HMMT-Nov 2023 guts, Problem 19

Suppose $a, b$, and $c$ are real numbers such that \[ \begin{aligned} a^{2}-b c & =14 \\ b^{2}-c a & =14, \text { and } \\ c^{2}-a b & =-3 \end{aligned} \] Compute $|a+b+c|$.

Solution: Solution: Subtracting the first two equations gives $(a-b)(a+b+c)=0$, so either $a=b$ or $a+b+c=0$. However, subtracting first and last equations gives $(a-c)(a+b+c)=17$, so $a+b+c \neq 0$. This means $a=b$. Now adding all three equations gives $(a-c)^{2}=25$, so $a-c= \pm 5$. Then $a+b+c= \pm \frac{17}{5}$. $\fbox{\frac{17}{5}}$.

Answer: \frac{17}{5}