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Revision as of 19:19, 3 July 2013

Problem

A polynomial

\[P(x) = c_{2004}x^{2004} + c_{2003}x^{2003} + ... + c_1x + c_0\]

has real coefficients with $c_{2004}\not = 0$ and $2004$ distinct complex zeroes $z_k = a_k + b_ki$, $1\leq k\leq 2004$ with $a_k$ and $b_k$ real, $a_1 = b_1 = 0$, and

\[\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}.\]

Which of the following quantities can be a nonzero number?

$\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}$

Solution

We have to evaluate the answer choices and use process of elimination:

  • $\mathrm{(A)}$: We are given that $a_1 = b_1 = 0$, so $z_1 = 0$. If one of the roots is zero, then $P(0) = c_0 = 0$.
  • $\mathrm{(B)}$: By Vieta's formulas, we know that $-\frac{c_{2003}}{c_{2004}}$ is the sum of all of the roots of $P(x)$. Since that is real, $\sum_{k = 1}^{2004}{b_k}=0=\sum_{k = 1}^{2004}{a_k}$, and $\frac{c_{2003}}{c_{2004}}=0$, so $c_{2003}=0$.
  • $\mathrm{(C)}$: All of the coefficients are real. For sake of contradiction suppose none of $b_{2\ldots 2004}$ are zero. Then for each complex root $z_i$, its complex conjugate $\overline{z_i} = a_i - b_ik$ is also a root. So the roots should pair up, but we have an odd number of imaginary roots! This gives us the contradiction, and therefore the product is equal to zero.
  • $\mathrm{(D)}$: We are given that $\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}$. Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.

There is, however, no reason to believe that $\boxed{\mathrm{E}}$ should be zero (in fact, that quantity is $P(1)$, and there is no evidence that $1$ is a root of $P(x)$).

See also

2004 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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