This question is inspired by two posed by P.Terzić (both given elegant synthetic proofs by F. Petrov). The starting point is a triangle $ABC$ and a triangle centre $G_1$. There are two classical ways to use this to generate a new triangle $A_1B_1C_1$ where the new vertices are
- the reflections of the original ones in $G_1$;
- the feet of the Cevians through this point.
We now add a second centre $G_2$ to the ingredients and consider the three circles through $AG_2 A_1$, $BG_2B_1$ and $CG_2C_1$.
These three circles have the one common point $G_2$, of course, and our question is to determine under which conditions they have a second one. More specifically, our question is whether the following dichotomy holds:
For a given (distinct) pair of triangle centres, then either the above result holds or the triangle is isosceles.
Remarks. 1. The results of Petrov display two situations where the first situation is true.
we are using the term “circle” in the generalised sense which includes lines. These occur in the specal case of an isosceles triangle (for an equilateral one we have three lines through the centre–the second point of intersection is at infinity).
we refer to the online Encyclopedia of Triangle Centers for the concepts and notations we are using.
since there are now over $40,000$ registered centres there are potentially that number squared such results. Hence a synthetic proof can’t be expected and one is forced to apply Tate’s maxim–think geometrically, prove algebraically.