Difference between revisions of "2008 AMC 12A Problems/Problem 23"
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The solutions of the equation <math>z^4+4z^3i-6z^2-4zi-i=0</math> are the vertices of a convex polygon in the complex plane. What is the area of the polygon? | The solutions of the equation <math>z^4+4z^3i-6z^2-4zi-i=0</math> are the vertices of a convex polygon in the complex plane. What is the area of the polygon? | ||
− | <math>\ | + | <math>\mathrm{(A)}\ 2^{\frac{5}{8}}\qquad\mathrm{(B)}\ 2^{\frac{3}{4}}\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 2^{\frac{5}{4}}\qquad\mathrm{(E)}\ 2^{\frac{3}{2}}</math> |
==Solution== | ==Solution== |
Revision as of 01:57, 26 April 2008
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:
We can apply a translation of
and a rotation of
(both operations preserve area) to simplify the problem:
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 $\frac{\left( 2 \cdot 2^{\frac{1}{8} \right)}^{2}}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D.$ (Error compiling LaTeX. Unknown error_msg)
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 |