Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #5
Grade 3 geometry-2d
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The map is already a diagram, so Tool #1 (Draw a Diagram) just asks us to use it — read the position of each point by counting blocks right and up from the corner. Because the streets form a grid, the shortest path between any two labeled points is simply (horizontal blocks between them) + (vertical blocks between them); there is no benefit to zig-zagging. Tool #7 (Identify Subproblems) then splits the loop into four independent legs F → A, A → B, B → C, C → F. Solve each tiny leg by counting, then add the four numbers.
Read the corners off the grid
Read each labeled point off the grid: F = (6, 5), A = (7, 3), B = (0, 0), C = (2, 4).
Describing where something is on the grid using right/up counts is Kindergarten position language.
K.G.A.1Draw A DiagramSet up the block-distance rule
On a grid the shortest leg is (horizontal blocks apart) + (vertical blocks apart); split the loop into four legs (Tool #7).
Adding side-distances to get a path length on a grid is the same idea as finding perimeter pieces in Grade 3.
On a map where the truck may drive only along the streets, the shortest route between two corners is exactly the number of columns they are apart plus the number of rows they are apart.
▸ Why?
Each block driven moves the truck one column sideways or one row up or down, so the route can be shortest only if it never wastes a block while still closing the full sideways gap and the full up-and-down gap between the two corners.
▸ Why?
Every sideways block crosses exactly one column line, so to get across all the columns that separate the two corners the route needs at least that many sideways blocks.
▸ Why?
In the same way, every up-or-down block crosses exactly one row line, so closing the up-and-down gap needs at least as many blocks as the corners are rows apart.
▸ Why?
A route that always heads toward the target, never doubling back, spends one block for each column of gap and one block for each row of gap and nothing more, so its length is just those two counts added together.
▸ Why?
The whole route is made of separate one-block pieces with no gaps and no overlaps, so its total length is found by adding up all of those pieces.
Measure leg F to A
Leg F → A: 1 block right and 2 blocks down, so 3 blocks.
Just counting blocks across and down and adding them is a Grade 2 addition word problem.
2.OA.A.1Draw A DiagramMeasure leg A to B
Leg A → B: 7 blocks left and 3 blocks down, so 10 blocks.
Counting blocks across and down, then adding, is still Grade 2 arithmetic.
2.OA.A.1Draw A DiagramMeasure leg B to C
Leg B → C: 2 blocks right and 4 blocks up, so 6 blocks.
Same idea: count across, count up, add.
2.OA.A.1Draw A DiagramMeasure leg C to F
Leg C → F: 4 blocks right and 1 block up, so 5 blocks.
One more across-and-up count, then add.
2.OA.A.1Draw A DiagramAdd the four legs
Add the four legs: 3 + 10 + 6 + 5 = 24 blocks, which matches choice (C).
Summing four small numbers fluently is Grade 3 add-within-1000.
3.NBT.A.2Identify SubproblemsThis AMC 8 problem only needs Grade 3 addition and the idea of "path length around a shape" you already know!
- Read the corners off the grid
- Set up the block-distance rule
- Measure leg F to A
- Measure leg A to B
- Measure leg B to C
- Measure leg C to F
- Add the four legs
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