How minimally can I cut a cake sector-wise to fit it into a slightly undersized square tin? 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.
 A: Here’s a solution using three pieces:
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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!)
A: 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.
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A: With John’s configuration, the maximum possible value of $\varepsilon$ is approximately $0.0291842223$:
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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.
A: Here’s a third distinct three-piece configuration, with the centres of the sectors at the vertices of an equilateral triangle:
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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.
