Geometric Mystery

Let the radius of the largest circle be $1$ . Then successive radii would be $\dfrac {1} { { \left( \sqrt{2} \right) }^{n} }$ , which means similar areas scaled to such radii would be in successive ratios of $\dfrac{1}{2}$ , starting with $1$ . The infinite sum $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...=1$ , so the blue areas have a total equal to that of the red one.

Let $R$ be the radius of the biggest circle and $S$ be the side length of the biggest inscribed square.

Then the diameter $2R$ is the length of the square's diagonal, so $S^2 + S^2 = (2R)^2 = 4R^2$ .

$S = R\sqrt{2}$

Clearly, the red area = $\pi R^2 - S^2 = \pi R^2 - 2R^2 = R^2(\pi - 2)$

For the second biggest circle, its diameter = $S = R\sqrt{2}$ , so its radius $r$ = $\dfrac{R}{\sqrt{2}}$ .

Again, for the second biggest inscribed square, the diameter equals the diagonal in length, so the side length $s$ can be calculated as:

$s^2 + s^2 = (2r)^2 = 4r^2$

$s = r\sqrt{2} = R$

Therefore, the area of the outermost blue area = $\pi r^2 - s^2 = \pi (\dfrac{R^2}{2}) - R^2 = (\dfrac{R^2}{2})(\pi - 2)$ .

In other words, the outermost blue area is half of the red area, and by evaluating the third biggest circle & square, we will get similar results that the area is progressing as half of the previous region.

Let $T$ denote for the total red area. Then the total blue area = $T[\dfrac{1}{2} + (\dfrac{1}{2})^2 + (\dfrac{1}{2})^3 + ... + (\dfrac{1}{2})^n]$

By using geometric progression, we will get:

$T[\dfrac{1}{2} + (\dfrac{1}{2})^2 + (\dfrac{1}{2})^3 + ... + (\dfrac{1}{2})^n] = T(\dfrac{\dfrac{1}{2}}{1- \dfrac{1}{2}}) = T$

As a result, the blue and the red regions have the same area.