Competition · AMC preparation · step 4 of 4
AMC 10 · 2024B · #10
Grade 8 geometry-2dPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Spot the similar triangles
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.
Parallel lines cut by transversals force matching alternate-interior angles — Grade 8 informal angle arguments make the AA similarity instant.
Parallel lines cut by a crossing line force matching angles, which makes the triangles the same shape.
▸ Why?
A line crossing two parallels makes equal angles with both of them.
▸ Why?
Triangles with identical angles have all their matching sides in one fixed ratio.
Find the similarity ratio
E is the midpoint, so AE = BC and the corresponding sides give AE:CB = 1:2, so every corresponding length — including AF:FC = 1:2.
Similar figures scale every length by the same factor — Grade 8 similarity. E being a midpoint pins the factor at 1/2.
8.G.A.4Identify SubproblemsTurn lengths into areas
Areas scale as the square of 1:2, so with [△ AEF] = x, [△ CBF] = 4x; same-height △ ABF splits by base to give [△ ABF] = 2x.
Ratio-of-areas = (ratio-of-sides)² for similar triangles, and same-height triangles split area by base — two Grade 8 similarity ideas in one step.
8.G.A.4Identify SubproblemsFind the triangle across the diagonal
The diagonal AC halves the parallelogram: [△ ABC] = [△ ABF] + [△ CBF] = 6x, so [△ ADC] = 6x too.
A diagonal of a parallelogram halves its area into two congruent triangles — a Grade 6 "area by composing" fact.
6.G.A.1Identify SubproblemsCompute the asked ratio
CDEF is [△ ADC] minus [△ AEF], so [CDEF] = 5x; against [△ CBF] = 4x the ratio is 5x : 4x.
Areas can be added and subtracted because the region CDEF is just △ ADC minus △ AEF — Grade 6 "compose and decompose polygons."
6.G.A.1Identify SubproblemsThis 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!
- Spot the similar triangles
- Find the similarity ratio
- Turn lengths into areas
- Find the triangle across the diagonal
- Compute the asked ratio
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