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Suppose you would like to cover the plane with congruent copies of Stars of David:

            StarDavid1
The goal is to overlap the least, to have a "thinnest" covering. One construction treats each Star of David as just an equilateral triangle, and ignores its other equilateral triangle, and then cover the plane with equilateral triangles. If my calculations are not in error, this leads to double-covering $3/9 = 1/3$ of the plane, something like this:
StarDavidTri
It seems unlikely this is the least double-covering fraction.

Q. What is the thinnest covering of the plane by congruent Stars of David?

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    You sure it's $2/3$? I seem to get only $1/3$. – 2017-01-06
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    Make an infinite column of stars one on top of each other so their horizontal edges coincide. This is a pattern with a zigzag boundary that can then be repeated to cover the plane without any more overlaps. I believe the double-covering fraction is $1/5$. – 2017-01-06
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    @Rahul: Nice idea! – 2017-01-06
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    @OscarLanzi: You are right---Thanks. Now corrected. – 2017-01-06

1 Answers 1

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Your method seems to give an overlap ratio of only $1/3$, but it is not the minimum.

Converting @Rahul's comment to an answer:

Let the outer vertices be $A, C, E, ..., K$ in rotational order, and the intervening inward-directed vertices be $B$ between $A$ and $C$, $D$ between $C$ and $E$, abd so on up to $L$. Draw the concave hexagon $AEFGKL$. This has its opposite sides congruent and parallel, so you can tile the plane with that hexagon from each star. Two points (not three) from each star protrude into adjacent hexagons giving an overlap fraction of $1/5$.

StarHex
(Figure added by JORourke.)

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    I'll accept this, as it is likely quite difficult to prove that this is the thinnest. – 2017-01-07