Harvard-MIT Mathematics Tournament

Solution:

Suppose $f$ has only one distinct root $r_{1}$. Then, if $x_{1}$ is a root of $f(f(x))$, it must be the case that $f(x_{1}) = r_{1}$. As a result, $f(f(x))$ would have at most two roots, thus not satisfying the problem condition. Hence $f$ has two distinct roots. Let them be $r_{1} \neq r_{2}$.

Since $f(f(x))$ has just three distinct roots, either $f(x) = r_{1}$ or $f(x) = r_{2}$ has one distinct root. Assume without loss of generality that $r_{1}$ has one distinct root. Then $f(x) = x^{2} + 6x + c = r_{1}$ has one root, so that $x^{2} + 6x + c - r_{1}$ is a square polynomial. Therefore, $c - r_{1} = 9$, so that $r_{1} = c - 9$. So $c - 9$ is a root of $f$. So $(c - 9)^{2} + 6(c - 9) + c = 0$, yielding $c^{2} - 11c + 27 = 0$, or $\left(c - \frac{11}{2}\right)^{2} = \frac{13}{2}$. This results to $c = \frac{11 \pm \sqrt{13}}{2}$.

If $c = \frac{11 - \sqrt{13}}{2}$, $f(x) = x^{2} + 6x + \frac{11 - \sqrt{13}}{2} = \left(x + \frac{7 + \sqrt{13}}{2}\right)\left(x + \frac{5 - \sqrt{13}}{2}\right)$. We know $f(x) = \frac{-7 - \sqrt{13}}{2}$ has a double root, $-3$. Now $\frac{-5 + \sqrt{13}}{2} > \frac{-7 - \sqrt{13}}{2}$ so the second root is above the vertex of the parabola, and is hit twice.

If $c = \frac{11 + \sqrt{13}}{2}$, $f(x) = x^{2} + 6x + \frac{11 + \sqrt{13}}{2} = \left(x + \frac{7 - \sqrt{13}}{2}\right)\left(x + \frac{5 + \sqrt{13}}{2}\right)$. We know $f(x) = \frac{-7 + \sqrt{13}}{2}$ has a double root, $-3$, and this is the value of $f$ at the vertex of the parabola, so it is its minimum value. Since $\frac{-5 - \sqrt{13}}{2} < \frac{-7 + \sqrt{13}}{2}$, $f(x) = \frac{-5 - \sqrt{13}}{2}$ has no solutions. So in this case, $f$ has only one real root.

So the answer is $c = \frac{11 - \sqrt{13}}{2}$.