Difference between revisions of "1997 AIME Problems/Problem 11"
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− | == Problem == | + | == Problem 11 == |
+ | Let <math>x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}</math>. What is the greatest integer that does not exceed <math>100x</math>? | ||
+ | |||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
+ | Note that | ||
+ | <math>\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}</math> by the cofunction identities.(We could have also written it as <math>\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=46}^{89} \sin n}{\sum_{n=1}^{44} \sin n} = \frac {\sin 89 + \sin 88 + \dots + \sin 46}{\sin 1 + \sin 2 + \dots + \sin 44}</math>.) | ||
+ | |||
+ | Now use the sum-product formula <math>\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)</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{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos 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\left(\frac{x}{2}\right) = \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}} \\ &=& \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}\\ | ||
+ | &=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44} | ||
+ | \end{eqnarray*}</cmath> | ||
+ | |||
+ | Using the identity <math>\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}</math> <math>\Longrightarrow \sin x + \cos x</math> <math>= \sin x + \sin (90-x)</math> <math>= 2 \sin 45 \cos (45-x)</math> <math>= \sqrt{2} \cos (45-x)</math>, that [[summation]] reduces to | ||
+ | |||
+ | <cmath>\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) | ||
+ | \end{eqnarray*} | ||
+ | </cmath> | ||
+ | |||
+ | This fraction is equivalent to <math>x</math>. Therefore, | ||
+ | <cmath>\begin{eqnarray*} | ||
+ | x &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\ | ||
+ | \frac {1}{\sqrt {2}} &=& x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\ | ||
+ | x &=& \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\ | ||
+ | \lfloor 100x \rfloor &=& \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\ | ||
+ | \end{eqnarray*}</cmath> | ||
+ | |||
+ | === Solution 3 === | ||
+ | A slight variant of the above solution, note that | ||
+ | |||
+ | <cmath>\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 | ||
+ | \end{eqnarray*}</cmath> | ||
+ | |||
+ | 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 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 <math>- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}</math> (by [[De Moivre's Theorem]] with geometric series) | ||
+ | |||
+ | <math>= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}</math> (after multiplying by [[complex conjugate]]) | ||
+ | |||
+ | <math>= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}</math> | ||
+ | |||
+ | <math>= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}</math> | ||
+ | |||
+ | Using the [[trigonometric identities|tangent half-angle formula]], this becomes <math>\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)</math>. | ||
+ | |||
+ | Dividing the two parts and multiplying each part by 4, the fraction is <math>\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}</math>. | ||
+ | |||
+ | 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 \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}</math>. | ||
+ | |||
+ | === 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Note that the <math>\cot(22.5^\circ) = \sqrt2 + 1</math> can be shown easily through half-angle formula. | ||
+ | |||
+ | |||
+ | |||
+ | === Solution 6 === | ||
+ | |||
+ | We write <math>x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}</math> since <math>\cos x = \sin (90^{\circ}-x).</math> Now we by the sine angle sum we know that <math>\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).</math> So the expression simplifies to <math>\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).</math> Therefore we have the equation <math>x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.</math> Finishing, we have <math>\lfloor 100x \rfloor = \boxed{241}.</math> | ||
+ | |||
+ | === Solution 7 === | ||
+ | |||
+ | We can pair the terms of the summations as below. | ||
+ | |||
+ | <cmath>\dfrac{(\cos{1} + \cos{44}) + (\cos{2} + \cos{43}) + (\cos{3} + \cos{42}) + \cdots + (\cos{22} + \cos{23})}{(\sin{1} + \sin{44}) + (\sin{2} + \sin{43}) + (\sin{3} + \sin{42}) + \cdots + (\sin{22} + \sin{23})}.</cmath> | ||
