πŸ‡ΊπŸ‡Έ AMC 8 ⇄ switch contest
1986 AJHSME

Problem 13

Problem 13 · 1986 AJHSME Hard
Geometry & Measurement perimeter-equals-bounding-box
Figure for AJHSME 1986 Problem 13
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Answer: C — 28.
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Hint 1 of 3
Only two lengths are labeled (8 and 6), yet the shape has six sides β€” so don't try to find each missing side. Instead ask: if you pushed the notched-in corner back out to make a full rectangle, would the total edge length change?
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Hint 2 of 3
Every horizontal piece on the staircase, added up, must still span the full width 8; every vertical piece must still span the full height 6. So the perimeter equals that of the surrounding 8-by-6 rectangle.
Still stuck? Show hint 3 →
Hint 3 of 3
Bounding rectangle: 8 wide, 6 tall β€” that alone fixes the perimeter even though individual side lengths are unknown.
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Approach: slide the step out β€” perimeter unchanged
  1. Look at all the horizontal edges: going across, they must total the full width, 8. Same for the vertical edges: top to bottom they total the full height, 6. Sliding the inward 'step' out to the corner just rearranges those pieces without adding or removing any length.
  2. So the perimeter is exactly that of the 8-by-6 bounding rectangle: 2(8 + 6) = 28.
  3. This is why the answer isn't 'cannot be determined' even though the step's individual sizes are hidden β€” for any such rectilinear staircase shape, the perimeter depends only on the overall width and height.
  4. Sanity check: the area would change if you moved the step, but the perimeter doesn't β€” a nice reminder that area and perimeter are independent.
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