Competition · AMC preparation · step 4 of 4

AMC 8 · 2019 · #24

Grade 8 geometry-2d

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

area-trianglesratio-proportionsimilar-triangles identify-subproblemsarea-difference ↑ Prerequisites: area-trianglesratio-proportion

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Inside △ ABC (area 360), point D sits on AC with AD:DC = 1:2, and E is the midpoint of BD. The line AE, when extended, hits side BC at a point F. Find the area of the small triangle △ EBF.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure has five named points and three concurrent cevians, so Tool #1 (Diagram) is essential — but we go a step further and put the diagram on a coordinate grid so that the unknown point F becomes computable, not guessed. Tool #7 (Subproblems) breaks the area question into a chain of three easier shared-height ratios (△ ABD from △ ABC, △ ABE from △ ABD, △ ABF from △ ABC). Tool #13 (Algebra) appears only briefly to pin down where line AE crosses BC, which gives BF:FC = 1:3. With those three ratios in hand the final answer is one subtraction. We deliberately avoid Menelaus's Theorem — it gives the same ratio in one line, but it is outside the CCSS K-8 toolkit our product targets.

1STEP 1

Put the figure on coordinates

Put the figure on a grid: B = (0, 0), C = (3, 0), A = (0, 3) — a right triangle whose ratios still match the problem.

B=(0,0), C=(3,0), A=(0,3), Area_coord(△ ABC)=9/2

2STEP 2

Locate D and E

Section and midpoint formulas place the inner points: D = (1, 2) and E = ( $\frac{1}{2}$ , 1).

D = A + 1/3(C - A) = (1, 2), E = 1/2(B + D) = (1/2, 1)

3STEP 3

Find where AE meets BC

Line AE (slope -4, so y = 3 - 4x) meets the x-axis at y = 0, giving F = ( $\frac{3}{4}$ , 0).

y - 3 = -4(x - 0) → y = 3 - 4x → 0 = 3 - 4x → x = 3/4

4STEP 4

Read off the BF ratio

On the x-axis BF = $\frac{3}{4}$ and FC = $\frac{9}{4}$ , so BF : FC = 1 : 3 and BF = $\frac{1}{4}$ of BC.

BF/BC = 3/4/3 = 1/4

5STEP 5

Chain the area ratios

Same-height area rule three times off 360: [△ ABD] = 120, [△ ABE] = 60, [△ ABF] = 90.

[△ ABD]=1/3 · 360=120, [△ ABE]=1/2 · 120=60, [△ ABF]=1/4 · 360=90

6STEP 6

Subtract to get the triangle

A, E, F are collinear, so [△ EBF] = [△ ABF] - [△ ABE] = 90 - 60 = 30, choice (B).

[△ EBF] = 90 - 60 = 30 → (B)

Answer

Each ratio in the chain is a clean unit fraction of 360:

\frac{1}{3}

\frac{1}{2}

\frac{1}{4}

, ending with 90 - 60 = 30. That is exactly choice (B), and it is plausibly small — △ EBF sits in a tight wedge near B, well under one tenth of the whole, and indeed

\frac{30}{360}

\frac{1}{12}

. As a sanity check, place the original coordinates from the asy diagram (B=(0,0), C=(3,0), A=(1.2,1.7)): the shoelace formula gives [△ EBF] ≈ 0.2125 while [△ ABC] ≈ 2.55, and

\frac{0.2125}{2.55}

× 360 = 30 on the nose.

💡Key takeaway

This AMC 8 problem only needs Grade 8 linear equations (to find one point) plus the Grade 6 "same-height triangles have areas in the same ratio as their bases" rule that you already know!

Put the figure on coordinates
Locate D and E
Find where AE meets BC
Read off the BF ratio
Chain the area ratios
Subtract to get the triangle

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