Really beautiful problem, but nearly impossible to solve under contest conditions. Here's what I got over the weekend with some hints (and a much nicer diagram!). For convenience, let $n = 100$ and let $\mathcal{X}$ denote the polygon $X_1X_2\dots X_n$. Furthermore, let $R_i = \overline{PP_i}\cap \overline{P_{i-1}P_{i+1}}$ and $T_i = \overline{P_iP_{i+2}}\cap \overline{P_{i-1}P_{i+1}}$ for each $i$. Note that $P_iR_iR_{i+1}P_{i+1}$ is cyclic with diameter $P_iP_{i+1}$ for each $i$, and $P$ is their common radical center. Hence inversion at $P$ with power $PP_i\cdot PR_i$ sends $\mathcal P$ to $\mathcal R$, which implies that $\mathcal R$ is cyclic with circumcircle $\Gamma$. Here's a diagram with $n = 7$: [asy][asy] size(370); defaultpen(fontsize(10pt)); pair P, O, P1, P2, P3, P4, P5, P6, P7; pair Q1, Q2, Q3, Q4, Q5, Q6, Q7; pair Q, R1, R2, R3, R4, R5, R6, R7; pair X, T1, T2, T3, T4, T5, T6, T7; pair D1, D2, D3, H; O = (0,0); P = 0.1*dir(110); P1 = dir(125.15); P3 = dir(220); P2 = intersectionpoint(P--(2*foot(P, P1, P3)-P), unitcircle); P4 = intersectionpoint(foot(P2, P, P3)--(4*foot(P2, P, P3)-3*P2), unitcircle); P5 = intersectionpoint(foot(P3, P, P4)--(4*foot(P3, P, P4)-3*P3), unitcircle); P6 = intersectionpoint(foot(P4, P, P5)--(4*foot(P4, P, P5)-3*P4), unitcircle); P7 = intersectionpoint(foot(P5, P, P6)--(4*foot(P5, P, P6)-3*P5), unitcircle); Q1 = extension(P2, P6, P3, P7); Q2 = extension(P3, P7, P4, P1); Q3 = extension(P4, P1, P5, P2); Q4 = extension(P5, P2, P6, P3); Q5 = extension(P6, P3, P7, P4); Q6 = extension(P7, P4, P1, P5); Q7 = extension(P1, P5, P2, P6); R1 = foot(P, P2, P7); R2 = foot(P, P3, P1); R3 = foot(P, P4, P2); R4 = foot(P, P5, P3); R5 = foot(P, P6, P4); R6 = foot(P, P7, P5); R7 = foot(P, P1, P6); T1 = extension(P1, R2, P2, R1); T2 = extension(P2, R3, P3, R2); T3 = extension(P3, R4, P4, R3); T4 = extension(P4, R5, P5, R4); T5 = extension(P5, R6, P6, R5); T6 = extension(P6, R7, P7, R6); T7 = extension(P7, R1, P1, R7); X = extension(Q1, R1, Q2, R2); Q = extension(Q1, Q1+P-R1, Q2, Q2+P-R2); D1 = foot(Q, T1, T7); D2 = foot(Q, T1, T2); D3 = foot(Q, T2, T3); H = orthocenter(Q, T1, T2); draw(P--P1^^P--P2^^P--P3^^P--P4^^P--P5^^P--P6^^P--P7, dashed+grey+linewidth(0.4)); draw(unitcircle, red); draw(P1--P2--P3--P4--P5--P6--P7--cycle, orange); draw(P1--P3--P5--P7--P2--P4--P6--cycle, grey+linewidth(0.4)); draw(P1--P4--P7--P3--P6--P2--P5--cycle, heavyred+linewidth(0.4)); draw(R1--R2--R3--R4--R5--R6--R7--cycle, lightblue+linewidth(0.3)); draw(circumcircle(Q1, Q2, Q3), heavygreen+linewidth(0.5)); draw(circumcircle(R1, R2, R3), heavycyan+linewidth(0.4)); draw(T1--H--T2, heavyred); draw(D1--D2--D3--cycle, lightblue); draw(Q--D2, dashed+magenta); dot("$P$", P, dir(60)); dot("$P_1$", P1, dir(P1)); dot("$P_2$", P2, dir(P2)); dot("$P_3$", P3, dir(P3)); dot("$P_4$", P4, dir(P4)); dot("$P_5$", P5, dir(P5)); dot("$P_6$", P6, dir(P6)); dot("$P_7$", P7, dir(P7)); dot("$Q_1$", Q1, dir(Q1)); dot("$Q_2$", Q2, dir(210)); dot("$Q_3$", Q3, dir(240)); dot("$Q_4$", Q4, dir(270)); dot("$Q_5$", Q5, dir(340)); dot("$Q_6$", Q6, dir(10)); dot("$Q_7$", Q7, dir(Q7)); dot("$R_1$", R1, dir(140)); dot("$R_2$", R2, dir(150)); dot("$R_3$", R3, dir(210)); dot("$R_4$", R4, dir(2400)); dot("$R_5$", R5, dir(0)); dot("$R_6$", R6, dir(90)); dot("$R_7$", R7, dir(90)); dot("$T_1$", T1, dir(T1)); dot("$T_2$", T2, dir(T2)); dot("$T_3$", T3, dir(T3)); dot("$T_4$", T4, dir(T4)); dot("$T_5$", T5, dir(T5)); dot("$T_6$", T6, dir(T6)); dot("$T_7$", T7, dir(T7)); dot("$Q$", Q, dir(70)); dot("$D_1$", D1, dir(110)); dot("$D_2$", D2, dir(195)); dot("$D_3$", D3, dir(245)); dot("$H$", H, dir(60)); [/asy][/asy] The key observation is that $\Gamma$ is the pedal circle of $P$ w.r.t. $\mathcal T$, so we can reflect $P$ across the center of $\Gamma$ to obtain point $Q$, the isogonal conjugate of $P$ w.r.t. $\mathcal T$. Let the $D_i$ be the feet from $Q$ to $\overline{T_{i-1}T_i}$ for each $i$, which also lie on $\Gamma$. Claim: Triangles $D_1D_2D_3$ and $P_1P_2P_3$ are homothetic. Proof. Since $P_1R_1R_2P_2$ and $D_1R_1D_2R_2$ are cyclic, lines $P_1P_2$ and $D_1D_2$ are both antiparallel to $\overline{R_1R_2}$ in $\angle P_1T_1P_2$, hence they are parallel. Similarly, we also have $\overline{P_2P_3}\parallel \overline{D_2D_3}$. Now since $PR_1P_2R_3$ is cyclic with diameter $\overline{PP_2}$, we can write \begin{align*} \angle T_2R_3R_1 &= \angle P_2R_3R_1 = \angle P_2PP_1 = 180^\circ - \angle PP_1P_2-\angle PP_2P_1 \\ &= 180^\circ - (90^\circ - \angle P_1P_2T_1) - (90^\circ - \angle P_2P_1T_1) \\ &= \angle P_1P_2T_1+\angle P_2P_1T_1 = \angle P_2T_1T_2, \end{align*}which implies $R_1T_1T_2R_3$ cyclic. Then $T_1T_2$ and $D_1D_3$ are both antiparallel to $R_1R_3$ in $\angle T_1P_2T_2$, which forces $\overline{T_1T_3}\parallel \overline{D_1D_3}$. This proves the claim. $\blacksquare$ Claim: We have $\overline{R_1R_2}\parallel \overline{Q_1Q_2}\perp \overline{QT_1}$. Proof. The parallel lines follow from the same antiparallel argument as before; $P$ and $Q$ being isogonal conjugates implies $\overline{QT_1}\perp \overline{R_1R_2}$. (This is the same as $\overline{AO}\perp \overline{EF}$, where $E$ and $F$ are the feet of the altitudes.) $\blacksquare$ Claim: [Main Step] We have $\overline{QQ_2}\perp \overline{T_1T_2}$. Proof. Let $H$ be the orthocenter of $\triangle QT_1T_2$; note that from the previous claim, $\triangle T_1HT_2$ and $\triangle P_1Q_2P_3$ are homothetic. We claim that the center of homothety is $D_2$. Indeed, since $\triangle D_1D_2D_3$ and $\triangle P_1P_2P_3$ are homothetic, we have $$\frac{D_2T_1}{D_2P_1} = \frac{D_1T_1}{D_1P_2} = \frac{D_3T_2}{D_3P_2} = \frac{D_2T_2}{D_2P_3},$$which is enough. Thus the homothety at $D_2$ sends $H$ to $Q_2$, so $\overline{QQ_2HD_2}\perp \overline{T_1T_2}$, as desired. $\blacksquare$ Repeating this argument, we see that $\triangle QQ_iQ_{i+1}$ and $\triangle PR_iR_{i+1}$ are homothetic for each $i$, so in fact we have $$Q_1Q_2\dots Q_n\sim R_1R_2\dots R_n.$$Since $\mathcal R$ is cyclic, we must have $\mathcal Q$ is cyclic, as needed. Remark: My original proof of the last step was a Law of Sines computation. If $E$ is the foot from $Q_2$ to $\overline{T_1T_2}$, one can write $$\frac{D_2T_1}{D_2T_2} = \frac{\tan \angle QT_2T_1}{\tan \angle QT_1T_2} = \frac{\tan \angle P_1P_2P_7}{\tan \angle P_3P_2P_4} = \frac{EP_1}{EP_3}$$after some angle chasing. So it suffices to show that $$\frac{P_1T_1}{P_3T_2} = \frac{\tan \angle P_1P_2P_7}{\tan \angle P_3P_2P_4},$$which is not too difficult with Law of Sines.

tastymath75025 wrote: Let $P_1P_2\dotsb P_{100}$ be a cyclic $100$-gon and let $P_i = P_{i+100}$ for all $i$. Define $Q_i$ as the intersection of diagonals $\overline{P_{i-2}P_{i+1}}$ and $\overline{P_{i-1}P_{i+2}}$ for all integers $i$. Suppose there exists a point $P$ satisfying $\overline{PP_i}\perp\overline{P_{i-1}P_{i+1}}$ for all integers $i$. Prove that the points $Q_1,Q_2,\dots, Q_{100}$ are concyclic. Michael Ren BTW, if $s \ge 0$ is an integer, and $Q_i$ is instead defined as $P_i P_{i+s} \cap P_{i+1} P_{i+s+1}$, the problem remains true. The given problem is the case $s=3$.

