AMC 10 · 2012 · #13

Grade 8 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basicindependent-eventslinear-equations-one-var complementary-countingconvert-to-algebraidentify-subproblems ↑ Prerequisites: probability-basic

📏 Medium solution 💡 3 insights

Problem

Four random rolls become the coefficients of two parabolas. Find the chance the curves meet.

Pick an answer.

(A)

$\frac{1}{2}$

(B)

$\frac{25}{36}$

(C)

$\frac{5}{6}$

(D)

$\frac{31}{36}$

(E)

How to solve

Strategy Change Focus / Count the Complement

The phrase "at least one point in common" is the standard Tool #16 trigger: the successes are messy and the failures are one clean condition, so chase the failures and subtract. Getting there needs three preparation moves. Tool #13 (Convert to Algebra) turns the geometric words "have a point in common" into a precise equation, and the translation must run both ways — a shared point forces a solution, and a solution hands back a genuine shared point. Tool #9 (Solve an Easier Related Problem) is the pivot the whole problem rests on: because both leading coefficients are 1, subtracting the two equations kills the x² terms and reduces a parabola-versus-parabola question to a single linear equation (a-c)x = d-b. Tool #7 (Identify Subproblems) then splits that linear equation into its only three behaviours (one solution / every solution / no solution), which is where the identical-curve case gets settled instead of quietly ignored. Tool #2 (Make a Systematic List) supplies the two dice counts, 6 matching ordered pairs out of 36 and 30 mismatching ones.

1STEP 1

Say what a common point means

A shared point solves one equation.

common point ⇔ ∃ x₀ ∈ R: x₀² + a x₀ + b = x₀² + c x₀ + d

2STEP 2

The x² terms cancel

The squared terms cancel at once.

(x²+ax+b) - (x²+cx+d) = (a-c)x + (b-d) ⟹ (a-c)x = d-b

3STEP 3

Classify the linear equation

The linear equation has three cases.

(a-c)x = d-b: a ≠ c → x = (d-b)/(a-c) (one point); a=c, b=d → same curve; a=c, b ≠ d → no solution

4STEP 4

Count the two dice conditions

Only one case gives no shared point.

P(a=c) = 6/36 = 1/6, P(b ≠ d) = 30/36 = 5/6

5STEP 5

Multiply the independent parts

Its two conditions are independent.

P(no common point) = 1/6 · 5/6 = 5/36 = 180/1296

6STEP 6

Subtract from one

Subtracting from one gives 31/36, choice (D).

1 - 5/36 = 36/36 - 5/36 = 31/36 = 1116/1296 → (D)

Answer

31/36

Failing needs two things at once — an exact match on (a,c) and a mismatch on (b,d) — so failure should be rare and the answer should sit just below 1. That alone rules out (A) 1/2, (B) 25/36, and (C) 5/6. It also cannot be (E) 1, since the explicit roll a=1, b=1, c=1, d=2 gives y = x²+x+1 and y = x²+x+2, two curves a constant distance 1 apart that never meet. The exact count settles it: 6 × 30 = 180 failing quadruples out of 1296, so (1296-180)/1296 = 1116/1296 = 31/36, matching the complement computation exactly. One case deserves a second look rather than a shrug: when a=c and b=d the two parabolas are the same curve. "At least one point in common" is true there — they share every point — so those 36 quadruples belong on the success side, which is exactly where the case analysis put them. Dropping them instead would give (1116-36)/1296 = 1080/1296 = 5/6, which is choice (C) — the trap this problem is built around.

💡Key takeaway

Both curves start with x², so subtracting them leaves a straight line, and a straight line misses zero only when it is perfectly flat and lifted off the axis — which happens only when a=c and b ≠ d.

Say what a common point means
The x² terms cancel
Classify the linear equation
Count the two dice conditions
Multiply the independent parts
Subtract from one