Find all real numbers p, q, and r that satisfy this system of equations:
1/p+q+r = p+1/q+r = p+q+1/r = 3
*** Adapted from a problem appearing at El Salvador Math Olympiad in 2017.
Solution set:
(1,1,1), (1/2,1/2,1/2), (3,3,-1/3), (3,-1/3,3), (-1/3,3,3)
Try summing all 3 equations:
(1/p)+(1/q)+(1/r) + 2(p+q+r) = 9 (doesn't seem to help much)
What if p=q=r=x? Then if 1/x + 2x = 3, 2x^2 - 3x + 1 = 0, x = {1,1/2}
Two solutions are: (1,1,1), (1/2,1/2,1/2)
Try subtracting one eqn from another:
p - 1/p = q - 1/q = r - 1/r
or
p - q = 1/p - 1/q
p - q = (q - p)/pq
(p - q) - (q - p)/pq = 0
(q - p)(1 + 1/pq) = 0
So either p = q (already dealt with this case)
or
(1 + 1/pq) = 0 --> pq = -1
For example p = x and q = -1/x
case one (x, x, -1/x) case two (x, -1/x, -1/x)
case one:
1/p+q+r = 1/x + x -1/x = 3 --> x = 3
case two:
1/p+q+r = 1/x - 1/x + x = 3 --> x = 3
But triplets such as (3,-1/3,-1/3) do not solve the original equations.
Three more solutions: (3,3,-1/3), (3,-1/3,3), (-1/3,3,3)
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Posted by Larry
on 2023-11-20 08:22:20 |
Solution