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
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.
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Posted by Charlie
on 2010-09-14 17:28:10 |
solution
