Difference between revisions of "2008 AMC 12A Problems/Problem 23"
(New page: ==Problem== 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>\textbf{(A)...) |
Nikoislife (talk | contribs) (→Solution) |
||
| (12 intermediate revisions by 7 users not shown) | |||
| Line 2: | Line 2: | ||
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== | ||
| + | Looking at the coefficients, we are immediately reminded of the binomial expansion of <math>{\left(x+1\right)}^{4}</math>. | ||
| + | |||
| + | Modifying this slightly, we can write the given equation as: | ||
| + | <cmath> {\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4} </cmath> | ||
| + | We can apply a translation of <math>-i</math> and a rotation of <math>-\frac{\pi}{4}</math> (both operations preserve area) to simplify the problem: | ||
| + | <cmath>z^{4}=2^{\frac{1}{2}}</cmath> | ||
| + | |||
| + | Because the roots of this equation are created by rotating <math>\frac{\pi}{2}</math> radians successively about the origin, the quadrilateral is a square. | ||
| + | |||
| + | We know that half the diagonal length of the square is <math>{\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}</math> | ||
| + | |||
| + | Therefore, the area of the square is | ||
| + | <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}} | |
Latest revision as of 00:38, 10 November 2019
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:
![\[{\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4}\]](http://latex.artofproblemsolving.com/e/1/0/e10a3fa3d0bae693db18fc3ac2f6ac67bbf43a78.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. 
