2015 AMC 12A Problem 12

Question

The parabolas

y = ax^2 - 2

and

y = 4 - bx^2

intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area

12.

What is

a + b?

1

1.5

2

2.5

3

The two

y

-intercepts are

-2

and

4;

the

x

-intercepts are symmetric about the

y

-axis

The kite’s diagonals are the vertical distance

6

and the horizontal spread of the

x

-intercepts

The

y

-intercepts of the two parabolas are

-2

and

4.

To intersect the

x

-axis, the first parabola opens upward and the second opens downward, so their

x

-intercepts are

\pm t

for some

t \gt 0.

The kite has one diagonal of length

4 - (-2) = 6

along the

y

-axis and the other of length

2t.

Its area is

\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12,

so

t = 2.

Thus the

x

-intercepts are

\pm 2.

For the first parabola,

0 = a(2)^2 - 2

gives

a = \dfrac{1}{2};

for the second,

0 = 4 - b(2)^2

gives

b = 1.

Therefore

a + b = 1.5.

Thus, the correct answer is B.

Problem 12 in other years