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2004 AMC 12A Problems/Problem 23

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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} ...

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 (ProblemsResources)
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
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