Let $\gamma_1, \gamma_2, \gamma_3$ be mutually externally tangent circles and $\Gamma_1, \Gamma_2, \Gamma_3$ also be mutually externally tangent circles. For each $1 \le i \le 3$, $\gamma_i$ and $\Gamma_{i+1}$ are externally tangent at $A_i$, $\gamma_i$ and $\Gamma_{i+2}$ are externally tangent at $B_i$, and $\gamma_i$ and $\Gamma_i$ do not meet. Show that the six points $A_1, A_2, A_3, B_1, B_2, B_3$ lie on either a line or a circle.
Although the problem look very hard, it's really easy
Let $A,B,C$ and $P,Q,R$ denote center of $\Gamma_1,\Gamma_2,\Gamma_3$ and $\gamma_1,\gamma_2,\gamma_3$ respectively.
WLOG that $\triangle{PQR}$ lie inside $\triangle{ABC}$
Let $(A)\cap (B)=T$, inversion center at $T$ with arbitrary radius $r>0$,
This is what we getSuppose $X\rightarrow X'$ for all object $X$
We get that $(A)'$ and $(B)'$ are two parallel lines, denoted them by $l_1$ and $l_2$ respectively.
Let $d$ denote distance between $l_1$ and $l_2$
Both $(R)'$ and $(C)'$ are two circles tangent to $l_1$ and $l_2$, so they have same radius.
And $(P)'$ is the circle tangent to $l_1,(R)',(C)'$ and similarly for $(Q)'$
Note that $(P)'$ also tangent with $(Q)'$, so both circles have diameter length $\frac{d}{2}$
Note that the six points we want to consider, become two tangent points of $(R)'$ with $l_1,l_2$ and four tangent points of $(P)'$ and $(Q)'$ with $(R)',l_1,l_2$
The rest is very easy, so I will left it to the reader.