The triangle on the left is equilateral. We produce the figure on the right by following a simple rule. We divide each side into three segments of equal length and add an equilateral triangle to the middle third segment as a base. The base of the new equilateral triangle is removed (see figure). If the original triangle has side $1$, what is the area of the resulting figure after an infinite number of applications of the rule?

enter image description here

I am getting the area of final figure as $$\sqrt{ \frac{3}{4} }+6\sqrt{\frac{3}{4}}\cdot\left(\frac{1}{3}\right)^2+12\sqrt{\frac{3}{4}}\cdot\left(\frac{1}{9}\right)^2\ldots$$

But my friend said the final figure is turning into a circle. I don't know how to finally reach to a logical answer and I am unable to sum this series also: $$\sqrt{ \frac{3}{4} }+6\sqrt{\frac{3}{4}}\cdot\left(\frac{1}{3}\right)^2+12\sqrt{\frac{3}{4}}\cdot\left(\frac{1}{9}\right)^2\ldots$$

  • $\begingroup$ Please type the text of the question instead of embedding it as an image, since images can not be searched. Also please make use of $\LaTeX$ formatting for your math. $\endgroup$ – MvG Jun 23 '14 at 15:15
  • $\begingroup$ I don't get it ,why do you want to search the question? and for LATEX I don't know really sorry for that. Here is the source :amburnchem.com/math/geometry.html set6 challenge Ques 7. $\endgroup$ – swapedoc Jun 23 '14 at 15:19
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    $\begingroup$ Hint: The area added is the sum of an infinite geometric series. An infinite geometric series with first term $k$ and common ratio $r$ has sum $\frac{k}{1-r}$. $\endgroup$ – André Nicolas Jun 23 '14 at 16:06
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    $\begingroup$ You may also find this interesting: en.wikipedia.org/wiki/Koch_snowflake $\endgroup$ – Jean-Claude Arbaut Jun 23 '14 at 16:28
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    $\begingroup$ The idea of Stack Exchange is that others can profit from your questions as well, which means they have to be able to find relevant questions using some kind of search, which means the questions should be searchable. I strongly suggest learning some basic MathJax, e.g. by reading the current (i.e. edited) source of this post, and/or the MathJax basic tutorial and quick reference. $\endgroup$ – MvG Jun 24 '14 at 8:56

Let the area of the original triangle be $a$. If the original side is $1$, then $a=\frac{\sqrt{3}}{4}$.

At the beginning, we have $3$ "sides," after the first stage we have $12$, after the second stage we have $48$, and so on. At the first stage we added $3$ triangles of area $\frac{a}{9}$, at the next stage $12$ with area $\frac{a}{9^2}$, and so on. So the total area is $$a\left(1+3\cdot\frac{1}{9}+12\cdot \frac{1}{9^2}+48\cdot\frac{1}{9^3}+\cdots\right).$$ Apart from the first term, this is a geometric series. The area is $$a+\frac{a}{3}\left(1+r+r^2+r^3+\cdots\right),$$ where $r=\frac{4}{9}$.

Thus the total area is $a+\frac{a}{3}\cdot \frac{1}{1-4/9}$.


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