There are three circles $w_1, w_2, w_3$. Let $w_{i+1} \cap w_{i+2} = A_i, B_i$, where $A_i$ lies insides of $w_i$. Let $\gamma$ be the circle that is inside $w_1,w_2,w_3$ and tangent to the three said circles at $T_1, T_2, T_3$. Let $T_iA_{i+1}T_{i+2}$'s circumcircle and $T_iA_{i+2}T_{i+1}$'s circumcircle meet at $S_i$. Prove that the circumcircles of $A_iB_iS_i$ meet at two points. ($1 \le i \le 3$, indices taken modulo $3$) If one of $A_i,B_i,S_i$ are collinear - the intersections of the other two circles lie on this line. Prove this as well.
Problem
Source: 2016 Korea Winter Camp 1st Test #7
Tags: geometry
IsoLyS
25.01.2016 12:05
Ten circles. Crazy. Step 1. $A_{1}, A_{2},S_{1},S_{2}$ are cyclic. Invert $A_{1},A_{2}, T_{1}, T_{2}, S_{1},S_{2}$ by $T_{3}$. Then - $A_{1}'A_{2}' \| T_{1}'T_{2}'$, since $\omega_{3}$ touches $\gamma$ at $T_{3}$ - $A_{1}',T_{2}',S_{2}'$ and $A_{2}',T_{1}',S_{1}'$ are collinear. - $T_{1}',T_{2}',S_{1}',S_{2}'$ are cyclic. So $S_{1}'S_{2}'$ is antiparallel to $T_{1}'T_{2}'$, thus also to $A_{1}'A_{2}'$. Hence $A_{1}'A_{2}'S_{1}'S_{2}'$ is cyclic, as desired. Step 2. $A_{1}B_{1},A_{2}B_{2},A_{3}B_{3}$ are concurrent at $P$, and it is also radical center of $A_iB_iS_i$. (trivial) Step 3. $A_{1}S_{1},A_{2}S_{2},A_{3}S_{3}$ are concurrent at $Q$, and it is also radical center of $A_iB_iS_i$. (trivial by Step 1) $P$ is different from $Q$ and are inside of the circles. Therefore 3 circles are coaxal, sharing $PQ$, as desired.
tbn456834678
26.05.2019 17:56
Inverting about $A_1$ gives a well-known configuration: Let $w_1$ be a circle and $w_2$, $w_3$, two lines intersecting $w_1$ at $A_3$, $B_3$, $A_2$ and $B_2$, respectively. Let $\{B_1\}=w_2 \cap w_3$ and $\gamma$ be the circle inside both the angle $\angle{A_2B_1A_3}$ and $w_1$, tangent to $w_1$, $w_2$ and $w_3$. Let $T_1$, $T_2$ and $T_3$ be the intersections of $\gamma$ with $w_1$, $w_2$ and $w_3$, respectively. The circumcircles of $ \bigtriangleup T_1A_3T_2$ and $ \bigtriangleup T_1A_2T_3$ intersect $T_2T_3$ and each other at $\{T_2,S_2\}$, $\{T_3,S_3\}$ and $\{T_1,S_1\}$, respectively. Prove that $B_1S_1$ and the circumcircles of $\bigtriangleup A_2B_2S_2$ and $ \bigtriangleup A_3B_3S_3$ meet at two points. It is well-known that $S_1$, $S_2$ and $S_3$ are the incircles of $ \bigtriangleup A_2B_1A_3$, $ \bigtriangleup A_2B_2A_3$ and $ \bigtriangleup A_2A_3B_3$. We need to prove that $B_1S_1$ is the radical axis of the circumcircles of $\bigtriangleup A_2B_2S_2$ and $ \bigtriangleup A_3B_3S_3$. It is clear that the power of $B_1$ wrt the two circles is the same, because $B_1A_2\cdot B_2B_1 =B_1A_3\cdot B_1B_3$. Furthermore, $A_2S_3S_2A_3$ lie on a circle centered at the midpoint of the arc $A_2A_3$ of $w_1$, therefore $S_1A_2\cdot S_1S_2=S_1S_3\cdot S_1A_3$, where the conclusion follows.