# Thinnest covering of the plane by regular pentagons

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."

• Suppose you line them up. (The nth pentagon has its base on [n,n+1].). How much space does that overuse? Gerhard "Minding Ones Pents And Gons" Paseman, 2017.11.08. – Gerhard Paseman Nov 9 '17 at 1:03
• @GerhardPaseman: This would not be the best way to do it. See my answer below and click on the link. You may also enjoy the animation auburn.edu/~kuperwl/pent_movie.mp4 showing a continuous transition between the densest packing and this covering. – Wlodek Kuperberg Nov 9 '17 at 2:34