πŸ‡ΊπŸ‡Έ AMC 8 ⇄ switch contest
1996 AJHSME

Problem 22

Problem 22 · 1996 AJHSME Hard
Geometry & Measurement shoelacelattice
Figure for AJHSME 1996 Problem 22
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Answer: B — 1/2.
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Hint 1 of 2
Look how nearly straight A, B, C are β€” B sits just a hair off the line from A to C, so this triangle is a razor-thin sliver. That's a hint the area will be tiny, not one of the bigger answer choices.
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Hint 2 of 2
Put A at the origin and read B and C off the grid as coordinates. Then a lattice-coordinate area formula (or counting boxes) pins down the sliver exactly.
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Approach: coordinates and the area formula
  1. Set A = (0,0) and read off B = (3,2), C = (4,3) from the grid.
  2. Shoelace formula: area = Β½ |0(2 βˆ’ 3) + 3(3 βˆ’ 0) + 4(0 βˆ’ 2)| = Β½ |9 βˆ’ 8| = 1/2.
  3. Sanity-check: the line from A through C has slope 3/4, and B = (3,2) would be on it only if 2 = 3Β·(3/4) = 2.25 β€” it's just 1/4 unit below, so the triangle is a thin sliver, exactly matching the tiny area 1/2 (and ruling out the larger choices).
Another way — Pick's theorem:
  1. The triangle has the 3 corners as its only boundary lattice points (no grid point lies on a side) and no interior lattice points: B = (3,2) just misses the line AC.
  2. Pick's theorem gives area = interior + boundary⁄2 βˆ’ 1 = 0 + 3⁄2 βˆ’ 1 = 1/2.
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