Work in the projective plane. By Brokard's theorem, $\triangle{PQR}$ is self-polar (w.r.t. $\omega$). Let line $MKR$ hit $\omega$ a second time at $K'$, and define the points $Z = KK\cap K'K'$, $F = OR\cap PQ$. Clearly $F,R$ are inverses, and since $R$ lies on the polar $KK'$ of $Z$, $Z$ lies on the polar $PQ$ of $R$. If $Z$ is a point at infinity (corresponding to $PQ$), the result is trivial, since $O_{KPQ},M,K,O$ are collinear in this case. Now suppose otherwise; then it suffices to show $ZK^2 = ZP\cdot ZQ$. We can easily compute $ZK^2 = ZO^2 - R_{\omega}^2 = (ZF^2+OF^2) - OR\cdot OF = ZF^2 + OF\cdot RF$ and $ZP\cdot ZQ = ZM^2 - MP^2$. On the other hand, $OZ\perp RM = KK'$ yields $OR^2 + ZM^2 = OM^2 + ZR^2 = (OF^2 + MF^2) + (ZF^2 + RF^2)$, so \begin{align*} ZP\cdot ZQ - ZK^2 &= (ZM^2 - ZF^2) - MP^2 - OF\cdot RF \\ &= (OF^2 + MF^2 + RF^2 - OR^2 ) - MP^2 - OF\cdot RF \\ &= (OF-OR)(OF+OR) - (MP-MF)(MP+MF) - RF\cdot OR \\ &= FR\cdot FO - FP\cdot FQ \\ &= 0, \end{align*}where we use $\triangle{FPR}\sim\triangle{FOQ}$ in the last step (which follows from the fact that $O$ is the orthocenter of $\triangle{PQR}$).

During the contest i found a very easy solution: Just take tangents from the Miquel point and apply Pascal theorem. Edit: I don't know your notations i am refering to the normal miquel point of $ABCD$ (here is the pic) \ i forgot to draw $KJ$ on the pic which is supposed to pass through R (the original).

paul1703 wrote: Just take tangents from the Miquel point and apply Pascal theorem. I assume you're basically saying this (using notations of my proof): Let the polar of $F$ be $URV$ with $UV$ a chord of $\omega$. By Brokard's theorem, $P' = KU \cap K'V$, $Q' = KV \cap K'U$, and $R = KK' \cap UV$ form a self-polar triangle; in particular, $P'Q'$ is the polar $PQ$ of $R$. But then $P'Q' = PQ\parallel UV$, so because $R$ is the midpoint of $UV$, $M$ must be the midpoint of $P'Q'$. Yet $O$ is the orthocenter of both $\triangle{PQR}$ and $\triangle{P'Q'R}$, so $\{P,Q\}=\{P',Q'\}$, whence the homothety centered at $K$ taking $URV$ to $P'MQ'$ sends $\omega$ to $(KPQ)$.

This problem is in fact a special case of Gergonne's solution to the Apollonius problem. Consider the circle $w$ along with the degenerate circles $P,Q$ with radius $0$. Note that $PQ$ is the Monge D'alembert line of these three circles. Since $PQR$ is self polar, $R$ is the pole of this line with respect to $w$. Now $P$ lies on the polar of $Q$ wrt $w$, and applying a homothety with scale factor $\frac{1}{2}$ centered at $Q$ maps this line to the radical axis of $Q,w$. Thus this radical axis passes through $M$. Similarly, the radical axis of $P,w$ passes through $M$, so $M$ is radical center of the three circles. Since $K$ is the intersection of $w$ and $RM$, the statement that $KPQ$ is tangent to $w$ is equivalent to Gergonne's solution in this case.

Work in the projective plane. Let $O$ be the center of $\omega$. Let $P_1, Q_1$ be the points on $\omega$ s.t. $OR \perp P_1Q_1$, and $R$ lies on $P_1Q_1$, and $P_1, Q_1$, are in a different order than $P, Q$ (this definition will be clearer after we note that $PQ || P_1Q_1$). Note that since $PQ$ is the polar of $R$, then $OR \perp PQ$, so $PQ || P_1Q_1$. Also, the tangents from $P_1, Q_1$ intersect on $PQ$. We'll now prove a few lemmas. Lemma 1: The tangents from $A$ and $C$ intersect on $PQ$. Proof: Use Pascal's Theorem on $AABCCD$. Lemma 2: $AP_1, CQ_1$ intersect on $PQ$. Proof: Let $AP_1\cap CQ_1 = X$ and $AQ_1\cap CP_1 = Y$. By Pascal's Theorem on $AAP_1CCQ_1$ and $P_1P_1AQ_1Q_1C$, we get that the $X, Y, AA \cap CC$, and $P_1P_1\cap Q_1Q_1$ are collinear. Since $AA \cap CC$ and $P_1P_1\cap Q_1Q_1$ both lie on $PQ$ (Lemma 1), $X$ and $Y$ both also lie on $PQ$. Lemma 3: $PP_1, QQ_1$ intersect on $\omega$. Proof: Let $PP_1 \cap \omega = P_2$. We will show that $Q_1P_2$ passes through $Q$. By Pascal's Theorem on $P_2Q_1CDAP_1$, we get that $P_2Q_1\cap DA$, and $Q_1C \cap AP_1$, and $AD\cap P_1P_2 = P$, are collinear. By Lemma 2, $Q_1C\cap AP_1$ lies on $PQ$, so $P_2Q_1\cap DA$ lies on $PQ$ also, and since $PQ \cap AD = Q$, $P_2Q_1$ passes through $Q$. Note that since $PQ \| P_1Q_1$, and $R$ is the midpoint of $P_1Q_1$ and $M$ is the midpoint of $PQ$, $K = PQ \cap P_1Q_1$. Now it is clear that a homothety centered at $K$ takes $KP_1Q_1$ to $KPQ$, so it maps the circumcircles to each other. Since both circles also pass through $K$, they are tangent at $K$.

