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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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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