2020 AMC 12B Problems/Problem 17

Problem

How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ , $b$ , $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$ )

$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution

We notice that $\frac{1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}$ , so in order for root $r$ to exist, $re^{i\frac{2\pi}{3}}$ must also be a root, meaning that 3 of the roots of the polynomial must be $r$ , $re^{i\frac{2\pi}{3}}$ , $re^{i\frac{4\pi}{3}}$ . However, since the polynomial is degree 5, there must be two additional roots, but in order for each of these roots to reach eachother by multiplying by 120 degrees, there must be three of them, but this clearly cannot be the case. This means that the polynomial is degree 5 with only the three previously determined roots in the form $(x-r)^m(x-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{4\pi}{3}})^p$ . Moreover, by Vieta's, we know that there is only one possible value for r as $r^5 = 2020$ . In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{i\frac{2\pi}{3}}$ and $re^{i\frac{2\pi}{3}}$ being conjugates. Since $m+n+p = 5$ as the polynomial is 5th degree, we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{\textbf{(C) } 2}$ .

-- Murtagh

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2020 AMC 12B Problems/Problem 17

Problem

Solution