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Arithmetic Progression Triangle (Posted on 2007-01-28) Difficulty: 3 of 5
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.

See The Solution Submitted by Bractals    
Rating: 2.5000 (2 votes)

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re: solution | Comment 4 of 5 |
(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
   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

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