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).

All three vertices of $\triangle ABC$ lie on the parabola defined by $y=x^2,$ with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64.$ What is the length of $BC?$

$4$

$6$

$8$

$10$

$16$

Answer: C

Concepts:parabola triangle area coordinate geometry

Difficulty rating: 1350

Solution:

Let the vertex in the first quadrant be $(x,x^2).$ By symmetry the base is $BC=2x$ and the height is $x^2,$ so $\tfrac12\cdot2x\cdot x^2=x^3=64.$ Thus $x=4$ and $BC=2x=8.$

Thus, the correct answer is C.

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