making leo.euler proud

isn't necessary, so

My problem! [asy][asy] // ELMO 2024/1 size(8cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen orangefill,reddraw,orangedraw,greendraw,lightbluedraw; orangefill = RGB(255,221,187); reddraw = RGB(255,85,85); orangedraw = RGB(255,136,0); greendraw = RGB(0,187,0); lightbluedraw = RGB(85,187,255); pair A,B,X,C,D,E,P,Q,R,S,K,L; path c,d,e,f; A = (-1,0); B = (9,0); X = (0,13); C = (B+X)/2; D = (2*A+X)/3; E = extension(A,C,B,D); c = circumcircle(A,D,E); d = circumcircle(B,C,E); P = intersectionpoints(c,A--B)[1]; Q = intersectionpoints(d,A--B)[0]; e = circumcircle(A,C,P); f = circumcircle(B,D,Q); R = intersectionpoints(e,A--X)[0]; S = intersectionpoints(f,B--X)[0]; K = intersectionpoints(f,A--X)[0]; L = intersectionpoints(e,B--X)[1]; filldraw(A--B--C--D--cycle,orangefill,orangedraw); draw(c,greendraw); draw(d,greendraw); draw(e,lightbluedraw); draw(f,lightbluedraw); draw(circumcircle(C,D,E),reddraw); draw(circumcircle(A,B,R),reddraw+dotted); draw(X--C,orangedraw); draw(X--D,orangedraw); draw(A--C,orangedraw); draw(B--D,orangedraw); dot("$A$",A,dir(235)); dot("$B$",B,dir(305)); dot("$X$",X,dir(95)); dot("$C$",C,1.2*dir(60)); dot("$D$",D,1.4*dir(165)); dot("$E$",E,2.4*dir(100)); dot("$P$",P,dir(280)); dot("$Q$",Q,dir(240)); dot("$R$",R,dir(130)); dot("$S$",S,1.2*dir(75)); dot("$K$",K,dir(145)); dot("$L$",L,dir(60)); [/asy][/asy] Let the circumcircle of $BDQ$ intersect $\overline{AD}$ again at $K$, and let the circumcircle of $ACP$ intersect $\overline{BC}$ again at $L$. Since $AC \cdot AE=AB \cdot AQ=AD \cdot AK$, we know that $CEDK$ is cyclic. Analogously, $DECL$ is cyclic, so $CEDKL$ is cyclic. Assume that $\overline{AD}$ and $\overline{BC}$ intersect at $X$; otherwise, we can finish by continuity. Since \[XA \cdot XR=XC \cdot XL=XD \cdot XK=XB \cdot XS,\]we obtain that $ARSB$ is cyclic, as desired.

Perform an inversion about $E$ of any radius let us keep our original point names in our inverted problem statement. inverted problem statement wrote: Suppose $ABCD$ is a convex quadrilateral who's diagonals intersect at $E$. Let the intersection of $\overline{DA}$, with $(BEA)$ not equal to $B$ be $Q$. Let $(BQD)$ intersect $(BCE)$, at $S$. Let $(APC)$ intersect $(ADE)$ at $R$. Prove $A$, $B$, $S$, and $R$ are concyclic. Note that $$C \in \text{Radical axis of } (AED) \text{and} (BEA), (DBQ) \text{and} (BEA)$$$$D \in \text{Radical axis of } (APC) \text{and} (BEA), (BCE) \text{and} (BEA)$$Thus $C$ lies on the radical axis of $(ADE)$, and $(DBQ)$, and $D$ lies on the radical axis of $(APC)$, and $(BCE)$. Let $(APC) \cap (BCE)$ not equal to $C$ be $H$, and $(ADE) \cap (DBQ)$ not equal to $D$ be $F$. Note that $C$, $H$, $F$, and $D$ are collinear. Since $\overline{CD}$ is the radical axis of $(ADE)$, and $(BQD)$, and is also the radical axis of $(BCE)$, and $(APC)$. Define $\overline{BS}\cap \overline{CD}=X$, $X$ lies on the radical of $(ADE)$, and $(BDQ)$, $(BCE)$, and $(APC)$, and $(BCE)$, and $(BDQ)$. Thus $$\mathrm{Pow}_{(BCE)}(X)=\mathrm{Pow}_{(BQD)}(X)=\mathrm{Pow}_{(ADE)}(X)$$Therefore $XS \cdot XB= XR \cdot XA$, or $ABSR$ is cyclic, and we conclude $\square$ I will unsplash you tiger

