2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 25 Hard

A quadratic polynomial with real coefficients and leading coefficient 1 is called 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) ?(2021 AMC Fall 10A, Question #25)

  • A.

    \frac{5}{16}

  • B.

    \frac{1}{2}

  • C.

    \frac{5}{8}

  • D.

    1

  • E.

    \frac{9}{8}

Answer:A

Solution 1:

Let r_{1} and r_{2} be the roots of \tilde{p}(x). Then, \tilde{p}(x)=\left(x-r_{1}\right)\left(x-r_{2}\right)=x^{2}-\left(r_{1}+r_{2}\right) x+r_{1} r_{2}. The solutions to \tilde{p}(\tilde{p}(x))=0 is the union of the solutions to \tilde{p}(x)-r_{1}=x^{2}-\left(r_{1}+r_{2}\right) x+\left(r_{1} r_{2}-r_{1}\right)=0 and \tilde{p}(x)-r_{2}=x^{2}-\left(r_{1}+r_{2}\right) x+\left(r_{1} r_{2}-r_{2}\right)=0 . Note that one of these two quadratics has one solution (a double root) and the other has two as there are exactly three solutions. WLOG, assume that the quadratic with one root is x^{2}-\left(r_{1}+r_{2}\right) x+\left(r_{1} r_{2}-r_{1}\right)=0. Then, the discriminant is 0 , so \left(r_{1}+r_{2}\right)^{2}=4 r_{1} r_{2}-4 r_{1}. Thus, r_{1}-r_{2}=\pm 2 \sqrt{-r_{1}}, but for x^{2}-\left(r_{1}+r_{2}\right) x+\left(r_{1} r_{2}-r_{2}\right)=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 2 r_{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)= (A) \frac{5}{16}. Remark * For x^{2}-\left(r_{1}+r_{2}\right) x+\left(r_{1} r_{2}-r_{2}\right)=0 to have two solutions, the discriminant \left(r_{1}+r_{2}\right)^{2}-4 r_{1} r_{2}+4 r_{2} must be positive. From here, we get that \left(r_{1}-r_{2}\right)^{2}>-4 r_{2}, so -4 r_{1}>-4 r_{2} \Longrightarrow r_{1}<r_{2}. Hence, r_{1}-r_{2} is negative, so r_{1}-r_{2}=-2 \sqrt{-r_{1}}.

Solution 2:

Let p(x)=(x-h)^{2}+k for some real constants h and k. Suppose that p(x) has real roots r and s. Since p(p(x))=0, we conclude that p(x)=r or p(x)=s. Without the loss of generality, we assume that p(x)=r has two real solutions and p(x)=s has one real solution. Therefore, we have k=s, from which p(x)=(x-h)^{2}+s. As p(s)=0, we expand the left side to obtain (s-h)^{2}+s=0, or s^{2}-(2 h-1) s+h^{2}=0 . \quad(\star) Since (\star) has real solutions for s, the discriminant is nonnegative: (2 h-1)^{2}-4 h^{2} \geq 0. We solve this inequality to get h \leq \frac{1}{4}. Either by the axis of symmetry or Vieta's Formulas, note that r+s=2 h. As we wish to maximize 2 h, we maximize h. Substituting h=\frac{1}{4} into (\star), we obtain s^{2}+\frac{1}{2} s+\frac{1}{16}=0. We factor the left side to get \left(s+\frac{1}{4}\right)^{2}=0, or s=-\frac{1}{4}. Finally, the unique such polynomial is \tilde{p}(x)=\left(x-\frac{1}{4}\right)^{2}-\frac{1}{4} from which \tilde{p}(1)= (A) \frac{5}{16}

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