Packing Squares with Side 1/n

up to 1/n

150

initial rectangle

5

6

×

47

60

packing excess =

5

6

×

47

60

+ 1 -

π

2

6

= -0.00784371092955134131

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

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25

26

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29

30

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39

40

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

n

2

=

π

2

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.