2011 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2011 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12A solutions, or check the answer key.

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Concepts:parabola basic probability casework

Difficulty rating: 1690

14.

Suppose $a$ and $b$ are single-digit positive integers chosen independently and at random. What is the probability that the point $(a, b)$ lies above the parabola $y = ax^2 - bx?$

$\dfrac{11}{81}$

$\dfrac{13}{81}$

$\dfrac{5}{27}$

$\dfrac{17}{81}$

$\dfrac{19}{81}$

Solution:

Substituting $x = a,$ $y = b,$ the point is above the parabola when $b \gt a^3 - ab,$ i.e. $b(a + 1) \gt a^3.$

For $a = 1:$ $b \gt \tfrac12,$ all $9$ values work. For $a = 2:$ $b \gt \tfrac83,$ so $b \ge 3,$ giving $7.$ For $a = 3:$ $b \gt \tfrac{27}{4} = 6.75,$ so $b \ge 7,$ giving $3.$ For $a \ge 4,$ no $b \le 9$ works.

The count is $9 + 7 + 3 = 19$ out of $81,$ so the probability is $\dfrac{19}{81}.$

Thus, the correct answer is E.

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