Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
m (→‎Solution 1)
Line 7: Line 7:
 
:<math>z^{1997}=1=1(\cos 0 + i \sin 0)</math>
 
:<math>z^{1997}=1=1(\cos 0 + i \sin 0)</math>
  
−
By [[De Moivre's Theorem]], we find that (<math>k \in \{0,1,\ldots,1996\}</math>)
+
Define <math>\theta = 2\pi/1997</math>. By [[De Moivre's Theorem]], we have
  
−
:<math>z=\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)</math>
+
:<math>z=\cos (k\theta) +i\sin(k\theta), \qquad k \in \{0,1,\ldots,1996\}</math>
  
−
Now, let <math>v</math> be the root corresponding to <math>\theta=\frac{2\pi m}{1997}</math>, and let <math>w</math> be the root corresponding to <math>\theta=\frac{2\pi n}{1997}</math>. The magnitude of <math>v+w</math> is therefore:
+
Now, let <math>v</math> be the root corresponding to <math>m\theta=2m\pi/1997</math>, and let <math>w</math> be the root corresponding to <math>n\theta=2n\pi/ 1997</math>. Then
−
<cmath>\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)}</cmath>
+
<cmath>\begin{align*}
 +
|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{align*}</cmath>
 +
The [[Trigonometric identities|cosine difference identity]] simplifies that to
 +
<cmath>|v+w|^2 = 2+2\cos((m-n)\theta) </cmath>
  
−
We need <cmath>\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}</cmath>The [[Trigonometric identities|cosine difference identity]] simplifies that to <cmath>\cos\left(\frac{2\pi m}{1997} - \frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}</cmath>Thus, <cmath>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \lfloor \frac{1997}{12} \rfloor =166</cmath>.
+
We need <cmath>\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}</cmath>Thus, <cmath>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \left\lfloor \frac{1997}{12} \right\rfloor =166</cmath>.
  
 
Therefore, <math>m</math> and <math>n</math> cannot be more than <math>166</math> away from each other.  This means that for a given value of <math>m</math>, there are <math>332</math> values for <math>n</math> that satisfy the inequality; <math>166</math> of them <math>> m</math>, and <math>166</math> of them <math>< m</math>.  Since <math>m</math> and <math>n</math> must be distinct, <math>n</math> can have <math>1996</math> possible values.  Therefore, the probability is <math>\frac{332}{1996}=\frac{83}{499}</math>.  The answer is then <math>499+83=\boxed{582}</math>.
 
Therefore, <math>m</math> and <math>n</math> cannot be more than <math>166</math> away from each other.  This means that for a given value of <math>m</math>, there are <math>332</math> values for <math>n</math> that satisfy the inequality; <math>166</math> of them <math>> m</math>, and <math>166</math> of them <math>< m</math>.  Since <math>m</math> and <math>n</math> must be distinct, <math>n</math> can have <math>1996</math> possible values.  Therefore, the probability is <math>\frac{332}{1996}=\frac{83}{499}</math>.  The answer is then <math>499+83=\boxed{582}</math>.

Revision as of 14:19, 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

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

Define $\theta = 2\pi/1997$. By De Moivre's Theorem, we have

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

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{align*} |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{align*} The cosine difference identity simplifies that to \[|v+w|^2 = 2+2\cos((m-n)\theta)\]

We need \[\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}\]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 $< m$. 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_14&oldid=163494"