Difference between revisions of "1971 AHSME Problems/Problem 22"
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\textbf{(E) }1 </math> | \textbf{(E) }1 </math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | <math>\boxed{\textbf{(A) }4}</math>. | + | Expanding the given expression yields <math>1+w-w^2-w-w^2+w^3+w^2+w^3-w^4=1-w^2+2w^3-w^4</math>. Recalling that <math>w^3=1</math>, we see that this expression equals <math>1-w^2+2-w=4-(1+w+w^2)</math>. By the properties of [[roots of unity]] <math>\neq 1</math>, <math>w^2+w+1=0</math>, so the given expression equals <math>\boxed{\textbf{(A) }4}</math>. |
| + | |||
| + | == Solution 2 (not recommended) == | ||
| + | Suppose <math>w=e^{i\tfrac{2\pi}3}=\cos(\tfrac{2\pi}3)+i\sin(\tfrac{2\pi}3) = \tfrac{-1+i\sqrt3}2</math>. Substituting this into the given expression, we can calculate the result: | ||
| + | \begin{align*} | ||
| + | (1-w+w^2)(1+w-w^2) &= (1-\frac{-1+i\sqrt3}2+(\frac{-1+i\sqrt3}2)^2)(1+\frac{-1+i\sqrt3}2-(\frac{-1+i\sqrt3}2)^2) \\ | ||
| + | &= (1-\frac{-1+i\sqrt3}{2}+\frac{1-3-2i\sqrt3}{4})(1+\frac{-1+i\sqrt3}2-\frac{1-3-2i\sqrt3}{4}) \\ | ||
| + | &= (1-\frac{-1+i\sqrt3}{2}+\frac{-1-i\sqrt3}{2})(1+\frac{-1+i\sqrt3}2-\frac{-1-i\sqrt3}{2}) \\ | ||
| + | &= (1-2(\frac{i\sqrt3}2))(1+2(\frac{i\sqrt3}2)) \\ | ||
| + | &= 1^2-(i\sqrt3)^2 \\ | ||
| + | &= 1+3 \\ | ||
| + | &= 4 | ||
| + | \end{align*} | ||
| + | Thus, our answer is <math>\boxed{\textbf{(A) }4}</math>. | ||
== See Also == | == See Also == | ||
{{AHSME 35p box|year=1971|num-b=21|num-a=23}} | {{AHSME 35p box|year=1971|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:27, 5 August 2024
Problem
If 
is one of the imaginary roots of the equation 
, then the product 
is equal to
Solution 1
Expanding the given expression yields 
. Recalling that 
, we see that this expression equals 
. By the properties of roots of unity 
, 
, so the given expression equals 
.
Solution 2 (not recommended)
Suppose 
. Substituting this into the given expression, we can calculate the result:
\begin{align*}
(1-w+w^2)(1+w-w^2) &= (1-\frac{-1+i\sqrt3}2+(\frac{-1+i\sqrt3}2)^2)(1+\frac{-1+i\sqrt3}2-(\frac{-1+i\sqrt3}2)^2) \\
&= (1-\frac{-1+i\sqrt3}{2}+\frac{1-3-2i\sqrt3}{4})(1+\frac{-1+i\sqrt3}2-\frac{1-3-2i\sqrt3}{4}) \\
&= (1-\frac{-1+i\sqrt3}{2}+\frac{-1-i\sqrt3}{2})(1+\frac{-1+i\sqrt3}2-\frac{-1-i\sqrt3}{2}) \\
&= (1-2(\frac{i\sqrt3}2))(1+2(\frac{i\sqrt3}2)) \\
&= 1^2-(i\sqrt3)^2 \\
&= 1+3 \\
&= 4
\end{align*}
Thus, our answer is .
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
