Let $(BQX)$ and $(CPX)$ intersect $AB,AC$ at $K,L$ respectively. Since $\angle KXE=\angle B=\angle KDE$ and $\angle DXL=\angle C=\angle DEL,$ we get $D,E,X,K,L$ are cyclic. Hence $AK.AB=AL.AC$ which gives that $A$ lies on the radical axis of $(BQXK),(CPXL)$ as desired.$\blacksquare$

Let $AB$ and $AC$ meet $(BQX)$ and $(CPX)$ again at $F$ and $G$. Claim: $XDEFG$ is cyclic We show that $F\in (XDE)$, $$\measuredangle DFX=\measuredangle BQX=\measuredangle DEX$$ By Reim's Theorem $BCFG$ is cyclic, so we are done by applying the Radical Axes Theorem on $(BCFG)$, $(BQXF)$, and $(CPXG)$.

India 1995.

Let $(BXC) \cap AB,AC$ at $Q',P'$.By it is known that, $BQ'P'C$ is cyclic.So,$\angle DEQ=180-\angle DQ'X=\angle PQE$ $\implies$ $BQ'XQ$ is cyclic and with the same way we can found $CP'XP$ is cyclic.Thus, radical axises of $(BQX),(CPX)$,and $BQ'P'C$ are concurrent.

When you do a four page coordinate bash :_(

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Let $K=\overline{AX}\cap\overline{BC}$. Then it suffices to show $KB\cdot KQ=KC\cdot KP$ which may be obtained by applying DDIT to $X$ and $BCED$, then projecting the resulting involution onto $\overline{BC}$ as $(\infty_{BC}K;BQ;CP)$.

Suppose that $AX$ intersects $BC, DE$ at $Z, T$ then $\dfrac{\overline{ZP}}{\overline{ZQ}} = \dfrac{\overline{TD}}{\overline{TE}} = \dfrac{\overline{ZB}}{\overline{ZC}}$. So $\overline{ZP} \cdot \overline{ZC} = \overline{ZQ} \cdot \overline{ZB}$ or $Z$ lies on radical axis of $(BQX)$ and $(CPX)$. Hence $Z \in XY$ or $A \in XY$

crazyeyemoody907 wrote: Let $K=\overline{AX}\cap\overline{BC}$. Then it suffices to show $KB\cdot KQ=KC\cdot KP$ which may be obtained by applying DDIT to $X$ and $BCED$, then projecting the resulting involution onto $\overline{BC}$ as $(\infty_{BC}K;BQ;CP)$.

Your solution is similar to me.(P.S. I AM WEAK) I didn't bring my cellphone to school ... My solution: (just for storage) : Let $T=AX \cap BC $,then we apply DDIT on quadrilateral $ADXE$ and $BC $ we have$ (BQ),(C,P),(T,\infty_{BC}) $are involution pairs. Then there exists a point $T'$ such that $TB \times TQ =TC\times TP =TT\times T \infty_{BC}$ So $T=T'$ and it lies on radical axis of $(BQX) and (CPX)$,so we are done.

This is not INMO 1995 problem, anyways this is easier than the P5. To prove: $\overline{A-X-Y}$ where $XY$ is radical axis. Let $AB\cap(BQX)=G$ and $AC \cap (CPX)=F$, we do this to use Reim's theorem to get $(BCFG)$ and then finish by radical axis on $(BCFG)$, $(BQXYG)$ and $(CPXBY)$ we prove that $\overline{A-X-Y}$. For that we need the following claims Claim I: $\odot(XDFGE)\implies \odot(BFGC)$ Proof: Reim's Theorem. Claim II: $\odot(XDFGE)$ Proof:\begin{align*} \measuredangle BQX=\measuredangle BGX&=\measuredangle DGX=\measuredangle DEX\\ \measuredangle CPX=\measuredangle CFX&=\measuredangle XFE=\measuredangle XDE \end{align*}

Rogestive8828 wrote: crazyeyemoody907 wrote: Let $K=\overline{AX}\cap\overline{BC}$. Then it suffices to show $KB\cdot KQ=KC\cdot KP$ which may be obtained by applying DDIT to $X$ and $BCED$, then projecting the resulting involution onto $\overline{BC}$ as $(\infty_{BC}K;BQ;CP)$.

