Perimeter Equals Area in a Triangle

Date: 4/2/96 at 11:19:9
From: Anonymous
Subject: Right triangles where area = perimeter

When will the area and perimeter of a right triangle be 
numerically equal?

Date: 4/3/96 at 12:54:16
From: Doctor Ken
Subject: Re: Right triangles where area = perimeter

Hello!

Well, let's see.  Let's say we know the lengths (x and y) of the 
two legs of the triangle.  Then the area of the triangle will be 
xy/2, and the perimeter will be x + y + Sqrt{x^2 + y^2}.  So 
we're looking for solutions to the equation

x * y
-----  =  x + y + Sqrt{x^2 + y^2}.
  2

Now how are we going to solve that thing?  Well, move the 
x and y over to the left side to get this:

x * y
-----  -  x - y   =   Sqrt{x^2 + y^2}.
  2

Now square both sides:

x^2 + 2xy - x^2 y + y^2 - xy^2 + (x^2 y^2)/4  =  x^2 + y^2

Cancel:

      2xy - x^2 y - xy^2 + (x^2 y^2)/4  =  0

So either x = 0 or y = 0 or we have

       2 - x - y + xy/4  =  0

             x(y/4 - 1)  =  y - 2
 
                             y - 2      4y - 8
                      x  =  -------  =  ------
                            y/4 - 1      y - 4


So I guess that's our answer.  Ether x or y is 0 (so we don't 
have a triangle at all), or x = (4y - 8)/(y - 4).

I encourage you to check out The Geometer's Sketchpad to 
see what intuition it can give you about this problem.  You 
can get a demo version from the page

http://mathforum.org/dynamic.html    

Have fun!

-Doctor Ken,  The Math Forum