Nice one!

Proof: Replace $P,Q$ by $E,F$. Let $\odot (NEF)$ intersect $BC$ at some point $L'$ by radical axis theorm (can be found on page 26 example 2.7 egmo proved on page 28.) we have that the radical axis of $\odot(NEF)$ and the given two circls are concurrent or that $KM,NL',AC$ are concurrent . But by pascals theorem on $BKMDNL$ (example 7.27 egmo page 135) we have that $BK \cap DN=A,KM \cap NL,BL \cap MD=C$ are collinear $\implies KM,NL,AC$ are concurrent $\implies L \equiv L'$ and we are done. $\blacksquare$.

Trivial! Sharygin 2020 CR P15 wrote: A circle passing through the vertices $B$ and $D$ of quadrilateral $ABCD$ meets $AB$, $BC$, $CD$, and $DA$ at points $K$, $L$, $M$, and $N$ respectively. A circle passing through $K$ and $M$ meets $AC$ at $P$ and $Q$. Prove that $L$, $N$, $P$, and $Q$ are concyclic. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -11.37, xmax = 11.37, ymin = -6.65, ymax = 6.65; /* image dimensions */ pen ffdxqq = rgb(1,0.8431372549019608,0); /* draw figures */ draw(circle((-0.07789654097849784,0.6489295730757243), 3.695012977275282), linewidth(1) + red); draw((-1.43,5.65)--(-3.71,-0.03), linewidth(1)); draw((-3.71,-0.03)--(3.572018589515622,0.07339465456700656), linewidth(1)); draw((0.75,4.25)--(0.844704973418243,-2.929048535837966), linewidth(1)); draw((-1.43,5.65)--(2.89,2.85), linewidth(1)); draw(circle((-2.191462782949008,-0.33954923437045126), 3.9904662982043706), linewidth(1) + ffdxqq); draw((-1.43,5.65)--(1.5307217335153656,-1.7780080285309827), linewidth(1)); draw(circle((0.9588267454729476,0.9035802942185787), 2.741893471682451), linewidth(0.8) + linetype("4 4") + green); /* dots and labels */ dot((-1.43,5.65),dotstyle); label("$A$", (-1.35,5.85), NE * labelscalefactor); dot((-3.71,-0.03),dotstyle); label("$B$", (-4.01,-0.39), NE * labelscalefactor); dot((0.81,0.03),dotstyle); label("$C$", (0.89,0.23), NE * labelscalefactor); dot((0.75,4.25),dotstyle); label("$D$", (0.83,4.45), NE * labelscalefactor); dot((2.89,2.85),dotstyle); label("$N$", (2.97,3.05), NE * labelscalefactor); dot((3.572018589515622,0.07339465456700656),dotstyle); label("$L$", (3.65,0.27), NE * labelscalefactor); dot((0.844704973418243,-2.929048535837966),dotstyle); label("$M$", (0.93,-2.73), NE * labelscalefactor); dot((-2.232533626375452,3.6507057027137866),dotstyle); label("$K$", (-2.19,3.23), NE * labelscalefactor); dot((-0.49,3.27),dotstyle); label("$P$", (-0.41,3.47), NE * labelscalefactor); dot((1.5307217335153656,-1.7780080285309827),dotstyle); label("$Q$", (1.61,-1.57), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Pascal's Theorem on the Hexagon $MDNLBK$ gives $MD\cap LB=C, DN\cap BK=A,NL\cap KM$ are collinear. So this implies $MK\cap NL\in AC$. Hence, $MK,NL,PQ$ are concurrent. Let this Concurrency Point be $T$. So, $$TL.TN=TM.TK=TQ.TP\implies L,N,P,Q\text{ are concyclic}. \blacksquare$$

Deserves the position of P1 Problem 15. A circle passing through the vertices $B$ and $D$ of quadrilateral $ABCD$ meets $AB, BC, CD,$ and $DA$ at points $K, L, M,$ and $N$ respectively. A circle passing through $K$ and $M$ meets $AC$ at $P$ and $Q$. Prove that $L, N, P,$ and $Q$ are concyclic. Solution. Proof. We begin by noting that $B,L,M,D,N,K$ all lie on a circle. Thus, by pascals theorem on hexagon $LBKMDN$ shows that $MK\cap NL$ lies on the line $CD$ and thus, $MK,NL,PQ$ are concurrent and let $X$ be this concurrency point. Now, we also know that $XK\cdot XM=XL\cdot XN$ because $L,M,N,K$ are concyclic points. Now, let $\omega_1,\omega_2$ denote the circumcircles of $\triangle NQL$ and cyclic quadrilateral $KPMO$. Clearly, $Q=\omega_1\cap\omega_2$ and thus, the radical axis of $\omega_1,\omega_2$ passes through $Q$. Also, power of $X$ with respect to $\omega_1$ and $\omega_2$ are $XL\cdot XN$ and $XM\cdot XK$ respectively, and they are the same as found earlier. Thus, $X$ also lies on the radical axis of $\omega_1,\omega_2$. Hence, $XQ$ is the radical axis of $\omega_1,\omega_2$. Now, as power of $P$ is zero with respect to $\omega_2$ and $P\in XQ\implies$ power of $P$ with respect to $\omega_1$ is also zero and hence, $P\in\omega_1\implies PLQN$ is a cyclic which completes the proof. Feedback: Better not say

Here is my solution why are problems 15-20 so easy this year ?

^Yeah & copied too. Cutoff is also rising dont worry

I solved this using barycentrics as well (I was practicing barycentric bashing for INMO by solving the sharygin problems ) Edit: INMO is Indian National Mathematical Olympiad

By Pascal on $BLNDMK,$ $X=\overline{NL}\cap\overline{KM}\cap\overline{AC}$ exists. Hence, $$XN\cdot XL=XK\cdot XM=XP\cdot XQ.$$$\square$

By Pascal on $BKMDNL$ we get that $KM \cap NL = X \in AC$, now, PoP finishes. The solution is way shorter than the question lol