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Integral Area

Problem

A rectangle measuring x by y has a unit square placed in the bottom right corner. The diagonal, d, joining the bottom left to the top right of the rectangle passes through the vertex of the square.


If the area of the rectangle is integer, what can you deduce about d?


Solution

By the Pythagorean Theorem, x2 + y2 = d2.

By similar triangles,
    (xminus1)/1 = 1/(yminus1)
    (xminus1)(yminus1) = 1
    xyminusxminusy+1 = 1
therefore xy = x + y

Squaring both sides, (xy)2 = x2 + y2 + 2(xy) = d2 + 2(xy)

therefore (xy)2 minus 2(xy) = d2
    (xy)2 minus 2(xy) + 1 = d2 + 1
    (xyminus1)2 = d2 + 1
    xyminus1 = plus or minusradical(d2 + 1)
therefore xy = 1 plus or minus radical(d2 + 1)

But as the area, xy greater than 0, we only need take the positive root.

    xy = 1 + radical(d2 + 1)

So for xy to be integer, d2+1 must be square. Let d2 + 1 = k2, where k greater than 1, hence d = radical(k2minus1).

Given that d = radical15, find the dimensions of the rectangle.
What if d = radical8?
For which values of d are both the area and the dimensions integer?

Problem ID: 223 (24 May 2005)     Difficulty: 3 Star

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