Difference between revisions of "2015 AIME I Problems/Problem 13"
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Recall that <math>\sin\alpha\cdot \sin(60^{\circ}-\alpha)\cdot \sin(60^{\circ}+\alpha)=\frac{1}{4}\cdot \sin3\alpha</math> | 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. | Since it is in csc, we can write in sin and then take reciprocal. | ||
− | We can group them | + | 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 |
− | So we take reciprocal, | + | <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 | ~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== |
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.