Problem 24 · 2019 AMC 8
Hard
Geometry & Measurement
area-fractionratio
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Answer: B — Area 30.
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Hint 1 of 2
Every quantity here is a ratio (1:2, midpoint, an area of 360), and area ratios don't care about the exact shape. So either chase the ratios directly with the "same height → areas split like the bases" rule, or drop in convenient coordinates and let the numbers do the work.
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Hint 2 of 2
If you go coordinate: place the triangle with easy numbers honoring AD:DC = 1:2, find E as a midpoint and F as the intersection of lines AE and BC, then compare area(EBF) to area(ABC).
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Approach: convenient coordinates, then compare to the whole
- Ratios are shape-independent, so pick easy coordinates honoring AD:DC = 1:2: A = (0, 0), B = (0, 1), C = (3, 0). Then D = (1, 0) and E = midpoint of BD = (0.5, 0.5).
- Line AE has slope 1, so y = x; line BC runs (0,1) to (3,0), so y = 1 − x/3. Setting equal: x = 1 − x/3 ⇒ x = 3/4, giving F = (3/4, 3/4).
- Shoelace on E(0.5, 0.5), B(0, 1), F(0.75, 0.75) gives area 1/8; area ABC = 3/2. So ▵EBF is (1/8) ÷ (3/2) = 1/12 of the whole.
- Therefore ▵EBF = (1/12) × 360 = 30.
- Why this transfers: when a problem gives only ratios and one total area, the answer is a fixed fraction of that total — computing on any convenient triangle gives the same fraction, so choose coordinates that make the arithmetic trivial.
Another way — pure area-ratio chain (no coordinates):
- Since AD:DC = 1:2, triangles ABD and CBD share the height from B and split as their bases: [ABD] = (1/3)(360) = 120.
- E is the midpoint of BD, so the median AE halves ▵ABD: [ABE] = 60.
- F lies on line BC, so ▵BEF and ▵ABF share base BF; their areas are in the ratio of the heights of E and A above line BC. Because E is the midpoint of BD and D is 2/3 of the way along CA, E's height above BC is 1/3 of A's, giving [BEF] : [ABF] = 1 : 3.
- With [ABF] = [ABE] + [BEF] = 60 + [BEF], set [BEF] / (60 + [BEF]) = 1/3 ⇒ 3[BEF] = 60 + [BEF] ⇒ [BEF] = 30.
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