2024 AMC 12A Problems/Problem 13

Problem

Let $P(x)$ be a cubic polynomial with complex coefficients whose leading coefficient is real. Suppose $P(x)$ has two real roots and one complex root $z$ . If $|z - 1| = 10$ and $P(1) = 3 + 4i$ , where $i = \sqrt{-1}$ , what is the maximum possible value of $|z|$ ?

$\textbf{(A)}~\sqrt{89} \qquad \textbf{(B)}~10 \qquad \textbf{(C)}~\sqrt{113} \qquad \textbf{(D)}~\sqrt{117} \qquad \textbf{(E)}~11$

Solution

Let $r_{1}$ and $r_{2}$ be the two real roots, and $a$ the leading coefficient of $P(x)$ . By linear factorization, $P(1) = 3 + 4i = a(1 - z)(1 - r_{1})(1 - r_{2}).$ Since $|P(1)| = |3 + 4i| = 5$ and $|1 - z| = |z - 1| = 10$ , it follows that $|a(1 - r_{1})(1 - r_{2})| = \tfrac{1}{2}$ , but since $a$ , $r_{1}$ , and $r_{2}$ are all real, we must have $a(1 - r_{1})(1 - r_{2}) \in \{-\tfrac{1}{2}, \tfrac{1}{2}\}$ and thus $(1 - z) \in \{(6 + 8i), (-6 - 8i)\}$ . This means $z \in \{(-5 - 8i), (7 + 8i)\}$ ; thus the two possible values of $|z|$ are $\sqrt{5^{2}+8^{2}}=\sqrt{89}$ and $\sqrt{7^{2}+8^{2}}=\sqrt{113}$ . The largest value is $\boxed{\textbf{(C)}~\sqrt{113}}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2024 AMC 12A Problems/Problem 13

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Solution

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