after getting x can only be 0 or 1, by symmetry, y & z is also bound to 0 or 1, giving 4 possible sets: {0,0,0}, {0,0,1}, {0,1,1} & {1,1,1}, and only one of them has a sum of 1
I worked it out a different way. I cubed both sides of x+y+z = 1 to get: (x+y+z)^3 = 1 x^3 + y^3 + z^3 + 3x^2 y + 3x^2 z + 3 y^2 x + 3y^2 z + 3z^2 x +3z^2 y + 6xyz = 1 Adding 2(x^3 + y^3 + z^3)=2 to both sides: 3x^3 + 3y^3 + 3z^3 + 3x^2 y + 3x^2 z + 3 y^2 x + 3y^2 z + 3z^2 x +3z^2 y + 6xyz = 3 or, after dividing by 3, x^3 + y^3 + z^3 + x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 2xyz = 1 Regrouping: (x^3 + y^2 x + z^2 x) + (x^2 y + y^3 + z^2 y) + (x^2 z + y^2 z + z^3) + 2xyz = 1 Factoring out the gcd: x(x^2 + y^2 + z^2) + y(x^2 + y^2 + z^2) + z(x^2 + y^2 + z^2) + 2xyz = 1 Using (x^2 + y^2 + z^2) = 1: x + y + z +2xyz = 1 Using x+y+z = 1: 2xyz = 0 One of x, y, or z is 0. Suppose z = 0: x + y = 1 and x^2 + y^2 = 1. A line and a circle can only intersect in at most two points, and clearly (1,0,0) and (0,1,0) are the solutions. By symmetry we get if z ≠ 0, that a third solution would be (0,0,1).
after getting x can only be 0 or 1, by symmetry, y & z is also bound to 0 or 1, giving 4 possible sets: {0,0,0}, {0,0,1}, {0,1,1} & {1,1,1}, and only one of them has a sum of 1
You are absolutely right, thank you for sharing.
Hi!
Can you make videos on Olympiad Combinatorics like problems from USAMO, Russian Olympiads etc?
Thanks a lot!
I worked it out a different way. I cubed both sides of x+y+z = 1 to get:
(x+y+z)^3 = 1
x^3 + y^3 + z^3 + 3x^2 y + 3x^2 z + 3 y^2 x + 3y^2 z + 3z^2 x +3z^2 y + 6xyz = 1
Adding 2(x^3 + y^3 + z^3)=2 to both sides:
3x^3 + 3y^3 + 3z^3 + 3x^2 y + 3x^2 z + 3 y^2 x + 3y^2 z + 3z^2 x +3z^2 y + 6xyz = 3 or, after dividing by 3,
x^3 + y^3 + z^3 + x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 2xyz = 1
Regrouping:
(x^3 + y^2 x + z^2 x) + (x^2 y + y^3 + z^2 y) + (x^2 z + y^2 z + z^3) + 2xyz = 1
Factoring out the gcd:
x(x^2 + y^2 + z^2) + y(x^2 + y^2 + z^2) + z(x^2 + y^2 + z^2) + 2xyz = 1
Using (x^2 + y^2 + z^2) = 1:
x + y + z +2xyz = 1
Using x+y+z = 1:
2xyz = 0
One of x, y, or z is 0. Suppose z = 0:
x + y = 1 and x^2 + y^2 = 1. A line and a circle can only intersect in at most two points, and clearly (1,0,0) and (0,1,0) are the solutions. By symmetry we get if z ≠ 0, that a third solution would be (0,0,1).
These are very clever substitutions you have used on the second paragraph of your solution. Thank you for sharing.