My problem I think this is the best geometry problem I've ever written, and I'm glad it's finally appeared. Since people seem to enjoy these things, I'll elaborate a bit on the history of this problem. Around two years ago, I solved 2017 Thailand TST P1, which about a pentagon with concurrent altitudes, and thought it was nice. I wondered what would happen if the pentagon were also constrained to be cyclic. After playing around in Geogebra, I found the following nice property: Let $ABCDE$ be a cyclic pentagon, and suppose that the altitudes from $A$ to $CD$, $B$ to $DE$, $C$ to $EA$, $D$ to $AB$, and $E$ to $BC$ concur. Then $AB\cap CD$, $BC\cap DE$, $CD\cap EA$, $DE\cap AB$, and $EA\cap BC$ are concyclic. I showed this to some 2018 ELMO PSC members, who agreed that it was a nice problem but found it too difficult for ELMO. Since I was close to graduating, I decided to wait it out and propose it to an official contest. My original solution involved inversion, so at the time I thought it would only be suitable for a contest like RMM. After a while, I noticed that the problem also held true for a cyclic $n$-gon if instead of the opposite side altitudes were dropped to the nearest diagonal. In fact, much more was true: Let $\ldots,P_{-2},P_{-1},P_0,P_1,P_2,\ldots$ be points on a circle, and let $k$ be an integer. For all integers $i$, let $Q_i=P_iP_{i+k}\cap P_{i+1}P_{i+k+1}$. If there exists a point $P$ with $PP_i\perp P_{i-1}P_{i+1}$ for all integers $i$, then $\ldots,Q_{-2},Q_{-1},Q_0,Q_1,Q_2,\ldots$ are concyclic. For a contest, it seemed better to have the $P_i$ be periodic modulo $n$ for some positive integer $n$ so that the problem could be stated using a cyclic $n$-gon. The previous version with the pentagon was just $n=5$ and $k=2$ with $P_1P_2P_3P_4P_5=ACEBD$. However, I could only solve $0\le k\le 3$, and it became clear that with my approach at the time the difficulty would only increase with $k$. I tried to induct on $k$ but could not find the right strengthening of the hypothesis for it to carry through to all $k$. I was very happy to be using induction on a pure geometry problem, and I hope that someone can complete it in this thread. On the bright side, I found a completely elementary solution for the $k=2$ case that was pretty nice, so I proposed it to IMO 2019 in the following form: Let $P_1P_2\cdots P_{2019}$ be a cyclic $2019$-gon. Define $Q_i=P_{i-1}P_{i+1}\cap P_iP_{i+2}$ for all $i$, where indices are taken modulo $2019$. Prove that if there exists a point $P$ satisfying $PP_i\perp P_{i-1}P_{i+1}$ for all $i$, then $Q_1Q_2\cdots Q_{2019}$ is cyclic. I estimated the $k=2$ version at a medium-hard difficulty of around G5 and the $k=3$ version at a hard difficulty of around G8. I suggested the $k=3$ version as an alternate problem, and I also included the unsolved general version in the hopes that it would intrigue the PSC enough to include some variant of the problem on the ISL. Sadly, it didn't, and they returned the problem to me after the IMO along with the comment "The problem was not popular among the PSC members." and a trig bash to the $k=2$ version. A week before the IMO last year, I remembered that this problem had a chance of appearing and became interested in solving the general version again. I used a completely different method that involved developing a lot of theory surrounding the configuration and was finally able to solve it after a year. I proposed the $k=3$ version to TST with a new year: Let $P_1P_2\cdots P_{2020}$ be a cyclic $2020$-gon. Define $Q_i=P_{i-2}P_{i+1}\cap P_{i-1}P_{i+2}$ for all $i$, where indices are taken modulo $2020$. Prove that if there exists a point $P$ satisfying $PP_i\perp P_{i-1}P_{i+1}$ for all $i$, then $Q_1Q_2\cdots Q_{2019}$ is cyclic. For "psychological reasons" $2020$ was changed to $100$ in the final draft. During the TST development discussions I also mentioned the following extremely aesthetic variant, which I would have loved to put on the test if it weren't completely impossible under contest conditions: Let $P_1P_2\cdots P_{2019}$ be a cyclic $2019$-gon. Define $Q_i=P_{i-20}P_{i+19}\cap P_{i+20}P_{i-19}$ for all $i$, where indices are taken modulo $2019$. Prove that if there exists a point $P$ satisfying $PP_i\perp P_{i-1}P_{i+1}$ for all $i$, then $Q_1Q_2\cdots Q_{2019}$ is cyclic. This was, of course, shot down for being too hard. Surprisingly, people thought that the $k=3$ version was a reasonable difficulty. In fact, multiple people reported it as medium-hard, though I'm guessing that none of them did the problem themselves and instead just read the solution, which seems short and not too difficult when condensed. The official solution I sent in is essentially the one in pinetree1's post above, which he found after I gave him some hints. The $k=2$ version has a similar, but much shorter, solution. In hindsight, I should have sent in the $k=2$ version again, but I was afraid that the IMO PSC's trig bash would make it less appealing to people. I had some reservations about putting the $k=3$ version on the test, but I thought that someone in the USA TST group would maybe be capable of solving it, especially with the easier P4 and P5. Unfortunately, I was wrong (see thread title). Anyway, I hope you guys enjoy solving this problem and the general version! I'll say some more about my solutions to these problems and what's really going on underneath the surface after someone else solves the general version and posts their solution in this thread. [asy][asy] size(9cm); pair O = origin; draw(circle(O,1)); real a = 16.502245059813934; real b = 60.51606646804328; real c = 96.13802261995933; real d = 123.69237966066693; pair P1 = dir(a); pair P2 = dir(b); pair P3 = dir(c); pair P4 = dir(d); pair P = extension(P2,foot(P2,P1,P3),P3,foot(P3,P2,P4)); pair P5 = 2*foot(O,P3,foot(P3,P,P4))-P3; pair P6 = 2*foot(O,P4,foot(P4,P,P5))-P4; pair P7 = 2*foot(O,P5,foot(P5,P,P6))-P5; pair P8 = 2*foot(O,P6,foot(P6,P,P7))-P6; pair P9 = 2*foot(O,P7,foot(P7,P,P8))-P7; pair P10 = 2*foot(O,P8,foot(P8,P,P9))-P8; pair P11 = 2*foot(O,P9,foot(P9,P,P10))-P9; pair[] A = {P1,P2,P3,P4,P5,P6,P7,P8,P9,P10,P11,P1,P2,P3,P4,P5,P6,P7,P8,P9,P10,P11}; for (int i = 0; i < 11 ; i = i+1) { for (int j = i+1; j < 11; j = j+1) { if (j-i != 1 && j-i != 2 && j-i != -10 && j-i != -9) draw(A[i]--A[j],gray); } } draw(P1--P2--P3--P4--P5--P6--P7--P8--P9--P10--P11--cycle); draw(P1--P3--P5--P7--P9--P11--P2--P4--P6--P8--P10--cycle); for (int p = 2; p <= 5; p = p+1) { draw(circumcircle(extension(A[0],A[p],A[1],A[p+1]),extension(A[1],A[p+1],A[2],A[p+2]),extension(A[2],A[p+2],A[3],A[p+3])),gray); for (int q = 0; q < 11; q = q+1) { dot(extension(A[q],A[p+q],A[q+1],A[p+q+1])); } } draw(circumcircle(extension(A[0],A[0]+A[0]*dir(90),A[1],A[1]+A[1]*dir(90)),extension(A[1],A[1]+A[1]*dir(90),A[2],A[2]+A[2]*dir(90)),extension(A[2],A[2]+A[2]*dir(90),A[3],A[3]+A[3]*dir(90))),gray); for (int i = 0; i < 11; i = i+1) { draw(extension(A[i],A[i]+A[i]*dir(90),A[i+1],A[i+1]+A[i+1]*dir(90))--extension(A[i+1],A[i+1]+A[i+1]*dir(90),A[i+2],A[i+2]+A[i+2]*dir(90)),gray); } for (int i = 0; i < 11; i = i+1) { dot(extension(A[i],A[i]+A[i]*dir(90),A[i+1],A[i+1]+A[i+1]*dir(90))); } dot(P1^^P2^^P3^^P4^^P5^^P6^^P7^^P8^^P9^^P10^^P11^^P); for (int i = 0; i < 11; i = i+1) { draw(P--A[i]); } [/asy][/asy]

Let $S_i = \overline{P_iP_i}\cap \overline{P_{i+1}P_{i+1}}$ for each $i$. Here's a proof that $\mathcal T$ and $\mathcal S$ are cyclic (carrying over my notation from above). [asy][asy] size(350); defaultpen(fontsize(10pt)); pair P, O, P1, P2, P3, P4, P5, P6, P7; pair Q1, Q2, Q3, Q4, Q5, Q6, Q7; pair R1, R2, R3, R4, R5, R6, R7; pair X, T1, T2, T3, T4, T5, T6, T7; pair S1, S2, S3, S4, S5, S6, S7; pair M1, M2, M3, M4, M5, M6, M7; pair D1, D2, D3, D4, D5, D6, D7; O = (0,0); P = 0.1*dir(110); P1 = dir(125.15); P3 = dir(220); P2 = intersectionpoint(P--(2*foot(P, P1, P3)-P), unitcircle); P4 = intersectionpoint(foot(P2, P, P3)--(4*foot(P2, P, P3)-3*P2), unitcircle); P5 = intersectionpoint(foot(P3, P, P4)--(4*foot(P3, P, P4)-3*P3), unitcircle); P6 = intersectionpoint(foot(P4, P, P5)--(4*foot(P4, P, P5)-3*P4), unitcircle); P7 = intersectionpoint(foot(P5, P, P6)--(4*foot(P5, P, P6)-3*P5), unitcircle); Q1 = extension(P2, P6, P3, P7); Q2 = extension(P3, P7, P4, P1); Q3 = extension(P4, P1, P5, P2); Q4 = extension(P5, P2, P6, P3); Q5 = extension(P6, P3, P7, P4); Q6 = extension(P7, P4, P1, P5); Q7 = extension(P1, P5, P2, P6); R1 = foot(P, P2, P7); R2 = foot(P, P3, P1); R3 = foot(P, P4, P2); R4 = foot(P, P5, P3); R5 = foot(P, P6, P4); R6 = foot(P, P7, P5); R7 = foot(P, P1, P6); T1 = extension(P1, R2, P2, R1); T2 = extension(P2, R3, P3, R2); T3 = extension(P3, R4, P4, R3); T4 = extension(P4, R5, P5, R4); T5 = extension(P5, R6, P6, R5); T6 = extension(P6, R7, P7, R6); T7 = extension(P7, R1, P1, R7); S1 = extension(P1, rotate(90, P1)*O, P2, rotate(90, P2)*O); S2 = extension(P2, rotate(90, P2)*O, P3, rotate(90, P3)*O); S3 = extension(P3, rotate(90, P3)*O, P4, rotate(90, P4)*O); S4 = extension(P4, rotate(90, P4)*O, P5, rotate(90, P5)*O); S5 = extension(P5, rotate(90, P5)*O, P6, rotate(90, P6)*O); S6 = extension(P6, rotate(90, P6)*O, P7, rotate(90, P7)*O); S7 = extension(P7, rotate(90, P7)*O, P1, rotate(90, P1)*O); M1 = (P1+P2)/2; M2 = (P2+P3)/2; M3 = (P3+P4)/2; M4 = (P4+P5)/2; M5 = (P5+P6)/2; M6 = (P6+P7)/2; M7 = (P7+P1)/2; D1 = foot(P, P1, P2); D2 = foot(P, P2, P3); D3 = foot(P, P3, P4); D4 = foot(P, P4, P5); D5 = foot(P, P5, P6); D6 = foot(P, P6, P7); D7 = foot(P, P7, P1); X = extension(Q1, R1, Q2, R2); draw(P--P1^^P--P2^^P--P3^^P--P4^^P--P5^^P--P6^^P--P7, dashed+grey+linewidth(0.4)); draw(unitcircle, grey); draw(P1--P2--P3--P4--P5--P6--P7--cycle); draw(P1--P3--P5--P7--P2--P4--P6--cycle, linewidth(0.3)); draw(P1--P4--P7--P3--P6--P2--P5--cycle, linewidth(0.4)); draw(R1--R2--R3--R4--R5--R6--R7--cycle, linewidth(0.3)); draw(circumcircle(Q1, Q2, Q3), grey); draw(circumcircle(R1, R2, R3), grey); draw(circumcircle(T1, T2, T3), dotted); draw(S1--S2--S3--S4--S5--S6--S7--cycle, linewidth(0.4)); draw(circumcircle(S1, S2, S3), dotted); draw(circumcircle(M1, M2, M3), dashed+linewidth(0.8)); dot("$P$", P, dir(60)); dot("$P_1$", P1, dir(P1)); dot("$P_2$", P2, dir(P2)); dot("$P_3$", P3, dir(P3)); dot("$P_4$", P4, dir(P4)); dot("$P_5$", P5, dir(P5)); dot("$P_6$", P6, dir(P6)); dot("$P_7$", P7, dir(P7)); dot("$Q_1$", Q1, dir(Q1)); dot("$Q_2$", Q2, dir(180)); dot("$Q_3$", Q3, dir(240)); dot("$Q_4$", Q4, dir(270)); dot("$Q_5$", Q5, dir(340)); dot("$Q_6$", Q6, dir(10)); dot("$Q_7$", Q7, dir(Q7)); dot("$R_1$", R1, dir(90)); dot("$R_2$", R2, dir(130)); dot("$R_3$", R3, dir(180)); dot("$R_4$", R4, dir(2400)); dot("$R_5$", R5, dir(0)); dot("$R_6$", R6, dir(90)); dot("$R_7$", R7, dir(90)); dot("$T_1$", T1, dir(T1)); dot("$T_2$", T2, dir(T2)); dot("$T_3$", T3, dir(T3)); dot("$T_4$", T4, dir(T4)); dot("$T_5$", T5, dir(T5)); dot("$T_6$", T6, dir(T6)); dot("$T_7$", T7, dir(T7)); dot("$S_1$", S1, dir(S1)); dot("$S_2$", S2, dir(S2)); dot("$S_3$", S3, dir(S3)); dot("$S_4$", S4, dir(S4)); dot("$S_5$", S5, dir(S5)); dot("$S_6$", S6, dir(S6)); dot("$S_7$", S7, dir(S7)); dot("$O$", O, dir(5)); dot(M1); dot(M2); dot(M3); dot(M4); dot(M5); dot(M6); dot(M7); dot(D1); dot(D2); dot(D3); dot(D4); dot(D5); dot(D6); dot(D7); [/asy][/asy] The key claim is that: Claim: $P$ and $O$ are isogonal conjugates w.r.t. $\mathcal P$. Proof. We have $\angle OP_2P_3 = 90^\circ - \angle P_2P_1P_3 = \angle P_1P_2P$, so $\overline{P_2P}$ and $\overline{P_2O}$ are isogonal in $\angle P_1P_2P_3$. Repeating this argument proves the claim. $\blacksquare$ Thus the pedal $n$-gons of $P$ and $O$ w.r.t. $\mathcal P$ are cyclic with circumcircle $\Gamma$. Inverting at $P$ with power $PR_i\cdot PP_i$ gives $\mathcal T$ cyclic, while inverting about the circumcircle gives $\mathcal S$ cyclic. So we're done!

