AMC 8 · 2019 · #24
Grade 8 geometry-2dProblem
In triangle △ABC, point D divides side AC so that AD:DC=1:2. Let E be the midpoint of BD and let F be the point of intersection of line BC and line AE. Given that the area of △ABC is 360, what is the area of △EBF?
Pick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $40$
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
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.
💡 Putting the polygon on a grid (Grade 6 standard) lets us replace "where is $F$?" with arithmetic.
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 $E = \tfrac{1}{2}(B + D) = (\tfrac{1}{2}, 1)$.
💡 Section formula and midpoint formula are coordinate-plane arithmetic — Grade 6 territory.
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 $F = (\tfrac{3}{4}, 0)$.
💡 Writing the slope-intercept equation and solving for $y = 0$ is Grade 8 linear-equation work.
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 $BF = \tfrac{1}{4} BC$.
💡 Comparing two lengths on the same number line is direct ratio reasoning — Grade 6.
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 $[\triangle ABF] = \tfrac{1}{4}\cdot 360 = 90$.
💡 When two triangles share a height, their areas are in the same ratio as their bases — Grade 6 triangle-area logic.
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 $\textbf{(B)}\ 30$.
💡 Decomposing a triangle into two smaller triangles by a cevian and adding/subtracting areas is core Grade 6 geometry.
6.G.A.3 Drop the figure onto coordinates that respect the ratios: $B = (0, 0)$, $C = (3, 6.NS.C.8 Locate $D$ and $E$ using the section/midpoint formulas. $D$ divides $\overline{A 8.EE.C.7 Find $F$ by intersecting line $\overleftrightarrow{AE}$ with the $x$-axis (which 6.RP.A.3 Read the key ratio straight off the $x$-axis. $F = (\tfrac{3}{4}, 0)$ and $C = ( 6.G.A.1 Use the shared-height area rule three times to chain ratios back to $\triangle A 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.1Find 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.3Draw 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.8Solve 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.3Use 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.7Solve 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!
⭐ 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!
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