Expert: Josh - 2/27/2005

Question
The side of a large rhombus is 3.  If rhombi are continually drawn within each other, with each rhombus half the size of the last, what is the sum of their perimeters?

Answer
Okay, a rhombus has four sides and all sides are equal in length. So, the largest one has a perimeter of 3*4=12...[#0]

Let us set up the problem as follows.
Use p(n) to denote the perimeter of the n th rhombus, where n=0 refers to the original (largest) rhombus.

So, we have p(0)=12 (as shown above).
Now, the next rhombus is indexed by n=1, where its side s(n)=s(n-1)/2,...[#1]
since its side is half as long as the previous rhombus.
So, its perimeter
p(n)=4*s(n)
   =4*s(n-1)/2 by substituting [#1] in the previous line
   =2*s(n-1)
But, p(n-1)=4*s(n-1) [in words, the perimeter of any given rhombus is four times the length of its side].

So, p(n)=0.5*p(n-1)...[#2]
This suggests that the perimeter is halved in relation to the previous rhombus.

Adding up the perimeters for all the rhombi, from n=0 (the largest one) to n=infinity, we have
Total_perimeter= p(0)+p(1)+p(2)+...+p(infinity)
              = p(0)+ p(0)/2+ p(0)/4 +...+p(0)/2^n
as n tends to infinity.

This is a geometric series (G.S.), with initial term a=p(0)=12, as indicated in [#0]. By inspection, the geometric ratio for the sequence 1, 1/2, 1/4, 1/8, 1/16,... is r=1/2.

Therefore, using the infinite sum formula for G.S.
where SUM_infinity = a/(1-r)
we have 12/(1-1/2)=24.

I don't know if you have learnt this in class. Let me know if anything bothers you.

Cheers.