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

The first graph peaks at

(a, b);

the second bottoms out at

(c, d)

Rewrite the intersection equation as

|x-a|+|x-c|=b-d

and use symmetry

At an intersection,

-|x-a|+b=|x-c|+d,

|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,

a+c=2+8=10.

Thus, the correct answer is C.

Problem 22 in other years