The infamous generalization of IMO 2009 Problem 2 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=291269 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=372184 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=439734

I didn't go through all the links but i think it's a new solution . Although it is not beautiful compared with those extraordinary solutions , you will see that it's a nice use of Ptolemy theorem . Let $ M,N,P $ be the midpoints of $ BE,CD,DE $ respectively , $O$ the circumcentre of triangle $ ABC $ , and the circumcircle of triangle $ MNP $ intersect $ XY$ the second time at $Q$ , we want to show that $ OQ \bot XY $ or equivalently $ OD^2 - OE^2 = QD^2 - QE^2 $ or $ AE\cdot CE - AD\cdot BD = 2PQ\cdot DE $ ( $ D-P-Q-E $ ) . Since $ \Delta QMN \sim \Delta AED $ and $ BD = 2MP ~,~ CE = 2 NP $ , we have $ AE\cdot CE - AD\cdot BD = 2 (\frac{ DE}{MN})( MQ\cdot NP - NQ\cdot MP ) $ $ = 2 (\frac{ DE}{MN})( MN\cdot PQ ) = 2PQ\cdot DE $ .

Here's a direct solution using the converse of Pascal's: Let the midpoints of $XY, DE, BE, CD$ be $M_1, M_2, M_3, M_4$ respectively. Let $AM_1\cap\Omega=A_1, BM_1\cap\Omega=BM_1, CM_1\cap\Omega=C_1$. Let $C_1A_1\cap XY=C_2, B_1A_a\cap XY=B_2, CB_2\cap BC_2=A_2$. Note that since $B_2, C_2, M$ are collinear, the converse of Pascal's on $A_1C_1CA_2BB_1$ implies $A_2$ lies on $\Omega$. Thus: \[\angle BAC=\angle BA_2C=\pi-\angle C_2BC-\angle B_2CB=\angle BC_2E+\angle CB_2D-\pi\] Note that by the Butterfly Theorem wrt $M_1$, we have $DM_1=M_1B_2, EM_1=M_1C_2$. Thus, from $SAS$ similarity we see that $M_1M_3||C_2B, M_1M_2||B_2C$. Thus: \[\angle BAC=\angle BC_2E+\angle CB_2D-\pi=\angle M_3M_1E+\angle M_4M_1D-\pi=\angle M_4M_1M_3\] Also since $M_2M_3||AB, M_2M_4||AC$ we have: \[\angle BAC=\angle M_4M_2M_3\] So thus $\angle M_4M_2M_3=\angle M_4M_1M_3\implies M_1, M_2, M_3, M_4$ are concyclic as required.

Using the above notations, we kinda get $\angle M_4M_2M_3 = \angle A$ as a beginning observation. Now, noting that the proof of the above uses parallel midlines, we also want to find some midlines to use for $\angle M_4M_1M_3$. This can be done with Butterfly Theorem. So we draw every lines needed to spam butterfly's theorem. (This is the motivation for $A_1, B_1, C_1, B_2, C_2$) Now by parallel midlines, we want the angle between $CB_2$ and $BC_2$ to be $\angle A$, so we take the intersection and claim that it lies on the circumcircle, which is simply the converse of Pascal. yay

My solution Denote the midpoints of $DE,BE,CD,XY$ by $P,Q,R,S$. $O_1$ is the circumcenter of $\triangle PRQ$. $O_2$ reflects with $O$ about $O_1$. So we need to prove that $O_2P \perp DE$ Take $U,V$ be the midpoints of $O_2C$ and $O_2B$. We have $O_1U = O_1V = \frac{R}{2}$ and $O_1R = O_1Q$. On the other hand, $\angle QO_1R =2 \angle QPR = 2 \angle A = \angle VO_1U \implies \angle QO_1V = \angle RO_1U$. So, $\triangle O_1QV = \triangle O_1RU$. It follows that $QV = RU$. Then, $O_2E = O_2D$ so $O_2P \perp DE$. Then $(O_1)$ passes through $S$, it means that $PQRS$ is cyclic

