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 "2019 AIME I Problems/Problem 8"

m (→‎Solution 1)
Line 64: Line 64:
 
~PCampbell
 
~PCampbell
  
 +
==Solution 5==
 +
Factor the first equation. <cmath>\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)</cmath>
 +
First of all, <math>\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x</math> because <math>\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x</math>
 +
We group the 1st, 3rd and 5th term and 2nd and 4th term. The 1st group: <cmath> \sin^8+\sin^4x\cos^4x+\cos^8x = (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x)
 +
=1+4\sin^4x\cos^4x-4\sin^2x\cos^2x</cmath> The 2nd group: <cmath>-\sin^6x\cos^2x-\sin^2x\cos^6x = -\sin^2x\cos^2x(\sin^4x+\cos^4x)=-\sin^2x\cos^2x(1-2\sin^2x\cos^2x) = -\sin^2x\cos^2x+2\sin^4x\cos^4x</cmath> Add the two together to make <cmath>1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x</cmath> Because this equals <math>\frac{11}{36}</math>, we have <cmath>5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0</cmath> Let <math>\sin^2x\cos^2x = a</math> so we get <cmath>5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}</cmath> Solving the quadratic gives us <cmath>a = \frac{1 \pm \frac{2}{3}}{2}</cmath> Because <math>\sin^2x\cos^2x \le \frac{1}{4}</math>, we finally get <math>a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}</math>.
 +
 +
Now from the second equation, <cmath>\sin^{12}x + \cos^{12}x = (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)=(1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)=(1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x)</cmath> Plug in <math>\sin^2x\cos^2x = \frac{1}{6}</math> to get <cmath>(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}</cmath>
 +
which yields the answer <math>\boxed{067}</math>
 +
 +
~ZericHang
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:57, 27 March 2019

Problem 8

Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$. Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

We can substitute $y = \sin^2{x}$. Since we know that $\cos^2{x}=1-\sin^2{x}$, we can do some simplification.

This yields $y^5+(1-y)^5=\frac{11}{36}$. From this, we can substitute again to get some cancellation through binomials. If we let $z=1/2-y$, we can simplify the equation to $(1/2+z)^5+(1/2-z)^5=\frac{11}{36}$. After using binomial theorem, this simplifies to $\frac{1}{16}(80z^4+40z^2+1)=11/36$. If we use the quadratic formula, we obtain the that $z^2=\frac{1}{12}$, so $z=\pm\frac{1}{2\sqrt{3}}$. By plugging z into $(1/2-z)^6+(1/2+z)^6$ (which is equal to $\sin^{12}{x}+\cos^{12}{x}$, we can either use binomial theorem or sum of cubes to simplify, and we end up with $\frac{13}{54}$. Therefore, the answer is $\boxed{067}$.

eric2020, inspired by Tommy2002

Solution 2

First, for simplicity, let $a=\sin{x}$ and $b=\cos{x}$. Note that $a^2+b^2=1$. We then bash the rest of the problem out. Take the tenth power of this expression and get $a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$. Note that we also have $\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)$. So, it suffices to compute $a^2b^2(a^8+b^8)$. Let $y=a^2b^2$. We have from cubing $a^2+b^2=1$ that $a^6+b^6+3a^2b^2(a^2+b^2)=1$ or $a^6+b^6=1-3y$. Next, using $\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$, we get $a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}$ or $y(1-3y)+2y^2=y-y^2=\frac{5}{36}$. Solving gives $y=\frac{5}{6}$ or $y=\frac{1}{6}$. Clearly $y=\frac{5}{6}$ is extraneous, so $y=\frac{1}{6}$. Now note that $a^4+b^4=(a^2+b^2)-2a^2b^2=\frac{2}{3}$, and $a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}$. Thus we finally get $a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}*\frac{1}{6}=\frac{13}{54}$, giving $\boxed{067}$.

-Emathmaster

Solution 3 (Newton Sums)

Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in solution 2. Let $\sin^2x$ and $\cos^2x$ be the roots of some polynomial $F(a)$. Then, $F(a)=a^2-a+b$ for some $b=\sin^2x\cdot\cos^2x$.

Let $S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k$. We want to find $S_6$. Clearly $S_1=1$ and $S_2=1-2b$. Newton sums tells us that $S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}$ where $k\ge 3$ for our polynomial $F(a)$.

Bashing, we have \[S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b\] \[S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1\] \[S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}\]

Thus \[5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0\] $b=\frac{1}{6} \text{ or } \frac{5}{6}$. Clearly, $\sin^2x\cdot\cos^2x\not=\frac{5}{6}$ so $\sin^2x\cdot\cos^2x=b=\frac{1}{6}$.

