Difference between revisions of "1985 AHSME Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | If < | + | If <cmath>x = \frac{-1+i\sqrt{3}}{2} \qquad\text{and}\qquad y=\frac{-1-i\sqrt{3}}{2},</cmath> where <math>i^2 = -1</math>, then which of the following is not correct? |
− | <math> \mathrm{(A)\ } x^5+y^5=-1 \qquad \mathrm{(B) \ }x^7+y^7=-1 \qquad \mathrm{(C) \ } x^9+y^9=-1 \qquad </math> | + | <math> \mathrm{(A)\ } x^5+y^5 = -1 \qquad \mathrm{(B) \ }x^7+y^7 = -1 \qquad \mathrm{(C) \ } x^9+y^9 = -1 \qquad </math> |
− | <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> |
− | ==Solution== | + | ==Solution 1== |
− | + | We can write <cmath>\begin{align*}&x = \cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right) = e^{\frac{2\pi}{3}i}, \text{ and} \\ &y = \cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right) = e^{-\frac{2\pi}{3}i},\end{align*}</cmath> which gives <cmath>\begin{align*}&x^k = e^{\frac{2\pi k}{3}i} = \cos\left(\frac{2\pi k}{3}\right)+i\sin\left(\frac{2\pi k}{3}\right), \text{ and} \\ &y^k = e^{-\frac{2\pi k}{3}i} = \cos\left(-\frac{2\pi k}{3}\right)+i\sin\left(-\frac{2\pi k}{3}\right) = \cos\left(\frac{2\pi k}{3}\right)-i\sin\left(\frac{2\pi k}{3}\right),\end{align*}</cmath> using the fact that <math>\cos</math> is an even function and <math>\sin</math> is an odd function. | |
− | + | Accordingly, <cmath>x^k+y^k = 2\cos\left(\frac{2\pi k}{3}\right),</cmath> and upon substituting the values <math>k = 5,7,9,11,13</math> from the answer choices, we find that <math>x^k+y^k = -1</math> for all such values except <math>k = 9</math>, where <math>x^9+y^9 = 2\cos(6\pi) = 2 \neq -1</math>. Thus the answer is <math>\boxed{\text{(C)} \ x^9+y^9 = -1}</math>. | |
− | + | ==Solution 2== | |
+ | Notice that <math>x+y = -1</math> and <math>xy = 1</math>, so <cmath>\begin{align*}1 &=(x+y)^2 \\ &= x^2+2xy+y^2 \\ &=x^2+y^2+2,\end{align*}</cmath> and hence <math>x^2+y^2=-1</math>. Similarly, <cmath>\begin{align*}x^3+y^3 &= (x+y)^3-3x^2y-3xy^2 \\ &= \left(-1\right)^3-3xy(x+y) \\ &= -1-3(1)\left(-1\right) \\ &=2,\end{align*}</cmath> and <cmath>\begin{align*}x^5+y^5 &= (x+y)^5-5x^4y-5xy^4-10x^3y^2-10x^2y^3 \\ &= \left(-1\right)^5-5xy\left(x^3+y^3+2xy\left(x+y\right)\right) \\ &= -1-5(1)\left(2+2\left(1\right)\left(-1\right)\right) \\ &=-1.\end{align*}</cmath> | ||
− | <math> = | + | Now let <math>z_n = x^{2n+1}+y^{2n+1}</math>. Then, using the results <math>x^2+y^2 = -1</math> and <math>xy = 1</math> from above, we obtain <cmath>\begin{align*}-z_n &= \left(x^{2n+1}+y^{2n+1}\right)\left(x^2+y^2\right) \\ &= x^{2n+3}+y^{2n+3}+x^2y^{2n+1}+x^{2n+1}y^2 \\ &= x^{2n+3}+y^{2n+3}+\left(xy\right)^2\left(x^{2n-1}+y^{2n-1}\right) \\ &= x^{2n+3}+y^{2n+3}+x^{2n-1}+y^{2n-1} \\ &= z_{n+1}+z_{n-1}.\end{align*}</cmath> |
− | + | Again from above, <math>z_1 = 2</math> and <math>z_2 = -1</math>, so <cmath>\begin{align*}1 &= -z_2 \\ &= z_3+z_1 \\ &= x^7+y^7+2,\end{align*}</cmath> giving <math>z_3 = x^7+y^7 = -1</math>. Similarly, <cmath>\begin{align*}1 &= -z_3 \\ &= z_4+z_2 \\ &= x^9+y^9-1,\end{align*}</cmath> giving <math>z_4 = x^9+y^9 = 2 \neq -1</math>, meaning that the answer must be <math>\text{(C)}</math>. To confirm this, we further note that <cmath>\begin{align*}-2 &= -z_4 \\ &= z_5+z_3 \\ &= x^{11}+y^{11}-1,\end{align*}</cmath> giving <math>z_5 = x^{11}+y^{11} = -1</math>, and finally <cmath>\begin{align*}1 &= -z_5 \\ &= z_6+z_4 \\ &= x^{13}+y^{13}+2,\end{align*}</cmath> giving <math>x^{13}+y^{13} = -1</math>, which shows that the only false statement is indeed <math>\boxed{\text{(C)} \ x^9+y^9 = -1}</math>. | |
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==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}} |
Latest revision as of 00:47, 20 March 2024
Contents
Problem
If where
, then which of the following is not correct?
Solution 1
We can write which gives
using the fact that
is an even function and
is an odd function.
Accordingly, and upon substituting the values
from the answer choices, we find that
for all such values except
, where
. Thus the answer is
.
Solution 2
Notice that and
, so
and hence
. Similarly,
and
Now let . Then, using the results
and
from above, we obtain
Again from above, and
, so
giving
. Similarly,
giving
, meaning that the answer must be
. To confirm this, we further note that
giving
, and finally
giving
, which shows that the only false statement is indeed
.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AHSME Problems and Solutions |
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