Let $\omega \cap MR=L \neq K$. Let the tangents from $K,L$ to $\omega$ intersect at $S$, let $OS \cap KL=T$. So $KL$ is the polar of $S$ wrt $\omega$. It is well-known that $\triangle PQR$ is self-polar wrt $\omega$. So $PQ$ is the polar of $R$. $R$ lies on $KL$, the polar of $S$, hence $S$ lies on $PQ$ the polar of $R$ (La Hire). $KS$ is tangent to $\omega$, now we'll prove it is tangent to $\odot KPQ$ too. So it suffices to prove $SP.SQ=SK^2$. Note that $R$ is the orthocentre of $\triangle OPQ$. It is well-known that half turn wrt $M$ sends $R$ to $O'$, the diametrically opposite point of $O$ in $\odot OPQ$. Since $\angle OTM=\angle OTO'=90^{\circ}$, we conclude $T$ lies on the circle with diameter $OO'$ meaning $OTPQ$ is cyclic. Hence $SP.SQ=ST.SO=SK^2$, so we are done. Note: It follows $\odot LPQ$ is tangent to $\omega$ as well.

MY SOLUTION: lamma:AM meet $\omega$at S then S is the second meetpoint of $\omega$ and (PQC) Assyme that circle $\omega$ andP,Q are fixed then R ,Mis fixed. LET K be the point st.PQK externally tangent to $\omega$ then use the lemma LET C=K, then S=C=K so A,C,M are colinear so R is the commom point of segement MR and $\omega$

Edit: woops

robinpark wrote: $M$ is still the midpoint of $PQ$. I am sorry to say, but this is false. It goes to the point at infinity, but any direction... My solution is the same as Orin The problem is just a consequence of working backwards.

Instead, we draw the circle through $P$ and $Q$ tangent to the given circumcircle at $T$. It suffices to prove $RT$ bisects $PQ$, because then $K=T$ and we are done. Equivalently, it suffices $RP$, $RT$, $RQ$ and $RI$ form a harmonic pencil, where $I$ is the point of infinity on $PQ$. We project this through the given circumcircle $w$. It now suffices to prove $P$, $Q$, $X$, and $O'$ form a harmonic pencil. Here $X$ is the pole of $RT$, the intersection of the circles' common tangent at $T$ with $PQ$. Also, $O'$ is the pole of $RI$, which means it is the projection of $O$ onto $PQ$ and lies on $OR$ ($PQR$ is self polar by Brokard). Now, $Q'$ is the image of $Q$ under inversion about $w$, and $P'$ is similar. We have the orthocenter of $OPQ$ is where $QP'$, $PQ'$, $OO'$ meet ($QP'$ and $OP$ are perpendicular since $P$ and $Q$ are on one another's polars, etc). Now, denote the circumcircle of $TPQ$ by $A$. Under an inversion about $w$, $A$ maps to $B$, the circle through $TP'Q'$. As inversion preserves tangency, we have $A$ and $B$ are tangent at $T$ and $TX$ is their radical axis. Consider the radical center of $A$, $B$, and the circle with diameter $PQ$. It lies on $XT$, $PQ$, and $P'Q'$. As $XT$ intersects $PQ$ at $X$, it must be $X$ and so $X$ lies on $P'Q'$. Now, $P'$, $Q'$, and $O'$ are the feet of the altitudes of the triangle, which concur. Because $X$ is the intersection of $PQ$ and $P'Q'$, by the Ceva-Menelaus configuration we have $X$, $P$, $O'$, and $Q$ are harmonic. This suffices to prove our result, and so we are done. EDIT: I should probably prove the true (and easy to prove) claim that $4$ concurrent lines form a harmonic pencil if and only if their collinear poles form a harmonic division. I think it is well known, (it is essentially "projecting through the circle") but just in case here is a proof. Call the points $P_1$, $P_2$, $P_3$, $P_4$, and $P$. Define $P_1'$ on ray $PP_1$ with $(PP_1')(PP_1)=r^2$ and similar. Note that $P_1P_2P_3P_4$ are a harmonic division iff $PP_1$, $PP_2$, $PP_3$, $PP_4$ are a harmonic pencil iff $PP_1'$, $PP_2'$, $PP_3'$, $PP_4'$ are a harmonic pencil. Of course, this is true iff the lines through $P_1'$ perpendicular to $PP_1'$, etcetera, are a harmonic pencil, because rotating all the lines in a pencil by $90$ degrees and centering it at a new point preserves if it is harmonic. So the claim is true.

