1951 AHSME Problems/Problem 9
Problem
An equilateral triangle is drawn with a side of length . A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:
Solution
The perimeter of the first triangle is . The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing .
The starting term is , and the common ratio is . Therefore, the sum is
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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