Indija mo 2011

We have three relations: $$\begin{array}{rl}a{\alpha }^2-b\alpha -c& =\lambda ,\\ b{\alpha }^2-c\alpha -a& =\lambda ,\\ c{\alpha }^2-a\alpha -b& =\lambda ,\end{array}$$ where $\lambda$ is the common value. Eliminating ${\alpha }^2$ from these, taking these equations pairwise, we get three relations: $$\begin{array}{rl}(ca-b^2)\alpha -(bc-a^2)& =\lambda (b-a),\\ (ab-c^2)\alpha -(ca-b^2)& =\lambda (c-b),\\ (bc-a^2)-(ab-c^2)& =\lambda (a-c).\end{array}$$ Adding these three, we get $$(ab+bc+ca-a^2-b^2-c^2)(\alpha -1)=0.$$ (Alternatively, multiplying above relations respectively by $b-c$, $c-a$ and $a-b$, and adding also leads to this.) Thus either $ab+bc+ca-a^2-b^2-c^2=0$ or $\alpha =1$. In the first case $$0=ab+bc+ca-a^2-b^2-c^2=\frac{1}{2}((a-b{)}^2+(b-c{)}^2+(c-a{)}^2)$$ shows that $a=b=c$. If $\alpha =1$, then we obtain $$a-b-c=b-c-a=c-a-b,$$ and once again we obtain $a=b=c$.