AMC 10 · 2010 · #22

Grade 7 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type
combinations-basicsystematic-enumeration easier-related-problem ↑ Prerequisites: combinations-basic
📏 Medium solution 💡 3 insights
Problem
Eight dots sit on a circle, and every pair of dots is joined by a chord, so the inside of the circle fills up with many crossing points. No three chords cross at the same inside point. Count the triangles whose three corners are all inside the circle — that is, triangles whose corners are chord-crossing points, not the dots on the circle.

Pick an answer.

(A)
28
(B)
56
(C)
70
(D)
84
(E)
140

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Counting the triangles head-on is hopeless — the inside of the circle is a tangle of crossing chords. The winning move is to change what we count (Tool #16). First draw the picture (Tool #1) to see that each triangle corner is a chord crossing, then break a triangle into its parts (Tool #7): three crossing chords with no shared endpoints, which pin down exactly six of the eight circle points. Next solve the tiny case of just six points (Tool #9) to see that six points always give exactly one interior triangle. That one-to-one match lets us stop chasing triangles and instead count six-point groups, a simple choose count (Tool #2). No algebra, no angle chasing — just a clean correspondence.

1STEP 1

See where the corners live

Its corners are not the dots on the circle but points where two chords cross inside, so a triangle is three pairwise-crossing chords.

interior vertex = crossing of two chords
2STEP 2

Three crossings need six points

Chords sharing an endpoint meet on the circle, not inside, so a triangle's three chords use 6 of the 8 dots.

3 chords × 2 endpoints = 6 distinct points
3STEP 3

Six points give exactly one triangle

Label six points 1 to 6 around the circle: only chords 1-4, 2-5, 3-6 cross pairwise, so six points give exactly one triangle.

{1-4, 2-5, 3-6} → exactly 1 triangle
4STEP 4

Count the point groups instead

Triangles and 6-point groups pair up one to one, so counting the ways to choose 6 of the 8 dots counts the triangles.

#triangles = #{ways to choose 6 of the 8 points}
5STEP 5

Choose 6 of 8

Keeping 6 is the same as dropping 2, so C(8, 6)=C(8, 2)=(8×7)/2=28 interior triangles — choice (A).

C(8, 6)=C(8, 2)=(8 × 7)/2=28 → (A)
Answer
28
Is 28 believable? With 8 points there are C(8, 4)=70 interior crossing points in all, and a triangle uses 3 of them, so a rough ceiling makes 28 triangles sensible while the bigger choices (56, 70, 84, 140) would need far more mutually-crossing chord triples than the geometry allows. The correspondence is airtight: no triangle can use fewer than 6 points (you can't make three non-touching chords with fewer) or more than 6 (three chords have only 6 endpoints), and each 6-point group yields exactly one triangle, so nothing is double-counted or missed. The count C(8, 6)=28 stands, choice (A).
💡Key takeaway

Every inside triangle is born from exactly six of the circle's points, so just count the ways to pick 6 out of 8: C(8, 6)=28.

  • See where the corners live
  • Three crossings need six points
  • Six points give exactly one triangle
  • Count the point groups instead
  • Choose 6 of 8

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