Your solution is similar to me.(P.S. I AM WEAK) I didn't bring my cellphone to school ... My solution: (just for storage) : Let $T=AX \cap BC $,then we apply DDIT on quadrilateral $ADXE$ and $BC $ we have$ (BQ),(C,P),(T,\infty_{BC}) $are involution pairs. Then there exists a point $T'$ such that $TB \times TQ =TC\times TP =TT\times T \infty_{BC}$ So $T=T'$ and it lies on radical axis of $(BQX) and (CPX)$,so we are done. Your solution is similar to mine. Let $T=\overline{AX}\cap\overline{BC}$, then we apply DDIT from $A$ to $\{\overline{DE},\overline{DP},\overline{QE},\overline{QP}\}$ and project on $\overline{BC}$ to get that $(B,Q),(C,P),(T,\infty_{\overline{BC}})$ are pairs under an involution. Thus $TB\cdot TQ=TC\cdot TP$ and $\overline{AXT}$ is the radical axis of $(BQX)$ and $(CPX)$.

Let $(BQX) $ intersects $AB$again in $M$ and $ (CPX) $ intersects $AC$ at point $N$ We see that $$\angle{XMD}=\angle{BMX}=\angle{XQP}=\angle{XED}$$ So $DXEM$ is cyclic in the same way we get $DXEN$ is cyclic so $DEMN$ is cyclic By reim's theorem we get that $MNCB$ is cyclic and because $NC$ and $MB$ intersect at $A$ so $A$ lies on radical axis between $w_1$ and $ w_2$ So easily we get that $A,X,Y$ are collinear

Let $(BXQ)\cap AB=\{S\}, (CPX)\cap AC=\{L\}$ Claim: $DXELS$ is cyclic Proof:We will use directed angles mod $\pi$ $$\measuredangle XDE=\measuredangle XPQ=\measuredangle XPC=\measuredangle XLC=\measuredangle XLE$$and similary for $S$ and we are done. Now we have $AD\cdot AS=AE\cdot AL$ and since $\frac{AD}{AE}=\frac{AB}{AC}$ we have that $AS\cdot AB=AL\cdot AC$ so $A$ has the same power wrt $(BXQ)$ and $(CXP)$ meaning it lies on their radical axis i.e $XY$ so we are done.

Let $(BQX)\cap AB,(CPX)\cap AC$ be $T,Z$ respectively and $T\neq B,Z\neq C$. If $TZCB$ were cyclic, $YX$ would pass through $A$ because of the radical axis theorem for $(CPX),(BQX),(TZCB)$. Proving this ends the question. $$\angle{XQP}=\alpha,\angle{XPQ}=\theta\Longrightarrow \angle{DTX}=\angle{BEX}=\alpha, \angle{XDE}=\angle{XZE}=\theta $$$$\Longrightarrow\hspace{1mm}\text{T,X,E,Z,D cyclic}$$from $\angle{DTZ}=\angle{DEZ}=\angle{ACB}\Longrightarrow\hspace {1mm}\text{TZCB cyclic}$ $\blacksquare$

