πŸ‡ΊπŸ‡Έ AMC 8 ⇄ switch contest
1993 AJHSME

Problem 17

Problem 17 · 1993 AJHSME Hard
Geometry & Measurement surface-areanet
Figure for AJHSME 1993 Problem 17
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Answer: B — 500.
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Hint 1 of 2
Each corner cut takes 5 off both ends of a side, so the base shrinks by 10 in each direction: 20βˆ’10 and 30βˆ’10. The 5-wide flaps that fold up become the box's height of 5.
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Hint 2 of 2
An open box has a bottom and 4 walls but NO lid. Add those five faces; don't include a top.
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Approach: unfold the box into bottom + four walls (no top)
  1. Cutting a 5Γ—5 square from each corner removes 5 from both ends of every side, so the base is (20βˆ’5βˆ’5) Γ— (30βˆ’5βˆ’5) = 10 Γ— 20 = 200. The flaps that fold upward are 5 tall, so the box height is 5.
  2. Four walls: two are 10Γ—5 and two are 20Γ—5, totaling 2(50) + 2(100) = 300. The interior is the bottom plus those four walls (no lid): 200 + 300 = 500.
  3. Trap to dodge: 'open box' means skip the top β€” count 5 faces, not 6. And remember the corner cuts subtract from both ends, so the base loses 2Γ—5 in each direction, not just 5. Choice E (1000) is what you'd get from the full original 20Γ—30 sheet's worth of double-counting.
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