Difference between revisions of "1997 AIME Problems/Problem 11"
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Although an exact value for <math>\cot (1/2^\circ)</math> in terms of radicals will be difficult, this is easily known: it is really large! | Although an exact value for <math>\cot (1/2^\circ)</math> in terms of radicals will be difficult, this is easily known: it is really large! | ||
− | So treat it as though it were <math>\infty</math>. The fraction is approximated by <math>\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \boxed{241}</math>. | + | So treat it as though it were <math>\infty</math>. The fraction is approximated by <math>\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}</math>. |
== See also == | == See also == |
Revision as of 20:29, 19 August 2012
Problem
Let . What is the greatest integer that does not exceed ?
Solution
Solution 1
Using the identity , that summation reduces to
\begin{eqnarray*}x &=& \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\ &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right) (Error compiling LaTeX. ! LaTeX Error: \begin{eqnarray*} on input line 20 ended by \end{document}.)
This fraction is equivalent to . Therefore,
Solution 2
A slight variant of the above solution, note that
\begin{eqnarray*} \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\\ &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\ \sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n (Error compiling LaTeX. ! LaTeX Error: \begin{eqnarray*} on input line 20 ended by \end{document}.)
This is the ratio we are looking for. reduces to , and .
Solution 3
Consider the sum . The fraction is given by the real part divided by the imaginary part.
The sum can be written (by De Moivre's Theorem with geometric series)
(after multiplying by complex conjugate)
Using the tangent half-angle formula, this becomes .
Dividing the two parts and multiplying each part by 4, the fraction is .
Although an exact value for in terms of radicals will be difficult, this is easily known: it is really large!
So treat it as though it were . The fraction is approximated by .
See also
1997 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 |