Dear Mathlinkers http://jl.ayme.pagesperso-orange.fr/Docs/A%20generalization%20of%20IMO%202009.pdf Sincerely Jean-Louis

Assume that $X$ is closer to $AB$.Let the midpoints of $XY$, $BE$, $CD$, and $DE$ be $U$, $N$, $P$, $T$. Let $XY = l$. Let $X_1$ be a point on $l$ such that $\overline{XX_1} \equiv \overline{EY}$. Let $Y_1$ be a point on $l$ such that $CY_1$ and $BX_1$ concur on $\Omega$ at point $A_1$. $$(XY_1;X_1Y)\stackrel{A_1}{=}(XC;BY)\stackrel{A}{=}(XE;DY)\implies \frac{YY_1}{X_1Y_1}=\frac{XD}{DE}\implies \frac{EY_1+EY}{DX_1+DY_1}=\frac{DX_1+EY}{EY_1+DY_1} \implies \overline{EY_1} \equiv \overline{DX_1} $$Now we have that $U$ is the midpoint of $EX_1$ and $DY_1$, so we have that $UP\mid\mid CY_1$ and $UN\mid\mid BX_1 \implies $ $$\measuredangle{NUP}=\measuredangle{(NU,UP)}=\measuredangle{(BX_1,CY_1)}=\measuredangle{BA_1C}=\measuredangle{BAC}=\measuredangle{(BA,AC)}=\measuredangle{(NT,TP)}=\measuredangle{NTP} \ \ \ \blacksquare$$

Let the midpoint of $BE,CF,XY,DE$ is $B_0,C_0,M,N$, respectively. It's easy to see (by angle chasing) $\angle B_0MC_0 = \angle BAC$, hence we only need to prove $\angle B_0NC_0 = \angle BAC$. Toss to complex plane, and let $N^*,B_0^*,C_0^*$ be the reflection of the circumcenter w.r.t $N,B_0,C_0$ (to eliminate fractions). Obviously, $\angle B_0NC_0 = \angle B_0^*N^*C_0^*$. Now, let $A,B,C,X,Y$ denoted as $a,b,c,x,y$. Then, $d=\frac{ab(x+y)-xy(a+b)}{ab-xy},e=\frac{ac(x+y)-xy(a+c)}{ac-xy}$. Then, $B_0^*=b+e = b+\frac{ac(x+y)-xy(a+c)}{ac-xy}$ and $C_0^*=c+d=c+\frac{ab(x+y)-xy(a+b)}{ab-xy}$. Also, $N^* = x+y$. Now, we wish to show $\frac{b+e-x-y}{c+d-x-y} / \frac{b-a}{c-a}$ is real. After some calculation this equals to $\frac{(ab-xy)(c-a)}{(ac-xy)(b-a)}$, which could be easily vertified is a real number. We are done.

[asy][asy] unitsize(2.5inches); pair A=dir(115); pair B=dir(220); pair C=dir(-40); pair X=dir(172); pair Y=dir(45); pair D=extension(A,B,X,Y); pair E=extension(A,C,X,Y); pair DD=(C+D)/2; pair EE=(B+E)/2; pair M=(X+Y)/2; pair N=(D+E)/2; pair D1=2*M-D; pair E1=2*M-E; pair P=2*foot(0,C,D1)-C; draw(circumcircle(A,B,C)); draw(X--Y); draw(B--A--C); draw(C--D); draw(B--E); draw(EE--M--DD); draw(B--P,dotted); draw(C--P,dotted); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$D$",D,dir(150)); dot("$E$",E,dir(75)); dot("$D'$",DD,dir(250)); dot("$E'$",EE,dir(300)); dot("$M$",M,dir(90)); dot("$N$",N,dir(N)); dot("$D_1$",D1,1.6*dir(D1)); dot("$E_1$",E1,dir(E1)); dot("$P$",P,dir(P)); [/asy][/asy] Let the midpoint of $BE$ be $E'$ and the midpoint of $CD$ be $D'$. Let $N$ be the midpoint of $DE$ and $M$ the midpoint of $XY$. Note that $E'N\parallel BA$ and $D'N\parallel CA$, so it suffices to show that $\angle E'MD'=\angle BAC$. To this end, let $D_1$ and $E_1$ be on $XY$ such that $BE_1\parallel E'M$ and $CD_1\parallel D'M$. Note that this definition immediately implies that $M$ is the midpoint of $DD_1$ and $EE_1$. It suffices to show that $BE_1$ and $CD_1$ concur on $(ABC)$. This follows by a sort of extended butterfly theorem, but not the standard butterfly theorem (maybe it does, but I can't see it). This is why it's important to know the proofs of these kind of theorems, since the proof can easily be repeated here and solves the problem. Let $P=BE_1\cap CD_1$, and let $\gamma$ be the unique conic passing through $PAXBC$. Our goal is to show that $Y\in\gamma$, because the only conic passing through $AXBCY$ is the circle $(ABC)$. We will show this by Desargues involution theorem on line $XY$ and quadrilateral $ABCP$. The theorem implies that there is a unique involution on line $XY$ swapping $(D_1,D)$, $(E_1,E)$, and $(X,\gamma\cap XY\setminus\{X\})$. The first two pairs imply that the involution is reflection in $M$, so we have that $\gamma\cap XY=\{X,Y\}$, completing the proof.

