My problem. [asy][asy] size(10cm); pair A = dir(180); pair O = origin; pair B = dir(0); pair X = dir(160); pair Z = dir(65); pair C = extension(A, X, B, Z); pair K = foot(C, A, B); pair Y = IP(C--K, unitcircle); pair T = extension(X, Z, A, B); pair L = extension(O, Y, X, Z); pair P = OP(Line(2*K-L, 2*L-K), unitcircle); pair Q = IP(Line(2*K-L, 2*L-K), unitcircle); pair S = circumcenter(X, Z, K); pair W = 2*S-Y; pair M = extension(K, L, O, W); draw(A--X--Y--Z--B--A); draw(K--Q, dotted); draw(M--K); draw(Y--T--X, dotted); draw(X--Z); draw(T--A, dotted); draw(arc(O,B,A)); draw(arc(O,A,B), dotted); draw(circumcircle(X, K, Z), dashed); draw(circumcircle(O, W, T), dotted); pair I = X+Z; draw(Y--K, dotted); draw(circumcircle(I, Y, M), dashed); draw(Y--W, dashed); draw(O--I); dot("$A$", A, dir(225)); dot("$O$", O, dir(-70)); dot("$B$", B, dir(-45)); dot("$X$", X, dir(X)); dot("$Z$", Z, dir(Z)); dot("$K$", K, dir(-70)); dot("$Y$", Y, dir(60)); dot("$T$", T, dir(225)); dot("$L$", L, 1.4*dir(280)); dot("$P$", P, dir(110)); dot("$Q$", Q, dir(250)); dot("$S$", S, dir(45)); dot("$W$", W, dir(45)); dot("$M$", M, dir(M)); dot("$I$", I, dir(160)); [/asy][/asy] Extend the semicircle to a circle $\Gamma$. Let line $LK$ meet $\Gamma$ again at two points $P$ and $Q$. Let $W$ be the point on ray $OM$ such that $OW \cdot OM = OA \cdot OB$. So points $P$, $Q$, $W$, $O$ are concyclic, say on $\gamma$. Now, $L$ is the radical center of $\gamma$, $\Gamma$, and the circumcircles of $XKOZ$, because lines $XZ$ and $PQ$ are radical axii. So, line $YO$ is the radical axis of $\Gamma$ and $\gamma$. Let $T$ denote the intersection of lines $XZ$ and $AB$. We have that $KO \cdot KT = KA \cdot KB = KP \cdot KQ$, so point $T$ also lies on $\gamma$. Also, according to $TA \cdot TB = TK \cdot TZ = TK \cdot TO$, we deduce that $\overline{TY}$ is tangent to $\Gamma$. Finally, let $S$ denote the midpoint of $\overline{YW}$. By a homothety of ratio $2$ at $W$, we have that the line passing through $S$ and the midpoint of $\overline{WT}$ is perpendicular to $\overline{YO}$. Moreover, $S$ lies on the perpendicular bisector of line $\overline{KO}$. Therefore, $S$ is the center of the circumcircle of quadrilateral $XKOZ$. Finally, the collinearity of $Y$, $S$, $W$ implies quadrilateral $YOMI$ is concyclic, since one can readily show that $OS \cdot OI = OW \cdot OM = OY^2$, hence $\angle OIM = \angle OWS = \angle OWY = \angle OYM$.

