AHSME 1951 Problem 9

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