AMC 10 · 2013 · #25
Grade 8 countingAll 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
Pick an answer.
AMC 10 2013 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: A regular octagon has all $20$ of its diagonals drawn. Count how many distinct points strictly inside the octagon (not on its boundary) are passed through by two or more diagonals. If several diagonals go through the same interior spot, that spot still counts as just one point.
Givens: The shape is a regular octagon: $8$ equal sides and $8$ equal angles, so it has strong rotational and mirror symmetry; All $20$ diagonals of the octagon are drawn; Only crossings strictly inside the octagon count, never on the boundary or at a vertex; If three or more diagonals pass through one interior point, that location is a single point; Answer choices: (A) $49$, (B) $65$, (C) $70$, (D) $96$, (E) $128$
Unknowns: The number of distinct interior points where two or more diagonals meet
Understand
Restated: A regular octagon has all $20$ of its diagonals drawn. Count how many distinct points strictly inside the octagon (not on its boundary) are passed through by two or more diagonals. If several diagonals go through the same interior spot, that spot still counts as just one point.
Givens: The shape is a regular octagon: $8$ equal sides and $8$ equal angles, so it has strong rotational and mirror symmetry; All $20$ diagonals of the octagon are drawn; Only crossings strictly inside the octagon count, never on the boundary or at a vertex; If three or more diagonals pass through one interior point, that location is a single point; Answer choices: (A) $49$, (B) $65$, (C) $70$, (D) $96$, (E) $128$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #2 Make a Systematic List
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 $\binom{8}{4}$. 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.
Execute — Answer: A
7.SP.C.8 Step 1 Count crossings by their four corner vertices
- Directly hunting for crossing points is messy, so change focus.
- Two diagonals that cross inside use four different vertices, and those four vertices form a quadrilateral whose two diagonals are exactly that crossing pair.
- Turn it around: pick any $4$ of the $8$ vertices, and they form one quadrilateral with exactly one interior crossing of its diagonals.
- So each set of $4$ vertices gives exactly one crossing point.
- If no three diagonals ever shared a point, the number of crossings would equal the number of ways to choose $4$ vertices.
💡 Two crossing diagonals always pin down four corners, so counting crossings is the same as counting four-corner groups.
6.EE.A.2 Step 2 Evaluate the 4-vertex count: 70
- Count the ways to choose $4$ vertices out of $8$.
- The choose formula gives $\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = \frac{1680}{24} = 70$.
- So there are $70$ quadrilaterals, giving a first count of $70$ crossing points — but this is exact only if every crossing is made by exactly one pair of diagonals.
💡 Choosing 4 things out of 8 is a fixed arithmetic value, and here it works out to 70.
8.G.A.1 Step 3 Symmetry makes some diagonals share a point
- The count $70$ assumes no interior point is used by more than two diagonals.
- But the octagon is regular, so rotating or reflecting it maps diagonals onto other diagonals.
- This symmetry forces several diagonals through the same interior point in two places.
- When $k$ diagonals meet at one point, that single point is counted once for every pair of diagonals through it, that is $\binom{k}{2}$ times instead of once.
- So we must find those spots and remove the extra counts.
💡 A regular shape's symmetry lines up extra diagonals through the same point, breaking the one-crossing-per-point assumption.
7.SP.C.8 Step 4 The center: 4 diagonals, over-counted by 5
- The four longest diagonals join opposite vertices, and all four pass through the center.
- Any $2$ of these $4$ diagonals form a quadrilateral that crosses at the center, so $\binom{4}{2} = 6$ of our $70$ quadrilaterals all point to that single center.
- It should count once, so we counted $6 - 1 = 5$ too many there.
💡 Four diagonals through the center make six pairwise crossings that are really one and the same point.
7.SP.C.8 Step 5 Eight triple points: over-counted by 16
- Away from the center, the octagon's symmetry creates a ring of $8$ special interior points where exactly $3$ diagonals meet at once, one in each of the $8$ symmetric sectors.
- At each such point the $3$ diagonals give $\binom{3}{2} = 3$ quadrilaterals all landing there, so it is counted $3$ times instead of once — that is $3 - 1 = 2$ too many.
- Across all $8$ points that is $8 \times 2 = 16$ extra counts.
💡 Each triple point is counted three times, so it carries two extra counts, and there are eight of them.
4.OA.A.3 Step 6 Subtract the extras to get 49
- Start from $70$, then remove the $5$ extra counts at the center and the $16$ extra counts from the eight triple points.
- Every other crossing uses exactly two diagonals and was already counted correctly once.
- So the number of distinct interior intersection points is $70 - 5 - 16 = 49$.
- The answer is $\textbf{(A)}\ 49$.
💡 The only fixes needed are removing the double- and triple-counts, and what is left is the true point total.
7.SP.C.8 Directly hunting for crossing points is messy, so change focus. Two diagonals th 6.EE.A.2 Count the ways to choose $4$ vertices out of $8$. The choose formula gives $\bin 8.G.A.1 The count $70$ assumes no interior point is used by more than two diagonals. But 7.SP.C.8 The four longest diagonals join opposite vertices, and all four pass through the 7.SP.C.8 Away from the center, the octagon's symmetry creates a ring of $8$ special inter 4.OA.A.3 Start from $70$, then remove the $5$ extra counts at the center and the $16$ ext Review
Reasonableness: The plain count $70$ is choice (C), exactly the trap for anyone who forgets that diagonals can share a point, which is a good sign our correction direction is right. Every correction only lowers the count, because concurrent points are over-counted and never under-counted, so the true answer must be below $70$, ruling out (D) $96$ and (E) $128$ at once. Removing $5 + 16 = 21$ extra counts lands on $49$, which is choice (A). The two kinds of concurrences — one $4$-fold center and eight $3$-fold points — are just what the octagon's $8$-fold symmetry predicts, so the size of the correction is believable.
Alternative: Instead of correcting a global count, use the symmetry to build one-eighth of the figure. The octagon splits into $8$ identical sectors; carefully draw the diagonals inside a single sector and count its interior crossings, then handle the center and the points shared on sector boundaries and multiply out. Adding the eight sectors' points plus the one shared center rebuilds the same total of $49$, confirming the answer without the choose-$4$ bookkeeping.
CCSS standards used (min grade 8)
7.SP.C.8Represent and count the outcomes of compound events using organized lists, tables, and tree diagrams (Counting by systematic enumeration the ways to choose 4 vertices, and the pairs of diagonals meeting at the center and at each triple point.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating the combination expression C(8,4) = (8x7x6x5)/(4x3x2x1) to get the first count of 70.)8.G.A.1Verify experimentally the properties of rotations and reflections of figures (Using the regular octagon's rotational and mirror symmetry to locate where 3 or 4 diagonals become concurrent.)4.OA.A.3Solve multistep problems using the four operations (Subtracting the over-counts (70 minus 5 minus 16) to reach the final total of 49.)
⭐ Count 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 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.
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