Difference between revisions of "1997 AIME Problems/Problem 11"
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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>. | 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>. | ||
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+ | === Solution 4 === | ||
+ | Consider the sum <math>\sum_{n = 1}^{44} \text{cis } n^\circ</math>. The fraction is given by the real part divided by the imaginary part. | ||
+ | |||
+ | The sum can be written as <math>\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)</math>. | ||
+ | Consider the rhombus <math>OABC</math> on the complex plane such that <math>O</math> is the origin, <math>A</math> represents <math>\text{cis } n^\circ</math>, <math>B</math> represents <math>\text{cis } n^\circ + \text{cis } 45-n^\circ</math> and <math>C</math> represents <math>\text{cis } n^\circ</math>. Simple geometry shows that <math>\angle BOA = 22.5-k^\circ</math>, so the angle that <math>\text{cis } n^\circ + \text{cis } 45-n^\circ</math> makes with the real axis is simply <math>22.5^\circ</math>. So <math>\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)</math> is the sum of collinear complex numbers, so the angle the sum makes with the real axis is <math>22.5^\circ</math>. So our answer is <math>\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}</math>. | ||
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+ | Note that the <math>\cot(22.5^\circ) = \sqrt2 + 1</math> can be shown easily through half-angle formula. | ||
== See also == | == See also == |
Revision as of 14:08, 3 September 2017
Problem 11
Let . What is the greatest integer that does not exceed ?
Contents
[hide]Solution
Solution 1
Using the identity , that summation reduces to
This fraction is equivalent to . Therefore,
Solution 2
A slight variant of the above solution, note that
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 .
Solution 4
Consider the sum . The fraction is given by the real part divided by the imaginary part.
The sum can be written as . Consider the rhombus on the complex plane such that is the origin, represents , represents and represents . Simple geometry shows that , so the angle that makes with the real axis is simply . So is the sum of collinear complex numbers, so the angle the sum makes with the real axis is . So our answer is .
Note that the can be shown easily through half-angle formula.
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 |
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