area of Koch snowflake (2006 UQ/QAMT PS Competition 11-12 p4)

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area of Koch snowflake (2006 UQ/QAMT PS Competition 11-12 p4)

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#1 h Jan 21, 2021, 6:53 AM • 2 Y

Y by Mango247, Mango247

The Koch snowflake is constructed as follows. Start with an equilateral triangle of side length $1$ . Now divide each side into $3$ equal parts and construct an equilateral triangle extending outwards on the middle section, and then delete its base. Repeatedly apply this subdivision step. The snowflake is the limiting shape obtained. What is its area?

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#2 h Jan 22, 2021, 6:59 PM

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Given the equilateral triangle, side $a$ and area $A_{0}=\frac{\sqrt{3}}{4}a^{2}$ .

Step 1: adding $3$ triangles, side of the added triangle $\frac{a}{3}$ , area $A_{1}$ .
Step 2: adding $3 \cdot 4$ triangles, side of the added triangle $\frac{a}{9}$ , area $A_{2}$ .
Step 3: adding $3 \cdot 4^{2}$ triangles, side of the added triangle $\frac{a}{27}$ , area $A_{3}$ .
and so on
Step k: adding $3 \cdot 4^{k-1}$ triangles, side of the added triangle $\frac{a}{3^{k}}$ , area $A_{k}$ .

Area $A_{k+1}=A_{k}+(3 \cdot 4^{k-1}) \cdot \frac{\sqrt{3}}{4}\frac{a^{2}}{(3^{k})^{2}}$ ,
$A_{k+1}=A_{k}+\frac{\sqrt{3}}{12}\frac{4^{k-1}}{9^{k-1}}a^{2}$ .
$A_{k+1}=A_{0}+\frac{\sqrt{3}}{12}\left(1+\frac{4}{9}+(\frac{4}{9})^{2}+(\frac{4}{9})^{3}+\cdots\right)a^{2}$ .

$A=A_{0}+\frac{\sqrt{3}}{12} \cdot \frac{9}{5} \cdot a^{2}=\frac{\sqrt{3}}{4}a^{2}+\frac{\sqrt{3}}{4}a^{2} \cdot \frac{3}{5}$ .
$A=\frac{\sqrt{3}}{4}a^{2} \cdot \frac{8}{5}$ .

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