Here is my solution from the contest. There is an easier synthetic solution but I didn't find it at the time. [asy][asy] size(10cm); import geometry; defaultpen(fontsize(10pt)); pair A,B,C,D,R,S,M,P,N,Q,T,U,X,Y; A=dir(135); D=dir(85); C=dir(-20); B=dir(200); R=(2*A+B)/3; S=(5*C+2*D)/7; P=extension(A,D,R,S); Q=extension(B,C,R,S); X=extension(A,B,C,D); Y=extension(A,D,C,B); M=2*foot(A,circumcenter(P,A,R),circumcenter(X,A,D))-A; N=2*foot(C,circumcenter(Q,C,S),circumcenter(X,C,B))-C; T=foot((0,0),X,Y); U=extension(R,S,X,Y); draw(unitcircle); draw(circle(X,A,D)); draw(circle(X,B,C)); draw(circle(X,R,S)); draw(circle(P,A,R)); draw(circle(S,C,Q)); draw(circle(T,U,P),blue+dotted); draw(circle(T,U,Q),blue+dotted); draw(circle(M,N,P),blue+dotted); draw(X--extension(M,P,X,Y),green+dotted); draw(P--extension(M,P,X,Y),green+dotted); draw(Q--extension(M,P,X,Y),green+dotted); draw(U--Q); draw(X--B--C--cycle); draw(D--Y--B); draw(X--Y); draw(C--Q); dot("$A$",A,SE); dot("$D$",B,SW); dot("$C$",C,dir(270)); dot("$B$",D,NE); dot("$S$",R,dir(270)); dot("$R$",S,dir(270)); dot("$P$",P,SW); dot("$Q$",Q,ESE); dot("$L$",X,dir(90)); dot("$K$",Y,dir(270)); dot("$T$",T,dir(T)); dot("$U$",U,dir(T)); dot("$M$",M,NE); dot("$N$",N,W); clip(currentpicture, (-4,-1.1)--(3,-1.1)--(3,3)--(-4,3)--cycle); [/asy][/asy] Note that $M=(LAB)\cap (LRS)$ and $N=(LCD)\cap (LRS)$. Let $T=(LAB)\cap (LCD)$, which is known to lie on on $\overline{KL}$ (radax), and let $U=\overline{RS}\cap \overline{KL}$. Then I will prove that $(TUMP)$, $(TUNQ)$, and $(MPNQ)$ exist, after which the problem finishes by radax. $\measuredangle TMP=\measuredangle TMA+\measuredangle AMP=\measuredangle TLA+\measuredangle ASP=\measuredangle TUP$, so $T,U,M,P$ are cyclic. Similarly $T,U,N,Q$ are cyclic. $\measuredangle PMN=\measuredangle PMS+\measuredangle SMN=\measuredangle PAS+\measuredangle SLN=\measuredangle BCD+\measuredangle DCN=\measuredangle RCN=\measuredangle PQN$, so $M,N,P,Q$ are cyclic, as desired.

Yay, one more my problem on the international contest! This year Ukraine first time took part in APMO, and accordingly first time our problems were sent to APMO and, luckily, successfully. I thought that this problem is of APMO style so I sent it. How did you like it?

mshtand1 wrote: Yay, one more my problem on the international contest! This year Ukraine first time took part in APMO, and accordingly first time our problems were sent to APMO and, luckily, successfully. I thought that this problem is of APMO style so I sent it. How did you like it? Congratz!! I personally enjoyed the problem (even though I wasn't a contestant at 2024 APMO). Wish you luck with your G1-G8 goal

