AIME 1997 · 第 14 题

Solution

Solution 1

$z^{1997}=1=1(\cos 0 + i \sin 0)$

Define $\theta = 2\pi/1997$. By De Moivre's Theorem the roots are given by

$z=\cos (k\theta) +i\sin(k\theta), \qquad k \in \{0,1,\ldots,1996\}$

Now, let $v$ be the root corresponding to $m\theta=2m\pi/1997$, and let $w$ be the root corresponding to $n\theta=2n\pi/ 1997$. Then

$$\begin{aligned} |v+w|^2 &= \left(\cos(m\theta) + \cos(n\theta)\right)^2 + \left(\sin(m\theta) + \sin(n\theta)\right)^2 \\ &= 2 + 2\cos\left(m\theta\right)\cos\left(n\theta\right) + 2\sin\left(m\theta\right)\sin\left(n\theta\right) \end{aligned}$$ The cosine difference identity simplifies that to

$$|v+w|^2 = 2+2\cos((m-n)\theta)$$ We need $|v+w|^2 \ge 2+\sqrt{3}$, which simplifies to

$$\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}.$$ Hence, $-\frac{\pi}{6} \le (m-n) \theta \le \frac{\pi}{6},$ or alternatively, $|m - n| \theta \le \frac{\pi}{6}.$

Recall that $\theta=\frac{2\pi}{1997}.$ Thus,

$$|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \left\lfloor \frac{1997}{12} \right\rfloor =166.$$ Therefore, $m$ and $n$ cannot be more than $166$ away from each other. This means that for a given value of $m$, there are $332$ values for $n$ that satisfy the inequality; $166$ of them $> m$, and $166$ of them $< m$ ($m \neq n$ since the problem says that $m$ and $n$ are distinct). Since $m$ and $n$ must be distinct, $n$ can have $1996$ total possible values for each value of $m$. Therefore, the probability is $\frac{332}{1996}=\frac{83}{499}$. The answer is then $499+83=\boxed{582}$.

Solution 2

The solutions of the equation $z^{1997} = 1$ are the $1997$th roots of unity and are equal to $\text{cis}(\theta_k)$, where $\theta_k = \tfrac {2\pi k}{1997}$ for $k = 0,1,\ldots,1996.$ Thus, they are located at uniform intervals on the unit circle in the complex plane.

The quantity $|v+w|$ is unchanged upon rotation around the origin, so, WLOG, we can assume $v=1$ after rotating the axis till $v$ lies on the real axis. Let $w=\text{cis}(\theta_k)$. Since $w\cdot \overline{w}=|w|^2=1$ and $w+\overline{w}=2\text{Re}(w) = 2\cos\theta_k$, we have

$$|v + w|^2 = (1+w)(1+\overline{w}) = 2+2\cos\theta_k$$ We want $|v + w|^2\ge 2 + \sqrt {3}.$ From what we just obtained, this is equivalent to

$$\cos\theta_k\ge \frac {\sqrt {3}}2 \qquad \Leftrightarrow \qquad -\frac {\pi}6\le \theta_k \le \frac {\pi}6$$ which is satisfied by $k = 166,165,\ldots, - 165, - 166$ (we don't include 0 because that corresponds to $v$). So out of the $1996$ possible $k$, $332$ work. Thus, $m/n = 332/1996 = 83/499.$ So our answer is $83 + 499 = \boxed{582}.$

Solution 3

We can solve a geometrical interpretation of this problem.

Without loss of generality, let $w = 1$. We are now looking for a point exactly one unit away from $w$ such that the point is at least $\sqrt{2 + \sqrt{3}}$ units away from the origin. Note that the "boundary" condition is when the point will be exactly $\sqrt{2+\sqrt{3}}$ units away from the origin; these points will be the intersections of the circle centered at $(1,0)$ with radius $1$ and the circle centered at $(0,0)$ with radius $\sqrt{2+\sqrt{3}}$. The equations of these circles are $(x-1)^2 + y^2 = 1$ and $x^2 + y^2 = 2 + \sqrt{3}$. Solving for $x$ yields $x = 1 + \frac{\sqrt{3}}{2}$. Clearly, this means that the real part of $v$ is greater than $\frac{\sqrt{3}}{2}$. Solving, we note that $332$ possible $v$s exist, meaning that $\frac{m}{n} = \frac{332}{1996} = \frac{83}{499}$. Therefore, the answer is $83 + 499 = \boxed{582}$.

