Label two of the triangles $A_1 A_2 A_3, B_1 B_2 B_3$ respectively. We will strengthen the conditions of the problem; in fact only one of the points in the third triangle, say $C$, is required. Note that if arc $A_i B_j$ is a minor arc (call such an arc a good arc) then we may draw rectangle $A_i B_j X Y$; if C is anywhere in minor arc $XY$then triangle $A_i B_j C$ is non-obtuse and we are done. Hence the problem reduces to covering the entire circle with good arcs. If $B_1$ lies in arc $A_1 A_2$, $B_2$ in $A_2 A_3$ and $B_3$ in $A_3 A_1$ then clearly $A_1 B_1, B_1 A_2, \ldots, B_3 A_1$ are 6 good arcs that cover the entire circle. Else say $B_2, B_3$ both lie in arc $A_2 A_3$, and $B_1$ in $A_1 A_2$. Then $A_1 B_1, B_1 A_2, A_2 B_3, B_2 A_3, B_3 A_1, A_3 B_1$ are 6 good arcs that cover the entire circle. Thus we are done. (Note that all arcs mentioned are minor arcs, either since they subtend sides of triangles opposite acute angles, or they are properly contained in such arcs.)