# What is the smallest square into which one can pack a trisected disc?

In his accepted answer to this question, 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.

$\hspace1.5in$

• I added a picture according to my understanding of the arrangement of the pieces. feel free to get rid of it if this is not what you mean. – achille hui Oct 16 '15 at 15:13
• Nice picture! Do I understand correctly that an inportant requirement is that you use exactly THREE pieces? – Vincent Oct 16 '15 at 15:17
• Thank you, @achillehui. That is exactly right. – John Bentin Oct 16 '15 at 15:55
• @Vincent: Yes. But, if this question proves too easy, please consider the case for four pieces. – John Bentin Oct 16 '15 at 16:01

The proposed solution is not optimal: one with a slightly larger value of $\varepsilon$ can be found, as follows. In the diagram, fix the three circles and slide the square upwards and rightwards along its $y=x$ diagonal, without rotation, until the upper left corner of the square meets the upper circle, and simultaneously the lower right corner meets the circle on the right. (The intersections will be respectively at the points $(a,\frac12\surd2+\frac12\surd6+a)$ and $(\frac12\surd2+\frac12\surd6+a,a)$, where $a=\frac14\surd2+\frac14\surd6-\frac12\surd\!\surd3$; however, knowing these values is not needed to follow the present argument.) Produce the lower side of the shifted square leftwards so that it forms a complete chord of the lower circle. This chord cuts off a segment of the disk which is congruent to the segment cut from the right-hand disk by the right-hand edge of the square. The remaining piece cut from the lower disk is a segment truncated by the produced lower edge of the square, perpendicular to its chord; and we correspondingly truncate, at its right-hand end, the corresponding segment of the upper disk that sits on the upper edge of the square, to make a matching piece congruent to the truncated segment from the lower disk.

The picture looks similar to the original one; but the cut-off segments have grown to fit into the upper left and lower right corners of the square, and the top segment has been truncated by a vertical cut which meets its circular arc above the diagonal line $y=x$. Because of this cut, the pieces are no longer jammed in the square: The top piece can be slid a little way to the right, and the right-hand piece can be slid upwards a bit, without touching each other and freeing them also from the large piece remaining at the bottom left corner. Moved thus, we can now slightly scale up the pieces, and they will still fit in the original square. Alternatively, we can reciprocally scale down the square, and the pieces from the unit disk will still fit.

I just come up with a idea, what about taking the unit disk as the limit of inscribed regular polygons sequence.

Consider a inscribed regular n-polygons , cut it into $2n$ equal pieces , every piece is a vertical triangle. And 2 pieces can form a rectangular , the lengths of sides are $sin(\frac {\pi}{2n})$ and $cos(\frac {\pi}{2n})$. Therefore , area of the rectangular is $sin(\frac {\pi}{2n})cos(\frac {\pi}{2n}) = \frac 12sin(\frac {\pi}n)$

Area of all rectangular is $\frac n2sin(\frac {\pi}{n})$.

Then let's put these rectangular in a square , i.e to find a suitable configuration.

I get stuck here , I don't know if this strategy is optimal. However , I find some thing interesting , use MATLAB to compute $\frac nr$ where $r = {\frac {cos(\frac {\pi}{2n})}{sin(\frac {\pi}{2n})}}$ , the ratio of the lengths of sides in rectangular. If $n\rightarrow \infty$ , the value of $\frac nr$ is approaching $\pi$ and $r$ is going to $\infty$

• I don't see how your rectangles, stacked in this way, can be seen as comprised of 3 pieces of the disc of the required form. – John Bentin Oct 16 '15 at 12:37

The solution is not optimal. There is a simple configuration where the side of the square is $a=\frac14(5 + \sqrt 7)\simeq 1.911437$. In the cartesian plane, the coordinates of the vertices of the square are $(0,\,0),$ $(a,\,0),$ $(a,\,a),$ and $(0,\,a)$, and those of the three centers of the circles are $(a-1,\, a-1),$ $(2a-1,\, \frac32),$ and $(\frac32,\, 2a-1)$.