Question 1078282
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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).
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It could be done shorter.


<pre>
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 = {{{(x+y+z)^2}}} = {{{x^2 + y^2 + z^2 + 2xy + 2xz + 2yz}}} = {{{(x^2+y^2+z^2)}}} + {{{2*(xy + xz + yz)}}}  ===>

{{{x^2 + y^2 + z^2}}} = {{{-2(xy+xz+yz)}}}  ===>

{{{(x^2 + y^2 + z^2)/(xy+xz+yz)}}} = -2.
</pre>

Solved.