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2008 AMC 12A Problems/Problem 23: Difference between revisions

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Modifying this slightly, we can write the given equation as:
Modifying this slightly, we can write the given equation as:
<cmath>  {\left(x+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cdot \text{cis}\, \frac {\pi}{4} </cmath>
<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:
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>
<cmath>z^{4}=2^{\frac{1}{2}}</cmath>
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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)}^{2}}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D. </math>
<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 $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?

$\mathrm{(A)}\ 2^{\frac{5}{8}}\qquad\mathrm{(B)}\ 2^{\frac{3}{4}}\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 2^{\frac{5}{4}}\qquad\mathrm{(E)}\ 2^{\frac{3}{2}}$

Solution

Looking at the coefficients, we are immediately reminded of the binomial expansion of ${\left(x+1\right)}^{4}$.

Modifying this slightly, we can write the given equation as: \[{\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4}\] We can apply a translation of $-i$ and a rotation of $-\frac{\pi}{4}$ (both operations preserve area) to simplify the problem: \[z^{4}=2^{\frac{1}{2}}\]

Because the roots of this equation are created by rotating $\frac{\pi}{2}$ radians successively about the origin, the quadrilateral is a square.

We know that half the diagonal length of the square is ${\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}$

Therefore, the area of the square is $\frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D.$

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

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