We use moving points. Fix $C,D,R,S$ and move $\overline{AB}$, keeping it parallel to a fixed line (this preserves the fact that $ABCD$ is cyclic). Then $P,M$ are fixed, $L$ is fixed while $K$ has degree $1$, and $Q$ has degree $1$. To find the degree of $N$, note that by Miquel's theorem we have $N=(LAB) \cap (LRS)$. An inversion at $L$ sends $(LRS)$ to a fixed line and sends $(LAB)$ to a variable line that stays parallel to a fixed line, so the image of $N$ under inversion moves with degree $1$ and hence $N$ has degree $2$. Furthermore, when $A=R$ we have $Q=N=R$, and likewise when $B=S$ we have $Q=N=S$ (assume these cases are distinct by continuity), so in fact $\deg \overline{QN} \leq 1+2-2=1$. Finally, when $A=B=L$ we have $N=L$, and since $K,Q,A,B$ are always collinear lines $\overline{KL}$ and $\overline{QN}$ coincide. Thus $X$ has degree at most $1+1-1=1$ and we want to show it lies on the fixed line $\overline{PM}$, so it suffices to check $2$ cases. The first case is simple: if $\overline{AB},\overline{CD},\overline{RS}$ are concurrent, all three lines $\overline{KL},\overline{QN},\overline{PM}$ share the point $K=Q=P$. The second case is more complicated: send $\overline{AB}$ to infinity. $\overline{KL}$ becomes the line through $L$ parallel to $\overline{CD}$ and $Q$ goes to $\infty_{\overline{RS}}$. $N$ gets sent to the intersection of the tangent to $(LCD)$ at $L$ and $(LRS)$: an inversion at $L$ sends $(LAB)$ to the line parallel through the (uninverted) $\overline{CD}$ passing through $L$, so $(LAB)$ should be the "line" through $L$ tangent to $(LCD)$ to begin with. Let $X=\overline{QN} \cap (LRS)$; I claim that this is the desired concurrency point. First, we have $\overline{LX} \parallel \overline{CD}$ because $\measuredangle XLC=\measuredangle XLS=\measuredangle NLR=\measuredangle LCD$, so $\overline{LX}=\overline{KL}$. Then let $P=\overline{CD} \cap \overline{RS}$; we have $\measuredangle MPD=\measuredangle MRD=\measuredangle MXL$. Since $\overline{PCD} \parallel \overline{XL}$ this implies $P,M,X$ are collinear, i.e. $\overline{PM},\overline{QN},\overline{KL}$ concurrent as desired. This finishes the problem. $\blacksquare$

ayeen_izady wrote: mshtand1 wrote: Yay, one more my problem on the international contest! This year Ukraine first time took part in APMO, and accordingly first time our problems were sent to APMO and, luckily, successfully. I thought that this problem is of APMO style so I sent it. How did you like it? Congratz!! I personally enjoyed the problem (even though I wasn't a contestant at 2024 APMO). Wish you luck with your G1-G8 goal I've actually already retired from creating high quality geometry problems. And the leaders' choices regarding the selection of geometry problems on past IMOs consistently decrease chances that I will regain motivation to come up with problems again. However, don't worry, I still have a plenty of unused ones to send to some international competitions, so you will still have a chance to enjoy more my problems, at least from Ukrainian contests

