AMC 8 · 2019 · #24

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

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

📏 Long solution 💡 4 insights 📊 Diagram

Problem

In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$ ?

$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

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

Givens: Area of $\triangle ABC = 360$; $D$ is on $\overline{AC}$ with $AD:DC = 1:2$ (so $AD = \tfrac{1}{3}AC$); $E$ is the midpoint of $\overline{BD}$ (so $BE = ED$); $F$ lies on $\overline{BC}$ on the line through $A$ and $E$; Answer choices: (A) $24$, (B) $30$, (C) $32$, (D) $36$, (E) $40$

Unknowns: The area of $\triangle EBF$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

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 ($\triangle ABD$ from $\triangle ABC$, $\triangle ABE$ from $\triangle ABD$, $\triangle ABF$ from $\triangle ABC$). Tool #13 (Algebra) appears only briefly to pin down where line $AE$ crosses $\overline{BC}$, 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.

Execute — Answer: B

#1 Draw a Diagram 6.G.A.3 Step 1

Drop the figure onto coordinates that respect the ratios: $B = (0, 0)$, $C = (3, 0)$, $A = (0, 3)$.
With this choice $\triangle ABC$ is a right triangle of area $\tfrac{1}{2}(3)(3) = \tfrac{9}{2}$.
Every area we compute in these coordinates will later be scaled by the factor $360 / \tfrac{9}{2} = 80$ to convert to the problem's units.

$$B=(0,0),\ C=(3,0),\ A=(0,3),\ \text{Area}_{\text{coord}}(\triangle ABC)=\tfrac{9}{2}$$

💡 Putting the polygon on a grid (Grade 6 standard) lets us replace "where is $F$?" with arithmetic.

#1 Draw a Diagram 6.NS.C.8 Step 2

Locate $D$ and $E$ using the section/midpoint formulas.
$D$ divides $\overline{AC}$ with $AD:DC = 1:2$, so $D = A + \tfrac{1}{3}(C - A) = (1, 2)$.
$E$ is the midpoint of $\overline{BD}$, so $E = \tfrac{1}{2}(B + D) = (\tfrac{1}{2}, 1)$.

$$D = A + \tfrac{1}{3}(C - A) = (1, 2),\quad E = \tfrac{1}{2}(B + D) = \left(\tfrac{1}{2}, 1\right)$$

💡 Section formula and midpoint formula are coordinate-plane arithmetic — Grade 6 territory.

#13 Convert to Algebra 8.EE.C.7 Step 3

Find $F$ by intersecting line $\overleftrightarrow{AE}$ with the $x$-axis (which is line $\overleftrightarrow{BC}$).
The line through $A=(0,3)$ and $E=(\tfrac{1}{2},1)$ has slope $\tfrac{1-3}{1/2-0} = -4$, so its equation is $y = 3 - 4x$.
Setting $y=0$ gives $x = \tfrac{3}{4}$, hence $F = (\tfrac{3}{4}, 0)$.

$$y - 3 = -4(x - 0)\;\Rightarrow\; y = 3 - 4x\;\Rightarrow\; 0 = 3 - 4x\;\Rightarrow\; x = \tfrac{3}{4}$$

💡 Writing the slope-intercept equation and solving for $y = 0$ is Grade 8 linear-equation work.

#7 Identify Subproblems 6.RP.A.3 Step 4

Read the key ratio straight off the $x$-axis.
$F = (\tfrac{3}{4}, 0)$ and $C = (3, 0)$, so $BF = \tfrac{3}{4}$ and $FC = 3 - \tfrac{3}{4} = \tfrac{9}{4}$, giving $BF : FC = 1 : 3$ and $BF = \tfrac{1}{4} BC$.

$$\dfrac{BF}{BC} = \dfrac{3/4}{3} = \dfrac{1}{4}$$

💡 Comparing two lengths on the same number line is direct ratio reasoning — Grade 6.

