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

Home > Shapes > Geometry
Random chord problem (Posted on 2009-05-11) Difficulty: 1 of 5
Pick two points at random on a circle and draw the chord connecting them.

Pick two more points and connect them with a second chord.

What is the probability that these chords intersect?

See The Solution Submitted by Jer    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips That can't be right (Hint, Spoiler) | Comment 2 of 17 |
If the first chord is a diameter, the chances that the chords cross is 50%.  If the first chord is less than a diameter, the chances that the chords cross is less than 50%.  So, averaging all possible chord lengths, the probability must be less than 50%.

/*********************************************/

I suggest solving by transforming the problem out.  Pick a circle with unit circumference, break it at one of the two points, and straighten it into a unit line.  Now the problem becomes:

   Pick one point on a unit length.
   Pick two more points. 
   What is the probability that they lie on opposite sides
     of the first point?

Well, if the first point is at x,
then the probability is 2x*(1-x)

And the puzzle solution
is the integral from 0 to 1 of (2x*(1-x))dx

Unfortunately, my calculus is a little rusty.  I'll do the integration later, unless somebody does it first.
 

Edited on May 11, 2009, 8:34 pm
  Posted by Steve Herman on 2009-05-11 12:39:41

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 (11)
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