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 $m + n + p$ ?

$\textbf{(A)}\ 602\qquad \textbf{(B)}\ 652\qquad \textbf{(C)}\ 702\qquad \textbf{(D)}\ 752 \qquad \textbf{(E)}\ 802$

Solution

$[asy] import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5; real f(real x){return 1/400*(x-125)*(x+rt);} real g(real x){return -f(100-x);} xaxis(-600,600,Ticks(k, 5),Arrows(6));yaxis(ymin,ymax,Ticks(k, 5),Arrows(6)); draw(graph(f,-450,300),red+linewidth(0.8),Arrows(6));draw(graph(g,-200,550),blue+linewidth(0.8),Arrows(6)); draw((50,ymax)--(50,ymin),linetype("4 4"),Arrows(4));dot((-rt,0));dot((100+rt,0));dot((-25,0));dot((125,0)); [/asy]$

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

$\begin{eqnarray*} p\left(\frac {75 + x_1}{2}\right) & = & a\left(\frac {75 - x_1}{2}\right)\left(\frac { - 75 + x_1}{2}\right) \\ & = & - \frac {a}{4}(x_1 - 75)^2 \\ - \frac {a}{4}(x_1 - 75)^2 & = & q\left(\frac {75 + x_1}{2}\right) \\ & = & - a\left(\frac {x_1 + 225}{2}\right)\left(\frac {3x_1 + 75}{2}\right) \\ & = & - \frac {a}{4}(x_1 + 225)(3x_1 + 75) \end{eqnarray*}$

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

$\begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\ & = & 3u^2 + 10u + 3 - (u^2 - 2u + 1) \\ & = & 2u^2 + 12u + 2 \\ & = & u^2 + 6u + 1 \end{eqnarray*}$

So $u = \frac { - 6\pm\sqrt {32}}{2} = - 3\pm2\sqrt2$ . Since $u < - 1$ , $u = - 3 - 2\sqrt2$ .

The answer is $r_4 - r_1 = (-r_1) - r_1 = - 150u = 450 + 300\sqrt {2}$ , and $450 + 300 + 2 = 752\ \mathbf{(D)}$ .

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2009 AMC 12A Problems/Problem 23

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