Let us construct a grid consisting of rectangles of height $h$ and width $w$. When we place a circle of radius $r$ over this grid, there is a certain amount of rectangles that completely lie within the circle.
When moving around the circle (note that this only has to happen on an interval $[0,w)\times[0,h)$ since we're looking at a grid), we will obtain a maximum number $N_\max$ of rectangles that fit into the circle.
Now the question arises: Is there always a placement of the circle possible such that the grid is symmetric in the two axes through the center of the circle which also has $N_\max$ rectangles inside of it?
Clarified: Assume we know $N_\max$. A figure consisting of $N_\max$ rectangles that fit inside our circle need not be symmetric in $x$ and $y$-direction. But can we find a figure consisting of the same amount $N_\max$ that is symmetric in $x$ and $y$-direction and fits in our circle.
Here we see a symmetric solution with $N_\max=11$, but can we always find a symmetric solution?
Best Answer
I have implemented an algorithm myself that counts the amount of rectangles ($w\times h$) inside a circle with radius $r$ with its centre at $(x,y)$.
I then took $(x,y)$ in the rectangle $[0,w]\times[0,h]$ and looked what their amount of rectangles was.
Making a contour plot of that gave me an insight in the behaviour of the amount of rectangles, which looks enormously interesting.
For a random $w=0.2785, h=0.0975, r=0.4728$, the result was:
This has a maximum of 17, but it is located just off-centre, and we see the amount is only 16 at $(\frac{w}{2},0)$, so there is no symmetric solution for this specific $w,h,r$.
It should be noted that I had to try a few random values before finding a non-symmetric case, I can't yet explain why symmetry is so common.
Below one can find the circle in an optimal setup
$w,h,r$ above with 17 rectangles within them">