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In geometrical terms, what is the smallest number $n$ such that a disc of unit radius can be cut into $n$ sectors that can be reassembled without overlapping to fit into a square of side $2-\varepsilon$, where $\varepsilon>0$ is as small as you like? The question could be set for arbitrary straight cuts; but I like to serve my guests with traditional sector-shaped pieces.

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Problems like this are very difficult to prove that a given solution in optimal. With some geometry one could calculate the smallest square that David Bevan's solution would support, but there might be another 5-piece solution lurking out there. –  Ross Millikan Nov 24 '11 at 15:12
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4 Answers

up vote 4 down vote accepted

With John’s configuration, the maximum possible value of $\varepsilon$ is approximately $0.0291842223$:

$\hspace{1.5in}$ solution

The pieces have angles approximately $228.8124035475^\circ$, $113.1341370910^\circ$ and $18.0534593614^\circ$.

Note that the large pieces are no longer positioned symmetrically.

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This is looks as though it could be the optimal solution: $n=3$ for sure, and this $\varepsilon$ seems hard to beat. But I'll wait a bit before accepting it officially. I'm not expecting any surprises, but then I didn't spot your solution either. –  John Bentin Nov 25 '11 at 18:14
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Here’s a third distinct three-piece configuration, with the centres of the sectors at the vertices of an equilateral triangle:

$\hspace{1.5in}$ 3 pieces

All the pieces have an angle greater than $\frac{\pi}{2}$ (the angles in the diagram are $149.5^\circ$, $110^\circ$ and $100.5^\circ$).

A solution with three congruent pieces doesn’t seem possible.

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When posting this question, I only knew that it could be done for large $n$. Since then, I found a way to do it (I think!) for $n=3$: Cut the disc into sectors of $248^\circ$, $104^\circ$, and $8^\circ$ (slight variations on these numbers are possible). Place the big sector symmetrically in one corner of the square, with the cut-out toward the opposite corner, and put the medium sector symmetrically in the cut-out so that the circular part of its boundary touches the two straight radial edges of the big sector. There is just enough room left to slip the remaining sliver in between a straight edge of the medium sector and the neighbouring side of the square. I hope that this is right, but I may have miscalculated. Any refutation would be stoically welcome. Confirmation is naturally welcome too.

$\hspace{1.5in}$ xxx

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You just beat me to a 3-piece solution! –  David Bevan Nov 24 '11 at 17:33
I've added a diagram (awaiting review). –  David Bevan Nov 24 '11 at 17:46
Let's call it a tie; and thank you for the diagram. It's interesting that our two solutions are so different. –  John Bentin Nov 24 '11 at 19:05
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Here’s a solution using three pieces:

$\hspace{1.5in}$ Cake in tin

Two pieces have angle $\pi-\alpha$ and one has angle $2\,\alpha\,$ for some small $\alpha$ $\;$($\alpha =\frac{\pi}{12}$ in the diagram).

(I eventually got there after starting with a 10-piece solution!)

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This looks like it requires a side of 2. The two big pieces are a full unit tall, no? –  Ross Millikan Nov 25 '11 at 15:01
@Ross: The upper piece is slightly to the right of the lower piece. So their combined height is less than 2. –  TonyK Nov 25 '11 at 17:25
@TonyK: Got it. Thanks –  Ross Millikan Nov 26 '11 at 5:51
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