Fractal Fraction

Geometry Level 3

Jim has a unit square.

  • On top of this unit square, he places a square with side length equal to 1 3 \frac13 the side length of the square below.
  • On top of that square, he places another square which again has side length equal to 1 3 \frac13 the side length of the square below, and so on.
  • He repeats this process until he has created an infinite stack of diminishing squares--a fractal.

What is the ratio of the perimeter of the fractal to the area of the fractal?

4:3 10:3 33:8 40:9 The perimeter is infinite, so it is not possible to give a ratio

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3 solutions

The Codfather
Jan 29, 2018

It is important to first note that the sum of the lengths of the top sides of the squares will always be 1. From there, it can be seen that the perimeter of the shape is equal to 4 + 2 3 \frac{2}{3} + 2 9 \frac{2}{9} ...

S = 2 3 \frac{2}{3} + 2 9 \frac{2}{9} + 2 27 \frac{2}{27} ...

3S = 2 + 2 3 \frac{2}{3} + 2 9 \frac{2}{9} + 2 27 \frac{2}{27} ...

2S = 2

S = 1

Therefore the perimeter of the fractal is equal to 4 + 1 = 5 .

The area of the fractal is 1 + 1 9 \frac{1}{9} + 1 81 \frac{1}{81} ...

Another series summation can be used to solve this.

S = 1 + 1 9 \frac{1}{9} + 1 81 \frac{1}{81} ...

9S = 9 + 1 + 1 9 \frac{1}{9} + 1 81 \frac{1}{81} ...

8S = 9

S = 9 8 \frac{9}{8}

So the ratio is 5 to 9 8 \frac{9}{8} . This can multiplied out by 8 to give a nice integer ratio, and arrive at the final answer of 40:9 .

Why are u taking 2/3 in perimeter plzz explain

Saksham Jain - 3 years, 4 months ago

That's the next smallest square, with side length 1/3. We already accounted for its top side, so the only sides remaining on the perimeter are the left and right sides. Since both are 1/3, the 2/3 represents the sum of the left and right side of every smaller square.

The Codfather - 3 years, 4 months ago

The area after n n interactions may be written as

A n = 1 + ( 1 3 ) 2 + ( 1 3 1 3 ) 2 + + ( 1 3 1 3 n times ) 2 = i = 0 n ( 1 3 ) 2 i = 1 ( 1 9 ) n + 1 1 1 9 = 9 ( 1 9 ) n 8 , A_n = 1 + \left( \frac{1}{3} \right)^{2} + \left( \frac{1}{3} \cdot \frac{1}{3} \right)^{2} + \cdots + \left( \overbrace{\frac{1}{3} \cdots \frac{1}{3}}^{n \; \text{times}} \right)^{2} = \sum_{i=0}^{n} \left( \frac{1}{3} \right)^{2i} = \frac{1 - \left( \frac{1}{9} \right)^{n+1}}{1 - \frac{1}{9}} = \frac{9 - \left( \frac{1}{9} \right)^n}{8},

which will clearly give us A n n 9 / 8 A_n \xrightarrow{n \rightarrow \infty} 9/8 . For the perimeter

P n = 4 + ( 1 3 + 3 3 ) + ( 1 3 1 3 + 3 3 1 3 ) + + ( 1 3 1 3 n times + 3 3 1 3 1 3 n 1 times ) = 4 + i = 0 n 1 ( 1 3 ) i i = 1 n ( 1 3 ) i , P_n = 4 + \left( - \frac{1}{3} + \frac{3}{3} \right) + \left( - \frac{1}{3} \cdot \frac{1}{3} + \frac{3}{3} \cdot \frac{1}{3} \right) + \cdots + \left( - \overbrace{\frac{1}{3} \cdots \frac{1}{3}}^{n \; \text{times}} + \frac{3}{3} \cdot \overbrace{\frac{1}{3} \cdots \frac{1}{3}}^{n-1 \; \text{times}} \right) = 4 + \sum_{i=0}^{n-1} \left( \frac{1}{3} \right)^{i} - \sum_{i=1}^{n} \left( \frac{1}{3} \right)^{i},

which actually shows us that

P n = 5 ( 1 3 ) n n 5. P_n = 5 - \left( \frac{1}{3} \right)^{n} \xrightarrow{n \rightarrow \infty} 5.

We want

( P n A n ) n = 5 9 / 8 = 40 9 . \left( \frac{P_n}{A_n} \right )_{n \rightarrow \infty} = \frac{5}{9/8} = \frac{40}{9}.

Rab Gani
Jan 30, 2018

The Perimeter = 3 + 4/3 + 4/9 + 4/27 + ..... = 3 + (4/3)/(1 – 1/3) = 5. The Area = 1 + 1/9 + 1/81 + .... = 1/(1 – 1/9) = 9/8. The ratio of P/A = 5/(9/8) = 40/9.

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