When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

Clearly, $\omega_i$ is determined by its center $O_i$, since we know that it passes through $A_i,A_{i+1}$. We also know that for all $i,\ O_i,A_{i+1},O_{i+1}$ are collinear. We must prove that in these conditions, $O_7=O_1$. Suppose we prove that for all $i$, the angles $\angle OA_iO_i$ and $\angle OA_iO_{i+2}$ are equal (where $O$ is the circumcenter of $A_iA_{i+1}A_{i+2}$). The result would then follow, since $O_i\mapsto O_{i+3}$ would be an involution on the perpendicular bisector $d_i$ of $A_iA_{i+1}$, mapping a point on this line to its harmonic conjugate wrt $O$ and the intersection of the tangents in $A_i,A_{i+2}$ to the circumcircle of $A_iA_{i+1}A_{i+2}$. Since $O_1\mapsto O_4\mapsto O_7$, we must have $O_7=O_1$, as desired. All that’s left to prove is: Let $O_1$ be any point on $d_1$, and put $O_2=O_1A_2\cap d_2,\ O_3=O_2A_3\cap d_3$. Then, $A_1O_1,\ A_1O_3$ make equal angles with $OA_1$. This, however, is a simple angle chase .

MithsApprentice wrote: Let $ A_1A_2A_3$ be a triangle and let $ \omega_1$ be a circle in its plane passing through $ A_1$ and $ A_2.$ Suppose there exist circles $ \omega_2, \omega_3, \dots, \omega_7$ such that for $ k = 2, 3, \dots, 7,$ $ \omega_k$ is externally tangent to $ \omega_{k - 1}$ and passes through $ A_k$ and $ A_{k + 1},$ where $ A_{n + 3} = A_{n}$ for all $ n \ge 1$. Prove that $ \omega_7 = \omega_1.$ Sorry to revive an old topic, but I have a rather different solution than the one posted previously:

E^(pi*i)=-1 wrote: Since $ \omega_k$ and $ \omega_{k + 1}$ are externally tangent at $ A_{k + 1}$, we see that $ A_{k + 1}$ is the midpoint of $ O_k$ and $ O_{k + 1}$. Why? darij

Oops, you're right, it's wrong. It seems like there should be a constant ratio between radii of circles through $ A_1$ and $ A_2$ and those through $ A_2$ and $ A_3$, though . . . maybe there is a way to multiply the coordinates by these factors such that it works out?

Hmm I think you can do this with just trivial angle chasing: $ A=\angle{A_{3}A_{1}A_{2}},B=\angle{A_{1}A_{2}A_{3}},C=\angle{A_{2}A_{3}A_{1}},X=\angle{O_{1}A_{1}A_{2}}$. Then, measures of the angles $ \angle{O_{i}A_{i}A_{i+1}}$ are easy to get, we find that $ \angle{O_{7}A_{1}A_{2}}=X$, and we also need $ O_{1},O_{7}$ on the same side of $ A_{1}A_{2}$ (which follows from the fact that we need $ A_{i}$ between $ O_{i+1}$ and $ O_{i+2}$, blah.

We let the radius of the circle $\omega_i$ by $r_i$.As usual denote the side opposite to vertex $A_i$ by $a_i$ for $i=1,2,3$. Let $O_1A_2A_1=\theta$,where $O_1$ is the centre of $\omega_1$. By applications of sine rule we may easily deduce the following relations: $\frac{r_1}{r_2}=-\frac{a_1}{a_3}\frac{\cos\theta}{\cos(A_2+\theta)}$ $\frac{r_2}{r_3}=-\frac{a_2}{a_1}\frac{\cos(A_2+\theta)}{\cos(A_2-A_3+\theta)}$ $\frac{r_3}{r_4}=-\frac{a_3}{a_2}\frac{\cos(A_2-A_3+\theta)}{\cos(A_1-A_2+A_3-\theta)}$ Multiplying these out we get $\frac{r_1}{r_4}=-\frac{\cos\theta}{\cos(A_1-A_2+A_3-\theta)}$. Similarly it is easy to observe that $\frac{r_4}{r_7}=-\frac{\cos(A_1-A_2+A_3-\theta)}{\cos(A_1-A_2+A_3-(A_1-A_2+A_3-\theta))}=-\frac{\cos(A_1-A_2+A_3-\theta)}{\cos\theta}$. Multiplying these equalities we obatain $r_7=r_1$.Also because $7 \equiv 1\pmod{3}$ we note that $\omega_7$ passes through $A_1$ and $A_2$.We thus obtain $\omega_7=\omega_1$.

