Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AIME I Problems/Problem 13"
(→‎Solution 5)
Line 90: Line 90:
 
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 three by three, having <math>(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})....(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})=(\frac{1}{4})^{15}\cdot \sin3^{\circ}\cdot \sin9^{\circ}....\sin87^{\circ}=(\frac{1}{4})^{20}\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}*\frac{\sqrt{2}}{2}\cdot sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}}</math>
 
  we can group them three by three, having <math>(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})....(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})=(\frac{1}{4})^{15}\cdot \sin3^{\circ}\cdot \sin9^{\circ}....\sin87^{\circ}=(\frac{1}{4})^{20}\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}*\frac{\sqrt{2}}{2}\cdot sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}}</math>
−
So we take reciprocal, the expression is <math>2^{\frac{89}{2}}</math>, the desired answer is <math>\2^{89}</math> leads to answer <math>\boxed{091}</math>
+
So we take reciprocal, the expression is <math>2^{\frac{89}{2}}</math>, the desired answer is <math>2^{89}</math> leads to answer <math>\boxed{091}</math>
  
 
~bluesoul
 
~bluesoul

Revision as of 01:32, 20 March 2022

Problem

With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$.

Solution 1

Let $x = \cos 1^\circ + i \sin 1^\circ$. Then from the identity \[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\] we deduce that (taking absolute values and noticing $|x| = 1$) \[|2\sin 1| = |x^2 - 1|.\] But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$, if we let our product be $M$ then \[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ\] \[= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|\] because $\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$, and so \[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = \boxed{91}$.

Solution 2

Let $p=\sin1\sin3\sin5...\sin89$

\[p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}\]

\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}\]

\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89)}}\]

\[=\sqrt{\frac{1}{2^{89}}\frac{\sin90\sin91\sin92\sin93...\sin177\sin178\sin179}{\cos1\cos2\cos3\cos4...\cos89}}\]

$=\sqrt{\frac{1}{2^{89}}}$ because of the identity $\sin(90+x)=\cos(x)$

we want $\frac{1}{p^2}=2^{89}$

Thus the answer is $2+89=\boxed{091}$

Solution 3

Similar to Solution $2$, so we use $\sin{2\theta}=2\sin\theta\cos\theta$ and we find that: \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*} Now we can cancel the sines of the multiples of $4$: \[1=2^{22}(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86))\] So $\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)=2^{-22}$ and we can apply the double-angle formula again: \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*} Of course, $\sin(45)=2^{-\frac{1}{2}}$ is missing, so we multiply it to both sides: \[2^{-22}\sin(45)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(45))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[\left(2^{-22}\right)\left(2^{-\frac{1}{2}}\right)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[2^{-\frac{45}{2}}=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] Now isolate the product of the sines: \[\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89)=2^{-\frac{89}{2}}\] And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: \[\csc^2(1)\csc^2(3)\csc^2(5)\csc^2(7)\cdots\csc^2(83)\csc^2(85)\csc^2(87)\csc^2(89)=\left(\frac{1}{2^{-\frac{89}{2}}}\right)^2=\left(2^{\frac{89}{2}}\right)^2=2^{89}\] The answer is therefore $m+n=(2)+(89)=\boxed{091}$.

Solution 4

Let $p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ$.

Then, $\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(2k-1)^\circ$.

Since $\sin\theta=\cos(90^{\circ}-\theta)$, we can multiply both sides by $\frac{\sqrt{2}}{2}$ to get $\sqrt{\frac{1}{2p}}=\prod_{k=1}^{23} \sin(2k-1)^\circ\cos(2k-1)^\circ$.

Using the double-angle identity $\sin2\theta=2\sin\theta\cos\theta$, we get $\sqrt{\frac{1}{2p}}=\frac{1}{2^{23}}\prod_{k=1}^{23} \sin(4k-2)^\circ$.

Note that the right-hand side is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \sin(4k)^\circ$, which is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} 2\sin(2k)^\circ\cos(2k)^\circ$, again, from using our double-angle identity.

Putting this back into our equation and simplifying gives us $\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ$.

Using the fact that $\sin\theta=\cos(90^{\circ}-\theta)$ again, our equation simplifies to $\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}$, and since $\sin90^\circ=1$, it follows that $2p = 2^{90}$, which implies $p=2^{89}$. Thus, $m+n=2+89=\boxed{091}$.

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 $x^n+1$ are $e^{\frac{(2k-1)\pi i}{n}}, k=1,2,...,n$, we have \[x^n + 1 = \prod_{k=1}^{n}(x-e^{\frac{(2k-1)\pi i}{n}})\] Let $x=1$, and take absolute value of both sides, \[2 = \prod_{k=1}^{n}|1-e^{\frac{(2k-1)\pi i}{n}}|= 2^n\prod_{k=1}^{n}|\sin\frac{(2k-1)\pi}{2n}|\] or, \[\prod_{k=1}^{n}|\sin\frac{(2k-1)\pi}{2n}| = 2^{-(n-1)}\] Let $n$ be even, then, \[\sin\frac{(2k-1)\pi}{2n} = \sin\left(\pi - \frac{(2k-1)\pi}{2n}\right) = \sin\left(\frac{(2(n-k+1)-1)\pi}{2n}\right)\] so, \[\prod_{k=1}^{n}\left|\sin\frac{(2k-1)\pi}{n}\right| = \prod_{k=1}^{\frac{n}{2}}\sin^2\frac{(2k-1)\pi}{2n}\] Set $n=90$ and we have \[\prod_{k=1}^{45}\sin^2\frac{(2k-1)\pi}{180} = 2^{-89}\], \[\prod_{k=1}^{45}\csc^2\frac{(2k-1)\pi}{180} = 2^{89}\] -Mathdummy

Solution 5

Recall that $\sin\alpha\cdot \sin(60^{\circ}-\alpha)\cdot \sin(60^{\circ}+\alpha)=\frac{1}{4}\cdot \sin3\alpha$ Since it is in csc, we can write in sin and then take reciprocal. 

we can group them three by three, having $(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})....(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})=(\frac{1}{4})^{15}\cdot \sin3^{\circ}\cdot \sin9^{\circ}....\sin87^{\circ}=(\frac{1}{4})^{20}\sin9^{\circ}\cdot \sin27^{\circ}\cdot \sin45^{\circ}\cdot \sin63^{\circ}\cdot \sin81^{\circ}=(\frac{1}{4})^{20}\cdot \frac{\sqrt{2}}{2}\cdot \sin9^{\circ}\cdot \cos9^{\circ}\cdot \sin27^{\circ}\cdot \cos27^{\circ}=(\frac{1}{4})^{21}*\frac{\sqrt{2}}{2}\cdot sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}}$

So we take reciprocal, the expression is $2^{\frac{89}{2}}$, the desired answer is $2^{89}$ leads to answer $\boxed{091}$

~bluesoul

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_13&oldid=172796"