2005 PMWC Problems/Problem T8

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Problem

An isosceles right triangle is removed from each corner of a square piece of paper so that a rectangle of unequal sides remains. If the sum of the areas of the cut-off pieces is and the lengths of the legs of the triangles cut off are integers, find the area of the rectangle.

Solution

Since the figure in the middle is a rectangle, the isosceles triangles on opposite vertices are congruent. Let $x$ be a leg of the first two, and $y$ the other two. The sum of the areas of the triangle is then $2\left(\frac{1}{2}x^2\right) + 2\left(\frac{1}{2}y^2\right) = x^2 + y^2 = 200$ . (Remember that the sides are of unequal lengths, so we exclude $x = y= 10$ ). Since squares $\equiv 0,1 \pmod{4}$ , we can reduce our search to even integers, and a short bit of trial and error yield $x = 2, y = 14$ works.

Using subtraction of areas or 45-45-90 triangles, we find that the area of the rectangle is $(x + y)^2 - x^2 - y^2 = 2xy$ ; so the area of the rectangle is $2(2)(14) = 56$ .

2005 PMWC (Problems)
Preceded by Problem T7	Followed by Problem T9
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10

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2005 PMWC Problems/Problem T8

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Problem

Solution

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