In his accepted answer to [this question](https://math.stackexchange.com/questions/84896/how-minimally-can-i-cut-a-cake-sector-wise-to-fit-it-into-a-slightly-undersized), David Bevan improved my answer to show that a unit disc can be cut into three sectors which fit into a square of side $2-\varepsilon$, where $\varepsilon\approx0.0291842223$. Now, suppose we drop the requirement that the pieces are sectors, and permit any three pieces whose perimeters are made up of straight lines and parts of the perimeter of the original disc---namely, three pieces formed by dissecting the disc with straight cuts. Allowing this, I can reduce the square to one of side $(1+\surd3)/\surd2$; that is, one of side $2-\varepsilon$, where $\varepsilon\approx0.068148347$. This can be done by cutting off two neighbouring segments of length $\surd2$ on the straight side, say from the left and bottom of the disc, rotating them $90^\circ$ respectively anticlockwise and clockwise, and shifting them so that their circular sides touch the rest of the disc respectively in the upper and lower parts of its top right-hand quadrant. [Anyone with the requisite skills (which I lack) is invited to paste in a diagram here.]

The question is: Is this solution optimal, or can one be found with a larger value of $\varepsilon$? It would be particularly interesting to see an asymmetric solution, like David's solution to the earlier question.

**EDIT**

My understanding of the arrangement of the three pieces - achille.

$\hspace1.5in$ [![3 pieces of a disc](https://i.stack.imgur.com/XLHjG.gif)](https://i.stack.imgur.com/XLHjG.gif)