Why do scary Korean geos, end up by having such a simple solution. Claim 1: $TPEQ$ is cyclic. Proof: Let $ET \cap \Omega=H$ and let $AH \cap PT=I$, notice by Reim's Theorem on $\Omega, (DES)$ we get $AH \parallel ST$, so by Reim's $I$ lies in $(APF)$, now $\angle TPQ=\angle HAQ=\angle TEQ$ finishes. Finishing: Now let $BD \cap CE=G$, by Reim's Theorem in $(TPQ), (TSPF)$ we get $SF \parallel QE$, so By Pascal in the Hexagon $ABEDCQ$ we get $RG \parallel SF \parallel QE$, hence by Reim's theorem we get $TRQP$ cyclic, but now using all the parallels we get on the arcs of $\Omega$ that $\widehat{BH}=\widehat{AQ}=\widehat{DE}$ so $BD \parallel EH$ so in fact by Reim's we have $UTED$ to be a Isosceles Trapezoid, now $\angle TPE=\angle BCE=\angle BDE=\angle DUT$ so $UGPT$ is cyclic. Joining both info's we get $PTRU$ cyclic as desired, thus we are done .

Easiest KMO P4 prob. Ever. XD The solution itself is ez but personally I think it's a really well built problem

Kscv wrote: Easiest KMO P4 prob. Ever. XD The solution itself is ez but personally I think it's a really well built problem I personally think that 2022 KMO P4 is easier

Similar to the above solution, but slightly different. All angles are directed modulo $180$. We note $\angle PTE = \angle STE - \angle STP = \angle SDE - \angle SFP = \angle AQE - \angle AQP = \angle PQE$. Thus, $P$, $T$, $Q$, $E$ are cyclic. Note that this implies $\angle QEF = \angle STP = \angle AFP$, so $QE$ is parallel to $AD$. Then, let $K = BQ\cap AC$ and $L = BD\cap CE$. By Pascal on $DBQECA$, we get that $KL\parallel AD$. Therefore, we can again conclude that $K$, $L$, $P$, $T$ are cyclic. Finally, we can see that $BU\cdot BL= BU\cdot BD\cdot \frac{BK}{BS} = BS\cdot BT\cdot \frac{BK}{BS}=BK\cdot BT$. Thus, $K$, $U$, $L$, $T$ are cyclic. Similarly, $K$, $R$, $L$, $T$ are also cyclic. Therefore, $K$, $L$, $P$, $T$, $R$, $U$ are concyclic.

How many points is too many? This is a competitor to IMO 2022/4. Let $CE \cap BD = X, BQ \cap AC = Y$ Firstly $TPQE$ is cyclic because $\measuredangle ETP = \measuredangle ETS - \measuredangle PTS = 180 - \measuredangle ADE - \measuredangle AFP = \measuredangle AQE - \measuredangle AQP = \measuredangle EQP$. So By Reim's, $EQ \parallel AD$, again by Reim's $BXYC$ is cyclic, again by reim's $PTYX$ is cyclic. Next define $TP \cap (DES) = T^{*}$, By Reim's, $EC \parallel T^{*}D$, $\measuredangle PXU = \measuredangle T^{*}DX = \measuredangle PTU$ so $TPXU$ is cyclic. Next $\measuredangle PRY = \measuredangle PTY$ so $PRTY$ is cyclic. Combining all these, we get that $PXYTUR$ is cyclic and we're done.

First, notice that $$\measuredangle ETP = \measuredangle ETS + \measuredangle STP = \measuredangle SDE + \measuredangle PFS = \measuredangle EQA + \measuredangle AQP = \measuredangle EQP,$$so $Q, T, P, E$ are concyclic. By Reim, $QE \parallel AD, TP \parallel BC$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-19.21261,8.10049); pair B = (-14.07612,-4.27064); pair C = (-6.15349,-4.24179); pair D = (-0.71696,2.28422); pair E = (-5.19668,12.96974); pair F = (-5.70353,3.85233); pair P = (-5.33857,10.41739); pair Q = (-9.46782,14.30213); pair R = (-10.96746,0.30793); pair S = (-11.60052,5.70674); pair U = (-4.71322,0.32340); pair T = (-10.44053,10.38185); pair Z = (-8.84090,-1.70190); pair K = (-12.55632,1.80957); pair L = (-5.93288,-0.27327); import graph; size(10cm); pen ccwwff = rgb(0.8,0.4,1); pen ffttww = rgb(1,0.2,0.4); pen qqzzqq = rgb(0,0.6,0); pen qqwwzz = rgb(0,0.4,0.6); draw(circle((-10.14710,4.61258), 9.71333), linewidth(1) + ccwwff); draw(A--B, linewidth(1)); draw(B--C, linewidth(1)); draw(C--D, linewidth(1)); draw(D--E, linewidth(1)); draw(E--A, linewidth(1)); draw(C--E, linewidth(1)); draw(B--D, linewidth(1)); draw(A--D, linewidth(1)); draw(circle((-11.98084,7.49397), 7.25715), linewidth(1) + ffttww); draw(A--C, linewidth(1)); draw(B--Q, linewidth(1)); draw(circle((-5.32956,6.63225), 6.33888), linewidth(1) + ffttww); draw(circle((-7.86769,7.26201), 4.04387), linewidth(1) + qqzzqq); draw(circle((-7.85241,5.19529), 5.79567), linewidth(1) + linetype("4 4") + qqwwzz); dot("$A$", A, dir((-10, 0))); dot("$B$", B, dir((-10, -10))); dot("$C$", C, dir((10, -10))); dot("$D$", D, dir((10, -10))); dot("$E$", E, dir((10, 10))); dot("$F$", F, dir((-10, 20))); dot("$P$", P, dir((-10, -10))); dot("$Q$", Q, dir((0, 10))); dot("$R$", R, dir((5, 10))); dot("$S$", S, dir((-15, -10))); dot("$U$", U, dir((-5, 10))); dot("$T$", T, dir((10, -10))); dot("$Z$", Z, dir((-3, -10))); dot("$K$", K, dir((-10, -10))); dot("$L$", L, dir((10, -10))); [/asy][/asy] Now, let $X = AQ \cap DE, Y = BQ \cap CE$. Both of them clearly lies on the radical axis of two circles $(AQPFR)$ and $(DETSU)$. Let $K = BQ \cap AC, L = CE \cap DB$. By Pascal's theorem on hexagon $ACEDBQ$ and $ACEQBD$, we get $X, Y, Z$ are collinear and $KL \parallel AD$. Therefore $Z$ lies on the radical axis, and $(A, R, U, D)$ are concylic. By Reim, we get that $T, K, R, L, U, P$ are concyclic. $\blacksquare$