#7 Identify Subproblems 6.G.A.1 Step 5

Use the shared-height area rule three times to chain ratios back to $\triangle ABC$.
(i) $\triangle ABD$ and $\triangle ABC$ share the altitude from $B$, with $AD = \tfrac{1}{3}AC$, so $[\triangle ABD] = \tfrac{1}{3}\cdot 360 = 120$.
(ii) $\triangle ABE$ and $\triangle ABD$ share the altitude from $A$, with $BE = \tfrac{1}{2}BD$, so $[\triangle ABE] = \tfrac{1}{2}\cdot 120 = 60$.
(iii) $\triangle ABF$ and $\triangle ABC$ share the altitude from $A$, with $BF = \tfrac{1}{4}BC$, so $[\triangle ABF] = \tfrac{1}{4}\cdot 360 = 90$.

$$[\triangle ABD]=\tfrac{1}{3}\cdot 360=120,\quad [\triangle ABE]=\tfrac{1}{2}\cdot 120=60,\quad [\triangle ABF]=\tfrac{1}{4}\cdot 360=90$$

💡 When two triangles share a height, their areas are in the same ratio as their bases — Grade 6 triangle-area logic.

#7 Identify Subproblems 6.G.A.1 Step 6

Subtract to isolate $\triangle EBF$.
Because $A$, $E$, $F$ are collinear, segment $\overline{AF}$ splits $\triangle ABF$ into $\triangle ABE$ and $\triangle EBF$.
So $[\triangle EBF] = [\triangle ABF] - [\triangle ABE] = 90 - 60 = 30$.
Answer choice $\textbf{(B)}\ 30$.

$$[\triangle EBF] = 90 - 60 = 30 \;\Rightarrow\; \textbf{(B)}$$

💡 Decomposing a triangle into two smaller triangles by a cevian and adding/subtracting areas is core Grade 6 geometry.

[1] #1 6.G.A.3 Drop the figure onto coordinates that respect the ratios: $B = (0, 0)$, $C = (3,

[2] #1 6.NS.C.8 Locate $D$ and $E$ using the section/midpoint formulas. $D$ divides $\overline{A

[3] #13 8.EE.C.7 Find $F$ by intersecting line $\overleftrightarrow{AE}$ with the $x$-axis (which

[4] #7 6.RP.A.3 Read the key ratio straight off the $x$-axis. $F = (\tfrac{3}{4}, 0)$ and $C = (

[5] #7 6.G.A.1 Use the shared-height area rule three times to chain ratios back to $\triangle A

[6] #7 6.G.A.1 Subtract to isolate $\triangle EBF$. Because $A$, $E$, $F$ are collinear, segmen

Review

Reasonableness: Each ratio in the chain is a clean unit fraction of $360$: $\tfrac{1}{3}, \tfrac{1}{2}, \tfrac{1}{4}$, ending with $90 - 60 = 30$. That is exactly choice (B), and it is plausibly small — $\triangle EBF$ sits in a tight wedge near $B$, well under one tenth of the whole, and indeed $30/360 = \tfrac{1}{12}$. 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 $[\triangle EBF] \approx 0.2125$ while $[\triangle ABC] \approx 2.55$, and $0.2125 / 2.55 \times 360 = 30$ on the nose.

Alternative: Tool #6 (Guess and Check) on the choices: every option (24, 30, 32, 36, 40) is consistent with the figure only if the supporting ratios match. Working backwards, $[\triangle EBF] = [\triangle ABF] - [\triangle ABE]$ where $[\triangle ABE] = 60$ is forced by the $1:2$ and midpoint conditions; the only choice that leaves a clean $\tfrac{1}{4}$ ratio for $[\triangle ABF]/[\triangle ABC]$ is $60 + 30 = 90 = \tfrac{1}{4}\cdot 360$, again pointing to (B).

CCSS standards used (min grade 8)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Applying the shared-height area rule three times to obtain $[\triangle ABD]=120,\ [\triangle ABE]=60,\ [\triangle ABF]=90$, and decomposing $\triangle ABF$ into $\triangle ABE + \triangle EBF$.)
6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Placing $\triangle ABC$ on a coordinate grid so that lengths and intersections become arithmetic.)
6.NS.C.8 Solve real-world problems by graphing points in all four quadrants (Computing $D = (1, 2)$ and $E = (\tfrac{1}{2}, 1)$ using the section and midpoint formulas in the coordinate plane.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Reading $BF:FC = 1:3$ off the $x$-axis and converting it to $BF/BC = 1/4$.)
8.EE.C.7 Solve linear equations in one variable (Setting up the line $y = 3 - 4x$ for $\overleftrightarrow{AE}$ and solving $0 = 3 - 4x$ to locate $F = (\tfrac{3}{4}, 0)$.)

⭐ 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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