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)
Problem
Source: 2022 Yasinsky Geometry Olympiad X-XI advanced p6,, Ukraine
Tags: geometry, concurrency, concurrent, mixtilinear incircle, mixtilinear
iammocnhan
09.11.2022 04:12
$\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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Hoanglahoang
09.11.2022 04:26
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
10.11.2022 10:52
Thank you for posting my problems.