AMC 8 · 2000 · #15

Grade 5 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

perimetersimilar-figuresfraction-multiplication identify-subproblemspattern-recognition ↑ Prerequisites: perimetermulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

📘 View easy version →

Problem

Three equilateral triangles ABC, ADE, and EFG are nested. D is the midpoint of AC and G is the midpoint of AE. Given AB = 4, find the perimeter of the seven-vertex outline ABCDEFG (the segments AB, BC, CD, DE, EF, FG, GA).

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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 △ ABC fixes △ ADE (because AD is half of AC), and the side of △ ADE fixes △ EFG (because EG is half of AE). Once every segment is labelled, the perimeter is one Grade 3 addition.

1STEP 1

△ ABC is equilateral, so AB = 4 makes every side equal: BC = 4 and AC = 4.

AB = BC = AC = 4

2STEP 2

D halves AC, so AD = 2; equilateral △ ADE gives DE = AE = 2, and the leftover CD = 2.

AD = DE = AE = 2, CD = AC - AD = 4 - 2 = 2

3STEP 3

G halves AE, so GE = 1; equilateral △ EFG gives EF = FG = 1, and the leftover GA = 1.

EF = FG = GE = 1, GA = AE - GE = 2 - 1 = 1

4STEP 4

Add the seven outline segments (AE stays interior): 4 + 4 + 2 + 2 + 1 + 1 + 1 = 15, choice (C).

AB + BC + CD + DE + EF + FG + GA = 4 + 4 + 2 + 2 + 1 + 1 + 1 = 15 → (C)

Answer

Quick size check: △ ABC alone has perimeter 4 + 4 + 4 = 12. The outline replaces one side (AC) of △ ABC with the detour C → D → E → F → G → 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 △ 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.

💡Key takeaway

Each triangle is half the size of the one before it (4 → 2 → 1). Label every segment on the outline, then add: 4 + 4 + 2 + 2 + 1 + 1 + 1 = 15, answer (C).

More like this