2000 AMC 8 Problem 15

Attempt Problem 15 of the 2000 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

Triangles ABC,ABC, ADE,ADE, and EFGEFG are all equilateral. Points DD and GG are midpoints of AC‾\overline{AC} and AE‾,\overline{AE}, respectively. If AB=4,AB = 4, what is the perimeter of figure ABCDEFG?ABCDEFG?
1212
1313
1515
1818
2121
Answer: C
Concepts:equilateral triangleperimeter
Small hint:
Use the midpoint information to find smaller equilateral side lengths
Big hint:
Add the outside boundary lengths only
Solution:
The large equilateral triangle has side length 4,4, the middle one has side length 2,2, and the smaller one has side length 1.1.
The perimeter is therefore CB+CD+DE+EF+FG+GA+AB=4+4+2+2+1+1+1=15. \begin{gather*} CB + CD + DE + EF \\ {}+ FG + GA + AB \\ = 4 + 4 + 2 + 2 + 1 \\ + 1 + 1 = 15. \end{gather*}
Thus, C is the correct answer.
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