2004 AMC 12B Problem 18

Attempt Problem 18 of the 2004 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 2004 AMC 12B solutions, or check the answer key.

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18.

Points

A

and

B

are on the parabola

y = 4x^2 + 7x - 1,

and the origin is the midpoint of

\overline{AB}.

What is the length of

AB?

2\sqrt{5}

5 + \dfrac{\sqrt{2}}{2}

5 + \sqrt{2}

7

5\sqrt{2}

B = (a, b),

then

A = (-a, -b)

Substitute both points into the parabola and subtract the equations

Let

B = (a, b)

and

A = (-a, -b).

Then

4a^2 + 7a - 1 = b

and

4a^2 - 7a - 1 = -b.

Subtracting gives

14a = 2b,

b = 7a.

Then

4a^2 + 7a - 1 = 7a

gives

a^2 = \dfrac14,

and

b^2 = 49a^2 = \dfrac{49}{4}.

AB = 2\sqrt{a^2 + b^2}

= 2\sqrt{\dfrac{50}{4}} = 5\sqrt{2}.

Thus, the correct answer is E.

Problem 18 in other years