Difference between revisions of "2015 AIME II Problems/Problem 13"
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If <math>n = 1</math>, <math>a_n = \sin(1) > 0</math>. Then if <math>n</math> satisfies <math>a_n < 0</math>, <math>n \ge 2</math>, and | If <math>n = 1</math>, <math>a_n = \sin(1) > 0</math>. Then if <math>n</math> satisfies <math>a_n < 0</math>, <math>n \ge 2</math>, and | ||
<cmath>a_n =\cfrac{1}{\sin{1}} \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)].</cmath> | <cmath>a_n =\cfrac{1}{\sin{1}} \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)].</cmath> | ||
− | Since <math>\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>. Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>. Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>. Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>. Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor | + | Since <math>\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>. Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>. Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>. Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>. Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 10:05, 27 March 2015
Contents
Problem
Define the sequence by , where represents radian measure. Find the index of the 100th term for which .
Solution 1
If , . Then if satisfies , , and Since is positive, it does not affect the sign of . Let . Now since and , is negative if and only if , or when . Since is irrational, there is always only one integer in the range, so there are values of such that at . Then the hundredth such value will be when and .
Solution 2
Notice that is the imaginary part of , by Euler's formula. Using the geometric series formula, we find that this sum is equal to We only need to look at the imaginary part, which is Since , , so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have . This only holds when is between and for integer [continuity proof here], and since this has exactly one integer solution for every such interval, the th such is .
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 |
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