6-periodic billiards trajectory in acute triangle

We can construct a 3-periodic billiards trajectory in an acute triangle in a classical geometric way, say taking the altitudes. Is there a similar way to construct a 6-periodic billiards?

• you could relect/mirror the triangle sequentially 6 times at a side and then connect two corresponding points, one at the original triangle and one at the 6th reflection. Feb 26 '19 at 5:44
• take a parallel to you 3-periodic solution and you get a 6-periodic solution. Feb 26 '19 at 5:48

Let your triangle be $$A,B,C$$. Let $$A'$$ be the reflection of $$A$$ across $$BC$$, $$C'$$ the reflection of $$C$$ across $$A'B$$, $$B'$$ the reflection of $$B$$ across $$A'C'$$, $$A''$$ the reflection of $$A'$$ across $$B'C'$$, $$C''$$ the reflection of $$C'$$ across $$A''B'$$. Then it turns out $$A'' C''$$ is parallel to $$AC$$. If possible, take a point $$p = t A + (1-t) C$$ of $$AC$$, $$0 < t < 1$$, and $$p'' = t A'' + (1-t) C''$$, such that the line $$p p''$$ is contained in the union of triangles $$ABC$$, $$A'BC$$, $$A'BC'$$, $$A'B'C'$$, $$A''B'C'$$, $$A''B'C''$$ (I don't know if this is guaranteed to exist for all acute triangles). Then we get a $$6$$-periodic trajectory $$p \to (t_1 B + (1-t_1) C) \to (t_2 A + (1-t_2) B) \to (t_3 A + (1-t_3) C) \to (t_4 B + (1-t_4) C) \to (t_5 A + (1-t_5) B) \to p$$ where $$t_1 B + (1-t_1) C$$, $$t_2 A' + (1-t_2) B$$, $$t_3 A' + (1-t_3) C'$$, $$t_4 B' + (1-t_4) C'$$, $$t_5 A'' + (1-t_5) B'$$ are on the line $$p p''$$.