All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: solution | Comment 4 of 6 |
(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

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information