Find two rectangles, with integral sides, such that the area of the first is three times the area of the second, and the perimeter of the second is three times the perimeter of the first.

There are 4 degrees of freedom so it should not be a surprise that there are multiple solutions.

I called the first A by B and the second C by D
then we have the system
AB = 3CD
3(A+B) = (C+D)

Solving for A gives
6A = (C+D)+√(Cē-106CD+Dē)

To find some solutions I let C=1 and did a quick table to find perfect-square discriminants. 
There are seven, but only 4 give integer values for B:
{A,B,C,D}={22,15,1,110}
{35,12,1,140}
{48,11,1,176}
{87,10,1,290}

I have no doubt that letting C=2,3,... will give more solutions but I will yield to a computer program to find more.

Edited on December 12, 2013, 3:53 pm

Posted by Jer on 2013-12-09 11:20:33