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birbal
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GMO 2006 question  
« on: Dec 4th, 2006, 8:21pm »
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This is a question from GMO 2006.
 
Given a square grid of size 6 x 6. The grid is divided into 9 rectangles with integer size lengths. Prove that at least two rectangles are congruent to each other.
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Re: GMO 2006 question  
« Reply #1 on: Dec 5th, 2006, 6:49am »
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Let's see...
What are the nine smallest rectangles we could possibly have?
To how many unit square do they add up to?
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birbal
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Re: GMO 2006 question  
« Reply #2 on: Dec 5th, 2006, 8:36pm »
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Ok... the nine smallest incongruent rectangles can be
1x1, 1x2, 1x3, 1x4, 1x5, 1x6, 2x2, 2x3, 2x4
 
and the total square unit they add up to are  
39 units. mm but the total area is 36 units. Cool!!
So this is not possible??  
I am not sure if this is proved. Roll Eyes
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Re: GMO 2006 question  
« Reply #3 on: Dec 6th, 2006, 6:02am »
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More formally:
1) Assume that no two rectangles are congruent.
2) The nine smallest incongruent rectangles by area are 1x1, 1x2, 1x3, 1x4, 2x2, 1x5, 1x6, 2x3, and 1x7.
3) These rectangles have a total area of 38 square units.
4) Therefore no set of nine incongruent rectangles can tile a 6x6 square, and premise (1) must be false.
5) Therefore its negation must be true: at least two rectangles are congruent.
Q.E.D.
 
(And if we consider 1x1 and 2x2 to be congruent, the situation only gets worse as the smallest rectangles are then 1x1, 1x2, 1x3, 1x4, 1x5, 1x6, 2x3, 1x7, and 1x8 for a total area of 40.)
 
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Re: GMO 2006 question  
« Reply #4 on: Dec 6th, 2006, 6:35am »
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on Dec 6th, 2006, 6:02am, SMQ wrote:
2) The nine smallest incongruent rectangles by area are 1x1, 1x2, 1x3, 1x4, 2x2, 1x5, 1x6, 2x3, and 1x7.
1x7 doesn't fit in a 6x6 square, so picking 2x4 makes more sense.
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Re: GMO 2006 question  
« Reply #5 on: Dec 6th, 2006, 8:37am »
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Well, that just means he proved the generalization where you're allowed to rearrange the pieces into rectangles.
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