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 | Comment 2 of 12 |
(PR)^2 is equal to (QP)^2+(QR)^2.

1. (QA)^2 = ((2/3)(QP))^2 + ((1/3)(QR))^2

2. (QB)^2 = ((1/3)(QP))^2 + ((2/3)(QR))^2

3. (QA)^2 + (QB)^2 = 5/9 (QP)^2 + 5/9 (QR)^2

4. (QA)^2 + (QB)^2 = k*(PR)^2

5. (QA)^2 + (QB)^2 = k * ((QP)^2 + (QR)^2)

If you distribute the k through, you get:

6. (QA)^2 + (QB)^2 = k*(QP)^2 + k*(QR)^2

7. From line 3, since both coefficients are equal, k = 5/9.

So the answer is, k = 5/9.

Another method is to realize that QP and QR will each be multiplied by 2/3 once, and 1/3 once. Since everything is squared, (2/3)^2 + (1/3)^2 = 4/9 + 1/9 = 5/9. Therefore, using either method, k = 5/9, or I'm an idiot and did something really easy, wrong.

Edited on February 3, 2006, 5:23 pm
  Posted by Justin on 2006-02-03 17:20:02

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 (3)
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