The perimeter of a rectangle in units equals its area in units squared. (4,4), (3,6), and (6,3) are three possible pairs of lengths for this rectangle.
Give another pair of positive integral sides for this rectangle or prove why there isn't another pair.
The answer is there are no other pairs.
Your constraint is equivalent to:
x * y = 2x + 2y
Solve for y:
xy - 2y = 2x
y(x-2) = 2x
y = 2x / (x-2)
Clearly, this works for the points given in the problem.
This is the equation for a hyperbola (I think). In any case... you can graph this equation, and it has a vertical asymptote and x=2, and a horizontal asymptote at y=2 (where the function "blows up").
Since we are restricted to the region of the function where both variable are positive, we can quickly check x = 1, where y = -2.... doesn't work.
x = 2 it blows up...
and x = 3, 4, and 6 we already have the solutions (given in the problem)
at x = 5, y = 10/3, not integral.... doesn't work...
so we are only concerned for x>=7.
Well for x>2, we have a continuously decreasing function with a horizontal asymptote of 2.
Since we've got the point (6,3) on this graph, the graph (as we move to the right) must continuously decrease approaching y=2.... Therefore, all other y values (for x>6) are NON-INTEGRAL between 2 and 3.
So.... no other positive pairs.
Edited on September 12, 2003, 2:28 pm
Solution (and spoiler) I think
