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

Home > Shapes > Geometry
Circular Logic III (Posted on 2006-08-09) Difficulty: 4 of 5
Suppose you had a square intersecting a circle like in Circular Logic, where one vertex is inside the circle and three are outside.

If the vertex inside the circle was not located at the center, but anywhere inside the circle, such that segments ending at that vertex were of length a and b and the circle had radius r, what would be the area of their intersection be?

Note: A geometric solution is expected, without using calculus.

No Solution Yet Submitted by Gamer    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): can this be simplified? -- spoiler Comment 3 of 3 |
(In reply to re: can this be simplified? -- spoiler by Bractals)

hmmmm...  that would be a whole different calculation. I can't imagine one formula would cover such different situations.

Also, Gamer didn't include a measure for the side of the square in the formulation of the puzzle.

  Posted by Charlie on 2006-08-09 13:46:54
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 (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information