definitely should have been swapped with problem 2 Sketch: Define $P_i^0$, which is the reflection of $P$ over the sides. These points form a cyclic polygon centered at $Q$ since $P$ and $Q$ are isogonal conjugates. By $\sqrt{bc}$ inversion, the lengths $PP_i^{2025}$ is equal to $PP_i^1\prod_j\frac{PP_j^1}{PP_j^0}$ and corresponding angles of $\mathcal P$ and the polygon formed by $P_i^0$ are also equal, so they are similar with spiral center at $P$. If the scale factor is $c_P$ and the similarity angle is $\theta_P$, the center is $(q-p)c_P\mathrm{cis}\theta_P+p$, so $O_PO_Q$ divided by $p-q$ is real if and only if $c_P\mathrm{cis}\theta_P+c_Q\mathrm{cis}\theta_Q$ is real. Note that $\frac{PP_j^1}{PP_j^0}$ only depends on $\angle PA_jA_{j+1}$, so $c_P=c_Q$ since $P$ and $Q$ are isogonal conjugates. Similarly, $\theta_P=-\theta_Q$, so the imaginary parts cancel out. Remark: instead of dealing with the lengths $PP_i^j$ and arguments of those segments separately, you can deal with them at the same time just by considering the complex number $p_i^j-p$.