This problem is a lot easier with ggb. Note that $M$ is the Miquel point of $ABRSLP$ and $N$ is the Miquel point of $CRSDLQ$. Let $E \ne R$ be $NRCQ \cap BRMP$. Let $Y$ be the Miquel point of $ABCD$. Claim: $NLRSM$ is cyclic. Proof: $N, M$ both lie on $(LRS)$. Claim: $QEPK$ is cyclic Proof: $\measuredangle QEP + \measuredangle QKP = \measuredangle QER + \measuredangle REP + \measuredangle QKP = \measuredangle QCR + \measuredangle RBP + \measuredangle QKP = \measuredangle KCB + \measuredangle CBK + \measuredangle CKB = 0^\circ$. Claim: $LBAMY$ and $NLYCD$ are cyclic. Proof: It remains to show $M$ lies on $(LBAY)$. $\measuredangle BMA + \measuredangle ALB = \measuredangle BMP + \measuredangle PMA + \measuredangle ALB = \measuredangle BRP + \measuredangle PSA + \measuredangle ALB = \measuredangle LRS + \measuredangle RSL + \measuredangle SLR = 0^\circ$. Claim: $NQMP$ is cyclic. Proof: $\measuredangle QNM + \measuredangle MPQ = \measuredangle QNS + \measuredangle SNM + \measuredangle MPQ = \measuredangle CDA + \measuredangle PRM + \measuredangle MBR = \measuredangle CDA + \measuredangle PBM + \measuredangle MBR = \measuredangle CDA + \measuredangle PBR = 0^\circ$ As such, $X$ lies on $ER$ as $X$ is the radical center of $(NQMP), (NRCQ), (BRMP)$. Claim: $NEXYM$ is cyclic Proof: $\measuredangle NEM + \measuredangle MXN = \measuredangle NER + \measuredangle REM + \measuredangle NXM = \measuredangle NQR + \measuredangle RPM + \measuredangle MXN = 0^\circ$. Then $\measuredangle NYM = \measuredangle NYL + \measuredangle LYM = \measuredangle NCL + \measuredangle LAM = \measuredangle NCR + \measuredangle SAM = \measuredangle NQR + \measuredangle SPM = \measuredangle NXM$. Now, let $F$ = $RS \cap LYK$. Claim: $NYFQ$ and $YMPF$ are cyclic Proof: $\measuredangle YMP + \measuredangle PFY = \measuredangle YMA + \measuredangle AMP + \measuredangle PFY = \measuredangle YLA + \measuredangle AMP + \measuredangle PFY = \measuredangle AMP + \measuredangle PSA = 0^\circ$, symmetry gives the other. Then by radical axis on $(NYPF), (NMPQ), (YMPF)$ the result follows.

Another pure projective problem on APMO? Outline of proof: Take a projective transformation that sends $A, B, C, D$ to a rectangle. Then show $MP // NQ$, so $K, L, X$ lie on the line at infinity. PS: I am not sure if this proof is correct, if it is wrong please help to explain why. thanks!

eulerleonhardfan wrote: Another pure projective problem on APMO? Outline of proof: Take a projective transformation that sends $A, B, C, D$ to a rectangle. Then show $MP // NQ$, so $K, L, X$ lie on the line at infinity. PS: I am not sure if this proof is correct, if it is wrong please help to explain why. thanks! Circles arent preserved under projective transformations

