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Source: 2016 Korea Winter Camp 1st Test #7

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#1 h Jan 25, 2016, 8:08 AM • 1 Y

Y by Adventure10

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.

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#2 h Jan 25, 2016, 10:05 AM • 1 Y

Y by Adventure10

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.

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#3 h May 26, 2019, 2:56 PM • 2 Y

Y by Adventure10, Mango247

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.

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