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Let $T_1=[P_1,P_2,P_3]$ be a triangle. Choose a point $P_4$ on $T_1$ such that the area of triangle $T_2=[P_2,P_3,P_4]$ is half the area of $T_1$. Repeat this process in the natural way. It is not too difficult to see the resulting sequence of triangles converges to a point.

One can ask:

What the area of the set of possible convergents as compared to the area of $T_1$?

A quick hack when $P_1=(1,1)$, $P_2=(2,0)$ and $P_3=(0,0)$ yields the following graph of 30,000 random convergents:

enter image description here

A reasonable estimate of the area of the triangle containing these convergents would be $2/9$. Experimentally, at least, this type of simulation quickly finds an answer to the first question.

Now suppose $P_{k+3}$ (and each following point) is chosen randomly either half way between $P_k$ and $P_{k+1}$ or one-third of the way from $P_k$ to $P_{k+2}$. Here is a graph of a few of these convergents:

enter image description here

What is the area of the set of possible convergents in this case?

Finally, suppose we choose the next point to lie half way on the left side three times as frequently as half way on the right side. Here is the resulting rather weird graph of 100,000 or so such convergents:

enter image description here

It seems to my untrained eyes as though asking questions about area no longer makes much sense, so I won't.

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  • $\begingroup$ Your second image looks like the Osgood curve, a curve of positive area. (It also looks like the Cesaro curve, a variant of the Koch snowflake curve, which has zero area.) $\endgroup$ – Akiva Weinberger Jan 22 '17 at 8:35
  • $\begingroup$ @AkivaWeinberger Yes, it does look similar to those curves. Thx! $\endgroup$ – O. S. Dawg Jan 24 '17 at 2:24

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