Difference between revisions of "2008 AMC 12A Problems/Problem 23"
Imagineny1 (talk | contribs) (→Solution) |
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Therefore, the area of the square is | Therefore, the area of the square is | ||
| − | <math> \frac{\left( 2 \cdot 2^{\frac{1}{8} \right)}^ | + | <math> \frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D. </math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2008|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:17, 2 May 2015
Problem
The solutions of the equation 
are the vertices of a convex polygon in the complex plane. What is the area of the polygon?
Solution
Looking at the coefficients, we are immediately reminded of the binomial expansion of 
.
Modifying this slightly, we can write the given equation as:
![\[{\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cdot \text{cis}\, \frac {\pi}{4}\]](http://latex.artofproblemsolving.com/d/5/7/d5730019e749fada542d3480cfd7661295d8d226.png)
We can apply a translation of 
and a rotation of 
(both operations preserve area) to simplify the problem:
![\[z^{4}=2^{\frac{1}{2}}\]](http://latex.artofproblemsolving.com/7/6/9/769db790032db3f7a51b9f866767cb64a99aa2b7.png)
Because the roots of this equation are created by rotating 
radians successively about the origin, the quadrilateral is a square.
We know that half the diagonal length of the square is 
Therefore, the area of the square is
See Also
| 2008 AMC 12A (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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
