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(2Rr))
x = sqrt(2r(2Rr))
Shouldn't it be?
a = sqrt(4r(2Rr))
x = sqrt(2r(R2r))
And what does that do for the values
of your sides?
Edited on January 31, 2007, 12:02 pm

Posted by Bractals
on 20070131 11:53:22 