Question

For distinct complex numbers $z_1, z_2, \ldots, z_{673},$ the polynomial $(x - z_1)^3 (x - z_2)^3 \cdots (x - z_{673})^3$ can be expressed as $x^{2019} + 20x^{2018} + 19x^{2017} + g(x),$ where $g(x)$ is a polynomial with complex coefficients and with degree at most $2016.$ The value of $\left| \sum_{1 \le j \lt k \le 673} z_j z_k \right|$ can be expressed in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

The polynomial's $2019$ roots are the numbers $z_j,$ each repeated three times. By Vieta's formulas, the coefficient of $x^{2018}$ is minus the sum of all roots: $3\sum_j z_j = -20,$ so $\sum_j z_j = -\frac{20}{3}.$

The coefficient of $x^{2017}$ is the sum over unordered pairs of roots. A pair may use two copies from one triple ( $\binom{3}{2} = 3$ pairs for each $j$ ) or copies from two different triples ( $3 \cdot 3 = 9$ pairs for each $j \lt k$ ). Writing $S = \sum_{j \lt k} z_j z_k,$ $19 = 3\sum_j z_j^2 + 9S = 3\left[\left(-\tfrac{20}{3}\right)^2 - 2S\right] + 9S = \frac{400}{3} + 3S,$ so $3S = 19 - \frac{400}{3} = -\frac{343}{3}$ and $S = -\frac{343}{9}.$

Hence $|S| = \frac{343}{9},$ which is in lowest terms, and $m + n = 343 + 9 = 352.$

2019 AIME I Problem 10

Solution:

Problem 10 in Other Years