Trig bash because Im bad and cant do synthetic. Let $\{F\} = (BQX) \cap AB$ and $\{G\} = (CPX) \cap AC$. From LOS in $\triangle{BXF}$ we have $\frac{FX}{\sin(\angle ABX)} = \frac{BX}{\sin(\angle BFX)} = \frac{BX}{\sin(\angle DEX)}$. Also apply LOS in $\triangle{CEX}$ and divide the relations to get: \begin{align*} \frac{\sin(\angle XGF)}{\sin(\angle XFG)} &= \frac{FX}{GX}\\ &= \frac{\sin(\angle ABX)}{\sin(\angle ACX)}\cdot\frac{BX}{CX}\cdot\frac{\sin(\angle XDE)}{\sin(\angle XED)}\\ &= \frac{\sin(\angle ABX)}{\sin(\angle ACX)}\cdot\frac{BX}{CX}\cdot\frac{EX}{DX} \end{align*} From LOS in $\triangle{BXA}$ we get $\frac{BX}{\sin(\angle BAX)}=\frac{AX}{\sin(\angle ABX)}$. Do the same in $\triangle{AXC}$ and divide the relations to get $\frac{\sin(\angle ABX)}{\sin(\angle ACX)}\cdot\frac{BX}{CX} = \frac{\sin(\angle BAX)}{\sin(\angle CAX)}$. So we finally get: $$\frac{\sin(\angle XGF)}{\sin(\angle XFG)} = \frac{\sin(\angle BAX)}{\sin(\angle CAX)}\cdot\frac{EX}{DX}$$ Now apply LOS in $\triangle{ADX}$ and $\triangle{AEX}$ like we did before and divide the relations to get: $$\frac{\sin(\angle ADX)}{\sin(\angle AEX)} = \frac{\sin(\angle BAX)}{\sin(\angle CAX)}\cdot\frac{EX}{DX}$$ So we actually have: $$\frac{\sin(\angle XGF)}{\sin(\angle XFG)}=\frac{\sin(\angle ADX)}{\sin(\angle AEX)}=\frac{\sin(\angle BDX)}{\sin(\angle CEX)}$$ Now denote $\angle XDE = x$ and $\angle XED = y$. We have: \begin{align*} \angle BDX + \angle CEX &= 360^{\circ} - \angle ADX - \angle AEX\\ &= 180^{\circ}+(180^{\circ}-x-y) - B - C\\ &= 180^{\circ} - (360^{\circ} - (180^{\circ}-x-y) - (180^{\circ} - B) - (180^{\circ} - C))\\ &= 180^{\circ} - (360^{\circ} - \angle QXP - \angle FXQ - \angle EXP)\\ &= 180^{\circ} - \angle FXE\\ &= \angle XGF + \angle XFG \end{align*} So we have that $\angle XGF = \angle BDX$ and $\angle XFG = \angle XEC$ (well known lemma). Finally, we have that: $$\angle BFG = \angle BFX + \angle XFG = y + \angle XEC = y + 180^{\circ} - C - y = 180^{\circ} - C$$ So $BFGC$ is cyclic and hence $A$ lies on the radical axis of $(BQX)$ and $(CPX)$ as desired.

crazyeyemoody907 wrote: Let $K=\overline{AX}\cap\overline{BC}$. Then it suffices to show $KB\cdot KQ=KC\cdot KP$ which may be obtained by applying DDIT to $X$ and $BCED$, then projecting the resulting involution onto $\overline{BC}$ as $(\infty_{BC}K;BQ;CP)$.

What's DDIT?

alba_tross1867 wrote: crazyeyemoody907 wrote: Let $K=\overline{AX}\cap\overline{BC}$. Then it suffices to show $KB\cdot KQ=KC\cdot KP$ which may be obtained by applying DDIT to $X$ and $BCED$, then projecting the resulting involution onto $\overline{BC}$ as $(\infty_{BC}K;BQ;CP)$.

What's DDIT? Dual Desargues Involution Theorem.

Let $AB$ and $AC$ meet $(BQX)$ and $(CPX)$ again at $U$ and $V$. CLAIM : $DUXVE$ is cyclic. $\angle QBU = \angle UXE $ but $\angle QBU=\angle EDA$ so $\angle UXE= 180^{\circ} - \angle UDE$ so $UXED$ is cyclic. Analogous we have $DXVE$ cyclic so $DUXVE$ is cyclic. So the claim is proved. By PoP we have $AD \cdot AU=AE \cdot AV$ and how $\frac{AB}{AD}=\frac{AC}{AE}$ we have $AB \cdot AU=AC \cdot AV$ so A lies on the radical axis of $(BQXU)$ and $(CPXV)$. So A lies on $XY$.So the problem is proved.

