Problem 23 · 2008 AMC 8
Hard
Geometry & Measurement
coordinate-bashshoelace
In square ABCE, AF = 2FE and CD = 2DE. What is the ratio of the area of ▵BFD to the area of square ABCE?
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Answer: C — 5/18.
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Hint 1 of 2
△BFD has no horizontal or vertical side, so it's awkward to measure head-on — but the three pieces left over (one at each used corner) are right triangles with easy legs.
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Hint 2 of 2
Subtract instead of build: △BFD = whole square − the three right triangles cut off in the corners. Pick side length 1 to make every leg a simple fraction.
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Approach: subtract three corner right triangles from the square
- Take side 1, so AF = 2/3, FE = 1/3, CD = 2/3, DE = 1/3. The three corner triangles cut off around △BFD are △ABF (legs 1, 2/3, area 1/3), △BCD (legs 1, 2/3, area 1/3), and △FED (legs 1/3, 1/3, area 1/18).
- △BFD = 1 − 1/3 − 1/3 − 1/18 = (18 − 6 − 6 − 1)/18 = 5/18.
- Why this transfers: a slanted triangle inside a rectangle is almost always easier as "rectangle minus the right triangles in the corners" — each corner triangle has axis-aligned legs you can read straight off the figure.
Another way — shoelace:
- Place E at the origin, side s. B = (s, s), F = (0, s/3), D = (s/3, 0).
- Area △BFD = ½ |s(s/3 − 0) + 0(0 − s) + (s/3)(s − s/3)| = ½(s²/3 + 2s²/9) = 5s²/18, so the ratio is 5/18.
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