# Domino's thin crust slicing area sans calculus (circle in squares)

If the circle above has radius 2, how to find the areas of A,B,C,D (the portions of the circle inside the squares) without using calculus? Notes:

• A obviously has area 1.

• Using calculus, I find B and D have area $\frac{1}{6} \left(-6+3 \sqrt{3}+2 \pi \right)$ and C has area $\frac{1}{3} \left(3-3 \sqrt{3}+\pi \right)$

• Since these answers are "complex", I suspect there's no easy way to find these areas without calculus. However, I also suspect I may be wrong about my suspicion.

An incorrect solution appears at: http://wordpress.barrycarter.info/index.php/2013/01/31/dominos-pizza-thin-crust-slicing/

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1) Derive formulae for areas such as these, using calculus if desired. 2) Publish them prominently. 3) Poof, future attempts to answer this question can be accomplished without calculus :P –  Michael Grant Apr 16 '13 at 14:58
I know you're joking, but how do you google something like this? I'm sure someone's done it, but I can't find the right terms to google. –  barrycarter Apr 16 '13 at 15:12

I derived the area of B without any calculus - well, that is, assuming that we could find the area of a circular sector without calculus. Anyway, draw radii to the intersections of the circular arc with the grid defining B. Note that the grid has side $1$ and the circle has radius of $2$.

The area of B is the area of the sector defined by the radii, minus the right triangle in sector A, plus the right triangle outside the sector in area B. Note that the angle subtended by the sector at the center of the circle is $\pi/6$. By simple geometry, you can see that the radius angled at $\pi/6$ from the vertical grid splits the grid boundary between A and B into lengths $1/\sqrt{3}$ and $1-1/\sqrt{3}$.

Given this, you may deduce the area of B to be

$$\frac{1}{2} 2^2 \frac{\pi}{6} + \frac{1}{2} \frac{(\sqrt{3}-1)^2}{\sqrt{3}} - \frac{1}{2} \frac{1}{\sqrt{3}} = \frac{\pi}{3} - 1 + \frac{\sqrt{3}}{2}$$

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It's a little simpler than that - area of B = area of sector (of angle $\pi/6$) + area of right triangle (sides $\sqrt3$ and $1$) - area of A, which gives your answer directly. Then the area of D follows, by symmetry, and the area of C can be found from area of quadrant - area A - area B - area D –  user36196 Apr 16 '13 at 16:15
@user36196: OK, I think I follow you. Sure, that works, too. –  Ron Gordon Apr 16 '13 at 16:22
OK, I got it. For some reason, I was looking for right triangles with dimension 1-2-Sqrt[5], which have no superspecial properties. –  barrycarter Apr 16 '13 at 17:58
@MichaelGrant: not sure if you are speaking to me, but it is a good idea to delete a comment when that comment refers to something that someone has fixed since the comment was posted. –  Ron Gordon Apr 16 '13 at 18:21