Let $O',Y$ be the circumcenter of $\Gamma,\Omega$ and $Z$ be the projection of $N$ on $MO$ and $C'$ lie on $\Gamma$ such that $CC' \parallel AB$. Clearly, $\overline{D-O'-O},~\odot(ZMXN),~O$ is the midpoint of $AY$, $O'$ is the midpoint of $RC'$.Consider \begin{align*} \odot(DNXO') &\iff OD \cdot OO' = ON \cdot OX \\& \iff OD \cdot OO' = OM \cdot OZ \\& \iff \odot(MZDO') \\& \iff O'Z \perp MO \iff O'N \perp OM \\& \iff O'N \perp LY \iff C'D \perp LY \end{align*}Let $Y'$ be the reflection of $C$ across $Y$ and $S$ is the intersection of $Y'P$ on $CR$. Hence, $\triangle CC'A \cup \{D\} \sim \triangle CSY' \cup \{Y\}$. Hence, $C'D \perp SY$ and it suffices to show that $S,Y,L$ are collinear. Claim: $\overline{Y-P-R}$. Proof: Let $Y'P \cap \Gamma = A'$ and $P'$ lie on $\Gamma$ such that $PP' \parallel CY'$. By angle chasing, $\square A'RCP'$ is a kite and $(A',C;R,P')=-1$. Then, project it to $CY'$ throught $P$. Let $L' = PR \cap AC, B'' = AY \cap CR$ and $LY \cap CR = S*$. $(A,C;Q,R)=(A,C;L,L') =(B'',C;S*,R)$ Hence, $AB'' \cap S*Q$ lies on $\Gamma$. Let $B' = AY \cap \Gamma$. It suffices to show that $AB' \cap SQ$ lies on $\Gamma$. Equivalent to $A'P, QB'$ and $CR$ are concurrent. Consider $\triangle CQB$, $P$ is $B-$Dumpty point. Let $T$ be the isogonal conjugate of $P$ which is $B-$Humpty point and $BT \cap CQ = T'$ which is the midpoint of $CQ$. Let $E$ be the foot of the altitude from $Q$ to $BC$, $\angle BAY = \theta, \angle PQR = \alpha$. By angle chaseing, $\angle CTT' = \angle BCT' = \angle CET$. Therefore, $\odot(CT'TE)$ and $\angle QET = 90^{\circ} - 2 \theta$. Notice that $A'C$ is the diameter of $\Gamma$. Let $G = B'Q \cap A'P$, By angle chasing, $\square BETQ \sim CB'GA'$ which mean $\overline{G-C-R}$. Therefore, $A'P, QB'$ and $CR$ are concurrent.

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As above shows, it's equivalent to prove $C'D \perp LY$. Here is another finishing (I'll just adopt above's diagram for laziness): Since $R$ is the midpoint of arc $PQ$, we have $\angle QAR= \angle RCP$, hence from $CR \perp AB$ we get $90^{\circ}= \angle QAC+ \angle RCA = \left( \angle QAC- \angle QAR \right) + \left( \angle RCA + \angle RCP \right) =\angle RAC + \angle ACP$. Because $YC \perp AC$, we have ghen $\angle YPC = \angle YCP = 90^{\circ} - \angle ACP =\angle CAR$, hence $Y,P,R$ are collinear. Further, $YPR \perp C'P$, so $YP$ is tangent to $(C',C'P)$ (which also passes through $Q$ by symmetry). Since $YP=YC$, $Y$ lies on the radical axis of $(C',C'P)$ and $(AC)$. But $LQ \cdot LP = LA \cdot LC$, so $L$ is also on the radical axis, therefore $C'D \perp YL$.

As in the figure above let $C'$ be the point diametrically opposite to $R$ in circle $\Gamma$ .It suffices to show that $O'N \perp OM$(on a point $X_1$) for then we would have that $OX_1\cdot OM=OD\cdot OO'=ON\cdot OX$ Clearly this is equivalent to showing that $C'D \perp YL$. (1) $C'P$ is tangent to $\Omega$ Some angle chasing gives that angles $\angle PBC=\angle PCA=\angle AQP=\angle CRC'=\angle CPC'$ Trivially $CC'$ is parallel to $AB$ (2) $PQ$ , $CC'$ concur on a point $Z$ on circle $\Omega$ trivial angle chasing: $\angle QPC=\angle QAC=\angle ZCA$ (3)Let $CP$ meet $AB$ on $P'$. $C',D,P'$ are collinear $AP=AP'$ because the bisector $CR$ of $PCQ$ is perpendicular to $AB$ and $AP'P=AQC=P'PA$ also $CC'=AP$ and the fact that $CC'=AP'$ is just what we need. Let $F$ be the pole of $PZ$ wrt to circle $\Omega$ then $L$ is the pole of $CZ$ and $CZ\perp LY$ . Clearly $FZ $ is parallel to $QC'$ and then triangles $P'QC'$ and $CZF$ are homothetic wrt $P$ and $P'C'$ is parallel to $CZ$ which is perpendicular to $LY$ done.

