AMC 8 · 2008 · #23

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglesarea-rectanglescoordinate-geometryfraction-arithmetic area-differenceidentify-subproblemscoordinate-geometry ↑ Prerequisites: area-trianglesarea-rectangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Square ABCE has point F on side AE with AF = 2 · FE, and point D on side CE with CD = 2 · DE. Find the ratio of the area of △ BFD to the area of square ABCE.

Pick an answer.

(A)

$\frac{1}{6}$

(B)

$\frac{2}{9}$

(C)

$\frac{5}{18}$

(D)

$\frac{1}{3}$

(E)

$\frac{7}{20}$

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is the natural entry: placing the square on a coordinate grid pins down every point in the figure. Tool #16 (Change Representation) lets us pick the side length, and choosing s=3 turns the 2{:}1 ratios into whole-number lengths so no fractions appear until the very last step. Tool #7 (Break into Subproblems) handles △ BFD indirectly: instead of computing its area directly, we cut the square into △ BFD plus three right triangles in the corners, find those three easy areas, and subtract.

1STEP 1

Put the square on a grid with side 3 and E at the origin: E(0,0), C(3,0), B(3,3), A(0,3), so [ABCE] = 9.

E=(0,0), C=(3,0), B=(3,3), A=(0,3), [ABCE]=9

2STEP 2

The 2-to-1 splits make each side into thirds, so F=(0,1) one unit up from E and D=(1,0) one unit right of E.

FE = 1, AF = 2 → F=(0,1); DE = 1, CD = 2 → D=(1,0)

3STEP 3

Cut the square into △ BFD plus three corner right triangles at A, C, E, so [△ BFD] = [ABCE] - [△ ABF] - [△ BCD] - [△ FED].

[ABCE] = [△ ABF] + [△ BCD] + [△ FED] + [△ BFD]

4STEP 4

Each corner triangle is half its two legs: [△ ABF] = 3, [△ BCD] = 3, and [△ FED] = $\frac{1}{2}$ .

[△ ABF] =

\frac{1}{2}

(3)(2) = 3, [△ BCD] =

\frac{1}{2}

(3)(2) = 3, [△ FED] =

\frac{1}{2}

(1)(1) =

\frac{1}{2}

5STEP 5

Subtract: [△ BFD] = 9 - 3 - 3 - $\frac{1}{2}$ = $\frac{5}{2}$ , so the ratio is $\frac{\frac{5}{2}}{9}$ = $\frac{5}{18}$ → (C).

[△ BFD] = 9 - 3 - 3 -

\frac{1}{2}

\frac{5}{2}

;

\frac{[△ BFD]}{[ABCE]}

\frac{\frac{5}{2}}{9}

\frac{5}{18}

→ (C)

Answer

\frac{5}{18}

Quick size check: the three corner triangles together have area 3 + 3 +

\frac{1}{2}

\frac{13}{2}

, which is a bit more than half of the square's area 9. That leaves

\frac{5}{2}

for △ BFD, a little less than half the square. The ratio

\frac{5}{18}

≈ 0.278 matches that: a touch over a quarter. Among the choices,

\frac{1}{6}

≈ 0.17 is too small,

\frac{1}{3}

≈ 0.33 is too big, and only

\frac{5}{18}

lands in the right window.

💡Key takeaway

Put the square on graph paper at side 3, slice off the three corner triangles, and what is left is △ BFD — a clean Grade 6 area-decomposition trick that turns a tricky ratio into simple subtraction.

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