Consider a regular hexagon with side length 1. Remove an equilateral triangle from the center of the hexagon, leaving 6 smaller equilateral triangles around it. Take each of these 6 smaller triangles and remove an equilateral triangle from their centers, leaving even smaller triangles around each one. Continue this process infinitely.
To calculate this area, let's consider the initial hexagon with side length 1. Its area can be divided into 6 equilateral triangles, each with side length 1. snowflake by haese mathematics
Since ( A_0 = \frac{\sqrt{3}}{4} ), the final area is: [ A_{\infty} = \frac{8}{5} \cdot \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{5} ] Consider a regular hexagon with side length 1
The correct value is ( 3√3 / 2 ) * ( 3 / 4) not (5√3 / 4) or anything... Continue this process infinitely
You're referring to the Snowflake problem from the Haese Mathematics series!
Since ( \frac{4}{3} > 1 ), as ( n \to \infty ): [ \lim_{n \to \infty} P_n = \infty ] The snowflake has an infinite perimeter.