Area of the overlap revisited

[BMAD]

Back around the time I first joined brainden I posted a favorite puzzle of mine:

It didn't take long for the brainden elite to crack it using calculus. Surprisingly and grateful though, I also saw the use of geometry (my favorite subject!!) to crack it too and as a result, I have been hooked to this site ever since.

I am now modifying the problem slightly. The challenge here of course is to not use calculus except to verify if needed. Here we go:

Suppose the area of the innermost intersection is Pi, can we extrapolate back to the circle's original radius?

Edited February 27, 2015 by BMAD

[gavinksong]

I mean, obviously the answer is yes. Like bonanova said, everything just scales.
Anyways, I hope you don't mind if I try my hand at a geometry solution as well.
I haven't read the previous solution, so apologies if mine is redundant.

 
Let's break down the picture. The red area is easy to calculate. It is the overlap between two arcs that together cover entire square - so adding two quarters of the area of the circle together should be larger than the area of the square by exactly the red area. It is π/2 - 1.

 

The blue area is also easy to calculate. It fits an equilateral triangle. Thus, adding two sixths of the area of a circle and subtracting the overlapping area in the shape of an equilateral triangle gives us the blue area. It is π/3 - √(3)/4.

 

There are really four blue areas and two red areas. Adding the blue areas and subtracting the red areas gives us the area of the square plus the centerpiece (can you see why?). Thus, the centerpiece's area is 4(π/3 - √(3)/4) - 2(π/2 - 1) - 1.

 

This simplifies to 1 - √3 - π/3.

[bonanova]

Re-reading the previous dialog makes me miss SuperPrismatic. And Cap'n Ed.

[BMAD]

They are greatly missed

[bonanova]

Distances scale as the square root of area.

Area scale = pi / (1 - sqrt(3) + pi/3 ) = k. 

 

New radius = sqrt(k)

[gavinksong]

[gavinksong]

I finally got around to reading the old geometry solution by CaptainEd.

It was quite confusing, but it seems our first two steps were exactly the same: solving for the red and blue areas.

However, while I calculate the result in just one more step, the old solution appears to involve at least half a dozen more constructions.

So it looks like I've found a much more trimmed-down proof.

 

That said, BMAD, why don't you mark this thread as solved?

The answer is clearly that the radius is √(π / (1 - √3 - π/3)), is it not?

[BMAD]

It is solved. People stop reading forums once they are marked that way I found, so I tend to draw out marking them as such right away.