Difference between revisions of "2015 AIME I Problems/Problem 13"
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==Problem== | ==Problem== | ||
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== | ||
| + | Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity | ||
| + | <cmath>\sin x = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath> | ||
| + | we deduce that (taking absolute values and noticing <math>|x| = 1</math>) | ||
| + | <cmath>|2\sin x| = |x^2 - 1|.</cmath> | ||
| + | But because <math>\csc</math> is the reciprocal of <math>\sin</math> and because <math>\sin z = \sin (180^\circ - z)</math>, if we let our product be <math>M</math> then | ||
| + | <cmath>\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ</cmath> | ||
| + | <cmath> = \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|</cmath> | ||
| + | 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> | ||
| + | It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</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}} | ||
Revision as of 17:57, 20 March 2015
Problem
With all angles measured in degrees, the product 
, where 
and 
are integers greater than 1. Find 
.
Solution
Let 
. Then from the identity
![\[\sin x = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\]](http://latex.artofproblemsolving.com/b/8/9/b89108c11dc58e7a8a3ff42ca4008f84f65987bd.png)
we deduce that (taking absolute values and noticing 
)
![\[|2\sin x| = |x^2 - 1|.\]](http://latex.artofproblemsolving.com/e/9/e/e9e3c2b7743abe88982f8473e1a30585a11c47aa.png)
But because 
is the reciprocal of 
and because 
, if we let our product be 
then
![\[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ\]](http://latex.artofproblemsolving.com/6/4/1/641ac31f4882b96463aaf3c5df622e79b1654f33.png)
![\[= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|\]](http://latex.artofproblemsolving.com/0/f/0/0f0c50deb3ec6dd601fd7d644416290c98fe749d.png)
because is positive in the first and second quadrants. Now, notice that 
are the roots of 
Hence, we can write 
, and so
![\[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.\]](http://latex.artofproblemsolving.com/8/4/1/8418b0ac667a85750f4bab65e2639648d7cc5c74.png)
It is easy to see that 
and that our 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. 
