AMC 12 A

Suppose $p(x) = (x - r)(x - s)$. Observe that $p(x)$ must have (two) real roots in order for $p(p(x))$ to have any roots at all. More specifically, if $y$ is a root of $p(p(x))$, then $p(y) = r$ or $p(y) = s$. That is, the equations
$$(x - r)(x - s) - r = 0 \quad \text{and} \quad (x - r)(x - s) - s = 0$$
together must have exactly three real roots among them. It follows that one of these two quadratics, say $(x - r)(x - s) - r$, must have discriminant zero.
Expansion yields $x^2 - (r+s)x + r(s-1) = 0$, so the discriminant $\Delta$ of this quadratic must satisfy
$$0 = \Delta = (r+s)^2 - 4r(s-1) = (r-s)^2 + 4r.$$
This implies that $r$ is negative, say $r = -r_0$, and that $s = r \pm \sqrt{-4r} = -r_0 \pm 2\sqrt{r_0}$. It follows that
$$r + s = 2(-r_0 \pm \sqrt{r_0}) \le 2(-r_0 + \sqrt{r_0}) \le 2 \cdot \frac{1}{4} = \frac{1}{2},$$
where the second inequality follows from the fact that $a - a^2 \le \frac{1}{4}$ for all real numbers $a$. Thus $r = -\frac{1}{4}$ and $s = \frac{3}{4}$, which works. In turn, $\tilde{p}(x) = (x + \frac{1}{4})(x - \frac{3}{4})$ and $\tilde{p}(1) = \frac{5}{4} \cdot \frac{1}{4} = \frac{5}{16}$.