Sangaku Iterations
Is it Wasan?
I happened on a sangaku problem posed by the Japanese mathematician Tumugu Sakuma (18191896) who worked at the later years of the Edu period of seclusion by sheer accident. An article in Mathematics magazine by Fukuzo Suzuki just followed the one on the Lights Out puzzle I've been reading. The author referred to a problem in a Japanese book People of Wasan on Record by A. Hirayama (1965) and noted some incorrect results. The problem in the article was a generalization of that in Hirayama's book and asked to find certain relationships in a configuration of an equilateral triangle whose side lines passed though the vertices of a given isosceles triangle. (In the book, the original sangaku required a right isosceles triangle.)
Somehow, I found both the problem and the solution unappealing. However, the problem did not fit the stereotype of the sangaku promoted by Tony Rothman, whose article in Scientific American caused much stir in the math education community. The problem did not have "circles within triangles, spheres within pyramids, ellipsoids surrounding spheres." For this reason alone I thought it worthy to be included in my collection.
But how does one construct an equilateral triangle with the side lines through the vertices of another triangle? A recollection flashed through my mind of another problem where a triangle was obtained as a limit of an iterative procedure. This was a trivial matter to modify the applet and the result is below.
For a given triangle, you can start iterations anywhere by clicking a mouse button. On each step, the iterations go from a point in a direction of a vertex, using all three vertices in a loop. If p_{0} is the starting point and v_{0} the first vertex, then the second iterate is chosen according to the formula
p_{1} = p_{0} + (v_{0}  p_{0})·Rn/(Rn + Rd) 
The secondd is computed analogously via
p_{2} = p_{1} + (v_{1}  p_{1})·Rn/(Rn + Rd). 
The subsequent iterates are calculated by the formula that forces equal sides at the limit:
(1)  p_{n+1} = p_{n} + (v_{n}  p_{n})/dist(p_{n}, v_{n})·(dist(p_{n}, p_{n1}) + dist(p_{n1}, p_{n2}))/2. 
As you can easily check this approach works for triangles not necessarily isosceles. However, in the presence of an obtuse angle, the iterations may not converge to a triangle, but to a selfintersecting equilateral hexagon resembling an arrow tip.
What if applet does not run? 
Why (1) leads to an equilateral triangle? Assuming the iterations converge, i.e.,
p_{3n} → A_{0}, p_{3n + 1} → B_{0}, p_{3n + 2} → C_{0}, 
and using a = B_{0}C_{0}, b = A_{0}C_{0}, and c = A_{0}B_{0}, (1) gives in the limit
a = (b + c)/2, b = (c + a)/2, c = (a + b)/2. 
This is a system of three linear equations with three quantities a, b, c and solutions that, because of the symmetry, are bound to satisfy
References
 F. Suzuki, An Equilateral Triangle with Sides through the Vertices of an Isosceles Triangle, Mathematics Magazine, Vol. 74, No. 4. (Oct., 2001), pp. 304310.
Sangaku

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