Replace $100$ with $n$. Let $R_i=\overline{PP_i}\cap\overline{P_{i-1}P_{i+1}}$ be the foot from $P$ to $\overline{P_{i-1}P_{i+1}}$, and let $S_i=\overline{P_{i-1}P_{i+1}}\cap\overline{P_iP_{i+2}}$. [asy][asy] size(12cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=deepcyan; pen sec2=heavygreen; pen tri=royalblue; pen tri2=heavycyan; pen qua=lightblue; pen qua2=mediumblue; pen fil=invisible; pen sfil=invisible; pen sfil2=invisible; pen tfil=invisible; pen tfil2=invisible; pen qfil=invisible; pen qfil2=invisible; real t=3.5; pair O,PP,RR,QQ; pair[] P,Q,R,SS,T; O=(0,0); PP=0.15*dir(110+t); P[0]=dir(90+t); P[1]=dir(57.4+t); for (int i = 2; i <= 8; i += 1) P[i]=reflect(0,O+P[i-1]-PP)*P[i-2]; for (int i = 0; i < 8; i += 1) Q[i]=extension(P[(i+6)%8],P[(i+1)%8],P[(i+7)%8],P[(i+2)%8]); for (int i = 0; i < 8; i += 1) R[i]=foot(PP,P[(i+7)%8],P[(i+1)%8]); RR=circumcenter(R[0],R[1],R[2]); QQ=2RR-PP; for (int i = 0; i < 8; i += 1) SS[i]=extension(P[(i+7)%8],P[(i+1)%8],P[i],P[(i+2)%8]); for (int i = 0; i < 8; i += 1) T[i]=foot(QQ,P[(i+7)%8],P[(i+1)%8]); for(int i = 0; i < 8; i += 1) { draw(P[(i-1)%8]--P[(i+1)%8],sec); draw(PP--P[i],sec+dashed); draw(P[(i-1)%8]--P[(i+2)%8],tri); draw(QQ--T[i],qua+dashed); } filldraw(circumcircle(T[0],T[1],T[2]),qfil2,qua2); filldraw(Q[0]--Q[1]--Q[2]--Q[3]--Q[4]--Q[5]--Q[6]--Q[7]--cycle,tfil2,tri2+linewidth(0.8)); filldraw(circumcircle(Q[0],Q[1],Q[2]),tfil2,tri2); filldraw(SS[0]--SS[1]--SS[2]--SS[3]--SS[4]--SS[5]--SS[6]--SS[7]--cycle,sfil2,sec2+linewidth(0.8)); filldraw(circle(O,1),fil,pri+linewidth(0.8)); filldraw(P[0]--P[1]--P[2]--P[3]--P[4]--P[5]--P[6]--P[7]--cycle,fil,pri+linewidth(0.8)); dot("$P$", PP, S); dot("$Q$", QQ, N); for (int i = 0; i < 8; i += 1) dot("$P_" + string(i+1) + "$", P[i], P[i]); dot("$Q_1$", Q[0], N); dot("$Q_2$", Q[1], dir(15)); dot("$Q_3$", Q[2], dir(-30)); dot("$Q_4$", Q[3], dir(-40)); dot("$Q_5$", Q[4], dir(240)); dot("$Q_6$", Q[5], dir(235)); dot("$Q_7$", Q[6], dir(185)); dot("$Q_8$", Q[7], N); dot("$R_1$", R[0], unit(P[1]-P[7])*dir(45)); dot("$R_2$", R[1], unit(P[2]-P[0])*dir(135)); dot("$R_3$", R[2], unit(P[3]-P[1])*dir(135)); dot("$R_4$", R[3], unit(P[4]-P[2])*dir(45)); dot("$R_5$", R[4], unit(P[5]-P[3])*dir(45)); dot("$R_6$", R[5], unit(P[6]-P[4])*dir(45)); dot("$R_7$", R[6], unit(P[7]-P[5])*dir(45)); dot("$R_8$", R[7], unit(P[0]-P[6])*dir(135)); dot("$S_1$", SS[0], unit(SS[0]-incenter(SS[0],SS[7],SS[1]))); dot("$S_2$", SS[1], unit(SS[1]-incenter(SS[1],SS[0],SS[2]))); dot("$S_3$", SS[2], unit(SS[2]-incenter(SS[2],SS[1],SS[3]))); dot("$S_4$", SS[3], unit(SS[3]-incenter(SS[3],SS[2],SS[4]))); dot("$S_5$", SS[4], unit(SS[4]-incenter(SS[4],SS[3],SS[5]))); dot("$S_6$", SS[5], unit(SS[5]-incenter(SS[5],SS[4],SS[6]))); dot("$S_7$", SS[6], unit(-SS[6]+incenter(SS[6],SS[5],SS[7]))); dot("$S_8$", SS[7], unit(-SS[7]+incenter(SS[7],SS[6],SS[0]))); dot("$T_1$", T[0], unit(P[1]-P[7])*dir(135)); dot("$T_2$", T[1], dir(100)); dot("$T_3$", T[2], unit(P[3]-P[1])*dir(45)); dot("$T_4$", T[3], unit(P[4]-P[2])*dir(90)); dot("$T_5$", T[4], unit(P[5]-P[3])*dir(90)); dot("$T_6$", T[5], unit(P[6]-P[4])*dir(135)); dot("$T_7$", T[6], unit(P[7]-P[5])*dir(135)); dot("$T_8$", T[7], unit(P[0]-P[6])*dir(90)); [/asy][/asy] Claim 1. $R_1R_2\cdots R_n$ is cyclic. Proof. Since $\angle P_iR_iP_{i+1}=\angle P_iR_{i+1}P_{i+1}=90^\circ$ for each $i$, we have $PP_i\cdot PR_i=PP_{i+1}\cdot PR_{i+1}$, meaning inversion at $P$ with radius $\sqrt{PP_i\cdot PR_i}$, which is fixed, sends the circumcircle of $P_1P_2\cdots P_n$ to that of $R_1R_2\cdots R_{100}$. $\blacksquare$ Claim 2. $\overline{Q_iQ_{i+1}}\parallel\overline{R_iR_{i+1}}$ for all $i$. Proof. By Reim's theorem on $(P_iR_iR_{i+1}P_{i+1})$ and $(P_{i-1}P_iP_{i+1}P_{i+2})$, $\overline{Q_iQ_{i+1}}=\overline{P_{i-1}P_{i+2}}\parallel\overline{R_iR_{i+1}}$. $\blacksquare$ Since $P$ has a pedal circle in $S_1S_2\cdots S_{100}$, the reflection $Q$ over the center of the pedal circle is the isogonal conjugate of $P$ in $S_1S_2\cdots S_{100}$. Let $T_i$ be the foot from $Q$ to $\overline{P_{i-1}P_{i+1}}$, so that there is a fixed circle through both $R_i$ and $T_i$ for all $i$. Claim 3. $\triangle P_{i-1}P_iP_{i+1}$ and $\triangle T_{i-1}T_iT_{i+1}$ are homothetic. Proof. By Reim's theorem on $(P_iR_iR_{i-1}P_{i-1})$ and $(R_iT_iT_{i-1}R_{i-1})$, we have $\overline{P_iP_{i-1}}\parallel\overline{T_iT_{i-1}}$, and similarly $\overline{P_iP_{i+1}}\parallel\overline{T_iT_{i+1}}$. Also $P_iS_{i-1}\cdot P_iR_{i-1}=P_iR_i\cdot P_iP=P_iS_{i+1}\cdot P_iR_{i+1}$, so $R_{i-1}S_{i-1}S_iR_{i+1}$ is cyclic, and $\overline{P_{i-1}P_{i+1}}\parallel\overline{R_{i-1}R_{i+1}}$ by Reim's theorem on $(R_{i-1}S_{i-1}S_iR_{i+1})$ and $(R_{i-1}T_{i-1}T_{i+1}R_{i+1})$. $\blacksquare$ Claim 4. $\overline{QS_i}\perp\overline{Q_iQ_{i+1}}$ for all $i$. Proof. Since $S_i$ is the orthocenter of $\triangle PP_iP_{i+1}$, we know $\overline{PS_i}\perp\overline{P_iP_{i+1}}$. Note that $\overline{S_iP}$ and $\overline{S_iQ}$ are isogonal wrt.\ $\angle P_iS_iP_{i+1}$, and $\overline{P_iP_{i+1}}$ and $\overline{P_{i-1}P_{i+2}}$ are isogonal, so $\overline{QS_i}\perp\overline{P_{i-1}P_{i+2}}$, as claimed. $\blacksquare$ Claim 5. $Q$, $Q_i$, $T_i$ are collinear for all $i$. Proof. By the parallel lines from Claim 3, \[\frac{T_iS_{i-1}}{S_{i-1}P_{i-1}}=\frac{T_iT_{i-1}}{P_iP_{i-1}}=\frac{T_iT_{i+1}}{P_iP_{i+1}}=\frac{T_iS_i}{S_iP_{i+2}}.\]Hence a homothety at $S$ sends $\triangle Q_iP_{i-1}P_{i+1}$ to $\triangle H_iS_{i-1}S_i$ for some point $H_i$ with $\overline{QS_i}\perp\overline{H_iS_{i-1}}$ and $\overline{QS_{i-1}}\perp\overline{H_iS_i}$. It follows that $H_i$ is the orthocenter of $\triangle Q_iS_{i-1}S_i$, so $H_i$ lies on $\overline{QT_i}$. This is sufficient, since $H_i\in\overline{T_iQ_i}$ by the homothety. $\blacksquare$ From this, $\triangle PR_iR_{i+1}$ and $\triangle QQ_iQ_{i+1}$ for each $i$, so $R_1R_2\cdots R_n\sim Q_1Q_2\cdots Q_n$. Since the former is cyclic, so is the latter, thus concluding the proof.

It can be proved via complex numbers that if ortogonality condition holds for $99$ points then it holds for all.

ABCDE wrote: the solution [which] seems short and not too difficult when condensed You're welcome

@v_Enhance so I guess this is worth 50Mohs?

I put 55 in my database. I think it's just too difficult to solve in the time limit.

wait evan update mohs on website

It suffices to prove that $Q_i,Q_{i+1},Q_{i+2},Q_{i+3}$ are concyclic for every $i$; by symmetry it is sufficient to prove $P_3Q_3\cdot P_3Q_4 = P_3Q_1\cdot P_3Q_2$. Let $E_i=PP_i\cap P_{i-1}P_{i+1}$. From the cyclic quads $PE_2Q_2E_3$ and $PE_3Q_3E_4$ we can see that $$ P_3Q_3\cdot P_3E_4 = P_3E_3 \cdot P_3P = P_3Q_2\cdot P_3E_2; $$so it is enough to prove that $Q_1Q_4\|E_2E_4$. Let $ \alpha = \angle{P_5P_3P}, \quad \beta = \angle{PP_3P_1}$, $x = \angle{P_4PQ_4} =\angle{P_3P_5P_4}=\angle{P_3P_2P_4} =\angle{Q_2PP_3}$ and $y = \angle{P_3PQ_3} =\angle{P_2P_4P_3} =\angle{P_2P_1P_3}=\angle{Q_1PP_2}$. From the quadrilateral $PQ_2P_3Q_3$ we can get $$ \tan\alpha \cdot \tan x = \frac{E_3Q_2\cdot E_3Q_3}{E_3P\cdot E_3P_3} = \tan\beta \cdot \tan y. $$ Let $PP_3=1$. From the right triangles $PP_3E_4$ and $PE_4Q_4$ etc. we obtain $P_3E_4 = \cos\alpha$, $E_4Q_4 = \sin\alpha\cdot\tan x$, $P_3E_2 = \cos\beta$, $E_2Q_1 = \sin\beta\cdot\tan y; $ $$ \frac{E_4Q_4}{P_3E_4} = \tan\alpha\cdot\tan x = \tan\beta\cdot\tan y = \frac{E_2Q_1}{P_3E_2}. $$That proves $Q_1Q_4\|E_2E_4$.

Géza Kós wrote: It suffices to prove that $Q_i,Q_{i+1},Q_{i+2},Q_{i+3}$ are concyclic for every $i$; by symmetry it is sufficient to prove $$ P_3Q_3\cdot P_3Q_4 = P_3Q_1\cdot P_3Q_2. $$ Let $E_i=PP_i\cap P_{i-1}P_{i+1}$. From the cyclic quads $PE_2Q_2E_3$ and $PE_3Q_3E_4$ we can see that $$ P_3Q_3\cdot P_3E_4 = P_3E_3 \cdot P_3P = P_3Q_2\cdot P_3E_2; $$so it is enough to prove that $Q_1Q_4\|E_2E_4$. Let $$ \alpha = \angle{P_5P_3P}, \quad \beta = \angle{PP_3P_1}, $$$$ x = \angle{P_4PQ_4} =\angle{P_3P_5P_4} =\angle{P_3P_2P_4} =\angle{Q_2PP_3},$$and $$ y = \angle{P_3PQ_3} =\angle{P_2P_4P_3} =\angle{P_2P_1P_3} =\angle{Q_1PP_2}. $$ From the quadrilateral $PQ_2P_3Q_3$ we can get $$ \tan\alpha \cdot \tan x = \frac{E_3Q_2\cdot E_3Q_3}{E_3P\cdot E_3P_3} = \tan\beta \cdot \tan y. $$ Let $PP_3=1$. From the right triangles $PP_3E_4$ and $PE_4Q_4$ etc. we obtain $$ P_3E_4 = \cos\alpha, \qquad E_4Q_4 = \sin\alpha\cdot\tan x, \qquad P_3E_2 = \cos\beta, \qquad E_2Q_1 = \sin\beta\cdot\tan y; $$$$ \frac{E_4Q_4}{P_3E_4} = \tan\alpha\cdot\tan x = \tan\beta\cdot\tan y = \frac{E_2Q_1}{P_3E_2}. $$That proves $Q_1Q_4\|E_2E_4$. This is a misread: the problem has $Q_i=\overline{P_{i-2}P_{i+1}}\cap\overline{P_{i-1}P_{i+2}}$ while you did $Q_i = \overline{P_{i-1}P_{i+1}}\cap\overline{P_iP_{i+2}}$.

May be, I just copied my notes from 2019, and this is not the same problem. EDIT: Indeed, my notes were written on the variant with $Q_i=P_{i-1}P_{i+1}\cap P_i\cap P_{i+2}$.

ABCDE wrote: Let $P_1P_2\cdots P_{2019}$ be a cyclic $2019$-gon. Define $Q_i=P_{i-1}P_{i+1}\cap P_iP_{i+2}$ for all $i$, where indices are taken modulo $2019$. Prove that if there exists a point $P$ satisfying $PP_i\perp P_{i-1}P_{i+1}$ for all $i$, then $Q_1Q_2\cdots Q_{2019}$ is cyclic. it's the statement proposed to IMO 2019 same as @above proof

redacted

This seems to be a bit different from the other solutions

Define $K_i=PP_i \cap P_{i-1}P_{i+1}$ and $H_i=P_iP_{i+2} \cap P_{i-1}P_{i+1}$. Since $P_iP_{i+1}K_iK_{i+1}$ is cyclic, an inversion at $P$ with power $PP_i \cdot PK_i$ gives that $K_1K_2 \cdots K_{100}$ is cyclic. Our goal is to show $Q_1Q_2 \cdots Q_{100}$ is homothetic with this polygon, which would imply it being cyclic. For this it is sufficient to show $\triangle K_3K_4K_5$ and $\triangle Q_3 Q_4 Q_5$ are homothetic. First, Reim on $P_3K_3P_4K_4$ gives that $K_3K_4$ is parallel to $P_2P_5$, and analogously $K_4K_5$ is parallel to $P_3P_6$, so it suffices to prove $K_3K_5$ is parallel to $Q_3Q_5$. Now notice that $\triangle P_4H_2H_5$ and the triangle determined by lines $P_3P_5, P_2P_3, P_5P_6$ are orthologic, since $PH_2 \perp P_2P_3$ and $PH_5 \perp P_5P_6$ follow from the fact that $H_2$ and $H_5$ are the orthocenters of $\triangle PP_2P_3$ and $\triangle PP_5P_6$ respectively. Since the perpendiculars from $P_3$ to $P_4H_2$ and from $P_5$ to $P_4H_5$ also intersect at $P$, the two centers of orthology coincide, so the two triangles are perspective. Let $l$ be the line through $P_2P_4 \cap P_5P_6$ and $P_2P_3 \cap P_4P_6$. The perspectivity above implies $P_3P_5 \in l$, and Pascal on $P_2P_4P_4P_6P_5P_3$ gives $P_4P_4 \cap P_3P_5 \in l$. Since lines $l$ and $P_3P_5$ cannot coincide this means $P_4P_4, P_3P_5$ and $H_2H_5$ are concurrent, say at $T$. Pascal on $P_3P_1P_4P_4P_2P_5$ gives $Q_3 \in H_2T$, and similarly $Q_5$ is on that line as well. So if we can show $H_2H_5$ being parallel to $K_3K_5$ we are done. But $K_4H_3 \cdot K_4P_3 = K_4P_4 \cdot K_4P = K_4H_4 \cdot K_4P_5$ implies $\frac{K_4H_4}{K_4P_3}=\frac{K_4H_3}{K_4P_5}$, and an easy angle chase gives $\triangle PH_2P_4 \sim \triangle PH_4P_3$ and $\triangle PH_5P_4 \sim \triangle PH_3P_5$, so $\frac{K_3H_2}{K_3P_4}=\frac{K_5H_5}{K_5P_4} \implies H_2H_5$ is parallel to $K_3K_5$, and we are done. $\square$

