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2023 AMC 8

Problem 19

Problem 19 · 2023 AMC 8 Medium
Geometry & Measurement area-fractionarea
Figure for AMC 8 2023 Problem 19
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Answer: C — 5 : 12.
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Hint 1 of 2
You never need the actual area formula. Lengths scale by 2:3, so areas scale by the square: (2/3)2 = 4/9. That one fact does all the work.
Still stuck? Show hint 2 →
Hint 2 of 2
Pick friendly numbers: let the inner triangle be 4 and the outer be 9 (ratio 4:9). The three trapezoids together fill the gap 9 − 4, then split it evenly into three.
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Approach: areas scale as side-length squared
  1. Don't reach for ½·base·height — the key is that for similar figures, doubling lengths quadruples area: area scales as the square of the length ratio. Side ratio 2:3 → area ratio 4:9.
  2. So set inner area = 4 and outer area = 9 (any pair in 4:9 works). The ring of three congruent trapezoids fills the leftover 9 − 4 = 5, so each trapezoid is 53.
  3. One trapezoid : inner triangle = 5/34 = 512, i.e. 5 : 12. Worth keeping: whenever you see a ratio of lengths, square it for the ratio of areas (and cube it for volumes) — then you can assign convenient numbers and skip the geometry.
Another way — solve for the trapezoid area (MAA):
  1. Let the inner triangle have area A. The outer triangle is (3/2)2 = 9/4 as large, so its area is 94A.
  2. Inner triangle plus three trapezoids fills the outer triangle: A + 3X = 94A, so 3X = 54A and X = 512A.
  3. Thus one trapezoid : inner triangle = 5 : 12.
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