Turtles All The Way Down!

A circle is inscribed in a square of side length 2. The region bound between the square and the circle is shaded. A copy of that figure is inscribed in the circle that is inscribed in the original square, and another copy inscribed in that circle ad infinitum...

What is the total shaded area?

@joe-shmo said
A circle is inscribed in a square of side length 2. The region bound between the square and the circle is shaded. A copy of that figure is inscribed in the circle that is inscribed in the original square, and another copy inscribed in that circle ad infinitum...

What is the total shaded area?

@shallow-blue said
These things are always fun, because at first glance it looks like you're going to have to go into an infinite regression and use differential equations or something like that, but if you look at it right you see that you'll just end up with a single equation where the same ratio figures on both sides.

In this case, the area of the square is 4 and that of the circle i ...[text shortened]... l that's left to do is find that proportion. Have patience, please, while I try to recall my trig...

Interestingly, it turns out that the square inscribed in a circle is exactly twice the area of the circumscribed square. That's because the circle has a radius of 1. If you draw it, you'll see that this means the outer square has a side of length 2, and the inner one a diagonal of length 2; which, by Pythagoras, means that its side is √2. Therefore, the area of the outer square is 4 and that of the inner is 2.

Plugging this into the previous equation, we get S = (4-π ) / (1-proportion[exscribed square/inscribed square]) = (4-π ) / (1-½ ) = (4-π ) / ½ = (4-π ) * 2 = 8 - 2π ≈ 1.7168. Which is slightly lower than the square root of three and slightly higher than the cube root of five, but I don't think either of those are significant as anything but another example of Guy's Strong Law.

More interestingly, as you draw it out, it becomes obvious that it doesn't matter a jot how the squares are rotated within their circles. The ratio remains the same. So the grey area is the same size if all squares are orthogonal, if they alternate orthogonal and diagonal, if they are completely random, or perhaps most æsthetically, if they go in a spiral. I should write some PostScript- or SVG-generating code to show those configurations.

@joe-shmo said
Sure...Take your time ( and the easy way out - What fun is it completing the infinite sum anyway? )

@shallow-blue said
Interestingly, it turns out that the square inscribed in a circle is exactly twice the area of the circumscribed square. That's because the circle has a radius of 1. If you draw it, you'll see that this means the outer square has a side of length 2, and the inner one a diagonal of length 2; which, by Pythagoras, means that its side is √2. Therefore, the area of the outer ...[text shortened]... go in a spiral. I should write some PostScript- or SVG-generating code to show those configurations.

@shallow-blue said
None! It's much more satisfying finding the higher solution. After all, repetitive drudge-work is what we have computers for. I'm a programmer, I should be above that. Discovering the easy way out is my raison d'être.

@joe-shmo said
I didn't mean actually calculate it! I just meant write the first couple terms down and recognize the geometric series and sum via formula!

But either way, yours is probably a maximal efficiency solution...if that's your thing...

😉