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        riddles >> hard >> Perfect Square
        
(Message started by: THUDandBLUNDER on Nov 5^th, 2003, 5:43am)

Title: Perfect Square
Post by THUDandBLUNDER on Nov 5^th, 2003, 5:43am

Find the shaded area.

Title: Re: Squares
Post by towr on Nov 5^th, 2003, 5:51am

My first intuition is to assign each side a variable, and solve it using gaussian elimination fluff ;)

[e]actually, there isn't a unique solution, since any integral multiple of the collection of squares would have the same properties.
But I suppose we can choose the smallest possible value[/e]

Title: Re: Perfect Square
Post by visitor on Nov 5^th, 2003, 10:32am

[hide]The square can be solved in two parts. First, the upper left hand corner forms a self-contained rectangle that is easily solved to give the mystery square a size of 38, when the smallest square (call it z) is 1. The rectangle then has dimensions of 94 by 111. Plug that into the whole image. If I assign x to the smallest of the remaining boxes and y to the box right above it, I end up with 41x+y=111z. and 4y-11x=94z. And because it's a square, you can also equate 26x+y+94z=3x+2y+111z. So z=1, x=2, and y=29. And the mystery square retains its size of 38.[/hide]

Title: Re: Perfect Square
Post by THUDandBLUNDER on Dec 10^th, 2003, 5:09am

Well done, visitor.

Now that you have the hang of it, what's the area of the large square in the top left-hand corner?

Title: Re: Perfect Square
Post by Barukh on Dec 10^th, 2003, 3:10pm

Is the whole thing still a square?

Title: Re: Perfect Square
Post by THUDandBLUNDER on Dec 10^th, 2003, 5:55pm

Quote:

Is the whole thing still a square?

Yes, just as in the first puzzle.

Title: Re: Perfect Square
Post by Barukh on Dec 11^th, 2003, 11:43am

OK, I think I know the answer: [smiley=blacksquare.gif][hide]50[/hide][smiley=blacksquare.gif].

The solution will follow...

Title: Re: Perfect Square
Post by Barukh on Dec 11^th, 2003, 2:25pm

Here's the solution:

[smiley=blacksquare.gif][hide]
First, I made an assumption – based on the relative sizes of squares – that Z < A < B < C < D.

I began by choosing Z, A, C as “independent” quantities. Then, we have T = C+Z, N=2C+Z, M = 3C+2Z etc. Closing on R, I got the following relation: 4A = 2C+Z, which showed that Z must be even. So, the smallest possible assignment is Z=2, A=3, C=5.

Next, I turned to the right-bottom part of the big square. Since S = 2*O+B = R+N+C-O, I get 3*O+B = 6C+3Z+3A, that is, B is a multiple of 3. This eliminates the aforementioned assignment for A and C, so I considered the next possibility A=4, C=7. Then, B=6, O=18, P=24, S=42, and the size of the big square is 112.

Finally, 2J+D = P+B = 30, and K+G = 2J+3D = 46, so D=8, J=11. The requirements of the problem (each square being of unique size) are fulfilled, and also are the assumptions. The rest is really straightforward.
[smiley=blacksquare.gif][/hide]

Title: Re: Perfect Square
Post by THUDandBLUNDER on Dec 11^th, 2003, 7:45pm

Well done, Barukh. :o

With all due respect, this puzzle is obviously not as difficult as I had assumed and moderators may feel free to move it where they think fit.

But finding such squares from scratch is difficult.
See, for example, http://mathworld.wolfram.com/PerfectSquareDissection.html