2009 AMC 12A Problem 23

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

Answer: D

Concepts:quadratic parabola symmetry

Difficulty rating: 2420

Small hint:

The map

(x, y) \mapsto (100 - x, -y)

is a

180^\circ

rotation about

(50, 0)

that sends the graph of

f

to the graph of

g

Big hint:

So the roots pair as

x_2 + x_3 = x_1 + x_4 = 100;

with

x_3 - x_2 = 150,

find

x_2, x_3

and use the vertex condition

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

\begin{aligned} 1 &= \frac{f(h)}{g(h)} \\ &= \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)}, \end{aligned}

which gives

h = -25 \pm 75\sqrt{2}.

Since

x_1 = 2h-125

must be less than

x_2=-25,

we need

h \lt 50,

so

h = -25-75\sqrt{2}.

Then

x_4 = 100 - x_1,

so

\begin{aligned} x_4 - x_1 &= 350 - 4h \\ &= 450 + 300\sqrt{2}. \end{aligned}

Hence

m + n + p = 450 + 300 + 2

= 752.

Thus, the correct answer is D.