Let $k$ be a circumcircle of triangle $ABC$ $(AC<BC)$. Also, let $CL$ be an angle bisector of angle $ACB$ $(L \in AB)$, $M$ be a midpoint of arc $AB$ of circle $k$ containing the point $C$, and let $I$ be an incenter of a triangle $ABC$. Circle $k$ cuts line $MI$ at point $K$ and circle with diameter $CI$ at $H$. If the circumcircle of triangle $CLK$ intersects $AB$ again at $T$, prove that $T$, $H$ and $C$ are collinear. .
Problem
Source: Bosnia and Herzegovina TST 2016 day 2 problem 2
Tags: geometry, incenter, circumcircle, arc midpoint, Cyclic, angle bisector
mihajlon
16.05.2016 16:01
Let $N$ be midpoint of arc $AB$ not containing $AB$. Knowing that if $A-$mixtilinear incircle touches $AC$ and $BC$ at $X$ and $Y$ it's well known that $KN$, $AB$ and $XY$ concur in this case at $T$, thus if $CT\cap k=\{ D^{'} \}$ by Line perpendicular to line determined by mixtilinear(own...) problem we know that $ID^{'}\perp CT$, but $\angle IDC=90^{\circ}$. $QED$
FabrizioFelen
16.05.2016 16:20
Let $T'=AB\cap CH$ and $N$ the midpoint of the arc $\overarc {AB}$, since $N$, $I$ and midpoint of $CI$ are collinear we get $\odot (CIH)$ and $\odot (IBA)$ are tangent $\Longrightarrow$ by radical axis in $\odot (ABC)$, $\odot (IBA)$ and $\odot (CHI)$ we get $CH$, $AB$ and the tangent to $\odot (IAB)$ by $I$ are concurrent $\Longrightarrow$ $T'I$ is tangent to $\odot (IAB)$ then $\measuredangle T'IL=90^{\circ}$ $\Longrightarrow$ since $MN$ is a diameter of $\odot (ABC)$ we get $\measuredangle IKN=90^{\circ}$, so from $\measuredangle CT'I=\measuredangle CIH=\measuredangle HKI$ $\Longrightarrow$ $HIKT'$ is cyclic hence $T'$, $K$ and $N$ are collinear $\Longrightarrow$ since $CBNK$ is cyclic we get $\measuredangle CLT'=\measuredangle CBN=\measuredangle T'KC$ $\Longrightarrow$ $T'CLK$ is cyclic $\Longrightarrow$ $T'=T$ hence $C$, $H$ and $T$ are collinear.
PROF65
16.05.2016 17:55
Let $D$ the intersection of $CI$ and $(ABC)$ which is the antipode of $M$ . $DKI$ and $CHI$ are right-angle triangles and have their centers on $CI$ so they have common tangent at $I$ say $d$.the radical 's theorem yields $KD,d,HC$ are concurrent at a point say $T'$ ;similarly the radical axes of $(IBA)-(ABC)-(CHI)$ ,which are $BA,CH,d $ concur at $T'$ it remains to prove that $T'=T$ i.e $T'$ is on $(KLC)$ indeed: $ \widehat{T'KC}=\widehat{T'KM}-\widehat{CKM}= \frac{\pi}{2}-\frac{\widehat{A}-\widehat{B}}{2}=\widehat{B}-\frac{\widehat{C}}{2} =\widehat{T'LC}$ i.e $KLT'A$ is cyclic . R HAS
AlastorMoody
10.07.2019 11:04
Let $\Delta DEF $ be the intouch triangle WRT $\Delta ABC$ with $I $ as incenter and $M,M'$ as midpoints of arc $BC $ not containing $A $ and arc $BAC $. Let $AI \cap BC $ $=$ $L $. Let $M'I $ $\cap $ $\odot (ABC) $ $=$ $T $. Let $\odot (AFE) $ $\cap $ $\odot (ABC) $ $=$ $K $. We want to show $GTLA$ cyclic. Obviously $\angle AKI=90^{\circ} $, so if $KI \cap \odot (ABC) $ $=$ $P $, then, $P $ is $A- $antipode. The line through $I $ perpendicular to $AI $ is ofcourse tangent to $\odot (AFE) $ and $\odot (IBC) $. Also, $T $ is the $A- $mixtilinear incircle touch point. By Radical Axes Theorem, $AK,$ $I-\text {tangent} $ and $BC $ are concurrent, say at $G $. Let $N $ be other tangent from $G $ to $\odot (BIC) $. So, from La Hire's Theorem $\implies $ $N $ $\in $ $M'I $. Since, $MT \perp IN $ $\implies $ $IT=NT $ $\implies $ $G \in TM $ . Also, $GKIT $ is cyclic. $$\angle MKI=\angle MAO=\angle AMO=\angle ITD=\angle DKI $$So $K,D,M$ collinear. Let $T_A$ be the $A-$extouch point, so by reflection argument, $T_A$ lies on $MK'$. $\sqrt{bc}$ inversion: $\overline{MT_AK'}$ $\mapsto$ $\odot (GTLA)$.
