2000 AMC 8 Problems/Problem 15
Contents
Problem
Triangles 
, 
, and 
are all equilateral. Points 
and 
are midpoints of 
and 
, respectively. If 
, what is the perimeter of figure 
?
![[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]](http://latex.artofproblemsolving.com/3/e/e/3ee31377617710d02efbfe48eb576a433c3499ec.png)
Solution 1
The large triangle has sides of length 
. The medium triangle has sides of length 
. The small triangle has sides of length 
. There are 
segment sizes, and all segments depicted are one of these lengths.
Starting at 
and going clockwise, the perimeter is:
, thus the answer is 
Solution 2
The perimeter of is the perimeter of the three triangles, minus segments 
and 
, which are on the interior of the figure. Because each of these segments is on two triangles, each segment must be subtracted two times.
As in solution 1, the sides of the triangles are 
and , and the perimeters of the triangles are thus 
and .
The perimeter of the three triangles is 
. Subtracting the two segments and two times, the perimeter of is 
, and the answer is .
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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