2015 AIME II Problems/Problem 13
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 .
Clearer Explanation for the End
Notice that we want
Notice that the only interval in which this is true, is where
Thus, we need So we have
But we want the 100th solution. Which means is in between So n = 628.
~mathboy282
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 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 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 by sum to product. 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 .
Solution 3
Similar to solution 2, we set a complex number . We start from instead of because starts from : be careful.
The sum of .
We are trying to make so that the imaginary part of this expression is negative.
The argument of is . The argument of , however, is a little more tricky. is on a circle centered on with radius . The change in angle due to is with respect to the center, but the angle that makes with the -axis is the change, due to Circle Theorems (this intercepted arc is the argument of ), because the - axis is tangent to the circle at the origin. So . Dividing by subtracts the latter argument from the former, so the angle of the quotient with the -axis is .
We want the argument of the whole expression . This translates into . also consists of points on the circle centered at , so we deal with this argument similarly: the argument of is twice the angle makes with the -axis. Since is always negative, , and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a angle with the -axis both ways.
So the argument of must be in the bound by doubling, namely the last negative before another rotation. Since there is always one in this category per rotation because is irrational, and the answer is .
Solution 4
By the product-to-sum formula, Thus, we can make the sum in the problem telescope: \begin{align*} a_n &= \sum_{k = 1}^n \sin k \\ &= \sum_{k = 1}^n \frac{\sin \frac{1}{2} \sin k}{\sin \frac{1}{2}} \\ &= \sum_{k = 1}^n \frac{\cos (k - \frac{1}{2}) - \cos (k + \frac{1}{2})}{2 \sin \frac{1}{2}} \\ &= \frac{(\cos \frac{1}{2} - \cos \frac{3}{2}) + (\cos \frac{3}{2} - \cos \frac{5}{2}) + \dots + (\cos \frac{2n - 1}{2} - \cos \frac{2n + 1}{2})}{2 \sin \frac{1}{2}} \\ &= \frac{\cos \frac{1}{2} - \cos \frac{2n + 1}{2}}{2 \sin \frac{1}{2}}. \end{align*}Then when This occurs if and only if for some integer Equivalently, In other words, The 100th index of this form is then
Video Solution
~MathProblemSolvingSkills
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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