Let $P', Q'$ be the inverses for $P, Q$ wrt $\odot ABCD$. Let $L = \odot OP'Q' \cap \odot OPQ$. Note that $ML \equiv MR \perp OL$ since $M$ is midpoint of $PQ$ and $R$ is orthocentre. Let $N = P'Q' \cap PQ \cap OL$. Note the polar of $N$ is therefore $KL$. It follows $NK$ is tangent to $\odot PQK, \odot ABK$ so done.

[asy][asy] size(10cm); draw(unitcircle); pointfontsize = 8; pointpen = black; pair U = Drawing("U", dir(55), dir(90)); pair V = Drawing("V", conj(U), dir(-90)); pair R = Drawing("R", (U+V)/2, dir(225)); pair S = Drawing("S", 1/R, dir(0)); pair K = Drawing("K'", dir(7), 1.4*dir(-20)); pair Q = Drawing("Q", extension(S,S+dir(90),U,K), dir(0)); pair P = Drawing("P", extension(S,S+dir(90),V,K), dir(0)); pair X = Drawing("X", 2/conj(V+K), 1.4*dir(-110)); pair T = Drawing("T", extension(U,X,P,Q), dir(0)); draw(P--T); draw(U--S--V, dotted); draw(P--V--U); draw(U--Q, dotted); draw(U--T, dotted); draw(X--K, dotted); draw(R--Q); pair W = Drawing("W", 2*foot(origin,U,T)-U, dir(-10)); pair M = Drawing("M", (P+Q)/2, dir(0)); draw(R--M, dashed); [/asy][/asy] By Brokard's Theorem, $\triangle PQR$ is self-polar. Let $\ell$ be the polar of $R$. Let $S$ be the inverse of $R$ about $\omega$ (hence on $\ell$) and set $U$, $V$ as the contact points of the tangents from $S$, with $U$ closer to $P$ than to $Q$. Define $K'$ to be the second intersection of $\omega$ of $\overline{PV}$. We will prove that $U$, $K'$ and $Q$ are collinear. Let $X$ be the intersection be the intersection of the tangents at $K'$ and $V$, and let $\overline{UX}$ meet $\omega$ and $\ell$ again at $W$ and $T$, respectively. Observe that $X$ is the pole of $\overline{VK'}$, whilst $R$ is the pole of line $\ell$. Because $\overline{UK'}$ and $\ell$ meet at $P$, we see that $\overline{RX}$ is the polar of $P$. That means $Q$, $R$, $X$ are collinear. Because $R$ is the midpoint of $\overline{UV}$, it follows that $Q$ is the midpoint of $\overline{ST}$ by homothety at $X$. Moreover, $UVWK'$ is a harmonic quadrilateral. Taking a perspectivity at $U$ onto $\ell$, noting again that $\overline{ST} \parallel \overline{UV}$, we find that $\overline{UK'}$ must bisect $\overline{ST}$. Hence $U$, $K'$, $Q$ are collinear as desired. Now to the original problem. Since $R$ is the midpoint of $\overline{UV}$, and $M$ is the midpoint of $\overline{PQ}$, it follows that $R$, $K'$, $M$ are collinear, whence $K \equiv K'$. So $K$ is the center of a homothety taking $\overline{PQ}$ to $\overline{VU}$. This implies that $(KPQ)$ and $(KUV)$ are tangent, as desired. $\blacksquare$ A word of motivation: there are only two common ways I know of to show that two circles are tangent, inversion and the homothety trick used above (see IMO 2011 \#6, APMO 2014 \#5). So constructing $U$ and $V$ is natural, and if your diagram is good then you discover that $U$, $R$, $V$ are collinear.

