Difference between revisions of "1997 AIME Problems/Problem 11"
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== Problem == | == Problem == | ||
− | Let <math>x=\frac{\ | + | 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__ | __TOC__ | ||
− | == 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}\\ | <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> | + | &=& \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 | 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 | ||
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<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)\\ | <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) | &=& \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> | </cmath> | ||
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
− | + | == Solution 3 == | |
A slight variant of the above solution, note that | A slight variant of the above solution, note that | ||
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\sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\\ | \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\\ | &=& \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</cmath> | + | \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>. | 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. | 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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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>. |
+ | |||
+ | == 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 | ||
+ | |||
+ | == Solution 8 (Integral Calculus, not rigorous) == | ||
+ | Because the numerator and denominator are both sums, we can replace each with <math>44</math> times the average (i.e. [[arithmetic mean]]) of the terms of the respective sums. These <math>44</math>s will cancel out, leaving simply the ratio of the averages of the respective sums. | ||
+ | |||
+ | Given that the sums span from <math>1^{\circ}\approx 0</math> radians to <math>44^{\circ}\approx \tfrac\pi4</math> radians, we would expect the average heights of cosine and sine on <math>[0,\tfrac\pi4]</math> to be very close if not equal to the given ratio. We can use definite [[integrals]] to calculate these averages. | ||
+ | |||
+ | The average height of <math>\cos x</math> on <math>[0,\tfrac\pi4]</math> is <math>\frac1{\pi/4}\int^{\pi/4}_0(\cos x) dx = \frac4\pi(\sin\frac\pi4-\sin0) = \frac4\pi\cdot\frac{\sqrt2}2</math>. | ||
+ | |||
+ | Similarly, the average height of <math>\sin x</math> on <math>[0,\tfrac\pi4]</math> is <math>\frac1{\pi/4}\int^{\pi/4}_0(\sin x) dx = \frac4\pi(-\cos\frac\pi4+\cos0) = \frac4\pi\cdot(1-\frac{\sqrt2}2)</math>. | ||
+ | |||
+ | Taking the ratio of these two values yields <math>\frac{\sqrt2/2}{1-\sqrt2/2}=\frac1{\sqrt2-1}\cdot\frac{\sqrt2+1}{\sqrt2+1}=\frac{\sqrt2+1}{2-1}=\sqrt2+1 \approx 1.41+1 = 2.41</math>. Thus, our answer is <math>\boxed{241}</math>. | ||
== See also == | == See also == | ||
− | {{AIME box|year=1997|num-b=10|num-a=12}} | + | 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]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:05, 8 September 2024
Problem
Let . What is the greatest integer that does not exceed ?
Contents
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
Solution 8 (Integral Calculus, not rigorous)
Because the numerator and denominator are both sums, we can replace each with times the average (i.e. arithmetic mean) of the terms of the respective sums. These s will cancel out, leaving simply the ratio of the averages of the respective sums.
Given that the sums span from radians to radians, we would expect the average heights of cosine and sine on to be very close if not equal to the given ratio. We can use definite integrals to calculate these averages.
The average height of on is .
Similarly, the average height of on is .
Taking the ratio of these two values yields . Thus, our answer is .
See also
Video Solution by Osman Nal: https://www.youtube.com/watch?v=o6udScV0F_o
1997 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 12 | |
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