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

Home > Shapes > Geometry
Three Lock Box (Posted on 2010-07-19) Difficulty: 3 of 5
You are asked to help design a Three Lock Box. Your job is to decide the locations of the three locks. The door is a unit square. To operate properly, each lock must be the same distance from the nearest door edge and also that same distance from each of the other two locks.

No Solution Yet Submitted by Larry    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): solution Comment 4 of 4 |
(In reply to re: solution by brianjn)

Maybe I can get this one right.

As each angle of an equilateral triangle is 60 degrees (pi/3 radians), we can determine the maximum the size of our triangle. Rotating it 15 degrees (pi/12 radians), that is, centering it so the bisector of the triangle is parallel to the diagonal of the square, we can calculate the distance the rotated triangle will be between the equidistant edges of the square. Let us assign d as the distance from the edge of the square and the length of the side of the triangle. Therefore, we have the two distances -- d and (1-2d) -- and the angles -- pi/2 and (pi/2 - pi/12 =) 5pi/12 -- needed to find the value of d.
Using the Law of Sines: 
     sin(5pi/12)/(1 - 2d) = sin(pi/2)/d
=> sin(5pi/12)/(1 - 2d) = 1/d
=> sin(5pi/12) = 1/d*(1 - 2d)
=> sin(5pi/12) = 1/d - 2
=> sin(5pi/12) + 2 = 1/d
=> d = 1/(sin(5pi/12) + 2)
=> d =~ 0.33716284848943383961467150932587

Edited on July 20, 2010, 2:13 pm
  Posted by Dej Mar on 2010-07-20 10:56:24

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

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

Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information