Beatiful results with incenter and angle chasing

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Beatiful results with incenter and angle chasing

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#1 h Aug 3, 2018, 5:25 PM • 4 Y

Y by nguyendangkhoa17112003, Adventure10, Mango247, ehuseyinyigit

Given $\Delta ABC$ inscribed $(O)$ and its incenter $I$ . Let $(I)$ touch $BC,CA,AB$ at $D,E,F$ . $AI \cap BC=J$ , $EF \cap BC=T$ . Let $N$ be the midpoint of arc $BAC$ .

a/ Let $NT \cap (O)=$ { $N,Q$ }. Prove $(T,Q,I,J)$ concyclic
b/ Let $A$ -symmedian intersect $(O)$ , $EF$ at $G,K$ . Prove $(T,K,I,G)$ concyclic

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#2 h Aug 3, 2018, 9:29 PM • 2 Y

Y by Adventure10, Mango247

Part a) Let $N'$ be the point diametrically opposed to $N$ on $(O)$ . Let $D'=\overline{QN'}\cap\overline{BC}$ . Note that the circle with diameter $\overline{TD'}$ is the $Q$ -Apollonius circle of $\bigtriangleup BQC$ , so $(T,D';B,C)=-1=(T,D,B,C)$ , so $D'=D$ . Let $T'$ be the inverse of $T$ wrt. incircle of $\bigtriangleup ABC$ . It is known it lies on segment $AD$ . Since $\angle TQD=90^\circ=\angle TT'D$ , quadrilateral $TQT'D$ is cyclic. Line $BC$ is the image of $(O)$ under the inversion with center $M$ and radius $\overline{MA}$ , hence, quadrilateral $QDJA$ must be cyclic and we conclude that $\angle QTI=\angle QDT'=\angle QDA=\angle QJA=\angle QJI$ the result follows.

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#3 h Aug 3, 2018, 9:54 PM • 2 Y

Y by Adventure10, Mango247

Part b). Quadrilateral $ABGC$ is harmonic. Note that $(NA,NJ;NB,NC)=-1$ , then $N,J,G$ are collinear. Let $J'=\overline{AI}\cap\overline{EF}$ . We get,
$\angle TKG=90^\circ-\angle BAN'+\angle BAG=\angle TJG$ thus, $TKJG$ is cyclic. Moreover, observe that $TQJ'N'$ is cyclic as well. So,
$\angle QTK=\angle QTJ'=\angle QN'J'=\angle QN'A=\angle QGA=\angle QGK$ hereby, $TQKG$ is cyclic. Thus, $TQKJG$ is a cyclic pentagon. Taking into account part a), we conclude that $T,Q,K,I,J,G$ lie on a same circle, which clearly implies the required assertion.

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#4 h Aug 4, 2018, 8:17 AM • 1 Y

Y by Adventure10

Does the red circle named ? I think this problem is beatiful

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#6 h Aug 4, 2018, 9:10 AM • 2 Y

Y by jbaca, Adventure10

My solution for Part a (a bit different from Jafet98)
We have $Q(TD,BC)=-1$ , so we obtained: $QD$ pass through midpoint of arc $BC$ and $\angle TQD=90^o$
$\angle AQD+ \angle AJB=90^o+\angle AQN+\angle AJB=180^o \Rightarrow (A,Q,D,J)$ concyclic
It's well known that $TI \perp AD$ so $\angle AIT=90^o - \angle DAI= 90^o - \angle ADB + AJB = \angle ADC - \angle AQN=\angle NQJ$ ( Hide )
So $T,Q,I,J$ concyclic

This post has been edited 1 time. Last edited by RopuToran, Aug 4, 2018, 9:11 AM

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#7 h Oct 9, 2018, 3:39 PM • 2 Y

Y by Adventure10, Mango247

Nice problem indeed! With the above post, $TQIJ$ cyclic can also be deduced from $TQI = DQA$ using the equal right angles without referencing $TI \perp AD$ .

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