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 $G_1$;
or
- 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.
in this situation one can proceed as follows. Wlog the triangle can be assumed to have vertices $(0,0)$, $(1,0)$ and $(p,q)$. Using the centre functions from the above source, one can compute the coordinates of all the relevant points explicitly in terms of $p$ and $q$. (This is tedious to impractial to do by hand but it is easy to write a small programme with Mathematics to compute them with the relevant centre functions as input).
Using the determinental formula for the equation of the circle through three given points and the fact that three circles through a given point meet at a second one if and only if their radical axes coincide, one can then restate the above question as one about the nature of the zeroes of an explicit function of $p$ and $q$.
The general problem seems quite intractable to me which is why I am posting it here. However, one can use the above method to explore its validity in certain concrete cases.
My question is a particular case of what I call the general Steiner-Lehmus principle--if you have a general symmetry condition which holds for isosceles triangles, then it characterises them. In this vague formulation, the principle is preposterous but, when used with common sense, it can suggest interesting problems and results. The most significant case is that of the equality of the lengths of two Cevians associated to a triangle (the classical--and notorious--Steiner-Lehmus theorem involves the incentre). Another significant and much studied example is that of the collinearity of three centres--again we have a dichotomy: either it always holds (e.g., the Euler line!) or it implies that the triangle is isosceles.
