BrainDen.com - Brain Teasers
• 0 ## Question This is a problem that I came up with over 10 years ago, and haven't really ever spent time solving it since then, because I got stuck and real life got in the way. Posting on this site has made me think of it again, and was wondering if anyone had some ideas.

Imagine a square, with a length and width of 1 unit. A square can be subdivided into two rectangles, each with an area of 1/2 of the square. This would be accomplished by drawing a line through the middle of the square.

A square can be subdivided into three rectangles many ways, each with an area of 1/3 of the square. The most obvious way is to draw two lines down the middle so that each subdividing rectangles has a length and width of 1/3 by 1 unit. However, there is another way to subdivide the square into two rectangles each with an area of 1/3. You can make one rectangle with the dimensions of 1/3 by 1 unit, and then draw a line perpendicular to the one to create that, to create two rectangles with dimensions of 1/2 by 2/3 units.

Let me be clear. There are more ways than this to subdivide a square into 3 parts, even 3 parts with equal area, but the purpose is to subdivide a square into equal area rectangles.

Also, a unique solution is based solely on the number of rectangles with specifc dimensions. Even if you can make the rectangles in the square look differently, if they have the same dimensions and quantity as another look, it is not a unique solution.

So, with 1 rectangle with an area of 1 (the square), there is only 1 unique solution.

With 2 rectangles each with an area of 1/2, there is only 1 unique solution.

With 3 rectangles each with an area of 1/3, there are only 2 unique solutions.

Can you continue with the reasoning on how many unique solutions there area for rectangles of areas 1/4, 1/5, 1/6, ...?

Moreover (and here's where I got stuck), can you come up with a function that describes this pattern? This isn't for everyone. I was unable to come up with a function that predicted the next number, both mathematical or conceptual. The issue is not double-counting some solutions (if you work through a few more of these, you'll see what I'm saying.)

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• 0 For those of you who prefer a visual representation, the first three iterations are below. ##### Share on other sites
• 0 I agree with the double counting thing. Not sure how to handle that. Are you looking at this recursively (i.e. breaking it a 5-piece down into 1x4 and 2x3, and then breaking those down more)? That was the first way I looked at it.

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• 0 The Solution is n-1 where n is the number of rectangles you are trying to create.

Edited by Doctor Dub

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• 0 Did you find 5 unique solutions for 4 rectangles?

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• 0

I get:

2^(n-2) where n is the number of rectangles you are trying to create.

Actually, I think it's better defined recursively, where f(2)=1 and then f(n) = sum(f(2)...f(n-1))+1, but that's equivalent to 2^(n-2), so I give the shorter answer.

I've drawn them out to 6 divisions, and am trying to get them into a decent picture to post.

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• 0 Did you find 5 unique solutions for 4 rectangles?

I only found 4 unique for dividing into 5 parts

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• 0 I agree with the double counting thing. Not sure how to handle that. Are you looking at this recursively (i.e. breaking it a 5-piece down into 1x4 and 2x3, and then breaking those down more)? That was the first way I looked at it.

That's how I looked at it too.

The Solution is n-1 where n is the number of rectangles you are trying to create.

Nope, sorry, doesn't fit with above, nor with below.

Did you find 5 unique solutions for 4 rectangles?

Yes, I did.

1) 4 of (1/4 by 1)

2) 2 of (1/4 by 1) and 2 of (1/2 by 1/2)

3) 1 of (1/4 by 1) and 3 of (3/4 by 1/3)

4) 1 of (1/4 by 1), 1 of (3/4 by 1/3) and 2 of (2/3 by 3/8)

5) 4 of (1/2 by 1/2)

Would post a picture, but no time.

I get 2^(n-2) where n is the number of rectangles you are trying to create.

Sorry. Not all powers of 2. See above.

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• 0 interesting problem, first post on this forum

F(n) = 1 + 4*F(n-1) for n>3

The +1 is for the all horizontal/all vertical configuration and the 4* is the four different flip orientations for the previous configurations found.

Edited by jack0

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• 0 if you go by the original post's exact wording, there is only 3 solutions for 4 rectangles. He said that a unique solution meant that the dimentions for each rectangle were different and not a variation of another solution. If this is unclear...i will draw it out.

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• 0 if you go by the original post's exact wording, there is only 3 solutions for 4 rectangles. He said that a unique solution meant that the dimentions for each rectangle were different and not a variation of another solution. If this is unclear...i will draw it out.

no you get five solutions for four rectangles

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• 0
interesting problem, first post on this forum

F(n) = 1 + 4*F(n-1) for n>3

The +1 is for the all horizontal/all vertical configuration and the 4* is the four different flip orientations for the previous configurations found.

jack0, I think your solution double-counts. Your equation gives f(4) = 9, I think we've established f(4) = 5. Not that I'm doing much better right now ##### Share on other sites
• 0 I'm trying to draw this bugger out. This is what I've got through 5. Am I missing anything?

no, your not missing anything, there are 11 for 5 divisions

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• 0

Well, I'm not going to have much more time for this today, but here's my next guess:

f(2)=1

f(n)=2*f(n-1)+1

If I'm wrong, you guys will have to fix it for me.

Another interesting question comes out of this. Suppose the original shape were a rectangle instead of a square. I think rect(n)=2*square(n), but I'm not sure. Can anyone confirm or deny? I think it would actually help in the recursion of the original problem if we knew.

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• 0 has anyone worked out 6 yet

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• 0 anyone want to draw out all the 6 division iterations? i'm getting 27 but i'm not sure if i missed any

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• 0 I'm only getting 22 arrangements for 6 rectangles.

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• 0 the catch in this problem is for n divisions, you can't just consider n-1 when doing recursion, because when you get into the higher n's, it also depends on recursions dealing with n-2, n-3, n-4 etc.

the key is finding a relation between the infinite recursions and the initial recursions n-1 and n-2

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• 0 I'm only getting 22 arrangements for 6 rectangles.

i know theres more than that

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• 0 nevermind i had some repeats there are 25 arangements for the 6

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• 0 Yeah, I forgot a few. I have 25 arrangements now too.

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• 0 Here are the 25 arrangements for 6 rectangles:

I put them into a sort of order. If you start with the yellow rectangles, then each row is basically the various arrangements of previous iterations. For example, if you first set one yellow rectangle (first row), then the rest of the unused space is filled with five rectangles, following the eleven arrangements. The picture isn't exactly to scale, I'm sure, but I tried my best.

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• 0 Another difficulty, in addition to the repeating issue, is that dividing a rectangle isn't necessarily the same as dividing a square. For instance, dividing a square into two sections yields only one unique solution. Dividing any other rectangle into 2 parts gives 2 solutions - cutting it the long way and the short way. This makes it more complicated, because you can't reduce it recursively to some rectangle and then say, "this rectangle must be subdivided n times, and I've solved the problem for dividing a square n times, so now I can just multiply that number by the number of ways of getting down to that n-division."

Long story short, I have no idea.

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• 0 Just had to say that the "The Death of Socrates " is a fantastic painting!! I love J.L. David!

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