AMC 10 · 2011 · #14

Grade 9 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

probability-basicfunction-evaluationfundamental-counting-principle convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights

Problem

Two random digits serve as both a point's coordinates and a curve's coefficients. Find the chance the point sits above the curve.

Pick an answer.

(A)

$\frac{11}{81}$

(B)

$\frac{13}{81}$

(C)

$\frac{5}{27}$

(D)

$\frac{17}{81}$

(E)

$\frac{19}{81}$

How to solve

Strategy Convert to Algebra

Nothing useful can be read off a picture here, because the parabola is different for every draw — there are 81 outcomes and 81 parabolas. What does not change is the algebraic test for "above": the point's height b must beat the curve's height at x = a. Writing that test out collapses the whole problem into one inequality, and the surprise is that b appears on both sides of it, since b is a coefficient of the curve as well as a coordinate of the point. Once b is gathered on one side, the inequality names a smallest allowed b for each a, and only counting remains. The one fragile point is the fraction a³/(a+1): rounding it by eye is how an off-by-one gets made, and the five answer choices are 11/81, 13/81, 15/81, 17/81, 19/81 — each one point apart in the count. So the plan converts that fraction into an exact whole-number cutoff before anything is counted.

1STEP 1

Substitute the point into its own curve

The point goes into its own curve.

y(a) = a · a² - b · a = a³ - ab ⟹ b ＞ a³ - ab

2STEP 2

Collect b on one side

Gathering gives one letter above a fraction.

b ＞ a³ - ab ⇔ b(a+1) ＞ a³ ⇔ b ＞ a³/(a+1)

3STEP 3

Turn the fraction into an exact cutoff

That fraction sharpens to a whole-number cutoff.

a³/(a+1) = (a²-a+1) - 1/(a+1), 0 ＜ 1/(a+1) ＜ 1 ⟹ b ＞ a³/(a+1) ⇔ b ≥ a²-a+1

4STEP 4

Rule out the large values of a

Large values of the first digit are ruled out.

a²-a+1 ≤ 9 ⇔ a(a-1) ≤ 8; a(a-1) = 0, 2, 6, 12, 20, … strictly increasing

5STEP 5

Count the b for each surviving a

Only 19 pairs survive.

#{b} = 9 - a(a-1): a = 1 → 9, a = 2 → 7, a = 3 → 3, total 9 + 7 + 3 = 19

6STEP 6

Divide by the whole grid

Dividing by the grid gives 19/81, choice (E).

P = 19/(9 · 9) = 19/81

Answer

19/81

Recount along the other axis, which regroups the same points in a way that would not copy a slip. Fix b instead of a; the condition b ≥ a²-a+1 says a(a-1) ≤ b-1, and a(a-1) runs 0, 2, 6, 12 for a = 1,2,3,4. So b = 1 and b = 2 admit only a = 1; b = 3,4,5,6 admit a = 1,2; and b = 7,8,9 admit a = 1,2,3. The tally is 1+1+2+2+2+2+3+3+3 = 19, matching the row count. Size check: 19/81 ≈ 0.23, and the favorable set is a thin wedge in the left columns of the 9 × 9 grid, since the cutoff a²-a+1 overshoots 9 the moment a = 4; roughly a quarter of the grid is the right order of magnitude. Finally 19 is prime and does not divide 81, so 19/81 is already in lowest terms and matches choice (E) as printed.

💡Key takeaway

When the point and the curve are built from the same two numbers, substituting puts the unknown on both sides: gather it onto one side, then turn any leftover fraction into an exact whole-number cutoff before you count anything.

Substitute the point into its own curve
Collect b on one side
Turn the fraction into an exact cutoff
Rule out the large values of a
Count the b for each surviving a
Divide by the whole grid