Here is a construction I came upon recently:

A'B'C' is the contact triangle. X(1) is the Incenter. A'',B'',C'' are the midpoints of the sides of the contact triangle. A''',B''',C''' are the midpoints of the segments AX(1), BX(1), CX(1). Then quadrilaterals A''A'''C''C''', B''B'''C''C''', A''A'''B''B''' are cyclic. Let Ob, Oa, Oc be the circumcenters of these 3 quadrilaterals. Finally by drawing circles with centers at Oa, Ob, Oc and passing through the incenter we get the 'inscribed-circumscribed' six point circle with its center at the midpoint of X(1) and X(3) - the point X(1385) in the ETC:

The inscribed-circumscribed circle

There is an even easier way to get a smaller concentric six point circle:
circumcircles of the quadrilaterals A''A'''C''C''', B''B'''C''C''', A''A'''B''B''' cut the sides of the triangle ABC at six concyclic points.

The mere fact that the point X(1385) is already described in the ETC supposedly makes the construction not particularly inspiring for a geometer. However as this 'X(1385)-circle' has never been previously mentioned (?) in the literature, there might still be a chance that it has some 'nice' properties of its own.

  • Specifically I am interested whether any Kimberling center belongs to this 'inscribed-circumscribed' circle? Or, alternatively, does its central function correspond to any known center?

1 Answer 1


After some symbol-bashing in Mathematica, here are barycentric coordinates of key points:

$$\begin{align} A' &= (0 : a + b - c : a - b + c) \\[0.75em] A'' := \tfrac12(B'+C') &= \left(a +\frac{(a-b+c)(a+b-c)}{-a+b+c} : b : c\right) \\[0.75em] A''' &= (2 a + b + c : b : c ) \\[0.75em] O_A &= (2 a^3 : a^3 - b^3 + c^3 + a^2 b - a^2 c - 3 a b^2 - a c^2 - b^2 c + b c^2 - 4 a b c \\ &\phantom{(2 a^3\;}\quad : a^3 + b^3 - c^3 - a^2 b + a^2 c - a b^2 - 3 a c^2 + b^2 c - b c^2 - 4 a b c) \end{align}$$ The coordinates of the points where, say, $\bigcirc O_A$ (the one passing through incenter $X_1$) meets the side-lines of the circle are a little messy, so I'll omit them. Nevertheless, the six prescribed points are indeed cyclic, and the equation of their common circle has this barycentric form (in $u:v:w$ coordinates): $$\begin{align} 0 &= u^2 b c (-a + b + c) \cos A + v^2 c a (a - b + c) \cos B + w^2 a b (a + b - c) \cos C \\[0.75em] &\quad+ v w (-a^3 + b^3 + c^3 - 2 a^2 b - 2 a^2 c - b^2 c - b c^2) \\[0.75em] &\quad+ w u (-b^3 + c^3 + a^3 - 2 b^2 c - 2 b^2 a - c^2 a - c a^2) \\[0.75em] &\quad+ u v (-c^3 + a^3 + b^3 - 2 c^2 a - 2 c^2 b - a^2 b - a b^2) \end{align}$$

Converting to trilinear form (in $\alpha:\beta:\gamma$ coordinates), the equation can be manipulated into this: $$(\lambda \alpha + \mu\beta+\nu\gamma)(a\alpha+b\beta+c\gamma)+\kappa(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$ where $$\begin{align} \lambda &:= (-a + b + c)\cos A \\ \mu &:= (\phantom{-}a-b+c)\cos B\\ \nu &:= (\phantom{-}a+b-c)\cos C\\ \kappa &:= -2(a+b+c) \end{align}$$

As $\lambda:\mu:\nu$ are the trilinear coordinates for Kimberling center $X(219)$ ("$X(8)$-Ceva Conjugate of $X(55)$"), the circle is the Central Circle of that point. This answers the "alternatively" aspect of OP's question.

I don't know what Kimberling centers may lie on the circle. I don't even know how one would go about searching for them. Someone with more fluency in triangle-center lore may have some insights.

Incidentally, for the $\bigcirc O_A$ passing through $B''$, $B'''$, $C''$, $C'''$, etc, the corresponding six-point circle has trilinear form $$(\lambda'\alpha+\mu'\beta+\nu'\gamma)(a\alpha+b\beta+c\gamma)+\kappa'(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$ where $$\begin{align} \lambda' &:= a (-a + b + c)^2 \\ \mu' &:= b (\phantom{-}a - b + c)^2 \\ \nu' &:= c (\phantom{-}a + b - c)^2 \\ \kappa' &:= -8 a b c \end{align}$$ This circle is therefore the Central Circle of Kimberling's $X(220)$ ("$X(9)$-Ceva conjugate of $X(55)$").

I have not investigated whether there is a general connection between OP's construction, Ceva-conjugates, and/or $X(55)$.

  • $\begingroup$ terrific! I missed the deadline for awarding this bounty. I believe it was stated somewhere in the rules that the earliest approved answer gets it automatically and it is clearly yours. $\endgroup$
    – A Z
    Sep 15, 2021 at 8:38

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