Question

Sean $z_1,$ $z_2,$ $z_3,$ $\ldots,$ $z_{12}$ los $12$ ceros del polinomio $z^{12} - 2^{36}.$ Para cada $j,$ sea $w_j$ uno de $z_j$ o $iz_j.$ Entonces el valor máximo posible de la parte real de $\sum_{j=1}^{12} w_j$ puede escribirse como $m + \sqrt{n},$ donde $m$ y $n$ son enteros positivos. Halla $m + n.$

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 $m + n.$

Respuesta

Solución:

Los ceros son $z_j = 8\left(\cos\frac{\pi j}{6} + i\sin\frac{\pi j}{6}\right)$ para $j = 1, \ldots, 12,$ y $\operatorname{Re}(iz_j) = -\operatorname{Im}(z_j).$ Como las elecciones son independientes, la parte real máxima de la suma es $\sum_j 8\max\left(\cos\frac{\pi j}{6},\, -\sin\frac{\pi j}{6}\right).$

Comparando los dos valores, $-\sin\frac{\pi j}{6}$ es mayor exactamente para $j = 5, \ldots, 10.$ Los cosenos conservados, para $j = 1, 2, 3, 4, 11, 12,$ suman $\begin{aligned} &\frac{\sqrt{3}}{2} + \frac{1}{2} + 0 - \frac{1}{2} \\ &\quad {}+ \frac{\sqrt{3}}{2} + 1 = 1 + \sqrt{3}, \end{aligned}$ y los valores $-\sin\frac{\pi j}{6}$ conservados, para $j = 5, \ldots, 10,$ suman $\begin{aligned} &-\frac{1}{2} + 0 + \frac{1}{2} + \frac{\sqrt{3}}{2} \\ &\quad {}+ 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}. \end{aligned}$

El máximo es $\begin{aligned} &8\left(2 + 2\sqrt{3}\right) \\ &= 16 + 16\sqrt{3} \\ &= 16 + \sqrt{768}, \end{aligned}$ así que $m + n = 16 + 768 = 784.$

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 $\begin{aligned} &\frac{\sqrt{3}}{2} + \frac{1}{2} + 0 - \frac{1}{2} \\ &\quad {}+ \frac{\sqrt{3}}{2} + 1 = 1 + \sqrt{3}, \end{aligned}$ and the values $-\sin\frac{\pi j}{6}$ kept, for $j = 5, \ldots, 10,$ sum to $\begin{aligned} &-\frac{1}{2} + 0 + \frac{1}{2} + \frac{\sqrt{3}}{2} \\ &\quad {}+ 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}. \end{aligned}$

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

2011 AIME II Problema 8

Solución:

El Problema 8 en otros años