Quite shrimple? Claim: $BC = CQ$, and similarly $BP = BC$. Proof. By directed lengths, we get that $BQ = b \cdot \frac{\frac{a}{2}}{a \cdot \frac{b}{b+c}} = \frac{b+c}{2}$. $\blacksquare$ Claim: $BQ, PC, AX$ concur on $(XPQ)$. Proof. Note that $\measuredangle (AX, QB) = \measuredangle CBQ = \measuredangle BQC = \measuredangle (BQ, QA)$. This implies that $W = BQ \cap AX$ lies on $(XPQ)$. By symmetry, we similarly get $W$ lies on $PC$. $\blacksquare$ As such, it follows that $\measuredangle QBM + \measuredangle BCP = \measuredangle QWP$ which implies that the circles $(BMQ), (CMP), (PXQ)$ concur at some point $D$. Now, by triangle Miquel on $BXC$ we get that $(XYZD)$ is cyclic. Claim: $AX, AD$ are tangents to $(XYZD)$. Proof. Since $\measuredangle MDQ = \measuredangle MBW = \measuredangle XWQ = \measuredangle XDQ$ we get that $M$ lies on $XD$. Note that since $\measuredangle AXD = \measuredangle AXM = \measuredangle DMB = \measuredangle DYB = \measuredangle DYX$, it follows that $AX$ is a tangent. By orthogonality it follows that $AD$ must also be a tangent. $\blacksquare$ As such, since \[ (BC;M\infty) \overset{X}= (YZ;DX) = -1 \]this implies $A$ lies on $YZ$.

My problem! [asy][asy] // ELMO 2024/5 size(9cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen greenfill,greendraw,lightbluedraw,bluedraw,purpledraw,pinkdraw; greenfill = RGB(204,255,204); greendraw = RGB(0,187,0); lightbluedraw = RGB(85,187,255); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); pair A,B,C,M,P,Q,X,Y,Z,R; path c,d,e; A = (-4/5,3*sqrt(7)/5); B = (-1,0); C = (1,0); M = (B+C)/2; c = circle(A,(distance(A,C)-distance(A,B))/2); P = intersectionpoints(c,A--2*A-B)[0]; Q = intersectionpoints(c,A--C)[0]; X = intersectionpoints(c,A--A+B-C)[0]; d = circumcircle(B,M,Q); e = circumcircle(C,M,P); Y = intersectionpoints(d,B--X)[0]; Z = intersectionpoints(e,C--X)[0]; R = intersectionpoints(d,e)[0]; filldraw(c,greenfill,greendraw); draw(d,purpledraw); draw(e,purpledraw); draw(circumcircle(X,Y,Z),pinkdraw); draw(P--A--B--C--A,bluedraw); draw(M--P,lightbluedraw); draw(A--X,lightbluedraw); draw(A--R,lightbluedraw); draw(A--Y,lightbluedraw); draw(B--X,lightbluedraw); draw(C--X,lightbluedraw); draw(M--X,lightbluedraw); dot("$A$",A,dir(40)); dot("$B$",B,dir(230)); dot("$C$",C,dir(290)); dot("$M$",M,dir(275)); dot("$P$",P,dir(110)); dot("$Q$",Q,1.7*dir(25)); dot("$X$",X,dir(130)); dot("$Y$",Y,dir(320)); dot("$Z$",Z,dir(190)); dot("$R$",R,1.6*dir(245)); [/asy][/asy] Let $\measuredangle$ denote directed angles modulo $180^\circ$. By Menelaus, we have \[\frac{BP}{PA} \cdot \frac{AQ}{QC} \cdot \frac{CM}{MB}=1,\]and since $BM=CM$ and $AP=AQ$, this gives $BP=CQ$. If we let $BP=CQ=x$ and $AP=AQ=y$, then we have $x-y=AB$ and $x+y=AC$, so $x=\tfrac{AB+AC}{2}=BC$. Let circle centered at $A$ through $X$, $P$, and $Q$ intersect $\overline{XM}$ again at $R$. Claim: $R$ lies on the circumcircles of $BMQ$ and $CMP$. Proof: We have \[\angle MRQ=180^\circ-\angle XRQ=\frac{\angle XAQ}{2}=\frac{180^\circ-\angle ACB}{2}=\angle MBQ,\]so $BMQR$ is cyclic. Analogously, $CMRP$ is cyclic. $\square$ Claim: $XYRZ$ is cyclic and its circumcircle is tangent to $\overline{AX}$. Proof: We have \[\measuredangle XYR=\measuredangle BYR=\measuredangle BMR=\measuredangle AXR,\]so the circumcircle of $XRY$ is tangent to $\overline{AX}$. Analogously, the circumcircle of $XRZ$ is tangent to $\overline{AX}$, so the circle through $X$ and $R$ tangent to $\overline{AX}$ passes through $Y$ and $Z$, as desired. $\square$ Let $\infty_{BC}$ be the point at infinity on $\overline{BC}$. We have $(X,R;Y,Z)\overset{X}{=}(\infty_{BC},M;B,C)=-1$, so $XRYZ$ is harmonic. Since $\overline{AX}$ is tangent to its circumcircle, we know that $\overline{AR}$ is tangent as well, so $A$, $Y$, and $Z$ are collinear by the symmedian configuration. $\blacksquare$

