Difference between revisions of "2019 AIME I Problems/Problem 8"
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− | ==Problem | + | ==Problem== |
Let <math>x</math> be a real number such that <math>\sin^{10}x+\cos^{10} x = \tfrac{11}{36}</math>. Then <math>\sin^{12}x+\cos^{12} x = \tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Let <math>x</math> be a real number such that <math>\sin^{10}x+\cos^{10} x = \tfrac{11}{36}</math>. Then <math>\sin^{12}x+\cos^{12} x = \tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
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We can substitute <math>y = \sin^2{x}</math>. Since we know that <math>\cos^2{x}=1-\sin^2{x}</math>, we can do some simplification. | We can substitute <math>y = \sin^2{x}</math>. Since we know that <math>\cos^2{x}=1-\sin^2{x}</math>, we can do some simplification. | ||
− | This yields <math>y^5+(1-y)^5=\frac{11}{36}</math>. From this, we can substitute again to get some cancellation through binomials. If we let <math>z=\frac{1}{2}-y</math>, we can simplify the equation to < | + | This yields <math>y^5+(1-y)^5=\frac{11}{36}</math>. From this, we can substitute again to get some cancellation through binomials. If we let <math>z=\frac{1}{2}-y</math>, we can simplify the equation to: |
+ | <cmath>\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5=\frac{11}{36}.</cmath> | ||
+ | After using binomial theorem, this simplifies to: | ||
+ | <cmath>\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}.</cmath> | ||
+ | If we use the quadratic formula, we obtain <math>z^2=\frac{1}{12}</math>, so <math>z=\pm\frac{1}{2\sqrt{3}}</math> (observe that either choice of <math>z</math> doesn't matter). Substituting <math>z,</math> we get: | ||
+ | |||
+ | <cmath>\sin^{12}{x}+\cos^{12}{x}=\left(\frac{1}{2}-z\right)^6+\left(\frac{1}{2}+z\right)^6=2z^6 + \frac{15z^4}{2} + \frac{15z^2}{8} + \frac{1}{32}=\frac{13}{54}.</cmath> | ||
+ | |||
+ | Therefore, the answer is <math>13+54=\boxed{067}</math>. | ||
-eric2020, inspired by Tommy2002 | -eric2020, inspired by Tommy2002 | ||
+ | ===Motivation=== | ||
+ | The motivation to substitute <math>z=\frac{1}{2}-y</math> comes so that after applying the binomial theorem to <math>y^5+(1-y)^5=\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5,</math> a lot of terms will cancel out. Note that all the terms with odd exponents in <math>\left(\frac{1}{2}+z\right)^5</math> will cancel out, while the terms with even exponents will be doubled. | ||
+ | '''mathboy282''' | ||
==Solution 2== | ==Solution 2== | ||
− | First, for simplicity, let <math>a=\sin{x}</math> and <math>b=\cos{x}</math>. Note that <math>a^2+b^2=1</math>. We then bash the rest of the problem out. Take the | + | First, for simplicity, let <math>a=\sin{x}</math> and <math>b=\cos{x}</math>. Note that <math>a^2+b^2=1</math>. We then bash the rest of the problem out. Take the fifth power of this expression and get <math>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</math>. Note that we also have <math>\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)</math>. So, it suffices to compute <math>a^2b^2(a^8+b^8)</math>. Let <math>y=a^2b^2</math>. We have from cubing <math>a^2+b^2=1</math> that <math>a^6+b^6+3a^2b^2(a^2+b^2)=1</math> or <math>a^6+b^6=1-3y</math>. Next, using <math>\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>, we get <math>a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}</math> or <math>y(1-3y)+2y^2=y-y^2=\frac{5}{36}</math>. Solving gives <math>y=\frac{5}{6}</math> or <math>y=\frac{1}{6}</math>. Clearly <math>y=\frac{5}{6}</math> is extraneous, so <math>y=\frac{1}{6}</math>. Now note that <math>a^4+b^4=(a^2+b^2)^2-2a^2b^2=\frac{2}{3}</math>, and <math>a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}</math>. Thus we finally get <math>a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}\cdot\frac{1}{6}=\frac{13}{54}</math>, giving <math>\boxed{067}</math>. |
'''- Emathmaster''' | '''- Emathmaster''' | ||
