Difference between revisions of "2000 AMC 8 Problems/Problem 15"
m (→Problem) |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Triangles <math> ABC </math>, <math> | + | 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> |
Latest revision as of 19:56, 6 November 2013
Contents
[hide]Problem
Triangles ,
, and
are all equilateral. Points
and
are midpoints of
and
, respectively. If
, what is the perimeter of figure
?
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.