Undergrad Smalles Number of Rectangles to make a perfect square?

[SiegeX]

This may seem like an innocuous question at first, but I'm not so sure that's really the case. The question is as follows:

If you have a rectangle of height X and length Y, what is the smallest number of those rectangles needed that when placed together create a perfect square? You are allowed to rotate the rectangles 90* in any pattern you desire. You are not allowed to trim the rectangles in any way.

The particular case I'm looking for is a rectangle with dimensions 63x88 but I'm curious about any old generic dimension. Thanks

 

[mathman]

If X/Y is rational, it can be done - I won't try to guess as to the smallest number. If X/Y is irrational, there is no solution.

 

[Ignea_unda]

I believe that there are 2 cases for the "rational" dimensions.

If x divides y, then the dimensions would be y by y for the smallest number with y/x rectangles.

Now if x doesn't divide y, then we must look at the gcd(x,y). Let us say this is equal to "t". Then the resulting square would be y/t by x/t.

This however would be only keeping the rectangles facing the same direction. I'm sure there are many more permutations that would use rotation, which I'm nowhere near qualified to figure out.

if I am wrong please let me know. I was just looking at this and thought that I had an answer to it. I would be glad to have a critique on this answer.

 

[SiegeX]

Thank you for the replies. Regarding the GCD(x,y), this is the same solution I came up with but then I thought to myself "what about rotation?" So I think that's really where the question lies now, could rotating the rectangles get you a smaller perfect square than no rotation at all?

 

[epkid08]

I think I have a solution:
First find the area of the triangle and multiply it by n, then set it equal to x^2 where x is a positive integer. Solve for x and you should have sqrt(n*(2772)). Now simplify, bring a 4 and a 9 out of the square root and you're left with 6sqrt(77n). Because this is as simplified as it gets, we know that the lowest number n can be where x is an integer is 77 itself. So I got 77 triangles, and it seems to work.

 

[n_bourbaki]

Think of it this way - perhaps it helps - if you are tiling with XxY rectangles, say N of them, then the cover an area of NXY square units. You require this to be a square.

 

[HallsofIvy]

I think I have a solution:
First find the area of the triangle and multiply it by n, then set it equal to x^2 where x is a positive integer. Solve for x and you should have sqrt(n*(2772)). Now simplify, bring a 4 and a 9 out of the square root and you're left with 6sqrt(77n). Because this is as simplified as it gets, we know that the lowest number n can be where x is an integer is 77 itself. So I got 77 triangles, and it seems to work.

I have no idea what you are talking about! What triangle? No one has mentioned a triangle. And where did you get the number 2772?

 

[epkid08]

Oh :zzz: my, I read the title wrong. 2772 is the area of a triangle whose height is 63 and length is 88. Although I think I'm right if the question was, "Smallest number of triangles..."