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

Home > Shapes > Geometry
Double Secant (Posted on 2018-02-15) Difficulty: 3 of 5
Two concentric circles of radius R and r, with R>r are both intersected by the same secant line. The points of intersection, in order, are A,B,C,D.

Prove AC*CD is constant.

No Solution Yet Submitted by Jer    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 3 of 3 |

Let O be the center of the two circles.
Let Q be the foot of the perpendicular
from O to the secant ABCD.

Let  x = |OQ|
      z = |QC|
      Z = |QA| = |QD|

Then z^2 = r^2 - x^2
        Z^2 = R^2 - x^2

     |AC| = |AQ| + |QC| = Z + z
     |CD| = |QD| - |QC| = Z - z

     |AC|*|CD| = Z^2 - z^2
               = (R^2 - x^2)-(r*2 - x^2)
               = R^2 - r^2

                 a constant


Note: If the secant passes through the
      center O, then the problem is trivial. 

  Posted by Bractals on 2018-02-20 19:07:58
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 (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information