IDMasterz wrote: Let $P', Q'$ be the inverses for $P, Q$ wrt $\odot ABCD$. Let $L = \odot OP'Q' \cap \odot OPQ$. Note that $ML \equiv MR \perp OL$ since $M$ is midpoint of $PQ$ and $R$ is orthocentre. Let $N = P'Q' \cap PQ \cap OL$. Note the polar of $N$ is therefore $KL$. It follows $NK$ is tangent to $\odot PQK, \odot ABK$ so done. Absolutely beautiful solution, but could you be any more terse? To write out the details... Because $L \in (OP'Q')$, we find that $\angle OLR = 90^{\circ}$. Also, the reflection of $R$ over the midpoint of $\overline{PQ}$ is the point diametrically opposite $O$ on $(OPQ)$, so the aforementioned right antgle gives $L$, $R$, $M$ are collinear. Now let $N$ denote $\overline{PQ} \cap \overline{P'Q'}$. By radical axis (or Miquel points, if you like) the point $N$ lies on $\overline{OL}$. Moreover, $L$ and $N$ are inverses to $\omega$. Thus it follows from $\overline{KL} \perp \overline{ON}$ that $\overline{KL}$ is the polar of $N$. Then $\overline{NK}$ is a tangent to $\omega$, and moreover, $NP \cdot NQ = NL \cdot NO = NK^2$ implies that $\overline{NK}$ is also tangent to $(PKQ)$.

Let O be center of circle w and G circumcenter of (QKP). By well-known fact QP is perpendicular to OR. So, if parallel from R to QP intersect circle w at points X and Y, then XR=YR. By Lemma: If two points A and B be points outside of circle w, such that, B lies on polar of A. Let take point T on a segment AB. And tangents from T meet circle w at points X and Y. Then AX and BY meet at circle w. Proof is here, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=592032 This lemma can be easily applied, since XY is polar of point T, where T belongs to the segment PQ. So we get QY and XP meet on circle w (X is closer to A and Y is closer to C). And since PQ is parallel to XY and points M, R are midpoints of the segments QP, XY we have MR, PX and QY are concurrent at point K. GM and OR are perpendicular to QP, so according to the circles (G) and (O), MG\RO=QP\XY and according to XYPQ, QP\XY=MK\KR, which yields MG\RO= MK\KR. So O, K and G are collinear. Which yields (G) and (O) are tangent at K.

Let $L$ be the second intersection point between $MR$ and $\omega$. Let $S$ be the intersection point between the tangents of $\omega$ at $K, L$. Then $KL$ is the polar of $S$ with respect to $\omega$. Let $T \equiv OS \cap KL$, so $T$ is the inverse of $S$. $SK$ is tangent to $\omega$, so it suffices to show $SK$ is also tangent to $(KPQ)$. $\triangle PQR$ is self-polar with respect to $\omega$, from Brocard, so $PQ$ is the polar of $R$. $R$ lies on $KL$, so $S$ lies on $PQ$ by La Hire's theorem. Again from Brocard, $R$ is the orthocenter of $\triangle OPQ$. Reflect $R$ in $M$ to $N$, the point diametrically opposite $O$ on $(OPQ)$. Hence $\angle OTN = \angle OTK = 90^{\circ}$. It follows that $PQOT$ is cyclic, so $SP \cdot SQ = ST \cdot SO$. Let $SO \cap \omega \equiv X,Y$. If $r$ is the radius of $\omega$, and as $S,T$ are inverses, we have: \begin{align*}ST \cdot TO = r^2 \implies SO(SO-ST) = r^2 &\implies SO^2 - r^2 = ST \cdot SO\\ &\implies (SO + r)(SO-r) = ST \cdot SO \\ &\implies SX \cdot SY = ST \cdot SO. \end{align*} But $SK^2 = SX \cdot SY$ from the tangent-secant theorem, so $SK^2 = ST \cdot SO = SP \cdot SQ$, and $SK$ is tangent to $(KPQ)$, as required.

