Difference between revisions of "2015 AIME II Problems/Problem 13"
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<cmath>a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath> | <cmath>a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath> | ||
Since <math>2\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>. | Since <math>2\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>. | ||
+ | ===Clearer Explanation for the End=== | ||
+ | Notice that we want <math>\cos(n+1/2)>\cos(1/2).</math> | ||
+ | |||
+ | Notice that the only interval in which this is true, is where <math>-1/2 < n + 1/2 < 1/2.</math> | ||
+ | |||
+ | Thus, we need <math>-1<n<0.</math> So we have <math>-1+2\pi k <n <0+2\pi k.</math> | ||
+ | |||
+ | But we want the 100th solution. Which means <math>n</math> is in between <math>200\pi - 1 < n < 200 \pi.</math> So n = 628. | ||
+ | |||
+ | ~mathboy282 | ||
==Solution 2== | ==Solution 2== | ||
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 | + | 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+1) - \sin n > \sin 1 \implies 2 \cos \left(n + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. 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== | ==Solution 3== | ||
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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 per rotation because <math>\pi</math> is irrational, <math>n_{100}\equiv z^{628}</math> and the answer is <math>\boxed{628}</math>. | 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 per rotation because <math>\pi</math> is irrational, <math>n_{100}\equiv z^{628}</math> and the answer is <math>\boxed{628}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | By the product-to-sum formula, | ||
+ | <cmath>\sin \frac{1}{2} \sin k = \frac{1}{2} \left[ \cos \left( k - \frac{1}{2} \right) - \cos \left( k + \frac{1}{2} \right) \right].</cmath>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 <math>a_n < 0</math> when <math>\cos \frac{1}{2} < \cos \frac{2n + 1}{2}.</math> This occurs if and only if | ||
+ | <cmath>2 \pi k - \frac{1}{2} < \frac{2n + 1}{2} < 2 \pi k + \frac{1}{2}</cmath>for some integer <math>k.</math> Equivalently, | ||
+ | <cmath>2 \pi k - 1 < n < 2 \pi k.</cmath>In other words, <math>n = \lfloor 2 \pi k \rfloor.</math> The 100th index of this form is then <math>\lfloor 2 \pi \cdot 100 \rfloor = \boxed{628}.</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/b1-cUUPjYNk | ||
+ | |||
+ | ~MathProblemSolvingSkills | ||
==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}} |
Latest revision as of 16:31, 7 July 2024
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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