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Weekly Problem 3 - 2007

Stage: 3 and 4 Challenge Level: Challenge Level:1

1:1

Paper - solution


Let the sheet of paper have length $x$ and width $y$. Then the uncovered area consists of two congruent rectangles of length $x-y$ and width $y-\frac{1}{2}x$. So the uncovered area is $2(x-y)(y-\frac{1}{2}x)$, that is, $(x-y)(2y-x)$.

The area covered twice is a rectangle of length $y-(x-y)$, that is, $2y-x$, and width $\frac{1}{2}x-(y-\frac{1}{2}x)$, that is, $(x-y)$. So the area covered twice is also $(x-y)(2y-x)$.

This problem is taken from the UKMT Mathematical Challenges.

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