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Difference between revisions of "2000 AMC 8 Problems/Problem 15"

(Created page with "Triangles <math> ABC </math>, <math>ABD</math>, and <math>EFG</math> are all equilateral. Points <math>D</math> and <math>G</math> are midpoints of <math> \overline{AC} </math> a...")
 
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Triangles <math> ABC </math>, <math>ABD</math>, and <math>EFG</math> are all equilateral. Points <math>D</math> and <math>G</math> are midpoints of <math> \overline{AC} </math> and <math> \overline{AE} </math>, respectively. If <math> AB = 4 </math>, what is the perimeter of figure <math> ABCDEFG </math>?
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==Problem==
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Triangles <math> ABC </math>, <math>ADE</math>, and <math>EFG</math> are all equilateral. Points <math>D</math> and <math>G</math> are midpoints of <math> \overline{AC} </math> and <math> \overline{AE} </math>, respectively. If <math> AB = 4 </math>, what is the perimeter of figure <math> ABCDEFG </math>?
  
 
<asy>
 
<asy>
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<math> \text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21 </math>
 
<math> \text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21 </math>
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==Solution 1==
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The large triangle <math>ABC</math> has sides of length <math>4</math>.  The medium triangle has sides of length <math>2</math>.  The small triangle has sides of length <math>1</math>.  There are <math>3</math> segment sizes, and all segments depicted are one of these lengths.
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Starting at <math>A</math> and going clockwise, the perimeter is:
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<math>AB + BC + CD + DE + EF + FG + GA</math>
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<math>4 + 4 + 2 + 2 + 1 + 1 + 1</math>
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<math>15</math>, thus the answer is <math>\boxed{C}</math>
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==Solution 2==
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The perimeter of <math>ABCDEFG</math> is the perimeter of the three triangles, minus segments <math>AD</math> and <math>EG</math>, which are on the interior of the figure.  Because each of these segments is on two triangles, each segment must be subtracted two times.
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As in solution 1, the sides of the triangles are <math>4, 2, </math> and <math>1</math>, and the perimeters of the triangles are thus <math>12, 6, </math> and <math>3</math>.
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The perimeter of the three triangles is <math>12 + 6 + 3 = 21</math>.  Subtracting the two segments <math>AD</math> and <math>EG</math> two times, the perimeter of <math>ABCDEFG</math> is <math>21 - 2 - 1 - 2 - 1 = 15</math>, and the answer is <math>\boxed{C}</math>.
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==See Also==
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{{AMC8 box|year=2000|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 19:56, 6 November 2013

Problem

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]

$\text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21$

Solution 1

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 $\boxed{C}$

Solution 2

The perimeter of $ABCDEFG$ is the perimeter of the three triangles, minus segments $AD$ and $EG$, 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 $4, 2,$ and $1$, and the perimeters of the triangles are thus $12, 6,$ and $3$.

The perimeter of the three triangles is $12 + 6 + 3 = 21$. Subtracting the two segments $AD$ and $EG$ two times, the perimeter of $ABCDEFG$ is $21 - 2 - 1 - 2 - 1 = 15$, and the answer is $\boxed{C}$.


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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