Question

设 $w$ 和 $z$ 是复数，满足 $|w| = 1$ 且 $|z| = 10$ 。令 $\theta = \arg\left(\tfrac{w-z}{z}\right)$ 。 $\tan^2 \theta$ 的最大可能值可写成 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质的正整数。求 $p + q$ 。（注意， $\arg(w)$ 对于 $w \ne 0$ 表示复平面中从 $0$ 指向 $w$ 的射线与正实轴所成角的大小。）

Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10.$ Let $\theta = \arg\left(\tfrac{w-z}{z}\right).$ The maximum possible value of $\tan^2 \theta$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$ (Note that $\arg(w),$ for $w \ne 0,$ denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.)

答案

答案：100

知识点：复数圆切线最优化

难度评级：2560

小提示：

$\frac{w-z}{z} = \frac{w}{z} - 1$ 位于半径为 $\frac{1}{10}$ 、圆心为 $-1$ 的圆上

$\frac{w-z}{z} = \frac{w}{z} - 1$ lies on the circle of radius $\frac{1}{10}$ centered at $-1$

大提示：

当从原点出发的射线与该圆相切时，辐角达到极值，所以它与实轴所成角的正弦为 $\frac{1}{10}$

The argument is extremal when the ray from the origin is tangent to that circle, so the sine of the angle with the real axis is $\frac{1}{10}$

解答：

因为 $\frac{w-z}{z} = \frac{w}{z} - 1$ ，而 $\frac{w}{z}$ 可以是任意模长为 $\frac{1}{10}$ 的复数，所以点 $\zeta = \frac{w-z}{z}$ 的轨迹是半径为 $\frac{1}{10}$ 、圆心为 $-1$ 的圆。

由于 $\tan^2\theta$ 在 $\theta$ 改变 $180^\circ$ 时不变，我们要求的是从原点指向该圆的射线与实轴所成的最大角 $\alpha$ 。极端位置的射线与圆相切，此时 $\sin \alpha = \frac{\frac{1}{10}}{1} = \frac{1}{10}$ 。

因此 $\begin{aligned} \tan^2\theta &= \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{\frac{1}{100}}{\frac{99}{100}} \\ &= \frac{1}{99} \end{aligned}\text{，}$ 所以 $p + q = 1 + 99 = 100$ 。

Since $\frac{w-z}{z} = \frac{w}{z} - 1,$ and $\frac{w}{z}$ can be any complex number of modulus $\frac{1}{10},$ the point $\zeta = \frac{w-z}{z}$ ranges over the circle of radius $\frac{1}{10}$ centered at $-1.$

Because $\tan^2\theta$ is unchanged when $\theta$ shifts by $180^\circ,$ we want the largest angle $\alpha$ that a ray from the origin to this circle makes with the real axis. The extreme rays are tangent to the circle, where $\sin \alpha = \frac{\frac{1}{10}}{1} = \frac{1}{10}.$

Then $\begin{aligned} \tan^2\theta &= \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{\frac{1}{100}}{\frac{99}{100}} \\ &= \frac{1}{99}, \end{aligned}$ so $p + q = 1 + 99 = 100.$

2014 AIME I 第 7 题

答案：100

难度评级：2560

小提示：

大提示：

解答：

其他年份的第 7 题