Let $P$ the midpoint of $XZ$ and $W=XZ\cap AB$ since $XKOZ$ is cyclic by inversion $\mathbf{I}$ in $\odot (AXYZB)$ we get $\mathbf{I}(W)=K$ $\Longrightarrow$ $WY$ is tangent to $\odot (AYB)$. Let $D$ the projection of $L$ in $WO$ $\Longrightarrow$ by angle-chasing we get $\measuredangle KYD$ $=$ $\measuredangle YDL$ $=$ $\measuredangle LDP$. On the other hand since $KY$ $\parallel$ $OM$ $\parallel$ $LD$ we get $\tfrac{KD}{DO}$ $=$ $\tfrac{YL}{LO}$ $=$ $\tfrac{YK}{MO}$ $\Longrightarrow$ $\triangle KYD$ $\sim$ $\triangle OMD$ $\Longrightarrow$ $\measuredangle OMD$ $=$ $\measuredangle KYD$ $=$ $\measuredangle DMO$ $\Longrightarrow$ $D$, $P$ and $M$ are collinear $\Longrightarrow$ $\measuredangle PMO$ $=$ $\measuredangle PDL$ $=$ $\measuredangle POL$ and since $WYPO$ is cyclic we get $\measuredangle PYO$ $=$ $\measuredangle PWO$ $=$ $\measuredangle POM$ $\Longrightarrow$ combining both results we get $\triangle YPO$ $\sim$ $\triangle OPM$ $\Longrightarrow$ $\measuredangle YPI$ $=$ $\measuredangle MPI$ $=$ $\measuredangle YOM$ $...(1)$ and $\tfrac{YO}{MO}$ $=$ $\tfrac{YP}{PO}$ $=$ $\tfrac{YP}{PI}$ $...(2)$ $\Longrightarrow$ by $(1)$ and $(2)$ we get $\triangle YPI$ $\sim$ $\triangle YOM$ $\Longrightarrow$ $\measuredangle YIO$ $=$ $\measuredangle YIP$ $=$ $\measuredangle YMO$ hence $YIMO$ is cyclic.

We use complex numbers with $(AXYZB)$ as the unit circle, $a=-1,b=1$. Invert in the unit circle, and denote inverses by $'$. The condition that $XKOZ$ is cyclic is equivalent to $K',X,Z$ collinear, and as $k'=\frac{2}{y+\overline{y}}$, \begin{align*}\frac{x-k'}{z-k'}=\overline{\left(\frac{x-k'}{z-k'}\right)}&\iff\frac{xy^2+x-2y}{y^2z+z-2y}=\frac{-2xyz+y^2z+z}{xy^2-2xyz+x}\\&\iff\frac{(y^2+1)(x-z)(xy^2-2xyz+x+y^2z-2y+z)}{x(y^2-2yz+1)(y^2z-2y+z)}=0.\end{align*}If $y=\sqrt{-1}$, then $K,O$ degenerate to the same point. Clearly $x\neq z$, so the condition is $$\boxed{xy^2-2xyz+x+y^2z-2y+z=0.}~(\star)$$Now $YOMI$ is cyclic iff $Y,I',M'$ are collinear. We have $i'=\frac{xz}{x+z}, m'=\frac{-2(x-y)(z-y)}{(y-1)(y+1)(x+z)}$, so \begin{align*}\frac{y-m'}{y-i'}-\overline{\left(\frac{y-m'}{y-i'}\right)}&=\frac{x y^3-3 x y+2 x z+y^3 z+2 y^2-3 y z}{(y-1) (y+1) (x y-x z+y z)} - \frac{3 x y^2-2 x y z-x-2 y^3+3 y^2 z-z}{(y-1)(y+1) (x-y+z)}\\&=-\frac{(2 x y-x z-y^2+2 y z) (x y^2-2 x y z+x+y^2 z-2 y+z)}{(y-1) (y+1) (x-y+z) (x y-x z+y z)}, \end{align*}which is zero by $(\star)$, and we're done.

By radical axes on circles $(AB), (XKOZ)$ and $(YKO)$, we see that $XZ$, $AB$ and the tangent at $Y$ to $(AB)$ concur at $T$. Let $TY$ meet the line through $M$ parallel to $AB$ at $Q$, then $Q,M,Y,O$ are cyclic on the circle with diameter $OQ$. Let $N,P$ be the midpoints of $OQ,OI$ respectively; then it suffices to show that $N$ lies on $TL$, so that $QI\parallel NP\perp IO \implies I$ lies on $(OQ)$ as well. By Menalaus on $\triangle YOQ$ we need $\frac{YT}{QT} \cdot \frac{QN}{ON} \cdot \frac{OL}{YL} =1$, or $\frac{YT}{QT}=\frac{YL}{OL}$. Since $KY\parallel OM\implies$ $\frac{YL}{OL}=\frac{KY}{OM}=\frac{YT}{QT}$, due to ratio of perpendiculars from $Y,Q$ to $AB$, we are done.