Surprised this was P5, here are my 2 solutions from in contest[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.5175300171852, xmax = 11.601175595567593, ymin = -8.956055083733087, ymax = 7.410766922176333; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw(circle((4.254919711426577,-5.20298347684431), 3.7435131156477155), linewidth(0.4)); draw((xmin, 1.54261979760901*xmin-5.28438719926584)--(xmax, 1.54261979760901*xmax-5.28438719926584), linewidth(0.4)); /* line */ draw((xmin, -12.941176470588207*xmin + 89.11235294117625)--(xmax, -12.941176470588207*xmax + 89.11235294117625), linewidth(0.4)); /* line */ draw((xmin, -0.10546875000000003*xmin-2.021171875)--(xmax, -0.10546875000000003*xmax-2.021171875), linewidth(0.4)); /* line */ draw((xmin, -0.4144990785383048*xmin-4.086126855675013)--(xmax, -0.4144990785383048*xmax-4.086126855675013) , linewidth(0.4)); /* line */ draw((xmin, -0.5939595367280669*xmin-2.1697333752991956)--(xmax, -0.5939595367280669*xmax-2.1697333752991956), linewidth(0.4)); /* line */ draw(circle((0.8426595582669799,-2.064783447542708), 1.1492779427133082), linewidth(0.4)); draw(circle((3 .1232469441580526,-4.984028973908837), 4.551537011214573), linewidth(0.4)); draw(circle((9.531527920699311,-6.701979989357918), 2.1432526154016927), linewidth(0.4) + ccqqqq); draw(circle((5.56367459938652,-6.623443014817272), 5.452620481752408), linewidth(0.4) + ccqqqq); draw((xmin, 0.4610732271996606*xmin + 1.764487879216972)--(xmax, 0.4610732271996606*xmax + 1.764487879216972), linewidth(0.4)); /* line */ draw(circle((-3.0156517025322014,2.874013441845172), 5.567963936631288), linewidth(0.4) + ccqqqq); draw(circle((-2.266863245489841,-0.9752515580494889), 4.428345185108061), linewidth(0.4) + ccqqqq); draw(circle((1.32468546885748,-6.539539363532021), 9.559742688704302), linewidth(0.4) + ccqqqq); draw(circle((0.25265982197443293,-4.214473444189836), 7.515905516100535), linewidth(0.4) + blue); draw((xmin, 0.7812249313369485*xmin-1.7515075390816968)--(xmax, 0.7812249313369485*xmax-1.7515075390816968), linewidth(0.4)); /* line */ draw((xmin, 50.522105600837584*xmin-548.0197493633044)--(xmax, 50.522105600837584*xmax-548.0197493633044), linewidth(0.4)); /* line */draw((xmin, 3.7301884156842817*xmin-34.137848032365724)--(xmax, 3.7301884156842817*xmax-34.137848032365724), linewidth(0.4)); /* line */ draw((-6.682046355909883,-1.3164247984001296)--(7.3929280424441135,-6.560833490453235), linewidth(0.4)); draw((7.609109322964804,-5.75443658216716)-- (-6.682046355909883,-1.3164247984001296), linewidth(0.4)); draw((7.609109322964804,-5.75443658216716)--(2.5519080152048264,2.9411043433041417), linewidth(0.4)); draw((7.3929280424441135,-6.560833490453235)--(2.551908015204826 4,2.9411043433041417), linewidth(0.4)); draw(circle((6.733111604383132,-0.9128314619886586), 5.6864122824040315), linewidth(0.4) + blue); draw(circle((-5.828642262913379,-23.797304463991193), 22.497072011369145), linewidth(0.4) + blue); draw((7.609109322964804,-5.75443658216716)--(10.678639181615239,-8.512412956497565), linewidth(0.4)); draw((1.4577758821827405,-3.035593262933805)--(7.609109322964804,-5.75443658216716), linewidth(0.4)); draw((7.3929280424441135,-6.560833490453235)--(10.749381590183365,-4.938357520361102), linewidth(0.4)); /* dots and labels */ dot((1.98,-2.23),dotstyle); label("$A$", (2.0650556867798113,-2.002520878910299), NE * labelscalefactor); dot((7.1,-2.77),dotstyle); label("$B$", (7.1974261712918555,-2.5228533704276503), NE * labelscalefactor); dot((7.44,-7.17),dotstyle); label("$C$", (7.528546847711987,-7.631572378052557), NE * labelscalefactor); dot((0.612257312 6215318,-4.339906947584976),dotstyle); label("$D$", (0.7169215042121316,-4.107502321866857), NE * labelscalefactor); dot((-0.30412344374858513,-1.9890963555421415),dotstyle); label("$P$", (-0.2054860943868071,-1.742354633151623), NE * labelscalefactor ); dot((1.4577758821827405,-3.035593262933805),dotstyle); label("$R$", (0.9534362730836544,-3.4925639228008962), NE * labelscalefactor); dot((7.3929280424441135,-6.560833490453235),linewidth(4pt) + dotstyle); label("$S$", (7.481243893937683,-6.378044103033482), NE * labelscalefactor); dot((10.6786391816 15239,-8.512412956497565),linewidth(4pt) + dotstyle); label("$Q$", (10.768799181251849,-8.317465207779975), NE * labelscalefactor); dot((1.0467378714360498,-0.9337698173412851),linewidth(4pt) + dotstyle); label("$M$", (1.1426480881808727,-0.7489926038912246), NE * labelscalefactor); dot((10.749381590183365,-4.938357 520361102),linewidth(4pt) + dotstyle); label("$N$", (10.839753611913306,-4.74609219781997), NE * labelscalefactor); dot((6.51740319958198,4.769488005409629),linewidth(4pt) + dotstyle); label("$K$", (6.606139249113048,4.9510133259124895), NE * labelscalefactor); dot((-6.682046355909883,-1.316424798 4001296),linewidth(4pt) + dotstyle); label("$L$", (-6.591384853917922,-1.1274162340856622), NE * labelscalefactor); dot((0.8596528747804995,2.1608508044634824),linewidth(4pt) + dotstyle); label("$T$", (0.9534362730836544,2.349350868325732), NE * labelscalefactor); dot((2.5519080152048264,2.9411043433041417),linewidth(4pt) + dotstyle) ; label("$E$", (2.656342608958618,3.1298496056017595), NE * labelscalefactor); dot((7.609109322964804,-5.75443658216716),linewidth(4pt) + dotstyle); label("$L'$", (7.694107185922054,-5.573893888870303), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $L' = (ABS) \cap (CQS)$ Claim 1: $LPL'Q$ is cyclic Proof: Miquel on $LPSC$ Let $E = KL \cap (KEMNSR)$ Claim 2: $MPEL,MQEL$ are cyclic Proof: $\measuredangle MEL = \measuredangle MSB = \measuredangle MPA = \measuredangle MPL$ similarly other follows Claim 3: $ELL'S$ is cyclic Proof: $\measuredangle LL'S = \measuredangle LL'Q - \measuredangle SL'Q = \measuredangle LPQ - \measuredangle SNQ = \measuredangle AKP = \measuredangle LES$ Claim 4 $MPL'SB$ is cyclic Proof: $MPSB$ is cyclic by miquel, now $\measuredangle PL’S = \measuredangle PL’Q - \measuredangle BCD = \measuredangle PBS$ so $(MPL’SB)$ is cyclic. Now by radax we're done. $\blacksquare$ Solution 2 Claim 1: $MPNQ$ is cyclic Proof: $\measuredangle MPQ = \measuredangle MPR = \measuredangle MRP + \measuredangle PAR = \measuredangle MNS + \measuredangle SNQ = \measuredangle MNQ $ Now extend $MP,NQ$ to meet $(MRSN)$ at $J,I$ then by Reims theorem we have $KJI $homeothetic to $LPQ,$ so done.

