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

Home > Shapes > Geometry
Inscribe Quadrilateral, Get Diophantus (Posted on 2007-04-09) Difficulty: 3 of 5
PQRS is a cyclic quadrilateral with PR perpendicular to QS while PR meets QS at H. T is the radius of the circumscribed circle.

Prove that:

HP2 + HQ2 + HR2 + HS2 = 4T2

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 1 of 1
 
Let O be the center of the circumscribed circle.
         HP^2 + HQ^2 + HR^2 + HS^2
       = PQ^2 + RS^2 
       = OP^2 + OQ^2 - 2(OP)(OQ)cos(POQ) +
         OR^2 + OS^2 - 2(OR)(OS)cos(ROS)
       = 4T^2 - 2T^2[cos(POQ) + cos(ROS)]
       = 4T^2 - 2T^2[cos(2PRQ) + cos(2RQS)] 
       = 4T^2 - 2T^2[cos(2HRQ) + cos(2RQH)]
       = 4T^2 - 2T^2[cos(2HRQ) + cos(2[90-HRQ])]
       = 4T^2 - 2T^2[cos(2HRQ) + cos(180 - 2HRQ)]
       = 4T^2 - 2T^2[cos(2HRQ) - cos(2HRQ)]
       = 4T^2
 

  Posted by Bractals on 2007-04-09 20:42:12
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 (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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