How minimally can I cut a cake sector-wise to fit it into a slightly undersized square tin?
Solution 1:
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!)
Solution 2:
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.