Miku3D
14.04.2020 14:52
Let $T’ = CH \cap AB$, $N$ be the other intersection of $CI$ with $k$, and $l$ be the line through $I$ perpendicular to $CN$
Claim: Lines $CH$, $l$, $AB$, and $KN$ are concurrent at $T’$
Proof:It is well known by the Incenter-Excenter Lemma that $N$ is the center of $(AIB)$.
Thus, $l$ is a common tangent to $(CHI)$ and $(AIB)$, so $l$ is their radical axis.
Therefore, by Radical Axis Theorem on circles $k$, $(CHI)$, and $(AIB)$, lines $CH$, $l$, and $AB$ are concurrent.
Clearly, $MN$ is a diameter of $k$.
Therefore, $\angle IKN = \angle MKN = 90^{\circ}$
Thus, $l$ is also tangent to $(IKN)$, so it’s also the radical axis of $(IKN)$ and $(CHI)$.
Therefore, by Radical Axis Theorem on circles $k$, $(CHI)$, and $(IKN)$, lines $CH$, $l$, and $KN$ are concurrent.
Therefore, all 4 lines are concurrent.
Since two of them intersect at $T’$, then they must all concur at $T’$
Thus, the Claim is proven $\blacksquare$
Now it suffices to show that quadrilateral $CLKT’$ is cyclic, which is equivalent to $NC \cdot NL = NT’ \cdot NK$
Note that $\angle NIT’ = 90^{\circ} = \angle NKI$, so $NK \cdot NT’ = NI^2$
Also, note that $\angle NCB = \angle NCA = \angle NBA = \angle NBL$, so $NB^2=NL \cdot NC$
Since $N$ is the center of $(AIB)$, then $NI=NB$
Therefore, $$NK \cdot NT’ = NI^2 = NB^2 = NL \cdot NC$$So quadrilateral $CLKT’$ is cyclic.
Thus, $T=T’$ and we have that points $T$,$H$,$C$ are collinear. $\blacksquare$
TheGKilOfQst
03.12.2021 03:49
I have a different solution using projective geometry. Lemma 1: If $T,A,X,P,B$ are collinear points such that $(T,X;A,B)=-1$ and $P$ is the midpoint of $AB$ then $XA.XB=XT.XP$. Proof: It´s easy to prove this using the fact $\frac{TA}{TB}=\frac{XA}{XB}$ and $PA=PB$. q.e.d Let $N$ the midpoint of arc $AB$ not containing $C$, $P$ the midpoint of $AB$, $T_1$ the intersection of $AB$ and a tangent line to $(ABI)$ at $I$ and $X= MI \cap AB$. By angle chasing, we can prove that $MA$ an $MB$ are tangent to $(ABI)$, then $MI$ is symmedian of $\triangle{ABI}$. So, $(T_1,X,A,B)=-1$. Now, since $MA=MB$ and $NA=NB$ $\Rightarrow$ $(M,N,A,B)=-1$. Projecting in $AB$ we get $T_1-K-N$ collinear. And, by radical center in $(ABI),(ABC),(HCI)$ $\Rightarrow$ $H-C-T_1$ are collinear. Futhermore, by Lemma 1, $XK.XM=XA.XB=XP.XT_1$. So, $T_1KPM$ is a cyclic quadrilateral. $\Rightarrow$ $\angle{KT_1P}=\angle{KMP}=\angle{KCN}$. Then, $T_1CLK$ is cyclic. So $T=T_1$ and we are done!