The solution I'll present is mainly based on two lemmas and then the use of homothety. Let me remark that the solution heavily uses the problem's symmetry with respect to $P$ and $Q$. Lemma 1: For every $i\in \{ 1, 2, \dots , n \}$ let $P'_i$ be the projection of $P$ on $A_iA_{i+1}$. The points $P'_1, P'_2, \dots , P'_n$ lie on a circle with center $M$, the midpoint of $\overline{PQ}$.' Proof:

Using a homothety with center $P$ and ratio $1/2$, we have that this circle is mapped into the circle that passes through $P'_1, P'_2, \dots , P'_n$ and has $M$, the midpoint of $\overline{PQ}$, as its center. From this lemma and the symmetry of the problem, we can state the following corollary: given the points $Q'_1, Q'_2, \dots , Q'_n$, defined as the projections of $Q$ on $A_{i}A_{i+1}$, they also lie on a circle with center $M$. More than that, like in the previous proof, taking $Y_1, Y_2, \dots Y_n$ like the points $X_1, X_2, \dots , X_n$, but in respect to $Q$, we conclude that the radii of these two circles equals $\frac{QX_i}{2}$ and $\frac{PY_i}{2}$, respectively. Nevertheless, it's easy to verify that the quadrilateral $PQY_iX_i$ is an isosceles trapezoid, which implies that its diagonals have the same measure. So, we have that the center of the two radii are the same, which implies that the two circles are the same. Lemma 2: The triangles $\triangle PP'_{i+1}P'_{i}$, $\triangle QQ'_{i}Q'_{i+1}$, and $\triangle OP_{i}P_{i+1}$ are similar. Proof:

Let's use lemma 1 corollary so can state that the projections of $P$ and $Q$ on $A_iA_{i+1}$ lies all in a circle with center $M$, the midpoint of $\overline{PQ}$. Now let us focus first on $P$ and establish a homothety that proves the existence of $O_P$. Hopefully, we can use this homothety to prove items b) and c) as well. We shall prove the following Lemma 3: Exists an homothety that, for every $i\in \{ 1, 2, \dots , n \}$, maps $\triangle QQ'_{i}Q'_{i+1}$ into $\triangle OP_{i}P_{i+1}$. Proof:

Let us call this homothety $\mathcal{H}_{P}$. By the symmetry of the problem, we can also state: that exists a homothety that, for every $i\in \{ 1, 2, \dots , n \}$, maps $\triangle PP'_{i}P'_{i+1}$ into $\triangle OQ_{i}Q_{i+1}$. Let us call this homothety $\mathcal{H}_{Q}$. a) As $Q'_{1}Q'_{2} \dots Q'_{n}$ is cyclic and is mapped into $P_{1}P_{2} \dots P_{n}$ by $\mathcal{H}_{P}$, we conclude that the the last polygon is also cyclic. By the symmetry of the problem, $Q_{1}Q_{2} \dots Q_{n}$ is cyclic as well. b) We know that \[ \mathcal{H}_{P}(M) = O_{P}, \mathcal{H}_{P}(Q) = O, \mathcal{H}_{Q}(M) = O_{Q}, \mathcal{H}_{Q}(P) = O. \] So, we have that $\overline{O_{P}O} \parallel \overline{MQ}$ and $\overline{O_{Q} O} \parallel \overline{MP}$. As $P$, $Q$, and $M$ are collinear, we conclude that $O_{P}$, $O_{Q}$ and $O$ are collinear as well. c) By what was proved in the previous item, we have that \[ \overline{O_{P}O_{Q}} \parallel \overline{PQ} \] Remark: It's possible to prove item a) only using angle-chasing and the first two lemmas.

Nice Problem! It is well known that isogonal conjugate of a point $P$ exists if and only if the "pedal polygon" of $P$ WRT $A_1A_2 \cdots A_n$ is also cyclic, morover the center of this circle is midpoint of $PQ$, say $R$. Call the foot of altitude from $P$ to $A_iA_{i+1}$; $B_i$, similarly for $Q$, define $C_i$. The following lemma is well known and is proved by angle chasing: Lemma 1: $\triangle PB_iB_{i+1} \sim \triangle QC_iC_{i+1} \sim \triangle OP_iP_{i+1}$, and in fact, $QC_iC_{i+1}$ is homothetic to $OP_iP_{i+1}$ The above lemma is sufficient to establish that $P_1P_2 \cdots P_n \cup O$ and $C_1C_2 \cdots C_n \cup Q$ are homothetic. (a) follows from $P_1P_2 \cdots P_n \cup O$ and $C_1C_2 \cdots C_n \cup Q$ are homothetic. (b,c) the homothety implies that $OO_P,OO_Q \parallel PQ$ which implies both the parts (b,c).

Strikingly similar to USA IMO TST 2025, perhaps?