How much paint does it take to color in the Koch Snowflake?

Algebra Level 2

What is the area enclosed by a Koch snowflake starting from an equilateral triangle with side length 1?

A. 1
B. $\frac{1}{2}$
C. $\frac{2\sqrt{3}}{5}$
D. $2 \frac{\sqrt{3}}{4}$
E. Area is infinite

A B C D E

1 solution

Zandra Vinegar Staff
Oct 12, 2015

The original equilateral triangle with side length 1 has area $\frac{\sqrt{3}}{4}$ , call that area $A$ . Every time new tips are added (one for each side in the existing figure from the previous stage), the triangles are $\frac{1}{9}$ the size of those added in the previous stage.

Every stage, there are 4 times the number of sides. Therefore, in total, $\frac{4}{9}$ again as much area is added each stage. In the first transition, the amount of area added (three triangles, for a total addition of $\frac{3\times 4}{9}A$ .

The sum of the infinite, geometric series: $\sum_{k=1}^{\infty}\frac{3 \times 4^{k-1}}{9^k} = \frac{\frac{3}{9}}{1-\frac{4}{9}} = \frac{3}{5}$

Therefore, the total amount of area is: $A(1 + \frac{3}{5}) = (\frac{\sqrt{3}}{4})(\frac{8}{5}) = \frac{2\sqrt{3}}{5}$

My aunt died when I was solving this problem, I received the news watching the gif while my mom was crying. I dedicate this problem to her... her name was Sofía.

Romeo Gomez - 4 years, 4 months ago

I am so sorry for the awful coincidence on your aunt ;(

I Love Brilliant - 9 months ago

number of sides on stage x = 4 $\times$ number of sides on stage x - 1

I believe, your assumption

in total, $\frac {4} {9}$ again as much area is added each stage

is wrong, because some sides are already present in previous stage. So while calculating how much area is added in the next stage, it is false when we consider all the sides in a certain stage.

Let $A_i$ = how much area has been added in terms of coefficient of $\frac {\sqrt{3}}{4}$ in figure i

Figure 2

3 triangles has been added

$A_2 = 3 * \left( \frac{1}{3} \right)^2$ which satisfies $\frac {3 \times 4^{k - 1}}{9^k}$ for $k = 1$

Figure 3

12 triangles has been added

$A_3 = 12 * \left(\frac{1}{9}\right)^2$ which satisfies $\frac {3 \times 4^{k - 1}}{9^k}$ for $k = 2$

Figure 4

36 triangles has been added

$A_4 = 36 * \left(\frac {1} {27}\right)^2\\ A_4 = 36 * \left(\frac {1} {729}\right)\\ A_4 = 36 * \left(\frac {1} {9}\right)^3$

which doesn't satisfy $\frac {3 \times 4^{k - 1}}{9^k}$ for $k = 3$

Please tell me if anything is wrong here or update your answer please. thanks

Abhisek Panigrahi - 3 years, 10 months ago

the number of new small squares replicated is 48 and not 36 hence \frac {\sqrt{3}}{4} (1+3*4/9+...)

Martin Anto - 1 year, 2 months ago

How much paint does it take to color in the Koch Snowflake?

1 solution

Figure 2

Figure 3

Figure 4

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