So it appears that this can be done with complex numbers with respect to the unit circle $\omega$. The nice thing about it is that we can work in terms of points $p,q,r$ without mentioning $a,b,c,d$. Let $X$ be a point on $\omega$. If we can show that $X \in MR \iff (XPQ)$ is tangent to $\omega$, we are done. Now, \begin{align*} X\in MR &\iff \frac{x-r}{\frac{1}{x}-\bar{r}} = \frac{r-m}{\bar{r}-\bar{m}} \\ &\iff \frac{x^2-xr}{1-x\bar{r}}=\frac{2r-p-q}{2\bar{r}-\bar{p}-\bar{q}} \\ &\iff (2\bar{r}-\bar{p}-\bar{q})x^2-r(2\bar{r}-\bar{p}-\bar{q})x = (2r-p-q)-\bar{r}(2r-p-q)x \\ &\iff (2\bar{r}-\bar{p}-\bar{q})x^2+[r(\bar{p}+\bar{q})-\bar{r}(p+q)]x + (p+q-2r)=0. \end{align*}If we let $T$ be the circumcentre of $(XPQ)$, then $(XPQ)$ is tangent to $\omega$ iff $O,X,T$ are collinear, or $\frac{t}{\overline{t}} = x^2$. \[ t= \left |\begin{matrix}1 & x & 1 \\ 1 & p & p \bar{p}\\ 1 & q & q\bar{q}\end{matrix}\right |\div\left |\begin{matrix}1 & x &\bar{x}\\ 1 & p &\bar{p}\\ 1 & q &\bar{q}\end{matrix}\right |. \]Note that the denominator is the negative of its conjugate, so \begin{align*} -x^2 &= \left |\begin{matrix}1 & x & 1 \\ 1 & p & p \bar{p}\\ 1 & q & q\bar{q}\end{matrix}\right |\div\left |\begin{matrix}1 & \bar{x} & 1\\ 1 & \bar{p} &p\bar{p}\\ 1 & \bar{q} &q\bar{q}\end{matrix}\right | = \frac{pq(\bar{q}-\bar{p})+x(p\bar{p}-q\bar{q})+(q-p)}{\bar{p}\bar{q}(q-p)+\bar{x}(p\bar{p}-q\bar{q})+(\bar{q}-\bar{p})}\\ &\iff pq(\bar{q}-\bar{p})+ x(p\bar{p}-q\bar{q})+(q-p) = -x^2\bar{p}\bar{q}(q-p)-x(p\bar{p}-q\bar{q})-x^2(\bar{q}-\bar{p})\\ &\iff x^2[\bar{p}\bar{q}(q-p)+(\bar{q}-\bar{p})]+2x(p\bar{p}-q\bar{q})+[pq(\bar{q}-\bar{p}+(q-p)]=0. \end{align*}It suffices to show that the two quadratic equations in $x$ that we have obtained have corresponding coefficients in the same ratio. $O$ is the orthocentre of $\triangle PQR$, so $R$ is the orthocentre of $\triangle OPQ$, so \[ r = \frac{(p\bar{q}+\bar{p}q)(q-p)}{\bar{p}q-p\bar{q}}. \]We compute \begin{align*} r(\bar{p}+\bar{q})-\bar{r}(p+q) &=\frac{(p\bar{q}+\bar{p}q)(q-p)(\bar{p}+\bar{q})}{\bar{p}-p\bar{q}} + \frac{(p\bar{q}+\bar{p}q)(\bar{q}-\bar{p})(p+q)}{\bar{p}q-p\bar{q}}\\ &= \frac{2(p\bar{q}+\bar{p}q)(q\bar{q}-p\bar{p})}{\bar{p}q-p\bar{q}}. \end{align*}So \[ \frac{ r(\bar{p}+\bar{q})-\bar{r}(p+q)}{2(p\bar{p}-q\bar{q})} = \frac{p\bar{q}+\bar{p}q}{p{q}-\bar{p}q}. \]If we can show that $\frac{2\bar{r}-\bar{p}-\bar{q}}{\bar{p}\bar{q}(q-p)+(\bar{q}-\bar{p})}=\frac{p\bar{q}+\bar{p}q}{p\bar{q}-\bar{p}q}$, we are done, as $\frac{p+q-2r}{pq(\bar{q}-\bar{p})+(q-p)}=\frac{p\bar{q}+\bar{p}q}{p\bar{q}-\bar{p}q}$ would then follow from conjugation. Cross-multiplying and substituting in the value of $\bar{r}$ gives \[ 2(p\bar{q}+\bar{p}q)(\bar{q}-\bar{p})-(\bar{p}+\bar{q})(p\bar{q}-\bar{p}q)=(p\bar{q}+\bar{p}q)[\bar{p}\bar{q}(q-p)+(\bar{q}-\bar{p})] \&\iff (p\bar{q}+\bar{p}q)[\bar{q}-\bar{p}+(p-q)\bar{p}\bar{q}-\bar{p}-\bar{q}]=-2\bar{p}q(\bar{p}+\bar{q}). \]Since $Q$ lies on the polar of $P$ with respect to $\omega$, $OP \perp QP'$ where $P'$ is the inverse of $P$ with respect to $\omega$. Thus \[ \frac{p}{\bar{p}} = -\frac{q-\frac{1}{\bar{p}}}{\bar{q}-\frac{1}{p}} \implies p\bar{q}+\bar{p}q=2. \]It now suffices to show that \[ \bar{q}-\bar{p}+(p-q)\bar{p}\bar{q} = (\bar{p}+\bar{q})(1-\bar{p}q) \&\iff (\bar{q}-\bar{p})(p\bar{q}+\bar{p}q) + 2(p-q)\bar{p}\bar{q}=(\bar{p}+\bar{q})(p\bar{q}-\bar{p}q), \]which is true upon expansion.

Nice problem. It is done by first noticing PQR is autopolar, and then finding two cyclic quads by spamming Power of a Point.

leminscate wrote: The nice thing about it is that we can work in terms of points $p,q,r$ without mentioning $a,b,c,d$. So too synthetically it's really a bit silly in my opinion; I think the problem should have just said "let $PQR$ be self-polar" or something like that, but alas...

