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

Home > Shapes
Fixed length from unit square. (Posted on 2009-10-01) Difficulty: 3 of 5
Given a unit square and a fixed length, r.

Construct the set of all points which are at distance r from some point on the square.

Find the area of this set.

Note: A square is composed of 4 line segments, not the interior. For some values of r the set will have a hole in it.

Example of the construction:

No Solution Yet Submitted by Jer    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 2 of 4 |

The area of the set is pi*r^2+4*r+1 minus the
area of the center hole.
For r < 1/2, 
   Area of center hole = (1-2*r)^2
For 1/2 <= r <= sqrt(2)/2,
   No center hole
For sqrt(2)/2 < r,
   Area of center hole 
          -- sqrt(r^2 - 1/4)
          |
          |
     =  4 | [sqrt(r^2 - x^2) - 1/2] dx
          |
          |
         -- 1/2 
     = 2*r^2*[arctan(t)-arctan(1/t)] - t + 1
       where t = 2*sqrt(r^2 - 1/4) and
       arctan returns radians.
 

  Posted by Bractals on 2009-10-01 14:41:27
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 (1)
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