Let $\ell$ be the radical axis of $\omega_1$ and $\omega_3$. Lemma 1: $\ell$ is $B_2D_2$. Proof: $B_2 \in \ell$, because $B_2A_1 \cdot B_2A_2 = B_2A_3 \cdot B_2A_4$. Let $X$ be the intersection of $B_2D_2$ and $A_2A_3$. Using the power of the point $X$ with the circles $\Omega_2$ and $\Omega_3$, we have $XA_2 \cdot XB_1 = XD_2 \cdot XB_2$ and $XA_3 \cdot XB_3 = XD_2 \cdot XB_2$. Therefore, $XA_2 \cdot XB_1 = XA_3 \cdot XB_3 \implies X \in \ell$, thus, $\ell$ is the line $B_2X$, i.e., the line $B_2D_2$.

This link is useful: https://artofproblemsolving.com/community/q1h1813119p12086815 With the above link, we know that the intersection point of the $B_iD_i$'s exists, call this point $P.$ Then from Lemma $1$ above, we know that $P$ has equal powers w.r.t. $\omega_1$ and $\omega_3$. With five similar results, we're done. $\square$

Why not ask for more? - Let $\Gamma_A = (A_i)$. The points $C_i$ too belong to a circle $\Gamma_C$. - The concurrency point of the problem is precisely the insimilicenter of $\Gamma_A$ and $\Gamma_C$.

All you need is that $C_1C_2C_3C_4C_5$ is cyclic and the concurrency point is going to be the centre of inversion that sends circle $(A_1A_2A_3A_4A_5)$ $\to$ $(C_1C_2C_3C_4C_5)$. Just notice that $C_i$ is the miguel point of quadrilateral $A_iA_{i+1}B_{i+2}A_{i+4}$ and that points $B_iC_iA_{i+4}C_{i+3}B_{i+2}$ all belong on a circle . The rest is simple angle chasing to show that $C_iC_{i+1}C_{i+2}C_{i+3}$ is cyclic.

mela_20-15 wrote: All you need is that $C_1C_2C_3C_4C_5$ is cyclic and the concurrency point is going to be the centre of inversion that sends circle $(A_1A_2A_3A_4A_5)$ $\to$ $(C_1C_2C_3C_4C_5)$. Just notice that $C_i$ is the miguel point of quadrilateral $A_iA_{i+1}B_{i+2}A_{i+4}$ and that points $B_iC_iA_{i+4}C_{i+3}B_{i+2}$ all belong on a circle . The rest is simple angle chasing to show that $C_iC_{i+1}C_{i+2}C_{i+3}$ is cyclic. I don't think $C_1C_2C_3C_4C_5$ being cyclic is enough for the problem. Why are $A_i$ and $C_i$ correlated though this inversion?

Quote: mela_20-15 wrote: All you need is that $C_1C_2C_3C_4C_5$ is cyclic and the concurrency point is going to be the centre of inversion that sends circle $(A_1A_2A_3A_4A_5)$ $\to$ $(C_1C_2C_3C_4C_5)$. Just notice that $C_i$ is the miguel point of quadrilateral $A_iA_{i+1}B_{i+2}A_{i+4}$ and that points $B_iC_iA_{i+4}C_{i+3}B_{i+2}$ all belong on a circle . The rest is simple angle chasing to show that $C_iC_{i+1}C_{i+2}C_{i+3}$ is cyclic. I don't think $C_1C_2C_3C_4C_5$ being cyclic is enough for the problem. Why are $A_i$ and $C_i$ correlated though this inversion? Well given that $C_1C_2C_3C_4$ is cyclic by means of angle chasing you show that $A_1C_1A_3C_3$ is cyclic and then $A_1C_1,A_2C_2,A_3C_3$ concur at a radical center.

Let $\omega_i=(A_iB_iB_{i+1})$ for all $i=1,2,3,4,5$. Since $A_1A_2A_3A_4A_5$ is cyclic, then $B_3= A_4A_5 \cap A_2A_3$ satisfies $B_3A_4.B_3A_5= B_3A_3.B_3A_2 \implies Pow_{\omega_2}B_3=Pow_{\omega_4}B_3 \implies B_3$ lies on the radical axis of $\omega_2, \omega_4 (\star)$ . Let $E_3=A_4C_4 \cap A_3C_3$ and define $E_1,E_2,E_4,E_5$ similarly. Furthermore, let $T= B_3D_3 \cap A_3A_4 \implies TA_4.TB_4=TB_3.TD_3=TA_3.TB_2 \implies Pow_{\omega_2}T=Pow_{\omega_4}T \implies T$ lies on the radical axis of $\omega_2,\omega_4$. From $(\star)$, $B_3T$ is the radical axis of $\omega_2,\omega_4$. Since $D_3 \in B_3T$, $B_3D_3$ is the radical axis of $\omega_2,\omega_4$. Similarly, $B_iD_i$ is the radical axis of $\omega_{i-1},\omega_{i+1}$ for all $i=1,2,3,4,5$, and $B_iD_i, A_iC_i, A_{i+1}C_{i+1}$ are concurrent by radical axis theorem on $\omega_{i-1},\omega_i,\omega_{i+1}$. Therefore, it is enough to prove that all $A_iC_i$ are concurrent.$(\spadesuit)$ Now, observe that $\angle A_1C_2B_2= \angle A_1C_2B_2+ \angle A_2C_2B_2= \angle A_1B_1A_2+ \angle A_4A_3B_1= 180º - \angle A_1B_4B_2 \implies B_4A_1C_2B_2$ is cyclic. Moreover, $\angle B_4C_5A_1= \angle B_4C_5A_5 + \angle A_5C_5A_1= \angle B_5A_4B_2 + \angle A_5B_5A_1= \angle B_5A_4B_2+ \angle A_4B_5B_2= 180º - \angle B_4B_2A_1 \implies A_1B_2B_4C_5$ is cyclic, so $A_1C_2B_2B_4C_5$ is cyclic $(\heartsuit)$. Notice that $\angle A_2A_5C_5= \angle C_5A_5A_4+ \angle A_4A_5A_2$, which from $(\heartsuit)$ is equals to $\angle C_5C_2B_2+ 180º - \angle A_2A_3A_4$. On the other hand, $\angle A_2C_2C_5= \angle A_2C_2B_2- \angle C_5C_2B_2= \angle A_2A_3A_4- \angle C_5C_2B_2$, so $\angle A_2C_2C_5+ \angle A_2A_5C_5= 180º \implies A_2C_2C_5A_5$ is cyclic. Similarly, $A_{i+2}C_{i+2}C_iA_i$ is cyclic for all $i=1,2,3,4,5$, and let these circles be $\Gamma_i$. Thus, by radical center theorem on $\Gamma_i, \omega_i, \omega_{i+1}$, we have that $A_iC_i, A_{i+1}C_{i+1}, A_{i+2}C_{i+2}$ are concurrent for all $i$, so all the $A_iC_i$ are concurrent $\implies$ from $(\spadesuit)$, we are done. $\blacksquare$