Two nondegenerate triangles 
and 
are cyclologic if the three circles
through 
,
,
and 
are concurrent. In this case, the circles
through 
,
,
and 
are also concurrent, and conversely. The two points
of concurrence are called the cyclologic centers
of the triangles (Rabinowitz and Suppa 2026, p. 105).
In the illustration above, the first point of concurrence
is denoted 
and the reciprocal point is denoted 
.
Let 
be any point inside 
. For 
equal to 13, 14, 15, or 16, let 
, 
, and 
be the Kimberling centers 
of 
, 
, and 
, respectively. Then 
and 
are cyclologic (Rabinowitz and Suppa 2026). The
four choices are the first Fermat point 
, second
Fermat point 
,
first isodynamic point 
, and second isodynamic
point 
,
respectively.
For 
,
an inversion centered at 
maps the three configurations to equilateral
triangles erected outwardly on the sides of the inverse
of 
.
Their circumcircles are concurrent at the first
Fermat point by Napoleon's theorem. Applying
the inversion again gives three concurrent circles
through 
,
,
and 
.
The assumption that 
is in the interior is essential
to the outward-orientation step: each first
isodynamic point lies inside the corresponding circumcircle,
while the opposite vertex lies outside it. Consequently,
the result for 
can fail when 
lies outside 
.
For 
,
the analogous inversion gives equilateral
triangles erected inwardly, and their circumcircles
concur at the second Fermat point.
