Double Quadratics (Fall 2021 AMC 12A #23)

Here's an interesting quadratic polynomial problem that utilizes more than one quadratic! Try it on your own first before following the solution.

Fall 2021 AMC 12A #23

Problem: 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 $\tilde{p}(1)$?

Solution: First, we can let the two roots of $p(x)$ be $r$ and $s$. We see that in order for $p(p(x))$ to equal zero, $p(x)$ must be equal to either $r$ or $s$. So, if we let $p(x) = (x-r)(x-s)$, we have that the solutions to $p(p(x))=0$ correspond to the solutions to $x^2-(r+s)x+rs = r$ and $x^2-(r+s)x+rs = s$.

If we let the three numbers that satisfy $p(p(x))=0$ be $m, n, q$, we can see that either one of the quadratic equations above has roots $m$ and $n$, and the other has a repeated root of $q$, or one has roots $m$ and $q$, and the other has roots $n$ and $q$. However, we see that the second case that one quadratic equation has roots $m$ and $q$ and the other has roots $n$ and $q$ is not possible, because the sum of the roots in both equations are the same: $r+s = m+q = n+q$, which means $m=n$ and only two numbers satisfy $p(p(x))=0$ instead of three.

So, we can let $x^2-(r+s)x+rs - r = (x-m)(x-n)$ and $x^2-(r+s)x+rs - s = (x-q)^2$. We now see that in order for the second quadratic $x^2-(r+s)x+rs - s$ to have a singular root, it's determinant must equal zero. We have $${b^2-4ac} = {(-(r+s))^2-4(rs-s)} = {(r^2+2rs+s^2)-4rs+4s} = {(r-s)^2+4s} = 0$$ (You can also solve with the quadratic equation or completing the square or solving equations of corresponding coefficients—there are many ways to proceed).

From here, you could square root both sides and try to maximize the sum of the roots $r+s$, but a more rigorous approach is to actually expand the final expression and solve for $s$ in terms of $r$.

Expanding, we have $s^2+(4-2r)s+r^2= 0$. Using the quadratic formula, we get $$s = \frac{(2r-4) \pm \sqrt{(4-2r)^2-4r^2}}{2} = \frac{(2r-4) \pm \sqrt{16-16r+4r^2-4r^2}}{2}= r - 2 \pm 2\sqrt{1-r}$$

Since we want to maximize the sum of the roots $r+s$, we can define $$f(r) = r + s = r + r - 2 \pm 2\sqrt{1-r} = 2 (r - 1 \pm \sqrt{1-r})$$ To find the maximum of $f(r)$, we can take the derivative $f'(r)$ and find where it equals zero since the derivative $f'(r)$ corresponds to the slope of the graph of $f(r)$, and $f(r)$ only reaches a maximum or minimum at a critical point when the slope is zero (when $f'(r)=0$) or when $f'(r)$ is undefined. Don't worry if you haven't learned calculus; you can test values of $r \leq 1$ (since $1-r$ must be $\geq 0$) such as $r = 1, \frac{3}{4}, \frac{1}{2}, \frac{1}{4}, 0$ and find where $f(r) = r+s$ appears to reach a maximum.

Taking the derivative, we find $f'(r) = 2(1 \pm \frac{-1}{2\sqrt{1-r}})$. (don't forget the chain rule!) Setting $f'(r) = 0$, we get $1 \pm \frac{-1}{2\sqrt{1-r}} = 0$ or $\pm \frac{1}{2\sqrt{1-r}} = 1$. Squaring both sides gives us $\frac{1}{4(1-r)} = 1$. Solving, we find that $1-r=\frac{1}{4}$, so $r = \frac{3}{4}$. We also need to consider when $f'(r)$ is undefined, which is when $\sqrt{1-r}=0$, giving us $r = 1$.

With the possible values of $r = \frac{3}{4}, 1$, we can plug them back into $f(r)$ to get:

• $r = \frac{3}{4}$: $f(r) = 2(\frac{3}{4} - 1 \pm \sqrt{\frac{1}{4}})$, which yields $f(r) = \frac{1}{2}, -\frac{3}{2}$.
• $r = 1$: $f(r) = 2(1-1 \pm 0)$, which yields $f(r) = 0$.

So, we can see that the sum of the roots is maximized when $f(r) = r+s = \frac{1}{2}$, where $r = \frac{3}{4}$ and $s = \frac{1}{2} - \frac{3}{4} = -\frac{1}{4}$.

Now, we can find $\tilde{p}(1) = (1)^2 - (r+s)(1) + rs = 1 - (\frac{1}{2}) + (\frac{3}{4})(-\frac{1}{4}) = 1 - \frac{1}{2} - \frac{3}{16} = \frac{5}{16}$.