I solved it with coordinate bash which was not that hard

Sorry but can you post the problem #2?

i solved this with some dirty calculations with LM and MO, n=but my friends say that it can be solved with pascal theorem...

Well, here's a "radical axis" solution: Let $N'$ be the midpoint of $AB$, and let $CQ$ meet $AM$ and $ON$ at $S, K$, respectively. First, note that $\angle QLO=90^{\circ}-\angle QBC=\angle OCQ \implies (LQCO) \text{ cyclic}$. Now since $(NKSL)$ is cyclic, we have $OC^2=OQ^2=OS\cdot OL=ON\cdot OK$, so the inversion WRT $\odot(O, OQ)$ sends $CKSQ$ to $\odot(OCNSL)$. Hence $OCNSL$ is cyclic. Now we have $\angle ACO=\angle NLO=90^{\circ}$, so by symmetry $ABCO$ is cyclic. Also, $\angle BPQ=\angle QCA=\angle QLN' \implies (PQLN') \text{ cyclic}$. Then we have the following claim. Claim. $X, Y, Z$ lies on the radical axis of $\odot(ABCO)$ and $\odot(PQLN')$. Proof. Let $l$ be the radical axis. Then, we have $XA\cdot XC=XP \cdot XQ \implies X \in l$, $YN' \cdot YL=YB\cdot YO \implies Y \in l$ (since $(N'LBO)$ are cyclic), $ZO\cdot ZC=ZQ \cdot ZL \implies Z\in l$. Hence we are done. $\blacksquare$

My solution at test. Rather complicated... 1. Let) $U, A' : AC$, $\Omega \cap BL$ $V= QM\cap AC$ $I=$ foot of perpendicular of $Z$ to $AM$ a= $LM$, b= $MO$. $BM=AA'\to B, L, A'$ collinear $\to$ as $A', N, M$ is collinear, $\angle A'QM=$ 90 $\to QM\bot BC$ $\angle BCO= 90-\angle CQL= 90-\angle CQA'= 90-\angle CAA'= 90-\angle C\to CO\bot BC,$ respectively $BO\bot AB.$ Also, $BC: AA'= 2: 1\to AU: UC= 1: 2$ 2. $B, M, C$ are harmonic, so by pencil at $L$, we can find that $Y, B, O, Z'$ are harmonic as well, and as a result, $L, M, O, I$ are harmonic ($Z'$ is a symmetrical point of $Z$ to $AM$) $\to \frac {LZ}{UZ}=\frac{LI}{LI+\frac{1}{3}AL}= \frac{LO\cdot \frac{LM}{LM-MO}}{\frac{1}{3}AL+LO\cdot \frac{LM}{LM-MO} }$ (because when A,B,C,D are harmonic ; $AD=AC\cdot \frac{BC}{AB-BC}$) $=\frac{(a+b)\cdot a}{\frac{1}{3}a\cdot (a-b)+(a+b)\cdot a}=\frac{3a+3b}{4a+2b}$ 3. $\frac{YN}{LY}=\frac{\frac{1}{2} BM}{BM\cdot \frac{OL}{MO}} +1=\frac{MO}{2OL}+1=\frac{2a+3b}{2a+2b}$ 4. by $\triangle BCU$ & $QMV$ menel $1=\frac{QU}{BQ} \cdot \frac{CV}{UV}$ $\to \frac{CV}{UV}=\frac{BQ}{QU}=\frac{BQ}{QL+\frac{1}{3} BL}=\frac{BM^2}{ML^2+\frac{1}{3} (BM^2+ ML^2)} =\frac{2LM\cdot MO}{ML^2+\frac{1}{3} (2LM\cdot MO+ML^2)} =\frac{6MO}{4ML+2MO}=\frac{3b}{2a+b}$ by the way, as $QP, QA', QC, QM$ are harmonic, $U, C, V, X$ are harmonic too. $\to XU=UV\cdot \frac{UC}{UC-CV}=\frac{UC}{1-\frac{2CV}{UV}}= UC\cdot \frac{1}{1-\frac{6b}{2a+b}}= UC\cdot \frac{2a+b}{2a-5b}$ $\to \frac{XU}{NX}=\frac{XU}{XU-\frac{1}{4}UC}=\frac{\frac{2a+b}{2a-5b}}{\frac{6a+9b}{8a-20b}}=\frac{8a+4b}{6a+9b}$ 5 ;so, as $\frac{LZ}{UZ} \cdot \frac{YN}{LY} \cdot \frac{XU}{NX}=\frac{(3a+3b)(2a+3b)(8a+4b)}{(4a+2b)(2a+2b)(6a+9b)}=1$ by $\triangle ULN$ & $XYZ$ menel we can find that problem holds.

