AMC 8 · 2018 · #20

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

similar-trianglesarea-trianglesratio-proportion area-differenceidentify-subproblems ↑ Prerequisites: similar-trianglesarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In $\triangle ABC,$ a point $E$ is on $\overline{AB}$ with $AE=1$ and $EB=2.$ Point $D$ is on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$ and point $F$ is on $\overline{BC}$ so that $\overline{EF} \parallel \overline{AC}.$ What is the ratio of the area of $CDEF$ to the area of $\triangle ABC?$

$\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}$

Pick an answer.

(A)

$\frac{4}{9}$

(B)

$\frac{1}{2}$

(C)

$\frac{5}{9}$

(D)

$\frac{3}{5}$

(E)

$\frac{2}{3}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: In $\triangle ABC$, point $E$ sits on side $\overline{AB}$ with $AE = 1$ and $EB = 2$. From $E$ we draw $\overline{DE} \parallel \overline{BC}$ (so $D$ lands on $\overline{AC}$) and $\overline{EF} \parallel \overline{AC}$ (so $F$ lands on $\overline{BC}$). The four points $C, D, E, F$ form a quadrilateral inside the triangle. What fraction of the whole triangle's area is the area of $CDEF$?

Givens: $E$ is on $\overline{AB}$ with $AE = 1$ and $EB = 2$, so $AB = 3$; $D$ is on $\overline{AC}$ and $\overline{DE} \parallel \overline{BC}$; $F$ is on $\overline{BC}$ and $\overline{EF} \parallel \overline{AC}$; Answer choices: (A) $\tfrac{4}{9}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{5}{9}$, (D) $\tfrac{3}{5}$, (E) $\tfrac{2}{3}$

Unknowns: The ratio $\dfrac{[CDEF]}{[\triangle ABC]}$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The figure is the heart of the problem, so Tool #1 (Draw a Diagram) comes first: sketch $\triangle ABC$, mark $E$ one-third of the way along $\overline{AB}$, then draw the two parallels. The picture immediately shows that $\triangle ABC$ is sliced into three pieces — a small triangle $\triangle ADE$ near vertex $A$, another small triangle $\triangle EBF$ near vertex $B$, and the leftover quadrilateral $CDEF$. That decomposition is exactly Tool #7 (Identify Subproblems): instead of attacking $[CDEF]$ head-on, compute the two corner triangles as fractions of $[\triangle ABC]$ and subtract from $1$. Each corner triangle is similar to $\triangle ABC$ (parallel sides force the same angles), so its area scales as the square of the side ratio — a single idea applied twice.

Execute — Answer: A

#1 Draw a Diagram 4.G.A.2 Step 1

Draw the figure.
Sketch $\triangle ABC$, put $E$ on $\overline{AB}$ so that $AE = 1$ and $EB = 2$ (so $E$ is one-third of the way from $A$ to $B$).
Draw $\overline{DE}$ parallel to $\overline{BC}$ and $\overline{EF}$ parallel to $\overline{AC}$.
The two new segments split $\triangle ABC$ into three regions: corner triangle $\triangle ADE$ at $A$, corner triangle $\triangle EBF$ at $B$, and the leftover quadrilateral $CDEF$.

$$[\triangle ABC] = [\triangle ADE] + [\triangle EBF] + [CDEF]$$

💡 Drawing the figure and labelling the parallel pairs is the Grade 4 skill of classifying figures by parallel sides — it makes the decomposition visible.

#1 Draw a Diagram 8.G.A.5 Step 2

Recognize the two corner triangles are similar to $\triangle ABC$.
Because $\overline{DE} \parallel \overline{BC}$, the angles of $\triangle ADE$ match those of $\triangle ABC$ (shared angle at $A$, corresponding angles at $D$ and $E$), so $\triangle ADE \sim \triangle ABC$ with scale factor $\tfrac{AE}{AB} = \tfrac{1}{3}$.
Likewise $\overline{EF} \parallel \overline{AC}$ makes $\triangle EBF \sim \triangle ABC$ with scale factor $\tfrac{EB}{AB} = \tfrac{2}{3}$.

$$\triangle ADE \sim \triangle ABC \;\text{with ratio}\; \tfrac{1}{3}, \quad \triangle EBF \sim \triangle ABC \;\text{with ratio}\; \tfrac{2}{3}$$

💡 A line parallel to one side of a triangle cuts off a smaller, same-shape copy — the Grade 8 fact about parallel lines and corresponding angles.

