Snowflakes can be constructed by starting with an equilateral triangle (Fig. $1$), then repeatedly

Question

Snowflakes can be constructed by starting with an equilateral triangle (Fig. $1$), then repeatedly altering each line segment of the resulting polygon as follows:
$1.$ Divide each outer line segments into three segments of equal length.
$2.$ Add an equilateral triangle that has the middle segment from step $1$ as its base.
$3.$ Remove the line segment that is the base of the triangle from step $2$.
$4.$ Repeat the above steps for a number of iterations, $n$.
The $1$st, $2$nd and $3$rd iteration produces the snowflakes in Fig. $2$, Fig. $3$ and Fig. $4$ respectively.
1. If ${{a}_{0}}$ denotes the area of the original triangle and the area of each new triangle added in the ${{n}^{th}}$ iteration is denoted by ${{a}_{n}}$, show that
  ${{a}_{n}}=\frac{1}{9}{{a}_{n-1}}$ for all positive integers $n$.[2]
2. Write down the number of sides in the polygons in Fig.$1$, Fig. $2$ and Fig. $3$, and deduce, with clear explanations, that the number of new triangles added in the ${{n}^{th}}$ iteration is ${{T}_{n}}=k\left( {{4}^{n}} \right)$ where $k$ is a constant to be determined.
  [2]
3. Find the total area of triangles added in the ${{n}^{th}}$ iteration, ${{A}_{n}}$in terms of ${{a}_{0}}$ and $n$.[2]
4. Show that the total area of the snowflake produced after the ${{n}^{th}}$ iteration is ${{a}_{0}}\left[ \frac{8}{5}-\frac{3}{5}{{\left( \frac{4}{9} \right)}^{n}} \right]$ $\mathrm{unit}{{\mathrm{s}}^{2}}$.[3]
  The snowflake formed when the above steps are followed indefinitely is called the Koch snowflake.
5. Determine the least number of iterations needed for the area of the snowflake to exceed $99\%$ of the area of a Koch snowflake.
  [3]

Accepted Answer

(i) $a_n=\frac{1}{9}a_{n-1}$ (ii) $k=\frac{3}{4}$ (iii) $A_n=\frac{3}{4}\left(\frac{4}{9}\right)^n a_0$ (iv) $S_n=a_0\left[\frac{8}{5}-\frac{3}{5}\left(\frac{4}{9}\right)^n\right]$ (v) $n=5$

2019 ASRJC P1 Q11

2019 ASRJC P1 Q11

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