Curvy Areas

Herschel, from the European School of Varese, sent us this solution:
 
First, notice that the picture is made of two halves, with semicircles of various sizes in each half of the picture. Each coloured shape is made from a larger semicircle with a smaller one cut out, and the "band" (coloured curvy shape) created this way is joined to one in the other half of the picture, making the wavy shapes.

To work out the areas of the various coloured bands, we must work out the area of each half of the shape. For simplicity, we'll take the smallest semicircles (the red and blue ones at either side of the picture) to have an area of 1 unit, so the succesively bigger semicircles have areas of $2^2, 3^2, 4^2 \text{ and }5^2$ (including the cut-out region).

Let's take the example of the green band. To work out the area of the top half, we will take away the area of the yellow semicircle (including the cut-out region) from the green semicircle, so the area of the top of the green band is $4^2-3^2 = 16 - 9 = 7$ units of area.
Likewise, we work out the area of the bottom half to be $2^2 - 1^2 = 4 - 1 = 3$.

The total area of the green shape is $7+3=10$ units (where, if you recall, 1 unit is the area of the top half of the red shape).

Now we'll generalise: we'll take a circle where the number of coloured bands is $x$ and we'll pick the $n^{th}$ band from the left (so in the previous example, $x=5$ and our for our green band, $n=4$).

So, we need to separately calculate the top and bottom halves.
The area of the top half will be $A_1 = n^2 - (n-1)^2$ (like the green area was $4^2 - 3^2$).
The area of the bottom half will be $A_2 = (x+1-n)^2 - (x-n)^2$ (that makes the green area $2^2 - 1^2$).

Add the two areas together and we get $A = A_1 + A_2 = [n^2 - (n-1)^2] + [(x+1-n)^2 - (x-n)^2]$

Now it's just a case of working through the algebra:
Expand the brackets: $A = [n^2 -n^2 +2n -1] + [x^2 +x -xn +x +1 -n -xn -n +n^2 - (x^2 -2xn +n^2)]$
Regrouping the like terms and cancelling out, we end up with:
$A = [2n-1] + [x^2 +2x -2xn +1 -2n +n^2 - x^2 +2xn -n^2]$

Cancelling the remaining parts we finally get: $A = 2x$

As you can see, the $n$ terms have been eliminated - all the bands will have an area of $2x$, where $x$ is the number of coloured wavy shapes. This agrees nicely with the area of the green shape we worked out above - the area was 10, and indeed, the number of colours in the example image is 5, so $2x=2 \times 5 = 10$.

So, to conclude, the area of each of the "waves" in a shape such as this is $2 \times$ [the number of wavy shapes]$\times$ [the area of the smallest semicircle].

All the waves in a given image will have the same area.

Throughout this process, we have defined 1 unit to be the area of the smallest semicircle. We could well have decided to define 1 unit to be the radius of the smallest semicircle; indeed, this would appear to be more obvious a unit to use. However, we would then have had to use the formula $A=\pi r^2$, which would unnecessarily clutter and complicate the already tedious algebra, only to get a similar answer. This answer would in fact be $x \times \pi$ square units, since $2x$ semicircles is $x$ whole circles, and as mentioned, the area of a circle = $\pi r^2$, so a radius of 1 means an area of $\pi$. Regardless of which units we choose, the fact remains that all the bands in a circle have exactly the same area, which is $\frac{1}{x}$ of the area of the complete circle (thus the green band in the example image takes up exactly 1/5 of the total area of the circle, as do the red, orange, yellow and blue bands).