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I originally found this question in James Stewart's Calculus, specifically in one of the Problems Plus sections.

The question asks how $3$ people can share a pizza while making just $2$ cuts, instead of the usual $3$. The basic idea is to divide a circle into $3$ equal pieces using $2$ parallel lines that are the same distance from the center, as in the picture below.

I am having trouble getting started on this question, mainly because I have no idea how to find the areas of the $3$ pieces. I can see that the cuts should not be too close to the edges or the center, as this would make middle piece too large or too small respectively. But I cannot see much more than this.

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  • $\begingroup$ Did it ask you where to cut, or might it have asked only how to prove that places exist where you can cut to achieve this result? $\endgroup$ – Michael Hardy Jul 13 '16 at 0:20
  • $\begingroup$ I don't have the book anymore, but I'm fairly certain it asks where to cut, as it is intended for an introductory course (or series of courses) in calculus. $\endgroup$ – cpiegore Jul 13 '16 at 0:26
  • $\begingroup$ Proving that such places exist might be less challenging to students in first-semester calculus than actually figuring out where they are. $\qquad$ $\endgroup$ – Michael Hardy Jul 13 '16 at 0:27
  • $\begingroup$ Regardless of what the book says, my question is where to cut. $\endgroup$ – cpiegore Jul 13 '16 at 0:29
  • $\begingroup$ Symmetry tells you you can focus on the top-right quadrant; you want an x-value such that (some integral from 0 to x) is 1/2 (some integral from x to 1). $\endgroup$ – Steve D Jul 13 '16 at 0:34
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Assuming that the pizza has unit radius, the area of the pizza is $\pi$ and each piece has to have area $\frac{\pi}{3}$. So it is enough to find the height of a circle segment such that its area is one third of the original circle, i.e. to solve

$$ \int_{-\sqrt{h(2-h)}}^{\sqrt{h(2-h)}}\sqrt{1-x^2}\,dx - 2(1-h)\sqrt{h(2-h)}=\frac{\pi}{3}$$ in terms of $h$, that with some substitution boils down to Kepler's equation$^{(*)}$, i.e. to a trascendental equation that, in general, cannot be solved in explicit terms, but is easy to solve numerically by Newton's method. In our case we get $h\approx \color{red}{0.735068}\approx\frac{4492}{6111}$, so the three slices have widths approximately proportional to $3:2:3$.


$(*)$: who guessed that planetary motion and pizza slicing are related?

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Picture an $x$-axis running at a right angle to the two vertical lines in your picture.

Draw the ray from the center to the point on the pizza in the upper right were your vertical line on the right intersects the boundary. Let $\theta$ be the angle that that ray makes with the $x$-axis.

Similarly draw a ray from the center to the lower right intersection of a vertical line with the boundary, corresponding to the angle $-\theta$.

Then the length of the intersection of that line with the pizza is $2\sin\theta$ (where the radius of the pizza is $1$).

The fraction of the pizza between those rays is $2\theta/(2\pi)$ of the whole pizza (since $2\theta$ is the angle between those bounding rays and $2\pi$ is the angle encompassing the whole pizza.

The part to the right of that vertical line is what you want to be $1/3$ of the pizza. That part is the whole part between those rays minus the part between those rays that is to the left of that line. The part between those rays to the left of that line is a triangle. The area of a triangle is $\frac 1 2\times\text{base}\times\text{height}$. Call that vertical part of length $2\sin\theta$ the base; then the height is $\cos\theta$. The area of the triangle is therefore $\frac 1 2 (2\sin\theta)(\cos\theta)$.

The area to the right of that vertical line is therefore \begin{align} & \left( \frac{2\theta}{2\pi}\times\text{area of the whole pizza} \right) - (\sin\theta\cos\theta) \\[10pt] = {} & \frac\theta\pi\cdot\pi - \sin\theta\cos\theta \\[10pt] = {} & \theta-\sin\theta\cos\theta\qquad = \theta - \frac 1 2 \sin(2\theta). \end{align}

You want to make that equal to $1/3$ of the whole pizza, thus to $(1/3)\pi$. You can do that by Newton's method.

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  • $\begingroup$ Overall nice answer, but it is a little hard to read. Please consider breaking up the big paragraph. $\endgroup$ – cpiegore Jul 13 '16 at 0:48

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