Solution 4

Since $z^{1997}=1$, the roots will have magnitude $1$. Thus, the roots can be written as $\cos(\theta)+i\sin(\theta)$ and $\cos(\omega)+i\sin(\omega)$ for some angles $\theta$ and $\omega$. We rewrite the requirement as $\sqrt{2+\sqrt3}\le|\cos(\theta)+\cos(\omega)+i\sin(\theta)+i\sin(\omega)|$, which can now be easily manipulated to $2+\sqrt{3}\le(\cos(\theta)+\cos(\omega))^2+(\sin(\theta)+\sin(\omega))^2$.

WLOG, let $\theta = 0$. Thus, our inequality becomes $2+\sqrt{3}\le(1+\cos(\omega))^2+(\sin(\omega))^2$, $2+\sqrt{3}\le2+2\cos(\omega)$, and finally $\cos(\omega)\ge\frac{\sqrt{3}}{2}$. Obviously, $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, and thus it follows that, on either side of a given point, $\frac{1997}{12}\approx166$ points will work. The probability is $\frac{166\times2}{1996} = \frac{83}{499}$, and thus our requested sum is $\boxed{582}$ ~SigmaPiE

Solution 5

Let $v = \operatorname{cis}\left(\frac{2\pi k}{1997}\right)$ and $w = \operatorname{cis}\left(\frac{2\pi c}{1997}\right)$, where $\operatorname{cis}\left(\theta\right) = \cos(\theta) + i\sin(\theta)$, $v \neq w$, and $k,c \in \{0,1,2,\dots,1996\}$

We are given

$$\sqrt{2+\sqrt{3}} \le |v+w|.$$ Squaring both sides gives

$$2+\sqrt{3} \le |v+w|^2.$$ Expanding $v+w$, we obtain

$$v+w = \cos\!\left(\tfrac{2\pi k}{1997}\right) + \cos\!\left(\tfrac{2\pi c}{1997}\right) + i\!\left(\sin\!\left(\tfrac{2\pi k}{1997}\right) + \sin\!\left(\tfrac{2\pi c}{1997}\right)\right).$$ Using sum-to-product identities,

$$v+w = 2\cos\!\left(\tfrac{2\pi(k+c)}{1997}\right)\cos\!\left(\tfrac{2\pi(k-c)}{1997}\right) + i\!\left(2\sin\!\left(\tfrac{2\pi(k+c)}{1997}\right)\cos\!\left(\tfrac{2\pi(k-c)}{1997}\right)\right).$$ Thus,

$$|v+w|^2 = 4\cos^2\!\left(\tfrac{2\pi(k-c)}{1997}\right) \left(\cos^2\!\left(\tfrac{2\pi(k+c)}{1997}\right) + \sin^2\!\left(\tfrac{2\pi(k+c)}{1997}\right)\right).$$ Since $\sin^2\theta+\cos^2\theta=1$, we have

$$|v+w|^2 = 4\cos^2\!\left(\tfrac{2\pi(k-c)}{1997}\right).$$ Substituting into the inequality gives

$$2+\sqrt{3} \le 4\cos^2\!\left(\tfrac{2\pi(k-c)}{1997}\right).$$ Taking square roots,

$$\sqrt{2+\sqrt{3}} \le 2\cos\!\left(\tfrac{2\pi(k-c)}{1997}\right) \quad \text{or} \quad-\sqrt{2+\sqrt{3}} \ge 2\cos\!\left(\tfrac{2\pi(k-c)}{1997}\right).$$ Dividing by $2$,

$$\sqrt{\frac{2+\sqrt{3}}{4}} \le \cos\!\left(\tfrac{2\pi(k-c)}{1997}\right) \quad \text{or} \quad -\sqrt{\frac{2+\sqrt{3}}{4}} \ge \cos\!\left(\tfrac{2\pi(k-c)}{1997}\right).$$ Noting that

$$\sqrt{\frac{2+\sqrt{3}}{4}} = \cos \frac{\pi}{12},$$ we conclude that

$$-\frac{\pi}{12} \le \frac{2\pi(k-c)}{1997} \le \frac{\pi}{12} \quad \text{or} \quad \frac{11\pi}{12} \le \frac{2\pi(k-c)}{1997} \le \frac{13\pi}{12}.$$ These two cases are symmetric, so we consider the first and double the result. Simplifying,

$$-\frac{1997}{24} \le k-c \le \frac{1997}{24}.$$ Since $k-c$ is an integer,

$$-83 \le k-c \le 83.$$ Because $k \neq c$, for each fixed $k$ there are exactly $166$ valid values of $c$. This works modulo $1997$ since the arguments lie on a circle (e.g. $\cos(\frac{\pi}{12})=\cos(\frac{-23\pi}{12})$).

Thus, the probability for one case is $\frac{166}{1996}$, and doubling gives $\frac{332}{1996} = \frac{83}{499}.$ The problem asks for $m+n$, so $83+499=\boxed{582}.$

~Voidling