Difference between revisions of "2005 PMWC Problems/Problem T8"

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== Problem ==
 
== Problem ==
An [[isosceles triangle|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.  
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An [[isosceles triangle|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 <math>200</math> and the lengths of the legs of the triangles cut off are integers, find the area of the rectangle.
  
 
== Solution ==
 
== Solution ==

Revision as of 11:04, 24 December 2019

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 $200$ and the lengths of the legs of the triangles cut off are integers, find the area of the rectangle.

Solution

2005 PMWC-T8.png

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 $2(2)(14) = 56$.

See also

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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