23
$\begingroup$

Suppose I have a convex set $S\subset \mathbb{R}^2$ of unit area. In fact, I have two congruent copies of $S$ which I would like to bundle together with a rubber band, i.e. take the convex hull $C$ of a disjoint union of these two copies. Assuming I do this in an area-minimizing way, what is the worst case scenario for the area $A$ of the excess space $C\setminus(S\sqcup S')$? Which shapes attain or approach this upper bound?

I can prove that one always has $A \le 1$, by inscribing the set in a rectangle with its diameter spanning the long axis, placing the two rectangles side-to-side, and shaving the excess off of the outer end of each rectangle.

It's easy to see that a circle forces $A=\frac{4-\pi}{\pi}\approx 0.2732$. I believe that a hexagon yields $A=1/3$, as realized by either of these configurations:

two hexagon configurations

Are there shapes that do worse? Better upper bounds on $A$? In the event that this question is resolved, what about the case of three dimensions or higher?

In the case when both copies have the same orientation, and are simply translations of one another, I can prove $A$ is equal to the excess area of the smallest parallelogram containing $S$. (As a consequence, better upper bounds on this excess area provide bounds on $A$.)

These questions seem adjacent to questions of packing density, as they approach such behavior in the limit as the number of copies ($2$ here) goes to infinity.

Update: Regular pentagons sharing an edge yield $A=\frac3{\sqrt{5}}-1\approx0.3416$. I haven't proved this is optimal, though it's superior to meeting at a vertex; it's possible some better pentagon packing yields a lower $A$ (would love to see one if so!).

Update 2020-10-28: After writing some code to compute excess areas for arbitrary convex polygons, I have become more optimistic that the pentagon (and some affine transformations thereof) are maximal for this problem; at the least, I do not think there are any local variations to the shape which make it harder to pack with itself. (I can also use this code to try various families of convex regions, and see if any exceed $0.3416$ - suggestions welcome.)

$\endgroup$
13
  • $\begingroup$ Unless I missed something, optimal circular sectors get you ~0.434. $\endgroup$ – Michael Biro Sep 26 '20 at 20:15
  • 1
    $\begingroup$ On should check the case of a Reuleaux triangle... $\endgroup$ – Jean Marie Sep 26 '20 at 21:04
  • 2
    $\begingroup$ Is there infinite friction between the objects? For example, if I had two rhombi and put them against each other sharing a side, one a translate of the other in a direction parallel to one of its sides, then a rubber band around that configuration has no excess area. But a real rubber band would apply force that would make them slide against each other until one was a translate of the other in a direction perpendicular to a side. Do we could that initial configuration with zero excess area or do we look at the excess area after it settles? $\endgroup$ – alex.jordan Sep 27 '20 at 2:49
  • 1
    $\begingroup$ The initial configuration, apologies for ambiguity. I suppose the title is not properly descriptive, as rubber bands would like to minimize perimeter (perhaps also an interesting question: minimize ratio of band perimeter to perimeter of $S$). Not sure if there's a good physical metaphor for what this post describes. $\endgroup$ – RavenclawPrefect Sep 27 '20 at 3:05
  • 2
    $\begingroup$ The excess area of the smallest parallelogram containing $S$ is at most $1$ and this bound is exact when $S$ is a triangle, see, for instance, Problem 61 in “Geometrical estimations and combinatorial geometry problems” by D.O. Shkyarskiy, N.N. Chentsov, I.M.Yaglom (Moskow, Nauka, 1971, in Russian). $\endgroup$ – Alex Ravsky Oct 23 '20 at 8:12
2
$\begingroup$

By a 1973 result of G. D. Chakerian (PDF link), we know that any convex set $S$ of unit area is contained in a convex quadrilateral of area at most $\sqrt{2}$. It is easy to see that in any convex quadrilateral, there is a side with two adjacent angles whose sum is at most $180$ degrees. (In fact, there are always at least two such sides.)

Then, by taking this quadrilateral and its $180^\circ$ rotation about the midpoint of such a side, we obtain a convex set of area $2\sqrt{2}$ containing two disjoint copies of $S$. So the excess area of the two copies in this configuration is at most $2\sqrt{2}-2\approx 0.8284$.

enter image description here

It is conjectured by W. Kuperberg that the minimal quadrilateral can have area at most $\frac{3}{\sqrt{5}}$, attained by the regular pentagon; if so, this would reduce the bound to $\frac{6}{\sqrt{5}}-2\approx 0.6833$.

I suspect that at least in the $\sqrt{2}$ case, one may be able to refine this result slightly and show that any convex shape taking up at least $\frac1{\sqrt{2}}-\epsilon$ of the quadrilateral can be packed more efficiently somehow, but my attempts thus far haven't proven fruitful.

$\endgroup$
1
$\begingroup$

For regular polygons sharing a side, the following results are obtained. In each case the convex hull area is reported, including the two unit polygons, so that the reader can see the overall size if the hull. The hull sizes are not all that different, but the pentagons do represent the worst case:

$\begin{array}{|c|c|} \hline Triangles & 2.00000 \\ \hline Squares & 2.00000\\ \hline Pentagons & 2.34164\\ \hline Hexagons & 2.33333\\ \hline Heptagons & 2.26829\\ \hline Octagons & 2.20711 \\ \hline Enneagons & 2.28686\\ \hline Decagons & 2.29443\\ \hline Hendecagons & 2.27374\\ \hline Dodecagons & 2.24402\\ \hline \end{array}$

Note that there is a minimum at each multiple of four sides. This may be interpreted geometrically: when the number of sides in each polygon is a multiple of four, there are sides oriented perpendicularly to the shared side, and these fit flush against a pair of parallel "bases" above and below the polygons. This flush fit tightens the convex hull and thus reduces its area. However, the effect of this fit diminishes with more sides as the perpendicularly oriented sides (like all sides) become smaller. The limiting value for infinitely many sides (circles), of course, is $1+4/\pi\approx 2.27324$.

For the physical process of wrapping a rubber band, one might suppose that a greater perimeter makes the wrapping harder. Let's look at this case. Again the polygons are normalized to unit area, and we get:

$\begin{array}{|c|c|} \hline Triangles & 6.07868 \\ \hline Squares & 6.00000\\ \hline Pentagons & 5.94984\\ \hline Hexagons & 5.87156\\ \hline Heptagons & 5.83719\\ \hline Octagons & 5.83809 \\ \hline Enneagons & 5.83607\\ \hline Decagons & 5.82418\\ \hline Hendecagons & 5.81672\\ \hline Dodecagons & 5.81701\\ \hline \end{array}$

Here the limiting perimeter for a pair of circles is $2(\pi+2)/\sqrt\pi\approx5.80167$, and the polygonal figures decrease monotonically towards this limit ... almost. Aside from the worst case and second worst case, triangles and squares respectively, we identify a maximum perimeter at every multiple of four sides, whereas the areas noted above are minimized. The same geometric factor is responsible: the same flush fit to a pair of bases that reduces the area also forces the perimeter to stretch around the base-fitting sides.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.