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

Home > Shapes > Geometry
Find GM, Get One Side? (Posted on 2007-06-11) Difficulty: 2 of 5
In triangle PQR; QR = p, PR = q and PQ = r with Angle QPR = 2* Angle PQR. The length of all the three sides of the triangle are different.

Is it always true that: p2 = q (q + r)?.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 1 of 4

From the law of sines,
    q          p           p              p
 -------- = -------- = --------- = ----------------
  sin(Q)     sin(P)     sin(2Q)     2 sin(Q)cos(Q)
                      or
           
 cos(Q) = p/2q
              
From the law of cosines,
 q^2 = p^2 + r^2 - 2pr cos(Q)
     = p^2 + r^2 - 2pr (p/2q)
          or
 p^2(q - r) = q(q + r)(q - r)
If q != r, then p^2 = q(q + r).
If q = r, then Q = R = 45 and P = 90
   and therefore p^2 = q^2 + r^2 = q^2 + qr
                     = q(q + r)
Thus, p^2 = q(q + r) whether q = r or not.
 

  Posted by Bractals on 2007-06-11 12:49:11
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 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information