By lengths, $AP=AQ=\frac{b-c}2$ so $BP=CQ=a$, which gives $P=(2a:c-b:0)$ and $Q=(2a:0:b-c)$. We get $X=(2a:b-c:c-b)$. The circumcircle of $BMQ$ is $-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{(2b-a)(b-c)}4x+\frac12a^2z\right)=0$ so line $BX$ intersects this at $(2a:y:c-b)$, implying $$y(-a^2(c-b)-2ac^2)-2ab^2(c-b)+\frac14(2a+c-b+y)((2b-a)(b-c)2a+2a^2(c-b))=0$$or $$y=\frac{(b-c)(c-2b)}{(b-2c)}.$$ Similarly, $CX$ intersects this at $(2a:b-c:z)$ where $z=\frac{(c-b)(b-2c)}{c-2b}$, so $A$, $Y$, and $Z$ collinear is equivalent to $\frac y{c-b}=\frac{b-c}z$, or $yz=-(b-c)^2$, which is true.

Nice problem! First we will prove that $BP=BC=CQ$. Note that by Menelaus's theorem we have \[ 1=\frac{BM}{MC} \cdot \frac{CQ}{QA} \cdot \frac{AP}{PB} = \frac{CQ}{PB},\]so $CQ=PB$. Also $CQ+PB=(CQ+QA)+(PB-PA)=AC+AB=2BC \Rightarrow CQ=PB=BC.$ Let $D=(BMQ) \cap (CMP)$. Now we will prove that $AD=AP=AQ$. Note that this condition is equivalent to condition that $A$ is center of circle $(PQD)$. Since $AP=AQ$, it's enough to prove that $\angle PAQ=2 \angle PDQ$. Note that $D$ is center of spiral similarity of segments $PQ$ and $CB$, so $\angle PQD=\angle BCD$ and $\angle PCD=\angle QBD$, so $\angle BDC=180^\circ-\angle DBC-\angle DCB=180^\circ - (\angle CBQ + \angle BCP)=180^\circ-(90^\circ - \frac{1}{2}\angle BCA + 90^\circ - \frac{1}{2}\angle CBA)=\frac{1}{2} (\angle ABC + \angle ACB)=\frac{1}{2}\angle QAP.$ So, really $\angle PAQ=2 \angle PDQ$ and $AP=AQ=AD$. Note that if $BQ \cap AX=E$, then $\angle AEQ=\angle CBQ=\angle CQB=\angle AQE$ and so $AQ=AE \Rightarrow E \in (PQD).$ Now note that since $AD=AE=AP=AQ=AX$, $PEQDX$ is cyclic and then $\angle AXD = \angle DQB = \angle DMB$ and so $M,D,X$ are collinear. By PoP we get $XY \cdot XB = XD\cdot XM = XC \cdot XZ$, so $BYZC$ is cyclic. We have $\angle DYX = \angle BMD = \angle CZD$, so $XYDZ$ is cyclic. Now $\angle AXZ = \angle XCB = \angle ZYX$, so $AX$ is tangent to circle $(XYDZ)$. Since $AD=AX$, line $AD$ is also tangent to circle $(XYDZ)$. By this reason it is enough to prove that $XYDZ$ is harmonic quadrilateral. Note that $\Delta XZD \sim \Delta XMC$ and $\Delta XDY \sim \Delta XBM$, so \[ \frac{XZ}{ZD}=\frac{XM}{MC}=\frac{XM}{MB}=\frac{XY}{YD} \Rightarrow XZ \cdot DY = XY \cdot ZD\]and so we are done!