(Solution from in contest) Let $\overline{MP}$, $\overline{NQ}$ meet $(LRS)$ at $P'$, $Q'$. Then, using cyclic quads $APSM$, we have $\overline{LP'}\parallel \overline{AB}$ and similarly $\overline{LQ'}\parallel\overline{CD}$. As $\overline{AB}$, $\overline{CD}$ are antiparallel wrt $\angle RLS$, $\overline{LP'}$ and $\overline{LQ'}$ are isogonal, and $\overline{P'Q'}\parallel\overline{RS}$. As those parallel lines mean triangles $KPQ$ and $LP'Q'$ are homothetic, the concurrency follows.

Really really beautiful problem. It's probably one of the most fun hard geometry problems I have ever solved. Pity I didn't try this problem as much as I would have liked in contest. I got caught up on Problem 3 which I thought I had a way to finish. Claim : $M$ lies on the circumcircle of $\triangle LAB$. Proof : This is a direct angle chase. Simply note that, \[\measuredangle BMA = \measuredangle BMP + \measuredangle PMA = \measuredangle BRP + \measuredangle PSA\]\[=\measuredangle LRS + \measuredangle RSL = \measuredangle RLS = \measuredangle BLA \]So it follows that $M$ indeed lies on $\triangle LAB$. Similarly, $N$ lies on $\triangle LCD$. Claim : Points $M$ and $N$ lie on the circumcircle of $\triangle LSR$. Proof : This should be true due to the spiral similarity but we prove it anyways. First observe that $N$ is the Miquel Point of $RSDC$ and $M$ is the Miquel Point of $ASRB$. Note that from this we can obtain that, \[\measuredangle LRN = \measuredangle CRN = \measuredangle DSN = \measuredangle LSN \]from which it is clear that $N$ lies on $(LSR)$. Similarly, $M$ also lies on $(LSR)$ which proves the claim. Now, we can also prove the following claim. Claim : Quadrilateral $MNQP$ is cyclic Proof : Again, we just angle chase. \[ \measuredangle QNM = \measuredangle QNR + \measuredangle RNM = \measuredangle DCR+ \measuredangle PSM = \measuredangle SAP + \measuredangle PSM \]\[ = \measuredangle SMP + \measuredangle PSM = \measuredangle SPM = \measuredangle QPM \] We are almost there. The final claim is as follows. Let $T$ be the second intersection of circles $(LAB)$ and $(LCD)$ i.e the Miquel Point of $ABCD$. Then, Claim : Quadrilateral $MNTX$ is cyclic. Proof : Note that, \[\measuredangle NTM = \measuredangle NTL + \measuredangle LTM = \measuredangle NCL + \measuredangle LAM = \measuredangle NCR + \measuredangle LAM \]\[ = \measuredangle NQR + \measuredangle SPM = \measuredangle XQP + \measuredangle QPX = \measuredangle QXP \]which implies that indeed $T$ lies on the circumcircle of $\triangle MNX$ as claimed. Now, what remains is a simple angle chase. It is well known that $T$ lies on $LK$ so we show $X-T-L$. Note that, \[\measuredangle NTX = \measuredangle NMX = \measuredangle XQP = \measuredangle XQR = \measuredangle NCR = \measuredangle NTL\]from which it is clear that points $X$ , $T$ and $L$ are collinear as claimed. So, we are done.

