Generalization of 2011 IMO Problem 6

by TelvCohl, Jul 14, 2017, 11:52 PM

Notation : Given a $\triangle ABC,$ a point $P$ $\in$ $\odot (ABC)$ and a line $\tau,$ let $\triangle A_{\tau}B_{\tau}C_{\tau}$ be the triangle determined by the reflection of $\tau$ in $BC,$ $CA,$ $AB,$ resp. and $P^{-1}$ be the point lying on $\odot (ABC)$ such that the Steiner line of $P^{-1}$ WRT $\triangle ABC$ passes through $P.$

Property : Given a $\triangle ABC$ and a line $\ell$ such that $\ell$ $\cap$ $\odot (ABC)$ $\neq$ $\varnothing .$ Then $\angle (\ell , \odot(ABC))$ $=$ $\angle (\odot (ABC), \odot (A_{\ell}B_{\ell}C_{\ell})).$

Proof :

Lemma 1 : Given a $\triangle ABC,$ a point $T$ $\in$ $\odot (ABC)$ and a line $\ell$ passing through $T.$ Then $T^{-1}$ $\in$ $\odot (A_{\ell}B_{\ell}C_{\ell}).$

Proof : Let $H$ be the orthocenter of $\triangle ABC$ and $X,Y,Z$ be the reflection of $T$ in $BC,$ $CA,$ $AB,$ respectively, then $H,X,Y,Z$ lie on the Steiner line $\mathbf{S}_T$ of $T$ WRT $\triangle ABC.$ From symmetry we get $\measuredangle A_{\ell}YT^{-1}$ $=$ $\measuredangle (HT, \ell)$ $=$ $\measuredangle A_{\ell}ZT^{-1},$ so $T^{-1}$ $\in$ $\odot (A_{\ell}YZ).$ Similarly, $T^{-1}$ lies on $\odot (B_{\ell}ZX),$ $\odot(C_{\ell}XY),$ so $T^{-1}$ is the Miquel point of the complete quadrilateral $\triangle A_{\ell}B_{\ell}C_{\ell}$ $\cup$ $\mathbf{S}_T$ $\Longrightarrow$ $T^{-1}$ $\in$ $\odot (A_{\ell}B_{\ell}C_{\ell}).$ $\qquad \blacksquare$

Corollary 1 (Generalization of 2011 IMO Problem 6) : Given a $\triangle ABC$ and a line $\ell.$ Let $U,V$ be the intersection of $\ell$ with $\odot (ABC).$ Then $\odot (ABC)$ $\cap$ $\odot (A_{\ell}B_{\ell}C_{\ell})$ $=$ $\{ U^{-1}, V^{-1} \}.$

Corollary 2 (2011 IMO Problem 6) : Given a $\triangle ABC$ and a point $T$ $\in$ $\odot (ABC).$ Let $\ell$ be the tangent of $\odot (ABC)$ at $T.$ Then $\odot (ABC)$ and $\odot (A_{\ell}B_{\ell}C_{\ell})$ are tangent to each other at $T^{-1}.$

Lemma 2 (well-known) : Given a $\triangle ABC$ and a line $\ell.$ Then the incenter of $\triangle A_{\ell}B_{\ell}C_{\ell}$ is the pole of the Simson line of $\triangle ABC$ with the direction $\parallel$ $\ell.$

Proof : Let $E,F$ be the intersection of $\ell$ with $CA,$ $AB,$ resp. and $Y,Z$ be the projection of $A_{\ell}$ on $CA,$ $AB,$ respectively. Clearly, $A$ is the incenter or excenter of $\triangle A_{\ell}EF,$ so $AA_{\ell}$ passes through the incenter of $\triangle A_{\ell}B_{\ell}C_{\ell}.$ Note that $YZ$ is the $A_{\ell}$ -midline of $\triangle A_{\ell}EF,$ so $AA_{\ell}$ passes through the pole of the Simson line of $\triangle ABC$ with the direction $\parallel$ $\ell.$ $\qquad \blacksquare$

Back to the main problem :

Let $T$ be one of the intersection of $\ell$ with $\odot (ABC)$ and $\varrho$ be the tangent of $\odot (ABC)$ at $T$ and $T_A, T_B, T_C$ be the reflection of $T$ in $BC,$ $CA,$ $AB,$ respectively. From Lemma 1 we get $T^{-1}$ lies on $\odot (A_{\ell}B_{\ell}C_{\ell}),$ $\odot (A_{\varrho}B_{\varrho}C_{\varrho}).$ Furthermore, by Corollary 2 we know $\odot (ABC), \odot (A_{\varrho}B_{\varrho}C_{\varrho})$ are tangent to each other at $T^{-1},$ so $\angle (\odot(ABC), \odot(A_{\ell}B_{\ell}C_{\ell})) = \angle (\odot(A_{\varrho}B_{\varrho}C_{\varrho}), \odot(A_{\ell}B_{\ell}C_{\ell})) \ . \qquad \qquad (\star)$
From the proof of Lemma 1 we get $\left\{\begin{array}{cc} \triangle T^{-1}T_CT_B \stackrel{+}{\sim} \triangle T^{-1}B_{\ell}C_{\ell} \stackrel{+}{\sim} \triangle T^{-1}B_{\varrho}C_{\varrho} \\\\ \triangle T^{-1}T_AT_C \stackrel{+}{\sim} \triangle T^{-1}C_{\ell}A_{\ell} \stackrel{+}{\sim} \triangle T^{-1}C_{\varrho}A_{\varrho} \\\\ \triangle T^{-1}T_BT_A \stackrel{+}{\sim} \triangle T^{-1}A_{\ell}B_{\ell} \stackrel{+}{\sim} \triangle T^{-1}A_{\varrho}B_{\varrho} \end{array}\right\| \Longrightarrow$ $T^{-1}$ is center of the spiral similarity of $\triangle A_{\ell}B_{\ell}C_{\ell},$ $\triangle A_{\varrho}B_{\varrho}C_{\varrho}.$ Let $I_{\ell}, I_{\varrho}$ be the incenter of $\triangle A_{\ell}B_{\ell}C_{\ell},$ $\triangle A_{\varrho}B_{\varrho}C_{\varrho},$ respectively, then combining $(\star)$ and Lemma 2 we conclude that $\angle (\ell, \odot (ABC)) = \angle (\ell, \varrho) = \angle (T^{-1}I_{\varrho}, T^{-1}I_{\ell}) = \angle (\odot (A_{\varrho}B_{\varrho}C_{\varrho}), \odot(A_{\ell}B_{\ell}C_{\ell})) \stackrel{(\star)}{=} \angle (\odot (ABC), \odot(A_{\ell}B_{\ell}C_{\ell})) \ . \qquad \qquad \blacksquare$

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Interesting!
The last step could also done with $U^{-1}OO' \sim OWU,$ where $O',W$ the circumcenter of $A_lB_lC_l, UVH.$
I have also another shorter solution without using lemma 1,2 :-D

This post has been edited 1 time. Last edited by TelvCohl, Dec 31, 2022, 4:36 AM
Reason: LaTEX

by duanby, Dec 29, 2022, 9:09 PM

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Generalization of 2011 IMO Problem 6

by TelvCohl, Jul 14, 2017, 11:52 PM

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