As nocky wocky is cringe, he wrote: Let the inverses of $P,Q$ WRT $(ABCD)$ be $P',Q'$. By Brokard's, $P$ lies on the polar of $Q$ WRT $(ABCD)$, so $PQ'\perp QQ'$ so $M$ is the center of $(PQ)$, which passes through $P',Q'$. Let $Z$ be the radical centre of $(OR),(ABCD),(PQ)$, which lies on $PQ$ as it is the radical axis of $(OR),(ABCD)$. Now, since $P,P'$ and $Q,Q'$ swap under inversion WRT $(ABCD)$, $(PQ)$ is fixed, so $(PQ)$ is orthogonal to $(ABCD)$. Then, the polar of $M$ in $(ABCD)$ is the radical axis of $(PQ),(ABCD)$, so $Z$ lies on the polar of $M$ in $(ABCD)$. By La Hire's, $M$ lies on the polar of $Z$ in $(ABCD)$. Let $Z'$ be the intersection of the tangent at $K$ to $(ABCD)$ with $PQ$. Notice that $PQ$ is the locus of points $X$ such that the polar of $X$ in $(ABCD)$ passes through $R$ by Brokard's, so the polar of $Z'$ passes through $R,K$, and thus $M$. Hence, $Z'=Z$, so $ZK$ is tangent to $(ABCD)$ and is the radical centre of $(ABCD),(OR),(PQ)$. Hence, $$ZP\times ZQ=ZP'\times ZQ'=\text{Pow}_{(OR)}(Z)=\text{Pow}_{(ABCD)}(Z)=ZK^2$$so $ZK$ is tangent to both $(ABCD),(KPQ)$, which are now tangent to each other.

trick, $(PQ), \omega$ are orthogonal with radical axis $\text{(the polar of } M) \ni S$, so $SK^2 = SP \cdot SQ \implies SK$ is tangent to $\omega$.

the G8 god strikes again

The first claim in my solution to this problem: https://artofproblemsolving.com/community/u934783h3059886p27591604

Let $R_1$ and $R_2$ be on $\omega$ such that $R$ is the midpoint of $R_1R_2$. Since $OR$ is perpendicular to $R_1R_2$, it follows by Brokard's theorem that $\overline{PQ} \parallel \overline{R_1R_2}$, so taking an obvious homothety it suffices to show that $\overline{R_1Q} \cap \overline{R_2P}$ lies on $\omega$. Take a homography sending $R$ to the center of $\omega$. Then $ABCD$ becomes a rectangle under this map, and the problem is trivialized.

Let line $(MR)$ cut $\omega$ again at a point $L$, and let $N$ be the midpoint of segment $KL$, so that $N$ is the point where line $(MR)$ meets the circle with diameter $OR$ a second time. Since $R$ is the orthocenter of $\triangle OPQ$ by Brokard's Theorem, we actually have that $N$ lies on the circumcircle of $\triangle OPQ$ (it's, with the standard $ABC$ notations, the intersection point of the circle with diameter $AH$ and line $(MH)$). We now let $T$ be the pole of the line $(MR)$ w.r.t $\omega$, so that $(TK)$ and $(TL)$ are both tangent to $\omega$. Since $R$ is the pole of line $(PQ)$ by Brokard's and $R$ lies on the polar of $T$, we have by La Hire's Theorem that $T$ lies on line $(PQ)$. Finally, we have $TK^2=TN\times TO=TQ\cdot TP$, so $T$ lies on the radical axis of $\omega$ and $(KPQ)$. Since $(TK)$ is already tangent to $\omega$, it must be the case that this radical axis is exactly the line $(TK)$, i.e. circle $(KPQ)$ is tangent to $\omega$. $\blacksquare$

Let $\overline{PK} \cap \omega = P' \neq K$ and $\overline{QK} \cap \omega = Q' \neq K$. Our goal is to show that the homothety sending $(KPQ) \to \omega$ is centered at $K$ which implies $P'Q' \parallel PQ$. This is equivalent to $P'Q' \cap PQ = \infty$ which is equivalent to showing that $-1 = (P, Q, M; PQ \cap P'Q' = \infty)$. Taking a homography sending $R$ to the center of $\omega$(while preserving the circle) sends $P$ and $Q$ to $\infty_P$ and $\infty_Q$. Since $-1 = (\infty_P, M; \infty_Q, \overline{\infty_P\infty_Q} \cap P'Q')$ we are done.

help cant homography Let $\overline{PK} \cap \omega = X$ and $\overline{QX} \cap \omega = Y$ - $(KPQ)$ being tangent to $(ABCD)$ is clearly equivalent to $\overline{PQ} \parallel \overline{XY}$, or if $\overline{PQ} \cap \overline{XY} = \infty_{XY}$ then $(P, Q; M, \infty_{XY}) = -1$. Now the problem is purely projective, so taking a homography sending $ABCD$ to a rectangle finishes.

