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, $\tilde{p}(x)=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$ . 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 (a double root) 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, it must be the case that $r_1-r_2=- 2\sqrt{-r_1}$ $*$ . It follows that the sum of the roots of $\tilde{p}(x)$ is $2r_1 + 2\sqrt{-r_1}$ , whose maximum value occurs when $r_1 = - \frac{1}{4}$ . Solving for $r_2$ yields $r_2 = \frac{3}{4}$ . Therefore, $\tilde{p}(x)=x^2 - \frac{1}{2} x - \frac{3}{16}$ , so $\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}$ .

$*$ For $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ to have two solutions, the discriminant $(r_1+r_2)^2-4r_1r_2+4r_2$ must be positive. From here, we get that $(r_1-r_2)^2>-4r_2$ , so $-4r_1>-4r_2 \rightarrow r_1<r_2$ . Hence, $r_1-r_2$ is negative, so $r_1-r_2=2\sqrt{-r_1}$ .

~ Leo.Euler

Solution 2 (Factored form)

The disrespectful function $p(x)$ has leading coefficient 1, so it can be written in factored form as $(x-r)(x-s)$ . Now the problem states that all $p(x)$ must satisfy $p(p(x)) = 0$ . Plugging our form in, we get: $((x-r)(x-s)-r)((x-r)(x-s)-s) = 0$ The roots of this equation are $(x-r)(x-s) = r, (x-r)(x-s) = s$ . By the fundamental theorem of algebra, each root must have two roots for a total of four possible values of x yet the problem states that this equation is satisfied by three values of x. Therefore one equation must give a double root. Without loss of generality, let the equation $(x-r)(x-s) = r$ be the equation that produces the double root. Expanding gives $x^2-(r+s)x+rs-r = 0$ . We know that if there is a double root to this equation, the discriminant must be equal to zero, so $(r+s)^2-4(rs-r) = 0 \implies r^2+2rs+s^2-4rs+4r = 0 \implies r^2-2rs+s^2+4r = 0$ .

From here two solutions can progress.

Solution 2.1 (Fastest)

We can rewrite $r^2-2rs+s^2+4r = 0$ as $(r-s)^2+4r = 0$ . Let's keep our eyes on the ball; we want to find the disrespectful quadratic that maximizes the sum of the roots, which is $r+s$ . Let this be equal to a new variable, $m$ , so that our problem is reduced to maximizing this variable. We can rewrite our equation in terms of m as $(m-2s)^2 + 4r = 0 \implies m^2 - 4ms+4s^2+4r = 0$ .