Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #23
Grade 7 probabilitycounting
Pick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count all the triangles
The sample space is every way to pick 3 of the 8 vertices, order aside — a combination giving 56 triangles.
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.8Make A Systematic ListSwitch to the complement
Counting 'at least one shared side' directly is messy, so flip to the complement: no two chosen vertices are adjacent.
Flipping a probability question from "at least one" to "none" is the complement rule from Grade 7 probability models.
7.SP.C.7Change Focus Count The ComplementCount the no-side triangles
List the triples with no two vertices adjacent (V₁ and V₈ count as adjacent); the circular non-adjacent formula confirms 16 of them.
Once you systematically list non-adjacent triples, the cyclic symmetry and the closed-form formula both confirm there are 16 such triangles.
Exactly 16 of the 56 possible triangles have no two of their three vertices sitting next to each other around the octagon.
▸ Why?
Organize the count by walking around the ring and recording the gaps between the chosen vertices; every 'no two adjacent' triangle falls into exactly one gap pattern, so adding up the patterns gives the full count of 16.
▸ Why?
Sorting each triangle's vertices around the ring and filing it under its gap pattern leaves no triangle out and counts none twice, so the sizes of the groups add back to the true total.
▸ Why?
The gap running from the highest-numbered chosen vertex back down to the lowest has to be counted too, because vertex V₈ is a neighbor of vertex V₁ — the ring of 8 vertices closes back on itself.
▸ Why?
We are allowed to count the valid triangles anchored at one starting vertex and then scale up, because turning the octagon one step lays it exactly onto itself and carries every valid triangle to another valid one.
Form the complement probability
Among the 56 triangles, 16 share no side, so the complement probability is .
Probability of an event is favorable count over total count — Grade 7 probability model.
7.SP.C.7Change Focus Count The ComplementSubtract from 1
Subtract from 1 to answer the original question: 1 − = , choice (D).
Subtracting a fraction from 1 and recognizing 5/7 in lowest terms is Grade 4 equivalent-fractions arithmetic.
4.NF.A.1Change Focus Count The ComplementThis 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!
- Count all the triangles
- Switch to the complement
- Count the no-side triangles
- Form the complement probability
- Subtract from 1
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