Let $Z$ and $S$ be the centers of $\Omega$ and $\Gamma$ respectively. Let $F$ be the foot of perpendicular from $C$ and let $Q'$ be the reflection of $Q$ over $F$. First, notice that $\triangle CQQ'$ is isosceles, and since $\triangle CQQ'\sim\triangle APQ'$, we see that the latter is also isosceles. As $RA$ internally bisects $\angle PAQ$, it follows that $AR\perp CQ'$, and hence $R$ is the orthocenter of $\triangle AQ'C$. As $O,D,S$ are collinear, to show that $S$ lies on $(DNX)$, it suffices to show \[OD\cdot OS=ON\cdot OX.\]To eliminate $X$, let $Y$ be the foot of perpendicular from $N$ to $OM$. Then $OD\cdot OS=OY\cdot OM$ if and only if $D,S,Y,M$ are concyclic or equivalently, $\angle SYM=90^\circ$. Thus all we have to show is that $Y,N,S$ are collinear, or equivalently, $NS\perp MO$. Now let $T$ be the midpoint of $Q'R$. Since $R$ is the orthocenter and $S$ is the center of $(ARC)$, it follows that $RTDS$ is a parallelogram, so $NS$ passes through $T$. Now homotheties centered at $A$ and $R$ with ratio 2 respectively show that $MO\parallel LZ$, and $NS\parallel Q'D$, we just have to show that $LZ\parallel Q'D$. Let $K$ be the Miquel point of cyclic quadrilateral $AQPC$. Clearly $K$ lies on $\overline{Q'L}$. Then \[DZ^2-Q'Z^2=DC^2-CZ^2-(Q'B^2-BZ^2)=DC^2-Q'B^2=DA^2-Q'B^2\]and \begin{align*} (LD^2-DA^2)-(LQ'^2-Q'B^2)&=LA\cdot LC-LQ'^2+Q'B^2\\ &=LK\cdot KQ'+Q'P\cdot Q'C-LQ'^2\\ &=LK\cdot LQ'+Q'K\cdot Q'L-LQ'^2 &=LQ'^2-LQ'^2=0 \end{align*}and we're done.

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Is there a way do draw this diagram (using only ruler and compass)?

hydo2332 wrote: Is there a way do draw this diagram (using only ruler and compass)? Why not, the students who would have written the actual test, must have drawn this by ruler and compass....

hydo2332 wrote: Is there a way do draw this diagram (using only ruler and compass)? I think that it is best to think about the problems with drawings by hand, but I also think that thinking about how the construction would be is a nice exercise (it has even helped me to solve some problems). yes it is possible to build it with ruler and compass, if you can't do it I recommend you look at the solutions above to help you, for example you would know that you only need the construction of $\text{B-Dumpty poin}$

By the given conditions, note that $\triangle ABC$ is isosceles. Let $O_1$ be the center of $\Omega$, and $O_2$ be the center of $\Gamma$. Draw $F$ to be the intersection of the line passing through $C$ parallel to $AB$ with $\Gamma$, so $AB \parallel CF$ and $F \in \Gamma$. Now, I claim $\overline{R O_2 F}$ collinear. This is true since $\measuredangle BKC = \measuredangle FCK = \measuredangle FCR = 90^{\circ}$, so $RF$ is a diameter of $\Gamma$. The condition $(X,N,D,O_2)$ concyclic is equivalent to $ON \cdot OX = OD \cdot OO_2$, but since $\measuredangle MXO = 90^{\circ}$, $ON \cdot OX = OH \cdot OM$ where $H$ is the foot from $N$ to $MO$. So, it suffices to show $OH \cdot OM = OD \cdot OO_2$. This is just proving $(H,M,D,O_2)$ concyclic. Note that $\measuredangle O_2 D M = 90^{\circ}$, so it is enough to prove $MH \perp HO_2$. But since $MH \perp HN$, it suffices to prove $\overline{HNO_2}$ collinear, or equivalently $MH \perp NO_2$. Now, since $M$ and $O$ are the midpoints of $AL$ and $AO_1$, respectively, it suffices to prove $LO_1 \perp NO_2$. Note that $N$ and $O_2$ are the midpoints of $RD$ and $RF$, respectively, Thus, $NO_2 \parallel DF$, so it suffices to prove $LO_1 \parallel DF$. Note that we can completely erase the point $O_2$ at this point. Let $\Pi$ be the circle $(AKC)$ with diameter $AC$, and $\Delta$ be the circle passing through $P$ and $Q$, centered at $F$. Then, $\mathrm{Pow}_\Pi L = LA \cdot LC = \mathrm{Pow}_\Gamma L = LQ \cdot LP = \mathrm{Pow}_\Delta L$, so $L$ lies on the radical axis of $\Pi$ and $\Delta$. Moreover, I claim that $\overline{RPO_1}$ collinear. First, $\angle RAC + \angle ACP = 90^{\circ}$, and $\angle RAC = \angle RFC = 90-\angle FRC = 90-\angle FPC$. So, $\angle ACP = \angle FPC$. Because $\angle O_1 P C = \angle O_1 C P = 90 - \angle PCA = \angle RAC = \widehat{RC}$ and $\angle RPQ = \widehat{RQ}$, $\angle QPF = \widehat{QF}$, $\angle FPC = \widehat{FC}$, we have $\angle RPQ + \angle QPF + \angle FPC + \angle CPO_1 = \widehat{RQ} + \widehat{QF} + \widehat{CF} + \widehat{RC} = 180^{\circ}$, so $\overline{RPO_1}$ collinear, as desired. Now, since $\overline{O_1 P R}$ collinear and $\measuredangle FPR = 90^{\circ}$, we have $\measuredangle O_1 P F = 90 ^{\circ}$, so $\mathrm{Pow}_\Delta O_1 = O_1 P^2 = O_1 C^2 = \mathrm{Pow}_\Pi O_1$ since $\measuredangle ACO_1 = 90^{\circ}$. Thus, $O_1$ also lies on the radical axis of $\Delta$ and $\Pi$. But this implies that $LO_1$ is the radical axis of $\Delta$ and $\Pi$, so $LO_1 \perp DF$, since $D$ is the center of $\Pi$ and $F$ is the center of $\Delta$, which is what we wanted to show.