The legendary problem...the true P6. I agree 100% with 55 MOHS. Denote $\mathcal P_Z$ the polygon (usually i do use that notation for the polars w.r.t. a circle however this time i'll use it for this as i thought i'd be cool) $Z_1Z_2....Z_{100}$, let $PP_i \cap P_{i-1}P_{i+1}=A_i$, let $P_{i-1}P_{i+1} \cap P_iP_{i+2}=L_i$ Claim 1: $\mathcal P_L$ and $\mathcal P_A$ are cyclic, and $P,O$ are isogonal in $\mathcal P_P$ Proof: Its easy to see why $P,O$ are isogonal in $\mathcal P_P$ just consider each pair of $\triangle P_{i-1}P_iP_{i+1}$ and remember that line connecting the center and the altitude are isogonal. Now note that $P_iA_iA_{i+1}P_{i+1}$ is cyclic so repeatings this over all $i$'s and then making $\sqrt{PA_i \cdot PP_i}$ inversion one gets that $\mathcal P_A \to \mathcal P_P$ hence $\mathcal P_A$ is cyclic, and by the same inversion $\mathcal P_L$ goes to the pedal circle of $P$ in $\mathcal P_P$ (as $P$ has isogonal conjugate) hence $\mathcal P_L$ is also cyclic. Now note that $P$ has pedal circle in $\mathcal P_L$ hence let $Q$ the isogonal conjugate of $P$ in $\mathcal P_L$ let $M_i$ a point in $L_{i-1}L_i$ such that $QM_i \perp L_{i-1}L_i$, now note that $\mathcal P_M$ and $\mathcal P_A$ have the same circle passing through all of their points which is the pedal circle of $P$ and $Q$ w.r.t. the circle passing through all the points in $\mathcal P_L$. Claim 2: $Q_iQ_{i+1} \parallel A_iA_{i+1} \perp QL_i$ and $QP_i \perp A_{i-1}A_{i+1}$ Proof: By Reim's on $(P_iA_iA_{i+1}P_{i+1})$ and $(P_{i-1}P_iP_{i+1}P_{i+2})$ we get $A_iA_{i+1} \parallel P_{i-1}P_{i+2}$ so $A_iA_{i+1} \parallel Q_iQ_{i+1}$ and by Reim's in $(QM_{i+1}L_iM_i)$ and the pedal circle of $P,Q$ w.r.t. $\mathcal P_L$ we have that $PL_i \perp A_iA_{i+1} \parallel Q_iQ_{i+1}$. Now by Reim's on $(QM_{i+1}P_iM_{i-1})$ and the pedal circle of $P,Q$ w.r.t. $\mathcal P_L$ we get that $QP_i \perp A_{i-1}A_{i+1}$. Claim 3: $M_i,Q_i,Q$ colinear. Proof: Let $H$ the ortocenter of $QL_{i-1}L_i$ then note that $\triangle P_{i-1}Q_iP_{i+1}$ is homothetic with $\triangle L_{i-1}HL_i$, now by Reim's in $(P_iP_{i+1}A_{i+1}A_i)$ and the pedal circle of $P,Q$ w.r.t. $\mathcal P_L$ we get that $P_iP_{i+1} \parallel M_iM_{i+1}$ now by Reim's in $(PA_iP_{i+1}A_{i+2})$ and the pedal circle of $P,Q$ w.r.t. $\mathcal P_L$ we hace that $PP_{i+1} \perp M_iM_{i+2}$ meaning that $P_iP_{i+2} \parallel M_iM_{i+2}$ hence $\triangle P_iP_{i+1}P_{i+2}$ is homothetic to $\triangle M_iM_{i+1}M_{i+2}$ and from here (and the parallels from all the claims) one gets the following ratios: $$\frac{M_{i+1}L_i}{P_iM_{i+1}}=\frac{M_iL_i}{P_{i+1}M_i}=\frac{M_{i+2}L_{i+1}}{P_{i+1}M_{i+2}}=\frac{M_{i+1}L_{i+1}}{P_{i+2}M_{i+1}} \implies M_{i+1},H,Q_{i+1} \; \text{colinear!}$$From here our claim holds. The Great Finish: Using Claim 3 we have that $\triangle Q_iQ_{i+1}Q_{i+2}$ is ortologic to $\triangle L_iP_{i+1}L_{i+1}$ meaning that $QP_{i+1} \perp Q_iQ_{i+2}$ so $A_iA_{i+2} \parallel Q_iQ_{i+2}$ meaning that $\triangle A_iA_{i+1}A_{i+2}$ is homothetic to $\triangle Q_iQ_{i+1}Q_{i+2}$ so that means $\mathcal P_A$ homothetic to $\mathcal P_Q$, as $\mathcal P_A$ is cyclic we get that $\mathcal P_Q$ is cyclic thus we are done Note: I remember once a time in the beginings of 2022 (or maybe the end of 2021) saying that i would never be able to solve problems like this one...well i proved myself wrong.

We will use complex numbers. Note that the perpendicularity condition implies $\vec{PP_{j-1}} \cdot \vec{PP_{j}} = \vec{PP_{j}} \cdot \vec{PP_{j+1}}$ for all $j$; thus we can take a real constant $k$ such that $\vec{PP_{j}} \cdot \vec{PP_{j+1}}=k$ for all $j$. We will compute $Q_i$ in terms of $P$,$P_i$, and $k$, and then show that, as $P_i$ varies on the circle, $Q_i$ is on a fixed circle; this will solve the problem. Let $P=p$, $P_{i-2},P_{i-1},P_i,P_{i+1},P_{i+2} = a,b,c,d,\varepsilon$; let the circumcircle of $P_1\dots P_{100}$ be the unit circle. Now $AC \perp BP$ implies $a=\frac{p-b}{c(\overline{p}-\frac{1}{b})}$ (because $ac = - \frac{a-c}{\overline{a-c}} = \frac{p-b}{\overline{p-b}}$); similarly $\varepsilon=\frac{p-d}{c(\overline{p}-\frac{1}{d})}$. Let $Q_i=q$. Then we can compute that \begin{align*}q &= \frac{ad(\varepsilon+b)-b\varepsilon(a+d)}{ad-b\varepsilon} = \frac{\frac{p-b}{c(\overline{p}-\frac{1}{b})}d(\frac{p-d}{c(\overline{p}-\frac{1}{d})}+b)-\frac{p-d}{c(\overline{p}-\frac{1}{d})}b(\frac{p-b}{c(\overline{p}-\frac{1}{b})}+d)}{\frac{p-b}{c(\overline{p}-\frac{1}{b})}d-\frac{p-d}{c(\overline{p}-\frac{1}{d})}b} \\ &= \frac{(p-b)d((p-d)+bc(\overline{p}-\frac{1}{d}))-(p-d)b((p-b)+dc(\overline{p}-\frac{1}{b}))}{c(p-b)(d\overline{p}-1)-c(p-d)(b\overline{p}-1)} \\ &= \frac{(p-b)(p-d)(d-b) + bcd\overline{p}((p-b)-(p-d)) - (p-b)bc + (p-d)dc}{c(d-b)(p\overline{p}-1)} \\ &= \frac{(p-b)(p-d) + bcd\overline{p} + (p-d-b)c}{c(p\overline{p}-1)}.\end{align*} Now let us compute $b+d$ and $bd$ in terms of $p$, $c$, and $k$. Because $PC \perp BD$, we can see, as earlier, that $bd = \frac{p-c}{\overline{p}-\frac{1}{c}}$. Computing $b+d$ is more difficult. The equation $\vec{PX} \cdot \vec{PC}$ for $x$ on the unit circle becomes $(p-x)(\overline{p}-\frac{1}{c}) + (\overline{p}-\frac{1}{x})(p-c) = 2k$, so $p(\overline{p}-\frac{1}{c}) + \overline{p}(p-c) - 2k = x(\overline{p}-\frac{1}{c}) + \frac{1}{x}(p-c)$. Multiplying by $x$, we get that $b$ and $d$ are the roots of the quadratic $x^2(\overline{p}-\frac{1}{c}) - x(p(\overline{p}-\frac{1}{c}) + \overline{p}(p-c) - 2k) + (p-c) = 0$, so, by Vieta, we have $b+d=\frac{p(\overline{p}-\frac{1}{c}) + \overline{p}(p-c) - 2k}{\overline{p}-\frac{1}{c}}$. Now \begin{align*}q(p\overline{p}-1) &= \frac{p^2+pc -(b+d)(p+c) + bd(c\overline{p}+1)}{c} \\ &= \frac{p^2+pc - (\frac{p(\overline{p}-\frac{1}{c}) + \overline{p}(p-c) - 2k}{\overline{p}-\frac{1}{c}})(p+c) + (\frac{p-c}{\overline{p}-\frac{1}{c}})(c\overline{p}+1)}{c} \\ &= \frac{(p^2+pc)(\overline{p}-\frac{1}{c}) - (p(\overline{p}-\frac{1}{c}) + \overline{p}(p-c) - 2k)(p+c) + (p-c)(c\overline{p}+1)}{c\overline{p}-1} \\ &= \frac{p(p+c)(\overline{p}-\frac{1}{c}) - p(p+c)(\overline{p}-\frac{1}{c}) - (p\overline{p}+c\overline{p})(p-c) + 2k(p+c) + (p-c)(c\overline{p}+1)}{c\overline{p}-1} \\ &= \frac{-(p\overline{p}-1)(p-c) + 2k(p+c)}{c\overline{p}-1}.\end{align*} Thus $q = \frac{(c-p) + f(p+c)}{c\overline{p}-1}$, where $f = \frac{2k}{p\overline{p}-1}$. We want to show that lies on a fixed circle as $c$ is variable on the unit circle. Now $q = \frac{(1+f)c+(f-1)p}{c\overline{p}-1} = \frac{1+f}{\overline{p}} + \frac{(f-1)p-\frac{1+f}{\overline{p}}}{c\overline{p}-1} = g + \frac{h}{c\overline{p}-1}$ for some constants $g$ and $h$. Now, as $c$ varies, $c\overline{p}-1$ varies on a circle (and note that $0$ does not lie on this circle, for else $|p|=1$, but then, following along in the computation of $Q$ as $AD \cap BE$, we would get $ad-b\varepsilon = 0$ since $p\overline{p}-1 = 0$, implying $AD$ and $BE$ are parallel or the same line, a contradiction). Then $\frac{1}{c\overline{p}-1}$ varies on the image of this circle inverted about the unit circle and then reflected over the $x$-axis; since the circle does not pass through $0$, this is a circle. Then $g+ \frac{h}{c\overline{p}-1}$ is the image of this circle under a (possibly degenerate) spiral similarity and then a (possibly degenerate) translation; this is still a circle (note that it's fine if the circle has radius $0$ in the case of $h=0$). Thus $q$ lies on a fixed circle. Thus we're done. QED.

Since currently there isn't a synthetic solution in the thread for the version proposed to IMO 2019, I will post my solution for the case $n=5, k=2$ of the general problem mentioned in #6 (which is generalizable for any $n$ and works for the IMO proposal as well). Henceforth, we will consider only the following problem: Special case of IMO 2019 proposal wrote: Let $ABCDE$ be a cyclic convex pentagon, such that there exists a point $P$ in its interior, satisfying $PA \perp BE$, $PB \perp AC$, etc. If $AC \cap BE=A_1$, $BD \cap AC=B_1$, $CE \cap BD=C_1$ and so on, then $A_1B_1C_1D_1E_1$ is cyclic.

VicKmath7 wrote: Since currently there isn't a synthetic solution for the version proposed to IMO 2019, I will post my solution for the case $n=5, k=2$ of the general problem mentioned in #6 (which is generalizable for any $n$ and works for the IMO proposal as well). Henceforth, we will consider only the following problem: Special case of IMO 2019 proposal wrote: Let $ABCDE$ be a cyclic convex pentagon, such that there exists a point $P$ in its interior, satisfying $PA \perp BE$, $PB \perp AC$, etc. If $AC \cap BE=A_1$, $BD \cap AC=B_1$, $CE \cap BD=C_1$ and so on, then $A_1B_1C_1D_1E_1$ is cyclic.

I think @ABCDE said there is a solution without inversion.

