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).
<pre>
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: 

{{{x/2012 = a - b}}}, {{{y/2012 = b - c}}}, and {{{z/2012 = c - a}}}

Add all those equations together:

{{{x/2012+y/2012+z/2012}}}{{{""=""}}}{{{a-b+b-c+c-a}}}

{{{x/2012+y/2012+z/2012}}}{{{""=""}}}{{{0}}}

Multiply thru by 2012:

{{{x+y+z}}}{{{""=""}}}{{{0}}}

So {{{z}}}{{{""=""}}}{{{-x-y}}}

We want to compute:

{{{(x^2 + y^2 + z^2)/(xy + yz + zx)}}}{{{""=""}}}

Substitute -x-y for z

{{{(x^2 + y^2 + (-x-y)^2)/(xy + y(-x-y) + (-y-x)x)}}}{{{""=""}}}

{{{(x^2 + y^2 + x^2+2xy+y^2)/(xy-xy-y^2-xy-x^2)}}}{{{""=""}}}

{{{(2x^2+2y^2+2xy)/(-x^2-y^2-xy)}}}{{{""=""}}}

{{{(2(x^2+y^2+xy))/(-(x^2+y^2+xy))}}}{{{""=""}}}

{{{(2(cross(x^2+y^2+xy)))/(-(cross(x^2+y^2+xy)))}}}{{{""=""}}}

{{{-2}}}

Edwin</pre>