Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #15
Grade 5 geometry-2d
Pick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Label the biggest triangle
△ ABC is equilateral, so AB = 4 makes every side equal: BC = 4 and AC = 4.
Equilateral means "equal sides," so one side tells you all three.
4.G.A.2Draw A DiagramLabel the middle triangle
D halves AC, so AD = 2; equilateral △ ADE gives DE = AE = 2, and the leftover CD = 2.
Half of 4 is 2, and equilateral spreads that 2 to all three sides of the second triangle.
5.NF.B.4Identify SubproblemsLabel the smallest triangle
G halves AE, so GE = 1; equilateral △ EFG gives EF = FG = 1, and the leftover GA = 1.
Halving again: 2 → 1. The same midpoint trick that worked for △ ADE works for △ EFG.
5.NF.B.4Identify SubproblemsAdd the seven segments
Add the seven outline segments (AE stays interior): 4 + 4 + 2 + 2 + 1 + 1 + 1 = 15, choice (C).
Perimeter is just the sum of the segments you trace around the outside — read each label and add.
The perimeter of figure ABCDEFG is the total length of its seven outline segments AB, BC, CD, DE, EF, FG, and GA, and the inner segment AE is not counted.
▸ Why?
Perimeter means the whole distance once around the boundary, and this boundary is made of exactly those seven segments joined end to end with no gaps or overlaps, so its length is their sum; AE runs through the inside, so it is not part of the trip around.
▸ Why?
The seven segment lengths come from a shrinking cascade: each triangle's side is half the side of the triangle before it, giving 4 and 4, then 2 and 2, then 1, 1, and 1.
▸ Why?
Triangle ABC is equilateral, so its three sides share one length, and that length is the given AB = 4; therefore BC and AC are also 4.
▸ Why?
Point D is the midpoint of AC, so the pieces AD and DC are equal and, with no gap or overlap, add back to AC = 4; each is 2, so CD = 2 and side DE of equilateral triangle ADE matches it at 2.
▸ Why?
AD and DC together remake the whole segment AC = 4 with nothing missing or doubled, and as equal halves each must be 2, which fixes CD = 2.
▸ Why?
Triangle ADE is equilateral, so DE equals AD, and AD is 2; therefore DE is 2 as well.
▸ Why?
Point G is the midpoint of AE, so the two halves are equal and add back to AE = 2; each is 1, so GA = 1 and the sides of equilateral triangle EFG match it at 1.
▸ Why?
AG and GE together remake the whole segment AE = 2 with nothing missing or doubled, and as equal halves each must be 1, which fixes GA = 1.
▸ Why?
Triangle EFG is equilateral, so EF and FG equal GE, and GE is 1; therefore EF and FG are 1 as well.
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).
- Label the biggest triangle
- Label the middle triangle
- Label the smallest triangle
- Add the seven segments
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