Problem 22 · 2018 AMC 8
Hard
Geometry & Measurement
area-fractionarea-decomposition
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Answer: B — Area 108.
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Hint 1 of 2
The crossing point F looks awkward, but AB (the full top side) is parallel to EC (half the bottom side). Parallel lines being crossed means the two triangles meeting at F are similar — and that similarity, with its 2 : 1 size ratio, locks down exactly where F sits on the diagonal.
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Hint 2 of 2
The technique: find the key ratio from an X-shaped pair of similar triangles, then chase areas as fractions of the square. Don't solve for the side length until the very end — carry s symbolically and let the 45 finish it.
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Approach: use the AF:FC ratio from similar triangles
- Let the square's side be s. Then AB = s and EC = s/2 (E is the midpoint of DC).
- Lines BE and AC cross at F, forming ▵AFB and ▵CFE as a bow-tie. Since AB ∥ EC, these triangles are similar with ratio AB : EC = 2 : 1, so AF : FC = 2 : 1 — F is 2/3 of the way from A down to C.
- That fixes F's height: being 2/3 of the way down the diagonal, F is 1/3 of the side above the base DC. So ▵CEF has base EC = s/2 and height s/3, giving area (1/2)(s/2)(s/3) = s2/12.
- AFED is the lower-left triangle ▵ACD minus ▵CEF: s2/2 − s2/12 = 5s2/12. Set equal to the given 45: 5s2/12 = 45.
- s2 = 108 — and s2 is the square's area, so no separate final step.
Another way — coordinates (let the side be 1, scale at the end):
- Place D = (0, 0), C = (1, 0), B = (1, 1), A = (0, 1) for a unit square, with E = (1/2, 0). Line AC goes from (0,1) to (1,0): y = 1 − x. Line BE goes from (1,1) to (1/2,0): solving, the two meet at F = (2/3, 1/3).
- Quadrilateral AFED has vertices A(0,1), F(2/3,1/3), E(1/2,0), D(0,0). The shoelace formula gives area 5/12 of the unit square.
- So AFED is 5/12 of the whole square. Since 5/12 of the area = 45, the area is 45 × 12/5 = 108 — the same ratio, reached with pure coordinates instead of similar triangles.
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