AMC 10 · 2011 · #22
Grade 7 countingPick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The words "ends of each diagonal differ" hide a clean shape, so Tool #1 (Draw a Diagram) is the first move: draw only the diagonals and follow them. They form a five-pointed star, and walking the star visits all five vertices and returns to the start — it is one closed loop (a ring) of 5. That turns the problem into "color a ring of 5 so neighbors differ, using 6 colors." Tool #2 (Make a Systematic List) counts the easy open-chain version with the multiplication counting principle. Tool #7 (Identify Subproblems) and Tool #16 (Count the Complement) then repair the one link the chain ignored by subtracting the bad cases, which shrinks the ring by one and gives a short recursion.
Turn the diagonals into a ring
Draw only the diagonals AC, CE, EB, BD, DA: walking A → C → E → B → D → A closes into one ring of 5 vertices, so sides never matter.
Every diagonal is one link of the star, and the star closes up into a single loop, so "diagonal ends differ" just means "ring neighbors differ."
7.SP.C.8Draw A DiagramCount the open chain first
Cut the link D - A and color the open chain A - C - E - B - D left to right: 6 choices for A, then 5 each, giving 3750.
Down a chain each new vertex dodges only its single left neighbor, so it always has 5 safe colors.
7.SP.C.8Make A Systematic ListSpot the one broken rule
That chain count ignored the closing diagonal D - A, so subtract the chains where A and D happen to share a color.
The open chain over-counts by allowing the two ends to clash, so subtract just the clashing ones.
The open chain over-counts by allowing the two ends to clash, so those must be subtracted off.
▸ Why?
Each link is coloured without regard to the far ones, so the chain count is a plain product.
▸ Why?
Counting everything then removing the rule-breakers is what keeps the tally honest.
Glue the ends: a smaller ring
Forcing A and D equal merges them into one vertex, so the chain closes into a 4-ring and the bad count is just the 4-ring count.
Forcing the two loose ends to match fuses them into one vertex, shrinking the ring by exactly one.
7.SP.C.8Change Focus Count The ComplementWork down to the 4-ring
Base case: the 3-ring is a triangle, 6 · 5 · 4 = 120. Then the 4-ring is 6 · 5³ - 120 = 630, the bad-chain count.
The same subtract-the-smaller-ring rule bottoms out at the triangle, whose count is a plain product.
6.EE.A.1Identify SubproblemsFinish the 5-ring
Chain minus bad: 3750 - 630 = 3120 valid colorings of pentagon ABCDE — choice (C).
One subtraction closes the loop: start from the free chain, remove the colorings the last diagonal forbids.
7.SP.C.8Identify SubproblemsHidden inside the pentagon, the diagonals form one five-point ring, so the puzzle is just "color a loop of 5 with 6 colors so neighbors differ": count the easy open chain (6 · 5⁴ = 3750), then subtract the 630 that break the last link to land on 3120.
- Turn the diagonals into a ring
- Count the open chain first
- Spot the one broken rule
- Glue the ends: a smaller ring
- Work down to the 4-ring
- Finish the 5-ring