New fastest p3 solve woo Denote by $P'$ and $Q'$ the inverses of $P$ and $Q$ with respect to the circumcircle, let $O$ be the circumcenter. Let $F$ be the pole of $MR$ with respect to the circumcircle, hence it also lies on $PQ$. Thus $KF$ is tangent to $(ABC)$. Notice that $OP'Q'R$ is cyclic by inversion. Moreover by Queue point properties we have that $(OP'Q'R), OF,$ and $MR$ concur. Inverting we find that $P'Q'$ passes through $F$. Now, $(PP'QQ')$ is the circle with diameter $PQ$. Since this contains $P$ and $P'$ it follows that the circle with diameter $PQ$ is orthogonal to $(O)$. Now $PQ$ is the radical axis of $(ABCD)$ and $(OP'Q'R)$, easily shown by say $ABRQ$ cyclic + invert and similarly so for $P$. This implies that $F$ has equal power to $(OP'Q'R)$ and $(PP'QQ')$, or $FK^2 = FP' \cdot FQ' = FP \cdot FQ.$ Yay.

Let $\overline{KP}\cap(ABCD)=P'$ and $\overline{KQ} \cap (ABCD)=Q'.$ We need to prove there is a homothety sending $KP'Q'$ to $KPQ$ so that the center of the circles are collinear with $K.$ Thus, we need to prove $P'Q'\|PQ.$ We prove this by showing that $$(P, Q ; M, \overline{PQ}\cap \overline{P'Q'}).$$Thus, we take a homography that sends $ABCD$ to a rectangle. $M$ doesn't have to be the midpoint of $PQ$ anymore, so redefine it as $\overline{PQ} \cap \overline{KR}.$ After the homography, $P$ and $Q$ are now points at infinity and $K$ is basically arbitrary. Note that $KP'$ is parralel to a pair of sides of the rectangle, and $KQ'$ is parralel to the other pair. Thus, $KP'\perp KQ',$ so $P'$ and $Q'$ are antipodes. Note that $R$ is the center of $(ABCD),$ so $P'RQ'$ is collinear. Project our desired cross ratio with perspectivity from $R$ onto $KQ'.$ Note that $Q$ goes to the point at infinity, $KR \cap PQ$ goes to $K,$ $P'Q' \cap PQ$ goes to $Q',$ and $P$ goes to the midpoint of $KQ'.$ Thus, this is a harmonic bundle, and we're done.$\blacksquare$

Solved with ihategeo_1969 and Om245. Longer solution that most of the above ones but it was a really enjoyable problem. We let $L$ denote the second intersection of $\overline{MR}$ with $\omega$ besides $K$. Further, let $S = \overline{MC} \cap \omega$ and $N = \overline{AS} \cap \overline{PQ}$. Let $O$ denote the center of $\omega$. Finally, let $M_q$ denote the Miquel Point of cyclic quadrilateral $ABCD$, which we know lies on $\overline{PQ}$. We first show the following known result. Claim (11MOPK31) : Point $S$ lies on circle $(APQ)$. Proof : Let $(\cdot)$ denote the point circle with center $\cdot$. Now, we define the function $f(\cdot) = Pow_{\omega}(\cdot) - Pow_{(M)}(\cdot)$, for any point $\cdot$. By Linearity of the Power of a Point, we know that $f$ is linear, and hence \begin{align*} MC \cdot MS &= f(M)\\ &= \frac{f(P)+f(Q)}{2}\\ &= \frac{PB\cdot PA - PM^2 + QD\cdot QA - QM^2}{2}\\ &= \frac{PM_q\cdot PQ + QM_q \cdot QP - PM^2 - QM^2 }{2}\\ &= \frac{PQ^2 - PM^2 - QM^2}{2}\\ &= PM^2 = QM^2 = PM\cdot QM \end{align*}Now, let $C'$ be the reflection of $C$ across $M$. It is immediate that $AQCP$ is cyclic, and due to the above length calculation, it follows that, \[MC' \cdot MS = MC \cdot MR = MP \cdot MQ\]and thus $SQC'P$ is also cyclic, which proves the claim. This will come in useful later. Now, we attack the bulk of the problem. First note that, \[\measuredangle NM_qC = \measuredangle PM_qC = \measuredangle ABC = \measuredangle ASC \]from which it follows that $NM_qCS$ is cyclic. Then, we have the following key claim. Claim : Points $M_q$ , $K$ , $L$ , $O$ and $N$ lie on the same circle. Proof : It is well known that points $M_q$ , $R$ and $O$ are collinear and $AOCM$ is cyclic (results of inversion at $M_q$), so \[RM_q \cdot RO = RA \cdot RC = RL \cdot RK\]so it follows that $M_qKOL$ is cyclic. Further, \[MM_q \cdot MN = MC \cdot MS = MK \cdot ML\]from which it follows that $M_qKLN$ is also cyclic, finishing the proof of the claim. Now, we are almost there. It is well known that $OR \perp PQ$. Note that this implies, \[\measuredangle NKL = \measuredangle NM_qL = 90 + \measuredangle OM_qL = 90 + \measuredangle OKL = 90 + \measuredangle KLO = 90 + \measuredangle KM_qO = \measuredangle KM_qM = \measuredangle KLN \]so we have $NK=NL$. Since $(NKL)$ passes through the center of $\omega$. This implies that $NL$ and $NK$ are tangents to $\omega$. Finally, we have that \[NK^2 = NS \cdot NA = NP \cdot NQ\]implying that $NK$ is also tangent to $(KPQ)$. Similarly we have that $NL$ is tangent to $(LPQ)$. Thus, it follows that circles $\omega$ and $(KPQ)$ are tangent at $K$ and $(LPQ)$ and $\omega$ are tangent at $L$ with the common tangents being $\overline{KN}$ and $\overline{LN}$ respectively.

