AMC8 2000 8, Problem 15

Triangles $ABC$, $ADE$, and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$, respectively. If $AB = 4$, what is the perimeter of figure $ABCDEFG$? [asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2); G = (4.5,sqrt(3)/2); draw(A--B--C--cycle); draw(D--EE--A); draw(EE--F--G); label("$A$",A,S); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,NE); label("$E$",EE,NE); label("$F$",F,SE); label("$G$",G,SE);[/asy]

Solution: The large triangle $ABC$ has sides of length $4$.  The medium triangle has sides of length $2$.  The small triangle has sides of length $1$.  There are $3$ segment sizes, and all segments depicted are one of these lengths. Starting at $A$ and going clockwise, the perimeter is: $AB + BC + CD + DE + EF + FG + GA$ $4 + 4 + 2 + 2 + 1 + 1 + 1$ $15$, thus the answer is $\fbox{15}$

Answer: 15