Let $\gamma_1$ and $\gamma_2$ be two circles tangent at point $T$, and let $\ell_1$ and $\ell_2$ be two lines through $T$. The lines $\ell_1$ and $\ell_2$ meet again $\gamma_1$ at points $A$ and $B$, respectively, and $\gamma_2$ at points $A_1$ and $B_1$, respectively. Let further $X$ be a point in the complement of $\gamma_1 \cup \gamma_2 \cup \ell_1 \cup \ell_2$. The circles $ATX$ and $BTX$ meet again $\gamma_2$ at points $A_2$ and $B_2$, respectively. Prove that the lines $TX$, $A_1B_2$ and $A_2B_1$ are concurrent. ***
Problem
Source: Romania TST 2 2010, Problem 3
Tags: geometry, circumcircle, power of a point, radical axis, geometry proposed
Luis González
26.08.2012 02:01
Inverting the figure with center $T,$ we get the following equivalent problem: $\gamma_1$ and $\gamma_2$ are two parallel lines. $T,X$ are two arbitrary points on its plane. $A,B$ are distinct points on $\gamma_1.$ $TA,TB$ cut $\gamma_2$ at $A_1,B_1$ and $XA,XB$ cut $\gamma_2$ at $A_2,B_2.$ Then we prove that $TX$ is the radical axis of $\odot(TA_1B_2)$ and $\odot(TB_1A_2).$ $XA$ cuts $\odot(TB_1A_2)$ again at $D$ and $XB$ cuts $\odot(TA_1B_2)$ again at $E.$ $\angle TDA_2=\angle TB_1A_2=\angle TBA$ and $\angle TEB=\angle TA_1B_2=\angle TAB$ $\Longrightarrow$ $D$ and $E$ lie on $\odot(TAB)$ $\Longrightarrow$ $\angle XDE=\angle XBA=\angle XB_2A_2$ $\Longrightarrow$ $D,E,A_2,B_2$ are concyclic $\Longrightarrow$ $XE \cdot XB_2=XD \cdot XA_2,$ i.e. $X$ has equal power WRT $\odot(TA_1B_2)$ and $\odot(TB_1A_2)$ $\Longrightarrow$ $TX$ is their radical axis.
ACCCGS8
28.08.2012 13:45
I performed the same inversion as in the above post. Let $XT$ meet $A_2B_2$ at $P$. We want to prove that $TP$ is the radical axis of the circumcircles of $A_2B_1T$ and $B_2A_1T$. Clearly $T$ lies on the radical axis. So we want to prove $P$ has equal powers w.r.t. both circumcircles i.e. $PB_1 \cdot PA_2 = PA_1 \cdot PB_2 \Leftrightarrow \frac{PA_2}{PA_1} = \frac{PB_2}{PB_1} \Leftrightarrow \frac{A_1A_2}{PA_1} =$ $ \frac{B_1B_2}{PB_1} \Leftrightarrow \frac{A_2A}{AX} \cdot \frac{XT}{TP} =$ $ \frac{B_2B}{BX} \cdot \frac{XT}{TP}$. The equality was reached via Menelaus. Now the last statement holds because $AB$ is parallel to $A_2B_2$.
junior2001
24.12.2014 20:04
any solution without inversion ?
IDMasterz
26.12.2014 19:50
^^Why, inversion is just similar triangles
Panoz93
27.07.2015 23:13
very easy for Romanian tst Consider $Y$ the intersection of lines $A_1 B_2$ and $A_2 B_1$. $Y \in TX$ is equivalent to proving that $Y$ belongs to the radical axis of the circumcircles of $ATX$ and $BTX$ . Consider $A_3,B_3 $ the intersection of lines $A_1 B_2$ and $A_2 B_1$ with the circumcircles of $BTX$ and $ATX$ respectively. Furthermore we need to prove $YA_2*YA_3=YB_3*YB_2$. Then $\angle BB_3 B_2=\angle B_1 TB_2=\angle B_1 A_1 B_2 \Leftrightarrow B_1 A_1 || BB_3 $ Similarly $B_1 A_1||AA_3$. Hence $ A,B,A_3,B_3$ lie all in a line parallel to $B_1 A_1$. Therefore $\angle A_3 B_3 A_1=180-\angle B_2 A_1B_1=180 -\angle B_2 A_2 B_1$ and $A_3,B_3,A_1,B_1$ are conclyclic as we wanted.
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