AMC 8 · 2018 · #23
Grade 7 probabilitycountingProblem
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
Pick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Label the $8$ vertices of a regular octagon $V_1, V_2, \ldots, V_8$ 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?
Givens: A regular octagon has $8$ vertices and $8$ sides; Two vertices form a side of the octagon if and only if they are adjacent in the cyclic order $V_1 V_2 \cdots V_8 V_1$; $3$ of the $8$ vertices are chosen at random to form a triangle; Answer choices: (A) $\tfrac{2}{7}$, (B) $\tfrac{5}{42}$, (C) $\tfrac{11}{14}$, (D) $\tfrac{5}{7}$, (E) $\tfrac{6}{7}$
Unknowns: The probability that the resulting triangle shares at least one of its $3$ sides with a side of the octagon
Understand
Restated: Label the $8$ vertices of a regular octagon $V_1, V_2, \ldots, V_8$ 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?
Givens: A regular octagon has $8$ vertices and $8$ sides; Two vertices form a side of the octagon if and only if they are adjacent in the cyclic order $V_1 V_2 \cdots V_8 V_1$; $3$ of the $8$ vertices are chosen at random to form a triangle; Answer choices: (A) $\tfrac{2}{7}$, (B) $\tfrac{5}{42}$, (C) $\tfrac{11}{14}$, (D) $\tfrac{5}{7}$, (E) $\tfrac{6}{7}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #2 Make a Systematic List
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.
Execute — Answer: D
7.SP.C.8 Step 1 - Count the size of the sample space.
- We pick $3$ vertices out of $8$, with order irrelevant, so the total number of triangles is the combination $\binom{8}{3}$.
💡 Choosing $3$ items from $8$ where order does not matter is a combination — the Grade 7 "organized list / tables for compound events" idea.
7.SP.C.7 Step 2 - Reframe the question using the complement.
- "At least one shared side" is hard to count directly because a triangle can share $1$ or $2$ sides.
- The opposite event — "NO shared side" — happens exactly when no two of the three chosen vertices are next to each other around the octagon.
- We will count those "no two adjacent" triples instead.
💡 Flipping a probability question from "at least one" to "none" is the complement rule from Grade 7 probability models.
7.SP.C.8 Step 3 - Systematically list all triples $\{V_a, V_b, V_c\}$ with $a < b < c$ that have NO two vertices adjacent (treating $V_1$ and $V_8$ as adjacent because the octagon is a cycle).
- Fix $a = 1$ and list valid $(b, c)$: $b$ cannot be $2$ (adjacent to $V_1$), and $c$ cannot be $8$ (adjacent to $V_1$) nor equal to $b+1$.
- The valid triples with $a=1$ are $\{1,3,5\}, \{1,3,6\}, \{1,3,7\}, \{1,4,6\}, \{1,4,7\}, \{1,5,7\}$ — that is $6$ triples.
- By cyclic symmetry, each vertex is the smallest member of the same number of "no two adjacent" triples, so the count $6$ from $a=1$ overcounts the full collection by a known factor.
- A cleaner way: use the standard formula for choosing $k$ non-adjacent items from $n$ arranged in a circle, $\dfrac{n}{n-k}\binom{n-k}{k}$, with $n=8$, $k=3$.
💡 Once you systematically list non-adjacent triples, the cyclic symmetry and the closed-form formula both confirm there are $16$ such triangles.
7.SP.C.7 Step 4 Form the complement probability: out of $56$ equally likely triangles, $16$ share NO side with the octagon.
💡 Probability of an event is favorable count over total count — Grade 7 probability model.
4.NF.A.1 Step 5 Subtract from $1$ to recover the original "at least one shared side" probability, and simplify the fraction.
💡 Subtracting a fraction from $1$ and recognizing $\tfrac{5}{7}$ in lowest terms is Grade 4 equivalent-fractions arithmetic.
7.SP.C.8 Count the size of the sample space. We pick $3$ vertices out of $8$, with order 7.SP.C.7 Reframe the question using the complement. "At least one shared side" is hard to 7.SP.C.8 Systematically list all triples $\{V_a, V_b, V_c\}$ with $a < b < c$ that have N 7.SP.C.7 Form the complement probability: out of $56$ equally likely triangles, $16$ shar 4.NF.A.1 Subtract from $1$ to recover the original "at least one shared side" probability Review
Reasonableness: The answer $\tfrac{5}{7} \approx 0.71$ 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: $5/7 + 2/7 = 1$, so the favorable and "no-side" cases partition the sample space exactly, with no double counting. Choice (D) matches.
Alternative: Tool #7 (Identify Subproblems) gives the direct casework the reference solution uses: Case A — triangle shares exactly $2$ sides $\Rightarrow$ $3$ consecutive vertices $\Rightarrow$ $8$ triangles. Case B — triangle shares exactly $1$ side $\Rightarrow$ pick a side ($8$ ways), then pick a third vertex that is NOT adjacent to either endpoint ($8 - 4 = 4$ ways) $\Rightarrow$ $32$ triangles. Total favorable $= 8 + 32 = 40$, probability $= 40/56 = 5/7$. Same answer (D), confirming the complement count of $16$.
CCSS standards used (min grade 7)
4.NF.A.1Explain why a fraction is equivalent to another fraction (Simplifying $\tfrac{16}{56} = \tfrac{2}{7}$ and recognizing $1 - \tfrac{2}{7} = \tfrac{5}{7}$ as a fraction in lowest terms.)7.SP.C.7Develop probability models and use them to find probabilities of events (Treating each of the $56$ triangles as equally likely and applying the complement rule $P(A) = 1 - P(\text{not } A)$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the total $\binom{8}{3} = 56$ triangles and the $16$ triangles with no two adjacent vertices via systematic enumeration.)
⭐ 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!
⭐ 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!
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