HMMT-Nov 2016 guts, Problem 27

Let $r_{1}, r_{2}, r_{3}, r_{4}$ be the four roots of the polynomial $x^{4}-4 x^{3}+8 x^{2}-7 x+3$. Find the value of \[ \frac{r_{1}^{2}}{r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}+\frac{r_{2}^{2}}{r_{1}^{2}+r_{3}^{2}+r_{4}^{2}}+\frac{r_{3}^{2}}{r_{1}^{2}+r_{2}^{2}+r_{4}^{2}}+\frac{r_{4}^{2}}{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}} \]

Solution: Add 1 to each fraction to get \[ \frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}{r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}+\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}{r_{1}^{2}+r_{3}^{2}+r_{4}^{2}}+\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}{r_{1}^{2}+r_{2}^{2}+r_{4}^{2}}+\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}} \] This seems like a difficult problem until one realizes that \[ r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}=\left(r_{1}+r_{2}+r_{3}+r_{4}\right)^{2}-2\left(r_{1} r_{2}+r_{1} r_{3}+r_{1} r_{4}+r_{2} r_{3}+r_{2} r_{4}+r_{3} r_{4}\right)=4^{2}-2 \cdot 8=0 \] Thus, our current expression is 0 . Noting that we added 4, the original value had to be -4 . $\fbox{-4}$.

Answer: -4