Let $(ACRP)$ intersect $BC$ again at $R_1 \ne C$ and let $(BDSQ)$ intersect $AD$ at $S_1 \ne D$. Claim: $CR_1S_1D$ is cyclic. Proof. Note that $BR_1 \cdot BC = BA \cdot BP = BE \cdot BD$ follows by PoP on $(ACRP), (AEDP)$ so it thus follows that $CR_1ED$ is cyclic. By symmetry, the result follows. $\blacksquare$ Now, let $AD \cap BC = Z$. We now have that \[ ZB \cdot ZS = ZS_1 \cdot ZD = ZR_1 \cdot ZC = ZA \cdot ZR \]by PoP on $(BDSQ), (CR_1S_1D), (ACRP)$ which finishes.

Let $AD$ intersect $(BQD)$ again at $F$, $BC$ intersect $(APC)$ again at $G$, and $AD$ and $BC$ intersect at $X$. Claim. $D$, $F$, $E$, $G$, $C$ are concyclic. Proof. By Power of a Point in $(BFDQ)$ and $(EBQC)$, $$AF \cdot AD = AB \cdot AQ = AE \cdot AC$$so $D$, $F$, $E$, $C$ are concyclic. By Power of a Point in $(APED)$ and $(APGC)$, $$BE \cdot BD = BP \cdot BA = BG \cdot BC$$so $E$, $D$, $G$, $C$ are concyclic, giving the claim. $\blacksquare$ It follows that $DFGC$ is cyclic, so by Power of a Point, $$XF \cdot XD = XG \cdot XC$$However, $DFBS$ and $AGCR$ are cyclic, so $XF \cdot XD = XB \cdot XS$ and $XG \cdot XC = XA \cdot XR$, ultimately giving $$XB \cdot XS = XA \cdot XR$$and the desired condition follows by the Converse of Power of a Point. $\blacksquare$

Very nice, just invert $3$ times (what i did in contest)

Radical axes spam. $\textbf{Claim 1:}$ $(CED),(APCR),\overline{BC}$ concur at a point $F$. $\textbf{Proof:}$ Let $(CED) \cap BC = F'$ then by radical axes theorem on $(CEDF');(APCR);(APED)$ we get $APCF'$ is cyclic which implies $F \equiv F'$. $\blacksquare$. Similarly define $H = (CED) \cap (BQDS) \cap AD$. Now let $U = AD \cap BC$; then by radical axes on $(CEDHF),(APCFR),(BQDHS)$ we get $U$ lies on radical axes of $(APCFR)$ and $(BQDHS)$, let $(ABR) \cap (BQDHS) = S'$ then by radical axes on $(ABRS'),(APCFR),(BQDHS)$ we get $S' \in BC$ which implies $S=S'$ so we're done. $\blacksquare$

I would like an explanation to the title

Another nice problem. The title is also erm pretty interesting. The solution I submitted goes something like this. Let $(BQD)$ intersect $\overline{AD}$ again at $M$ and $(APC)$ intersect $\overline{BC}$ again at $N$. We now show the following. Claim : Points $C$ , $D$ , $M$ , $N$ and $E$ are concyclic. Proof : Simply note that, \[AD \cdot AM = AQ \cdot AB = AE\cdot AC\]So, $(DMCE)$ must be cyclic. Similarly we can show that $(DNCE)$ is also cyclic which proves the claim. Now, let $X = \overline{AD} \cap \overline{BC}$. It is easy to see that, \[XR \cdot XA = XC \cdot CN = XD \cdot XM = XS \cdot XB\]which implies that points $A$ , $B$ , $S$ and $R$ are indeed concyclic as desired. I'm not exactly sure if it's necessary but we separately show the case where $ \overline{AD} \cap \overline{BC}$ does not exist (i.e $\overline{AD} \parallel \overline{BC}$) separately. First, note that due to the given parallel lines we immediately have that $ANCR$ and $MNCD$ are isosceles trapezoids. Thus, $MN = DC$ and $AN=RC$. Further, \[\measuredangle AMN = \measuredangle ANC + \measuredangle CNM = \measuredangle NCR + \measuredangle DCN = \measuredangle DCR \]so $\triangle ANM \cong \triangle DCR$ and in particular, this implies that $AM=DR$. Again, due to the parallel lines $BMDS$ must be an isosceles trapezoid and so $BM=DS$. Also, \[\measuredangle BMA = \measuredangle BSD = \measuredangle RDS \]so $\triangle ABM \cong \triangle DRS$. This allows us to conclude that $AB=RS$ from which it follows that $ABRS$ is an isosceles trapezoid and hence cyclic, as desired.

