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From the analysis of the passage from one stage to the next, I made some observations:
The total area of the white triangles (that is the triangles removed) is given by the series: $$\left({1\over 4}\right) + 3\left({1\over 4}\right)^2 + 3^2\left({1\over 4}\right)^3 +... 3^{n-1}\left({1\over4}\right)^n = 1 - \left({3\over 4}\right)^n$$ and this is as expected from the calculation of the total area of the red triangles.
In order to calculate the dimension $d$ from the formula $n=m^d$ , where $n$ is the number of self similar pieces and $m$ is the magnification factor, I see that for this problem $n=3$ and $m = 2$. So, $3 = 2^d$ which gives $\log 3 = \log\left(2^d\right)$ and hence $$d = {\log 3\over \log 2} =\log_2 3\approx 1.58 $$ I see that the dimension $d$ is between $1$ and $2$, as expected.
Sierpinski Triangle
Age 16 to 18
Challenge Level 
Thank you Jeremy from Drexel University, Philadelphia, USA and Andrei, from Tudor Vianu National College, Bucharest, Romania for two more excellent solutions.
To solve this problem, first I made a table, and I filled it with the properties of the figures in the problem.
| Stage |
Number of
red triangles
|
Area
of red
triangles
|
Number of
white triangles
|
Area
of white
triangles
|
| $0$ | $1$ | $1$ | $0$ | $1-1$ |
| $1$ | $3$ | $3\over 4$ | $1$ | $1-{3\over 4}$ |
| $2$ | $3 \times3$ | ${3\over 4}\times {3\over 4}$ | $1+3$ | $1-\left({3\over 4}\right)^2$ |
From the analysis of the passage from one stage to the next, I made some observations:
- The number of red triangles at each stage is multiplied by three to give the number of red triangles at the next stage.
- The total area of the red triangles at each stage is multiplied by 3/4 to give the total area of the red triangles at the next stage.
- For each red triangle at stage $n$ one additional white triangle appears at stage $n+1$. Equivalently the number of white triangles added at each stage is three times the number of white triangles added at the previous stage.
- The total area of all the white triangles is [1 - (total area of red triangles)].
In the table below, I summarised the results obtained from
these observations, and I have also calculated the limits for $n
\to \infty$.[Jeremy gave a proof by induction for each of these
formulae]
| Stage |
Number
ofred triangles
|
Area
of red
triangles
|
Number
of white
triangles
|
Area
of white
triangles
|
| $n$ | $3^n$ | $\left(\frac{3}{4}\right)^n$ | $1 + 3 +3^2 ... +3^{n-1}= {3^n -1\over 2} $ | $1-\left(\frac{3}{4}\right)^n$ |
| $n \to \infty$ | $\infty$ | $0$ | $\infty$ | $1$ |
The total area of the white triangles (that is the triangles removed) is given by the series: $$\left({1\over 4}\right) + 3\left({1\over 4}\right)^2 + 3^2\left({1\over 4}\right)^3 +... 3^{n-1}\left({1\over4}\right)^n = 1 - \left({3\over 4}\right)^n$$ and this is as expected from the calculation of the total area of the red triangles.
In order to calculate the dimension $d$ from the formula $n=m^d$ , where $n$ is the number of self similar pieces and $m$ is the magnification factor, I see that for this problem $n=3$ and $m = 2$. So, $3 = 2^d$ which gives $\log 3 = \log\left(2^d\right)$ and hence $$d = {\log 3\over \log 2} =\log_2 3\approx 1.58 $$ I see that the dimension $d$ is between $1$ and $2$, as expected.
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