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:

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 3n 3/4n
1 + 3 +32 ... +3n-1 = 3n -1
2

1-3/4n
\n® ¥ ¥ 0 ¥ 1

The total area of the white triangles (that is the triangles removed) is given by the series:
( 1
4
 ) + 3( 1
4
 )2 + 32( 1
4
 )3 +... 3n-1( 1
4
 )n = 1 - ( 3
4
 )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=md , 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 = 2d which gives log3 = log(2d) and hence
d = log3
log2
 =log2 3 » 1.58
I see that the dimension d is between 1 and 2, as expected.