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Difference between revisions of "1985 AHSME Problems/Problem 23"
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<math> \mathrm{(D) \  } x^{11}+y^{11}=-1 \qquad \mathrm{(E) \  }x^{13}+y^{13}=-1 </math>
 
<math> \mathrm{(D) \  } x^{11}+y^{11}=-1 \qquad \mathrm{(E) \  }x^{13}+y^{13}=-1 </math>
  
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==Solution==
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==Solution 1==
 
Notice that <math> x+y=-1 </math> and <math> xy=1 </math>. We have <math> 1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1 </math>.  
 
Notice that <math> x+y=-1 </math> and <math> xy=1 </math>. We have <math> 1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1 </math>.  
  
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<math> 1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1 </math>, and this is true for all other answer choices.
 
<math> 1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1 </math>, and this is true for all other answer choices.
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==Solution 2==
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Note that <math>x=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})</math> and that $y=\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3})=\cos(\frac{2\pi}{3})-i\sin(\frac{2\pi}{3})
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=22|num-a=24}}
 
{{AHSME box|year=1985|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:35, 5 November 2013

Problem

If $x=\frac{-1+i\sqrt{3}}{2}$ and $y=\frac{-1-i\sqrt{3}}{2}$, where $i^2=-1$, then which of the following is not correct?

$\mathrm{(A)\ } x^5+y^5=-1 \qquad \mathrm{(B) \ }x^7+y^7=-1 \qquad \mathrm{(C) \ } x^9+y^9=-1 \qquad$

$\mathrm{(D) \ } x^{11}+y^{11}=-1 \qquad \mathrm{(E) \ }x^{13}+y^{13}=-1$

Solution 1

Notice that $x+y=-1$ and $xy=1$. We have $1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1$.

We also have $x^3+y^3=(x+y)^3-3x^2y-3xy^2=(-1)^3-3xy(x+y)=-1-3(1)(-1)=2$.

Finally, $x^5+y^5=(x+y)^5-5x^4y-5xy^4-10x^3y^2-10x^2y^3$

$=(-1)^5-5xy(x^3+y^3+2xy(x+y))=-1-5(1)(2+2(1)(-1))=-1$.


Let $x^{2n+1}+y^{2n+1}=z$. We have

$-z=(x^{2n+1}+y^{2n+1})(x^2+y^2)$

$-z=x^{2n+3}+y^{2n+3}+x^2y^{2n+1}+x^{2n+1}y^2$

$-z=x^{2n+3}+y^{2n+3}+(xy)^2(x^{2n-1}+y^{2n-1})$

$-z=x^{2n+3}+y^{2n+3}+x^{2n-1}+y^{2n-1}$.


For $n=2, z=-1$, so

$1=x^7+y^7+x^3+y^3=x^7+y^7+2\implies x^7+y^7=-1$.


Therefore, for $n=3, z=-1$, and

$1=x^9+y^9+x^5+y^5=x^9+y^9-1\implies x^9+y^9=2$, so $x^9+y^9\not=-1, \boxed{\text{C}}$.


As a check, we have $z=2$ for $n=4$, and

$-2=x^{11}+y^{11}+x^7+y^7=x^{11}+y^{11}-1\implies x^{11}+y^{11}=-1$.


Finally, for $n=5$ we have $z=-1$, and

$1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1$, and this is true for all other answer choices.

Solution 2

Note that $x=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})$ and that $y=\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3})=\cos(\frac{2\pi}{3})-i\sin(\frac{2\pi}{3})

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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