Note $S_4=\frac{7}{18}$. Solving for $S_6$, we get $S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}$. Finally, $13+54=\boxed{067}$.

Solution 4 (Not Fun)

We let $a = \sin^2(x)$ and $b = \cos^2(x)$, so we have $a+b=1$ and $a^5 + b^5 = \frac{11}{36}$. Noticing that $ab$ might be a useful value to find, we let $c = ab$. Then we can work our way up to find $c$. \[a + b = 1\] \[(a+b)(a+b) = 1(a+b)\] \[a^2 + 2ab + b^2 = 1\] \[a^2 + b^2 = -2c + 1\] \[(a+b)(a^2 + b^2) = -2c + 1\] \[a^3 + ab^2 + a^2b + b^3 = -2c + 1\] \[a^3 + b^3 + ab(a + b) = -2c + 1\] \[a^3 + b^3 + c = -2c + 1\] \[a^3 + b^3 = -3c + 1\] \[(a + b)(a^3 + b^3) = -3c + 1\] \[a^4 + ab^3 + a^3b + b^4 = -3c + 1\] \[a^4 + b^4 + ab(a^2 + b^2) = -3c + 1\] \[a^4 + b^4 + c(-2c + 1) = -3c + 1\] \[a^4 + b^4 - 2c^2 + c = -3c + 1\] \[a^4 + b^4 = 2c^2 - 4c + 1\] \[a^5 + ab^4 + a^4b + b^5 = 2c^2 - 4c + 1\] \[a^5 + b^5 + ab(a^3 + b^3) = 2c^2 - 4c + 1\] \[\frac{11}{36} + c(-3c + 1) = 2c^2 - 4c + 1\] \[\frac{11}{36} = 5c^2 - 5c + 1\] \[5c^2 - 5c + \frac{25}{36} = 0\] using quadform you get $c = \frac{1}{6}$ or $c = \frac{5}{6}$. Since $c = \sin^2(x)\cos^2(x) = (\sin(x)\cos(x))^2 = (\frac{\sin(2x)}{2})^2$, and since $\sin(2x)$ can't exceed 1, $c$ can't exceed $(\frac{1}{2})^2 = \frac{1}{4}$. Clearly, $c = \frac{1}{6}$. And finally, \[a^5 + b^5 = (a + b)(a^5 + b^5)\] \[a^5 + b^5 = a^6 + ab^5 + a^5b + b^6\] \[\frac{11}{36} = a^6 + b^6 + c(a^4 + b^4)\] looking back to previous results, we see that $a^4 + b^4 = 2c^2 - 4c + 1 = \frac{14}{36}$ (it's easier not to simplify the fraction). \[\frac{11}{36} = a^6 + b^6 + c(\frac{14}{36}) = a^6 + b^6 + \frac{14}{216}\] \[a^6 + b^6 = \frac{11}{36} - \frac{14}{216} = \frac{13}{54}\] which yields the answer $\boxed{067}$.

~PCampbell

Solution 5

Factor the first equation. \[\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)\] First of all, $\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x$ because $\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x$ We group the 1st, 3rd and 5th term and 2nd and 4th term. The 1st group: \[\sin^8+\sin^4x\cos^4x+\cos^8x = (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x) =1+4\sin^4x\cos^4x-4\sin^2x\cos^2x\] The 2nd group: \[-\sin^6x\cos^2x-\sin^2x\cos^6x = -\sin^2x\cos^2x(\sin^4x+\cos^4x)=-\sin^2x\cos^2x(1-2\sin^2x\cos^2x) = -\sin^2x\cos^2x+2\sin^4x\cos^4x\] Add the two together to make \[1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x\] Because this equals $\frac{11}{36}$, we have \[5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0\] Let $\sin^2x\cos^2x = a$ so we get \[5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}\] Solving the quadratic gives us \[a = \frac{1 \pm \frac{2}{3}}{2}\] Because $\sin^2x\cos^2x \le \frac{1}{4}$, we finally get $a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}$.

Now from the second equation, \[\sin^{12}x + \cos^{12}x = (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)=(1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)=(1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x)\] Plug in $\sin^2x\cos^2x = \frac{1}{6}$ to get \[(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}\] which yields the answer $\boxed{067}$

~ZericHang

See Also

2019 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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=2019_AIME_I_Problems/Problem_8&oldid=104981"