does anybody know any similar problems? I liked that very much!

First, if $A_1 \dots A_n$ is cyclic in radius $R$, then pick coordinates so that $A_i = (R\cos 2\theta_i, R\sin 2\theta_i)$ with $0 < \theta_1 < \dots < \theta_n < \pi$. Then for $1 \le i < j \le n$ we have \begin{align*} A_i A_j &= 2R \sin(\theta_j - \theta_i) \\ &= 2R (\sin \theta_j \cos \theta_i - \sin \theta_i \cos \theta_j) \end{align*}and thus we may take the choice \[ (b_j, c_j) = \left( \sqrt{2R} \sin \theta_j, \sqrt{2R} \cos \theta_j \right) \]which works great! The converse direction is by Ptolemy theorem; for any $i < j < k < \ell$ a calculation gives \[ A_i A_j \cdot A_k A_\ell = A_i A_k \cdot A_j A_\ell \]as desired.

Does anyone have a solution for this using complex numbers? This seems like a good candidate for it, given that all points lie on a circle, and the given condition for the assigned real numbers is a determinant.

Note that due to Ptolemy's inequality on $A_iA_jA_kA_l$, where $i<j<k<l$, we have that \[A_iA_j \cdot A_kA_l + A_jA_k \cdot A_iA_l \geq A_iA_k\cdot A_jA_l\]Substituting in our values, \[(b_jc_i-b_ic_j)\cdot(b_lc_k-b_kc_l)+(b_kc_j-b_jc_k)\cdot (b_lc_i-b_ic_l) \geq (b_kc_i-b_ic_k)\cdot (b_lc_j-b_jc_l)\]Expanding \[(b_jb_lc_ic_k-b_jb_kc_ic_l-b_ib_lc_jc_k+b_ib_kc_jc_l)+(b_kb_lc_ic_j-b_ib_kc_jc_l-b_jb_lc_ic_k+b_ib_jc_kc_l)\geq (b_kb_lc_ic_j-b_jb_kc_ic_l-b_ib_lc_jc_k+b_ib_jc_kc_l)\]Cancelling some terms on the LHS, we have \[(-b_jb_kc_ic_l-b_ib_lc_jc_k)+(b_kb_lc_ic_j+b_ib_jc_kc_l)\geq (b_kb_lc_ic_j-b_jb_kc_ic_l-b_ib_lc_jc_k+b_ib_jc_kc_l)\]This is equality, so these values exist iff we are at the equality case of Ptolemy's, so for all $i<j<k<l$ we have that $A_iA_jA_kA_l$ are concyclic. Thus we clearly have that for all $n\geq 4$, $A_n\in (A_1A_2A_3)$, so all the terms lie on the same circle and are concyclic. $\blacksquare$ Now for the construction for concyclic $A_1\ldots A_n$. Assume all points lie on a unit circle, write $A_i = (\cos(\theta_i),\sin(\theta_i)$. Thus \[A_iA_j^2 = (\cos(\theta_i)-\cos(\theta_j))^2+ (\sin(\theta_i)-\sin(\theta_j))^2 = 2-2\cos(\theta_i)\cos(\theta_j) -2\sin(\theta_i)\sin(\theta_j) = 2 -2\cos(\theta_j-\theta_i)\]We may now take $(b_i,c_i) =( \sqrt{2}\sin(\frac{1}{2} \theta_i), \sqrt{2}\cos(\frac12 \theta_j))$ Note that \[b_jc_i-b_ic_j = 2\left( \sin(\frac12 \theta_j)\cos(\frac12 \theta_i) - \sin(\frac12 \theta i)\cos(\frac12 \theta_j) \right) = 2\sin (\frac{\theta_j-\theta_1}{2}) = \sqrt{4 \sin^2(\frac{\theta_j-\theta_1}{2})}\]By double angle formula, we have $\sin^2(x) = \frac{1-cos(2x)}{2}$, using this here gives us that it becomes \[= \sqrt{2-2\cos(\theta_j-\theta_i))}\]Thus, we have $A_iA_j^2 = b_jc_i-b_ic_j$, so we have verified that our construction works! $\blacksquare$

Suppose we can assign such pairs, then we claim that $A_1A_2A_3A_4$ is cyclic. We can later generalize this to all four points, proving that the whole polygon is cyclic. Note that \begin{align*} &~~~~(A_1A_2)(A_3A_4)+(A_2A_3)(A_4A_1)\\ &=(b_2c_1-b_1c_2)(b_4c_3-b_3c_4)+(b_3c_2-b_2c_3)(b_4c_1-b_1c_4)\\ &=b_2b_4c_1c_3-b_2b_3c_1c_4-b_1b_4c_2c_3+b_1b_3c_2c_4+b_3b_4c_1c_2-b_2b_4c_1c_3-b_1b_3c_2c_4+b_1b_2c_3c_4\\ &=-b_2b_3c_1c_4-b_1b_4c_2c_3+b_3b_4c_1c_2+b_1b_2c_3c_4\\ &=(b_3c_1-b_1c_3)(b_4c_2-b_2c_4)\\ &=(A_1A_3)(A_2A_4) \end{align*}as desired. Now, suppose the polygon is cyclic, then let $A_k=r\cos(\theta_k)+ri\sin(\theta_k)$. Then, \begin{align*} A_jA_k^2 &= \left(r\cos(\theta_j)-r\cos(\theta_k)\right)^2+\left(r\sin(\theta_j)-r\sin(\theta_k)\right)^2\\ &= r^2(\cos(\theta_k)^2+\cos(\theta_j)^2-2\cos(\theta_j)\cos(\theta_k)+\sin(\theta_k)^2+\sin(\theta_j)^2-2\sin(\theta_j)\sin(\theta_k))\\ &= 2r^2(1-\cos(\theta_j)\cos(\theta_k)-\sin(\theta_j)\sin(\theta_k))\\ &= 2r^2(1-\cos(\theta_j-\theta_k))\\ &= 2r^2\left(2\sin^2\left(\frac{\theta_j-\theta_k}{2}\right)\right) \end{align*}so we have that \begin{align*} A_jA_k &= 2r\sin\left(\frac{\theta_j-\theta_k}{2}\right)\\ &= 2r\left(\sin\left(\frac{\theta_j}{2}\right)\cos\left(\frac{\theta_k}{2}\right)-\sin\left(\frac{\theta_k}{2}\right)\cos\left(\frac{\theta_j}{2}\right)\right) \end{align*}and we're done.

wow very funny

For the if direction, verify that $$(b_2c_1-b_1c_2)(b_4c_3-b_3c_4) + (b_4c_1-b_1c_4)(b_3c_2-b_2c_3) = (b_3c_1-b_1c_3)(b_4c_2-b_2c_4).$$So Ptolemy's theorem finishes. For the only if direction, notice that by scaling, we may assume that the circumcircle has radius exactly $1$. Then for each $A_i$ with argument $\theta_i$, take $b_i = \sqrt 2 \sin\left(\frac{\theta_i}2\right)$ and $c_i = \sqrt 2 \cos\left(\frac{\theta_i}2\right).$