Let $QN\cap (LDCN)=F$ and $PM\cap (LABM)=E$. $M$ is the miquel point of $ABRS$ and $N$ is the miquel point of $CDSR$. Hence both $M$ and $N$ lye on $(LSR)$. \[180-\angle FLS=180-\angle FLD=\angle DNF=180-\angle QND=180-\angle QSD=\angle LSQ\]Thus, $LF\parallel \overline{PQRS}$. Similarily, \[180-\angle ELS=180-\angle ELA=180-\angle EMA=\angle AMP=180-\angle PSA=\angle ASR=\angle LSR \]So $LE\parallel \overline{PQRS}$. These two give that $L,E,F$ are collinear and $\overline{LEF}\parallel \overline{PQRS}$. \[\angle NMP=\angle NMS+\angle SMP=180-\angle SRN+\angle SAP=\angle NRQ+\angle DCB=\angle NCQ+\angle DCB=180-\angle RQN\]This yields $P,Q,M,N$ are cyclic. We get \[\angle EFN=\angle EFQ=\angle PQF=\angle PQX=\angle XMN=\angle EMN\]Hence $M,N,E,F$ are cyclic. We have $Pow(X,(LAB))=XE.XM=XF.XN=Pow(X,(LCD))$ thus, $X$ lies on the radical axis of $(LAB)$ and $(LCD)$ which is $KL$ as desired.$\blacksquare$

Solved with ggb at 3am. $M$ is the miquel point of the complete quadrilateral $ABRSPL$, which means $LMSR$ is cyclic. Likewise, $N$ is the miquel point of the complete quadrilateral $CDSRQL$ so $NLSR$ is cyclic. Therefore, $NLMSR$ is cyclic. Now, we claim that $PMNQ$ is cyclic. To that end, we have $\angle NMP = \angle NMS + \angle SMP = \angle NRQ + \angle PAS = \angle NRQ + \angle DCB = \angle NRQ + \angle QNR = 180 - \angle RQN = 180 - \angle PQN$. Let lines $PM, QN$ intersect $(NLMSR)$ for the second time at $F,G$ respectively. We will show that triangles $LFG$ and $KPQ$ are homothetic, which will finish the problem. We have $\angle FGX = \angle FMN = \angle PQX$ so $FG \parallel PQ$. Similarly, $\angle LGX = \angle LSN = \angle DQX \implies LG \parallel KQ$ and $\angle FLA = \angle FGS = \angle PMS = \angle PAS = \angle GAL \implies LF \parallel KP $. Thus the claim is proven and we're done.

