Packing Circles
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A 10x10 square can obviously hold 100 unit circles (diameter=1) when arranged
in rows and columns. How many additional circles can it hold if the circles
are packed closer, in a hexagonal arrangement?
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What are the dimensions of the smallest area rectangle which can hold an
additional 10 unit circles when the arrangement changes from rows and columns
to hexagonal?
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What are the dimensions of the smallest area rectangle which can hold
100 unit circles in a hexagonal arrangement?
Source: Original.
Solutions were received from Henry Bottomley, Doug Dickson,
Philippe Fondanaiche, and Glenn Simon.
Philippe pointed out there are several other web sites which investigate
packing circles more closely. One such is
Erich Friedman's,
which has links to more.
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106 circles can fit in a 10x10 square. 8 rows of row/column circles
(which take up 8 units of space) can
be replaced by 9 hexagonally packed rows (which take up 1+8*sqrt(3)/2, or
7.928 units of space.) This leaves 2 more units of space in the square which
can be filled with two complete rows of 10 circles. Altogether, there are
4 rows with 9 circles and 7 rows with 10 circles. [Most people, myself
included, simply packed all the rows as close together as possible and
found the answer of 105 circles. Thanks to Philippe for finding an even
better answer!]
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Henry Bottmley sent:
A 10.866x1.866 rectangle =(10+sqrt(3)/2)x(1+sqrt(3)/2) =20.276 has space
for one row of ten circles in rectangular formation, but two rows of ten in
hexagonal formation.
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Philippe found the smallest rectangle to hold 10 circles above the original
100:
The smallest area rectangle which can pack 110 unit
circles, is equal to 22.5 * [ 1 + 2*sqrt(3) ] = 100.4423...
This configuration is obtained by packing in a hexagonal arrangement 22 rows
of 5 unit circles each.
The length of the rectangle is equal to 22*1 + 0.5(radius of a circle
representing the shift due to the hexagonal arrangement) = 22.5 while its
width is defined by 0.5 + 4*0.5*sqrt(3) + 0.5 = 1 + 2*sqrt(3). This result
corresponds to the integer n=5 such as n minimizes the quantity
[INT(110/n)+0.5]*[1+0.5*(n-1)*sqrt(3)] where INT is the rounded up integer of
110/n.
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Henry Bottmley's solution:
A 7.062x13 rectangle =(1+7*sqrt(3)/2)x13 =91.808 has space for 8 rows of
alternately 13 and 12 circles making 100 circles in total.
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Philippe's slightly better solution:
With the same approach, the smallest area rectangle containing 100 unit
circles is equal to 20.5*[ 1 + 2*sqrt(3) ] = 91.51408.. that is to say 20
rows of 5 unit circles each.
Mail to Ken