Question

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 $m + n + p?$

$602$

$652$

$702$

$752$

$802$

Solution:

Because $g(x) = -f(100 - x),$ the graphs of $f$ and $g$ are reflections of each other through the point $(50, 0),$ so the four intercepts pair up with $x_2 + x_3 = x_1 + x_4 = 100.$

With $x_3 - x_2 = 150,$ we get $x_2 = -25$ and $x_3 = 125.$

Take $x_1, x_3$ as the roots of $f,$ whose vertex has $x$ -coordinate $h = \dfrac{x_1 + x_3}{2},$ so $x_1 = 2h - 125.$ The condition that the vertex of $f$ lies on the graph of $g$ gives $1 = \frac{f(h)}{g(h)} = \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)},$ which solves to $h = -25 - 75\sqrt{2}.$

Then $x_4 = 100 - x_1,$ so $x_4 - x_1 = 350 - 4h = 450 + 300\sqrt{2}.$ Hence $m + n + p = 450 + 300 + 2 = 752.$

Thus, the correct answer is D.

2009 AMC 12A Problem 23

Solution:

Problem 23 in Other Years