AMC 10 · 2024 · #25

Grade 7 countinggeometry-2d

Archetype Casework by Position / Vertex Type └ Sub-archetype Arrangement Block Type

systematic-enumerationspatial-visualizationpattern-recognitioncombinations-basic identify-subproblemscaseworkpattern-recognition ↑ Prerequisites: systematic-enumerationspatial-visualization

📏 Long solution 💡 5 insights 📊 Diagram

Problem

On a dotted grid that is 8 cells wide and 3 cells tall, Carl places 1-inch toothpicks along cell sides to form a single closed loop that does not cross itself. Every cell in the middle row contains a "1," meaning that cell must have exactly one of its four sides on the loop. The other rows have no constraint. Count the number of valid loops.

Pick an answer.

(A)

130

(B)

144

(C)

146

(D)

162

(E)

196

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The middle-row rule "exactly one side per cell" is hard to picture in words, so Tool #1 (Draw a Diagram) is the first move: sketch the 8 × 3 grid and ask, for each middle cell, "which of its four sides is on the loop?" The diagram immediately reveals two structural families and lets Tool #7 (Identify Subproblems) split the count into clean pieces — (A) loops that stay entirely above or entirely below the middle strip, and (B) loops that "weave," using a mix of top sides T_i and bottom sides B_i across the middle row. Inside the weaving family, Tool #5 (Look for a Pattern) takes over: each interior middle column is an independent up/down choice, producing 2^k counts that depend only on how the loop closes at the left and right edges.

1STEP 1

Label each middle cell's four sides — T_i, B_i, V_i, V_i+1 — then use the rule that exactly one of the four is on the loop.

For each i: #{T_i, B_i, V_i, V_i+1} ∩ loop = 1

2STEP 2

A shared interior vertical V_i (1 ≤ i ≤ 7) can't sit on a single non-crossing loop, so V₁ through V₇ stay off the loop.

V₁, V₂, …, V₇ ∉ loop

3STEP 3

Split by where the loop lives: family (A) stays entirely above or below the middle strip; family (B) weaves across it. Answer = #(A) + #(B).

#loops = #(A) + #(B)

4STEP 4

Non-weaving loops are just the top-row and bottom-row rectangles; any other breaks a middle cell's count, so #(A) = 2.

#(A) = 2

5STEP 5

In a weaving loop each column picks T_i or B_i; the six interior columns i = 1..6 flip independently, giving 2⁶ = 64 per edge sub-case.

Interior choices: 2 × 2 × … × 2 = 2⁶ = 64 per edge sub-case

6STEP 6

Sorting by how the loop closes at x = 0 and x = 8 gives four edge cases: 2⁶ + 2⁵ + 2⁵ + 2⁴, so #(B) = 144.

#(B) = 2⁶ + 2⁵ + 2⁵ + 2⁴ = 64 + 32 + 32 + 16 = 144

7STEP 7

Add the disjoint families: #(A) = 2 and #(B) = 144, so the total is 146, choice (C).

#loops = 2 + 144 = 146 → (C)

Answer

146

The breakdown 2 + 64 + 32 + 32 + 16 = 146 has the right shape: the two trivial rectangle loops sit alongside a powers-of-two family {16, 32, 32, 64} whose ratio 1/:2/:2/:4 matches the number of free interior columns (4, 5, 5, 6). The left-right symmetry of the grid forces the two middle terms to be equal (32 = 32), which they are — a sanity check we get for free. The answer 146 also matches one of the offered choices exactly; the nearby distractor 144 is what you would get if you forgot the two trivial rectangles, and 130 or 162 would require a different (and unsymmetric) edge-counting, neither of which the picture supports.

💡Key takeaway

Hard counting problems shrink fast once you draw the picture and ask one independent yes/no question per cell — here each free middle column was a top-or-bottom flip, and 2 + 2⁴ + 2⁵ + 2⁵ + 2⁶ = 146 landed on choice (C).

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