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Coaxial circles related to Gergon point
Headhunter   1
N Apr 3, 2025 by internationalnick123456
Source: I tried but can't find the source...
Hi, everyone.

In $\triangle$$ABC$, $Ge$ is the Gergon point and the incircle $(I)$ touch $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively.
Let the circumcircles of $\triangle IDGe$, $\triangle IEGe$, $\triangle IFGe$ be $O_{1}$ , $O_{2}$ , $O_{3}$ respectively.

Reflect $O_{1}$ in $ID$ and then we get the circle $O'_{1}$
Reflect $O_{2}$ in $IE$ and then the circle $O'_{2}$
Reflect $O_{3}$ in $IF$ and then the circle $O'_{3}$

Prove that $O'_{1}$ , $O'_{2}$ , $O'_{3}$ are coaxial.
1 reply
Headhunter
Apr 3, 2025
internationalnick123456
Apr 3, 2025
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Coaxial circles related to Gergon point
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Headhunter
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Headhunter
#1 h  Apr 3, 2025, 2:48 AM
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Hi, everyone.

In $\triangle$$ABC$, $Ge$ is the Gergon point and the incircle $(I)$ touch $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively.
Let the circumcircles of $\triangle IDGe$, $\triangle IEGe$, $\triangle IFGe$ be $O_{1}$ , $O_{2}$ , $O_{3}$ respectively.

Reflect $O_{1}$ in $ID$ and then we get the circle $O'_{1}$
Reflect $O_{2}$ in $IE$ and then the circle $O'_{2}$
Reflect $O_{3}$ in $IF$ and then the circle $O'_{3}$

Prove that $O'_{1}$ , $O'_{2}$ , $O'_{3}$ are coaxial.
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internationalnick123456
124 posts
internationalnick123456
#2 h  Apr 3, 2025, 4:02 PM
Y by
For convenience, we will solve the following problem:
Quote:
Let $ABC$ be a triangle with circumcenter $O$ and Lemoine point $L$. Let $A_1,B_1,C_1$ be the reflections of $L$ wrt $OA,OB,OC$. Prove that the circles $(OAA_1), (OBB_1), (OCC_1)$ are coaxial.
We make use of the following lemma:
Lemma. Let $ABC$ be a triangle with a pair of isogonal conjugate points $P,Q$. Let $X,Y,Z$ be midpoints of sides $BC,CA,AB$, respectively. Then the Euler-Poncelet point of $A,B,C,P$ is the Anti-Steiner point of $OQ$ wrt $\triangle XYZ$.
Proof. See Proposition 20 in the below attached document.

Back to the problem, by considering the homothety $\mathcal{H}(L, 1/2)$, we only need to prove that the Euler circles of the triangles $ALO$, $BLO$, and $CLO$ are concurrent at a point (other than the midpoint of $OL$).
Indeed, let $S$ be the Euler-Poncelet point of $A, B, C, L$, and let $X, Y, Z$ be the midpoints of $BC, CA, AB$. According to the lemma, $S$ is the Anti-Steiner point of Euler line with respect to triangle $XYZ$.
Let $M$ be the reflection of the circumcenter of $XYZ$ across $YZ$. Then $SM$ being the reflection of $OG$ across $BC$, hence,
\[ (SM, SY) \equiv (SM, YZ) + (YZ, YS) \equiv (YZ, OG) + (OG, XY) + \frac{\pi}{2} \equiv (BC, BA) + \frac{\pi}{2} \pmod \pi \]Additionally, let $D$ be the midpoint of $AL$. Since $S$ lies on the Euler circle of triangle $ALC$, , we have
\[ (SY, SD) \equiv (AL, AC) \pmod \pi \]Therefore,
\[ (SM, SD) \equiv (BC, BA) + \frac{\pi}{2} + (AL, AC) \equiv (AC, AO) + (AL, AC) \equiv (AL, AO) \pmod \pi\]This implies that $S$ lies on the Euler circle of $\triangle AOL$. By similar reasoning for the other two triangles, the result follows.
Attachments:
Nine_Point_Circle_Pedal_Circle_and_Cevia.pdf (397kb)
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