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

Home > Just Math
Trisected Triangle (Posted on 2006-02-03) Difficulty: 3 of 5
A right triangle PQR, has its hypotenuse, PR, trisected at points A and B. Two lines, QA and QB are then drawn and k is such that QA^2 + QB^2 = (PR^2) * k. Find the value of k.

No Solution Yet Submitted by Chris, PhD    
Rating: 4.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Simplify by Going Complex | Comment 3 of 12 |

k has the value 5/9 independent of the shape of the right triangle.

A simple demonstration results when the triangle is placed in the complex plane, inscribed in the unit circle with its hypotenuse PR the diameter along the real axis. Q is then exp(it) for some angle of t radians in the interval (0,pi), A is at 1/3, and B is at -1/3.  Thus QA^2=[exp(it)-1/3][exp(-it)-1/3] and QB^2=[exp(it)+1/3][exp(-it)+1/3] so that QA^2+QB^2=2 2/9=20/9. Since PR^2=4, we then have k=5/9.

Edited on February 3, 2006, 5:47 pm
  Posted by Richard on 2006-02-03 17:45:59

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 (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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