Non POP sol outline Main claim: BDR is similar to ACS, which clearly finishes the problem Proof: Try to use SAS, by using spiral sim at C and D to convert ratios to remove R,S , followed by a short trig bash.

No spiral similarity solution??

until now

recommended by bjump, thanks! (: + diagram If we let the intersection of $(QBD)$ with $\overline{AD}$ be $F$ and similarly, label the intersection of $(PAC)$ with $\overline{BC}$ $G$, then we have \[AE\cdot AC = AB\cdot AQ = AF\cdot AD\]therefore, $D$, $F$, $E$, and $C$ are concyclic. Similarly, $D$, $E$, $G$, and $C$ are concyclic, so we can obtain that $DFEGC$ is cyclic. Now label the intersection of $\overline{AD}$ and $\overline{BC}$ $H$, so \[HF\cdot HD = HB\cdot HS\]and \[HG\cdot HC = HA\cdot HR.\]Because $HF\cdot HD = HG\cdot HC$, we have $HB\cdot HS=HA\cdot HR$, so $A$, $B$, $S$, and $R$ are cyclic, as desired. $\square$

$P$ is the center of spiral similarity from $ED$ to $CR$, so if these points are $p,e,d,c,r$ in complex numbers, $r=\frac{(c-p)(d-p)}{(e-p)}+p=\frac{ep-cp-dp+cd}{e-p}$. By symmetry, the function $f(z)=\frac{z(e-c-d)+cd}{e-z}$ sends $p,q$ to $r,s$ respectively. As $z$ moves on $\overline{AB}$, since $z\to e-z\to \frac{\text{constant}}{e-z}\to \text{constant}+\frac{\text{constant}}{e-z}=f(z)$ is a composition of a reflection, inversion+reflection, and translation, $f(z)$ moves on a circle or line. For $z=a$, if $R'=f(a)$, the spiral similarity at $A$ sending $DE$ to $CR'$ is a dilation, so $R'\in \overline{AD}$ and $\overline{CR'}\parallel \overline{BD}$. If $T=\overline{AD}\cap\overline{BC}$, then $\frac{TA}{TR'}=\frac{TA\cdot TB}{TC\cdot TD}$. Thus if $S'$ is defined symmetrically to $S$, since $R,R',S,S'$ are cyclic (none are the same because $f$ is injective), $$TA\cdot TR=TR\cdot TR'\cdot \frac{TA\cdot TB}{TC\cdot TD}=TS\cdot TS'\cdot \frac{TA\cdot TB}{TC\cdot TD}=TB\cdot TS,$$as desired.

