No teams got this problem correct at the tournament :).

$a,b,c,d,x,y,z$ are real numbers such that

\begin{eqnarray*}ax+by+cz &=& d,\\ bx+cy+dz &=& a,\\ cx+dy+az &=& b,\\ dx+ay+bz &=& c. \end{eqnarray*}

If $a\neq c$, find all possible values of $y$.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119313[/url]

question 10 someone posted it...

[hide]Subtracting the third equation from the first, we get
\[(a-c)x+(b-d)y+(c-a)z = (d-b) \ \Rightarrow \ (a-c)(x-z) = (d-b)(y+1).\]
Subtracting the fourth equation from the second, we get
\[(b-d)x+(c-a)y+(d-b)z = (a-c) \ \Rightarrow \ (b-d)(x-z) = (a-c)(y+1).\]

The product of these two equations is
\[(a-c)(b-d)(x-z)^{2}= (a-c)(d-b)(y+1)^{2}.\]
Assuming $a \neq c$, we can divide both sides by $a-c$.  Re-arranging, we get
\[(b-d)[(x-z)^{2}+(y+1)^{2}] = 0.\]

If the first factor is zero, then $b = d$.  Then $(b-d)(x-z) = (a-c)(y+1) = 0$ and $a \neq c$, so $y =-1$.  If the second factor is zero, then again $y$ must be $-1$.  So the only possible value of $y$ is $-1$.

There are many possible solutions when $y =-1$.  For example, take $x = y = z =-1$, and $a$, $b$, $c$, $d$ any real numbers such that $a+b+c+d = 0$ (and $a \neq c$).[/hide]

Nice problem!