Proof without words: Twelve corners

Here we are again dealing with a proof without words and the circular number $\pi$ . What is the area of a regular dodecagon that has a circumcircle of radius $r=1$ ? The puzzle you have in front of you provides an answer in an amazingly simple way: Put the pieces together so that they form three squares of equal size with an edge length of 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:

We can easily calculate the area of the regular dodecagon. However, it is not immediately obvious why this results in such a beautiful whole number. Nevertheless, we want to do this here shortly. 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 which each have an acute angle of $\alpha/2=\pi/12$ (they meet 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 — i.e. $a=\sin(\alpha/2)$ — and $h$ is the height — i.e. $h=\cos(\alpha/2).$ Thus we get

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

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 $\pi/3=60^\circ$ , i.e. exactly twice $\alpha$ . Thus the sine of $\alpha$ is exactly $1/2$ , because this is the ratio in which an angle bisector in an equilateral triangle intersects the opposite side.

So the sector with the peak 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? Now — this has to do with the nice 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 that is the reason in the above calculation.

However, if you are looking for a more descriptive reason, we can only refer you 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 that have an area that is a rational number for a radius of $r=1$ . However, the area of each such polygon is always a algebraic number. A famous theorem of the mathematician Carl Friedrich Gauss states that such a regular $n$ -gon is constructible with compass and ruler exactly when 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 are (these are such Prime numbers 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 $\alpha=2\pi/n$ is exactly for these values $n$ a so-called constructible number — that is, one that 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