AMC 10 · 2024 · #10

Grade 8 geometry-2d

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

similar-trianglesarea-trianglesratio-proportioncoordinate-geometry identify-subproblemscoordinate-geometryarea-difference ↑ Prerequisites: similar-trianglesarea-trianglesratio-proportion

📏 Medium solution 💡 3 insights

Problem

ABCD is a parallelogram. E is the midpoint of side AD, and F is the point where line EB crosses diagonal AC. Find the ratio area(CDEF) : area(△ CFB).

Pick an answer.

(A)

5:4

(B)

4:3

(C)

3:2

(D)

5:3

(E)

2:1

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The setup is purely geometric — a parallelogram, a midpoint, a diagonal, and one intersection point. Tool #1 (Draw a Diagram) puts every length and triangle on paper so similar-triangle pairs can be spotted at a glance. Tool #7 (Identify Subproblems) then breaks the answer into bite-sized pieces: (a) find the similarity ratio of △ AEF and △ CBF, (b) use that ratio to express the areas of △ AEF, △ CBF, △ ABF in terms of one variable x, (c) use the diagonal to get the area of △ ADC, (d) subtract to get area(CDEF), then form the requested ratio.

1STEP 1

Parallel sides AD ∥ BC make alternate angles equal, and the vertical angle at F matches too, so three equal angles give △ AEF ∼ △ CBF by AA.

△ AEF ∼ △ CBF

2STEP 2

E is the midpoint, so AE = $\frac{1}{2}$ BC and the corresponding sides give AE:CB = 1:2, so every corresponding length — including AF:FC = 1:2.

\frac{AE}{CB}

\frac{1}{2}

\frac{AF}{FC}

\frac{1}{2}

3STEP 3

Areas scale as the square of 1:2, so with [△ AEF] = x, [△ CBF] = 4x; same-height △ ABF splits by base to give [△ ABF] = 2x.

[△ AEF] = x, [△ CBF] = (

\frac{1}{2}

)⁻² · x = 4x, [△ ABF] =

\frac{1}{2}

[△ CBF] = 2x

4STEP 4

The diagonal AC halves the parallelogram: [△ ABC] = [△ ABF] + [△ CBF] = 6x, so [△ ADC] = 6x too.

[△ ABC] = 2x + 4x = 6x = [△ ADC]

5STEP 5

CDEF is [△ ADC] minus [△ AEF], so [CDEF] = 5x; against [△ CBF] = 4x the ratio is 5x : 4x.

\frac{[CDEF]}{([△ CBF])}

\frac{5x}{4x}

\frac{5}{4}

→ (A)

Answer

5:4

Sanity check the parts: the parallelogram has total area 12x (6x + 6x), and the four named pieces inside it — △ AEF (x), △ CBF (4x), △ ABF (2x), and quadrilateral CDEF (5x) — sum to x + 4x + 2x + 5x = 12x, accounting for the whole parallelogram exactly. The ratio 5:4 is also the only choice strictly between 1:1 and 3:2, matching the geometric intuition that CDEF is just a bit bigger than △ CBF.

💡Key takeaway

This AMC 10 problem only needs Grade 8 similar-triangle reasoning — the midpoint forces a 1:2 side ratio, which becomes a 1:4 area ratio, and the rest is adding and subtracting triangle areas!

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