Heavy config geo involving mixtilinear

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Heavy config geo involving mixtilinear

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Source: Bulgaria Spring Mathematical Competition 2025 12.4

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#1 h Mar 30, 2025, 1:23 PM • 1 Y

Y by cubres

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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#2 h Mar 30, 2025, 3:03 PM • 2 Y

Y by AlexCenteno2007, cubres

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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#3 h Mar 30, 2025, 6:00 PM • 3 Y

Y by ehuseyinyigit, VicKmath7, cubres

problem gives me vibe

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#4 h Mar 31, 2025, 2:33 PM • 4 Y

Y by VicKmath7, Assassino9931, bo18, cubres

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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