1951 AMC 12 Problem 9

Question

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:

Infinite

\left(5\dfrac14\right)a

2a

6a

\left(4\dfrac12\right)a

Answer: D

Concepts:geometric sequence perimeter equilateral triangle

Difficulty rating: 1400

Small hint:

Each midpoint triangle has half the side length of the preceding triangle

Big hint:

The perimeters form

3a+\frac{3a}{2}+\frac{3a}{4}+\cdots

Solution:

The first perimeter is

3a,

and every later perimeter is half the preceding one. Hence the total is

\begin{aligned} &3a\left(1+\frac12+\frac14+\cdots\right)\\ &\quad=3a\cdot\frac{1}{1-\frac12}\\ &\quad=6a. \end{aligned}

Thus, the correct answer is D.