Here's a solution using moving points.

[asy][asy] unitsize(100); pair A = dir(110), B = dir(210), C = dir(330), X = dir(160), Y = dir(30); pair D = extension(A, B, X, Y), E = extension(A, C, X, Y); pair K = (B+E)/2, L = (C+D)/2, M = (D+E)/2, N = (X+Y)/2; pair R = orthocenter(A, D, E), T = -A, S = (R+T)/2; draw(D--R--E^^K--S--L^^B--T--C, gray(0.5)); draw(S--N^^A--R, gray(0.5)); draw(R--T, gray(0.5)+dashed); draw(unitcircle); draw(A--B--C--cycle); draw(X--Y); draw(B--E^^C--D); draw(K--M--L); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$T$", T, dir(T)); dot("$D$", D, dir(dir(A-D)+dir(X-D))); dot("$E$", E, dir(dir(A-E)+dir(Y-E))); dot("$K$", K, rotate(90) * dir(E-B)); dot("$L$", L, rotate(90) * dir(C-D)); dot("$M$", M, dir(150)); dot("$N$", N, dir(N)); dot("$R$", R, dir(50)); dot("$S$", S, dir(220)); [/asy][/asy] Let $K$, $L$, $M$, $N$ be the midpoints of $\overline{BE}$, $\overline{CD}$, $\overline{DE}$, $\overline{XY}$, $R$ be the orthocenter of $\triangle ADE$, $T$ be the antipode of $A$ on $(ABC)$, and $S$ be the midpoint of $\overline{RT}$. Claim. $\angle SKM = \angle SLM = 90^\circ$. Proof. Since $\overline{BT}$ and $\overline{ER}$ are perpendicular to $\overline{AB}$, $\overline{KS}$ is perpendicular to $\overline{AB}$ as well. As $\overline{AB} \parallel \overline{MK}$, we obtain $\angle SKM = 90^\circ$. Likewise $\angle SLM = 90^\circ$. $\square$ Claim. $\angle SNM = 90^\circ$. Proof. Since $A$ and $T$ are antipodes on $(ABC)$, the midpoint of $\overline{AT}$ is equidistant from $X$ and $Y$. Now $\overline{AR} \perp \overline{XY}$, so the midpoint of $\overline{RT}$ is also equidistant from $X$ and $Y$, which means $\angle SNM = 90^\circ$. $\square$ It follows that $K$, $L$, $M$, $N$, $S$ are concyclic.

