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Difference between revisions of "2015 AIME II Problems/Problem 13"

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==Solution 2==
 
==Solution 2==
  
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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> <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>
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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> Multiplying the denominator by its conjugate, 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>.
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==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 16:06, 28 March 2015

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 \[\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}\] Multiplying the denominator by its conjugate, 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 $\lfloor 200\pi \rfloor = \boxed{628}$.

See also

2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AIME Problems and Solutions

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