Maximum volume convex body coverable by a unit square - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T04:04:24Z http://mathoverflow.net/feeds/question/96102 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96102/maximum-volume-convex-body-coverable-by-a-unit-square Maximum volume convex body coverable by a unit square Joseph O'Rourke 2012-05-06T01:24:03Z 2012-05-06T16:41:51Z <p>Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, <a href="http://mathoverflow.net/questions/95867/" rel="nofollow">Covering a Cube with a Square</a>.)</p> <p>The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \sqrt{\pi}} \approx 0.094$.</p> <p>In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in <em><a href="http://gfalop.org/" rel="nofollow">Geometric Folding Algorithms: Linkages, Origami, Polyhedra</a></em>, Demaine, O'Rourke, Cambridge, 2007.]: <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://cs.smith.edu/~orourke/MathOverflow/MaxVol.jpg" alt="Max Volume Polyhedron"><br /> The seeming randomness of this shape discouraged my further pursuit.</p> <p>The case $k=2$ resembles the famous <a href="http://en.wikipedia.org/wiki/Paper_bag_problem" rel="nofollow">tea-bag problem</a>, but here I am restricting attention to convex bodies.</p> <p>As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a <a href="http://en.wikipedia.org/wiki/Geodesic_dome" rel="nofollow">geodesic dome</a>.</p> <p>I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$?</p> <p><b>Update</b>. Here is my interpretation of Gerhard's suggestion for $k=2$: <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://cs.smith.edu/~orourke/MathOverflow/SquareBox221.jpg" alt="Square to Box"></p>