TITLE: 2-layer tilings with a center-of-gravity constraint
QUESTION [10 upvotes]: I've encountered a tiling problem with a physical constraint that
might place it outside the literature on tiling.
"Tiling" is a bit of a misnomer; it is a special type of cover.
All the tiles are identical (congruent), convex shapes $S$.
Layer 1, $L_1$, is some infinite pattern of copies of $S$
arranged on the plane with pairwise disjoint interiors.
Layer 2, $L_2$, is the exact same pattern as $L_1$,
but after some rigid motion of the whole arrangement
(translation, rotation, perhaps reflection).
Together $L_1$ and $L_2$ cover the plane:
$L_1 \cup L_2 = \mathbb{R}^2$.
But I want the center of gravity (c.g.) of each tile of $L_2$ to be on top
of a point of some tile in $L_1$.
Consider this last requirement an abstraction of a balance or
support constraint.
Finally, I would like to minimize double coverage, i.e.,
$L_1 \cap L_2$.
Below are some examples.
Two penny packings of the plane (a), with $L_2$ shifted
by the radius of the disk, satisfy the constraints.
But if I've calculated correctly, 81% of the plane is
doubly covered—not very good.
The staggered squares in (b) improve the double coverage to $\frac{1}{3} =$ 33%.
Here the c.g. of each $L_2$ square sits on the meeting of two corners
of squares from $L_1$.
One can improve this tiling by clipping off a corner (1/8-th)
of each square, as in (c). Now the c.g. of each $L_2$ tile
sits over an interior point of an $L_1$ tile.
I calculate this reduces double coverage to $\frac{2}{9} =$ 22%.

         

Because the c.g. in (c) sits over an interior point, there is room
for improvement. Below in (d) I clip off a tiny portion of the opposite
corner (shown in green) to move the c.g. to the boundary, resulting in a tiny
improvement to 21.8%.

     

I have no reason to believe this is an optimal, or even a good tiling
under these constraints.
It seems there should be some fundamental positive lower bound
to the double coverage, but I am not seeing an argument to
establish such a bound.

Update1. Improved by Yoav Kallus to 19.8% double-coverage with a simpler construction!
So the outstanding issue is: Lower bound??
Update2. Surely Yoav's new 12.5% tiling made of overlapping equilateral triangles
is the optimal.

REPLY [11 votes]: The pentagon $(0,0)(0,1)(1/2,1+x)(1,1),(1,0)$ with $x=(\sqrt{21}-3)/4$ gives a double coverage fraction of $x/2=0.1978\ldots$.

UPDATE: The triangle can get you $1/8=0.125$