Each of x, y, and z is a real number that satisfies this system of equations:
• 1/x+y+z=3
• x+1/y+z=3
• x+y+1/z=3
Determine all possible solutions to the above system of equations.
Note: Adapted from a problem that appeared on 2017 El Salvador Math Olympiad.
I get 5 solutions (3 of which are the same except for swithing the order of x,y,z
By inspection, (1,1,1) is a solution
Case one: Suppose x=y=z;
1/x+2x=3; 2x^2  3x + 1 = 0
(x1)(2x1)=0; so (1/2,1/2,1/2) is also a solution
Case two: Suppose 2 are equal but the third is different
wlog, 1/x+x+z=3
1/x+x+z=3
x+1/x+z=3
x+x+1/z=3
x+1/x = 2x; 1/x = x; x = ±1; 1 does not solve 3rd eqn.
eqn 3: 1/z = 32x; z=1/(32x); also z = 3x1/x
1/(32x) = 3x1/x
1 = (32x) (3x1/x)
1 = 9  3x  3/x  6x + 2x^2 + 2
2x^3  9x^2 + 10x  3 = 0
(x1)(2x^2  7x + 3) = 0
det: 49  24 = 25
quadratic's roots are (7 ± 5) / 4 = {3, 1/2}
The other root is x=y=1, we already have (1,1,1)
The last root is x=y=1/2, we already have (1/2,1/2,1/2)
try x=y=3
1/3 + 3 + (1/3) = 3
2*3 + (3) = 3 (3,3,1/3)(3,1/3,3)(1/3,3,3) are solutions
Case three: Suppose all different
Add 3 equations
(1/x) + (1/y) + (1/z) + 2(x+y+z) = 9
not sure this helps
3z = 1/x+y = x+1/y
y + xy^2 = x^2y + x
y + xy^2  x^2y  x = 0
(yx) + xy(yx) = 0
(yx)(1+xy) = 0 note (yx) not zero, since assuming all different
(1+xy) = 0 x = 1/y
(1+xz) = 0 x = 1/z
(1+yz) = 0
Therefore y=z which is a contradiction.
So no solutions for case three.
Summary, I get 5 solutions for (x,y,z):
(1,1,1)
(1/2,1/2,1/2)
(3,3,1/3)
(3,1/3,3)
(1/3,3,3)

Posted by Larry
on 20230918 09:58:32 