Area of the overlap

[BMAD]

Suppose you have a unit circle. You cut the circle into four equal parts. Where the radii of one of the four pieces exists (0,0) to (0,1) and (0,0) to (1,0). Putting one of the four pieces on each of the following coordinates, (0,0) , (0,1) , (1,1) , (1,0) and fitting them together so that the radii produce a square causes the interior to have overlapping arcs. There is a piece, in the innermost center, where all four arcs overlap. What is the area of the overlapped center?

[superprismatic]

(pi/3)+1-sqrt(3) which is the definite integral of 4*sqrt(2*x-x^2)-2 from 1-sqrt(3)/2 to 1/2.

[CaptainEd]

The problem as I understand it is to give the area of X if all the circles have radius 1.

.

1 - sqrt(3) + pi/3, same as Superprismatic, only days later...

First we need to learn the areas of some outside areas, A, C, and E

Starting with A, we see that B+A is one fourth of the entire circle (pi/4), and B+2A is an entire square of side 1. So
A = 1 - pi/4, and
B=(pi/2 - 1)

Next, we need G, the area of overlap of side-by-side squares.

Draw a triangle from the intersection of the circles to the two centers (T). Note the little wedges (W) that are left over.
Since T is an equilateral triangle of side 1, its area is (sqrt(3)/4).
Also, its angle is 60 degrees, so T+W is exactly one-sixth of the circle, hence T + W = pi/6.
So W = pi/6 - sqrt(3)/4 = (2*pi - 3*sqrt(3))/12
And G = 2T+4W = (4*pi-3*sqrt(3))/6

Now we can calculate E, using a rectangle around two squares.
Call this rectangle R
R = 2 * 3 = 6
It contains two circles (eachk of area pi), but an overlap of G, and four A corners and 2 E triangles.

R = 6 = 2*pi - G + 4*A + 2*E
2*E = 6 - 2*pi + G - 4*A
= 6 - 2*pi + (4*pi-3*sqrt(3))/6 - 4*(1-pi/4)
= 6 - 2*pi + (4*pi-3*sqrt(3))/6 - 4 + pi

E = 1 - pi/6 - sqrt(3)/4

Finally to calculate X. First consider the square circumscribed around the 4 circles. Area is 9.
Remove the 4 A triangles and the 4 E triangles to get the area that is covered in circles (Y).
Y = 9 - 4*A - 4*E =
Y = 9 - 4*(1-pi/4) - 4*(1-pi/6-sqrt(3)/4)
= 1 + sqrt(3) + 5*pi/3

But Y can be built up from superimposing the circles, as long as we remove double counted areas.

area of two adjacent circles = 2*pi - G
= 2*pi - (4*pi-3*sqrt(3))/6
= 4*pi/3 + sqrt(3)/2

add another circle, remove another G,
= 4*pi/3 + sqrt(3)/2 + pi - (4*pi-3*sqrt(3))/6)
= 5*pi/3 + sqrt(3)

But when we added the third circle, we double counted a tiny C section covered by the opposite two squares, so we'll subtract that out
=5*pi/3 + sqrt(3) - C

Now we add the fourth circle, remove two more Gs, but now we've gone too far and removed a C and the X, so add them back in (the C happily cancels out)

=5*pi/3 + sqrt(3) - C + pi - 2*(4*pi-3*sqrt(3))/6) + C + X

= 4*pi/3 + 2*sqrt(3) + X.

This quantity is also equal to Y, so
1 + sqrt(3) +5*pi/3 = 4*pi/3 + 2*sqrt(3) + X

X = 1 - sqrt(3) + pi/3

Edit: Sorry folks, I don't know why my spoilers didn't hide... :-(

Edit: correct sign on sqrt(3)

Edited February 20, 2013 by CaptainEd

[Yoruichi-san]

Lol nice. Hooray for calculus ;P. I had to explain a similar problem to a kid I was tutoring for SAT II Math and I couldn't use calc...it was a pain (the explaining, not the kid).

[bonanova]

I learned calculus once. I remember using the word integral correctly in a sentence.

For most of my career I used algebra and computers to solve problems. Since retirement,

I differentiate, but only in the logical sense.

I am in awe that SP did what he did; I understand what the Cap'n did.

[CaptainEd]

Glad to please.

For that matter, I'd like to know more about how SP did what he did. My calculus, too, is out in the shed, gathering rust, plant food, and spider webs. But *maybe* I could follow it.

[superprismatic]

Just like Bonanova, I spent most of my career using algebra

and computers to do my job. There were a handful of times
when I had to optimize some function, which I did using a
little differential calculus. So, when I saw this problem,
I decided to try to rub off a little rust from my calculus
and derive the definite integral which solves the problem.
It took longer than I expected. Anyway, here are the
particulars:

I refer to the Captain's first diagram which delineates
the area he calls X. The leftmost point of X is the leftmost
point in common with the two circles centered at (1,1) (whose
equation is (x-1)²+(y-1)²=1) and (1,0) (whose equation is
(x-1)²+y²=1). Solving these simultaneously gives the leftmost
point of X as (1-sqrt(3)/2,1/2). Because of symmetry, I only
needed to integrate to get the left half of X, then double it.
So, I only needed to integrate from x=1-sqrt(3)/2 to x=1/2.
Now for an x in this range, the top curve is from the circle
centered at (1,0). Using its equation, I get that
y=sqrt(2x-x²). Similarly, the bottom curve's equation gives
y=1-sqrt(2x-x²). So, the function to integrate is the
difference of the two which is 2sqrt(2x-x²)-1. As I must
double this to get the full area of X, the thing to integrate
is 4sqrt(2x-x²)-2 from x=1-sqrt(3)/2 to x=1/2.
This is where my real troubles began. In my student
days, I could integrate away like nobody's business. I found,
to my horror, that all that ability had atrophied! After a
few unsuccessful hours, I gave up and decided to install a
math manipulation program on my machine and, thereby, cheat.
I looked around for a good open source program and found
wxMaxima which was intuitively easy to use (I didn't need
to resort to reading documentation!). Within minutes, I had
the integration done.

I really enjoyed the Captain's derivation. It never
occurred to me to try doing it that way. Nice going there,
Captain!

[Yoruichi-san]

Yeah the integral is pretty ugly...I think you have to do it by parts and the second part will end up being a hyperbolic sine :/.

Coming up with the integral is the fun part, IMO...in my college days we had a school-wide license for Mathematica ;P.

[CaptainEd]

Thank you, SP and Y-san, I remember enough about calculus to know what you're saying, and even remember the notion of integration by parts. And I've heard of Mathematica. I can spell wxMaxima. I'll bet it took me more hours to do the geometry than it took you to do the calculus. (My geometry is stored in the mudroom right outside the shed)

Interesting puzzle, BMAD, thanks.