First, notice that $P$ and $Q$ both lie on line through $M$ parallel to the angle bisector of $\angle BAC$. Thus, if $P_1 = 2P-B$ and $Q_1=2Q-C$, then $AP_1=AC$ and $AQ_1=AB$. Thus, $BP = CQ = \tfrac{AB+AC}2 = BC$ Now, let $XM$ intersect $\odot(XPQ)$ at $T$. We have $$\angle QTM = 180^\circ - \angle QTX = \frac{\angle QAX}2 = 90^\circ - \frac{\angle C}2 = \angle QBM,$$so $T\in\odot(BMQ)$. Similarly, $T\in\odot(CMP)$. By power of point from $X$, we get $BCYZ$ concyclic. Moreover, by Miquel's theorem, we get $XYZT$ concyclic. Finally, notice that since $XM$ is median of $\triangle XBC$, we get that $XT$ is symmedian of $\triangle XYZ$. Thus, $XYZT$ is harmonic quadriltaral. Since we have $XA$ tangent to $\odot(XYZ)$ and $AX=AT$, it follows that $A\in YZ$, done.

Nice Problem! Looks like I over-complicated lol. 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Let $BW \cap AC = Q'$ and $CW \cap AB = P'$ $\textbf{Claim 1:}$ $P' \equiv P$ and $Q' \equiv Q$. $\textbf{Proof:}$ By reflection $\measuredangle Q'CB = 90 - \frac{C}{2}$ so $BC = CQ' = BP'$; $$AP = BC-AB = AC-BC = AQ$$and by converse of menelaus : $$\frac{BM}{MC} \cdot \frac{BP}{AP} \cdot \frac{AQ}{CQ} = 1$$, we get $M,P,Q$ are collinear which implies the claim as there exists an unique choice of $P,Q$. $\textbf{Claim 2:}$ $AW \parallel BC$, $AW = AP=AQ$ $\textbf{Proof:}$ let $M_A$ denote the arc midpoint of arc $BC$, by ptolemy $M_AB(AB+AC) = AM_A(BC)$ so $2M_AB=AM_A$ which implies $M_AI_A=M_AI=AI$ by incenter-excenter lemma.Let $AI \cap BC = D$, then it is well known that $$-1=(BA,BD;BI,BI_A) = (AD;II_A) = \frac{IA}{ID} \cdot \frac{I_AD}{I_AA} = \frac{3 \cdot ID}{I_AD}$$, therefore $AD = I_AD$, now reflect $I_A$ about $B,C$ to get $I_A'$ and $I_A''$ then $\overline{AI_A'I_A''}$ is collinear and parallel to $BC$ as $I_AD=AD$ so we conclude $AW \parallel BC$, now for $AW=AP$ consider homothety of half at $I_A$ to get $AW = 2MD = BC-AC=AP=AQ$ Let $L = CW \cap MM_A$, let $D \in (ABC)$ such that $AD \parallel BC$, let $T = PQ \cap AD$, let $S = AD \cap MM_A$. $\textbf{Claim 3:}$ $L \in (CMPY)$ and $\overline{LPX}$ are collinear. $\textbf{Proof:}$ $$\measuredangle CLM = \measuredangle