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 "2015 AIME II Problems/Problem 13"

m (→‎Solution 2)
(→‎Solution 2)
Line 13: Line 13:
 
Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath>
 
Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath>
 
Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin n + \sin 1 < \sin (n+1) \implies \sin (n+1) - \sin n > \sin 1</math>. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.
 
Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin n + \sin 1 < \sin (n+1) \implies \sin (n+1) - \sin n > \sin 1</math>. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.
 +
 +
==Solution 3==
 +
 +
Similar to solution 2, we set a complex number <math>z=\cos 1+i\sin 1</math>. We start from <math>z</math> instead of <math>1</math> because <math>k</math> starts from <math>1</math>: be careful.
 +
 +
The sum of <math>z+z^2+z^3+z^4+z^5\dots=\frac{z-z^{n+1}}{1-z}=\frac{z}{z-1}\left(z^n-1\right)</math>.
 +
 +
We are trying to make <math>n</math> so that the imaginary part of this expression is negative.
 +
 +
The argument of <math>z</math> is <math>1</math>. The argument of <math>z-1</math>, however, is a little more tricky. <math>z-1</math> is on a circle centered on <math>(-1,0)</math> with radius <math>1</math>. The change in angle due to <math>z</math> is <math>1</math> with respect to the center, but the angle that <math>z-1</math> makes with the <math>y</math>-axis is <math>half</math> the change, due to [[Power of a Point Theorem]], because the <math>y</math>- axis is tangent to the circle at the origin. So <math>\text{arg}(z-1)=\frac{\pi+1}{2}</math>. Dividing <math>z</math> by <math>z-1</math> subtracts the latter argument from the former, so the angle of the quotient with the <math>x</math>-axis is <math>\frac{1-\pi}{2}</math>.
 +
 +
We want the argument of the whole expression <math>-\pi<\theta<0</math>. This translates into <math>\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}</math>. <math>z^n-1</math> also consists of points on the circle centered at <math>(-1,0)</math>, so we deal with this argument similarly: the argument of <math>z^n</math> is twice the angle <math>z^n-1</math> makes with the <math>y</math>-axis. Since <math>z^n-1</math> is always negative, <math>\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}</math>, and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a <math>\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}</math> angle with the <math>y</math>-axis both ways.
 +
 +
So the argument of <math>z^n</math> must be in the bound <math>-1<\theta<0</math> by doubling, namely the last <math>z^n</math> negative before another rotation. Since there is always one <math>z^n</math> in this category because <math>\pi</math> is irrational, <math>n_{100}\equiv z^{628}</math> and the answer is <math>\boxed{628}</math>.
  
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=12|num-a=14}}
 
{{AIME box|year=2015|n=II|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:06, 29 February 2016

Problem

Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$, where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$.

Solution 1

If $n = 1$, $a_n = \sin(1) > 0$. Then if $n$ satisfies $a_n < 0$, $n \ge 2$, and \[a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $\sin 1$ is positive, it does not affect the sign of $a_n$. Let $b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)$. Now since $\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ and $\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$, $b_n$ is negative if and only if $\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)$, or when $n \in [2k\pi - 1, 2k\pi]$. Since $\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\pi, 4\pi, \cdots$. Then the hundredth such value will be when $k = 100$ and $n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}$.

Solution 2

Notice that $a_n$ is the imaginary part of $\sum_{k=1}^n e^{ik}$, by Euler's formula. Using the geometric series formula, we find that this sum is equal to \[\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}\] We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have \[\frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}\] We only need to look at the imaginary part, which is \[\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}\] Since $\cos 1 < 1$, $2-2 \cos 1 > 0$, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have $\sin n + \sin 1 < \sin (n+1) \implies \sin (n+1) - \sin n > \sin 1$. This only holds when $n$ is between $2\pi k - 1$ and $2\pi k$ for integer $k$ [continuity proof here], and since this has exactly one integer solution for every such interval, the $100$th such $n$ is $\lfloor 200\pi \rfloor = \boxed{628}$.

Solution 3

Similar to solution 2, we set a complex number $z=\cos 1+i\sin 1$. We start from $z$ instead of $1$ because $k$ starts from $1$: be careful.

The sum of $z+z^2+z^3+z^4+z^5\dots=\frac{z-z^{n+1}}{1-z}=\frac{z}{z-1}\left(z^n-1\right)$.

We are trying to make $n$ so that the imaginary part of this expression is negative.

The argument of $z$ is $1$. The argument of $z-1$, however, is a little more tricky. $z-1$ is on a circle centered on $(-1,0)$ with radius $1$. The change in angle due to $z$ is $1$ with respect to the center, but the angle that $z-1$ makes with the $y$-axis is $half$ the change, due to Power of a Point Theorem, because the $y$- axis is tangent to the circle at the origin. So $\text{arg}(z-1)=\frac{\pi+1}{2}$. Dividing $z$ by $z-1$ subtracts the latter argument from the former, so the angle of the quotient with the $x$-axis is $\frac{1-\pi}{2}$.

We want the argument of the whole expression $-\pi<\theta<0$. This translates into $\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}$. $z^n-1$ also consists of points on the circle centered at $(-1,0)$, so we deal with this argument similarly: the argument of $z^n$ is twice the angle $z^n-1$ makes with the $y$-axis. Since $z^n-1$ is always negative, $\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}$, and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a $\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}$ angle with the $y$-axis both ways.

So the argument of $z^n$ must be in the bound $-1<\theta<0$ by doubling, namely the last $z^n$ negative before another rotation. Since there is always one $z^n$ in this category because $\pi$ is irrational, $n_{100}\equiv z^{628}$ and the answer is $\boxed{628}$.

See also

2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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=2015_AIME_II_Problems/Problem_13&oldid=77124"