#7 Identify Subproblems 7.G.A.1 Step 3

Convert each side-ratio into an area-ratio.
When you scale a figure by factor $k$, its area scales by $k^2$ (one factor of $k$ for the base, one for the height).
So $\triangle ADE$ takes up $\bigl(\tfrac{1}{3}\bigr)^2 = \tfrac{1}{9}$ of $[\triangle ABC]$, and $\triangle EBF$ takes up $\bigl(\tfrac{2}{3}\bigr)^2 = \tfrac{4}{9}$ of $[\triangle ABC]$.

$$\dfrac{[\triangle ADE]}{[\triangle ABC]} = \left(\tfrac{1}{3}\right)^2 = \tfrac{1}{9}, \quad \dfrac{[\triangle EBF]}{[\triangle ABC]} = \left(\tfrac{2}{3}\right)^2 = \tfrac{4}{9}$$

💡 On a scale drawing with side-scale $k$, areas scale by $k^2$ — the Grade 7 standard for scale drawings.

#7 Identify Subproblems 5.NF.A.1 Step 4

Subtract the two corner pieces from the whole.
The quadrilateral $CDEF$ is everything left over, so its area as a fraction of $[\triangle ABC]$ is $1 - \tfrac{1}{9} - \tfrac{4}{9}$.

$$\dfrac{[CDEF]}{[\triangle ABC]} = 1 - \tfrac{1}{9} - \tfrac{4}{9} = \tfrac{9 - 1 - 4}{9} = \tfrac{4}{9} \;\Rightarrow\; \textbf{(A)}$$

💡 Adding and subtracting fractions with the same denominator $9$ is the Grade 5 fraction-arithmetic skill that finishes the problem.

[1] #1 4.G.A.2 Draw the figure. Sketch $\triangle ABC$, put $E$ on $\overline{AB}$ so that $AE

[2] #1 8.G.A.5 Recognize the two corner triangles are similar to $\triangle ABC$. Because $\ove

[3] #7 7.G.A.1 Convert each side-ratio into an area-ratio. When you scale a figure by factor $k

[4] #7 5.NF.A.1 Subtract the two corner pieces from the whole. The quadrilateral $CDEF$ is every

Review

Reasonableness: The two corner pieces use up $\tfrac{1}{9} + \tfrac{4}{9} = \tfrac{5}{9}$ of the triangle, leaving $\tfrac{4}{9}$ for $CDEF$ — a clean fraction that matches choice (A). A sanity check: since $E$ sits closer to $A$ than to $B$, the corner near $B$ should be the bigger triangle ($\tfrac{4}{9}$ vs $\tfrac{1}{9}$), which is exactly what the picture shows. Also, $\tfrac{4}{9} < \tfrac{1}{2}$, consistent with the visual that $CDEF$ takes up a bit less than half the triangle.

Alternative: Tool #13 (Convert to Algebra) with coordinates: place $A = (0,0)$, $B = (3,0)$, $C = (0,3)$ so $[\triangle ABC] = \tfrac{9}{2}$. Then $E = (1,0)$; the line through $E$ parallel to $\overline{BC}$ (slope $-1$) meets $\overline{AC}$ at $D = (0,1)$; the line through $E$ parallel to $\overline{AC}$ (the $y$-axis is $\overline{AC}$, slope vertical) meets $\overline{BC}$ at $F = (1,2)$. Shoelace on $C(0,3), D(0,1), E(1,0), F(1,2)$ gives area $2$, and $\tfrac{2}{9/2} = \tfrac{4}{9}$ — the same answer.

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Recognizing the two parallel pairs ($\overline{DE} \parallel \overline{BC}$ and $\overline{EF} \parallel \overline{AC}$) and using them to decompose $\triangle ABC$ into two corner triangles plus the parallelogram $CDEF$.)
8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (including the angle-angle criterion for similarity) (Justifying $\triangle ADE \sim \triangle ABC$ and $\triangle EBF \sim \triangle ABC$ via the parallel-line, corresponding-angle argument that gives the AA similarity criterion.)
7.G.A.1 Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing (Converting the side-length scale factors $\tfrac{1}{3}$ and $\tfrac{2}{3}$ into area scale factors $\tfrac{1}{9}$ and $\tfrac{4}{9}$ (areas scale as the square of the side ratio).)
5.NF.A.1 Add and subtract fractions with unlike denominators (Combining the corner-triangle area fractions: $1 - \tfrac{1}{9} - \tfrac{4}{9} = \tfrac{4}{9}$.)

⭐ This AMC 8 problem only needs Grade 8 similar-triangles reasoning — parallel lines make smaller copies whose areas shrink by the square of the side ratio — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

More like this