C/2 = \measuredangle (90- A/2 - B/2) = \measuredangle BPC - \measuredangle APQ = \measuredangle CPM $$. $$90^{\circ}= \measuredangle CPL = \measuredangle (90-B/2) + \measuredangle (B/2) = \measuredangle CPB + \measuredangle BPX = \measuredangle CPX $$ $\textbf{Claim 4:}$ $APQSM_{BC}$ is cyclic where $M_{BC}$ is arc midpoint of $BC$ containing $A$. $\textbf{Proof:}$ $APQM_{BC}$ is cyclic as $M_{BC}QC \equiv M_{BC}PB$, notice that $\overline{MPQ}$ is simson line of $M_{BC}$ as $\measuredangle AQM_{BC} = 180-\measuredangle (90-A/2)-\measuredangle A/2 = 90 $ so $(APQ)$ is circle with diameter $M_{BC}A$ which implies $S \in (APQ)$ as well. $\textbf{Claim 5:}$ $PXYT$ is cyclic. $\textbf{Proof:}$ $$\measuredangle TPY = \measuredangle MPY = \measuredangle MCY = \measuredangle MCX = \measuredangle TXY $$ $\textbf{Claim 6:}$ $STYZM$ is cyclic. $\textbf{Proof:}$ It suffices to prove that $STYM$ is cyclic and the conclusion follows by symmetry. $$\measuredangle MYT = \measuredangle MYP - \measuredangle PYT = 180- \measuredangle MCP $$$$- \measuredangle PXA = 180-\measuredangle (90-B/2) - \measuredangle (B/2) = 90^{\circ} = \measuredangle MST $$ $\textbf{Claim 7:}$ $BCYZ$ is cyclic. $\textbf{Proof:}$ $$\measuredangle YSW = \measuredangle YST = \measuredangle YMP = \measuredangle YCW$$so $CSWY$ is cyclic, and similarly $SQZB$ is cyclic, so $XC \cdot XY = XS \cdot XW =XB \cdot XZ$ $\textbf{Claim 8:}$ $AX$ is tangent to $(XYZ)$ $\textbf{Proof:}$ $$\measuredangle AXY = \measuredangle YCB = \measuredangle XZY$$ Now we note that $S,T$ are inverses WRT $(APQ)$ because $A,S,T$ are collinear and $S \in (APQ)$ so $$AS \cdot AT = AX^2$$so $A$ lies on radical axes of $(STYZM)$ and $(XYZ)$, that is $YZ$, hence we're done. $\blacksquare$

Very nice, just add the reflection $T$ of $X$ over $A,$ the reflection of $A$ over $T,$ the midpoint of $AT,$ the reflection of $B$ over $C,$ the intersections of the perpendicular bisector of $BC$ with lines $BQ$ and $CP,$ the foot from $A$ to $BC,$ the intouch point, the extouch point, the incenter, the orthocenter, the centroid, the circumcenter and the nagel point and now it is trivial (what I did in contest)

