AMC 10 · 2013 · #25
Grade 8 countingPick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Chasing scattered crossing points one by one is hard, so Tool #16 (Change Focus) reframes the count: every interior crossing is made by two diagonals, which together use exactly four vertices, and those four vertices form one quadrilateral. So instead of counting points, count choices of 4 vertices, giving a clean C(8, 4). That count is only exact when no three diagonals share a point, so Tool #1 (Draw a Diagram) and the octagon's symmetry reveal where diagonals become concurrent. Tool #7 (Identify Subproblems) splits the fix into two pieces — the single center and a ring of triple points — and Tool #2 (Make a Systematic List) tallies the over-counts to subtract.
Count crossings by their four corner vertices
Two diagonals crossing inside use four vertices that form one quadrilateral, so count choices of 4 vertices instead of chasing points.
Two crossing diagonals always pin down four corners, so counting crossings is the same as counting four-corner groups.
Two crossing diagonals always pin down four corners, so counting crossings is counting four-corner groups.
▸ Why?
Each group of four corners gives exactly one crossing, so groups and crossings pair off.
▸ Why?
Counting the corners without regard to order keeps each group from being tallied more than once.
Evaluate the 4-vertex count: 70
C(8, 4) = 1680/24 = 70, so a first count of 70 crossing points — exact only if no point hosts three diagonals.
Choosing 4 things out of 8 is a fixed arithmetic value, and here it works out to 70.
6.EE.A.2Make A Systematic ListSymmetry makes some diagonals share a point
But the regular octagon's symmetry lines extra diagonals up on one spot, and a point holding k diagonals is counted C(k, 2) times, not once.
A regular shape's symmetry lines up extra diagonals through the same point, breaking the one-crossing-per-point assumption.
8.G.A.1Draw A DiagramThe center: 4 diagonals, over-counted by 5
The 4 longest diagonals all pass through the center, so C(4, 2) = 6 quadrilaterals point at one spot: 5 counts are extra.
Four diagonals through the center make six pairwise crossings that are really one and the same point.
7.SP.C.8Identify SubproblemsEight triple points: over-counted by 16
A ring of 8 points each hold exactly 3 diagonals, counted C(3, 2) = 3 times instead of once, so 8 × 2 = 16 counts are extra.
Each triple point is counted three times, so it carries two extra counts, and there are eight of them.
7.SP.C.8Identify SubproblemsSubtract the extras to get 49
Every other crossing was already counted once correctly, so 70 - 5 - 16 = 49 distinct interior points. The answer is (A).
The only fixes needed are removing the double- and triple-counts, and what is left is the true point total.
4.OA.A.3Make A Systematic ListCount each crossing by the four corners that make it — that gives 70 — then subtract the extra counts wherever the octagon's symmetry pushes three or four diagonals through the same point.
- Count crossings by their four corner vertices
- Evaluate the 4-vertex count: 70
- Symmetry makes some diagonals share a point
- The center: 4 diagonals, over-counted by 5
- Eight triple points: over-counted by 16
- Subtract the extras to get 49
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