Apply complex numbers on $\Omega$ as the unit circle centered at $O=0$. Let the midpoints of $\overline{XY}$, $\overline{BE}$, $\overline{CD}$, $\overline{DE}$ be $Z$, $F$, $G$, $H$, respectively. BY chord intersection formula $d=\frac{xy(a+b)-ab(x+y)}{xy-ab}$ and $e=\frac{xy(a+c)-ac(x+y)}{xy-ac}$, $f=\frac{b+e}{2}$, $g=\frac{c+d}{2}$, $h=\frac{d+e}{2}$, $z=\frac{x+y}{2}$. Then $ZFGH$ is cyclic if and only if $(Z,H;F,G)=\frac{(z-f)(h-g)}{(h-f)(z-g)}=\frac{(c-x)(c-y)(xy-ab)^2}{(b-x)(b-y)(xy-ac)^2}\in \mathbb{R}$, which is true by conjugating (it took me 15 minutes at most to compute $(Z,H;F,G)$ on paper). We're done. $\square$

We use the method of moving points. Let $P$ be such that $PEDB$ is a parallelogram and $T$ be the point so that $AT$ and $AD$ are isotomic wrt $\angle XAY$. By homothety at $D$, the problem is equivalent to showing $PCTE$ concyclic. Observe that as $E$ moves, the perpendicular bisector of $ET$ is a moving line with $\deg 1$, the perpendicular bisector of $PT$ is a moving line with $\deg 1$, and the perpendicular bisector of $CT$ is a moving line with $\deg 2$, thus the desired concyclicity is a $\deg 4$ condition, and we need to check $5$ values of $E$. For $E\in\{X,Y,D\}$ the result is obvious. For $E=P_\infty$, $E=P=P_\infty$ so the result is again obvious. Finally, we check $E=AA\cap XY$, which will finish because of continuity in the case $AX=AY$. In this case, we need to show $A,T,P,E$ concyclic. Let line $BP$ intersect $(AXY)$ again at $K$. Clearly $DTKB$ is an isosceles trapezoid, so $TKPE$ is an isosceles trapezoid. Then, the result follows from the observation that \[\measuredangle KPE = \measuredangle KBA=\measuredangle KAE.\]

Fix $B$, $C$, $X$, and $Y$ in that order. Let $A$ vary around $\Omega$. Now, we see that $D$ and $E$ vary projectively about $XY$ with respect to $A$. The midpoints of $DE$, $BE$, and $CD$ do, too, so we need to show three points of $A$ on the circumcircle that satisfy that $BE$, $CD$, $XY$, and $DE$'s midpoints are concyclic. Note that as $A$ approaches infinitely close to $X$ (e.g. $A=X$), we have a homothety of $\Omega$ centered about $X$ with ratio $\frac12$ that gives us the circumcircle of the midpoints of $BE$, $CD$, $XY$, and $DE$. Similarly, as $A$ approaches infinitely close to $Y$ (e.g. $A=Y$), we have a homothety of $\Omega$ centered about $Y$ with ratio $\frac12$ that gives us the circumcircle of the midpoints of $BE$, $CD$, $XY$, and $DE$. Finally, with $A$ being the point on $XY$ such that the midpoints of $DE$ and $XY$ coincide, we see that this is true, too, because all triangles have circumcircles. As a result, our condition is true and we are done.

Ptolemy sinus lemma

Fix $XY, BC$ and $\Omega$ and move $A$ in $\Omega$ such that $\text{deg} A=2$ so now from this $\text{deg} D=1$ and $\text{deg} E=1$, now let $M,N,P,Q$ the midpoints of $XY, DE, BE, CD$ respectivily, note that $P$ moves on the B-midbase of $\triangle XBY$ and $Q$ moves on the C-midbase of $\triangle XCY$ so from here and the degrees of $D,E$ we have that $\text{deg} P=1$ and $\text{deg} Q=1$, now let $Q'$ a point in the C-midbase of $\triangle XCY$ such that $\angle PMQ'=\angle BAC$ now that $P \mapsto Q'$ is projective since its just rotation over a fixed point by a fixed angle so the degrees are preserved so $\text{deg} Q'=1$ hence $Q=Q'$ is a $1+1+1=3$ degree condition. Cases 1,2: $A=X,Y$ Both cases are trivial since they are just homothety with scale factor of $0.5$ Case 3: $A$ is such that $M=N$ By the midbases we have $\angle BAC=\angle PMQ$ but this means that $Q$ satisfies $Q'$ condition so $Q=Q'$ as desired. Hence $Q=Q'$ holds always so $MNPQ$ is cyclic as desired thus we are done .

