IMO Shortlist 2010 G7
 Thanic Nur Samin
 Posts: 176
 Joined: Sun Dec 01, 2013 11:02 am
IMO Shortlist 2010 G7
Three circular arcs $\gamma_1, \gamma_2,$ and $\gamma_3$ connect the points $A$ and $C.$ These arcs lie in the same halfplane defined by line $AC$ in such a way that arc $\gamma_2$ lies between the arcs $\gamma_1$ and $\gamma_3.$ Point $B$ lies on the segment $AC.$ Let $h_1, h_2$, and $h_3$ be three rays starting at $B,$ lying in the same halfplane, $h_2$ being between $h_1$ and $h_3.$ For $i, j = 1, 2, 3,$ denote by $V_{ij}$ the point of intersection of $h_i$ and $\gamma_j$ (see the Figure below). Denote by $\widehat{V_{ij}V_{kj}}\widehat{V_{kl}V_{il}}$ the curved quadrilateral, whose sides are the segments $V_{ij}V_{il},$ $V_{kj}V_{kl}$ and arcs $V_{ij}V_{kj}$ and $V_{il}V_{kl}.$ We say that this quadrilateral is $circumscribed$ if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ are circumscribed, then the curved quadrilateral $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$ is circumscribed, too.
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Re: IMO Shortlist 2010 G7
Lemma : Let two circular arcs $\alpha $ & $\beta$ connect pionts $A,B$. If two circle $ \varpi_1$ & $ \varpi_2$ are tangent to $\alpha $ & $\beta$ ,then their external center of simillitude lies in $AB$.
Proof :Let an external common tangent $l$ of $\alpha $ & $\beta$ meet $AB$ at piont $M$.Then the invertion with center $M$ ,taking $\alpha $ to itself ,takes $\beta$ to itself.Thus it takes $ \varpi_1$ to a circle tangent to $l$ , $\alpha $ & $\beta$ ,but there can be atmost two circles tangent to two circular arcs and a line and lying in the same side of the line . So the invertion takes $ \varpi_1$ to $ \varpi_2$,so $M$ is their external center of simillitude.
Solution :Let $O_1,O_2,O_3$ be the incenters of $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ , respectively. Let $(O_4) $ be the circle tangent to $h_2 ,\gamma _2,\gamma _3$ such that $(O_3),(O_4)$ lies in the sane side of $h_2$. Let $O_1O_3 \cap h_2=X.O_2O_4 \cap h_2=Y$.$O_1O_3 \cap AC=Z$. Then $X$ & $Z$ are the internal and external center of simillitude of $(O_1),(O_3)$.So $(BA,h_2 ;BO_1,BO_3)=1$. Similerly $(BA,h_2 ;BO_2,BO_4)=1$ .As $B,O_1,O_2$ are collinear so $B,O_3,O_4$ are collinear implying that $(O_4)$ is tangent to $h_3$ thus it is the incircle of $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$.
Proof :Let an external common tangent $l$ of $\alpha $ & $\beta$ meet $AB$ at piont $M$.Then the invertion with center $M$ ,taking $\alpha $ to itself ,takes $\beta$ to itself.Thus it takes $ \varpi_1$ to a circle tangent to $l$ , $\alpha $ & $\beta$ ,but there can be atmost two circles tangent to two circular arcs and a line and lying in the same side of the line . So the invertion takes $ \varpi_1$ to $ \varpi_2$,so $M$ is their external center of simillitude.
Solution :Let $O_1,O_2,O_3$ be the incenters of $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ , respectively. Let $(O_4) $ be the circle tangent to $h_2 ,\gamma _2,\gamma _3$ such that $(O_3),(O_4)$ lies in the sane side of $h_2$. Let $O_1O_3 \cap h_2=X.O_2O_4 \cap h_2=Y$.$O_1O_3 \cap AC=Z$. Then $X$ & $Z$ are the internal and external center of simillitude of $(O_1),(O_3)$.So $(BA,h_2 ;BO_1,BO_3)=1$. Similerly $(BA,h_2 ;BO_2,BO_4)=1$ .As $B,O_1,O_2$ are collinear so $B,O_3,O_4$ are collinear implying that $(O_4)$ is tangent to $h_3$ thus it is the incircle of $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$.
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 Albert Einstein