TITLE: Thinnest covering of the plane by regular pentagons
QUESTION [9 upvotes]: Q. Is it known what is the thinnest covering of the infinite plane by regular pentagons?

By covering I mean every point of the plane is covered.
By thinnest I mean the proportion of the plane covered more than once is minimal among all coverings.
This seems like it must be known, but I cannot find it, perhaps because I 
don't know the correct terminology.
This is a natural attempt:

          


If I've calculated correctly, this
covering doubly covers about $38\%$ of the plane:
$$\tfrac{1}{2}
   \left(3-\sqrt{5}\right) \approx 0.382 \;.$$
I am interested because the above covering can be achieved by "rolling"
a dodecahedron, and I'd like to know if there is a thinner cover
which might not be "rollable."

REPLY [7 votes]: The thinnest known covering of the plane with congruent regular pentagons is shown in my answer to:  Terrible tilers for covering the plane. What you see there is probably not "rollable". The covering is of the "double-lattice" type and is known to be the thinnest among double lattice coverings with regular pentagons (to be published). Also, it is conjectured to be the thinnest one among all coverings with congruent regular pentagons. The conjecture is still open.
By the way, Joe, your "natural attempt" is of a double-lattice type also, generated by a trapezoid contained in the pentagon. The trapezoid is a p-hexagon too, but not of maximum area, and the larger the p-hexagon, the thinner the covering generated by it.