Question

Let $z_1,$ $z_2,$ $z_3,$ $\ldots,$ $z_{12}$ be the $12$ zeroes of the polynomial $z^{12} - 2^{36}.$ For each $j,$ let $w_j$ be one of $z_j$ or $iz_j.$ Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m + \sqrt{n},$ where $m$ and $n$ are positive integers. Find $m + n.$

答案

Solution:

The zeroes are $z_j = 8\left(\cos\frac{\pi j}{6} + i\sin\frac{\pi j}{6}\right)$ for $j = 1, \ldots, 12,$ and $\operatorname{Re}(iz_j) = -\operatorname{Im}(z_j).$ Since the choices are independent, the maximum real part of the sum is $\sum_j 8\max\left(\cos\frac{\pi j}{6},\, -\sin\frac{\pi j}{6}\right).$

Comparing the two values, $-\sin\frac{\pi j}{6}$ is larger exactly for $j = 5, \ldots, 10.$ The cosines kept, for $j = 1, 2, 3, 4, 11, 12,$ sum to $\frac{\sqrt{3}}{2} + \frac{1}{2} + 0 - \frac{1}{2} + \frac{\sqrt{3}}{2} + 1 = 1 + \sqrt{3},$ and the values $-\sin\frac{\pi j}{6}$ kept, for $j = 5, \ldots, 10,$ sum to $-\frac{1}{2} + 0 + \frac{1}{2} + \frac{\sqrt{3}}{2} + 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}.$

The maximum is $8\left(2 + 2\sqrt{3}\right) = 16 + 16\sqrt{3} = 16 + \sqrt{768},$ so $m + n = 16 + 768 = 784.$

2011 AIME II Problem 8

Solution:

Problem 8 in Other Years