Fitting squares

Q

QwertyKnight

New member
Joined
May 3, 2015
Messages
7
Is it possible to fit smallers squares into a unit square with the following areas: sum from i=2 to infinite(1/i^2), and if its possible how could i do that?
I know that the sum of those squares are less than one but i didnt manage to find such an algorithm(for how i should place the smaller squares) that would provide us the anwser. Any idea?
 
Yes, you can most definitely fit an infinite number of non-overlapping successively smaller squares following your indicated pattern inside a unit square. That's what the summation you did proved. Compare the areas of the sum of the infinite squares to the area of the unit square:

\(\displaystyle \displaystyle \sum_{i=2}^{\infty} \: i^{-2}=\dfrac{\pi^2}{6}-1 \approx 0.6449 \le 1\)

Although this proves you can fit all the squares inside, it, as you correctly identified, doesn't tell you how to arrange the squares. But the reality is that it doesn't matter in this case, because there are infinitely many ways to arrange the smaller squares so that they all fit. One such way is to place the first square, with area 1/4, in the lower left corner of the unit square. Then place the second square, with area 1/9, so that the lower left corner of it is at the same point as the upper right corner of the first square. Repeat indefinitely.

See if you can find some others, and hopefully in the process intuit out why there must be infinitely many arrangements. In fact, I believe (although I'm not 100% sure) that there is an infinite number of ways to arrange any number (finite or infinite) of non-overlapping squares of any size inside a unit square, so long as the squares' areas sum to 1 or less, and the biggest square has an area of 1/4 or less. Maybe play around with that too and see what comes of it. I think it will further your understanding of the ideas at play here.
 
Actually I don't see why would the area inequality prove the statement.If I'd just start putting squares next to each other I'd get parts of the harmonic series(1/n), which is not limited. I tried to divide the elements of the original series into some others that has an upper limit(for the sides of the squares). I've also tried many different techniques to position the squares "correctly". If we could fill the square with such shapes that allows us the "every" one of those shapes we could place "every" square then that would be also a proof.(example: prove the statement with rectangles with the sides of: (1 & 1/2), (1/2 & 1/3), ..., (1/n & 1/(n+1)) (n approaches infinity))
 
Last edited:
QwertyKnight said:
Is it possible to fit smallers squares into a unit square with the following areas: sum from i=2 to infinite(1/i^2),
and if its possible how could i do that? I know that the sum of those squares are less than one but i didnt manage to find such an algorithm
(for how i should place the smaller squares) that would provide us the anwser. Any idea?
Click to expand...

Let \(\displaystyle \ S_2 \ \) be the square with side length of 1/2.

And in general, let \(\displaystyle \ S_n \ \) be the square with side length of 1/n, for n a positive integer.

Place these squares flush against one of the interior sides of the unit square.

_______________________
\(\displaystyle S_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ S_3\)
\(\displaystyle S_4S_5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ S_7S_6\)
\(\displaystyle S_8S_9S_{10}S_{11}\ \ \ \ \ S_{15}S_{14}S_{13}S_{12}\)
\(\displaystyle \cdot\cdot\cdot \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot\cdot\cdot \)


The side lengths along the left-hand side form an infinite geometric sequence. Its corresponding series adds up to 1,
the side length of the unit square.

The side lengths along the right-hand side form an infinite geometric sequence. Its corresponding series adds up to 2/3,
less than the side length of the unit square.
 
ksdhart2 said:
... there are infinitely many ways to arrange the smaller squares so that they all fit. One such way is to place the first square, with area 1/4, in the lower left corner of the unit square. Then place the second square, with area 1/9, so that the lower left corner of it is at the same point as the upper right corner of the first square. Repeat indefinitely.
Click to expand...

I tried this, but part of the third square fell outside the unit square.

If the lower-left corner of the first square is at the origin; the coordinates at its upper-right are (1/2,1/2).

Placing the second square as described, its upper-right corner is at (5/6,5/6).

The upper-right corner of the third square is at (13/12,13/12).
 
Otis said:
I tried this, but part of the third square fell outside the unit square.

If the lower-left corner of the first square is at the origin; the coordinates at its upper-right are (1/2,1/2).

Placing the second square as described, its upper-right corner is at (5/6,5/6).

The upper-right corner of the third square is at (13/12,13/12).
Click to expand...

Ah, yes, both you and the OP are correct that my outlined method doesn't work. Sorry about that. What happened was I got confused and thought each side length of the squares was 1/4, 1/9, 1/16, etc. rather than the area of each square following that sequence, meaning that the sides are really 1/2, 1/3, 1/4, etc.

I still believe that the infinite sequence of squares can be fit inside the unit square, and I still believe it can be done in infinitely many ways - my suggested way just wasn't one of them. The additional rectangles exercise as proposed by the OP is proving interesting to me, because I'm not certain as of yet if it's possible. My gut instinct is to say no, but I'm not ready to rule it out the possibility just yet.
 
ksdhart2 said:
I still believe that the infinite sequence of squares can be fit inside the unit square, and I still believe it can be done in infinitely many ways.
Click to expand...

It can be done; lookagain posted one arrangement. I was considering circular and spiral arrangements, but I ran out of time.

Not sure about the infinite-arrangements hypothesis ... :cool:
 
You must log in or register to reply here.
Top