AMC 8 · 2018 · #23

Grade 7 probabilitycounting

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

probability-basiccombinations-basiccomplementary-countingsystematic-enumeration complementary-countingcasework ↑ Prerequisites: combinations-basicfraction-arithmetic

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Label the 8 vertices of a regular octagon V₁, V₂, …, V₈ in order around the shape. Pick 3 of these vertices uniformly at random and connect them to form a triangle. What is the probability that at least one side of the triangle coincides with one of the 8 sides of the octagon?

Pick an answer.

(A)

$\frac{2}{7}$

(B)

$\frac{5}{42}$

(C)

$\frac{11}{14}$

(D)

$\frac{5}{7}$

(E)

$\frac{6}{7}$

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

The phrase "at least one side" is the classic trigger for Tool #16 (Complement). Counting triangles that share at least one side splits into messy overlapping cases (one side vs. two sides shared), but the opposite event — triangles where NO two chosen vertices are adjacent — is a single clean condition. Tool #2 (Systematic List) then lets us count those "no two adjacent" triples by fixing the smallest vertex and listing valid gaps, which is much easier than juggling inclusion-exclusion. Finally we subtract from 1 to get the answer the problem actually asks for.

1STEP 1

The sample space is every way to pick 3 of the 8 vertices, order aside — a combination giving 56 triangles.

C(8, 3) =

\frac{(8 × 7 × 6)}{(3 × 2 × 1)}

= 56

2STEP 2

Counting 'at least one shared side' directly is messy, so flip to the complement: no two chosen vertices are adjacent.

P(at least one) = 1 - P(none)

3STEP 3

List the triples with no two vertices adjacent (V₁ and V₈ count as adjacent); the circular non-adjacent formula confirms 16 of them.

\frac{8}{(8-3)}

C(8-3, 3) =

\frac{8}{5}

C(5, 3) =

\frac{8}{5}

× 10 = 16

4STEP 4

Among the 56 triangles, 16 share no side, so the complement probability is $\frac{2}{7}$ .

P(no shared side) =

\frac{16}{56}

\frac{2}{7}

5STEP 5

Subtract from 1 to answer the original question: 1 − $\frac{2}{7}$ = $\frac{5}{7}$ , choice (D).

P(at least one shared side) = 1 -

\frac{2}{7}

\frac{5}{7}

→ (D)

Answer

\frac{5}{7}

The answer

\frac{5}{7}

≈ 0.71 should feel large — most triangles drawn from an octagon's vertices do touch a side of the octagon, since you only avoid that by picking three pairwise non-adjacent vertices, which is restrictive. A quick cross-check:

\frac{5}{7}

\frac{2}{7}

= 1, so the favorable and "no-side" cases partition the sample space exactly, with no double counting. Choice (D) matches.

💡Key takeaway

This AMC 8 problem only needs Grade 7 probability models you already know — flip an 'at least one' question into a 'none' question, count both, and subtract!

More like this