Here I have constructed the orthocenter H of triangle ABC.
I then constructed the orthocenter of triangles HBC, HAB, and HAC.
Notice that the only points that appear on this drawing are A, B, C, and H.
The reason is simple. H is the common intersection of the three lines containing the altitudes.
Now let's look at the orthocenter of triangle HAB. We will need to construct the three altitudes for triangle HAB.
If we do this same procedure for triangle HAC and HBC we get that B and A are their orthocenters respectively.
The blue circle is the circumcircle of triangle ABC, the other three circles are the circumcircles of the other triangles.
One interesting conjecture is that the three purple circles always intersect at H. Proving this is very easy. The definition of a circumcircle is a circle that is the circumscribed circle of a triangle. In other words, all three vertices of a triangle, lie on the circle. Well since H is a vertex of all three triangles (besides triangle ABC) then the three circumcircles should intersect.
Notice that when H and C are at the same point it looks like three of the circles have disappeared. In actuality only two have disappeared. The circle that is to the right is the circumcircle of triangle ABC and HAB. The two circles that have "disappeared" are the circumcircles for triangle HAC and HBC since triangles HAC and HBC are no more than two straight lines.