2016 AMC 10B Problem 9

Attempt Problem 9 of the 2016 AMC 10B 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 2016 AMC 10B solutions, or check the answer key.

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9.

All three vertices of ABC\bigtriangleup ABC lie on the parabola defined by y=x2,y=x^2, with AA at the origin and BC\overline{BC} parallel to the xx-axis. The area of the triangle is 64.64. What is the length of BC?BC?

 4 \ 4

 6 \ 6

 8 \ 8

 10 \ 10

 16 \ 16

Answer: C
Concepts:parabolatriangle areacoordinate geometry
Difficulty rating: 1280
Solution:

Let the points be A=(0,0),A=(0,0),B=(x1,y1),B=(x_1,y_1),C=(x2,y2).C=(x_2,y_2).

Then, since BCBC is parallel with the xx-axis, we know y1=y2,y_1=y_2, which we will let be y.y. Then, x12=y=x22,x_1^2 = y=x_2^2, so x12=x22.x_1^2=x_2^2 . This implies that either x1=x2x_1=-x_2 or x1=x2.x_1 = x_2. The second option cannot happen since that would set two points as the same, which would create an area of 0.0. As such, let x=x1=x2.x= -x_1=x_2. Then y=x2y=x^2 also.

Then, the points are A=(0,0),A=(0,0),B=(x,x2),B=(-x,x^2),C=(x,x2).C=(x,x^2). With a base of BC,BC, the length is 2x2x and the height is x2.x^2. This would make the area x3=64.x^3 = 64. Therefore, x=4,x=4, so BC=24=8.BC = 2\cdot 4 = 8.

Thus, the correct answer is C .

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