Difference between revisions of "2015 AIME I Problems/Problem 13"
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With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>. | With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>. | ||
− | + | ==Solution 1== | |
− | |||
Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity | Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity | ||
<cmath>\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath> | <cmath>\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath> | ||
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It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>. | It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>. | ||
− | + | ==Solution 2== | |
Let <math>p=\sin1\sin3\sin5...\sin89</math> | Let <math>p=\sin1\sin3\sin5...\sin89</math> | ||
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Thus the answer is <math>2+89=\boxed{091}</math> | Thus the answer is <math>2+89=\boxed{091}</math> | ||
− | + | == Solution 3 == | |
Similar to Solution <math>2</math>, so we use <math>\sin{2\theta}=2\sin\theta\cos\theta</math> and we find that: | Similar to Solution <math>2</math>, so we use <math>\sin{2\theta}=2\sin\theta\cos\theta</math> and we find that: | ||
<cmath>\begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\ | <cmath>\begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\ | ||
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The answer is therefore <math>m+n=(2)+(89)=\boxed{091}</math>. | The answer is therefore <math>m+n=(2)+(89)=\boxed{091}</math>. | ||
− | + | ==Solution 4== | |
Let <math>p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ</math>. | Let <math>p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ</math>. | ||
Then, <math>\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(2k-1)^\circ</math>. | Then, <math>\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(2k-1)^\circ</math>. | ||
− | Since <math>\sin\theta=\cos( | + | Since <math>\sin\theta=\cos(90^{\circ}-\theta)</math>, we can multiply both sides by <math>\frac{\sqrt{2}}{2}</math> to get <math>\sqrt{\frac{1}{2p}}=\prod_{k=1}^{23} \sin(2k-1)^\circ\cos(2k-1)^\circ</math>. |
Using the double-angle identity <math>\sin2\theta=2\sin\theta\cos\theta</math>, we get <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{23}}\prod_{k=1}^{23} \sin(4k-2)^\circ</math>. | Using the double-angle identity <math>\sin2\theta=2\sin\theta\cos\theta</math>, we get <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{23}}\prod_{k=1}^{23} \sin(4k-2)^\circ</math>. | ||
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Putting this back into our equation and simplifying gives us <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ</math>. | Putting this back into our equation and simplifying gives us <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ</math>. | ||
− | Using the fact that <math>\sin\theta=\cos( | + | Using the fact that <math>\sin\theta=\cos(90^{\circ}-\theta)</math> again, our equation simplifies to <math>\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}</math>, and since <math>\sin90^\circ=1</math>, it follows that <math>2p = 2^{90}</math>, which implies <math>p=2^{89}</math>. Thus, <math>m+n=2+89=\boxed{091}</math>. |
+ | ==Solution 5== | ||
+ | Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think). | ||
+ | |||
+ | Recall that the roots of <math>x^n+1</math> are <math>e^{\frac{(2k-1)\pi i}{n}}, k=1,2,...,n</math>, we have | ||
+ | <cmath> x^n + 1 = \prod_{k=1}^{n}(x-e^{\frac{(2k-1)\pi i}{n}})</cmath> | ||
+ | Let <math>x=1</math>, and take absolute value of both sides, | ||
+ | <cmath>2 = \prod_{k=1}^{n}|1-e^{\frac{(2k-1)\pi i}{n}}|= 2^n\prod_{k=1}^{n}|\sin\frac{(2k-1)\pi}{2n}| </cmath> | ||
+ | or, | ||
+ | <cmath> \prod_{k=1}^{n}|\sin\frac{(2k-1)\pi}{2n}| = 2^{-(n-1)}</cmath> | ||
+ | Let <math>n</math> be even, then, | ||
+ | <cmath> \sin\frac{(2k-1)\pi}{2n} = \sin\left(\pi - \frac{(2k-1)\pi}{2n}\right) = \sin\left(\frac{(2(n-k+1)-1)\pi}{2n}\right) </cmath> | ||
+ | so, | ||
+ | <cmath> \prod_{k=1}^{n}\left|\sin\frac{(2k-1)\pi}{n}\right| = \prod_{k=1}^{\frac{n}{2}}\sin^2\frac{(2k-1)\pi}{2n}</cmath> | ||
+ | Set <math>n=90</math> and we have | ||
+ | <cmath>\prod_{k=1}^{45}\sin^2\frac{(2k-1)\pi}{180} = 2^{-89}</cmath>, | ||
