concurrency wanted, mixtilinear incircle related

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Source: 2022 Yasinsky Geometry Olympiad X-XI advanced p6,, Ukraine

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#1 h Nov 8, 2022, 7:04 PM

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Let $\omega$ be the circumscribed circle of the triangle $ABC$ , in which $AC< AB$ , $K$ is the center of the arc $BAC$ , $KW$ is the diameter of the circle $\omega$ . The circle $\gamma$ is inscribed in the curvilinear triangle formed by the segments $BC$ , $AB$ and the arc $AC$ of the circle $\omega$ . It turned out that circle $\gamma$ also touches $KW$ at point $F$ . Let $I$ be the center of the triangle $ABC$ , $M$ is the midpoint of the smaller arc $AK$ , and $T$ is the second intersection point of $MI$ with the circle $\omega$ . Prove that lines $FI$ , $TW$ and $BC$ intersect at one point.

(Mykhailo Sydorenko)

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#2 h Nov 9, 2022, 1:12 AM • 1 Y

Y by StarLex1

$\gamma$ touches $BC,\omega$ respectively at $E,Z$
Since $\gamma$ is the $B$ -mixtilinear circle of $\Delta ABC$ , we have $W,E,Z$ are colinear.
Therefore, $\Delta WBE \sim \Delta WZB \Rightarrow WB^2=\overline{WZ} \cdot \overline{WE}$
Also, since $WF$ is the tangent line of $\omega$ , we have $WF^2=\overline{WZ} \cdot \overline{WE}$
$\Rightarrow WB=WZ$
Moreover, $W$ is the center of $(IBC)$ , hence $B,F,I,C$ are concyclic, and $W$ is the center of $(BFIC)$ .
$\Rightarrow \widehat{FBI}=\widehat{FCI}$
$\Rightarrow 2\widehat{FBI}=\widehat{FBI}+\widehat{FCI}=\widehat{FBC}-\dfrac{\widehat{B}}{2}+\dfrac{\widehat{C}}{2}-\widehat{FCB}$
Since $KW$ is the perdipencular bisector of $BC$ , we have $\widehat{FBC}=\widehat{FCB}$
$\Rightarrow 2\widehat{FBI}=\dfrac{\widehat{C}-\widehat{B}}{2}=\widehat{KWA}$
$\Rightarrow \widehat{FBI}=\widehat{KWM}$
Since $W$ is the center of $(BFIC)$ , we have $\widehat{WFI}=90^o-\widehat{FBI}$
$\Rightarrow \widehat{WFI}=\widehat{WKM} \Rightarrow \widehat{WFI}=180^o-\widehat{WTM}$
$\Rightarrow W,F,I,T$ are concyclic.
Therefore, by radical axis for $(WFIT),(BFIC),(BCTW)$ we have $FI,BC,WT$ are concurrent.

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#3 h Nov 9, 2022, 1:26 AM • 1 Y

Y by StarLex1

Claim 1: $I,F,B,C$ are concyclic:
Proof: Let $J$ be the center of $\gamma$ .
$X$ is the tangential point of $\gamma$ with $BC$ .
We have: $\overline{JI}.\overline{JB} = JX^2 = JF^2$
$\Rightarrow$ $FJ$ is a tangent of $(IFB)$
$\Rightarrow (FJ,FB) \equiv (IF,IB)\pmod \pi$ .
Moreover, since $FJ\perp KW$ and $KW\perp BC\Rightarrow FJ\parallel BC$
$\Rightarrow (FJ,FB)\equiv (BC,BF)\equiv (CF,CB)\pmod \pi$
$\Rightarrow (IF,IB)\equiv (CF,CB)\pmod \pi$ , or $I,F,B,C$ are concyclic.
Claim 2: $\frac{IC}{IB} = \frac{TC}{TB}$
Proof: $BI,CI$ intersect $(O)$ again at $P,Q$ .
We have: $PQ\perp AI$ and $AI\perp AK\Rightarrow PQ\parallel AK$
$\Rightarrow AKPQ$ is an isosceles trapezoid
$\Rightarrow MP=MQ$ .
Also: $I(TQ,CM) = I(MB,PT)\Rightarrow (TQ,CM) = (MB,PT)\Rightarrow (CM,TQ) = (BM,TP)$
$\Rightarrow \frac{TC}{TM} : \frac{QC}{QM} = \frac{TB}{TM} : \frac{PB}{PM}$
$\Rightarrow \frac{TC}{TB} = \frac{CQ}{BP} = \frac{IC}{IB}$ .
Now, back to our problem: We have $I,F,B,C$ are concyclic and $FB=FC\Rightarrow IF$ is the external bisector of $\angle BIC$
$IF$ intersects $BC$ at $N\Rightarrow \frac{NB}{NC} = \frac{IB}{IC}$
$\Rightarrow \frac{NB}{NC} = \frac{TB}{TC}\Rightarrow TN$ is the external bisector of $\angle BTC$
$\Rightarrow N,T,W$ are collinear. (Q. E. D)

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MikeSO

#4 h Nov 10, 2022, 7:52 AM • 1 Y

Y by Mango247

Thank you for posting my problems.

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