2010 AMC 12B Problem 23

Below is the professionally curated solution for Problem 23 of the 2010 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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Concepts:polynomial completing the square symmetry (algebra)

Difficulty rating: 2420

23.

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 $Q(x)?$

$-100$

$-82$

$-73$

$-64$

$0$

Solution:

Write $P(x)=(x-h_1)^2-k_1^2$ and $Q(x)=(x-h_2)^2-k_2^2,$ with 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.

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