My first APMO p5 !(also solved without geogebra ) The idea is to simplify the problem:more precisely eliminating the annoying points $M$ and $N$ And the magical idea is to consider $X'$ and $Y'$ such that $\overline{L-X'-Y'}\parallel PQ$ and $X'\in (LAB)$ and $Y'\in (LCD)$ Claim:$\overline{P-M-X'}$ and $\overline{Q-N-Y'}$ Proof:$M$ is miquel for $SABR$ so $PMBR$ cyclic and since $LMBX'$ cyclic by reim it is easy to get the first conclusion. The second one is similar:$QNCR, LNCY'$ are cyclic and finish again by reim. Now with this claim we eliminated the points M and N and we want to show that $PX'\cap QY'\cap KL \neq \emptyset$ For this we do in this way:WLOG $D-C-K$ collinear in this order(other case similar) and let $K_1=LK\cap PQ$.We will do trig bash. By homothety stuff it is enough to prove that $\frac{LY'}{LX'}=\frac{K_1Q}{K_1P}$.Denote by $t$ the angle $\angle LRS$ By law of sines in LBX' and LCY' we can see that $\frac{LY'}{LX'}=\frac{CD}{AB}\cdot \frac{sin(B+t)}{sin (A+t)}$ so we want to prove that $\frac{CD}{AB}\cdot \frac{sin(B+t)}{sin (A+t)}=\frac{K_1Q}{K_1P}$ Now by law of sines in $KK_1Q$ and $KK_1P$ we want to show that $\frac{CD}{AB}=\frac{K_1Q}{K_1P}:\frac{KQ}{KP}$ but the latter is just $\frac{sin \angle K_1KQ}{sin \angle K_1KP}$ and now angle chase and law of sines solves the problem (basicaly now we are working only with $A,B,C,D,K,L$)

Too easy for APMO 5? First, notice that $L, M, S, R, N$ are concyclic because $N$ is the Miquel point of $SRCD$ and $M$ is the Miquel point of $SRBA$. Then we may prove that $M, N, P, Q$ are concyclic by angle chasing: \[\measuredangle NMP = \measuredangle NMS + \measuredangle SMP = \measuredangle NRQ + \measuredangle DAB = \measuredangle NCQ + \measuredangle QCB = \measuredangle NQP\]Let $(LMSRN)$ intersect line $KL$ again at $T$. The claim is that $T, M, P, K$ are concyclic. Indeed, this is again by angle chasing: \[\measuredangle KTM = \measuredangle LTM = \measuredangle LRM = \measuredangle BRM = \measuredangle BPM = \measuredangle KPM\]Similarly, we get that $T, N, Q, K$ are concyclic. Now applying the Radical Axis theorem on $(MNPQ), (TMPK), (TNQK)$ gives us the desired result.

A solution via concyclic chasing. 1. Let $LM$ meet $AB$ at $E$ and $LN$ meet $CD$ at $F$. 2. By angle chasing, we have $L,M,N,S,R$ are concyclic, implying $M,N,P,Q$ are also concyclic. 3. More angle chasing, we have $M,A,D,E$ are concyclic, so are $N,B,C,F$. 4. The above facts imply $M,P,E,Q,F,N$ are concyclic. 5. Applying Pascal to $MPFNQE$, we are done.

Similar to #18. We check that $KL$ and $KX$ are the same line. By the spiral similarity configuration, we have $\Delta MAB \sim \Delta MSR$, so \[ \frac{\sin \angle KPX}{\sin \angle XPQ} = \frac{MA}{MS} = \frac{AB}{SR} \]and similarly $ \frac{\sin \angle KQX}{\sin \angle XQP} = \frac{CD}{SR}$ so by Trig Ceva on $\Delta KPQ$ we have \[ \frac{\sin \angle PKX}{\sin \angle XKQ} =\frac{\sin \angle KPX}{\sin \angle XPQ} \cdot \frac{\sin \angle XQP}{\sin \angle KQX} = \frac{AB}{SR} \cdot \frac{SR}{CD} = \frac{AB}{CD}. \] The other ratio can be computed easily: by Ratio Lemma and similar triangles we have \[ \frac{\sin \angle PKL}{\sin \angle LKQ} = \frac{BL}{LC} \cdot \frac{KC}{KB} = \frac{BL}{LC} \cdot \frac{AC}{BD} = \frac{LB}{BD} \cdot \frac{AC}{LC} = \frac{LA}{AC} \cdot \frac{AC}{LC} = \frac{LA}{LC} = \frac{AB}{CD} \]which solves the problem (ignoring sign issues).