+ | <cmath>\prod_{k=1}^{45}\csc^2\frac{(2k-1)\pi}{180} = 2^{89}</cmath> | ||
+ | -Mathdummy | ||
+ | ==Solution 6== | ||
+ | Recall that <math>\sin\alpha\cdot \sin(60^{\circ}-\alpha)\cdot \sin(60^{\circ}+\alpha)=\frac{1}{4}\cdot \sin3\alpha</math> | ||
+ | Since it is in csc, we can write in sin and then take reciprocal. | ||
+ | We can group them by threes, <math>P=(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})\cdots(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})</math>. Thus | ||
+ | <cmath>\begin{align*} | ||
+ | P &=\frac{1}{4^{15}}\cdot \sin3^{\circ}\cdot \sin9^{\circ}\cdots\sin87^{\circ}\\ | ||
+ | &=\frac{1}{4^{20}}\cdot \sin9^{\circ}\cdot \sin27^{\circ}\cdot \sin45^{\circ}\cdot \sin63^{\circ}\cdot \sin81^{\circ}\\ | ||
+ | &=\frac{1}{4^{20}}\cdot \frac{\sqrt{2}}{2}\cdot \sin9^{\circ}\cdot \cos9^{\circ}\cdot \sin27^{\circ}\cdot \cos27^{\circ}\\ | ||
+ | &=\frac{1}{4^{21}}\cdot \frac{\sqrt{2}}{2}\cdot \sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}} | ||
+ | \end{align*}</cmath> | ||
+ | So we take reciprocal, <math>\frac 1P=2^{\frac{89}{2}}</math>, the desired answer is <math>\frac{1}{P^2}=2^{89}</math> leads to answer <math>\boxed{091}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | We have | ||
+ | |||
+ | <cmath>\prod_{k=1}^{45} \csc^2(2k-1)^\circ = \left(\frac{1}{\sin1^\circ \cdot \sin3^\circ \cdots \sin89^\circ}\right)^2.</cmath> | ||
+ | |||
+ | Multiplying by <math>\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}</math> gives | ||
+ | |||
+ | <cmath>\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin88^\circ \cdot \sin89^\circ}\right)^2</cmath> | ||
+ | |||
+ | <cmath> = \left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin 45^\circ \cdot \cos 44^\circ \cdot \cos 43^\circ \cdots \cos1^\circ}\right)^2.</cmath> | ||
+ | |||
+ | Using <math>\sin\alpha \cos\alpha = \frac{1}{2}\sin{2\alpha}</math> gives | ||
+ | |||
+ | <cmath> \left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\frac{1}{2} \sin2^\circ \cdot \frac{1}{2} \sin4^\circ \cdots \frac{1}{2} \sin88^\circ \cdot \sin45^\circ}\right) ^2 </cmath> | ||
+ | |||
+ | <cmath> = \left(\frac{1}{(\frac{1}{2})^{44} \cdot \frac{\sqrt{2}}{2}}\right)^2 </cmath> | ||
+ | |||
+ | <cmath> = 2^{89}. </cmath> | ||
+ | |||
+ | Thus, the answer is <math>2+89 = \boxed{091}.</math> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=12|num-a=14}} | {{AIME box|year=2015|n=I|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 16:26, 16 January 2024
Contents
Problem
With all angles measured in degrees, the product , where and are integers greater than 1. Find .
Solution 1
Let . Then from the identity we deduce that (taking absolute values and noticing ) But because is the reciprocal of and because , if we let our product be then because is positive in the first and second quadrants. Now, notice that are the roots of Hence, we can write , and so It is easy to see that and that our answer is .
Solution 2
Let
because of the identity
we want
Thus the answer is
Solution 3
Similar to Solution , so we use and we find that: Now we can cancel the sines of the multiples of : So and we can apply the double-angle formula again: Of course, is missing, so we multiply it to both sides: Now isolate the product of the sines: And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: The answer is therefore .
Solution 4
Let .
Then, .
Since , we can multiply both sides by to get .
Using the double-angle identity , we get .
Note that the right-hand side is equal to , which is equal to , again, from using our double-angle identity.
Putting this back into our equation and simplifying gives us .
Using the fact that again, our equation simplifies to , and since , it follows that , which implies . Thus, .
Solution 5
Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think).
Recall that the roots of are , we have Let , and take absolute value of both sides, or, Let be even, then, so, Set and we have , -Mathdummy
Solution 6
Recall that Since it is in csc, we can write in sin and then take reciprocal. We can group them by threes, . Thus So we take reciprocal, , the desired answer is leads to answer
~bluesoul
Solution 7
We have
Multiplying by gives
Using gives
Thus, the answer is
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.