Difference between revisions of "1997 AIME Problems/Problem 14"

Revision as of 18:36, 4 July 2013

Problem

Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Solution 1

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

By De Moivre's Theorem, we find that ( $k \in \{0,1,\ldots,1996\}$ )

$z=\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)$

Now, let $v$ be the root corresponding to $\theta=\frac{2\pi m}{1997}$ , and let $w$ be the root corresponding to $\theta=\frac{2\pi n}{1997}$ . The magnitude of $v+w$ is therefore:

$\sqrt{\left(\cos\left(\frac{2\pi m}{1997}\right) + \cos\left(\frac{2\pi n}{1997}\right)\right)^2 + \left(\sin\left(\frac{2\pi m}{1997}\right) + \sin\left(\frac{2\pi n}{1997}\right)\right)^2}$

$=\sqrt{2 + 2\cos\left(\frac{2\pi m}{1997}\right)\cos\left(\frac{2\pi n}{1997}\right) + 2\sin\left(\frac{2\pi m}{1997}\right)\sin\left(\frac{2\pi n}{1997}\right)}$

We need $\cos \left(\frac{2\pi m}{1997}\right)\cos \left(\frac{2\pi n}{1997}\right) + \sin \left(\frac{2\pi m}{1997}\right)\sin \left(\frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}$ . The cosine difference identity simplifies that to $\cos\left(\frac{2\pi m}{1997} - \frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}$ . Thus, $|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \lfloor \frac{1997}{12} \rfloor =166$ .

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 $< m$ . Since $m$ and $n$ must be distinct, $n$ can have $1996$ possible values. 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 $\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right)$ for $k = 0,1,\ldots,1996.$ They are also located at the vertices of a regular $1997$ -gon that is centered at the origin in the complex plane.

Without loss of generality, let $v = 1.$ Then $\begin{eqnarray*} |v + w|^2 & = & |\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right) + 1|^2 \\ & = & \left|\left[\cos\left(\frac {2\pi k}{1997}\right) + 1\right] + i\sin\left(\frac {2\pi k}{1997}\right)\right|^2 \\ & = & \cos^2\left(\frac {2\pi k}{1997}\right) + 2\cos\left(\frac {2\pi k}{1997}\right) + 1 + \sin^2\left(\frac {2\pi k}{1997}\right) \\ & = & 2 + 2\cos\left(\frac {2\pi k}{1997}\right) \end{eqnarray*}$

We want $|v + w|^2\ge 2 + \sqrt {3}.$ From what we just obtained, this is equivalent to $\cos\left(\frac {2\pi k}{1997}\right)\ge \frac {\sqrt {3}}2.$ This occurs when $\frac {\pi}6\ge \frac {2\pi k}{1997}\ge - \frac {\pi}6$ 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 $83 + 499 = \boxed{582}.$

Revision as of 23:58, 23 March 2011 (view source) Danielguo94 (talk \| contribs) (→‎Solution 2) ← Older edit		Revision as of 18:36, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
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1997 AIME (Problems • Answer Key • Resources)
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