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

Home > Just Math
Product = Sum + 2 (Posted on 2010-09-14) Difficulty: 3 of 5
Determine all possible triplet(s) (p, q, r) of positive real numbers, with p ≤ q ≤ r, that satisfy the following system of equations:

p*q + q*r + p*r = 12 , and:
p*q*r = p + q + r + 2

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 3

Starting with p*q*r = p + q + r + 2:

We need not consider any set where p>2, as 3*3*r = 9*r > 8+r = 3+3+r+2 for r>=3.

For any given p and q, r = (p + q + 2) / (p * q - 1), and for any given p this is monotonically decreasing as q increases, once q >= 2:

p  q  r(calculated)
1  1 inf
1  1  0
1  2  5
1  3  3
1  4  2.333333
1  5  2
1  6  1.8
1  7  1.666667
1  8  1.571429
2  2  2
2  3  1.4
2  4  1.142857
2  5  1
2  6  .9090909
2  7  .8461539
2  8  .8
2  9  .7647059

Beyond this, the r values are fractional, and certainly less than q.

The only candidates are (1,2,5), (1,3,3) and (2,2,2).

1*2 + 2*5 + 1*5 = 17
1*3 + 3*3 + 1*3 = 15
2*2 + 2*2 + 2*2 = 12

So (2,2,2) is the only triplet that satisfies both conditions.


  Posted by Charlie on 2010-09-14 17:28:10
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 (7)
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