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 $\beta = \measuredangle A_1 A_2 A_3$, $\gamma = \measuredangle A_2 A_3 A_1$ and $\alpha = \measuredangle A_1 A_2 A_3$. [asy][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)), invisible, deepgreen); filldraw(circle(O2,abs(A2-O2)), invisible, deepgreen); filldraw(circle(O3,abs(A3-O3)), invisible, deepgreen); filldraw(A1--A2--A3--cycle, invisible, 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); [/asy][/asy] Initially, we set $\theta = \measuredangle O_1 A_2 A_1$. Then we compute \begin{align*} \measuredangle O_1 A_2 A_1 &= \theta \\ \measuredangle O_2 A_3 A_2 &= -\beta-\theta \\ \measuredangle O_3 A_1 A_3 &= \beta-\gamma+\theta \\ \measuredangle O_4 A_2 A_1 &= (\gamma-\beta-\alpha)-\theta \\ \end{align*}and repeating the same calculation another round gives \[ \measuredangle O_7 A_2 A_1 = k-(k-\theta) = \theta \]with $k = \gamma-\beta-\alpha$. This implies $O_7 = O_1$, so $\omega_7 = \omega_1$.

I think you can do this with moving points, but here is the angle chase approach. Let $O_i$ be the center of $\omega_i$ for $i=1,2,\dots,7$. Observe \[\measuredangle O_7A_1A_2=-\measuredangle A_2A_1A_3-\measuredangle A_3A_1O_6=-\measuredangle A_2A_1A_3-\measuredangle O_6A_3A_1=-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2+\measuredangle A_2A_3O_5=\]\[-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2+\measuredangle O_5A_2A_3=-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2-\measuredangle A_3A_2A_1-\measuredangle A_1A_2O_4.\]A similar angle chase finishes.

Let $O_i$ be the center of $\omega_i$. Also let $\theta_1=\measuredangle A_1A_2A_3,\theta_2=\measuredangle A_2A_3A_1,\theta_3=\measuredangle A_3A_1A_2$. Observe that $\omega_1$ is uniquely determined by $\measuredangle O_1A_2A_1$, and $\omega_7$ is uniquely determined by $\measuredangle O_7A_2A_1$, so if we can show those are equal we're done. Now, let $\measuredangle O_1A_2A_1=\theta$. By basic angle chasing (since $O_i,A_{i+1},O_{i+1}$ are collinear), we can find: \begin{align*} \measuredangle O_1A_2A_1&=\theta\\ \measuredangle O_2A_3A_2&=-\theta_1-\theta\\ \measuredangle O_3A_1A_2&=\theta_1-\theta_2+\theta\\ \measuredangle O_4A_2A_1&=-\theta_1+\theta_2-\theta_3-\theta\\ \measuredangle O_5A_3A_2&=\theta_1-\theta_2+\theta_3-\theta_1+\theta\\ \measuredangle O_6A_1A_2&=-\theta_1+\theta_2-\theta_3+\theta_1-\theta_2-\theta\\ \measuredangle O_7A_2A_1&=\theta_1-\theta_2+\theta_3-\theta_1+\theta_2-\theta_3+\theta=\theta. \end{align*}hence $\omega_1=\omega_7$ as desired. $\blacksquare$