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Another solution i found recently using pascal let) $W=AA'\cap PC$ $T$=point on $AM$ s.t. $A'T//ZO$ $K=CP\cap AM$ by pascal on $CCAA'PQ \to W,Y,X$ collinear $WTS) WYZ$ collinear $ETS) \frac{WA'}{YL}=\frac{ZA'}{ZL} \frac{WA'}{YL}=\frac{AM}{OL}$ (similarity of $\triangle YLO$ and $\triangle WAM$) $=\frac{TO}{OL}=\frac{ZA'}{ZL}$ (similarity of $\triangle ALA'$ and $\triangle OLZ$) ...done

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Let $Q'$ be the point where the $C-$symmedian meet the circumcircle of $\triangle AMC$ for the second time. Let $F$ be the midpoint of $CQ'$. Now it is well-known that $F'$ is the $C-$dumpty pointof $\triangle AMC$. Now we have $$\angle LMF=90^{\circ}-\angle CMF=90^{\circ}-\angle ACF=90^{\circ}-\angle MCL=\angle MLC=\angle BLM$$Hence $MF\|BL$, a homothety centered at $C$ shows that $Q'$ lies on $BL$ as desired. $\blacksquare$ CLAIM 2. $Q,L,N,O,C$ are concyclic. Proof. Notice that $\angle QLO=\angle CLO$ and $OQ=OC$, so by fact 5 we have $O\in (QLC)$. From CLAIM 1, $$\angle NLC=\angle LCB=\angle ACQ=\angle NCQ=\angle NQC$$as desired. $\blacksquare$ CLAIM 3. Let $PQ$ and $BO$ intersect at $D$, then $D$ lies on the circle $(QLNOC)$ as well. Proof. This is just angle chasing. From CLAIM 2, $\angle ABO=\angle ACO=\angle NLO-90^{\circ}$ $$\angle QDO=\angle PBD+\angle BPD=90^{\circ}+\angle ACQ=180^{\circ}-\angle QCO$$as desired. $\blacksquare$ We finish the problem by directly applying Pascal's theorem to $QDONCL$.

a simple solution: let $N'$ be the midpoint of $AB$ and let $K$ be the midpoint of $BM$ easily get $O$ is the antipode of $A$ in $(ABC)$ claim(1):$NCOXQL$ is cyclic proof: $\angle OCN=90=\angle OLN$ note that $\angle CLO=\angle CNO=\angle BN'O=\angle BLO=\angle QLO$ combined with $OQ=OC$ we have $OQLC$ is cyclic $2MX.MC=MB.MC=MA.MO=2ML.MO$ $\blacksquare$ claim(2):$PN'QL$ is cyclic proof: since $\angle N'XC=90=\angle CPN'$ so $PN'XC$ is cyclic $BX.BC=BQ.BL=BN ' . BP$ $\blacksquare$ note that $XY$ is the radical axis of $(ABCO) , (PN'QL)$ $QL$ is the radical axis of $(LQOCN) , (PN'QL)$ $XY$ is the radical axis of $(ABCO) , (LQOCN)$ and we win

It has been more than a year since I solved this question in the test. You simply let Q' be the intersection of BL and the circumcenter of AMC, and let W be the intersection of AQ' and CP. With Pascal's on QQ'ACCP, we obtain Z, W, X collinear. Then simple length observation gives ZP:ZQ'=YL:WQ', implying that Z, Y, W collinear. QED.

Can we prove this with pascal and La Hire?

My solution: Let T be the intersection of XQ and circumcircle of BQC, and let R be the intersection of YB and circumcircle of BQC, and let S be the intersection of CZ and circumcircle of BQC. Let BC and ST meet at W. By Pascal on CCQBST, X,W,Z are collinear. It's easy to see that C,Q,Y are collinear. and well known Z,T,R are collinear (With pappus hexagon theorem). So, by Pascal on QBSTRC, Y,W,Z are collinear as required. please see this solution is true or not.

Another way to finish after the pascal of chrono 223's post about his another solution Since $WA'C$ and $YLO$ are homothetic, we get the desired result

Point $D$ ($AMCD$ is rectangular) kills this problem