Problem 5 · 2025 AMC 8
Stretch
Geometry & Measurement
taxicab-distancegrid
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Answer: C — 24 blocks.
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Hint 1 of 2
You can't cut diagonally across blocks. So however Betty zigzags within one leg, the distance is the same — what two things does it depend on?
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Hint 2 of 2
Each leg's length is just its sideways blocks plus its up-and-down blocks (taxicab distance). Read those off the map for F→A, A→B, B→C, C→F and add.
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Approach: taxicab / Manhattan distance per leg
- Key idea: on a street grid you can't cut diagonally, so the distance for any leg is just (horizontal blocks) + (vertical blocks) — and as long as you never backtrack, the exact path doesn't matter, only where you start and end. (This is the taxicab / Manhattan distance.)
- Read the four legs off the map: F→A = 1 + 2 = 3, A→B = 7 + 3 = 10, B→C = 2 + 4 = 6, C→F = 4 + 1 = 5.
- Total: 3 + 10 + 6 + 5 = 24 blocks.
- Why this transfers: on any grid where you only move along streets, the shortest distance between two points is fixed — just add the sideways and up-down gaps. The wiggly path you choose is irrelevant.
Another way — C is already on the way back (MAA):
- Notice C lies on a shortest path from B back to F, so visiting C costs nothing extra. The problem reduces to F → A → B → F.
- F→A = 3, A→B = 10, B→F (through C) = 6 + 5 = 11. Total: 3 + 10 + 11 = 24 blocks.
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