AMC 8 2016 Problem 22: Geometry Solution

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Answer: C — Area 3.

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Hint 1 of 3

The figure is left-right symmetric, so the two wings have equal area — find ONE wing and double. The wing isn't a clean triangle, but it's a big triangle with a small triangle bitten out where the two slant lines cross.

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Hint 2 of 3

The crossing point H comes from similar triangles: ▵BCH and ▵EFH share the vertical angle at H and have parallel bases BC = 1 (top) and EF = 3 (bottom), so they're similar in ratio 1 : 3. Their heights split the rectangle's height 4 in that same 1 : 3 ratio.

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Hint 3 of 3

Once H's height is known, one wing = (a triangle with base on the top edge) minus (the little triangle above H).

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Approach: use the symmetry, find the crossing height by similar triangles, compute one wing and double

Let H be where the two inner slants CF and BE cross. ▵BCH ~ ▵EFH (vertical angle at H; parallel bases BC = 1 and EF = 3), so their heights are in ratio 1 : 3 and together span the height 4 — giving ▵BCH a height of 1.
Area ▵BCH = (1/2)(BC)(height) = (1/2)(1)(1) = 1/2.
▵BCE has base BC = 1 along the top and reaches the full height 4 down to E, so its area = (1/2)(1)(4) = 2.
One wing = ▵BCE − the bitten-out ▵BCH = 2 − 1/2 = 3/2. By symmetry the two wings total 2 × 3/2 = 3.
Why this transfers: when two cevians/slants cross, the crossing point's position comes free from a similar-triangle ratio of the two parallel bases — no coordinates needed.

Another way — coordinates + shoelace (trust-the-arithmetic check):

Put E = (0, 0), F = (3, 0), D = (0, 4), A = (3, 4); then C = (1, 4) and B = (2, 4). Inner slants: line BE is y = 2x, line CF is y = −2x + 6, crossing at H = (1.5, 3).
Left wing is ▵ with vertices C(1, 4), E(0, 0), H(1.5, 3). Shoelace area = (1/2)|1(0 − 3) + 0(3 − 4) + 1.5(4 − 0)| = (1/2)|−3 + 6| = 3/2.
Double for both wings: 2 × 3/2 = 3.