Interesting problem. We will aply an inversion on $\Gamma$ with center $A_1$ and ratio $R$ (it's not important). Define $Inv _{\Gamma} (\omega _i)=\Omega _i$ and $Inv _{\Gamma} (A_i)=B_i$. Note that $\Omega _i$ is a line for $i=1, 3, 4, 6, 7$ because $\omega _i$ pass through $A_1$. Also note that $\Omega _i$ and $\Omega _{i+1}$ are parallels for $i=3, 6$ because $\Omega _i$ and $\Omega _{i+1}$ are tangents in $A_1$. We want that $\Omega _1=\Omega _7$, it's sufficient to prove that $\Omega _1 \parallel \Omega _6$. Now we will use directed angles. Let $\ell=B_2B_3$: $$\measuredangle (\Omega _1,\ell)=\measuredangle(\ell,\Omega _3)=\measuredangle(\ell,\Omega _4)=\measuredangle (\Omega _6,\ell).$$ So we have that $\Omega _1$ and $\Omega _6$ are parallels as desired.

Not sure if this works. Through some manipulations we find that it suffices to show $\angle O_1A_2O_4=\angle O_7A_1O_4$, but this is true because $$\angle O_1A_2A_4=\angle O_5A_2O_2=\angle O_5A_3O_2=\angle O_3A_3O_6=\angle O_3A_1O_6=\angle O_7A_1O_4$$

wait i think this works only when using directed angles (because $A_1O_4$ bisecting $\angle O_1A_1O_7$ is possible)

Let $O_i$ be the center of $\omega_i$. Define $\angle A_3A_1A_2=\alpha,\angle A_1A_2A_3=\beta,\angle A_2A_3A_1=\gamma.$ Define $\angle A_iA_{i+1}O_i=\angle A_{i+1}A_iO_i=\theta_i.$ It suffices to show $\theta_1=\theta_7.$ In fact, note that \begin{align*} \theta_2 &= 180^\circ-\beta-\theta_1 \\ \theta_3 &= 180^\circ-\gamma-\theta_2 \\ &= \beta-\gamma +\theta_1 \\ \theta_4 &= 180^\circ - \alpha -\theta_3 \\ &= 180^\circ - \alpha - \beta + \gamma - \theta_1 \\ &= 2\gamma - \theta_1 \end{align*}Note that this process is the exact same for $\theta_4\to \theta_7$, and since function $f(k)=2\gamma -k$ is cyclic with order $2$, we are done.

Let $\theta=\angle A_1O_1A_2$, where $O_i$ is the center of $\omega_i$. Furthermore, let $\alpha, \beta, \gamma$ be the angles of the triangle. Then, check that we have the sequence $$\theta \to 2\beta - \theta \to \theta+2\gamma-2\beta \to 2\alpha-2\gamma+2\beta-\theta=360^\circ-4\beta-\theta.$$Upon applying another sequence of three transformations, this returns the identity $\theta$, so the centers of $\omega_1$ and $\omega_7$ coincide, which suffices.

Define $O_{k}$ as the centers of each respective circle. Let $\alpha = \angle{O_{1}A_{2}A_{1}}$. \begin{align*} \angle{A_{2}O_{4}A_{1}} &= 180-\angle{A_{1}}-\angle{O_{3}A_{1}A_{3}} \\ &= 180-\angle{A_{1}}-(180-\angle{A_{3}-\angle{A_{2}A_{3}O_{2}}}) \\ &= \angle{A_{3}}+\angle{A_{2}A_{3}O_{2}}-\angle{A_{1}} \\ &= \angle{A_{3}}+180-\angle{A_{2}}-\alpha-\angle{A_{1}} \end{align*}We can apply this transformation twice to get that, \[\angle{A_{2}A_{1}O_{7}} = \alpha\implies \omega_{1} = \omega_{7}\]

