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Coaxial circles related to Gergon point
Headhunter   1
N 35 minutes ago 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
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internationalnick123456
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Heavy config geo involving mixtilinear
Assassino9931   3
N Mar 31, 2025 by africanboy
Source: Bulgaria Spring Mathematical Competition 2025 12.4
Let $ABC$ be an acute-angled triangle \( ABC \) with \( AC > BC \) and incenter \( I \). Let \( \omega \) be the mixtilinear circle at vertex \( C \), i.e. the circle internally tangent to the circumcircle of \( \triangle ABC \) and also tangent to lines \( AC \) and \( BC \). A circle \( \Gamma \) passes through points \( A \) and \( B \) and is tangent to \( \omega \) at point \( T \), with \( C \notin \Gamma \) and \( I \) being inside \( \triangle ATB \). Prove that:
$$\angle CTB + \angle ATI = 180^\circ + \angle BAI - \angle ABI.$$
3 replies
Assassino9931
Mar 30, 2025
africanboy
Mar 31, 2025
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Heavy config geo involving mixtilinear
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geometry
mixtilinear
tangent circles
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Source: Bulgaria Spring Mathematical Competition 2025 12.4
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Assassino9931
1220 posts
Assassino9931
#1 h  Mar 30, 2025, 1:23 PM
Y by
Let $ABC$ be an acute-angled triangle \( ABC \) with \( AC > BC \) and incenter \( I \). Let \( \omega \) be the mixtilinear circle at vertex \( C \), i.e. the circle internally tangent to the circumcircle of \( \triangle ABC \) and also tangent to lines \( AC \) and \( BC \). A circle \( \Gamma \) passes through points \( A \) and \( B \) and is tangent to \( \omega \) at point \( T \), with \( C \notin \Gamma \) and \( I \) being inside \( \triangle ATB \). Prove that:
$$\angle CTB + \angle ATI = 180^\circ + \angle BAI - \angle ABI.$$
This post has been edited 2 times. Last edited by Assassino9931, Mar 30, 2025, 2:03 PM
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VicKmath7
1386 posts
VicKmath7
#2 h  Mar 30, 2025, 3:03 PM
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Let the mixtilinear incircle touch $CA, CB$ at $E, F$ and $(ABC)$ at $T_c$; let $CI \cap AB, (ABC)=D, M$. It's known that $I$ is the midpoint of $EF$, so $TI$ and $TC$ are isogonal in $\angle ETF$ and thus the angle equality rewrites as $\angle ATE-\angle BTF=\frac{\beta}{2}-\frac{\alpha}{2}$. Due to the tangency we have $\angle ATE=\angle ABT-\angle TFE=\angle ABT-\angle TT_cE$ and similarly $\angle BTF=\angle TAB-\angle TT_cF$. Thus, the angle equality rewrites as $\angle TBA-\angle TAB=\frac{\beta}{2}-\frac{\alpha}{2}+\angle TT_cE-\angle TT_cF$. It's known that $T_cE$ bisects $\angle AT_cC=\beta$ and $T_cF$ bisects $\angle BT_cC=\alpha$, so $$\angle TT_cE-\angle TT_cF=\angle CT_cE+\angle CT_cT-(\angle CT_cF-\angle CT_cT)=\frac{\beta}{2}-\frac{\alpha}{2}+2\angle CT_cT.$$Thus, we need $\angle TBA-\angle TAB=\beta-\alpha+2\angle CT_cT$.

Observe that by radical axes for $(ABC), (ABT), (TEF)$, the tangents at $T, T_c$ to the mixtilinear incircle meet at a point $S$ on $AB$. Let $EF \cap AB=X$, $IT_c \cap AB=U$ and $CT_c \cap EF=V$; it's well-known (follows by Pascal) that $X \in MT_c$ and $IT_c, MT_c$ are internal and external angle bisectors of $\angle AT_cB$. Thus, the circle $UT_cXT$ centered at $S$ is Apollonius circle of $AB$ ($T$ lies on it as $ST_c=ST$). Moreover, $\angle T_cVX=\angle T_cUX=90^{\circ}-\frac{\gamma}{2}+\angle BCT_c$, so $V$ also lies on this Apollonius circle.

Finally, observe that $\angle TSV=2\angle CT_cT$ and $\angle TSA=\angle TBA-\angle TAB$ (as $TX$ is external angle bisector of $\angle ATB$ and $\angle TXA=\frac{\angle TBA-\angle TAB}{2}$), so the angle equality rewrites as $\angle VSA=\beta-\alpha$, or $UV \parallel CI$. But by power of point at $M$ we have that $CDT_cX$ is cyclic, so $\angle UVT_c=\angle UXT_c=\angle DCT_c$, which finishes the problem.

Edit: @2below Yeah I know I overkilled it, but at least it was fun lol. I have done too much config geo, unfortunately.
This post has been edited 2 times. Last edited by VicKmath7, Mar 31, 2025, 3:14 PM
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Assassino9931
1220 posts
Assassino9931
#3 h  Mar 30, 2025, 6:00 PM • 2 Y
Y by ehuseyinyigit, VicKmath7
problem gives me vibe
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africanboy
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africanboy
#4 h  Mar 31, 2025, 2:33 PM • 3 Y
Y by VicKmath7, Assassino9931, bo18
Too easy for 12.4.

Let \(CT \) intersect \( \omega \) at \(J \) and let the mixtilinear incircle touch \(CA, CB \) at \(D, E \), respectively . It's known that \(I \) is the midpoint of \(EF \). Also it's clear that \(DJET \) is harmonic.

\( \angle CTB + \angle ATI = \angle TAB + \angle TEJ + \angle ATI = \angle TAI + \frac{\alpha}{2} + \angle TEJ + \angle ATI = 180^\circ - \angle AIT + \frac{\alpha}{2} + \angle TEJ = 180^\circ - \angle AID - \angle DIT + \frac{\alpha}{2} + \angle TEJ = 180^\circ - \frac{\beta}{2} + \frac{\alpha}{2} = 180^\circ + \angle BAI - \angle ABI \)

We used that \( \angle AID = \angle IDC - \angle CAI = \frac{\alpha}{2} + \frac{\beta}{2} - \frac{\alpha}{2} = \frac{\beta}{2} \).
We also used the well-known property of the harmonic quadrilateral \(DJET \) , which states that \( \angle DIT = \angle TEJ \).
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