2012 AMC 12B Problem 10

Attempt Problem 10 of the 2012 AMC 12B 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 2012 AMC 12B solutions, or check the answer key.

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10.

What is the area of the polygon whose vertices are the points of intersection of the curves

x^2 + y^2 = 25

and

(x - 4)^2 + 9y^2 = 81?

24

27

36

37.5

42

Solve the first equation for

y^2

and substitute into the second

The three intersection points form a triangle; use one vertical side as the base

From

x^2+y^2=25

we get

y^2=25-x^2.

Substituting into

(x-4)^2+9y^2=81

gives

x^2+x-20=0,

x=4

x=-5.

The intersection points are

(-5,0),

(4,3),

and

(4,-3).

The vertical side from

(4,3)

(4,-3)

has length

6,

and the horizontal distance to

(-5,0)

9,

so the area is

\tfrac12\cdot6\cdot9=27.

Thus, the correct answer is B.

Problem 10 in other years