Let $AD$ meet $BC$ at $F$, $(APC)$ meet $BC$ the second time at $M$ and $(BQD)$ meet $AD$ the second time at $N$. We need to show that $F$ lies on the radical axis of $(APC)$ and $(BQD)$. From PoP we have: $$BE\cdot BD=BP\cdot BA=BC\cdot BM$$ Hence $M$ lies on $(DEC)$ and similarly $N$ also lies on $(DEC)$. Therefore we have $FN\cdot FD=FM\cdot FC$ so $F$ lies on the desired radical axis. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.9890374450015588, xmax = 29.178181259936437, ymin = -7.132634314161692, ymax = 11.782627064724863; /* image dimensions */ pen qqzzff = rgb(0.,0.6,1.); pen ffwwqq = rgb(1.,0.4,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((7.97838255842413,-5.567651168482594)--(17.97838255842413,-5.567651168482594)--(15.97838255842413,0.4323488315174062)--(10.05323069776398,1.0798317405114535)--cycle, linewidth(0.8) + qqzzff); /* draw figures */ draw((7.97838255842413,-5.567651168482594)--(17.97838255842413,-5.567651168482594), linewidth(0.8) + qqzzff); draw((17.97838255842413,-5.567651168482594)--(15.97838255842413,0.4323488315174062), linewidth(0.8) + qqzzff); draw((15.97838255842413,0.4323488315174062)--(10.05323069776398,1.0798317405114535), linewidth(0.8) + qqzzff); draw((10.05323069776398,1.0798317405114535)--(7.97838255842413,-5.567651168482594), linewidth(0.8) + qqzzff); draw((7.97838255842413,-5.567651168482594)--(15.97838255842413,0.4323488315174062), linewidth(0.8)); draw((10.05323069776398,1.0798317405114535)--(17.97838255842413,-5.567651168482594), linewidth(0.8)); draw(circle((9.791757766574651,-2.4861037591506636), 3.57550892622229), linewidth(0.8)); draw(circle((16.278011170543703,-2.801108297776072), 3.2473100117142333), linewidth(0.8)); draw(circle((9.791757766574648,0.3478485539833815), 6.187201840252989), linewidth(0.8) + ffwwqq); draw((10.05323069776398,1.0798317405114535)--(11.665278282857873,6.244575399199787), linewidth(0.8)); draw(circle((16.278011170543696,0.4530982789995032), 6.256251814471539), linewidth(0.8) + ffwwqq); draw((14.025858612358784,6.289920669713442)--(15.97838255842413,0.4323488315174062), linewidth(0.8)); draw(circle((13.091386378694224,1.4477247580232377), 3.060348871273118), linewidth(0.8) + qqwuqq); draw((11.665278282857873,6.244575399199787)--(12.814096669251525,9.925206499035212), linewidth(0.8)); draw((12.814096669251525,9.925206499035212)--(14.025858612358784,6.289920669713442), linewidth(0.8)); /* dots and labels */ dot((7.97838255842413,-5.567651168482594),dotstyle); label("$A$", (7.38780015145917,-6.073379676944045), NE * labelscalefactor); dot((17.97838255842413,-5.567651168482594),dotstyle); label("$B$", (18.15328605624258,-6.159849443247526), NE * labelscalefactor); dot((15.97838255842413,0.4323488315174062),dotstyle); label("$C$", (15.018757027741387,0.7577318610309858), NE * labelscalefactor); dot((10.05323069776398,1.0798317405114535),dotstyle); label("$D$", (10.371007088929273,1.2981679004277447), NE * labelscalefactor); dot((13.257788337815448,-1.608096833939105),linewidth(4.pt) + dotstyle); label("$E$", (13.138039610640671,-1.0365157897662531), NE * labelscalefactor); dot((11.60513297472517,-5.567651168482593),linewidth(4.pt) + dotstyle); label("$P$", (11.62481870032975,-6.073379676944045), NE * labelscalefactor); dot((14.577639782663258,-5.567651168482593),linewidth(4.pt) + dotstyle); label("$Q$", (13.981119832099612,-6.030144793792304), NE * labelscalefactor); dot((11.665278282857873,6.244575399199787),linewidth(4.pt) + dotstyle); label("$R$", (11.192469868812344,6.57282364494011), NE * labelscalefactor); dot((14.025858612358784,6.289920669713442),linewidth(4.pt) + dotstyle); label("$S$", (14.197294247858316,6.61605852809185), NE * labelscalefactor); dot((14.791757766574648,3.9922232070658485),linewidth(4.pt) + dotstyle); label("$M$", (15.018757027741387,4.173287630018501), NE * labelscalefactor); dot((10.801917396793561,3.4785046561947133),linewidth(4.pt) + dotstyle); label("$N$", (10.284537322625791,3.6977039153493534), NE * labelscalefactor); dot((12.814096669251525,9.925206499035212),linewidth(4.pt) + dotstyle); label("$F$", (12.705690779123264,10.161318946534587), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $(APC)\cap BC=X\neq C$ and $(BQD)\cap AD=Y\neq D$. Claim: $BPXC$ is cyclic and analogously $AQCY$. Proof. Angle chase! We get \[\measuredangle PDB=\measuredangle PDE=\measuredangle PAE=\measuredangle PAC=\measuredangle PXC=\measuredangle PXB\]as desired. $\blacksquare$ Claim: $CXDY$ is cyclic. Proof. First, note \[\measuredangle CXD=\measuredangle BXD =\measuredangle BPD=\measuredangle APD=\measuredangle AED=\measuredangle CED\]so $CEDX$ is cyclic. Similarly, $CEDY$ is cyclic so $CXDY$ is cyclic. $\blacksquare$ Radical axis on $(CXDY), (BDY), (AXC)$ gives the radical axis of $(BDY)$ and $(AXC)$ passes through $BC \cap AD$. Then, radical axis on $(BYDS)$, $(AXCR)$, and $(ABS)$ gives that $(ABC)$ passes through $R$ because of the above concurrence.

