A $$6 \times 6 \text{ square}$$ is dissected into $$9 \text{ rectangles}$$ by lines parallel to its sides
such that all these rectangles have only integer sides. Prove that $$ \text{ there
are always two congruent rectangles}$$.
Congruent Rectangles
- Fatin Farhan
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- asif e elahi
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Re: Congruent Rectangles
Let the rectangle can be dissected into $9$ incongruent rectangles.Let the area of these rectangles are $x_{1},x_{2}.......x_{9}$ and $x_{1}\leq x_{2}\leq .......\leq x_{9}$
Then $x_{1}\geq 1\times 1=1$
$x_{2}\geq 2\times 1=2$
$x_{3}\geq 3\times 1=3$
$x_{4}\geq 4\times 1=4$
$x_{5}\geq 2\times 2=4$
$x_{6}\geq 5\times 1=5$
$x_{7}\geq 6\times 1=6$
$x_{8}\geq 3\times 2=6$
$x_{9}\geq 7\times 1=7$
So $36=x_{1}+x_{2}.......+x_{9}\geq 38$
This is not true.So there are always $2$ congruent rectangles.
Then $x_{1}\geq 1\times 1=1$
$x_{2}\geq 2\times 1=2$
$x_{3}\geq 3\times 1=3$
$x_{4}\geq 4\times 1=4$
$x_{5}\geq 2\times 2=4$
$x_{6}\geq 5\times 1=5$
$x_{7}\geq 6\times 1=6$
$x_{8}\geq 3\times 2=6$
$x_{9}\geq 7\times 1=7$
So $36=x_{1}+x_{2}.......+x_{9}\geq 38$
This is not true.So there are always $2$ congruent rectangles.