When reading this sol please draw a diagram on paper :pleading_face: (just do a pentagon, start by drawing $P$ and the pedal polygon/circle first) Note that the condition is equivalent to $P$ and $Q$ being isogonal conjugates. Thus $P$ and $Q$ have pedal circles. Claim. $\measuredangle PP_i^{n+1} P_{i+1}^{n+1} = \measuredangle PP_i^n P_{i+1}^n.$ Proof. Take a homothety with scale factor $2$ at $P$ sending $P_j^{k}$ to $P_j^{\ast k}$. Then we have $P_{i+1}^n-P_i^{\ast(n+1)}-P_{i+1}^{\ast(n+1)}$ by perpendicularity spam, and furthermore one obtains \[ \measuredangle PP_i^{n+1} P_{i+1}^{n+1} = \measuredangle PP_i^{\ast(n+1)} P_{i+1}^{\ast(n+1)} = \measuredangle PP_i^{\ast(n+1)} P_{i+1}^{n} = \measuredangle PP_i^n P_{i+1}^n, \]where the last equality holds because $PP_i^nP_{i+1}^nP_i^{\ast(n+1)}$ is cyclic. $\square$ One can similarly angle chase the claim $\measuredangle PP_{i+1}^{n+1}P_i^{n+1} = \measuredangle PP_{i+2}^nP_{i+1}^n$ by doing the exact same thing as above. Now define $P^0_i$ to be the reflection of $P$ over $A_iA_{i-1}$. Note that $P^0_i$ are cyclic because $P$ has a pedal circle. The above claims still hold. Hence inducting down we have \[ \measuredangle PP_i^{2025}P_{i+1}^{2025} = \measuredangle PP_i^{0}P_{i+1}^0 \]and \[ \measuredangle PP_{i+1}^{2025}P_{i}^{2025} = \measuredangle PP_{i+2026}^{0}P_{i+2025}^0 = \measuredangle PP_{i+1}^{0}P_{i}^{0}, \]hence it's easy to verify that $\triangle PP_i^{2025}P_{i+1}^{2025} \overset{+}{\sim} \triangle PP_i^0P_{i+1}^0$. By combining a bunch of positive similarities, one concludes that $P_1^0P_2^0\dots P_{2025}^0 \sim \mathcal P$, hence $\mathcal P$ is cyclic. Similarly $\mathcal Q$ is cyclic. For part b, note the length equalities \[ PP_{i}^{n+1} = \frac{PP_i^n}{2 \sin (\angle PP_{i+1}^n P_i^n)} = \frac{PP_{i+1}^n}{2 \sin (\angle PP_{i}^nP_{i+1}^n)}. \]The same length equalities hold for $Q$ and $Q_j^k$ instead of $P$. One determines then, by inducting the claims in part a and just expanding the sine product in the denominator, that \[ \frac{PP_i^{2025}}{PP_i^0} = \frac{1}{2^{2025} \prod_{k=1}^{2025} \sin(\angle PP_{k+1}^0P_k^0)} = \frac{1}{2^{2025} \prod_{k=1}^{2025} \sin(\angle PP_{k}^0P_{k+1}^0)}. \]The same equalities hold for $Q$. However one can check that \[ \measuredangle QQ_k^0 Q_{k+1}^0 = \measuredangle QA_{k-1}A_k = \measuredangle PA_{k-1}A_{k-2} = \measuredangle PP_{k-1}P_{k-2}, \]but this is just a 'reflection' and a 'constant shift', so the sine products for $Q$ are actually equal. Hence we obtain \[ \frac{PP_i^{2025}}{PP_i^0} =\frac{QQ_i^{2025}}{QQ_i^0}, \]so the ratios of similarity of $(\mathcal P)$ and $(\mathcal Q)$ to the pedal circle of $P$ are the same. Now WLOG we may assume the $A_i$ are labeled clockwise. We consider \[ \sum_{i = 0}^{2024} \angle P_1^iPP_1^{i+1}. \]Using the claims in part a and a LOT of straightforward angle chasing one verifies that this sum becomes \[ \sum_{i=0}^{2024} \angle P_i^0PA_i. \]This gives us the rotation angle of the spiral similarity sending the $P_i^0$ to $\mathcal P$. However one notices that $\angle Q_i^0QA_i = 90^\circ - \angle QA_iA_{i-1} = 90^\circ - \angle PA_iA_{i+1} = \angle P_{i+1}^0PA_i$. Thus the corresponding spiral similarity for $Q$ has the negative angle, since this every angle that isn't accounted for by $\angle P_i^0PA_i$. Since the ratios are the same, and $\angle O_PPQ = \angle PQO_Q$, it's easy to see $O_P$ and $O_Q$ form an isosceles trapezoid with $P$ and $Q$, hence we are done. Some motivational remarks. The solution I gave is not magical. Firstly $P$ and $Q$ isogonal conjugates is given, so one immediately goes "yeah there's probably a pedal circle here." So you construct the pedal circle. The next thing to notice is that the circumcenter construction is kinda annoying, so you use antipodes instead; turns out what you get going from pedal polygon to polygon to $P^1$ to $P^2$ at that point is the same thing. The angle chases thus must work. There should be a gut feeling after doing the first angle chase that the angles "cycle around" as you construct more and more layers. I wasn't sure if it was going to work out to be exactly 2025, but it'd be kinda stupid if it didn't, so I trusted the process. This gives you part a). Part b) was just "well if it's a spiral similarity, it's probably the case that $(\mathcal P)$ and $(\mathcal Q)$ are closely related. So you just straight up compute the length ratio and the angle! It dies.

Very nice, here's my solution: Let $n = 2025$. Let $P_i^0,Q_i^0$ be the reflections of $P,Q$ in $A_{i-1}A_i$. Let $\mathcal{P}^i,\mathcal{Q}^i$ be the $P^i_{1}P^i_{2}\dotsm P^i_n, Q^i_{1}Q^i_{2}\dotsm Q^i_n$. You can angle chase to get that $(P,\mathcal{P}^0)$ and $(P,\mathcal{P})$ are similar figures and similarly for $Q$. Hence, as $\mathcal{P}^0$ is cyclic with circumcenter $Q$, we get that $\mathcal{P}$ is cyclic and $\triangle PP_1^0P_1^n\overset{+}{\sim} \triangle PQO_P$ and similar for $Q$. Observe that inverting $\mathcal{P}^0$ at $P$ gives a polygon similar to $\mathcal{Q}^0$ by angle chase. Now consider the transformation of inverting at $P$ and then compose it with a spiral similarity which itself sends $P$ to $Q$ and the inverted $\mathcal{P}^0$ to $\mathcal{Q}$. By induction, this sends $\mathcal{P}^i$ to $\mathcal{Q}^{n-i}$. So, this sends $P_1^0,P_1^n$ to $Q_1^n,Q_1^0$. Hence, $\triangle PQO_P\overset{+}{\sim} \triangle \triangle PP_1^0P_1^n\overset{-}{\sim} \triangle QQ_1^0Q_1^n \overset{+}{\sim} \triangle QPO_Q$ giving that $PQO_QO_P$ is an isosceles trapezium and we are done!