I guess I'm sort of posting this just so I can have better retention of the solution, though I didn't actually solve. It certainly is very beautiful and just to help myself learn I'll try to section this off with motivation for each part. Below is vEnhance's diagram from OTIS. (gah its squished!) [asy][asy] size(16cm); pair P_1 = dir(178); pair P_2 = dir(136); pair P_3 = dir(82); pair P_4 = dir(32); pair K_2 = foot(P_2, P_1, P_3); pair K_3 = foot(P_3, P_2, P_4); pair P = extension(P_2, K_2, P_3, K_3); pair K_4 = foot(P_3, P, P_4); pair P_5 = -P_3+2*foot(origin, P_3, K_4); pair K_5 = foot(P_4, P, P_5); pair P_6 = -P_4+2*foot(origin, P_4, K_5); pair K_1 = foot(P_2, P, P_1); pair P_0 = -P_2+2*foot(origin, P_2, K_1); fill(P--P_1--P_2--cycle, opacity(0.1)+paleblue); fill(P--P_2--P_3--cycle, opacity(0.1)+lightgreen); fill(P--P_3--P_4--cycle, opacity(0.1)+paleblue); fill(P--P_4--P_5--cycle, opacity(0.1)+lightgreen); draw(unitcircle, blue); draw(P_0--P_1--P_2--P_3--P_4--P_5--P_6, blue); draw(P--P_1, red); draw(P--P_2, red); draw(P--P_3, red); draw(P--P_4, red); draw(P--P_5, red); draw(P_1--P_3--P_5, deepgreen); draw(P_0--P_2--P_4--P_6, deepgreen); draw(P_0--P_3, orange); draw(P_1--P_4, orange); draw(P_2--P_5, orange); draw(P_3--P_6, orange); pair H_1 = orthocenter(P, P_1, P_2); pair H_2 = orthocenter(P, P_2, P_3); pair H_3 = orthocenter(P, P_3, P_4); pair H_4 = orthocenter(P, P_4, P_5); pair Q_2 = extension(P_0, P_3, P_1, P_4); pair Q_3 = extension(P_1, P_4, P_2, P_5); pair Q_4 = extension(P_2, P_5, P_3, P_6); pair N = circumcenter(K_1, K_2, K_3); draw(CP(N, K_1), deepgreen); pair E = 2*N-P; pair E_2 = foot(E, P_1, P_3); pair E_3 = foot(E, P_2, P_4); pair E_4 = foot(E, P_3, P_5); draw(E_2--E_3--E_4, lightblue+1.5); draw(K_2--K_3--K_4, orange+1.5); draw(P--H_1--E, deepcyan); draw(P--H_2--E, deepcyan); draw(P--H_3--E, deepcyan); draw(P--E); draw(E_2--E, brown); draw(E_3--E, brown); draw(E_4--E, brown); pair F_1 = extension(E, H_1, P_0, P_3); pair F_2 = extension(E, H_2, P_1, P_4); pair F_3 = extension(E, H_3, P_2, P_5); draw(E_2--E_4, deepgreen+1.5); dot("$P_1$", P_1, dir(P_1)); dot("$P_2$", P_2, dir(P_2)); dot("$P_3$", P_3, dir(P_3)); dot("$P_4$", P_4, dir(P_4)); dot("$K_2$", K_2, dir(K_2)); dot("$K_3$", K_3, dir(K_3)); dot("$P$", P, dir(P)); dot("$K_4$", K_4, dir(K_4)); dot("$P_5$", P_5, dir(P_5)); dot("$K_5$", K_5, dir(K_5)); dot(P_6); dot("$K_1$", K_1, dir(K_1)); dot(P_0); dot("$H_1$", H_1, dir(H_1)); dot("$H_2$", H_2, dir(H_2)); dot("$H_3$", H_3, dir(H_3)); dot("$H_4$", H_4, dir(H_4)); dot("$Q_2$", Q_2, dir(Q_2)); dot("$Q_3$", Q_3, dir(Q_3)); dot("$Q_4$", Q_4, dir(Q_4)); dot(N); dot("$E$", E, dir(E)); dot("$E_2$", E_2, dir(E_2)); dot("$E_3$", E_3, dir(E_3)); dot("$E_4$", E_4, dir(E_4)); dot(F_1); dot(F_2); dot(F_3); /* TSQ Source: !size(16cm); P_1 = dir 178 P_2 = dir 136 P_3 = dir 82 P_4 = dir 32 K_2 = foot P_2 P_1 P_3 K_3 = foot P_3 P_2 P_4 P = extension P_2 K_2 P_3 K_3 K_4 = foot P_3 P P_4 P_5 = -P_3+2*foot(origin, P_3, K_4) K_5 = foot P_4 P P_5 P_6 .= -P_4+2*foot(origin, P_4, K_5) K_1 = foot P_2 P P_1 P_0 .= -P_2+2*foot(origin, P_2, K_1) !fill(P--P_1--P_2--cycle, opacity(0.1)+paleblue); !fill(P--P_2--P_3--cycle, opacity(0.1)+lightgreen); !fill(P--P_3--P_4--cycle, opacity(0.1)+paleblue); !fill(P--P_4--P_5--cycle, opacity(0.1)+lightgreen); unitcircle blue P_0--P_1--P_2--P_3--P_4--P_5--P_6 blue P--P_1 red P--P_2 red P--P_3 red P--P_4 red P--P_5 red P_1--P_3--P_5 deepgreen P_0--P_2--P_4--P_6 deepgreen P_0--P_3 orange P_1--P_4 orange P_2--P_5 orange P_3--P_6 orange H_1 = orthocenter P P_1 P_2 H_2 = orthocenter P P_2 P_3 H_3 = orthocenter P P_3 P_4 H_4 = orthocenter P P_4 P_5 Q_2 = extension P_0 P_3 P_1 P_4 Q_3 = extension P_1 P_4 P_2 P_5 Q_4 = extension P_2 P_5 P_3 P_6 N .= circumcenter K_1 K_2 K_3 CP N K_1 deepgreen E = 2*N-P E_2 = foot E P_1 P_3 E_3 = foot E P_2 P_4 E_4 = foot E P_3 P_5 E_2--E_3--E_4 lightblue+1.5 K_2--K_3--K_4 orange+1.5 P--H_1--E deepcyan P--H_2--E deepcyan P--H_3--E deepcyan P--E E_2--E brown E_3--E brown E_4--E brown F_1 .= extension E H_1 P_0 P_3 F_2 .= extension E H_2 P_1 P_4 F_3 .= extension E H_3 P_2 P_5 E_2--E_4 deepgreen+1.5 */ [/asy][/asy] 1. Recognizing the configuration. I was able to get this part, at least. There isn't really much to it, but it's very important to draw $H_{\bullet}$ (which, for some reason, I neglected; doing so would have likely advanced me much farther in the question, making me very sad). Without this construction, you can still find that $P$ essentially gets the pedal polygons $K_{2i}$ and $K_{2i+1}$ for two cyclic $50$-gons. This is clearly not very nice to work with. If you have a brain (not me), then you construct $H_1H_2\dots H_{100}$, and realize (after either angle chase, which is not clean, and inversion---which I missed since I'm stupid---which is clean) that $K_{\bullet}$ are cyclic. So guess what? By a well-known configuration, it follows that there should exist a point $E$ which is the reflection of $P$ across the center of $K_{\bullet}$ which is the isogonal conjugate of $P$. Specifically, we can also define the pedal polygon $E_{\bullet}$ of $E$, and then the circle passing through $E_{\bullet}$ also passes through $K_{\bullet}$. 2. Guessing the center of homothety. For this step, I guessed the point $E$ first, and I was clearly wrong. Unfortunately, this led me to give up on the question. But homothety in the first place is motivated by Reim's to find $K_2K_3\parallel Q_2Q_3$ which coincides with $P_1P_4$. When I discovered this, I immediately thought of homothety. At the same time, proving a local statement about $Q_{\bullet}$ seemed impossible (it wasn't) and so I was motivated to instead prove something like $Q_1Q_3\parallel E_1E_3$. This was the extent of my progress. (how many points?) I'm still figuring out how to motivate the collinearity claim $E_3Q_3E$. I think I knew the whole time that more direct proofs that $K_{\bullet}$ and $Q_{\bullet}$ were homothetic would fail, so the main point, I guess, is to include more points in homotheties. Specifically, I should consider the following: Under homothety from $K_{\bullet}$ to $Q_{\bullet}$, where do other points go? I need to Transform the whole diagram so I don't miss things. In this sense, I suppose, homothety should be treated more like inversion. In any case, it's not unreasonable to conjecture that $P$ is sent to $E$ (I'm still a little shaky on this foundation, but we'll have to practice more, I suppose!). Then perpendicularity gives the collinear claim. 3. Finishing with (much) easier synthetic observations. I won't deny that Evan's diagram is definitely assisting me here, but I still want to provide motivation for the way it's easy to reverse-engineer from here. First, it's very nice that the diagram is so smooth and symmetric; I suppose I should be more clever in my construction of diagrams. This step alone would greatly help in understanding problems. (Note: After this point, it's all my work---admittedly, though, it's made much easier with Evan's diagram. Just wanted to put that out there! I'm not completely useless!) Second, we employ linearity to give nice claims. We have to do some work, though. Since $H_{\bullet}P$ and $H_{\bullet}E$ are isogonal, this implies $H_iE\perp P_{i-1}P_{i+2}$. Let's use $\triangle EH_2H_3$ as the reference triangle. If we move $Q_3$ along $E_3E$ to the orthocenter of the reference triangle, we should expect $P_2$ to move to $H_2$ and $P_4$ to $H_3$. That means it suffices to show \[\frac{E_3H_2}{E_3P_2}=\frac{E_3H_3}{E_3P_4}.\]It actually just suffices to show that $E_2E_3E_4$ is homothetic to $P_2P_3P_4$ then. Oh wait, oops. In the midst of typing this I found an astonishing finish: reflect across the center of $K_{\bullet}$, notice that $E_2E_4$ is sent to $E_2'E_4'$ where $E_2'$ actually lies on $PK_2$. Well, that means we can use Reim's one last time. Since $E_2'E_4'K_2K_4$ is cyclic and $P_2P_4K_4K_2$ is cyclic, we get $E_2'E_4'\parallel P_2P_4$. End proof. Everything bolded is something insanely important which I should remember. There's one last thing I'd like to add: I feel like it's important to have an explicit set of skills to apply when stuck. I guess Evan calls these hard skills, while the bolded things in my analysis are what he would call soft skills. Whew!

I was able to complete all up until the second claim by myself. I needed hints to prove the second claim tho . Anyways, great problem (new favorite), took me on an 8 hour journey!!! Define $R_i=PP_i\cap P_{i-1}P_{i+1}$ and let $H_i=P_{i-1}P_{i+1}\cap P_iP_{i+2}$ be the orthocenter of $PP_iP_{i+1}$. Observe the following $\mathcal R$ is cyclic by inversion at $P$. Denote its circumcircle by $(\mathcal R)$. Let $S_i=(\mathcal R)\cap P_{i-1}P_{i+1}\neq R_i$. Then, the perpendiculars from $S_i$ to $P_{i-1}P_{i+1}$ all pass through the reflection of $P$ around the center of $(\mathcal R)$ by homothety By reim on $(\mathcal R)$ and $(P_iR_iP_{i+1}R_{i+1})$ we have $P_{i}P_{i+1}\parallel S_iS_{i+1}$. Then, we obtain $\mathcal P$ and $\mathcal S$ are homothetic from Reim on $(\mathcal R)$ and $(H_iR_iH_{i+1}R_{i+2})$ to get $P_{i-1}P_{i+1}\parallel S_{i-1}S_{i-2}$. The quadrilateral is cyclic since \[P_iH_{i-1}\cdot P_iR_{i-1}=P_iR_i\cdot P_i\cdot P_iP=P_i\cdot H_{i+1}\cdot P_i\cdot H_{i+2}\] Let $S$ be the reflection of $P$ across the center of $(\mathcal R)$, so, as established above, $\mathcal S$ is the pedal polygon of $S$. Claim: $SH_i \perp P_{i-1}P_{i+2}$ Proof. Let the intersections of the two above lines be $X_i$. Then, \begin{align*} & \measuredangle S_{i+1}P_{i+2}X_i=\measuredangle H_iP_{i+2}P_{i-1}=\measuredangle H_iR_{i+1}R_i=\measuredangle S_{i+1}R_{i+1}R_i \\ & =\measuredangle S_{i+1}S_iR_i=\measuredangle S_{i+1}S_iH_i=\measuredangle S_{i+1}SH_i=\measuredangle S_{i+1}SX_i \end{align*}Where we used $R_iR_{i+1}\parallel P_{i-1}P_{i+2}$ from Reim and $SS_iH_iS_{i+1}$ cyclic from right angles. Then, $X_iSP_{i+2}S_{i+1}$ cyclic and as $\angle P_{i+2}S_{i+1}S=90^{\circ}$ the result follows. $\blacksquare$ We now make the following key claim. Claim: [Impossible -- needed hints] $S_i,Q_i,S$ are collinear. Proof. First, using the last bullet above, we have \[\frac{S_iH_i}{H_iP_{i+1}}=\frac{H_iS_{i+1}}{H_iP_i}=\frac {H_{i-1}S_{i-1}}{H_{i-1}P_i}=\frac{S_iH_{i-1}}{S_iP_{i-1}}\]Hence there is a homothety centered at $S_i$ taking $H_{i-1}H_i$ to $P_{i-1}P_{i+1}$. In this homothety $S$ is sent to $S'_i$ such that $P_{i+1}S'_i\parallel SH_i\perp P_{i-1}Q_{i}$ and $P_{i-1}S'_i\parallel SH_{i-1}\perp P_{i+1}Q_{i}$. Thus $Q_i$ is the orthocenter of $P_{i-1}P_{i+1}S'_i$ so $QS'_i\perp P_{i-1}P_{i+1}$. But by the definition of homothety, $S_i,S,S'_i$ are collinear as and this line is also perpendicular to $P_{i-1}P_{i+1}$ as desired. $\blacksquare$ Now, if $P_{i-1}P_{i+2}\cap SH_i=F_i$, then $S_iQ_iF_iH_i$ and $H_iF_iS_{i+1}Q_{i+1}$ are cyclic by right angles. Thus $SQ_i\cdot SS_i=SF_i\cdot SH_i=SQ_{i+1}\cdot SS_{i+1}$. Repeating, we get that an inversion at $S$ sends $\mathcal S$ to $\mathcal Q$ which finishes!! Remark: How would you be expected to get this in contest. Geogebra helped motivate lot of the constructions and i dont see how this is humanly possible on paper and under time pressure. Speaking of time pressure, this took me $8$ hours--$5$ spent on the ``impossible'' claim and the other $3$ on the rest of the problem. Remark: Claim $2$ might be one of the hardest claims I've ever attempted (don't judge difficulty by solution length ), only getting it after struggling for longer than the oallotted time for an olympiad and getting hints on the length chase. That being said, this claim is so beautiful, making this my new favorite problem!

