AMC 8 · 2000 · #15
Grade 5 geometry-2dProblem
Triangles ABC, ADE, and EFG are all equilateral. Points D and G are midpoints of AC and AE, respectively. If AB=4, what is the perimeter of figure ABCDEFG?
Pick an answer.
AMC 8 2000 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: Three equilateral triangles $ABC$, $ADE$, and $EFG$ are nested. $D$ is the midpoint of $\overline{AC}$ and $G$ is the midpoint of $\overline{AE}$. Given $AB = 4$, find the perimeter of the seven-vertex outline $ABCDEFG$ (the segments $AB$, $BC$, $CD$, $DE$, $EF$, $FG$, $GA$).
Givens: $\triangle ABC$, $\triangle ADE$, $\triangle EFG$ are equilateral; $D$ is the midpoint of $\overline{AC}$, so $AD = DC = \tfrac{1}{2} AC$; $G$ is the midpoint of $\overline{AE}$, so $AG = GE = \tfrac{1}{2} AE$; $AB = 4$; Answer choices: (A) $12$, (B) $13$, (C) $15$, (D) $18$, (E) $21$
Unknowns: The perimeter of figure $ABCDEFG$ = $AB + BC + CD + DE + EF + FG + GA$
Understand
Restated: Three equilateral triangles $ABC$, $ADE$, and $EFG$ are nested. $D$ is the midpoint of $\overline{AC}$ and $G$ is the midpoint of $\overline{AE}$. Given $AB = 4$, find the perimeter of the seven-vertex outline $ABCDEFG$ (the segments $AB$, $BC$, $CD$, $DE$, $EF$, $FG$, $GA$).
Givens: $\triangle ABC$, $\triangle ADE$, $\triangle EFG$ are equilateral; $D$ is the midpoint of $\overline{AC}$, so $AD = DC = \tfrac{1}{2} AC$; $G$ is the midpoint of $\overline{AE}$, so $AG = GE = \tfrac{1}{2} AE$; $AB = 4$; Answer choices: (A) $12$, (B) $13$, (C) $15$, (D) $18$, (E) $21$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Break the Problem into Smaller Parts
The figure already shows three equilateral triangles linked by midpoints, so Tool #1 (Draw a Diagram) lets us label each segment with its length and read the perimeter straight off the picture — no algebra needed. Tool #7 (Break into Smaller Parts) splits the work into a triangle-by-triangle cascade: the side of $\triangle ABC$ fixes $\triangle ADE$ (because $AD$ is half of $AC$), and the side of $\triangle ADE$ fixes $\triangle EFG$ (because $EG$ is half of $AE$). Once every segment is labelled, the perimeter is one Grade 3 addition.
Execute — Answer: C
4.G.A.2 Step 1 - Label $\triangle ABC$.
- Since $\triangle ABC$ is equilateral and $AB = 4$, every side of $\triangle ABC$ is $4$.
- In particular $BC = 4$ and $AC = 4$.
💡 Equilateral means "equal sides," so one side tells you all three.
5.NF.B.4 Step 2 - Move to $\triangle ADE$ (smaller part #1).
- $D$ is the midpoint of $\overline{AC}$, so $AD = \tfrac{1}{2} \cdot 4 = 2$.
- Since $\triangle ADE$ is also equilateral, every side equals $2$.
- That fixes $AD = DE = AE = 2$, and the leftover piece $CD$ on the outline is the other half of $AC$, also $2$.
💡 Half of $4$ is $2$, and equilateral spreads that $2$ to all three sides of the second triangle.
5.NF.B.4 Step 3 - Move to $\triangle EFG$ (smaller part #2).
- $G$ is the midpoint of $\overline{AE}$, so $GE = \tfrac{1}{2} \cdot 2 = 1$.
- Since $\triangle EFG$ is equilateral, every side equals $1$.
- That gives $EF = FG = GE = 1$, and the leftover piece $GA$ on the outline is the other half of $AE$, also $1$.
💡 Halving again: $2 \to 1$. The same midpoint trick that worked for $\triangle ADE$ works for $\triangle EFG$.
3.MD.D.8 Step 4 - Walk the outline $A \to B \to C \to D \to E \to F \to G \to A$ and add the seven labelled segments.
- The segment $AE$ is *not* on the outline (it is interior to $\triangle ADE$), so we use $CD = 2$ and $GA = 1$ in its place.
💡 Perimeter is just the sum of the segments you trace around the outside — read each label and add.
4.G.A.2 Label $\triangle ABC$. Since $\triangle ABC$ is equilateral and $AB = 4$, every 5.NF.B.4 Move to $\triangle ADE$ (smaller part #1). $D$ is the midpoint of $\overline{AC} 5.NF.B.4 Move to $\triangle EFG$ (smaller part #2). $G$ is the midpoint of $\overline{AE} 3.MD.D.8 Walk the outline $A \to B \to C \to D \to E \to F \to G \to A$ and add the seven Review
Reasonableness: Quick size check: $\triangle ABC$ alone has perimeter $4 + 4 + 4 = 12$. The outline replaces one side ($AC$) of $\triangle ABC$ with the detour $C \to D \to E \to F \to G \to A$, whose length is $2 + 2 + 1 + 1 + 1 = 7$. So the perimeter is $12 - 4 + 7 = 15$ — matches (C). The trap choice (A) $12$ is the perimeter of $\triangle ABC$ by itself, forgetting the detour. (E) $21$ is the sum of all sides of all three triangles ($12 + 6 + 3$), which double-counts the interior segments $AE$ and the two midpoint halves.
Alternative: Tool #5 (Look for a Pattern): the three triangles have sides $4, 2, 1$ — each one is half the previous. On the outline, the two sides of each triangle that stay on the boundary contribute $4 + 4 = 8$ from $\triangle ABC$ (the $AC$ side is replaced by the next triangle), $2 + 2 = 4$ from $\triangle ADE$ (the $AE$ side is replaced), and all three sides $1 + 1 + 1 = 3$ from $\triangle EFG$ (it is the last triangle, so nothing replaces $AE$ on its end — the half $GA$ closes the loop). Sum: $8 + 4 + 3 = 15$, again (C).
CCSS standards used (min grade 5)
4.G.A.2Classify two-dimensional figures based on properties of their sides (Using the equilateral property of $\triangle ABC$, $\triangle ADE$, $\triangle EFG$ to conclude all three sides of each triangle share the same length.)5.NF.B.4Apply and extend previous understandings of multiplication to multiply a fraction by a whole number (Taking half of a length at each midpoint — $\tfrac{1}{2} \cdot 4 = 2$ and $\tfrac{1}{2} \cdot 2 = 1$ — to size $\triangle ADE$ and $\triangle EFG$.)3.MD.D.8Solve real world and mathematical problems involving perimeters of polygons (Adding the seven outline segments $4 + 4 + 2 + 2 + 1 + 1 + 1 = 15$ to get the perimeter of figure $ABCDEFG$.)
⭐ Each triangle is half the size of the one before it ($4 \to 2 \to 1$). Label every segment on the outline, then add: $4 + 4 + 2 + 2 + 1 + 1 + 1 = 15$, answer (C).
⭐ Each triangle is half the size of the one before it ($4 \to 2 \to 1$). Label every segment on the outline, then add: $4 + 4 + 2 + 2 + 1 + 1 + 1 = 15$, answer (C).
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