Isn't it the 2024 Brazil Olympic Revenge P3??

My problem! As you may imagine, I did not come up with this statement with $O(n^2)$ points on the spot. Initially I was exploring the idea of "repeatedly taking perpendicular bisectors" for the case $n = 3$. This turns out to yield a very rich configuration that I synthesized in the following problem. Prequel wrote: Let $P$ and $Q$ isogonal conjugates inside triangle $\Delta$. The perpendicular bisectors of the segments joining $P$ to the vertices of $\Delta$ form triangle $\mathcal{P}_1$. The perpendicular bisectors of the segments joining $P$ to the vertices of $\mathcal{P}_1$ form triangle $\mathcal{P}_2$. Similarly, construct $\mathcal{Q}_1$ and $\mathcal{Q}_2$. Let $O$ be the circumcenter of $\Delta$. Prove that the circumcenter of $\triangle OPQ$ is the radical center of the circumcircles of triangles $\Delta$, $\mathcal{P}_2$ and $\mathcal{Q}_2$. I was really happy with this question, and was looking forward to proposing it to the TST. However, pieater314159 pointed out that the configuration was reminiscent of RMM Shortlist 2023 G3 which in the above notation states the following. RMM Shortlist 2023 G3 wrote: Show that the perpendicular bisector of $PQ$ is the radical axis of the circumcircles of $\Delta$ and $\mathcal{P}_1$ and $\mathcal{Q}_1$. After thinking about this, I decided that students who had seen it would have an "unfair" advantage and decided to not propose my problem to any high stakes competition. Nevertheless, I really recommend trying the prequel (especially if you enjoyed G3). I currently only know one solution, and would be interested in seeing whether there are other ways to think about it. I've created a new thread for it: https://artofproblemsolving.com/community/c6h3479111_geometry Eventually I realized there was nothing special about the case $n=3$. After refining my thoughts, I proposed the following statement to the TST. Original Proposal wrote: Let $\Omega$ be a convex 2025-gon, and $P$ and $Q$ isogonal conjugates inside it. Construct the following sequences of polygons: $\mathcal{P}_1=\Omega$, and for $i\geq 2$, $\mathcal{P}_i$ is the 2025-gon formed by the perpendicular bisectors of the segments joining $P$ to each of the 2025 vertices of $\mathcal{P}_{i-1}$. $\mathcal{Q}_1=\Omega$, and for $i\geq 2$, $\mathcal{Q}_i$ is the 2025-gon formed by the perpendicular bisectors of the segments joining $Q$ to each of the 2025 vertices of $\mathcal{Q}_{i-1}$. Prove $\mathcal{P}_{2025}, \mathcal{Q}_{2025}$ are cyclic, and the line through their circumcenters is parallel to $PQ$. Here is the solution I included in my submission. The official solution packet features nice figures that accompany the explanation.

I think it is very hard to anticipate the difficulty of an ad hoc problem like this one. In my proposal I described it as a medium geometry problem that could be unexpectedly hard. The reviewers seemed to agree that the difficulty was somewhere between IMO 2/3, and considering the unusually hard P1, I was not that surprised to see my problem used as P3.