Why did the asymptote take more time than solving the problem? Here's a short solution, same as others I think. Solution: Consider $U \coloneqq AB \cap \odot(XQB) \ne B$ and $V \coloneqq AC \cap \odot(XPC) \ne C$. Here's the central claim of the problem. [asy][asy] import olympiad; import geometry; size(9cm); // defaultpen(fontsize(10pt)); pair A = (-0.36,0.93); pair B = (-0.86,-0.51); pair C = (0.86,-0.51); pair D = (-0.59, 0.27); pair E = extension(D,D+(C-B),A,C); pair X = (-0.13, -0.12); pair P = extension(D,X,B,C); pair Q = extension(E,X,B,C); pair U = intersectionpoints(circumcircle(D,X,E),circumcircle(B,Q,X))[0]; pair V = intersectionpoints(circumcircle(D,X,E), circumcircle(X,P,C))[0]; pair Y = intersectionpoints(circumcircle(X,P,C), circumcircle(X,Q,B))[0]; draw(A--B--C--A, red); draw(A--X, magenta+dashed); draw(D--E, red); draw(D--P, red); draw(E--Q, red); draw(circumcircle(P,X,C), orange); draw(circumcircle(Q,X,B), orange); draw(circumcircle(D,X,E), deepgreen+dashed); draw(circumcircle(B,C,U), blue+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$X$", X, dir(270)*1.8); dot("$Y$", Y, dir(E)*1.8); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$U$", U, dir(110)); dot("$V$", V, dir(90)); [/asy][/asy] Claim: $DXEUV$ and $UVBC$ are cyclic quadrilaterals. Proof: By symmetry, it suffices to show that $V \in \odot(DXE)$. This is true since \[\measuredangle XDE = \measuredangle XPQ = \measuredangle XVC = \measuredangle XVE. \]For $UVBC$, observe \begin{align*} \measuredangle UVC &= \measuredangle UVX + \measuredangle XVC \\ &= \measuredangle BDX + \measuredangle QPX \\ &= \measuredangle BDP + \measuredangle BPD \\ &= \measuredangle DBP = \measuredangle UBC. \end{align*}This completes the proof. $\square$ Since the pairwise radical axis of $\odot(XPCV)$, $\odot(XQBU)$ and $\odot(UVBC)$ concur, it must follow that $A-X-Y$ are collinear. $\blacksquare$

USEMO derusting, 38 minute solve WLOG $AB<AC$. Let $(XBP)$ intersect $AB$ again at $F$, and $(XQC)$ intersect $AC$ again at $G$. Now $\angle XDE = \angle XQP = \angle XGE $ which means that $XEGD$ is cyclic. Similarly $XFED$ is cyclic. Therefore $XEGDF$ is cyclic. Now observe $$\angle BFG + \angle C = 180^\circ-\angle DFG+ \angle C = 180^\circ - \angle DEG +\angle C = 180^\circ - \angle C + \angle C = 180^\circ$$therefore $BFEC$ is cyclic and since $A$ is the radical center of $(PBFYX)$, $(QCGYX)$, and $(BCFG)$. Then $A$ lies on the radical axis of $(PBFYX)$ and $(QCGYX)$ therefore $A$, $X$, and $Y$ are collinear.

Let $(BQX)\cap AB=S$ and $(CPX)\cap AC=S$ Claim: Points $D$, $T$, $X$ and $E$ are concyclic Proof: Since $\angle XSB=\angle XQP=\angle XED$ which implies the points $D$, $E$, $X$ and $S$ are concyclic. Similarly $\angle XTC=\angle XPQ=\angle XDE$ gives the points $D$, $T$, $X$ and $E$ are concyclic. Thus, points $D$, $E$, $X$, $T$ and $S$ are concyclic. Claim: Points $S$, $T$, $B$ and $C$ are concyclic Proof: Observe that from the circumcirle $(DEXTS)$ and $DE\parallel BC$, we have $$\dfrac{AS}{AT}=\dfrac{AE}{AD}=\dfrac{AC}{AB}$$which means points $S$, $T$, $B$ and $C$ are concyclic (also can be shown via Reim's Theorem). Thus, point $A$ is on radical axis of $(BQX)$ and $(CPX)$ and $A\in XY$ as desired.

Let $\overline{AX} \cap \overline{BC}=T$. Apply DDIT from $X$ to $DECB$ and we get $(\overline{XD};\overline{XC})$; $(\overline{XE};\overline{XB})$; $(\overline{XA};\overline{X \infty})$ are pairs under some involution. Projecting this onto $\overline{BC}$ we get that $(P,C)$; $(Q,B)$; $(T,\infty)$ are pairs under some involution. This gives us that $T$ is the center of this involution (inversion) and so $TP \cdot TC=TQ \cdot TB$ and we are done. Remark: This is just the buffed down version of IMO $2019/2$.