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We know that <math>a_0=2</math>, <math>a_1=1</math>, <math>a_5=\frac{11}{36}</math>. | We know that <math>a_0=2</math>, <math>a_1=1</math>, <math>a_5=\frac{11}{36}</math>. | ||
Solving the rest of the recursion gives | Solving the rest of the recursion gives | ||
− | + | ||
− | < | + | <cmath>a_2=1+2q</cmath> |
− | < | + | <cmath>a_3=1+3q</cmath> |
− | < | + | <cmath>a_4=1+4q+2q^2</cmath> |
− | < | + | <cmath>a_5=1+5q+5q^2=\frac{11}{36}</cmath> |
− | < | + | <cmath>a_6=1+6q+9q^2+2q^3</cmath> |
− | + | ||
Solving for <math>q</math> in the expression for <math>a_5</math> gives us <math>q^2+q+\frac{5}{36}=0</math>, so <math>q=-\frac{5}{6}, -\frac{1}{6}</math>. Since <math>q=-(\sin^2 x)(\cos^2 x)</math>, we know that the minimum value it can attain is <math>-\frac{1}{4}</math> by AM-GM, so <math>q</math> cannot be <math>-\frac{5}{6}</math>. | Solving for <math>q</math> in the expression for <math>a_5</math> gives us <math>q^2+q+\frac{5}{36}=0</math>, so <math>q=-\frac{5}{6}, -\frac{1}{6}</math>. Since <math>q=-(\sin^2 x)(\cos^2 x)</math>, we know that the minimum value it can attain is <math>-\frac{1}{4}</math> by AM-GM, so <math>q</math> cannot be <math>-\frac{5}{6}</math>. | ||
− | Plugging in the value of <math>q</math> into the expression for <math>a_6</math>, we get <math>a_6=1-1+\frac{1}{4}-\frac{1}{108}=\frac{26}{108}=\frac{13}{54}</math>. Our final answer is then <math>13+54=\boxed{ | + | Plugging in the value of <math>q</math> into the expression for <math>a_6</math>, we get <math>a_6=1-1+\frac{1}{4}-\frac{1}{108}=\frac{26}{108}=\frac{13}{54}</math>. Our final answer is then <math>13+54=\boxed{067}</math> |
-Natmath | -Natmath | ||
+ | |||
+ | ==Solution 6== | ||
+ | Let <math>m=\sin^2 x</math> and <math>n=\cos^2 x</math>, then <math>m+n=1</math> and <math>m^5+n^5=\frac{11}{36}</math> | ||
+ | |||
+ | <math>m^6+n^6=(m^5+n^5)(m+n)-mn(m^4+n^4)=(m^5+n^5)-mn(m^4+n^4)</math> | ||
+ | |||
+ | Now factoring <math>m^5+n^5</math> as solution 4 yields <math>m^5+n^5=(m+n)(m^4-m^3n+m^2n^2-mn^3+n^4)</math> | ||
+ | <math>=m^4+n^4-mn(m^2-mn+n^2)=m^4+n^4-mn[(m+n)^2-3mn]=m^4+n^4-mn(1-3mn)</math>. | ||
+ | |||
+ | Since <math>(m+n)^4=m^4+4m^3n+6m^2n^2+4mn^3+n^4</math>, <math>m^4+n^4=(m+n)^4-2mn(2m^2+3mn+2n^2)=1-2mn(2m^2+3mn+2n^2)</math>. | ||
+ | |||
+ | Notice that <math>2m^2+3mn+2n^2</math> can be rewritten as <math>[\sqrt{2}(a+b)]^2-mn=2-mn</math>. Thus,<math>m^4+n^4=1-2mn(2-mn)</math> and <math>m^5+n^5=1-2mn(2-mn)-mn(1-3mn)=1-5mn+5(mn)^2=\frac{11}{36}</math>. As in solution 4, we get <math>mn=\frac{1}{6}</math> and <math>m^4+n^4=1-2*\frac{1}{6}(2-\frac{1}{6})=\frac{7}{18}</math> | ||
+ | |||
+ | Substitute <math>m^4+n^4=\frac{7}{18}</math> and <math>mn=\frac{1}{6}</math>, then | ||
+ | <math>m^6+n^6=\frac{11}{36}-\frac{1}{6}*\frac{7}{18}=\frac{13}{54}</math>, and the desired answer is <math>\boxed{067}</math> | ||
+ | |||
+ | ==Solution 7 (Official MAA)== | ||
+ | Let <math>c=\sin^2x\cdot\cos^2x,</math> and let <math>S(n)=\sin^{2n}x+\cos^{2n}x.</math> Then for <math>n\ge 1</math> | ||
+ | <cmath>\begin{align*} | ||
+ | S(n)&=(\sin^{2n}x+\cos^{2n}x)\cdot(\sin^2x+\cos^2x)\ | ||
+ | &=\sin^{2n+2}x+\cos^{2n+2}x+\sin^2x\cdot\cos^2x(\sin^{2n-2}x+\cos^{2n-2}x)\ | ||
+ | &=S(n+1)+cS(n-1). | ||
+ | \end{align*}</cmath> | ||
+ | Because <math>S(0)=2</math> and <math>S(1)=1,</math> it follows that <math>S(2)=1-2c, S(3)=1-3c,S(4)=2c^2-4c+1,</math> and <math>\tfrac{11}{36}=S(5)=5c^2-5c+1.</math> Hence <math>c=\tfrac16</math> or <math>\tfrac56,</math> and because <math>4c=\sin^2{2x},</math> the only possible value of <math>c</math> is <math>\tfrac16.</math> Therefore <cmath>S(6)=S(5)-cS(4)=\frac{11}{36}-\frac16\left(2\left(\frac16\right)^2-4\left(\frac16\right)+1\right)=\frac{13}{54}.</cmath> The requested sum is <math>13+54=67.</math> | ||
+ | |||
+ | ==Solution 8 (Recursion)== | ||
+ | Let <math>a_n=\sin^nx+\cos^nx</math> for non-negative integers <math>n</math>. Then <math>a_0=2</math> and <math>a_2=1</math>. In addition,<cmath>a_n=\sin^nx+\cos^nx=\left(\sin^{n-2}x+\cos^{n-2}x\right)\left(\sin^2x+\cos^2x\right)-\sin^2x\cos^2x\left(\sin^{n-4}x+\cos^{n-4}x\right)=a_{n-2}-Xa_{n-4},</cmath>where <math>X=\sin^2x\cos^2x</math>. So we can compute | ||
