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

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=== Solution 2 ===
 
=== Solution 2 ===
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The solutions of the equation <math>z^{1997} = 1</math> are the <math>1997</math>th [[roots of unity]] and are equal to <math>\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right)</math> for <math>k = 0,1,\ldots,1996.</math> They are also located at the vertices of a [[regular polygon|regular]] <math>1997</math>-gon that is centered at the origin in the complex plane.
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The solutions of the equation <math>z^{1997} = 1</math> are the <math>1997</math>th [[roots of unity]] and are equal to <math>\text{cis}(\theta_k)</math>, where <math>\theta_k = \tfrac {2\pi k}{1997}</math> for <math>k = 0,1,\ldots,1996.</math> Thus, they are located at uniform intervals on the unit circle in the complex plane.  
  
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[[Without loss of generality]], let <math>v = 1.</math> Then
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The quantity <math>|v+w|</math> is unchanged upon rotation around the origin, so, WLOG, we can assume <math>v=1</math> after rotating the axis till <math>v</math> lies on the real axis. Let <math>w=\text{cis}(\theta_k)</math>. Since <math>w\cdot \overline{w}=|w|^2=1</math> and <math>w+\overline{w}=2\text{Re}(w) = 2\cos\theta_k</math>, we have <cmath>|v + w|^2 = (1+w)(1+\overline{w}) = 2+2\cos\theta_k</cmath>
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<cmath>
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We want <math>|v + w|^2\ge 2 + \sqrt {3}.</math>  From what we just obtained, this is equivalent to <cmath>\cos\theta_k\ge \frac {\sqrt {3}}2 \qquad \Leftrightarrow \qquad -\frac {\pi}6\le \theta_k \le  \frac {\pi}6</cmath> which is satisfied by <math>k = 166,165,\ldots, - 165, - 166</math> (we don't include 0 because that corresponds to <math>v</math>).  So out of the <math>1996</math> possible <math>k</math>, <math>332</math> work.  Thus, <math>m/n = 332/1996 = 83/499.</math>  So our answer is <math>83 + 499 = \boxed{582}.</math>
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\begin{eqnarray*} |v + w|^2 & = & |\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right) + 1|^2 \\
 
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& = & \left|\left[\cos\left(\frac {2\pi k}{1997}\right) + 1\right] + i\sin\left(\frac {2\pi k}{1997}\right)\right|^2 \\
 
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& = & \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) \\
 
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& = & 2 + 2\cos\left(\frac {2\pi k}{1997}\right) \end{eqnarray*}
 
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</cmath>
 
−
 
 
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We want <math>|v + w|^2\ge 2 + \sqrt {3}.</math>  From what we just obtained, this is equivalent to <math>\cos\left(\frac {2\pi k}{1997}\right)\ge \frac {\sqrt {3}}2.</math>  This occurs when <math>\frac {\pi}6\ge \frac {2\pi k}{1997}\ge - \frac {\pi}6</math> which is satisfied by <math>k = 166,165,\ldots, - 165, - 166</math> (we don't include 0 because that corresponds to <math>v</math>).  So out of the <math>1996</math> possible <math>k</math>, <math>332</math> work.  Thus, <math>m/n = 332/1996 = 83/499.</math>  So our answer is <math>83 + 499 = \boxed{582}.</math>
 
  
 
=== Solution 3 ===
 
=== Solution 3 ===

Revision as of 12:09, 14 October 2021

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 $\text{cis}(\theta_k)$, 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 $u = 1$. We are now looking for a point exactly one unit away from $u$ 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 $(x-1)^2 = 1$ and $x^2 + y^2 = 2 + \sqrt{3}$. Solving for $x$ yields $x = \frac{\sqrt{3}}{2}$. 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 $83 + 499 = \boxed{582}$.

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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