AMC 8 2000 Problem 15
AMC8 2000 年第 15 题
三角形★★★☆☆
Triangles , , and are all equilateral. Points and are midpoints of and , respectively. If AB = 4 , what is the perimeter of figure ?
三角形 ABC，ADE 和 EFG 都是等边三角形。点 D 和 G 分别是形 ABCDEFG 的周长是多少？
和
- A.12
- B.13
- C.15
- D.18
- E.21
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
The large triangle 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 and going clockwise, the perimeter is: AB + BC + CD + DE + EF + FG + GA4 + 4 + 2 + 2 + 1 + 1 + 115, thus the answer is
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 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 and two times, the perimeter of is 21 - 2 - 1 - 2 - 1 = 15, and the answer is .
First, remove segments , and . Notice that segments , and can fit into , and , respectively. This means we only need to account for the perimeters of triangles , and , whose perimeters are 12, and 3 respectively. Adding these perimeters gives us 12+3=15, or answer choice .
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