2016 AMC 12B Problem 6

Attempt Problem 6 of the 2016 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 2016 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

All three vertices of △ABC\triangle 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?
44
66
88
1010
1616
Answer: C
Concepts:parabolatriangle areacoordinate geometry
Small hint:
Let a top vertex be (x,x2);(x,x^2); by symmetry the other is (−x,x2)(-x,x^2)
Big hint:
The base is BC=2xBC=2x and the height is x2,x^2, so 12(2x)(x2)=64\tfrac12(2x)(x^2)=64
Solution:
Let the vertex in the first quadrant be (x,x2).(x,x^2). By symmetry the base is BC=2xBC=2x and the height is x2,x^2, so 12⋅2x⋅x2=x3=64.\tfrac12\cdot2x\cdot x^2=x^3=64. Thus x=4x=4 and BC=2x=8.BC=2x=8.
Thus, the correct answer is C.
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