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It is an unsolved problem to determine the "thinnest" $2$-fold covering of the plane by disks. The $2$-fold coverage problem by disks is to find the minimum number of congruent (unit-radius) disks such that each point of the plane is covered by at least $2$ disks. Here is a paper from 1997 on the topic, including their Fig.1 showing the thinnest $1$-fold covering and a potential thinnest $2$-fold covering:

Bezdek, András, and Włodzimierz Kuperberg. "Circle covering with a margin." Periodica Mathematica Hungarica 34.1-2 (1997): 3-16. journal link


BezdekKuperbergFig1


And here is a paper from just a few days ago on this topic, which says that thinnest $k$-fold disk coverage for $k>1$ is "hopelessly difficult":

Jingchao Chen. "Two-Fold Circle-Covering of the Plane under Congruent Voronoi Polygon Conditions." 3 Oct 2014. arXiv Abstract link

My question is:

Q. For which shapes $K$ (planar convex bodies) is the thinnest $2$-fold covering (by congruent copies of $K$) known?

I would allow $K$ to be rigidly transformed: translated, rotated. Certainly if $K$ tiles the plane, the question is settled. So for all triangles, and for all quadrilaterals (see this earlier MO question), matters are settled. Some pentagons tile the plane. How about for $K$ a regular pentagon? Etc. (Added:). I am especially interested to learn if some thinnest $2$-fold covering is known for a shape that does not perfectly tile the plane.

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    $\begingroup$ There are must be nice non-convex polygons which do not tile the plane but do give perfect $2$-coverings ( $2/3$ of a hexagon, chosen wisely?) If you could tile one of those with a convex non-plane tiler you would be set. $\endgroup$ – Aaron Meyerowitz Oct 10 '14 at 2:57
  • $\begingroup$ My example does tile $2/3$ of a hexagon but not in the way you imagined... $\endgroup$ – Noam D. Elkies Oct 14 '14 at 18:14
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    $\begingroup$ this is about teh case when only translations are allowed: link.springer.com/article/10.1007/s00493-012-2860-3 $\endgroup$ – Dima Pasechnik Oct 14 '14 at 20:06
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[Edited to add a two-parameter family of hexagons]

Here's a family of convex pentagons $K_\epsilon$, each of which does not tile the plane but does have a perfect $2$-covering by rotated translates.

Start with a regular hexagon $\,\widetilde{\!H}$. Its six shorter diagonals form a "star of David" with a smaller hexagon $H$ at its center whose area is $1/3$ the area of $\,\widetilde{\!H}$. Now tile the plane with parallel copies of $\,\widetilde{\!H}$, and remove the corresponding copy of $H$ from the center of each translate. The resulting region, call it $R$, is shown shaded in the first part of the image below (with $\,\widetilde{\!H}$ and its translates outlined in blue, and the star of David in light blue). This picture shows how $R$ is also the hexagonal tiling by copies of $H$ with every third hexagon removed. Hence three parallel copies of $R$ constitute a perfect $2$-covering of the plane.

Now each copy of $\,\widetilde{\!H} - H$ is tiled by six rotations of a pentagon $K_0$ with angles $120^\circ, 90^\circ, 120^\circ, 90^\circ, 120^\circ$. So $K_0$ has a perfect $2$-covering. This is not yet our pentagon: it tiles the plane, because three copies fill $H$ exactly. But we can deform $K_0$ by rotating its shortest two sides by $\epsilon$ degrees to a pentagon $K_\epsilon$, with angles $(120+\epsilon)^\circ, (90-\epsilon)^\circ, 120^\circ, (90+\epsilon)^\circ, (120-\epsilon)^\circ$, six copies of which still tile $\,\widetilde{\!H} - H$ as shown in the second part of the image. For small but nonzero $\epsilon$ this $K_\epsilon$ cannot tile the plane, not even with reflections allowed $-$ a contradiction is soon reached starting with a neighborhood of a $(120\pm\epsilon)^\circ$ angle. But $K_\epsilon$ still perfectly $2$-covers the plane via $\,\widetilde{\!H}-H$ and $R$, so we have our family of convex pentagons with a minimal tiling thickness of $2$.

P.S. I later noticed a further modifiction to a two-parameter family of convex hexagons $K = K_{\epsilon,\delta}$ that do not tile the plane but do tile $R$ and thus perfectly $2$-cover the plane. Deform each edge of $\,\widetilde{\!H}$ to the same Z-shaped path ("Z-shaped": centrally-symmetric concatenation of three line segments). Call the resulting polygon $\,\widetilde{\!H}'$; it still tiles the plane. Do not change $H$. Then $\,\widetilde{\!H}' - H$ is tiled by six rotations of the same convex hexagon $K$, as seen in the first part of the next image:

Hence $K$ tiles $R$, as seen in the second part, and therefore $K$ perfectly $2$-covers the plane. Compared with the pentagons $K_\epsilon$, this variation has the advantage not just of an extra parameter but also that it's much easier to check there's no perfect $1$-covering: If a convex hexagon tiles the plane then we can assume that any two adjacent tiles must share a complete edge, and that no more than three tiles meet in every corner; and that's soon seen to be impossible here.

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  • $\begingroup$ This is an exceptionally clever 2-tiling by an irregular pentagon. And I appreciate your lucid explanation and the precisely informative figures. Thanks! $\endgroup$ – Joseph O'Rourke Oct 14 '14 at 22:00
  • $\begingroup$ You're welcome, and thank you! I noticed later that there's a further generalization to a 2-parameter family of convex hexagons; see the new pictures. $\endgroup$ – Noam D. Elkies Oct 23 '14 at 22:55
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    $\begingroup$ Those hexagon 2-covers are pleasingly intricate, Noam! :-) $\endgroup$ – Joseph O'Rourke Oct 24 '14 at 1:39

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