Let the line passing through $R$ and parallel to $PQ$ intersect $\omega$ at $X, Y$. We only need to prove $PX, QY$ and $\omega$ are concurrent, the rest is just homothety and obvious Brokard. Consider the homography that preserves $\omega$ and sends $ABCD$ to a rectangle. Then $P$, $Q$ becomes point of infinities and $R$ becomes the center of $\omega$ and $XY$ is the diameter of $\omega$. Since $P, Q$ are point of infinities $PX$ and $QY$ are parallel to $AB$ and $BC$, respectively $\implies$ $PX \cap QY$ lies on $\omega$. $\blacksquare$

Invert from $Q$ with radius $\sqrt{QC.QB}$. Let $S$ be the miquel point where $P$ swaps with under this inversion. $M^*$ is on $PQ$ such that $QS=SM^*$. Let the tangents from $S$ to $\omega$ be $T_1,T_2$. We will show that $Q,M^*,R^*,T_1,T_2$ are cyclic which is enough since $K$ swaps with $T_1$ or $T_2$. $S$ lies on the polar of $R$ according to $\omega$ hence by La Hire, we get that $R$ lies on the polar of $S$ thus, $R,T_1,T_2$ are collinear. If $O$ is the circumenter of $\omega,$ then $\overline{ORS}$ is the perpendicular bisector of both $T_1T_2$ and $QM^*$ hence $QM^*T_1T_2$ is an isosceles trapezoid. Also $RT_1.RT_2=RB.RD=RR^*.RQ$ so $R^*$ also lie on $(QT_1T_2)$. These yield $Q,M^*,T_1,T_2,R^*$ are cyclic.$\blacksquare$

Let $PK \cap \omega$ be $E$ and $QK \cap \omega$ be $F$. It suffices to show that $EF || PQ$ or that $(P,Q;M,EF \cap PQ)=-1$ so now the problem is purely projective. Now taking a homography sending $R$ to centre of $\omega$ finishes.

let $MR\cap\omega=K,L$ let $N$ be the midpoint of $KL$ and let $X$ be pole of $MR$ wrt $\omega$ so $XK$ tangent to $\omega$. by self polar orthogonality $(PQ)$ orthogonal to $\omega$ so $MK\cdot ML=MP^2$ and we have $XP\cdot XQ-XK^2=XM^2-XK^2-MP^2=NM^2-NK^2-MP^2=MK\cdot ML-MP^2=0$ so $XK$ tangent to $(KPQ)$ done

Just forget about $ABCD$. Invert around $\omega$ and denote by $X'$ the inverse of $X$. Let $\ell$ be the tangent to $\omega$ at $K$. By pole-polar duality, we have that $RM'$, $\ell$ and $PQ$ are concurrent at some point $T$. Also note that $M'$ is the $O$-Humpty point in $POQ$ so $QRM'P$ is cyclic (orthocenter config). Notice that $Q'$ is the foot from $P$ in $\triangle POQ$ and similarly for $P'$. By radical center we get that $P'Q'$, $RM'$ and $PQ$ are concurrent at the same point $T$. Now, it's easy to see that $OM'KR'T$ is cyclic and its circumcircle has diameter $OT$. Therefore \begin{align*} TK^2&=TO^2-OK^2\\ &=OM'^2+TM'^2-OM'\cdot OM\\ &=TM^2-MM'^2-OM'\cdot MM'\\ &=TM^2-MM'\cdot MO\\ &=TM^2-MP^2\\ &=\text{Pow}(T,(PQQ'P'))\\ &=TP\cdot TQ \end{align*} So we are done. $\blacksquare$