AMC 8 · 2000 · #15
Easy mode Grade 5Problem
Picture three equilateral triangles attached one after another, each smaller than the last. (Equilateral means all three sides have the same length.)
- The biggest triangle is △ABC. Its side AB has length 4.
- The middle triangle is △ADE. Point D sits at the middle of side AC.
- The smallest triangle is △EFG. Point G sits at the middle of side AE.
Together these three triangles form one connected shape called ABCDEFG.
What is the perimeter of the shape ABCDEFG (the total length all the way around its outside)?
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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