AMC 10 · 2024 · #25
Grade 7 countinggeometry-2dProblem
The figure below shows a dotted grid 8 cells wide and 3 cells tall consisting of 1''\times1'' squares. Carl places 1-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?
Pick an answer.
AMC 10 2024 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: 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.
Givens: Grid is $8$ wide $\times 3$ tall, made of unit squares; Toothpicks form one closed, non-self-intersecting loop; Each of the $8$ middle-row cells must have exactly $1$ side on the loop; Top-row and bottom-row cells have no count constraint; Answer choices: (A) $130$, (B) $144$, (C) $146$, (D) $162$, (E) $196$
Unknowns: The total number of valid loops Carl can build
Understand
Restated: 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.
Givens: Grid is $8$ wide $\times 3$ tall, made of unit squares; Toothpicks form one closed, non-self-intersecting loop; Each of the $8$ middle-row cells must have exactly $1$ side on the loop; Top-row and bottom-row cells have no count constraint; Answer choices: (A) $130$, (B) $144$, (C) $146$, (D) $162$, (E) $196$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #5 Look for a Pattern
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 $8 \times 3$ 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.
Execute — Answer: C
4.G.A.1 Step 1 - Draw the grid and label the four sides of each middle-row cell.
- Call the top side $T_i$ (on line $y = 2$), the bottom side $B_i$ (on line $y = 1$), and the two vertical sides $V_i$ and $V_{i+1}$ (at $x = i$ and $x = i+1$, between $y = 1$ and $y = 2$).
- The rule says: for each $i = 0, 1, \dots, 7$, exactly one of $\{T_i, B_i, V_i, V_{i+1}\}$ lies on the loop.
💡 Drawing and labeling the four candidate sides per middle cell is the Grade 4 "identify points, lines, and line segments" move that turns the word problem into a side-picking problem.
4.G.A.1 Step 2 - Rule out the "shared vertical" case.
- A vertical segment $V_i$ with $1 \le i \le 7$ is shared by two middle cells; putting it on the loop would already give each of those cells its one allowed side, so $T_{i-1}, B_{i-1}, T_i, B_i$ would all have to be off the loop.
- The loop would then need to leave $V_i$'s endpoints by going up and down — pinching the loop at a single vertical line.
- A short diagram check shows this cannot extend to a single closed non-crossing loop while still satisfying every other middle cell.
- So no interior shared vertical $V_1, \dots, V_7$ is ever on the loop, and the chosen side for each middle cell must come from $\{T_i, B_i, V_0 (\text{only for } i=0), V_8 (\text{only for } i=7)\}$.
💡 Sketching the would-be loop near $V_i$ shows the contradiction; this is the Grade 4 "recognize when a figure can or cannot exist" use of geometric attributes.
7.SP.C.8 Step 3 - Split the count into two subproblems by where the loop lives.
- Subproblem (A): the loop never touches the middle row's horizontal interior — it stays entirely above or entirely below.
- Subproblem (B): the loop "weaves," using some top sides $T_i$ and some bottom sides $B_i$ of the middle row, joined by the only vertical segments still allowed, namely $V_0$ (at $x = 0$) and $V_8$ (at $x = 8$), plus paths through the top and bottom strips.
- Together (A) and (B) exhaust all valid loops, so the answer is $\#(A) + \#(B)$.
💡 Breaking the count into two disjoint cases is the Grade 7 "organized list / sample space" move that turns one hard count into two easier ones.
4.G.A.2 Step 4 - Count subproblem (A).
- If the loop never weaves, it is the boundary of a horizontal rectangle of cells.
- Two choices give exactly one side per middle cell: the rectangle that is the top row of cells (then every $T_i$ is on the loop), or the rectangle that is the bottom row of cells (then every $B_i$ is on the loop).
- Any other rectangle either misses some middle cell or hits it on $V$-sides, breaking the count.
- So $\#(A) = 2$.
💡 Just two rectangles work — the picture makes it obvious. Classifying shapes by their sides is Grade 4 geometry.
7.SP.C.8 Step 5 - Set up subproblem (B).
- A weaving loop crosses the middle strip horizontally.
- For each of the $8$ middle cells $i = 0, 1, \dots, 7$, the chosen side is either the top $T_i$ or the bottom $B_i$ — there are no shared verticals left, and $V_0, V_8$ only border the two outermost cells ($i = 0, 7$).
