Difference between revisions of "2015 AIME I Problems/Problem 13"
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== Solution 3 == | == 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*}\ | + | <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\ | + | &=(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}(\ | + | &=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}(\ | + | &=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>: | Now we can cancel the sines of the multiples of <math>4</math>: | ||
− | <cmath>1=2^{22}(\ | + | <cmath>1=2^{22}(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86))</cmath> |
− | So <math>\ | + | 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}&=\ | + | <cmath>\begin{align*}2^{-22}&=(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)\\ |
− | &=(2\ | + | &=(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\ | + | &=(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}(\ | + | &=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}(\ | + | &=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>\ | + | Of course, <math>\sin(45)=2^{-\frac{1}{2}}</math> is missing, so we multiply it to both sides: |
− | <cmath>2^{-22}\ | + | <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>(2^{-22})(2^{-\frac{1}{2}})=2^{22}(\ | + | <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}(\ | + | <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: | Now isolate the product of the sines: | ||
− | <cmath>\ | + | <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: | 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^ | + | <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}}})^2=\left(2^{\frac{89}{2}})^2=2^{89}</cmath> |
The answer is therefore <math>m+n=(2)+(89)=\boxed{091}</math>. | The answer is therefore <math>m+n=(2)+(89)=\boxed{091}</math>. | ||
Revision as of 18:12, 21 March 2015
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:
\[\csc^2(1)\csc^2(3)\csc^2(5)\csc^2(7)\cdots\csc^2(83)\csc^2(85)\csc^2(87)\csc^2(89)=\left(\frac{1}{2^{\frac{89}{2}}})^2=\left(2^{\frac{89}{2}})^2=2^{89}\] (Error compiling LaTeX. Unknown error_msg)
The answer is therefore .
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.