+ | \begin{align*} | ||
+ | a_4&=1-2X\ | ||
+ | a_6&=1-3X\ | ||
+ | a_8&=1-4X+2X^2\ | ||
+ | a_{10}&=1-5X+5X^2=\frac{11}{36} | ||
+ | \end{align*}so <math>X=\frac{1}{6},\frac{5}{6}</math>. But by the sin double angle formula, <math>\sin^2x\cos^2x=\frac{1}{4}\sin^22x\leq\frac{1}{4}</math>, so <math>X=\frac{1}{6}</math>. Then<cmath>a_{12}=a_{10}-Xa_8=\frac{11}{36}-\frac{1}{6}\cdot\frac{7}{18}=\frac{13}{54}</cmath>so the answer is <math>\boxed{067}</math> as desired. | ||
+ | |||
+ | A quick note: this solution uses recursion and is similar to the solution 7 above. It was from trumpeter, posted in the AoPS Forums, Contest Discussion. | ||
+ | |||
+ | ==Video Solution By The Power Of Logic== | ||
+ | https://youtu.be/TWQn4DvBATc | ||
+ | |||
+ | ~ Hayabusa1 | ||
==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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 12:19, 1 February 2024
Contents
[hide]Problem
Let be a real number such that . Then where and are relatively prime positive integers. Find .
Solution 1
We can substitute . Since we know that , we can do some simplification.
This yields . From this, we can substitute again to get some cancellation through binomials. If we let , we can simplify the equation to: After using binomial theorem, this simplifies to: If we use the quadratic formula, we obtain , so (observe that either choice of doesn't matter). Substituting we get:
Therefore, the answer is .
-eric2020, inspired by Tommy2002
Motivation
The motivation to substitute comes so that after applying the binomial theorem to a lot of terms will cancel out. Note that all the terms with odd exponents in will cancel out, while the terms with even exponents will be doubled. mathboy282
Solution 2
First, for simplicity, let and . Note that . We then bash the rest of the problem out. Take the fifth power of this expression and get . Note that we also have . So, it suffices to compute . Let . We have from cubing that or . Next, using , we get or . Solving gives or . Clearly is extraneous, so . Now note that , and . Thus we finally get , giving .
- 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 . Let and be the roots of some polynomial . Then, by Vieta, for some .
Let . We want to find . Clearly and . Newton sums tells us that where for our polynomial .
Bashing, we have
Thus . Clearly, so .
Note . Solving for , we get . Finally, .
Solution 4
Factor the first equation. First of all, because We group the first, third, and fifth term and second and fourth term. The first group: The second group: Add the two together to make Because this equals , we have Let so we get Solving the quadratic gives us Because , we finally get .
Now from the second equation, Plug in to get which yields the answer
~ZericHang
Solution 5
Define the recursion We know that the characteristic equation of must have 2 roots, so we can recursively define as . is simply the sum of the roots of the characteristic equation, which is . is the product of the roots, which is . This value is not trivial and we have to solve for it. We know that , , . Solving the rest of the recursion gives
Solving for in the expression for gives us , so . Since , we know that the minimum value it can attain is by AM-GM, so cannot be .
Plugging in the value of into the expression for , we get . Our final answer is then
-Natmath
Solution 6
Let and , then and
Now factoring as solution 4 yields .
Since , .
Notice that can be rewritten as . Thus, and . As in solution 4, we get and
Substitute and , then , and the desired answer is
Solution 7 (Official MAA)
Let and let Then for Because and it follows that and Hence or and because the only possible value of is Therefore The requested sum is
Solution 8 (Recursion)
Let for non-negative integers . Then and . In addition,where . So we can compute
A quick note: this solution uses recursion and is similar to the solution 7 above. It was from trumpeter, posted in the AoPS Forums, Contest Discussion.
Video Solution By The Power Of Logic
~ Hayabusa1
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.