Given lengths R and r with R > 2r. Construct a triangle with side lengths in arithmetic progression and R and r its circumradius and inradius respectively.

(In reply to solution by Art M)

You have an algebra problem at this step:

   More simplification: 

   a^2 - 4x^2 = 12r^2 
   a^2 - x^2 = 6rR 

   And: 

   a = sqrt(4r(2R-r)) 
   x = sqrt(2r(2R-r))

Shouldn't  it be? 

   a = sqrt(4r(2R-r)) 
   x = sqrt(2r(R-2r))

And what does that do for the values
of your sides? 

Edited on January 31, 2007, 12:02 pm

Posted by Bractals on 2007-01-31 11:53:22