We will prove the following. Let $P_1P_2\cdots{}P_{100}$ be the cyclic $100$-gon and let $P_k=P_{k+100}$ for all integers $k$. For all integers $n$ and $k$, define $Q_{n,k}=\overline{P_kP_{k+n}}\cap\overline{P_{k+1}P_{k+1+n}}$. Then, there exist coaxial circles $\cdots,\Gamma_{-1},\Gamma_0,\Gamma_1,\cdots$ such that $Q_{n,k}$ lies on $\Gamma_n$ for all integers $n$ and $k$. Let $O$ be the center of $\Gamma$ the circumcircle of $P_1P_2\cdots{}P_{100}$. Let $F_k$ be the foot from $P$ to $\overline{P_kP_{k+1}}$ for all integers $k$ and let $M_k$ be the foot from $O$ to $\overline{P_kP_{k+1}}$ for all integers $k$. Claim $1$. We have that $F_1,F_2,\cdots,F_{100},M_1,M_2,\cdots,M_{100}$ all lie on a circle. Proof. For all integers $k$, we have that $\overline{OP_k}$ and $\overline{PP_k}$ are isogonal in $\angle{}P_{k-1}P_kP_{k+1}$ since $\overline{PP_k}\perp\overline{P_{k-1}P_{k+1}}$, implying that $F_{k-1},M_{k-1},F_k,$ and $M_k$ lie on a circle centered at the midpoint of $\overline{OP}$. This implies the claim. Inverting $M_1,M_2,\cdots,M_{100}$ about $\Gamma$ gives that if $T_k$ is the intersection of the tangents to $\Gamma$ at $P_k$ and $P_{k+1}$ for all integers $k$, then $T_1,T_2,\cdots,T_{100}$ are concyclic. As $T_k=Q_{0,k}$ for all integers $k$ and $Q_{-n,k}=Q_{n,k-n}$ for all integers $k$ and $n$, it suffices to prove the statement for positive integers $n$ and verify that $\Gamma_0,\Gamma_1,\cdots$ are coaxial. Claim $2$. Let $n$ be an integer at least $4$. Let $X_1,X_2,\cdots,X_n,Z_1,Z_2,\cdots,Z_n$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,\cdots,n-1$. Then, let $A=\overline{X_1X_{n-1}}\cap\overline{X_2X_n},B=\overline{Z_1Z_{n-1}}\cap\overline{Z_2Z_n},C=\overline{X_1Z_1}\cap\overline{X_2Z_2},$ and $D=\overline{X_{n-1}Z_{n-1}}\cap\overline{X_nZ_n}$. If $Y_1,Y_2,\cdots,Y_n,W_1,W_2,\cdots,W_n$ are coconic, then $A,B,C,$ and $D$ are collinear. Proof. First, we prove a few statements. Lemma $2.1.1$. Let $X_1,X_2,X_3,X_4,Z_1,Z_2,Z_3,$ and $Z_4$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,3$. Then, let $A=\overline{X_1X_3}\cap\overline{X_2X_4},B=\overline{Z_1Z_3}\cap\overline{Z_2Z_4},C=\overline{X_1Z_1}\cap\overline{X_2Z_2},$ and $D=\overline{X_3Z_3}\cap\overline{X_4Z_4}$. If $Y_1,Y_2,Y_3,W_1,W_2,$ and $W_3$ are coconic, then $A,B,C,$ and $D$ are collinear. Proof. By Desargues's Theorem on $\triangle{}X_1AX_2$ and $\triangle{}Z_1BZ_2$ and Desargues's Theorem on $\triangle{}X_3AX_4$ and $\triangle{}Z_3BZ_4$ it suffices that $\overline{X_1X_3}\cap\overline{Z_1Z_3},\overline{X_2X_4}\cap\overline{Z_2Z_4},\overline{X_1X_2}\cap\overline{Z_1Z_2},$ and $\overline{X_3X_4}\cap\overline{Z_3Z_4}$ are collinear. We will prove that $\overline{X_1X_2}\cap\overline{Z_1Z_2},\overline{X_3X_4}\cap\overline{Z_3Z_4}$, and $\overline{X_1X_3}\cap\overline{Z_1Z_3}$ are collinear. The proof that $\overline{X_1X_2}\cap\overline{Z_1Z_2},\overline{X_3X_4}\cap\overline{Z_3Z_4},$ and $\overline{X_2X_4}\cap\overline{Z_2Z_4}$ are collinear is analogous. Define $Y$ as the intersection of $\overline{Y_2Y_3}$ and the tangent to $\gamma$ at $X_1$ and define $W$ as the intersection of $\overline{W_2W_3}$ and the tangent to $\gamma$ at $Z_1$. Also, define $E=\overline{Y_1Y_2}\cap\overline{W_2W_3}$ and $F=\overline{Y_2Y_3}\cap\overline{W_1W_2}$. Taking a point line duality around $\gamma$ gives that it suffices that $\overline{YW},\overline{Y_1W_1},$ and $\overline{Y_3W_3}$ concur. Then, by Brianchon's Theorem on $EY_1YYFW_1W$ we see that $\overline{EF},\overline{YW},$ and $\overline{Y_1W_1}$ concur and by Pascal's Theorem on $Y_1Y_2Y_3W_3W_2W_1$ we see that $\overline{EF},\overline{Y_1W_1},$ and $\overline{Y_3W_3}$ concur, so $\overline{YW},\overline{Y_1W_1},$ and $\overline{Y_3W_3}$ concur, proving the lemma. Lemma $2.1.2$. Let $X_1,X_2,X_3,X_4,Z_1,Z_2,Z_3,$ and $Z_4$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,3$. Then, let $A=\overline{X_1X_2}\cap\overline{X_3X_4},B=\overline{Z_1Z_2}\cap\overline{Z_3Z_4},C=\overline{X_1Z_1}\cap\overline{X_3Z_3},$ and $D=\overline{X_2Z_2}\cap\overline{X_4Z_4}$. If $Y_1,Y_2,Y_3,W_1,W_2,$ and $W_3$ are coconic, then $A,B,C,$ and $D$ are collinear. Proof. By Desargues's Theorem on $\triangle{}X_1AX_3$ and $\triangle{}Z_1BZ_3$ and Desargues's Theorem on $\triangle{}X_2AX_4$ and $\triangle{}Z_2BZ_4$ it suffices that $\overline{X_1X_2}\cap\overline{Z_1Z_2},\overline{X_3X_4}\cap\overline{Z_3Z_4},\overline{X_1X_3}\cap\overline{Z_1Z_3},$ and $\overline{X_2X_4}\cap\overline{Z_2Z_4}$ are collinear. This is true by the same argument as in lemma $2.1.1$. Claim $2.2$. Let $n$ be an integer at least $4$. Let $X_1,X_2,\cdots,X_n,Z_1,Z_2,\cdots,Z_n$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,\cdots,n-1$. Then, let $Y$ be the intersection of the tangents to $\gamma$ at $X_2$ and $X_{n-1}$ and let $W$ be the intersection of the tangents to $\gamma$ at $Z_2$ and $Z_{n-1}$. If $Y_1,Y_2,\cdots,Y_n,W_1,W_2,\cdots,W_{n-1},$ and $W_n$ lie on a conic $\omega$, then $Y_1,Y_{n-1},Y,W_1,W_{n-1},$ and $W$ are coconic. Proof. We will induct on $n$. For $n=4$ this is true by definition. Now, assume that it is true for $n=j$ for some integer $j\ge4$. We will prove that it is true for $n=j+1$. First, by taking a point line duality around $\omega$, we see that the intersections of the tangents to $\omega$ at $Y_k$ and $Y_{k+1}$ for $k=1,2,\cdots,j-1$ and the intersection of the tangents to $\omega$ at $W_k$ and $W_{k+1}$ for $k=1,2,\cdots,j-1$ are all coconic. Therefore, we see by the inductive hypothesis that the intersections of the tangents to $\omega$ at $Y_k$ and $Y_{k'}$ and the intersections of the tangents to $\omega$ at $Y_k$ and $Y_{k'}$ for $(k,k')=(1,2),(2,j-1),(j-1,j)$ are coconic. Then, by lemma $2.1.2$ with $\left(X_1,X_2,X_3,X_4,Z_1,Z_2,Z_3,Z_4\right)\rightarrow\left(Y_1,Y_2,Y_{j-1},Y_j,W_1,W_2,W_{j-1},W_j\right)$ we see that if $C=\overline{Y_1W_1}\cap\overline{Y_{j-1}W_{j-1}}$ and $D=\overline{Y_2W_2}\cap\overline{Y_jW_j}$, then $C$ and $D$ lie on $\overline{YW}$. By lemma $2.1.2$ with $\left(X_1,X_2,X_3,X_4,Z_1,Z_2,Z_3,Z_4\right)\rightarrow\left(Y_1,Y_2,Y_{j-1},Y_j,W_j,W_{j-1},W_2,W_1\right)$ we see that if $E=\overline{Y_1W_j}\cap\overline{Y_{j-1}W_2}$ and $F=\overline{Y_2W_{j-1}}\cap\overline{Y_jW_1}$, then $E$ and $F$ lie on $\overline{YW}$. Let $Y'_k$ be the intersection of the tangents to $\omega$ at $Y_k$ and $Y_{k+j-2}$ and let $W'_k$ be the intersection of the tangents to $\omega$ at $W_k$ and $W_{k+j-2}$ for $k=1,2$. Note by Brokard's Theorem that $C$ lies on $\overline{Y'_1W'_1}$, that $D$ lies on $\overline{Y'_2W'_2}$, that $E$ lies on $\overline{Y'_1W'_2}$, and that $F$ lies on $\overline{Y'_2W'_1}$. By Desargues's Involution Theorem on $Y'_1Y'_2W'_1W'_2$ with line $\overline{YW}$ we see that there is an involution swapping $C$ and $D$, swapping $E$ and $F$, and swapping $Y$ and $W$. Therefore, by the converse of Desargues's Involution Theorem on $Y_1Y_jW_1W_j$ line $\overline{YW}$ we see that $Y_1,Y_j,Y,W_1,W_j,$ and $W$ are coconic, completing the induction. Then, the claim follows from lemma $2.1.1$ with $\left(X_1,X_2,X_3,X_4,Z_1,Z_2,Z_3,Z_4\right)\rightarrow\left(X_1,X_2,X_{j-1},X_j,Z_1,Z_2,Z_{j-1},Z_j\right)$. Claim $3$. Let $n$ be an integer at least $4$. Let $X_1,X_2,\cdots,X_n,Z_1,Z_2,\cdots,Z_n$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,\cdots,n-1$. Then, let $A=\overline{X_1X_{n-1}}\cap\overline{X_2X_n}$ and $B=\overline{Z_1Z_{n-1}}\cap\overline{Z_2Z_n}$. If $Y_1,Y_2,\cdots,Y_n,W_1,W_2,\cdots,W_n$ are on a conic $\omega$, then $A,B,$ and the four intersections of $\gamma$ and $\omega$ are coconic. Proof. Define $C=\overline{X_1Z_1}\cap\overline{X_2Z_2}$ and $D=\overline{X_{n-1}Z_{n-1}}\cap\overline{X_nZ_n}$, so by claim $2$, we have that $A,B,C,$ and $D$ are collinear. By Brokard's Theorem we see that $A$ is on $\overline{Y_1Y_{n-1}}$, that $B$ is on $\overline{W_1W_{n-1}}$, that $C$ is on $\overline{Y_1W_1}$, and that $D$ is on $\overline{Y_{n-1}W_{n-1}}$. Then, by Desargues's Involution Theorem on $Y_1Y_{n-1}W_1W_{n-1}$ with circumconic $\omega$ and line $\overline{AB}$, there exists an involution swapping $A$ and $B$, swapping $C$ and $D$, and swapping the two intersections of $\overline{AB}$ with $\omega$. Also, Desargues's Involution Theorem on $X_1X_{n-1}Z_1Z_{n-1}$ with circumconic $\gamma$ gives that there exists an involution swapping $A$ and $B$, swapping $C$ and $D$, and swapping the two intersections of $\overline{AB}$ with $\gamma$. These involutions must be the same, so there exists an involution swapping $A$ and $B$, swapping the two intersections of $\overline{AB}$ with $\gamma$, and swapping the two intersections of $\overline{AB}$ with $\omega$. This implies the claim by the converse of Desargues's Involution Theorem on the quadrilateral formed by the four intersections of $\gamma$ and $\omega$ with circumconics $\gamma$ and $\omega$. Claim $3$ implies the following. Let $n$ be an integer at least $4$ and let $\gamma$ and $\omega$ be conics. Let $X_1,X_2,\cdots,X_n$ be points on $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ for $k=1,2,\cdots,n-1$ such that $Y_1,Y_2,\cdots,Y_{n-1}$ are on $\omega$. Let $A=\overline{X_1X_{n-1}}\cap\overline{X_2X_n}$. As $X_1,X_2,\cdots,X_n$ vary, the point $A$ lies on a fixed conic through the four intersection points of $\gamma$ and $\omega$. By taking $\left(X_1,X_2,\cdots,X_n\right)=\left(P_{k+1},P_{k+2},\cdots,P_{k+n}\right)$ for all integers $k$, we see that there exists a conic $\Gamma_{n-2}$ through the four intersections of $\Gamma_0$ and $\Gamma=\Gamma_1$ such that $Q_{n-2,k}$ lies on $\Gamma_{n-2}$ for all integers $n\ge4$, meaning that $\Gamma_{n-2}$ is a circle coaxial with $\Gamma_0$ and $\Gamma_1$, so we are done.