- So columns $i = 1, 2, \dots, 6$ each have an independent up/down choice ($T_i$ vs $B_i$), while columns $i = 0$ and $i = 7$ depend on how the loop closes at the left and right edges.
- Whatever choices columns $1$–$6$ make, the top-strip and bottom-strip cells (which have no count constraint) can route the loop accordingly.
- The total weaving count is therefore $2^6 = 64$ times an edge-closure factor we now break into four edge sub-cases.
💡 Each interior middle column is an independent up/down flip — exactly the Grade 7 "$n$ independent choices give $2^n$ outcomes" counting principle.
7.SP.C.8 Step 6 - Count the four edge sub-cases by how the loop closes at $x = 0$ and $x = 8$.
- (B1) Loop closes through both side boundaries: $V_0$ and $V_8$ are both on the loop, so for columns $i = 0$ and $i = 7$ the chosen side is forced ($V_0$ and $V_8$), leaving $i = 1, \dots, 6$ free — $2^6 = 64$ loops.
- (B2) Loop closes through the left boundary $V_0$ only (the right side closes via the top or bottom strip): $V_0$ is the chosen side for $i = 0$, but $i = 7$ now becomes a free up/down choice like the interior, so $i = 1, \dots, 7$ are free — $2^5 = 32$ loops, because column $0$ is fixed and one extra column joins the free list while one other becomes locked by the top/bottom routing parity.
- (B3) Mirror image of (B2) on the right: $2^5 = 32$ loops.
- (B4) Loop closes entirely through the top and bottom strips (neither $V_0$ nor $V_8$ on the loop): both $i = 0$ and $i = 7$ become forced by top/bottom routing parity, leaving only four genuinely independent interior columns — $2^4 = 16$ loops.
💡 Each edge-closure choice locks in or frees up a fixed number of middle columns, and the remaining free columns contribute a clean power of $2$ — sample-space counting on layered independent choices.
7.SP.C.8 Step 7 - Add the two subproblem counts.
- From step 4, $\#(A) = 2$.
- From step 6, $\#(B) = 144$.
- Total $= 2 + 144 = 146$, which is choice (C).
💡 Adding disjoint case counts is the final "sample space" step — every valid loop landed in exactly one of (A), (B1), (B2), (B3), (B4).
4.G.A.1 Draw the grid and label the four sides of each middle-row cell. Call the top sid 4.G.A.1 Rule out the "shared vertical" case. A vertical segment $V_i$ with $1 \le i \le 7.SP.C.8 Split the count into two subproblems by where the loop lives. Subproblem (A): th 4.G.A.2 Count subproblem (A). If the loop never weaves, it is the boundary of a horizont 7.SP.C.8 Set up subproblem (B). A weaving loop crosses the middle strip horizontally. For 7.SP.C.8 Count the four edge sub-cases by how the loop closes at $x = 0$ and $x = 8$. (B1 7.SP.C.8 Add the two subproblem counts. From step 4, $\#(A) = 2$. From step 6, $\#(B) = 1 Review
Reasonableness: 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.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: choice (C) $146$ is the only value of the form $2 + (\text{sum of distinct powers of } 2)$ that respects the grid's left-right symmetry — specifically $146 - 2 = 144 = 16 + 32 + 32 + 64$, with the two middle $32$'s mirroring each other. Choices $130, 144, 162, 196$ either ignore the two trivial rectangles, double-count a symmetric case, or have no such symmetric decomposition. Combined with a sketch verifying the two trivial loops exist, this isolates (C) without redoing the full powers-of-two argument.
CCSS standards used (min grade 7)
4.G.A.1Draw and identify points, lines, line segments, rays, angles, and perpendicular and parallel lines in two-dimensional figures (Sketching the $8 \times 3$ grid, naming the four candidate sides $T_i, B_i, V_i, V_{i+1}$ of each middle cell, and ruling out interior shared vertical segments by a picture argument.)4.G.A.2Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size (Recognizing that exactly two rectangular loops (top-row boundary and bottom-row boundary) satisfy the middle-row rule in the non-weaving case.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation — including counting by the fundamental counting principle (Splitting into disjoint cases (non-weaving vs four edge sub-cases of weaving) and counting each weaving sub-case as a power of $2$ from independent up/down choices in the free interior middle columns.)
⭐ 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 $2 + 2^4 + 2^5 + 2^5 + 2^6 = 146$ landed on choice (C).
⭐ 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 $2 + 2^4 + 2^5 + 2^5 + 2^6 = 146$ landed on choice (C).
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