Mock AIME I 2015 Problems/Problem 11

Solution

For convenience, let's use $a, b, c$ instead of $\alpha, \beta, \gamma$ . Define a polynomial $P(x)$ such that $P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc$ . Let $j = ab + ac + bc$ and $k = -abc$ . Then, our polynomial becomes $P(x) = x^3 - (a+b+c)x^2 + jx + k$ . Note that we want to compute $-\frac{j}{k}$ .

From the given information, we know that the coefficient of the $x^2$ term is $6$ , and we also know that $P(-1) = -33$ , or in other words, $-j + k = -26$ . By Newton's Sums (since we are given $a^3 + b^3 + c^3$ ), we also find that $6j + k = 43$ . Solving this system, we find that $(j, k) \in (\frac{69}{7}, -\frac{113}{7})$ . Thus, $\frac{j}{-k} = \frac{69}{113}$ , so our final answer is $69 + 113 = \boxed{182}$ .

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Mock AIME I 2015 Problems/Problem 11

Solution