Question

令求不超过 $100x$ 的最大整数。 $x = \frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle\sum_{n=1}^{44} \sin n^\circ}.$

Let $x = \frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle\sum_{n=1}^{44} \sin n^\circ}.$ What is the greatest integer that does not exceed $100x?$

答案

答案：241

知识点：三角恒等式裂项相消

难度评级：2710

解答：

分子分母同乘 $2\sin\frac{1}{2}^\circ$ 。由于 $2\cos n^\circ \sin\frac{1}{2}^\circ$ $= \sin\left(n + \frac{1}{2}\right)^\circ$ $- \sin\left(n - \frac{1}{2}\right)^\circ$ ，且 $2\sin n^\circ \sin\frac{1}{2}^\circ$ $= \cos\left(n - \frac{1}{2}\right)^\circ$ $- \cos\left(n + \frac{1}{2}\right)^\circ$ ，两边的求和都会望远镜消去：最后一步使用和差化积公式。 $\begin{aligned} x &= \frac{\sin 44.5^\circ - \sin 0.5^\circ}{\cos 0.5^\circ - \cos 44.5^\circ} \\ &= \frac{2\cos 22.5^\circ \sin 22^\circ}{2\sin 22.5^\circ \sin 22^\circ} \\ &= \cot 22.5^\circ, \end{aligned}$

由半角公式， $\cot 22.5^\circ = \frac{1 + \cos 45^\circ}{\sin 45^\circ} = \sqrt{2} + 1$ 。因此 $100x = 100\sqrt{2} + 100$ $= 241.42\ldots$ ，不超过它的最大整数是 $241$ 。

Multiply numerator and denominator by $2\sin\frac{1}{2}^\circ.$ Since $2\cos n^\circ \sin\frac{1}{2}^\circ$ $= \sin\left(n + \frac{1}{2}\right)^\circ$ $- \sin\left(n - \frac{1}{2}\right)^\circ$ and $2\sin n^\circ \sin\frac{1}{2}^\circ$ $= \cos\left(n - \frac{1}{2}\right)^\circ$ $- \cos\left(n + \frac{1}{2}\right)^\circ,$ both sums telescope: $\begin{aligned} x &= \frac{\sin 44.5^\circ - \sin 0.5^\circ}{\cos 0.5^\circ - \cos 44.5^\circ} \\ &= \frac{2\cos 22.5^\circ \sin 22^\circ}{2\sin 22.5^\circ \sin 22^\circ} \\ &= \cot 22.5^\circ, \end{aligned}$ using the sum-to-product identities in the last step.

By the half-angle formula, $\cot 22.5^\circ = \frac{1 + \cos 45^\circ}{\sin 45^\circ} = \sqrt{2} + 1.$ Hence $100x = 100\sqrt{2} + 100$ $= 241.42\ldots,$ and the greatest integer not exceeding it is $241.$

1997 AIME 第 11 题

答案：241

难度评级：2710

解答：

其他年份的第 11 题