In a given rectangle, equal circles are arranged .
The ratios
-
length of the rectangle to the diameter of each circle,and
-
breadth of rectangle to diameter,
are both natural numbers.
All the circles are arranged such that no more circles can be placed and no circle overlaps another or exits the rectangle.
To be simple, all the circles are arranged like this:
In such a case, does the total area of circles in the rectangle change if circles of a different size are arranged the same way in the same rectangle? (again, the above mentioned ratios must be integers)
Will the same apply to spheres in a cuboid?
I would also like to know if there are any articles on this topic on the internet.
Best Answer
Note that in your arrangement each circle can be inscribed in a square. That packing has density $\frac \pi 4 \approx 0.785$. As long as you stay to the rectangular array the density will not change, but that is not the most efficient packing of circles. You can do better with a triangular packing, which has a limiting density of $\frac {\pi \sqrt 3}6 \approx 0.9069$ For finite rectangles there is a contest between the denser triangular packing and the area lost at the boundary. You might look at the examples at packomania. The first square where there is a packing known better than a square array is $49$ circles, where you can pack $49$ circles with radius $0.0716926817$ in a unit square, while square packing only allows a radius of $\frac 1{14}\approx 0.0714825$