
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,
(x
1)/1 = 1/(y
1)
(x
1)(y
1) = 1
xy
x
y+1 = 1
xy = x + y
Squaring both sides, (xy)2 = x2 + y2 + 2(xy) = d2 + 2(xy)
(xy)2
2(xy) = d2
(xy)2
2(xy) + 1 = d2 + 1
(xy
1)2 = d2 + 1
xy
1 = ![]()
(d2 + 1)
xy = 1
(d2 + 1)
But as the area, xy
0, we only need take the positive root.
xy = 1 +
(d2 + 1)
So for xy to be integer, d2+1 must be square. Let d2 + 1 = k2, where k
1, hence d =
(k2
1).
Given that d =
15, find the dimensions of the rectangle.
What if d =
8?
For which values of d are both the area and the dimensions integer?
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