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'Sierpinski Triangle' printed from https://nrich.maths.org/
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.