How many cookies in the cookie box? -- Tiling stars With holiday season coming up I decided to make some cinnamon stars. That was fun (and the result tasty), but my inner nerd cringed when I put the first tray of stars in the box and they would not fit in one layer:



Almost! Is there a way they could have fit? How well can we tile stars, anyway? Given that these are regular six-pointed stars, we could certainly use the well-known hexagon tilings as an approximation, like so:


Messed up the one to the upper right, whoops.

But is this optimal? There's plenty of room between the tips.

For this consideration, let us restrict ourselves to rectangular boxes and six-pointed, regular stars, i.e. there are thirty degrees (or $\frac{\pi}{6}$) between every tips and its neighbour nooks. The stars are characterised by the inner radius $r_i$ and outer radius $r_o$:


[source]

Note that we have hexagons for $r_i = \frac{\sqrt{3}}{2} \cdot r_o$ and hexagrams for $r_i = \frac{1}{\sqrt{3}} \cdot r_o$. I think it's reasonable to consider these the extremes (for cookies) and restrict ourselves to the range in between, i.e. $\frac{r_i}{r_0} \in \Bigl[\frac{1}{\sqrt{3}}, \frac{\sqrt{3}}{2}\Bigr]$. 

My cookies have $r_i \approx 17\mathrm{mm}$ and $r_o \approx 25\mathrm{mm}$ ignoring imperfections -- I was going for taste, not form for once!

What is an optimal tiling for stars as characterised above? If there is no static best tiling, is there an algorithm to find a good one efficiently?
Let me answer your question partially for the hexagram case.

You can make the following tiling

$\hskip1.2in$

By this you will cover 12/14=6/7 of the plane (count the triangles in the dashed quadrilateral).

Is this optimal? I would think so. Although I am not giving a proof I will provide some arguments. One can ask, how good we can fill the space (triangle) in between the pointy spikes. In the above tiling we fill half of it. Can we do better?

$\hskip20mm$

It is possible that two hexagram will intersect this space, but then they cover very little of its area (without proof). If there is only one hexagram intersecting I assume that its tip touches the concave corner of the other hexagram as depicted in the picture. If this would not be the case we can improve by moving the intersecting hexagram to this corner (again no proof here). Under these assumptions it is not hard to see that the case where there is side-to-side contact maximizes the intersection. If you do the math, then you will find out that the area of the intersection equals 
$$\frac{\sec ^2(x)}{2 \sqrt{3} \tan (x)+2}.$$

The plot of this function looks like this and shows that our intuition was right.

$\hskip30mm$