Consider if we extend the internal tangent of all of the pairs of circles $\omega_{k-1}$ and $\omega_k$ for $k = 2, 3, \dots, 8$. Name these $\ell_1, \ell_2, \dots \ell_7$. Since $\ell_x$ is just $\ell_{x-1}$ reflected over the perpendicular bisecter of some side, and because the internal tangent is not dependent on the circle (it is in fact what makes the next circle), we just have to prove that $\ell_1 = \ell_7$. Call $\angle A_2A_1A_3$ $a$, $\angle A_3A_2A_1$ $b$, and $\angle A_1A_3A_2$ $c$ We know that the angle formed by $A_1$, $A_2$, and $\ell_1$ is less than $a$ (otherwise $\omega_2$ does not exist). Thus we call this $x$ and we know that the angle formed by $\ell_1$, $A_1$, and $A_3$ is $a-x$ we can see that this is equal to the angle formed by $A_1$, $A_3$, and $\ell_2$, so we know that the angle formed by $\ell_2$, $A_3$, $A_2$ is $b-a+x$. Continuing this pattern for subsequent angles we find that the next angles are $180-2b-x$, $a+2b+x - 180$, $180 - a - b - x$, and $x$. Thus we find that $\ell_7 = \ell_1$ and we are done. $\blacksquare$

For $1 \le i \le 7$, let $O_i$ be the center of $\omega_i$ and $\theta_i = \measuredangle O_iA_iA_{i + 1} = \measuredangle A_iA_{i + 1}O_i$. Then for $2 \le i \le 7$, the tangency condition tells us that $O_{i - 1}, A_i$, and $O_i$ are collinear, so $$\theta_i = \measuredangle O_iA_iA_{i + 1} = \measuredangle O_{i - 1} A_iA_{i + 1} = \measuredangle O_{i - 1}A_iA_{i - 1} + \measuredangle A_{i - 1}A_iA_{i + 1} = -\theta_{i - 1} + \measuredangle A_{i - 1}A_iA_{i + 1}.$$Therefore, applying this identity in succession, we have $$\theta_7 = \theta_1 + \sum_{i = 2}^7 (-1)^{7 - i} \measuredangle A_{i - 1}A_iA_{i + 1} = \theta_1 + \sum_{i = 2}^4 ((-1)^{7 - i} + (-1)^{4 - i})(\measuredangle A_{i -1}A_iA_{i + 1}) = \theta_1.$$Thus, $\measuredangle O_1A_1A_2 = \measuredangle O_7A_1A_2$ and $\measuredangle A_1A_2O_1 = \measuredangle A_1A_2O_7$. This is enough to imply that $O_1 = O_7$ since we would otherwise have $\angle O_1A_1A_2 + \angle O_7A_1A_2 = 180^{\circ}$ (using non-directed angles), which is impossible since $\angle O_1A_1A_2, \angle O_7A_1A_2 < 90^{\circ}$. So, we are done.

Cute problem. Let $O_i$ denote the center of each circle $\omega_i$. Let $\ell_1 , \ell_2$ and $\ell_3$ denote respectively the perpendicular bisectors of segments $BC , AB$ and $AC$. Note that for all $k = 2,3, \dots, 7$, the center $O_k$ of $\omega_k$ lies on the line $\ell_{k-1 \pmod{3}}$. First, note that since $O_k$ is a center of a circle passing through points $A_{k+1}$ and $A_{k}$, it must lie on $\ell_{k \pmod{3}}$. Further, since it is externally tangent to $\omega_{k-1}$, which clearly passes through $A_k$ as well, $O_k$ also lies on $\overline{O_kA_{k \pmod{3}}}$. Now, with this information in hand, it suffices to show that point $O_1 , A_1$ and $O_6$ are collinear since then $O_7 = O_1$ and thus $\omega_7 = \omega_1$ as desired. To see why, simply note that \[\measuredangle O_1A_1O_4 = \measuredangle O_4A_2O_1 = \measuredangle O_5A_2O_2 = \measuredangle O_2A_3O_5 = \measuredangle O_3A_3O_6 = \measuredangle O_6A_1O_3\]which implies that $O_1 , A_1$ and $O_6$ are collinear finishing the proof.