+ | |||
+ | From here, we use the cosine and sine subtraction formulas as shown. | ||
+ | |||
+ | \begin{align*} | ||
+ | &\dfrac{(\cos(45-44) + \cos(45-1)) + (\cos(45-43) + \cos(45-2))+ \cdots (\cos(45-23) + \cos(45-22))}{(\sin(45-44) + \sin(45-1)) + (\sin(45-43) + \sin(45-2))+ \cdots (\sin(45-23) + \sin(45-22))} \\ | ||
+ | &= \dfrac{(\cos{45}\cos{44} +\sin{45}\sin{44} + \cos{45}\cos{1} + \sin{45}\sin{1})+ \cdots + (\cos{45}\cos{23} + \sin{45}\sin{23} + \cos{45}\cos{22} + \sin{45}\sin{22})}{(\sin{45}\cos{44}-\cos{45}\sin{44} + \sin{45}\cos{1} - \cos{45}\sin{1}) + \cdots + (\sin{45}\cos{23} -\cos{45}\sin{23} + \sin{45}\cos{22} - \cos{45}\sin{22})} \\ | ||
+ | &= \dfrac{\sqrt{2}/2(\cos{44} + \sin{44} + \cos{1}+\sin{1} +\cos{43} + \sin{43} + \cos{2} + \sin{2} + \cdots + \cos{23} + \sin{23} + \cos{22} + \sin{22})}{\sqrt{2}/2(\cos{44} -\sin{44} +\cos{1} - \sin{1} + \cos{43}-\sin{43} + \cos{2} -\sin{2} +\cdots + \cos{23}-\sin{23} + \cos{22} - \sin{22})} \\ | ||
+ | &=\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ}. | ||
+ | \end{align*} | ||
+ | |||
+ | Since all of these operations have been performed to ensure equality, we can set the final line of the above mess equal to our original expression. | ||
+ | |||
+ | <cmath>\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ} = \frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}.</cmath> | ||
+ | |||
+ | For the sake of clarity, let <math>\sum\limits_{n=1}^{44} \cos n^\circ = C</math> and <math>\sum\limits_{n=1}^{44} \sin n^\circ = S</math>. Then, we have | ||
+ | |||
+ | <cmath>\dfrac{C+S}{C-S} = \dfrac{C}{S} \implies CS+S^2 = C^2-CS.</cmath> | ||
+ | |||
+ | Finishing, we have <math>S^2+2CS=C^2</math>. Adding <math>C^2</math> to both sides gives <math>(C+S)^2 = 2C^2</math>, or <math>C+S = \pm C\sqrt{2}</math>. Taking the positive case gives <math>S= C(\sqrt{2}-1)</math>. Finally, | ||
+ | |||
+ | <cmath>x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ} = \dfrac{C}{S} = \dfrac{1}{\sqrt{2}-1} = \sqrt{2} +1 \implies \lfloor{100x\rfloor} = \boxed{241}.</cmath> | ||
+ | |||
+ | ~NTfish | ||
+ | |||
== See also == | == See also == | ||
− | + | Video Solution by Osman Nal: | |
+ | https://www.youtube.com/watch?v=o6udScV0F_o | ||
+ | {{AIME box|year=1997|num-b=10|num-a=12}} | ||
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:16, 18 June 2024
Problem 11
Let . What is the greatest integer that does not exceed
?
Contents
Solution
Solution 1
Note that
by the cofunction identities.(We could have also written it as
.)
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.
Solution 6
We write since
Now we by the sine angle sum we know that
So the expression simplifies to
Therefore we have the equation
Finishing, we have
Solution 7
We can pair the terms of the summations as below.
From here, we use the cosine and sine subtraction formulas as shown.
\begin{align*} &\dfrac{(\cos(45-44) + \cos(45-1)) + (\cos(45-43) + \cos(45-2))+ \cdots (\cos(45-23) + \cos(45-22))}{(\sin(45-44) + \sin(45-1)) + (\sin(45-43) + \sin(45-2))+ \cdots (\sin(45-23) + \sin(45-22))} \\ &= \dfrac{(\cos{45}\cos{44} +\sin{45}\sin{44} + \cos{45}\cos{1} + \sin{45}\sin{1})+ \cdots + (\cos{45}\cos{23} + \sin{45}\sin{23} + \cos{45}\cos{22} + \sin{45}\sin{22})}{(\sin{45}\cos{44}-\cos{45}\sin{44} + \sin{45}\cos{1} - \cos{45}\sin{1}) + \cdots + (\sin{45}\cos{23} -\cos{45}\sin{23} + \sin{45}\cos{22} - \cos{45}\sin{22})} \\ &= \dfrac{\sqrt{2}/2(\cos{44} + \sin{44} + \cos{1}+\sin{1} +\cos{43} + \sin{43} + \cos{2} + \sin{2} + \cdots + \cos{23} + \sin{23} + \cos{22} + \sin{22})}{\sqrt{2}/2(\cos{44} -\sin{44} +\cos{1} - \sin{1} + \cos{43}-\sin{43} + \cos{2} -\sin{2} +\cdots + \cos{23}-\sin{23} + \cos{22} - \sin{22})} \\ &=\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ}. \end{align*}
Since all of these operations have been performed to ensure equality, we can set the final line of the above mess equal to our original expression.
For the sake of clarity, let and
. Then, we have
Finishing, we have . Adding
to both sides gives
, or
. Taking the positive case gives
. Finally,
~NTfish
See also
Video Solution by Osman Nal: https://www.youtube.com/watch?v=o6udScV0F_o
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 | ||
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