WOLFRAM|DEMONSTRATIONS PROJECT

Packing Squares with Side 1/n

​
up to 1/n
150
initial rectangle
5
6
×
47
60
packing excess =
5
6
×
47
60
+ 1 -
2
π
6
= -0.00784371092955134131
A finite volume of potatoes will fit in a finite sack. This seemingly simple statement leads to a family of very difficult questions, sometimes called potato sack problems.
Consider squares with sides
1
2
,
1
3
,
1
4
, …,
1
n
. What is the smallest rectangle that can contain the squares as
n∞
? One bound is
∞
∑
n=2
1
2
n
=
2
π
6
-1
, but no one has found a packing for a rectangle of that area. In 1968, Meir and Moser showed that a square of size
5
6
×
5
6
was enough. The current record is held by Marc Paulhus, who developed the packing algorithm used for this Demonstration.