1999 AMC 12 Problem 22

Attempt Problem 22 of the 1999 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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22.

The graphs of $y = -|x - a| + b$ and $y = |x - c| + d$ intersect at points $(2, 5)$ and $(8, 3).$ Find $a + c.$

$7$

$8$

$10$

$13$

$18$

Answer: C

Concepts:absolute value coordinate geometry midpoint

Difficulty rating: 1740

Solution:

At an intersection, $-|x-a|+b=|x-c|+d,$ or $|x-a|+|x-c|=b-d.$ Because there are two isolated intersections, they lie on opposite sides of the interval with endpoints $a$ and $c.$ The two solutions are symmetric about $x=\tfrac{a+c}{2}.$ Their $x$ -coordinates are $2$ and $8,$ so $a+c=2+8=10.$

Thus, the correct answer is C.

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