Difference between revisions of "2015 AIME I Problems/Problem 13"
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because <math>\sin</math> is positive in the first and second quadrants. Now, notice that <math>x^2, x^6, x^{10}, \dots, x^{358}</math> are the roots of <math>z^{90} + 1 = 0.</math> Hence, we can write <math>(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1</math>, and so | because <math>\sin</math> is positive in the first and second quadrants. Now, notice that <math>x^2, x^6, x^{10}, \dots, x^{358}</math> are the roots of <math>z^{90} + 1 = 0.</math> Hence, we can write <math>(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1</math>, and so | ||
<cmath>\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.</cmath> | <cmath>\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.</cmath> | ||
− | It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{ | + | It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>. |
===Solution 2=== | ===Solution 2=== | ||
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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>. | 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 | + | 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>. |
Note that the right-hand side is equal to <math>\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \sin(4k)^\circ</math>, which is equal to <math>\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} 2\sin(2k)^\circ\cos(2k)^\circ</math>, again, from using our double-angle identity. | Note that the right-hand side is equal to <math>\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \sin(4k)^\circ</math>, which is equal to <math>\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} 2\sin(2k)^\circ\cos(2k)^\circ</math>, again, from using our double-angle identity. | ||
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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>. | 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 | ||
+ | |||
==See Also== | ==See Also== |
Revision as of 10:47, 18 February 2021
Contents
Problem
With all angles measured in degrees, the product , where and are integers greater than 1. Find .
Solution
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
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.