2010 AMC 12B Problem 23

Question

Monic quadratic polynomials

P(x)

and

Q(x)

have the property that

P(Q(x))

has zeros at

x=-23,

-21,

-17,

and

-15,

and

Q(P(x))

has zeros at

x=-59,

-57,

-51,

and

-49.

What is the sum of the minimum values of

P(x)

and

Q(x)?

-100

-82

-73

-64

0

Answer: A

Concepts:polynomial completing the square symmetry (algebra)

Difficulty rating: 2420

Small hint:

Write

P(x)=(x-h_1)^2-k_1^2

and

Q(x)=(x-h_2)^2-k_2^2;

the zeros of

P(Q(x))

are symmetric about

x=h_2

Big hint:

The minimum values are

-k_1^2

and

-k_2^2;

find

k_1,k_2

from the spacings of the given zeros

Solution:

If

P

had only one real root, then

P(Q(x))=0

would have at most two real solutions, not four. Thus

P

has two distinct real roots, and the same argument applies to

Q.

Write

P(x)=(x-h_1)^2-k_1^2

and

Q(x)=(x-h_2)^2-k_2^2,

with

k_1,k_2\gt0

and minimum values

-k_1^2

and

-k_2^2.

The zeros of

P(Q(x))

occur where

Q(x)=h_1\pm k_1;

their four solutions are symmetric about

h_2,

so

h_2

is the average

\tfrac{-23-21-17-15}{4}=-19.

Then

Q(-15)-Q(-17)

=(16-k_2^2)-(4-k_2^2)

=12,

and this difference equals

2k_1,

so

k_1=6.

Symmetrically,

h_1=\tfrac{-59-57-51-49}{4}=-54,

and

P(-49)-P(-51)

=(25-k_1^2)-(9-k_1^2)

=16=2k_2,

so

k_2=8.

The sum of the minimum values is

-k_1^2-k_2^2=-36-64=-100.

Thus, the correct answer is A.