# Find the attractor of the sequence of orthic triangles (formed by joining the feet of altitudes of the previous triangle).

Triangle 1 (see the picture) is given. Find the point toward which the vertices of triangle n -> infinity converge, assuming that triangle n is constructed by uniting the feet of the altitudes of triangle n-1.

Sequence of triangles formed by the above mentioned rule.

For the definition of "foot of an altitude" please see: Perpendicular Foot

• Can you explain, perhaps in a separate paragraph from the question, the rule for constructing additional triangles? An algorithm for triangle $n$, in terms on $n-1$, would fine. Basically I have no idea what you mean by "uniting the feet of the altitudes". Commented Jan 18, 2019 at 21:32
• @ShapeOfMatter: I believe it is quite clear: if $T_n$ is the triangle at the $n$-th step, $T_{n+1}$ is the orthic triangle of $T_{n}$. Commented Jan 18, 2019 at 21:37

By denoting as $$T_n$$ the triangle at the $$n$$-th iteration we can easily describe the angles of $$T_{n+1}$$, orthic triangle of $$T_n$$, in terms of the angles of $$T_n$$. We may check that the area and the perimeter of $$T_n$$ converge to zero, but the "shape" of $$T_n$$ (i.e. the triple of the angles) does not converge, in the general case.
Actually it is known that such iteration is usually chaotic, and not difficult to prove: assuming that our sequence is convergent to a point $$P$$, from some $$n$$ onward the orthocenter of $$T_n$$ has to lie in the interior of $$T_n$$, meaning that $$T_n$$ is acute-angled for any $$n$$ sufficiently large. On the other hand the shape of $$T_n$$ changes according to $$(A,B,C)\to (\pi-2A,\pi-2B,\pi-2C)$$ and almost surely the map sending $$x$$ into $$-2x\pmod{\pi}$$ is not convergent.