Difference between revisions of "1997 AIME Problems/Problem 11"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
+ | Note that | ||
+ | <math>\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}</math> | ||
+ | |||
+ | Now use the sum-product formula <math>\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})</math> | ||
+ | We want to pair up <math>1, 44</math>, <math>2, 43</math>, <math>3, 42</math>, etc. from the numerator and <math>46, 89</math>, <math>47, 88</math>, <math>48, 87</math> etc. from the denominator. Then we get: | ||
+ | <cmath>\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}</cmath> | ||
+ | |||
+ | To calculate this number, use the half angle formula. Since <math>\cos(\frac{x}{2}) = \pm \sqrt{\frac{\cos x + 1}{2}}</math>, then our number becomes: <cmath>\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}</cmath> in which we drop the negative roots (as it is clear cosine of <math>22.5</math> and <math>67.5</math> are positive). We can easily simplify this: | ||
+ | |||
+ | <cmath>\begin{eqnarray*} | ||
+ | \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &=& \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &=& \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &=& \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &=& \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &=& \sqrt{2}+1 | ||
+ | \end{eqnarray*}</cmath> | ||
+ | |||
+ | And hence our answer is <math>\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
<cmath>\begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\ | <cmath>\begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\ | ||
&=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44} | &=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44} | ||
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
− | === Solution | + | === Solution 3 === |
A slight variant of the above solution, note that | A slight variant of the above solution, note that | ||
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This is the [[ratio]] we are looking for. <math>x</math> reduces to <math>\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1</math>, and <math>\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}</math>. | This is the [[ratio]] we are looking for. <math>x</math> reduces to <math>\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1</math>, and <math>\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}</math>. | ||
− | === Solution | + | === 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. | 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. | ||
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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>. | ||
− | === Solution | + | === Solution 5 === |
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. | 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. | ||
Revision as of 21:41, 4 August 2019
Problem 11
Let . What is the greatest integer that does not exceed
?
Contents
Solution
Solution 1
Note that
Now use the sum-product formula
We want to pair up
,
,
, etc. from the numerator and
,
,
etc. from the denominator. Then we get:
To calculate this number, use the half angle formula. Since , then our number becomes:
in which we drop the negative roots (as it is clear cosine of
and
are positive). We can easily simplify this:
And hence our answer is
Solution 2
Using the identity
, that summation reduces to
This fraction is equivalent to . Therefore,
Solution 3
A slight variant of the above solution, note that
This is the ratio we are looking for. reduces to
, and
.
Solution 4
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 5
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.