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Difference between revisions of "1999 AIME Problems/Problem 11"

(→‎Alternate Solution)
(→‎Alternate Solution)
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We note that <math>\sin x = \mbox{Im } e^{ix}</math>. We thus have that
 
We note that <math>\sin x = \mbox{Im } e^{ix}</math>. We thus have that
−
<math>\sum_{k = 1}^{35} \sin 5k</math>
+
<cmath>\begin{align*}
−
<math>= \sum_{k = 1}^{35} \mbox{Im } e^{5ki} </math>
+
\sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\
−
<math>= \mbox{Im } \sum_{k = 1}^{35} e^{5ki} </math>
+
&= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\
−
<math>= \mbox{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}} </math>
+
&= \mbox{Im } \frac{e^{5i}(1 - e^{175i})}{1 - e^{5i}}\
−
<math>= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5} </math>
+
&= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\
−
<math>= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)((1 - \cos 5) + i \sin 5)}{(1 - \cos 5)^2 + \sin^2 5} </math>
+
&= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\
−
<math>= \frac{2 \sin 5}{2 - 2 \cos 5} = \frac{\sin 5}{1 - \cos 5}</math>
+
&= \frac{2 \sin 5}{2 - 2 \cos 5}\
−
<math>= \frac{\sin 175}{1 + \cos 175} = \tan \frac{175}{2}.</math>  
+
&= \frac{\sin 5}{1 - \cos 5}\
−
 
+
&= \frac{\sin 175}{1 + \cos 175} \
−
 
+
&= \tan \frac{175}{2}.\
 +
\end{align*}</cmath>
 
The desired answer is thus <math>175 + 2 = \boxed{177}</math>.
 
The desired answer is thus <math>175 + 2 = \boxed{177}</math>.
−
 
−
Comment: I think it should be <math> \mbox{Im } \sum_{k = 1}^{35} e^{5ki} = \mbox{Im } \frac{e^{5i}(1 - e^{175i})}{1 - e^{5i}} </math>
 
  
 
== See also ==
 
== See also ==

Revision as of 11:10, 12 October 2020

Problem

Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$

Solution

Let $s = \sum_{k=1}^{35}\sin 5k = \sin 5 + \sin 10 + \ldots + \sin 175$. We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identity $\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))$, we can rewrite $s$ as

\[s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 = \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\] \[s = \frac{0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 - \cos 20 + \ldots - \cos 170 + \cos 165 - \cos 175+ \cos 170 - \cos 180)}{\sin 5}\]

This telescopes to $s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}$. Manipulating this to use the identity $\tan x = \frac{1 - \cos 2x}{\sin 2x}$, we get $s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2}$, and our answer is $\boxed{177}$.

Alternate Solution

We note that $\sin x = \mbox{Im } e^{ix}$. We thus have that \begin{align*} \sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\\ &= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\\ &= \mbox{Im } \frac{e^{5i}(1 - e^{175i})}{1 - e^{5i}}\\ &= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\ &= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\ &= \frac{2 \sin 5}{2 - 2 \cos 5}\\ &= \frac{\sin 5}{1 - \cos 5}\\ &= \frac{\sin 175}{1 + \cos 175} \\ &= \tan \frac{175}{2}.\\ \end{align*} The desired answer is thus $175 + 2 = \boxed{177}$.

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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