Let $R=XZ\cap AB$. Note athat $XKOZ$ being cyclic implies that $AX$ and $BZ$ intersect on $KY$, thus by Brokard we have that $RY$ is tangent to circle $(AXYZB)$. Let $\omega$ denote the circle $(KXZO)$ and let $I'$ be the circumcenter of $\omega$. Invert with center $O$ and radius $OA$. $K$ goes to $R$ and $L$ goes to the $L'$ intersection of $OY$ with $\omega$. $M$ goes to $S$, the intersection of line $OM$ with the circle $(ORL')$. Because $I$ goes to $I"$ we are left to prove that $Y, I'$ and $S$ are colinear. Let $Q$ be the midpoint of $RS$ and the circumcenter of $(ORL'S)$. Because OL' is the radical axis of $(ORL'S)$ and $\omega$ we have that $QI'$ is perpendicular to $OY$, thus parallel to $RY$, hence $\overline{OI'}$ is the $S$-midline in $\triangle{RSY}$ and because the bisector of $OK$ intersect $YS$ exactly at the midpoint of $YS$, we have that $I'$ is the midpoint of $YS$.

Let XZ-OK=T, reflection of Y-OT=D, G-midpoint XZ, D-G=F, M-G=N (XOZ) cuts AD tat T, AG-(O)=H. Because GOM=YOM-YOG=YDO-YDG=GDO so OM tangent (OT). Note that OFIH cyclic, OHN=OEY=OIN so OHIN cyclic then NOFH cyclic. So MOD+MID=YDO+NOF=YHF+NOF=180 so q.e.d

Nice problem! Although this should be 2/5 level in my opinion. [asy][asy] unitsize(2.5cm); defaultpen(fontsize(10pt)); pair A = (-1,0); pair B = (1,0); pair O = (0,0); pair X = dir(165); pair Y = dir(112); pair K = foot(Y,A,B); pair T = (-1/abs(K-O),0); pair Z = 2*foot(O,T,X)-X; pair L = extension(O,Y,X,Z); pair M = extension(K,L,O,(0,1)); pair S = circumcenter(Y,O,M); pair P = 2*S-T; pair I = 2*foot(O,X,Z)-O; fill(Y--K--T--cycle,palered); fill(M--O--P--cycle,palered); draw(arc((0,0),1,0,180)); draw(B--T--P); draw(Y--O); draw(K--M); draw(K--Y--T,red); draw(O--M--P--cycle,red); draw(S--(O+Y)/2,red); draw(S--(O+M)/2,red); dot("$A$",A,dir(270)); dot("$B$",B,dir(270)); dot("$T$",T,dir(270)); dot("$O$",O,dir(270)); dot("$K$",K,dir(270)); dot("$X$",X,NW); dot("$Y$",Y,N); dot("$Z$",Z,SW); dot("$L$",L,2*N); dot("$M$",M,N); dot("$S$",S,NW); dot("$P$",P,N); dot("$I$",I,N); [/asy][/asy] By radical axis on $\odot(YKO)$, $\odot(XKOZ)$ and $\odot(AB)$, we get $YY, XZ, AB$ are concurrent at $T$. Now let $P$ be the point such that $MP\parallel AB$ and $OP\parallel YT$, then $\triangle KYT$ and $\triangle MOP$ are homothetic with center $L$ therefore $P\in XZ$. Let $S$ be the midpoint of $PT$, then observe that $S$ is the circumcenter of $\triangle YOM$. As $S\in XZ$, we are done.