Difference between revisions of "2015 AIME I Problems/Problem 13"
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==Solution 1== | ==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 | + | <cmath>\sin 1 = \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>) | we deduce that (taking absolute values and noticing <math>|x| = 1</math>) | ||
− | <cmath>|2\sin | + | <cmath>|2\sin 1| = |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 | 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}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ</cmath> | ||
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Let <math>p=\sin1\sin3\sin5...\sin89</math> | Let <math>p=\sin1\sin3\sin5...\sin89</math> | ||
− | <cmath>=\sqrt{\sin1\sin3\sin5...\sin177\sin179}</cmath> | + | <cmath>p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}</cmath> |
<cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}</cmath> | <cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}</cmath> | ||
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we want <math>\frac{1}{p^2}=2^{89}</math> | we want <math>\frac{1}{p^2}=2^{89}</math> | ||
− | Thus the answer is <math>2+89=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: | ||
+ | <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))\\ | ||
+ | &=(2\sin(2)\sin(88))(2\sin(4))\sin(86))(2\sin(6)\sin(84))(2\sin(8)\sin(82))\cdots(2\sin(42)\sin(48))(2\sin(44)\sin(46))\\ | ||
+ | &=2^{22}(\sin(2)\sin(88)\sin(4)\sin(86)\sin(6)\sin(84)\sin(8)\sin(82)\cdots\sin(42)\sin(48)\sin(44)\sin(46))\\ | ||
+ | &=2^{22}(\sin(2)\sin(4)\sin(6)\sin(8)\cdots\sin(82)\sin(84)\sin(86)\sin(88))\end{align*}</cmath> | ||
+ | Now we can cancel the sines of the multiples of <math>4</math>: | ||
+ | <cmath>1=2^{22}(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86))</cmath> | ||
+ | So <math>\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)=2^{-22}</math> and we can apply the double-angle formula again: | ||
+ | <cmath>\begin{align*}2^{-22}&=(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)\\ | ||
+ | &=(2\sin(1)\cos(1))(2\sin(3)\cos(3))(2\sin(5)\cos(5))(2\sin(7)\cos(7))\cdots(2\sin(41)\cos(41))(2\sin(43)\cos(43))\\ | ||
+ | &=(2\sin(1)\sin(89))(2\sin(3)\sin(87))(2\sin(5)\sin(85))(2\sin(7)\sin(87))\cdots(2\sin(41)\sin(49))(2\sin(43)\sin(47))\\ | ||
+ | &=2^{22}(\sin(1)\sin(89)\sin(3)\sin(87)\sin(5)\sin(85)\sin(7)\sin(83)\cdots\sin(41)\sin(49)\sin(43)\sin(47))\\ | ||
+ | &=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\end{align*}</cmath> | ||
+ | Of course, <math>\sin(45)=2^{-\frac{1}{2}}</math> is missing, so we multiply it to both sides: | ||
+ | <cmath>2^{-22}\sin(45)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(45))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))</cmath> | ||
+ | <cmath>\left(2^{-22}\right)\left(2^{-\frac{1}{2}}\right)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))</cmath> | ||
+ | <cmath>2^{-\frac{45}{2}}=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))</cmath> | ||
+ | Now isolate the product of the sines: | ||
+ | <cmath>\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89)=2^{-\frac{89}{2}}</cmath> | ||
+ | And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: | ||
+ | <cmath>\csc^2(1)\csc^2(3)\csc^2(5)\csc^2(7)\cdots\csc^2(83)\csc^2(85)\csc^2(87)\csc^2(89)=\left(\frac{1}{2^{-\frac{89}{2}}}\right)^2=\left(2^{\frac{89}{2}}\right)^2=2^{89}</cmath> | ||
+ | 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>. | ||
+ | |||
+ | Then, <math>\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(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 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. | ||
+ | |||
+ | 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(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 17:26, 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.