Let $\measuredangle$ denote directed angles modulo $180^\circ$. Let $K$, $L$, and $M$ be the midpoints of segments $BE$, $CD$, and $DE$, respectively, and let the circumcircle $\omega$ of $KLM$ intersect $\overline{DE}$ again at $N$. Let a homothety of scale factor $2$ centered at $D$ and $E$ map $\omega$ to $\omega_D$ and $\omega_E$, respectively, and let $D'$ and $E'$ be the centers of $\omega_D$ and $\omega_E$. Claim: $ND'=NE'$. Proof: Notice that the reflections of $D$ and $E$ over $N$ lie on $\omega_D$ and $\omega_E$ respectively, so $N$ lies on the radical axis of $\omega_D$ and $\omega_E$. Since $\omega_D$ and $\omega_E$ have twice the radius of $\omega$, we have $ND'=NE'$ by symmetry. $\square$ Claim: $OD'=OE'$. Proof: We claim that $OBE' \cong OCD'$. Indeed, we have $OB=OC$, $BE'=CD'$, and, noticing that $\measuredangle BAC=\measuredangle KML$ because $\overline{MK} \parallel \overline{AB}$ and $\overline{ML} \parallel \overline{AC}$, \begin{align*} &\measuredangle OBE'=\measuredangle OBA+\measuredangle ABE'=\measuredangle BCA+90^\circ+\measuredangle MKO_1=\measuredangle BCA+\measuredangle MLK=\measuredangle BCA+\measuredangle CAB+\measuredangle MLK+\measuredangle KML \\ &=\measuredangle CBA+\measuredangle MKL=\measuredangle OCA+\measuredangle MLO_1=\measuredangle OCA+\measuredangle ECD'=OCD', \end{align*}so $OBE' \cong OCD'$ by SAS, as desired. $\square$ Also, notice that $D'$ and $E'$ are the reflections of $D$ and $E$ over the center of $\omega$, so $DED'E'$ is a parallelogram. Thus, the perpendicular bisector $\overline{ON}$ of $\overline{D'E'}$ is perpendicular to $\overline{DE}$, so $N$ is the midpoint of $\overline{XY}$, as desired. $\square$

Fix $X,B,C,Y$ and move $A$ on $\Omega$ where $\text{deg} A=2.$ Let $M,N,P,Q$ be the midpoints of $XY,DE,BE,CD,$ respectively. Note that $P,Q$ moves on the midline $B,C$ of $\triangle BXY, \triangle CXY,$ respectively. Note that $\angle BAC=\angle PNQ$ due to the homotheties at $D$ and $E.$ Thus, if we show $\angle BAC=PMQ$ we're done. Let $\angle BAC=\angle PMQ'.$ There is a map from $A\mapsto Q'$ because $P\mapsto Q'$ because it's just rotating about $M$ by $\angle BAC,$ which is fixed. Thus, we just need to check $Q'=Q$ for three cases and we're done. Case $1,2, A=X,Y$ Trivial because it's just a homothety of scale $0.5.$ Case $3, M=N.$ $\angle PMQ=\angle PNQ,$ because $M=N,$ duh. Thus, we're done.$\blacksquare$

Here's another pure synth (not even proj ;-. Haven't double checked this as I did this a while ago. Define $F$ as the reflection of $D$ over $P$. Define $M$ as the reflection of $D$ over the midpoint of $XY$. Note that $BDFE$ is a cyclic quadrilateral and that $F$ lies on the line $\ell$ through $B$ parallel to $XY$. It remains to show that $(EFMC)$ is cyclic by a homothety. Note that $DBMZ$ is cyclic and thus $EFMZ$ is an isoceles trapezoid and is thus cyclic. Let $Z = \ell \cap (ABC)$. It remains to show that quadrilateral $CEFZ$ is cyclic, which follows as $\measuredangle ECZ = \measuredangle ACZ = \measuredangle DBZ = \measuredangle EFZ$.