Square And Circle And Square And Circle

Let a square's side length be $x$ . Then, the first inscribed circle within it has radius $x/2$ . Then, the blue area between the first square and the first circle is given by $x^2 - \pi (\frac{x}{2})^2 = x^2(1 - \pi/4)$ . Furthermore, the next square inscribed within this first circle has side length $x/\sqrt{2}$ , and the circle within this square has radius $x/(2\sqrt{2})$ , so the area between the second square and the second circle is $(x/\sqrt{2})^2 - \pi (x/(2\sqrt{2}))^2 = \frac{1}{2}x^2(1 - \pi/4)$ , precisely half the area from the square before. Thus, we have a convergent geometric series; taking the initial square to have side length $s$ , we see that the entire blue area is equal to: $s^2 \left(1 - \frac{\pi}{4} \right) \sum_{i = 0}^{\infty} \left(\frac{1}{2} \right)^i = 2s^2(1 - \frac{\pi}{4})$ using the fact that $\sum_{i = 0}^{\infty} \left(\frac{1}{2} \right)^i = \frac{1}{1 - \frac{1}{2}} = 2$ .

A similar pattern occurs in the yellow areas. Let the radius of a circle be $r$ . The area between this circle and the first square inscribed in it (which side length $r\sqrt{2}$ ) is $\pi r^2 - (r\sqrt{2})^2 = r^2(\pi - 2)$ . Then, the next circle has radius $r/\sqrt{2}$ , and the square within it has side length $r\sqrt{2}/\sqrt{2} = r$ , so the area between these is $\pi(r/\sqrt{2})^2 - r^2 = \frac{1}{2}( r^2(\pi - 2))$ , so again we have a geometric series where each successive term is half the previous. Taking the first circle to have radius $s/2$ (since the first square has side length $s$ ), the yellow area is equal to: $s^2 \left(\frac{\pi}{4} - \frac{1}{2} \right) \sum_{i = 0}^{\infty} \left(\frac{1}{2} \right)^i = 2s^2 \left(\frac{\pi}{4} - \frac{1}{2} \right)$

Thus, which area is larger comes down to which of $1 - \frac{\pi}{4}$ and $\frac{\pi}{4} - \frac{1}{2}$ is larger. A little work by hand reveals that it is the latter, so it is the yellow area which is larger.