2021 AMC Fall 10A Problem 25

Let $p(x) = (x - r)(x - s)$ with real roots $r, s$ (if the roots were not real, $p(x) = \text{root}$ would have no real solutions). Then $p(p(x)) = 0$ means $p(x) = r$ or $p(x) = s$. Each of these is a quadratic equation with $0$, $1$, or $2$ real solutions, and the solution sets are disjoint since $r \ne s$ (if $r = s$, only one equation appears and it has at most $2$ solutions). Exactly three solutions in total therefore means one equation has $2$ solutions and the other has exactly $1$, a double root.

A monic quadratic $p(x) = k$ has a double root exactly when $k$ equals the minimum value of $p$, which is attained at the vertex $x = \tfrac{r + s}{2}$:
$$p\!\left(\tfrac{r+s}{2}\right) = \left(\tfrac{s - r}{2}\right)\left(\tfrac{r - s}{2}\right) = -\frac{(r - s)^2}{4}.$$
So (say) $r = -\dfrac{(r - s)^2}{4}$, i.e. $(r - s)^2 = -4r$; in particular $r \le 0$, and then $s > r$ is automatically above the minimum, so $p(x) = s$ has two solutions.

Maximize $r + s$. Put $u = \sqrt{-r} \ge 0$, so $r = -u^2$ and $s = r + 2u = -u^2 + 2u$ (taking $s > r$). Then
$$r + s = -2u^2 + 2u = -2\left(u - \tfrac12\right)^2 + \tfrac12,$$
maximized at $u = \tfrac12$ with value $\tfrac12$. This gives $r = -\tfrac14$ and $s = \tfrac34$, so
$$\tilde p(x) = \left(x + \tfrac14\right)\left(x - \tfrac34\right), \qquad \tilde p(1) = \tfrac54 \cdot \tfrac14 = \tfrac{5}{16}.$$

Check: $\tilde p$ has vertex at $x = \tfrac14$ with value $-\tfrac14 = r$, as required.

The answer is $\boxed{\textbf{(A)}\ \frac{5}{16}}$.

Composing a quadratic with itself doubles the degree, but $p(p(x)) = 0$ is just two ordinary quadratic equations, $p(x) = r$ and $p(x) = s$. An odd number of solutions is a fingerprint of tangency: one of the horizontal lines $y = r$, $y = s$ must pass through the vertex. That single condition ties $r$ and $s$ together, turning "maximize the sum of the roots" into maximizing a one-variable quadratic.

Work with the sum $\sigma = r + s$ and product $\pi = rs$ directly: the minimum value of $p$ is $\pi - \sigma^2/4$, and setting it equal to the root $r$ with $s = \sigma - r$ gives $r^2 - (\sigma - 1) r + \sigma^2/4 = 0$. A real $r$ exists only if the discriminant $(\sigma - 1)^2 - \sigma^2 = 1 - 2\sigma$ is nonnegative, so $\sigma \le \tfrac12$, with equality at $r = \tfrac{\sigma - 1}{2} = -\tfrac14$.

• Maximizing the sum of the four solutions of p(p(x)) = 0, or evaluating p at 1 for a polynomial found from a wrong tangency root (using the larger root as the vertex value).
• Assuming a double root of $p$ itself ($r = s$) can give three solutions; then $p(p(x)) = 0$ has at most two real solutions.