2021 Fall AMC 10A Problems/Problem 25

Problem

A quadratic polynomial with real coefficients and leading coefficient $1$ is called $\emph{disrespectful}$ if the equation $p(p(x))=0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$ ?

$\textbf{(A) } \frac{5}{16} \qquad\textbf{(B) } \frac{1}{2} \qquad\textbf{(C) } \frac{5}{8} \qquad\textbf{(D) } 1 \qquad\textbf{(E) } \frac{9}{8}$

Solution 1

Let $r_1$ and $r_2$ be the roots of $\tilde{p}(x)$ . Then, $\tilde{p}(x)=x^2-(r_1+r_2)x+r_1r_2$ . The solutions to $\tilde{p}(\tilde{p}(x))=0$ is the union of the solutions to $x^2-(r_1+r_2)x+(r_1r_2-r_1)=0$ and $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ . It follows that one of these two quadratics has one solution and the other has two. WLOG, assume that the quadratic with one root is $x^2-(r_1+r_2)x+(r_1r_2-r_1)=0$ . Then, the discriminant is $0$ , so $(r_1+r_2)^2 = 4r_1r_2 - 4r_1$ . Thus, $r_1-r_2=\pm 2\sqrt{-r_1}$ , but for $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ to have two solutions, $r_1-r_2=- 2\sqrt{-r_1}$ . It follows that the sum of the roots of $\tilde{p}(x)$ is $2r_1 + 2\sqrt{-r_1}$ , and its maximum value occurs when $r_1 = - \frac{1}{4}$ . Therefore, $\tilde{p}(x)=x^2 - \frac{1}{2} x - \frac{3}{16}$ , so $\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}$ .

~ Leo.Euler

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2021 Fall AMC 10A Problems/Problem 25

Problem

Solution 1