Fractals and Visualising Solids | FIO

Question 2

Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Solution

Understand the Question

The Sierpinski Triangle is a fractal created by starting with an equilateral triangle and iteratively removing the central inverted triangle from each solid triangle.
At each step:
- Every remaining solid triangle is split into $3$ smaller solid triangles, and $1$ central triangular hole is removed.
- The remaining triangles multiply by $3$ at each step ( $3^n$ at Step $n$ ).
- The number of holes accumulates: at each step $n$ , $3^{n-1}$ new holes are added.

(i) Find the number of holes at each step of the shape sequence that leads to the Sierpinski Triangle.

Step 1 · Count the Number of Holes at Each Step

Step 0: Starting with one solid triangle (no holes):

$\text{Number of holes at Step 0} = 0$

Diagram 6

Step 1: Removing the middle triangle creates $1$ hole:

$\text{Number of holes at Step 1} = 1$

Diagram 2

Step 2: A hole is removed from each of the $3$ remaining triangles, adding $3$ new holes:

\begin{aligned} \text{Number of holes at Step 2} &= (\text{holes from Step 1}) + (\text{new holes added}) \\ &= 1 + 3 \\ &= 4 \end{aligned}

Diagram 3

Step 3: Middle parts are removed from each of the $9$ triangles, adding $9$ new holes:

\begin{aligned} \text{Number of holes at Step 3} &= (\text{holes from Step 2}) + (\text{new holes added}) \\ &= 4 + 9 \\ &= 13 \end{aligned}

Diagram 4

Step 4: Middle parts are removed from each of the $27$ triangles, adding $27$ new holes:

\begin{aligned} \text{Number of holes at Step 4} &= (\text{holes from Step 3}) + (\text{new holes added}) \\ &= 13 + 27 \\ &= 40 \end{aligned}

Diagram 5

Pattern: At step $n$ (for $n \ge 1$ ), $3^{n-1}$ new holes are added: $\text{Number of holes at Step } n = 3^0 + 3^1 + 3^2 + \dots + 3^{n-1}$

Answer

(i) Step 0: $0$ , Step 1: $1$ , Step 2: $4$ , Step 3: $13$ , Step 4: $40$ , and Step $n$ : $3^0 + 3^1 + \dots + 3^{n-1}$

(ii) Find the number of triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Step 1 · Count the Remaining Triangles at Each Step

Step 0: Starting with $1$ large solid triangle:

$\text{Number of remaining triangles at Step 0} = 1 = 3^0$

Step 1: Removing the central triangle leaves $3$ smaller triangles:

$\text{Number of remaining triangles at Step 1} = 3 = 3^1$

Diagram 7

Step 2: Each of the $3$ triangles is divided into $3$ smaller triangles:

\begin{aligned} \text{Number of remaining triangles at Step 2} &= 3 \times (\text{triangles from Step 1}) \\ &= 3 \times 3 \\ &= 9 = 3^2 \end{aligned}

Diagram 8

Step 3: Each of the $9$ triangles is divided into $3$ smaller triangles:

\begin{aligned} \text{Number of remaining triangles at Step 3} &= 3 \times (\text{triangles from Step 2}) \\ &= 3 \times 9 \\ &= 27 = 3^3 \end{aligned}

Diagram 9

Step 4: Each of the $27$ triangles is divided into $3$ smaller triangles:

\begin{aligned} \text{Number of remaining triangles at Step 4} &= 3 \times (\text{triangles from Step 3}) \\ &= 3 \times 27 \\ &= 81 = 3^4 \end{aligned}

Diagram 10

Pattern: At Step $n$ , the number of remaining triangles is: $\text{Number of remaining triangles at Step } n = 3^n$

Answer

(ii) Step 0: $1$ , Step 1: $3$ , Step 2: $9$ , Step 3: $27$ , Step 4: $81$ , and Step $n$ : $3^n$

Common Mistakes

Total vs. New Holes: Confusing the new holes added at Step $n$ ( $3^{n-1}$ ) with the total cumulative holes present at Step $n$ ( $3^0 + 3^1 + \dots + 3^{n-1} = \dfrac{3^n - 1}{2}$ ).
Step 0 Indexing: Forgetting that Step 0 has $3^0 = 1$ triangle and $0$ holes, leading to an off-by-one exponent error.

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