Problem: Let $A_1 A_2 A_3$ be a triangle,
and let $\omega_1$ be a circle in its plane
passing through $A_1$ and $A_2$.
Suppose there exists circles $\omega_2$, $\omega_3$, \dots, $\omega_7$
such that for $k = 2, 3, \dots, 7,$
circle $\omega_k$ is externally tangent to $\omega_{k-1}$ and passes
through $A_k$ and $A_{k+1}$ (indices mod $3$).
Prove that $\omega_7 = \omega_1$.

Solution: The idea is to keep track of the subtended arc
$\widehat{A_i A_{i+1}}$ of $\omega_i$ for each $i$.
To this end, let $\beta = \dang A_1 A_2 A_3$,
$\gamma = \dang A_2 A_3 A_1$
and $\alpha = \dang A_1 A_2 A_3$.

\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(9pt));
pair A2 = (0,0);
pair O1 = (-8,0);
pair O2 = (5,0);
pair A1 = abs(A2-O1)*dir(285)+O1;
pair A3 = abs(A2-O2)*dir(245)+O2;
pair O3 = extension(O2, A3, midpoint(A1--A3), midpoint(A1--A3)+dir(A1-A3)*dir(90));

filldraw(circle(O1,abs(A1-O1)), opacity(0.1)+lightgreen, deepgreen);
filldraw(circle(O2,abs(A2-O2)), opacity(0.1)+lightgreen, deepgreen);
filldraw(circle(O3,abs(A3-O3)), opacity(0.1)+lightgreen, deepgreen);
filldraw(A1--A2--A3--cycle, opacity(0.1)+orange, red);

pair J = (0,-3.2);
draw(O1--O2--O3, deepcyan+dotted);

dot("$O_1$", O1, dir(90), deepcyan);
dot("$O_2$", O2, dir(90), deepcyan);
dot("$O_3$", O3, dir(-90), deepcyan);
dot("$A_1$", A1, dir(A1-J), red);
dot("$A_2$", A2, dir(A2-J), red);
dot("$A_3$", A3, dir(A3-J), red);

label("$\alpha$", A1, 2.4*dir(J-A1), red);
label("$\beta$",  A2, 2*dir(J-A2), red);
label("$\gamma$", A3, 2*dir(J-A3), red);
\end{asy}
\end{center}

Initially, we set $\theta = \dang O_1 A_2 A_1$.
Then we compute
\begin{align*}
  \dang O_1 A_2 A_1 &= \theta \\
  \dang O_2 A_3 A_2 &= -\beta-\theta \\
  \dang O_3 A_1 A_3 &= \beta-\gamma+\theta \\
  \dang O_4 A_2 A_1 &= (\gamma-\beta-\alpha)-\theta \\
\end{align*}
and repeating the same calculation another round gives
\[ \dang O_7 A_2 A_1 = k-(k-\theta) = \theta \]
with $k = \gamma-\beta-\alpha$.
This implies $O_7 = O_1$, so $\omega_7 = \omega_1$.