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Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

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.