πŸ‡ΊπŸ‡Έ AMC 8 ⇄ switch contest
1985 AJHSME

Problem 4

Problem 4 · 1985 AJHSME Medium
Geometry & Measurement bounding-rect-minus-notch
Figure for AJHSME 1985 Problem 4
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Answer: C — 46.
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Hint 1 of 2
The shape has one corner pushed IN (the bite near E and F). Instead of chopping it into pieces, imagine filling that bite to make a full rectangle β€” then take the bite back out. What is the full rectangle?
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Hint 2 of 2
This is the bounding-box trick: enclose any rectilinear shape in the smallest rectangle that contains it, then subtract the rectangular pieces that aren't part of the shape. One subtraction beats slicing into several rectangles.
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Approach: bounding rectangle minus the notch
  1. Fill the bite to complete the full rectangle: it is 6 wide (the top AB) by 9 tall (the right side BC), so its area is 6 Γ— 9 = 54.
  2. The bite that was removed is itself a small rectangle. Its width is 6 βˆ’ 4 = 2 (top width minus bottom width DC) and its height is 9 βˆ’ 5 = 4 (right side minus left side AF), so its area is 2 Γ— 4 = 8.
  3. Polygon = full rectangle βˆ’ bite = 54 βˆ’ 8 = 46.
  4. Why this transfers: any L-shape, T-shape, or staircase is just a rectangle with rectangular bites taken out β€” find the missing side lengths by subtracting the parts you know, then subtract areas.
Another way — split into two rectangles:
  1. Cut horizontally at the level of F/E. Top piece: 6 wide Γ— 5 tall = 30. Bottom piece: 4 wide Γ— (9 βˆ’ 5) = 4 Γ— 4 = 16.
  2. 30 + 16 = 46 β€” same answer, built up instead of cut down.
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