A 6 x 6 square is dissected into 9 rectangles by lines parallel to its sides such that all these rectangles have only integer sides. Prove that there are always 2 congruent rectangles.
Prove step by step please.
2006-12-04 03:44:18 · 2 answers · asked by Naval Architect 5 in Science & Mathematics ➔ Mathematics
I have a rather ugly brute force proof of this although the question should be 'at least 2 congruent rectangles'
Clearly the sum of the areas of the 9 rectangles will still equal the area of the square.
There are 21 different sized rectangles (including squares which special cases of rectangles) that can fit into the square, ie 1x1, 1x2....5x6, 6x6
For there to be no congruent rectangles, we must have 9 distinct sized rectanges.
The smallest 9 (by area) of these rectangles are:
(1x1) = 1, (2x1) = 2, (3x1) = 3, (4x1) = 4, (5x1) = 5, (6x1) = 6, (4x2) = 8, (3x3) = 9, (5x2) = 10
Add these areas gives a total area of 48 so if we cannot fit the smallest 9 distinct rectangles in the 6 x 6 = 36 area of the square, we cannot fit any 9 distinct rectangles in.
Therefore we must have at least 2 congruent rectangles.
2006-12-04 20:30:09 · answer #1 · answered by Status: Paranoia 4 · 1⤊ 1⤋
Well, there are only three ways to split six units into the sum of three positive integers: (a) 2 + 2 + 2 (b) 1 + 2 + 3 (c) 1 + 1 + 4. That leaves only six dissections to consider, since the remaining three are mirror images of one of the other six.
That should be enough of a hint for you to work this out. (Sorry, but, for your own good, I'm omitting the step-by-step analysis. You'll probably get one from another poster anyway, but I can't in good conscience be a party to this.)
2006-12-04 11:53:45 · answer #2 · answered by Anonymous · 1⤊ 1⤋