Let $AD\cap BC=F$, $(BDQ)\cap AD=N$ and $(ACP)\cap BC=M$. Claim: $D$,$E$,$C$,$M$ and $N$ are cyclic. Since $BP\cdot BA=BE\cdot BD=BC\cdot BM$, the points $C$, $D$, $E$ and $M$ are concyclic. On the other hand, notice that $AQ\cdot AB=AE\cdot AC=AD\cdot AN$ implies the points $C$, $D$, $E$ and $N$ are concyclic. Thus, $D$,$E$,$C$,$M$ and $N$ are cyclic. Claim: $A$,$B$,$R$ and $S$ are cyclic. Since $XK\cdot XD=XL\cdot XC=XR\cdot XA=XS\cdot XB$, we get the desired concyclicity.

write solution before another olympiad speedrun

OronSH wrote: Very nice, just invert $3$ times (what i did in contest) Here's that solution, for completeness: First, let $DE$ meet $CR$ at $A'$. Then we see $P$ is the Miquel point of $\overline{ADR},\overline{ACE},\overline{A'DE},\overline{A'CR}$. Consider Miquel inversion $\Phi_1$ at $P$, swapping pairs $(A,A'),(C,D),(E,R)$. Since $B=DE\cap AP$, we have $B'=\Phi_1(B)=(ACPR)\cap A'P$. Next, since $Q=(BCE)\cap AP$, we have $Q'=\Phi_1(Q)=(B'DR)\cap A'P$. The conclusion, which may be rewritten as $\overline{BC},(ABR),(BDQ)$ concurring, becomes $(B'DP),(A'B'E),(B'CQ')$ sharing a second point. Now delete points $A,B,Q,S$ from the diagram. We now have points $A',B',C,D,E,P,Q',R$ satisfying \[\overline{A'B'PQ'},\overline{A'CR},\overline{A'DE},(A'CEP),(A'DPR),(B'DQ'R),(B'CPR).\] Next, consider inversion $\Phi_2$ at $B'$. Let this rename points such that $\Phi_2(A',C,D,E,P,Q',R,\infty)=D,E,G,H,A,B,C,F$, respectively. The problem becomes RELSMO 2024/2: Quote: In $\triangle ABC$ let $D,E$ be points on $AB,AC$ respectively. The circumcircle of $\triangle CDE$ intersects $AB$ at $F$. The circumcircle of $\triangle ACD$ intersects $BC$ at $G$. The circumcircle of $DFG$ intersects the circumcircle of $ADE$ at $H$. Prove that $AG,BE,DH$ concur. To solve this problem, consider inversion $\Phi_3$ at $D$. Let this keep points named the same, but with $\Phi_3(\infty)=D$. In our new problem, we have $\triangle CAF$ with points $B,D$ on $AF$, point $E$ on $CF$, and point $G$ on $AC$ such that $ACDE$ and $BCDG$ are cyclic, and we want to show $H=AE\cap FG$ lies on the radical axis of $(ADG)$ and $(BDE)$. To finish, the main idea is to consider $(CEG)$. Radical Axis theorem on $(BDE),(CEG),(BCDG)$ implies $AE$ is the radical axis of $(BDE)$ and $(CEG)$. Radical Axis theorem on $(ADG),(CEG),(ACDE)$ implies $FG$ is the radical axis of $(ADG)$ and $(CDG)$. Finally, Radical Axis theorem on $(ADG),(BDE),(CEG)$ implies that $AE\cap FG=H$ lies on the radical axis of $(ADG)$ and $(BDE)$, as desired.