h}{2},\]

where a is the length of the base side — i.e. a=\sin(\alpha/2) — and h is the height — i.e. h=\cos(\alpha/2). This gives us

    \[A_\Delta=\frac{\sin(\alpha/2)\cos(\alpha/2)}{2}.\]

Now we can use the addition theorem for the sine and in this way we get

    \[A_\Delta=\frac{2\sin(\alpha/2)\cos(\alpha/2)}{4}=\frac{\sin(2\alpha/2)}{4}.\]

But what is the sine of the angle \alpha? To do this, observe that the interior angle at each corner of an equilateral triangle is exactly \pi/3=60^\circ, exactly twice \alpha. Thus the sine of \alpha is exactly 1/2, because this is the ratio at which an angle bisector in an equilateral triangle intersects the opposite side.

So the sector with the apex of the angle \alpha has exactly the area A_{S}=2A_\Delta=2\cdot1/2\cdot1/4=1/4.

So if we take all twelve sectors together, we get exactly an area of 12\cdot 1/4=3 for the dodecagon.

But why is this area such a beautiful number? Well — this has to do with the beautiful properties of the angle \alpha=\pi/3=60^\circ, because via this we get the value \sin(\alpha)=\sin(30^\circ)=1/4. At least this is the reason in the above calculation.

Figure 1: The “Twelve Corners” exhibit

However, if you are looking for a more descriptive reason, we can only refer here to the explanation provided by the exhibit itself. It should be noted at this point that the square and the dodecagon are the only regular polygons which, for a radius of r=1, have an area which is a rational number. However, the area of any such polygon is always an algebraic number. A famous theorem of the mathematician Carl Friedrich Gauss states that such a regular n-gon can be constructed with compass and ruler exactly if the number n can be written as n=2^k p_1\cdots p_l, where k\geq 0 is an integer and p_1,\ldots,p_l are so-called Fermat primes (those primes which can be written as p=2^{2^e}+1, named after the mathematician Pierre de Fermat). This is due to the fact that \sin(\alpha) with \alpha=2\pi/n exactly for these values n is a so-called constructible number — i.e. one which can be represented as an expression with only integers, +, - and square roots \sqrt{\cdot}.


Literature

[1] https://de.wikipedia.org/wiki/Regelmaßiges_Polygon

[2] https://de.wikipedia.org/wiki/Konstruierbares_Polygon

[3] https://de.wikipedia.org/wiki/Zwölfeck#Regelmäßiges_Zwölfeck