Team Selection Test

Let $$\begin{array}{rlrl}a{x}^2+bx+c& =x_1,& a{x}^2+cx+b& =x_2\\ b{x}^2+cx+a& =x_3,& b{x}^2+ax+c& =x_4\\ c{x}^2+ax+b& =x_5,& c{x}^2+bx+a& =x_6\end{array}$$ Then we get $$\begin{array}{rl}(b-c)(x-1)& =x_1-x_2,(c-a)(x-1)=x_3-x_4,(a-b)(x-1)=x_5-x_6\\ (b-c)(x^2-1)& =x_4-x_5,(c-a)(x^2-1)=x_6-x_1,(a-b)(x^2-1)=x_2-x_3\\ (b-c)(x^2-x)& =x_3-x_6,(c-a)(x^2-x)=x_5-x_2,(a-b)(x^2-x)=x_1-x_4.\end{array}$$ For $x=0$, the set $\{x_1,x_2,x_3,x_4,x_5,x_6\}=\{c,b,a,c,b,a\}$ can not include three identical elements. Similarly, for $x=-1$, the set $\{x_1,x_2,x_3,x_4,x_5,x_6\}=\{a+c-b,a-c+b,b-c+a,b-a+c,c-a+b,c-b+a\}$ can not include three identical elements since if two of $a+b-c$, $b+c-a$, $c+a-b$ are equal, then two of $a$, $b$, $c$ are equal which is not possible. Consider the case $x\ne -1,0,1$. Since $a$, $b$, $c$ are distinct, using the equations above, we get $x_u\ne x_v$ for any odd $u$ and even $v$. Therefore the sets $S_1=\{x_1,x_3,x_5\}$ and $S_2=\{x_2,x_4,x_6\}$ should be disjoint. Therefore if there are three identical elements in $S_1\cup S_2$, then either $x_1=x_3=x_5$ or $x_2=x_4=x_6$. W.L.O.G. assume that $x_1=x_3=x_5$. In this case, we have $$a{x}^2+bx+c=b{x}^2+cx+a=c{x}^2+ax+b.$$

$x=1$ is a common root of the three 2-nd order polynomials given above, and hence we obtain that $$(x-1)(x-t_1)=(x-1)(x-t_2)=0$$ where $t_1=\frac{c-a}{a-b}$, $t_2=\frac{a-b}{b-c}$. There must be another common root different from $1$, and hence $$\frac{c-a}{a-b}=\frac{a-b}{b-c}$$ which is equivalent to $$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$$ which is not possible since $a$, $b$, $c$ are distinct. Therefore we get $x=1$.