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.

The following formulas will help:

R²=(abc)²/p(p-2a)(p-2b)(p-2c)
r²=p(p-2a)(p-2b)(p-2c)/4p²

where (a,b,c) are the sides, p=a+b+c is the perimeter of the triangle.

If the sides of the triangle are (a-x,a,a+x), the equations will be simplified:

12r²=a²-4x²
3R²=(a²-x²)²/(a²-4x²)

More simplification:

a²-4x²=12r²
a²-x² =6rR

And:

a=√(4r(2R-r))
x=√(2r(2R-r))

The sides of the triangle would be:

(√2-1)x, √2x, (√2+1)x.
Edited on January 30, 2007, 11:36 pm

Posted by Art M on 2007-01-30 23:35:09