1951 AHSME Problems/Problem 9

Problem

An equilateral triangle is drawn with a side of length $a$. 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:

$\textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$

Solution

The perimeter of the first triangle is $3a$. 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 $3a+\frac{3a}{2}+\frac{3a}{4}+\cdots$.

The starting term is $3a$, and the common ratio is $1/2$. Therefore, the sum is $\frac{3a}{1-\frac{1}{2}}=\boxed{\textbf{(D)}\ 6a}$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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