AMC 10 · 2010 · #22
Grade 7 countingEight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
Pick an answer.
AMC 10 2010 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: 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.
Givens: 8 points lie on a circle, and every pair is joined by a chord (all $\binom{8}{2}=28$ chords are drawn); No three chords meet at a single point inside the circle, so each interior crossing comes from exactly one pair of chords; We only want triangles whose three vertices are interior crossing points, not the 8 points on the circle; Answer choices: (A) 28, (B) 56, (C) 70, (D) 84, (E) 140
Unknowns: How many triangles have all three vertices strictly inside the circle
Understand
Restated: 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.
Givens: 8 points lie on a circle, and every pair is joined by a chord (all $\binom{8}{2}=28$ chords are drawn); No three chords meet at a single point inside the circle, so each interior crossing comes from exactly one pair of chords; We only want triangles whose three vertices are interior crossing points, not the 8 points on the circle; Answer choices: (A) 28, (B) 56, (C) 70, (D) 84, (E) 140
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #9 Solve an Easier Related Problem, #2 Make a Systematic List
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.
Execute — Answer: A
7.G.A.2 Step 1 See where the corners live
- Draw the circle with the 8 dots and sketch a few chords.
- A triangle 'with all three vertices in the interior' does **not** use the dots on the circle as its corners.
- Each of its three corners is a point where two chords cross **inside** the circle.
- So a triangle here is really three chords that cross one another pairwise, fencing off a little triangle in the middle.
💡 The triangle's corners are chord crossings, so a triangle is just three chords that cross each other pairwise.
7.SP.C.8 Step 2 Three crossings need six points
- Each inside crossing is made by two chords, and two chords use $2+2=4$ endpoints on the circle.
- A triangle has three corners, so it is built from three chords — one for each pair of crossing sides.
- For all three chords to actually cross inside, no two of them may share an endpoint (chords that share an endpoint meet **on** the circle, not inside).
- Three chords with no shared endpoint use $3\times 2 = 6$ different dots, so every interior triangle uses exactly 6 of the 8 points.
💡 A triangle is three chords that never share an endpoint, so it always uses exactly six of the circle's points.
7.G.A.2 Step 3 Six points give exactly one triangle
- Zoom in on just 6 of the points and label them $1$ through $6$ going around the circle.
- Which three chords cross each other pairwise inside?
- Only the three 'long' chords joining opposite points: $1\text{-}4$, $2\text{-}5$, and $3\text{-}6$.
- These three cross one another and fence off a single triangle in the middle.
- Any other pairing of the six points leaves at least two chords that never cross inside, so no triangle forms.
- So each group of 6 points produces **exactly one** interior triangle.
💡 Pick six points and there is only one way to draw three mutually crossing chords, so six points always mean one triangle.
7.SP.C.8 Step 4 Count the point groups instead
- Step 3 says triangles and 6-point groups come in perfect pairs: every interior triangle points back to the 6 circle points that source its corners, and every choice of 6 points gives back exactly one triangle.
- So instead of hunting triangles in the tangle of chords, just count how many ways we can **choose 6 of the 8 points**.
- That count is the number of triangles.
💡 Because each triangle matches exactly one group of six points, counting the groups counts the triangles.
3.OA.A.1 Step 5 Choose 6 of 8
- Choosing which 6 points to keep is the same as choosing which 2 points to leave out, so the count is $\binom{8}{6}=\binom{8}{2}$.
- There are $8$ ways to pick the first left-out point and $7$ for the second, and each pair is counted twice because order doesn't matter, giving $\tfrac{8\times 7}{2}=28$.
- So there are $28$ interior triangles, which is choice **(A)**.
💡 Choosing 6 to keep is the same as choosing 2 to drop, and there are $8\times 7/2 = 28$ such pairs.
7.G.A.2 Draw the circle with the 8 dots and sketch a few chords. A triangle 'with all th 7.SP.C.8 Each inside crossing is made by two chords, and two chords use $2+2=4$ endpoints 7.G.A.2 Zoom in on just 6 of the points and label them $1$ through $6$ going around the 7.SP.C.8 Step 3 says triangles and 6-point groups come in perfect pairs: every interior t 3.OA.A.1 Choosing which 6 points to keep is the same as choosing which 2 points to leave Review
Reasonableness: Is $28$ believable? With 8 points there are $\binom{8}{4}=70$ 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 $\binom{8}{6}=28$ stands, choice (A).
Alternative: Instead of reframing, count directly through the crossing points. Each interior crossing is set by 4 of the circle's points, and a triangle is three such crossings. Track how the three corners of one triangle share their source points — they overlap so that the whole triangle uses just 6 points, each point feeding 2 of the 3 chords. Following that overlap forces the same triangle-to-six-points match, so the total is again $\binom{8}{6}=28$. Same answer, reached by bookkeeping crossings rather than changing focus.
CCSS standards used (min grade 7)
7.G.A.2Draw geometric shapes with given conditions including triangles (Drawing the circle, chords, and the small interior triangle, and checking that on any 6 points exactly one set of three mutually crossing chords (1-4, 2-5, 3-6) makes a triangle.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Reasoning that three non-touching chords use exactly 6 endpoints, then organizing the count as a one-to-one match between interior triangles and 6-point groups.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Computing the number of ways to choose 6 of 8 points as $\binom{8}{2}=\tfrac{8\times 7}{2}=28$.)
⭐ Every inside triangle is born from exactly six of the circle's points, so just count the ways to pick 6 out of 8: $\binom{8}{6}=28$.
⭐ Every inside triangle is born from exactly six of the circle's points, so just count the ways to pick 6 out of 8: $\binom{8}{6}=28$.
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