During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

Difference between revisions of "1985 AHSME Problems/Problem 23"

(Created page with "==Problem== If <math> x=\frac{-1+i\sqrt{3}}{2} </math> and <math> y=\frac{-1-i\sqrt{3}}{2} </math>, where <math> i^2=-1 </math>, then which of the following is ''not'' correct? ...")
 
Line 48: Line 48:
 
==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}}

Revision as of 12:01, 5 July 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

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.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS