based projective enjoyer vs cringe transform user

I thought Rijul Sir would have killed this too by providing all solutions. Most people applied Pascal thrice, but here's my solution using inversion and radax:

Proposed by Rijul Saini (I think?).

Redefining $J$ as the second intersection point of $KL$ and $\omega$ and applying Pascal theorem on $KKJDAB$ and $JJKBCD$ gets the result that $KB$ and $JD$ are concurrent on $MN$ and the tangent line from $J$ to $\omega$ passes through point $N$.

Solved with kingu who had the really bright idea of using homography. I wonder why all these really artificial constraints are added related to the positions of $M,N$ and $K,J$. We solve the following equivalent problem instead. Quote: Let $ABCD$ be a convex cyclic quadrilateral with circumcircle $\omega$. Let $BA$ produced beyond $A$ meet $CD$ produced beyond $D$, at $P$. Let $K$ be an arbitrary point on the circle $\omega$, and $J$ the second intersection of line $KP$ and $\omega$. Let $M$ and $n$ be the intersections of the tangents to $\omega$ at $K$ and $J$ and lines $AD$ and $BC$ respectively. Prove that points $P$ , $M$ and $N$ are collinear. We almost cheat and cheese this via a homography. Note that the problem statement and conclusion are all purely projective. We consider the homography $\mathcal{H}$ sending $ABCD$ to a rectangle and preserving $\omega$. Now, $P$ goes to $\overline{AB} \cap \overline{CD} = P_{\infty}$. Then $K$ is still an arbitrary point on $\omega$. Note that $KJ \parallel AB$ since lines $AB$ , $CD$ and $KJ$ concur at $P$. $M$ and $N$ conserve their definitions since they are defined by only intersections. We are then left with the following picture, [asy][asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,D,K,J,M,N; A= (1.5,1); B = (1.5,-1); C = (-1.5,-1); D = (-1.5,1); J = (1.76,0.4); line p = parallel(J,line(A,B)); K = intersectionpoints(circle(A,B,C),p)[0]; line tK = tangent(circle(A,B,C),K); line tJ = tangent(circle(A,B,C),J); M = intersectionpoint(tK,line(A,D)); N = intersectionpoint(tJ,line(B,C)); filldraw((path)(A--B--C--D--cycle), white+0.1*pri, pri); filldraw(circumcircle(A,B,C), tfil, tri); draw(K--J); draw(A--M); draw(B--N); draw(M--N); draw(K--M); draw(J--N); draw(A--J); draw(K--B); draw(A--K,dashed); draw(B--J,dashed); dot("$A$",A,dir(60)); dot("$B$",B,dir(300)); dot("$C$",C,dir(210)); dot("$D$",D,dir(130)); dot("$K$",K,dir(0)); dot("$J$",J,dir(30)); dot("$M$",M,dir(60)); dot("$N$",N,dir(0)); [/asy][/asy] We wish to show that $MN \parallel AB$ as well (since we needed to prove $P$ , $M$ and $N$ collinear). For this, we simply make the following observation. Claim : Triangles $KMA$ and $JNB$ are congruent. Proof : Note that $ABKJ$ is an isoceles trapezoid due to being cyclic, and $AB \parallel KJ$, so we clearly have $AK = BJ$. Further, \[\measuredangle MAK = 90 + \measuredangle BAK = 90 + \measuredangle JBA = \measuredangle JBN\]and also since $AN \perp KJ \perp AM$, \[\measuredangle KMA = 90 + \measuredangle MKJ = 90 + \measuredangle KAJ = 90 + \measuredangle KBJ = 90 + \measuredangle KJN = \measuredangle BNJ \]from which it follows that $\triangle AMK \cong \triangle BNJ$, as we wished to show. Now, from this claim, $BN=AM$ from which it follows that $ABNM$ must also be a rectangle, and thus $MN \parallel AB$ as desired. Note that here $K$ is not necessarily inside arc $AB$ and the problem holds just fine. It turns out that as $K$ varies on $\omega$, $M$ varies along $\overline{AD}$ (which is obvious) but never lies inside the segment $AD$.

Take a homography sending $AC \cap BD$ to the center of $\omega$ which sends $ABCD$ to a rectangle. Then $AB \parallel CD \implies L$ is the point at infinity along line $AB$. Then it follows that $MN \parallel AB$ since $M$, $N$ lie on the line containing $L_\infty$ and by symmetry it also follows that $KJ \parallel MN \implies K - J - L_\infty$.

This problem is funny tbh. Basically let $M$ drop its second tangent to $\omega$ which hits it at $K'$, and redefine $LK \cap \omega=J$ and $LK' \cap \omega=J'$. Note that $-1=(A, D; K, K') \overset{L}{=} (B, C; J, J')$ so if we let the tangents from $J,J'$ to $\omega$ hit at $N'$, we have $N'$ lies on $BC$, plus by Brokard and Pascal considering $KK'J'J$ we have that $LM$ is the polar of $K'J \cap KJ'$ and that therefore $N'$ lies on $LM$ as well, but then this gives $LM \cap BC=N'$ so $N=N'$ and thus we are done .

dolphinday wrote: Take a homography sending $AC \cap BD$ to the center of $\omega$ which sends $ABCD$ to a rectangle. Then $AB \parallel CD \implies L$ is the point at infinity along line $AB$. Then it follows that $MN \parallel AB$ since $M$, $N$ lie on the line containing $L_\infty$ and by symmetry it also follows that $KJ \parallel MN \implies K - J - L_\infty$. Many contestants lost points if they used homography, but did not specify the location of tangency points along the circle.

D_S wrote: dolphinday wrote: Take a homography sending $AC \cap BD$ to the center of $\omega$ which sends $ABCD$ to a rectangle. Then $AB \parallel CD \implies L$ is the point at infinity along line $AB$. Then it follows that $MN \parallel AB$ since $M$, $N$ lie on the line containing $L_\infty$ and by symmetry it also follows that $KJ \parallel MN \implies K - J - L_\infty$. Many contestants lost points if they used homography, but did not specify the location of tangency points along the circle. Yes, kingu mentioned this to me so I'll try to patch my sol.

Define $K'$ as the second intersection of $LJ$ with $w$. Define $M'=K'K'\cap AD$. We need to show that $L$, $M'$, and $N$ are collinear. Define $Z=DJ\cap K'B$. Pascal on $JJABCK'$ gives $N$, $Z$, and $L$ are collinear. Pascal on $K'K'BADJ$ gives $M'$, $Z$, and $L$ are collinear. The result follows.