2021 AMC Fall 10A Problem 25

1. Break $p(p(x)) = 0$ apart. Write $p(x) = (x - r)(x - s)$, where $r$ and $s$ are the roots, with $r$ the smaller one. The roots must be real and different (otherwise there are too few solutions). Now $p(\text{something}) = 0$ exactly when "something" is $r$ or $s$. So
$$p(p(x)) = 0 \quad \text{means} \quad p(x) = r \quad \text{or} \quad p(x) = s.$$
2. How many solutions can each have? Each of these is a quadratic equation, with $0$, $1$ or $2$ solutions. A value of $x$ cannot solve both, since $r \ne s$. To get exactly $3$ in total, one equation has $2$ solutions and the other has exactly $1$.
3. When does $p(x) = k$ have exactly one solution? The graph of $p$ is a U-shaped parabola. A horizontal line $y = k$ touches it at exactly one point only when $k$ is the lowest value of $p$, at the bottom of the U.
4. Find the lowest value. The bottom of the U is halfway between the roots, at $x = \frac{r+s}{2}$. There
$$p\left(\tfrac{r+s}{2}\right) = \left(\tfrac{s - r}{2}\right)\left(\tfrac{r - s}{2}\right) = -\frac{(s - r)^2}{4}.$$
This is negative, so the root equal to it must be the smaller root $r$ (and then $s$, being larger, gives $2$ solutions). The condition is
$$r = -\frac{(s - r)^2}{4}.$$
5. Use one variable. Let $u = \frac{s - r}{2}$, half the distance between the roots. Then $r = -u^2$ and $s = r + 2u = -u^2 + 2u$. The sum of the roots is
$$r + s = -2u^2 + 2u.$$
6. Make the sum as large as possible. Complete the square: $-2u^2 + 2u = -2\left(u - \frac12\right)^2 + \frac12$. The largest value is $\frac12$, at $u = \frac12$. (Check some values: $u = 0.4$ gives $0.48$, $u = 0.5$ gives $0.5$, $u = 0.6$ gives $0.48$.)
7. Find the polynomial. With $u = \frac12$: $r = -\frac14$ and $s = -\frac14 + 1 = \frac34$. So
$$\tilde p(x) = \left(x + \tfrac14\right)\left(x - \tfrac34\right).$$
Check: the bottom of the U is at $x = \frac14$, where $\tilde p = \frac12 \times \left(-\frac12\right) = -\frac14 = r$. Good.
8. Answer the question.
$$\tilde p(1) = \left(1 + \tfrac14\right)\left(1 - \tfrac34\right) = \tfrac54 \times \tfrac14 = \tfrac{5}{16}.$$

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

$p(p(x)) = 0$ looks like a hard fourth-degree equation, but it is just two ordinary quadratics: $p(x) = r$ and $p(x) = s$. An odd number of solutions means a horizontal line just touches the parabola at its bottom point. That one condition links $r$ and $s$, and "largest sum of the roots" becomes finding the top of a simple quadratic in one variable.

Work with the sum $S = r + s$. The lowest value of $p$ is $rs - \frac{S^2}{4}$. Setting it equal to $r$, and writing $s = S - r$, gives
$$r^2 - (S - 1)\,r + \frac{S^2}{4} = 0.$$
For a real $r$ to exist, the discriminant must not be negative: $(S - 1)^2 - S^2 = 1 - 2S \ge 0$. So $S \le \frac12$, and at $S = \frac12$ the double root is $r = \frac{S - 1}{2} = -\frac14$, giving $s = \frac34$ as before.

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