Let $(BQX)$ and $(CPX)$ intersect $AB$ and $AC$ at $M;N$,respectively. Claim 1.:$DEXN$ is cyclic. Proof:$$\measuredangle BPD=\alpha\Rightarrow \measuredangle PDE=\measuredangle XPE=180-\alpha$$$$\measuredangle DPC=\measuredangle XPC=\measuredangle XNC=\measuredangle XNE=180-\alpha$$$$\measuredangle XPE=\measuredangle XNE$$. Do this for $\measuredangle CQE=\beta$ to get $DEXM$ is cyclic. $DEXN$ and $DEXM$ is cyclic$\Rightarrow DEXNM$ is cyclic. Claim 2:$BMNC$ is cyclic Proof:$\measuredangle BMN=\measuredangle BMX+\measuredangle XMN=\measuredangle BQX+\measuredangle XEN=\measuredangle CQX+\measuredangle XEC=\measuredangle CQE+\measuredangle QEC=-\measuredangle ECQ=\measuredangle QCE=\measuredangle BCN$ We got cyclic quadrilaterals $BMNC;BXYM;CXYN$, so their radical axises $BM;CN;XY$ concur at some point. But we know that $BM$ and $CN$ concur at $A$, so that means line $XY$ passes through $A$

Let $(BQX)$ meet $AB$ again at $F$ and $(CPX)$ meet $(AC)$ again at $G$. By Reim's theorem, $DXEF$ and $DXEG$ are cyclic. Thus, $DEGF$ is cyclic, so by Reim's theorem again, $BFGC$ is cyclic. Thus, $A$ must be the radical center of $(BQXF)$, $(CPXG)$, and $(BFGC)$, so it lies on $XY$.

Easy let the circumcircles meet $AB$ at $H$ and $AC$ at $F$ then $X$ $D$ $H$ and $F$ are concyclic (that follows directly from the parallelism and the $2$ concyclicities then the statement is equivalent to show that $AH$.$AB$=$AF$.$AC$ but we already have that $AH$.$AD$=$AF$.$AE$ but since $DE\parallel BC$ $$\dfrac{AD}{AE}=\dfrac{AB}{AC}$$and hence $A$ lies on the radical axcis of $(HXQB)$ and $(FXPC)$ and hence the conclusion $\blacksquare$

Did this one while ago but I am posting it now for storage. , Let $\odot(BQXY)=\omega_B, \odot(CPXY)=\omega_C$ , $\omega_B \cap AB=\{K\} , \omega_C \cap AC=\{L\}$ Claim: Points $K,D,L$ and $E$ are concyclic Proof: Let $\odot (XED)=\Gamma$ $\angle DEX \equiv DEQ \stackrel{DE \parallel BC}{=} \angle EQC \equiv XQC =180-\angle XQB \stackrel{\omega_b}{=}\angle BKX=180-\angle DKX \implies \angle DEX=180-\angle DKX \implies$ $\angle DEX+\angle DKX=180 \implies$ Points $X,K,E$ and $D$ are concyclic. $\implies K \in \Gamma$ Simmilarly: $\angle EDX \equiv EDB \stackrel{ED \parallel BC}{=}180-\angle DPC \equiv 180-\angle XPC \stackrel{\omega_C}{=} \angle XLC \equiv \angle XLE\equiv \angle ELX \implies$ $ \angle EDX=\angle ELX \implies$ Points $E,L,D$ and $X$ are concyclic. $\implies L \in \Gamma$ So since $K,L \in \Gamma \implies$ Points $K,D,L$ and $E$ are concyclic $\square$. Claim: Points $B,K,L$ and $C$ are concyclic Proof: $180-\angle BKL=\angle AKL \equiv \angle DXL \stackrel{\Gamma}{=} \angle DEL \equiv \angle DEA \equiv \angle AED \stackrel{DE \parallel BC}{=} \angle ACB \equiv \angle LCB \implies 180-\angle BKL=\angle LCB \implies$ $ \angle BKL+\angle LCB=180 \implies$ Points $B,K,L$ and $C$ are concyclic $\square$ Claim: Points $\overline{A-X-Y}$ are collinear Proof: By using Power of the Point Theorem we get: Points $B,K,L$ and $C$ are concyclic $\implies AK \cdot AB=AL \cdot AC \implies Pow(A,\omega_B)=Pow(A, \omega_C) \implies$ $A$ lies on the radical axis of $\omega_B$ and $\omega_C$ but $\omega_B \cap \omega_C=\{X,Y \} \implies$ Points $\overline{A-X-Y}$ are collinear $\blacksquare$