Similarly to others we can prove that lines BQ and CP meet at a point D at the circumcircle of XPQ. Here is another way to finish Let T be the intersection of PQ with AX. Let A' and X' be the reflections of A and X through D, respectively Let W be the reflection of X through A'. After a few computations we can get A'M perpendicular to BC. Thus DX'CB is a trapezoid. Note that XYQT and XZTP are cyclic Then angles CZT, XPT, XDQ and the supplementary of DX'C are equal. Hence TZCX' is cyclic Then XZ * XC = XT * XX' = XA * XW and thus angles XAZ and WCX are equal Likewise we obtain angles XAY and WBX are equal. Since XBCW is a trapezoid we get that angles WCX and WBX are equal, thus angles XAZ and XAY are equal. Conclusion follows Note: The computations I did to get A'M perpendicular to BC involve the hypothesis AC + AB = 2BC. Thus if the tangency point of BC with the incircle of ABC divides BC in two segments of lenghts 2b < 2c, then AB = 3b + c, AC = b + 3c, AA' = 2c - 2b, and from there we may compute AM and the altitude from A to BC

Any explanation to the title ?

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Let me also add one sol. without details) One can easily observe that $PB=BC=CQ$(by Menelaus th). Let $BQ \cap PC=T$, by cheva we have $X - A - T$ and simply we have $(XPQT)$ is cyclic. Next step is proving $(BYZC)$ is cyclic. Take the miquel point of $BMCTPQ$ and call that point $R$. By angle chasing we have $\angle XRT=\angle TRM=90$, so $X - R - M$. Then by PoP $(BYZC)$ is cyclic, then $AX$ is tangent to $(XYZ)$. Since $AX=AQ$, it is enough to prove that $\angle AQZ=\angle ZYQ$(because by proving this, we will prove that $A$ lies on radical axes of circles $(XYZ)$ and $(YZQ)$, which means $A - Y - Z$). Let $PM \cap XT=F$ and $PM \cap CX=G$, then $(PFZX)$ and $(YZQG)$ are cyclic, and rest follows from angle chasing, so we are done:).

CyclicISLscelesTrapezoid wrote: My problem! Let $\measuredangle$ denote directed angles modulo $180^\circ$. By Menelaus, we have \[\frac{BP}{PA} \cdot \frac{AQ}{QC} \cdot \frac{CM}{MB}=1,\]and since $BM=CM$ and $AP=AQ$, this gives $BP=CQ$. If we let $BP=CQ=x$ and $AP=AQ=y$, then we have $x-y=AB$ and $x+y=AC$, so $x=\tfrac{AB+AC}{2}=BC$. Let $T$ be the reflection of $X$ over $A$. Notice that $APT \sim BPC$ and $AQT \sim CQB$ by SAS, so $\overline{BQ}$ and $\overline{CP}$ intersect at $T$. Let the circumcircles of $BMQ$ and $CMP$ intersect again at $R$. Claim: $R$ lies on the circle centered at $A$ through $P$, $Q$, $X$, and $T$. Proof: We have \[\measuredangle PRQ=\measuredangle PRM+\measuredangle MRQ=\measuredangle PCM+\measuredangle MBQ=\measuredangle CTB=\measuredangle PTQ,\]so $PTQR$ is cyclic, as desired. Since $\measuredangle BMR=\measuredangle BQR=\measuredangle TQR=\measuredangle TXR$, we obtain that $X$, $R$, and $M$ are collinear. Claim: $XYRZ$ is cyclic and its circumcircle is tangent to $\overline{AX}$. Proof: We have \[\measuredangle XYR=\measuredangle BYR=\measuredangle BMR=\measuredangle AXR,\]so the circumcircle of $XRY$ is tangent to $\overline{AX}$. Analogously, the circumcircle of $XRZ$ is tangent to $\overline{AX}$, so the circle through $X$ and $R$ tangent to $\overline{AX}$ passes through $Y$ and $Z$, as desired. Let $\infty_{BC}$ be the point at infinity on $\overline{BC}$. We have $(X,R;Y,Z)\overset{X}{=}(\infty_{BC},M;B,C)=-1$, so $XRYZ$ is harmonic. Since $\overline{AX}$ is tangent to its circumcircle, we know that $\overline{AR}$ is tangent as well, so $A$, $Y$, and $Z$ are collinear by the symmedian configuration. ADMITS *throws water balloon* i didnt see anyone throw water balloons at MOP this year, can someone confirm that tiger was hit with a water balloon?

