Question

设 $A$ 和 $B$ 是半径为 $2$ 的半圆弧的两个端点。六个等距点 $C_1$ 、 $C_2$ 、 $\ldots$ 、 $C_6$ 将该半圆弧分成七段全等的弧。画出所有形如 $\overline{AC_i}$ 或 $\overline{BC_i}$ 的弦。令 $n$ 为这十二条弦长的乘积。求 $n$ 除以 $1000$ 的余数。

Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2.$ The arc is divided into seven congruent arcs by six equally spaced points $C_1,$ $C_2,$ $\ldots,$ $C_6.$ All chords of the form $\overline{AC_i}$ or $\overline{BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the remainder when $n$ is divided by $1000.$

答案

答案：672

知识点：单位根复数弦

难度评级：3160

小提示：

在复平面中表示圆，取 $A = -2$ 、 $B = 2$ ，且 $C_i = 2\omega^i$ ，其中 $\omega = e^{\frac{\pi \mathrm{i}}{7}}$

Model the circle in the complex plane with $A = -2,$ $B = 2,$ and $C_i = 2\omega^i$ where $\omega = e^{\frac{\pi \mathrm{i}}{7}}$

大提示：

$AC_i \cdot BC_i = 4\,|\omega^{2i} - 1|$ ，且 $\omega^{2i}$ 正好是非平凡的 $7$ 次单位根；它们到 $1$ 的距离乘积为 $7$

$AC_i \cdot BC_i = 4\,|\omega^{2i} - 1|,$ and the $\omega^{2i}$ are exactly the nontrivial $7$ th roots of unity, whose product of distances to $1$ is $7$

解答：

在复平面中放置该圆，圆心为 $0$ ，取 $A = -2$ 、 $B = 2$ ，且对 $i = 1, \ldots, 6$ ，令 $C_i = 2\omega^i$ ，其中 $\omega = e^{\frac{\pi \mathrm{i}}{7}}$ 。则 $AC_i = 2\,|\omega^i + 1|$ ， $BC_i = 2\,|\omega^i - 1|$ ，所以 $AC_i \cdot BC_i = 4\,\bigl|\omega^{2i} - 1\bigr|\text{。}$

当 $i$ 取 $1, \ldots, 6$ 时， $\omega^{2i}$ 遍历所有六个非平凡的 $7$ 次单位根。令 $\zeta = \omega^2$ 。由于 $\prod_{j=1}^{6} (x - \zeta^j)$ $= 1 + x + \cdots + x^6$ ，代入 $x = 1$ 得 $\prod_{j=1}^{6} \bigl|1 - \zeta^j\bigr| = 7$ 。因此 $\begin{aligned} n &= \prod_{i=1}^{6} 4\,\bigl|\omega^{2i} - 1\bigr| \\ &= 4^6 \cdot 7 = 28672 \end{aligned}\text{。}$

$n$ 除以 $1000$ 的余数为 $672$ 。

Put the circle in the complex plane with center $0,$ $A = -2,$ $B = 2,$ and $C_i = 2\omega^i$ for $i = 1, \ldots, 6,$ where $\omega = e^{\frac{\pi \mathrm{i}}{7}}.$ Then $AC_i = 2\,|\omega^i + 1|$ and $BC_i = 2\,|\omega^i - 1|,$ so $AC_i \cdot BC_i = 4\,\bigl|\omega^{2i} - 1\bigr|.$

As $i$ runs over $1, \ldots, 6,$ the numbers $\omega^{2i}$ run over all six nontrivial $7$ th roots of unity. Let $\zeta = \omega^2.$ Since $\prod_{j=1}^{6} (x - \zeta^j)$ $= 1 + x + \cdots + x^6,$ plugging in $x = 1$ gives $\prod_{j=1}^{6} \bigl|1 - \zeta^j\bigr| = 7.$ Therefore $\begin{aligned} n &= \prod_{i=1}^{6} 4\,\bigl|\omega^{2i} - 1\bigr| \\ &= 4^6 \cdot 7 = 28672. \end{aligned}$

The remainder when $n$ is divided by $1000$ is $672.$

2009 AIME II 第 13 题

答案：672

难度评级：3160

小提示：

大提示：

解答：

其他年份的第 13 题