/puzzle/ - Puzzles and recreational math

Let ABC be the vertices of the central triangle of side L. Because the branches are symmetric about the medians of the central triangle, we just need to find r such that the branches reaches the medians. We observe just the branch with root in B. Let Pn be the nearest point of the new trianles of the n-th generation in our branch to the median, so that P0=B.

Observe in the picture, that after n=2, all the Pn's are aligned perpendicular to the median. After P2, we can use the geometric series to find out the distance from P2 to the next Pn's.

It's even easier if you use this point (let's call it D) instead. The distance between D and the median is (l+rl+r^2l)*sin30=l/2*(1+r+r^2). We can use the geometric series to get the distance from D to Pn. r must be such that r^2l+r^3l+r^4l+…=l/2*(1+r+r^2). Or:

\sum(r^k)-1-r=(1+r+r^2)/2

The solution of this is r=(\sqrt{5}-1)/2, approximately 0.61803.