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1951 AMC 12 Question 9

AMC12 · 1951 AMC 12

Question 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:
  1. Infinite
  2. $\left(5\dfrac14\right)a$
  3. $2a$
  4. $6a$
  5. $\left(4\dfrac12\right)a$