¶ 1951 AHSME Problem 9
Problem:
An equilateral triangle is drawn with a side of length . A new equilateral triangle is formed by joining the mid-points of the sides of the first one. Then a third equilateral triangle is formed by joining the mid-points 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:
Answer Choices:
A. Infinite
B.
C.
D.
E.
Solution:
Let be the perimeter of the -th triangle, with the first triangle having a perimeter of . The problem states that each subsequent perimeter is half of the previous one, so the common ratio is .
The sum of all the perimeters, , forms an infinite geometric series. We can find this sum using the formula , where is the first term of the series.
