SOLUTION: Let x = 2012(a - b), y = 2012(b - c), and z = 2012(c - a), where a,b,c are real numbers, and assume xy + yz + zx is not = 0. Compute (x^2 + y^2 + z^2)/(xy + yz + zx).

Question 1078282: Let x = 2012(a - b), y = 2012(b - c), and z = 2012(c - a), where a,b,c are real numbers, and assume
xy + yz + zx is not = 0. Compute (x^2 + y^2 + z^2)/(xy + yz + zx).
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20064) (Show Source): You can put this solution on YOUR website!
Let x = 2012(a - b), y = 2012(b - c), and z = 2012(c - a), where a,b,c are real numbers, and assume
xy + yz + zx is not = 0. Compute (x^2 + y^2 + z^2)/(xy + yz + zx).

The 2012 was just put in there to complicate matters. It could have been any number whatever. Most likely this problem was made up 5 years ago in in 2012. Let's just divide everything through by 2012: , , and Add all those equations together: Multiply thru by 2012: So We want to compute: Substitute -x-y for z Edwin

Answer by ikleyn(52879) (Show Source): You can put this solution on YOUR website!
.
Let x = 2012(a - b), y = 2012(b - c), and z = 2012(c - a), where a,b,c are real numbers, and assume
xy + yz + zx is not = 0. Compute (x^2 + y^2 + z^2)/(xy + yz + zx).
~~~~~~~~~~~~~~~~~~~~~

It could be done shorter.

x = 2012(a - b), y = 2012(b - c), and z = 2012(c - a) ===> x + y + z = 2012*(a-b+b-c+c-a) = 0. Hence 0 = = = + ===> = ===> = -2.

Solved.

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