2015 AMC 12A Problem 12

Attempt Problem 12 of the 2015 AMC 12A 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 2015 AMC 12A solutions, or check the answer key.

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12.

The parabolas y=ax22y = ax^2 - 2 and y=4bx2y = 4 - bx^2 intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area 12.12. What is a+b?a + b?

11

1.51.5

22

2.52.5

33

Answer: B
Concepts:parabolakitecoordinate geometryarea
Difficulty rating: 1630
Solution:

The yy-intercepts of the two parabolas are 2-2 and 4.4. To intersect the xx-axis, the first parabola opens upward and the second opens downward, so their xx-intercepts are ±t\pm t for some t>0.t \gt 0.

The kite has one diagonal of length 4(2)=64 - (-2) = 6 along the yy-axis and the other of length 2t.2t. Its area is 1262t=6t=12,\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12, so t=2.t = 2.

Thus the xx-intercepts are ±2.\pm 2. For the first parabola, 0=a(2)220 = a(2)^2 - 2 gives a=12;a = \dfrac{1}{2}; for the second, 0=4b(2)20 = 4 - b(2)^2 gives b=1.b = 1. Therefore a+b=1.5.a + b = 1.5.

Thus, the correct answer is B.

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