Define $K$ and $L$ such that $LB=LC$ and $KB=KC$, $A,K,L$ lie on the same side of $BC$, $\angle BLC=\angle ABC$ and $\angle BKC=\angle ACB$. Claim: $CQ=BC=BP$. Proof. By menelaus on $ABC$ and $P,Q,M$, \[1=\frac{BM}{MC}\cdot \frac{CQ}{QA}\cdot \frac{AP}{PB}=\frac{CQ}{PB}\]so $CQ=PB=y$ since $BM=MC$, $AP=AQ=x$. Now, $AB=BP-AP=y-x$ and $AC=y+x$ so $AB+AC=2y=2BC$ so $y=BC$. $\blacksquare$ Claim: $K$ lies on $XP, BQ$, and $(PMC)$. Proof. Since $\angle QBA=90^{\circ} - \angle \frac{\angle C}{2}$ from $CQ=CB$, $K$ lies on $BQ$ as $\angle KBC=90^{\circ} - \frac{\angle C}{2}$. Now, $\angle MQC=\frac{\angle A}{2}$ as $AP=AQ$. Thus $\angle CMP = \angle CMQ = 180^{\circ} - \frac{\angle A}{2}-\angle C$. As $\angle PCM=90^{\circ} - \frac{\angle B}{2}$, \[\angle MPC=180^{\circ} - \angle CMP - \angle PCM = \frac{\angle C}{2}\]But since $\angle MKC = \frac{\angle C}{2}$, $K$ lies on $(CMP)$. Now, as $A$ is the circumcenter of $XPQ$ and $\angle XAQ=180^{\circ} - \angle C$, we get $\angle XPQ=90^{\circ} - \frac{\angle C}{2}$. But \[\angle MPK=180^{\circ} - \angle KMC = 180 ^{\circ} - \left(90^{\circ} - \frac{\angle C}{2}\right) = 180 ^{\circ} - \angle XPQ,\]so $K$ lies on $XP$. $\blacksquare$ Similar results hold for $L$, so since \[\measuredangle QLK=\measuredangle QLM=\measuredangle QBM=\measuredangle KBC=\measuredangle MCK=\measuredangle MPK=\measuredangle QPK,\]$PKQL$ is cyclic, so by radical axis, $X$ lies on the radical axis of $(PCM)$, $(BMQ)$, $(PKQL)$. Now, invert at $X$ with radius $\sqrt{\operatorname{pow}_{(PKQL)}X}$ which fixes $(PCM)$, $(BMQ)$, $(PKQL)$. It sends $Y$ to $B$, $Z$ to $C$, and $(XPQ)$ to $KL$, the perpendicular bisector of $BC$. As $A$ is the circumcenter of $XPQ$, $A$ goes to $X'$, the reflection of $X$ over $KL$. Thus $AYZ$ inverts to $XBCX'$, an isosceles trapezoid, so inverting back we conclude.

Let $E=\overline{XQ}\cap (BMQ)$ , $F=\overline{XP}\cap (CMP)$ note that $\overline{PQ}\parallel \overline{AD}$ where $D$ is the intersection of $\angle BAC$ bisector and $BC$ from $\frac{CQ}{CA}=\frac{CM}{CD}$ we can get $CQ=BC=BP$ some angle chase shows that $M$, $E$, $F$ are collinear, $BYZC$ is cyclic and $\overline{BEF} \perp \overline{BC}$ Now let $R$ be the reflection of $X$ over $\overline {ME}$, abviously $XBCR$ is cyclic. $\angle QER=\angle XER=2\angle XEF=2\angle QBM=180-\angle QAX$ so $XA.XR=XY.XB$ To finish note that he inversion centered at $X$ with radius $\sqrt{XY.XB}$ sends $(XBCR)$ to a line $\overline{YZA}$ as needed.