A 10x10 square can obviously hold 100 unit circles (diameter=1) when arranged in rows and columns. What is the maximum number of non-overlapping unit circles a 10x10 square can hold if the circles are packed closer together?

(In reply to hmmm... by Thalamus)

Thought, thanks for chiming in, and I wish Vee-Liem had pointed out what the problem was.

In point of fact, Vee-Liem was correct in questioning my calculation.  It should read using Pythagorean theorem we calculate:
x² + (1/2)² = 1
and the vertical distance we're interested in is twice x (or 4x if like the previous problem you want the distance between odd rows) .

Now the correct calculation gives us 2x ≈ .866205 (not .894427 as I said previously).  This allows us to pack a little closer.

A corrected table shows the following vertical coordinates of the rows:

row  vertical coordinate
 1   0.5
 2   1.366025
 3   2.232051
 4   3.098076
 5   3.964102
 6   4.830127
 7   5.696152
 8   6.562178
 9   7.428203
10   8.294229
11   9.160254

Making all the odd rows a 10-circle row, and all the even rows a 9-circle row, this will fit 105 circles as before.  But you'll notice a lot more "white space" at the bottom now.  You'll also notice though that row 9 now is less than 7.5.  This means that we don't have to "pack" a 9-circle row as row 10 anymore.

Row 10 can be a 10-circle row, pushing row 11 down a bit, but still fitting in the square.

So, by doing this, the vertical coordinates of rows 10 and 11 will be 8.428203 and 9.428203 respectively, and both rows are 10-circle rows.

So we get one additional circle in there, for a total of 106 circles.

Edited on July 30, 2004, 5:41 pm

Posted by Thalamus on 2004-07-30 13:03:09