Replace $100$ with general $n \geq 4$. Denote $\mathcal{M}$ as the polygon with vertices $M_i$ for some variable $M$, and $O_\Gamma$ the circumcenter of some circle $\Gamma$. Then also denote $A^{BC}$ as the projection of any vertex $A$ onto some line $BC$. Claim: $P_iR_iR_{i+k}P_{i+k}$ is cyclic. Let $R_i = P_iP \cap P_{i+1}P_{i-1}$. Then notice that $\measuredangle P_iR_iP_{i+1} = \measuredangle P_iR_{i+1}P_{i+1} = 90^\circ$ so $P_iR_iR_{i+1}P_{i+1}$ is cyclic. So then by PoP we get that $PR_i \cdot PP_i$ is fixed over all $i$ which implies the claim. Claim: $\mathcal R$ is cyclic. Inverting with radius $\sqrt{PR_i \cdot PP_i}$ swaps $P_i$ and $R_i$. Since $\mathcal{R}^\ast = \mathcal P$ is cyclic, we find that $\mathcal{R}$ is cyclic as well. Claim: $\overline{R_iR_{i+1}} \parallel \overline{Q_iQ_{i+1}}$ Now since $\measuredangle P_iP_{i+1}P_{i-1} = \measuredangle{P_iR_{i+1}R_i} = \measuredangle{P_iP_{i+2}P_{i-1}}$ it follows that $\overline{R_iR_{i+1}} \parallel \overline{Q_iQ_{i+1}}$. Claim: $T_i$ is the foot of the altitude from the isogonal conjugate of $P$ wrt $\mathcal R$, to $P_{i-1}P_{i+1}$. Let $S_i = P_iP_{i+2} \cap P_{i+1}P_{i-1}$, and let $R_iS_i \cap (\mathcal R) \neq R_i = T_i$. Then note that $P$ has an isogonal conjugate wrt $\mathcal S$ since the $P$-pedal circle is cyclic, which we will call $Q$. Then notice that $Q$ is the reflection of $P$ over $O_\mathcal{R}$. Then since $T_i$ is the reflection of $R_i$ over $O_{\mathcal R}^{R_iT_i}$ and $P_iO_{\mathcal R}^{R_iT_i}$ passes through $O_{\mathcal R}$, we have $T_i = Q^{S_iS_{i+1}}$). Claim: $\mathcal T$ is similar to $\mathcal P$. We first show that $T_{i}T_{i+k} \parallel P_iP_{i+k}$ which follows by Reim's on $(P_iP_{i+k}R_{i+k}R_i)$ and $(T_iT_{i+k}R_{i+k}R_i)$, and $P_i - T_i - Q$ so our claim is proven. Claim: $R_iR_{i+1} \perp QS_i$ First notice that $P$ is the orthocenter of $\triangle P_iP_{i+1}S_i$ and $S_iP$ is isogonal to $S_iQ$ wrt $\angle P_{i+1}S_iP_i$, so $S_iQ$ passes through $O_{(P_iP_{i+1}S_i)}$. Lemma: In $\triangle ABC$, $AO \perp EF$, where $E$ and $F$ are the feet of the altitudes from $B$ and $C$. Proof: $\angle OAB = 90^\circ - C$, and $\angle AFE = \angle C$ so $\angle AOF = 90^\circ$. Applying this lemma to $\triangle P_iP_{i+1}S_i$ gives us the desired. Claim: $Q \in \overline{T_iQ_i}$.

Note that we have $\triangle T_iS_{i-1}T_{i-1} \sim P_{i-1}S_{i-1}P_i$ since $P_iP_{i+1} \parallel T_iT_{i+1}$, so $S_i$ is the center of negative homothety sending $T_iT_{i+1}$ to $P_iP_{i+1}$. From this, we get that $\frac{T_iS_i}{S_iP_{i+1}} = \frac{T_iT_{i+1}}{P_iP_{i+1}} = \frac{T_{i+1}T_{i+2}}{P_{i+1}P_{i+2}} = \frac{T_{i+1}S_{i+1}}{S_{i+1}P_{i+2}}$ so $T_i$ is the center of homothety sending $S_{i-1}S_{i}$ to $P_iP_{i+2}$. Let $U_i$ be a point so that there exists a homothety at $T_i$ sending $\triangle S_{i}S_{i-1}U_i \to \triangle P_{i-1}P_iQ_i$. We first notice that by the homothety we have $S_{i-1}U_i \parallel Q_{i-1}Q_i \parallel R_iR_{i+1} \perp QS_i$ so $S_{i-1}U_i \perp QS_i$. Analogously we have $S_{i}U_i \perp QS_{i-1}$, so $U_i$ is the orthocenter of $QS_iS_{i-1}$. This implies that $Q - U_i - T_i$, however $U_i - T_i - Q_i$ by the homothety, so our claim is proven. Final Claim: $\mathcal Q \sim \mathcal P$ We have $QQ_i \perp P_{i-1}P_{i} \perp PR_i$ which implies that there is a homothety sending $\mathcal Q$ to $\mathcal R$ which also sends $Q$ to $P$ evidently, so we are done as this implies $\mathcal Q$ cyclic. Extra comments: this problem took me 10-11 hours over 2 days, I thoroughly enjoyed it even though i got stuck a ton

There is a further generalization, which we will prove. Let $m$ be a positive integer. Also, let $P_1P_2\cdots{}P_{100}$ be the cyclic $100$-gon and let $P_k=P_{k+100}$ for all integers $k$. For all integers $n$ and $k$ with $n\ge0$, define $Q_{n,k}=\overline{P_kP_{k+n}}\cap\overline{P_{k+m}P_{k+m+n}}$, with $\overline{P_aP_a}$ being the tangent to $\left(P_1P_2\cdots{}P_{100}\right)$ at $P_a$ for all integers $a$. Then, there exist conics $\Gamma_0,\Gamma_1,\Gamma_2,\cdots$ such that $Q_{n,k}$ lies on $\Gamma_n$ for all integers $n$ and $k$. Furthermore, there exist $4$ points which these conics all pass through.

. Let the second to last paragraph in the quote be claim $4$. Claim $5$. Let $n$ be an integer at least $2$. Let $X_1,X_2,\cdots,X_n$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ for $k=1,2,\cdots,n-1$. Then, let $Y$ be the intersection of the tangents to $\gamma$ at $X_1$ and $X_n$. Suppose that $Y_1,Y_2,\cdots,Y_{n-1}$ lie on a conic $\omega$. Then, as $X_1,X_2,\cdots,X_n$ vary while fixing $\gamma$ and $\omega$, we have that $Y$ lies on a fixed conic. Proof. When $n=2$ the conic is $\omega$. Otherwise, let $Y_0$ be the intersection of the tangent to $\gamma$ at $X_1$ with $\omega$ other than $Y_1$ and let $Y_n$ be the intersection of the tangent to $\gamma$ at $X_n$ with $\omega$ other than $Y_{n-1}$. Then, let $A=\overline{Y_0Y_{n-1}}\cap\overline{Y_1Y_n}$. The locus of $A$ when $X_1,X_2,\cdots,X_n$ vary while fixing $\gamma$ and $\omega$ is a conic $\Omega$ by the second to last paragraph in the quote with $n$ going to $n+1$, with $X_1,X_2,\cdots,X_n$ going to $Y_0,Y_2,\cdots,Y_n$, with $\gamma$ going to $\omega$, and with $\omega$ going to the dual of $\gamma$ over $\omega$. Then, by claim $2$ with $n$ going to $n+1$, with $X_1,X_2,\cdots,X_n,Z_1,Z_2,\cdots,Z_n$ going to $Y_0,Y_1,\cdots,Y_n,Y_0,Y_1,\cdots,Y_n$, with $\gamma$ going to $\omega$, and with $\omega$ going to the dual of $\gamma$ over $\omega$, we see that if $C$ is the intersection of the tangents to $\omega$ at $Y_0$ and $Y_1$ at $\omega$ and $D$ is the intersection of the tangents to $\omega$ at $Y_{n-1}$ and $Y_n$, the tangent to $\Omega$ at $A$ is $\overline{CD}$, which is the polar of $Y$ with respect to $\omega$. Therefore, we see that $Y$ lies on the dual of $\Omega$ over $\omega$, proving the claim. Claim $5$ implies that there exists a conic $\Gamma_0$ through $Q_{0,k}$ for all integers $k$. Claim $6$. Let $a$ and $b$ be positive integers and let $n=2a+b+1$. Let $X_1,X_2,\cdots,X_n,Z_1,Z_2,\cdots,Z_n$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,\cdots,n-1$. Suppose that $Y_1,Y_2,\cdots,Y_n,W_1,W_2,\cdots,W_{n-1},$ and $W_n$ lie on a conic $\omega$. Then, let $Y'_1$ be the intersection of the tangents to $\gamma$ at $X_1$ and $X_{a+1}$, let $Y'_2$ be the intersection of the tangents to $\gamma$ at $X_{n-a}$ and $X_n$, and let $Y$ be the intersection of the tangents to $\gamma$ at $X_{a+1}$ and $X_{n-a}$. Also, let $W'_1$ be the intersection of the tangents to $\gamma$ at $Z_1$ and $Z_{a+1}$, let $W'_2$ be the intersection of the tangents to $\gamma$ at $Z_{n-a}$ and $Z_n$, and let $W$ be the intersection of the tangents to $\gamma$ at $Z_{a+1}$ and $Z_{n-a}$. Then, we have that $Y'_1,Y'_2,Y,W'_1,W'_2,$ and $W$ are coconic. Proof. We will define points $\cdots,X_{-1},X_0,X_1,\cdots$ and $\cdots,Z_{-1},Z_0,Z_1,\cdots$ on $\gamma$ such that the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at a point $Y_k$ on $\omega$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at a point $W_k$ on $\omega$ for all integers $k$. We will induct on $a$ and $b$. The base case, where $a=b$, is true by claim $5$. Now, assume that $a\ne1$ and the claim is true when $(a,b)$ is replaced with $\left(\min(a,b),|a-b|\right)$. We will show that the claim is true, which would complete the induction since replacing $(a,b)$ with $\left(\min(a,b),|a-b|\right)$ repeatedly for positive integers $a$ and $b$ will always result in $a=b$ eventually since otherwise the quantity $a+b$ strictly decreases after an iteration. We will define $S_k$ as the intersection of the tangents to $\gamma$ at $X_k$ and $X_{b+k}$ and $T_k$ as the intersection of the tangents to $\gamma$ at $Z_k$ and $Z_{b+k}$ for all integers $k$, so that $Y=S_{a+1}$ and $W=T_{a+1}$. Note that there exists a conic $\gamma_b$ which passes through $\cdots,S_{-1},S_0,S_1,\cdots$ and $\cdots,T_{-1},T_0,T_1,\cdots$ by claim $5$. Then, let $U_k$ be the intersection of the tangents to $\gamma_b$ at $S_k$ and $S_{k+1}$ and let $V_k$ be the intersection of the tangents to $\gamma_b$ at $T_k$ and $T_{k+1}$. We will show that $\cdots,U_{-1},U_0,U_1,\cdots$ and $\cdots,V_{-1},V_0,V_1,\cdots$ all lie on a conic. Taking a point line duality around $\gamma_b$ gives that we want a conic tangent to $\overline{S_kS_{k+1}}$ and $\overline{T_kT_{k+1}}$ for all integers $k$. Then, taking a point line duality around $\gamma$ gives that we want that there exists a conic passing through $\overline{X_kX_{k+b}}\cap\overline{X_{k+1}X_{k+1+b}}$ and $\overline{W_kW_{k+b}}\cap\overline{W_{k+1}W_{k+1+b}}$ for all integers $k$, which is true by claim $4$. Define $A=\overline{YW}\cap\overline{S_1T_1},B=\overline{YW}\cap\overline{S_{n-b}T_{n-b}},C=\overline{YW}\cap\overline{S_1T_{n-b}},$ and $D=\overline{YW}\cap\overline{S_{n-b}T_1}$. If $a<b$, then by the inductive hypothesis, we see that the poles of $\overline{S_{1-b}S_{a+1-b}},\overline{S_1S_{a+1}},\overline{S_1S_{a+1-b}},\overline{T_{1-b}T_{a+1-b}},\overline{T_1T_{a+1}},$ and $\overline{T_1T_{a+1-b}}$ with respect to $\gamma_b$ are coconic. Then, since $Y'_1=\overline{S_{1-b}S_1}\cap\overline{S_{a+1-b}S_{a+1}}$ and $W'_1=\overline{T_{1-b}T_1}\cap\overline{T_{a+1-b}T_{a+1}}$, by lemma $2.1.1$, we see that $A$ lies on $\overline{Y'_1W'_1}$. By taking the relabeling sending $X_k$ to $X_{n+1-k}$ for all integers $k$ or the relabeling sending $Y_k$ to $Y_{n+1-k}$ for all integers $k$ or both and applying the same reasoning, we see that $B$ lies on $\overline{Y'_2W'_2}$, that $C$ lies on $\overline{Y'_1W'_2}$, and that $D$ lies on $\overline{Y'_2W'_1}$. If $a>b$, then by the inductive hypothesis, we see that the poles of $\overline{S_{1-b}S_1},\overline{S_{a+1-b}S_{a+1}},\overline{S_1S_{a+1-b}},\overline{T_{1-b}T_1},\overline{T_{a+1-b}T_{a+1}},$ and $\overline{T_1T_{a+1-b}}$ with respect to $\gamma_b$ are coconic. Then, since $Y'_1=\overline{S_{1-b}S_1}\cap\overline{S_{a+1-b}S_{a+1}}$ and $W'_1=\overline{T_{1-b}T_1}\cap\overline{T_{a+1-b}T_{a+1}}$, by lemma $2.1.2$, we see that $A$ lies on $\overline{Y'_1W'_1}$. By taking the relabeling sending $X_k$ to $X_{n+1-k}$ for all integers $k$ or the relabeling sending $Y_k$ to $Y_{n+1-k}$ for all integers $k$ or both and applying the same reasoning, we see that $B$ lies on $\overline{Y'_2W'_2}$, that $C$ lies on $\overline{Y'_1W'_2}$, and that $D$ lies on $\overline{Y'_2W'_1}$. Now, by Desargues's Involution Theorem on $S_1S_{n-b}T_1T_{n-b}$ with circumconic $\gamma_b$ and line $\overline{YW}$, we get that there exists an involution swapping $A$ and $B$, swapping $C$ and $D$, and swapping $Y$ and $W$. Therefore, by the converse of Desargues's Involution Theorem on $Y'_1Y'_2W'_1W'_2$ with line $\overline{YW}$, we see that $Y'_1,Y'_2,Y,W'_1,W'_2,$ and $W$ are coconic, proving the claim. Claim $7$. Let $a$ and $b$ be positive integers and let $n=2a+b+1$. Let $X_1,X_2,\cdots,X_n,Z_1,Z_2,\cdots,Z_n$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,\cdots,n-1$. Suppose that $Y_1,Y_2,\cdots,Y_n,W_1,W_2,\cdots,W_{n-1},$ and $W_n$ lie on a conic $\omega$. Then, let $A=\overline{X_1X_{1+a+b}}\cap\overline{X_{n-a-b}X_n}$ and $B=\overline{Z_1Z_{1+a+b}}\cap\overline{Z_{n-a-b}Z_n}$. We will define points $\cdots,X_{-1},X_0,X_1,\cdots$ and $\cdots,Z_{-1},Z_0,Z_1,\cdots$ on $\gamma$ such that the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at a point $Y_k$ on $\omega$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at a point $W_k$ on $\omega$ for all integers $k$. We will define $S_k$ as the intersection of the tangents to $\gamma$ at $X_k$ and $X_{a+k}$ and $T_k$ as the intersection of the tangents to $\gamma$ at $Z_k$ and $Z_{a+k}$ for all integers $k$, so that there exists a conic $\gamma_a$ which passes through $\cdots,S_{-1},S_0,S_1,\cdots$ and $\cdots,T_{-1},T_0,T_1,\cdots$ by claim $5$. Then, we have that $A,B,$ and the four intersections of $\gamma$ and $\gamma_a$ are coconic. Proof. Let $Y'_1$ be the intersection of the tangents to $\gamma$ at $X_1$ and $X_{a+1}$, let $Y'_2$ be the intersection of the tangents to $\gamma$ at $X_{n-a}$ and $X_n$, and let $Y$ be the intersection of the tangents to $\gamma$ at $X_{a+1}$ and $X_{n-a}$. Also, let $W'_1$ be the intersection of the tangents to $\gamma$ at $Z_1$ and $Z_{a+1}$, let $W'_2$ be the intersection of the tangents to $\gamma$ at $Z_{n-a}$ and $Z_n$, and let $W$ be the intersection of the tangents to $\gamma$ at $Z_{a+1}$ and $Z_{n-a}$. By claim $6$ we have that $Y'_1,Y'_2,Y,W'_1,W'_2,$ and $W$ are coconic. Define $C=\overline{X_1Z_1}\cap\overline{X_{a+1}Z_{a+1}}$ and $D=\overline{X_{n-a}Z_{n-a}}\cap\overline{X_nZ_n}$. Then, by lemma $2.1.1$ we see that $A,B,C,$ and $D$ are collinear. Also, we see that $A$ lies on $\overline{Y'_1Y'_2}$, that $B$ lies on $\overline{W'_1W'_2}$, that $C$ lies on $\overline{Y'_1W'_1}$, and that $D$ lies on $\overline{Y'_2W'_2}$. Now, by Desargues's Involution Theorem on $X_1X_{n-a}Z_1Z_{n-a}$ with circumconic $\gamma$ and line $\overline{AB}$ we see that there exists an involution swapping $A$ and $B$, swapping $C$ and $D$, and swapping the two intersections of $\overline{AB}$ with $\gamma$. Then, by Desargues's Involution Theorem on $Y'_1Y'_2W'_1W'_2$ with circumconic $\gamma_a$ and line $\overline{AB}$ there exists an involution swapping $A$ and $B$, swapping $C$ and $D$, and swapping the two intersections of $\overline{AB}$ with $\gamma_a$. These involutions must be the same, so there exists an involution swapping $A$ and $B$, swapping the two intersections of $\overline{AB}$ with $\gamma$, and swapping the two intersections of $\overline{AB}$ with $\gamma_a$. This implies the claim by the converse of Desargues's Involution Theorem on the quadrilateral formed by the four intersections of $\gamma$ and $\gamma_a$ with circumconics $\gamma$ and $\gamma_a$. Claim $8$. Let $a$ and $b$ be positive integers and let $n=2a+b+1$. Let $X_1,X_2,\cdots,X_n,Z_1,Z_2,\cdots,Z_n$ be points on a conic $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at $W_k$ for $k=1,2,\cdots,n-1$. Suppose that $Y_1,Y_2,\cdots,Y_n,W_1,W_2,\cdots,W_{n-1},$ and $W_n$ lie on a conic $\omega$. Then, let $A=\overline{X_1X_{1+a}}\cap\overline{X_{n-a}X_n}$ and $B=\overline{Z_1Z_{1+a}}\cap\overline{Z_{n-a}Z_n}$. We will define points $\cdots,X_{-1},X_0,X_1,\cdots$ and $\cdots,Z_{-1},Z_0,Z_1,\cdots$ on $\gamma$ such that the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at a point $Y_k$ on $\omega$ and the tangents to $\gamma$ at $Z_k$ and $Z_{k+1}$ intersect at a point $W_k$ on $\omega$ for all integers $k$. We will define $S_k$ as the intersection of the tangents to $\gamma$ at $X_k$ and $X_{a+b+k}$ and $T_k$ as the intersection of the tangents to $\gamma$ at $Z_k$ and $Z_{a+b+k}$ for all integers $k$, so that there exists a conic $\gamma_{a+b}$ which passes through $\cdots,S_{-1},S_0,S_1,\cdots$ and $\cdots,T_{-1},T_0,T_1,\cdots$ by claim $5$. Then, we have that $A,B,$ and the four intersections of $\gamma$ and $\gamma_{a+b}$ are coconic. Proof. Let $Y'_1$ be the intersection of the tangents to $\gamma$ at $X_1$ and $X_{a+1}$, let $Y'_2$ be the intersection of the tangents to $\gamma$ at $X_{n-a}$ and $X_n$, and let $Y$ be the intersection of the tangents to $\gamma$ at $X_{a+1}$ and $X_{n-a}$. Also, let $W'_1$ be the intersection of the tangents to $\gamma$ at $Z_1$ and $Z_{a+1}$, let $W'_2$ be the intersection of the tangents to $\gamma$ at $Z_{n-a}$ and $Z_n$, and let $W$ be the intersection of the tangents to $\gamma$ at $Z_{a+1}$ and $Z_{n-a}$. By claim $6$ we have that $Y'_1,Y'_2,Y,W'_1,W'_2,$ and $W$ are coconic. Let $Y'_3$ be the intersection of the tangents to $\gamma$ at $X_1$ and $X_{a+b+1}$, let $Y'_4$ be the intersection of the tangents to $\gamma$ at $X_{n-a-b}$ and $X_n$, let $W'_3$ be the intersection of the tangents to $\gamma$ at $Z_1$ and $Z_{a+b+1}$, and let $W'_4$ be the intersection of the tangents to $\gamma$ at $Z_{n-a-b}$ and $Z_n$. Define $C=\overline{X_1Z_1}\cap\overline{X_{a+b+1}Z_{a+b+1}}$ and $D=\overline{X_{n-a-b}Z_{n-a-b}}\cap\overline{X_nZ_n}$. Then, by lemma $2.1.2$ we see that $A,B,C,$ and $D$ are collinear. Also, we see that $A$ lies on $\overline{Y'_3Y'_4}$, that $B$ lies on $\overline{W'_3W'_4}$, that $C$ lies on $\overline{Y'_3W'_3}$, and that $D$ lies on $\overline{Y'_4W'_4}$. Now, by Desargues's Involution Theorem on $X_1X_{n-a-b}Z_1Z_{n-a-b}$ with circumconic $\gamma$ and line $\overline{AB}$ we see that there exists an involution swapping $A$ and $B$, swapping $C$ and $D$, and swapping the two intersections of $\overline{AB}$ with $\gamma$. Then, by Desargues's Involution Theorem on $Y'_3Y'_4W'_3W'_4$ with circumconic $\gamma_{a+b}$ and line $\overline{AB}$ there exists an involution swapping $A$ and $B$, swapping $C$ and $D$, and swapping the two intersections of $\overline{AB}$ with $\gamma_{a+b}$. These involutions must be the same, so there exists an involution swapping $A$ and $B$, swapping the two intersections of $\overline{AB}$ with $\gamma$, and swapping the two intersections of $\overline{AB}$ with $\gamma_{a+b}$. This implies the claim by the converse of Desargues's Involution Theorem on the quadrilateral formed by the four intersections of $\gamma$ and $\gamma_{a+b}$ with circumconics $\gamma$ and $\gamma_{a+b}$. Claims $5,7,$ and $8$ imply the following. Let $m$ be a positive integer, let $n$ be an integer at least $m+2$, and let $\gamma$ and $\omega$ be conics. Let $X_1,X_2,\cdots,X_n$ be points on $\gamma$ and let the tangents to $\gamma$ at $X_k$ and $X_{k+1}$ intersect at $Y_k$ for $k=1,2,\cdots,n-1$ such that $Y_1,Y_2,\cdots,Y_{n-1}$ are on $\omega$. Let $A=\overline{X_1X_{n-m}}\cap\overline{X_{m+1}X_n}$ and let $B$ be the intersection of the tangents to $\gamma$ at $X_1$ and $X_{m+1}$. As $X_1,X_2,\cdots,X_n$ vary, the point $B$ lies on a fixed conic $\gamma_m$ and the point $A$ lies on a fixed conic through the four intersection points of $\gamma$ and $\gamma_m$. By taking $\left(X_1,X_2,\cdots,X_n\right)=\left(P_{k+1},P_{k+2},\cdots,P_{k+n}\right)$ for all integers $k$, we see that there exists a conic $\Gamma_{n-m-1}$ through the four intersections of $\Gamma_0$ and $\left(P_1P_2\cdots{}P_{100}\right)=\Gamma_m$ such that $Q_{n-m-1,k}$ lies on $\Gamma_{n-m-1}$ for all integers $n\ge{}m+2$, so we are done.

