2009 AMC 12A Problems/Problem 23

Problem

Functions $f$ and $g$ are quadratic, $g(x) = - f(100 - x)$ , and the graph of $g$ contains the vertex of the graph of $f$ . The four $x$ -intercepts on the two graphs have $x$ -coordinates $x_1$ , $x_2$ , $x_3$ , and $x_4$ , in increasing order, and $x_3 - x_2 = 150$ . The value of $x_4 - x_1$ is $m + n\sqrt p$ , where $m$ , $n$ , and $p$ are positive integers, and $p$ is not divisible by the square of any prime. What is $m + n + p$ ?

$import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5;real f(real x){return 1...$

The two quadratics are $180^{\circ}$ rotations of each other about $(50,0)$ . Since we are only dealing with differences of roots, we can translate them to be symmetric about $(0,0)$ . Now $x_3 = - x_2 = 75$ and $x_4 = - x_1$ . Say our translated versions of $f$ and $g$ are $p$ and $q$ , respectively, so that $p(x) = - q( - x)$ . Let $x_3 = 75$ be a root of $p$ and $x_2 = - 75$ a root of $q$ by symmetry. Note that since they each contain each other's vertex, $x_1$ , $x_2$ , $x_3$ , and $x_4$ must be roots of alternating polynomials, so $x_1$ is a root of $p$ and $x_4$ a root of $q$

$p(x) = a(x - 75)(x - x_1) \\q(x) = - a(x + 75)(x + x_1)$

The vertex of $p(x)$ is half the sum of its roots, or $\frac {75 + x_1}{2}$ . We are told that the vertex of one quadratic lies on the other, so

$\begin{eqnarray*} p\left(\frac {75 + x_1}{2}\right) & = & a\left(\frac {75 - x_1}{2}\right)\left(\frac { - 75 + x_1}{...$

Let $x_1 = 75u$ and divide through by $75^2$ , since this is a timed competition and it will drastically simplify computations. We know $u < - 1$ and that $(u - 1)^2 = (3u + 1)(u + 3)$ , or

$\begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\& = & 3u^2 + 10u + 3 - (u^2 - 2u + 1) \\& = &...$