Solved with mcmp. Solution will be similar to ones above. Fantastic and difficult problem. The problem holds for a general polygon with at least $7$ sides; diagram is presented as a $7$-gon. Let the polygon $P_1P_2 \cdots P_{100}$ be denoted as $\mathcal{P}$. We start by defining $Y_n = P_nP_n \cap P_{n+1}P_{n+1}$, $X_n = P_nP_{n+2} \cap P_{n-1}P_{n+1}$, $F_n = PP_{n} \cap P_{n-1}P_{n+1}$, and $Q_n = P_{n-1}P_{n+1} \cap P_{n-1}P_{n+2}$, with indices taken $\bmod 100$. Let these respectively form the polygons $\mathcal{Y}$, $\mathcal{X}$, $\mathcal{F}$ and $\mathcal{Q}$. Claim: $\mathcal{F}$ is cyclic. From $\angle P_2F_2P_1 = \angle P_2F_1P_1 = 90$, we get that $P_1F_1F_2P_2$ cyclic. Repeating this argument gives $P_nF_nF_{n+1}P_{n+1}$ cyclic, all of them with radical center $P$. Therefore inversion at $P$ with radius $\sqrt{PF_n \cdot PP_n}$ switches the vertices of $\mathcal{P}$ and $\mathcal{F}$, implying $\mathcal{F}$ is cyclic. Note also that the right angles also give that $X_n$ is the orthocenter of $PP_nP_{n+1}$. Claim: $P$ and $O$ are isogonal conjugates in $\mathcal{P}$. This is from angle chase. $\angle{P'P_2P_1}=90-\angle{P_2P'X1}=90-\angle{P2P1X3} = \frac{180-\angle{P_2OP_3}}{2} = \angle{OP_2P_3}$. Repeating this around the polygon gives the conclusion. This implies that the pedal circle of $\mathcal{P}$ (through the midpoints of the sides and the feet from $P$) exists. Claim: $\mathcal{X}$ and $\mathcal{Y}$ are cyclic. Applying the same inversion as before, vertices of $\mathcal{X}$ is sent to the feet of $P$ onto the sides of $\mathcal{P}$. Inverting about the circumcircle of $\mathcal{P}$ takes the vertices of $\mathcal{Y}$ to the midpoints of the sides of $\mathcal{P}$. These belong to the pedal circle, therefore $\mathcal{X}$ and $\mathcal{Y}$ are cyclic. Now observe that $P$ has a pedal circle in $\mathcal{X}$, therefore construct its isogonal conjugate $P'$. Let the feet from $P'$ (or second intersections of pedal circle with sides) be labeled with the corresponding $G$, and its polygon as $\mathcal{G}$. Claim: $\mathcal{G}$ is homothetic to $\mathcal{P}$. Consider Reim's in the lines $X_1P_1$ and $X_1P_2$ with the cyclic quadrilaterals $P_1F_1F_2P_2$ and $F_1F_2G_1G_2$ which then implies that $P_1P_2 \parallel G_1G_2$. Similarly $P_2P_3 \parallel G_2G_3$. Now consider Reim's in lines $P_2X_1$ and $P_2X_2$, cyclic quads $F_1F_3G_1G_3$ and $F_1X_1X_2F_3$ (from $P_2X_2 \cdot P_2F_3 = P_2F_2 \cdot P_2P = P_2X_1 \cdot P_2F_1$), which gives that $G_1G_3$ and $P_1P_3$ are parallel. This gives that $P_1P_2P_3$ and $G_1G_2G_3$ are homothetic, which after repeating this argument around the polygons gives the conclusion. Claim: The sides of $\mathcal{F}$ are parallel to the sides of $\mathcal{Q}$. Consider Reim's in lines $X_2P_2$ and $X_2P_3$. From cyclics $P_2P_3P_1P_4$ and $P_2P_3F_2F_3$, $F_2F_3 \parallel P_1P_4 = Q_2Q_3$. Repeating around the polygon gives the conclusion. Claim: $P'X_n \perp F_nF_{n+1}$. Observe in $\angle X_1X_2X_3$ that $X_2P'$ and $X_2P$ are isogonal. Now observe that $G_2G_3$ and $F_2F_3$ are antiparallel, therefore when translated to pass through $X_2$ are isogonal too. Now by isogonals, from $PX_2 \perp P_2P_3 \parallel G_2G_3$, $P'X_2 \perp F_2F_3$. Claim:$P'$, $Q_n$, $G_n$ collinear. Construct the orthocenter $H$ of $P'X_2X_3$, and note that $X_2HX_3$ and $P_2Q_3P_4$ are homothetic by the parallel lines. But now note that $\frac{G_3X_2}{G_3P_2} = \frac{G_2X_2}{G_2P_3} = \frac{G_4X_3}{G_4P_3} = \frac{G_3X_3}{G_3P_4}$, which gives that $G_3$ is the homothety center, and thus $G_3$, $H$, $Q_3$ and also $P'$ are collinear. This finishes as now $P'Q_nQ_{n+1}$ and $PF_{n}F_{n+1}$ have corresponding parallel sides, which, when repeated, gives that $\mathcal{F}$ is homothetic to $\mathcal{Q}$, thus $\mathcal{Q}$ is cyclic.