Sketch: ACE, BDF share same 9point circle by midlines and lengths (PoP combined with weird length condition), reduces to proving ACE, BDF, XYZ centroids collinear by homothety at the center of the common 9pc, easy finish with vectors/complex

Let $M_A$ be the midpoint of $CE$, $M_B$ the midpoint of $DF$, and so on. Note that $XY$ passes through $M_C$ and $M_F$ by trapezoid properties, and so on. The key claim is this: CLAIM: All $M_P$ lie on a circle. Proof: $M_EX \cdot M_BX = \frac{1}{4} CD \cdot AF = \frac{1}{4} AB \cdot DE = M_FX \cdot M_CX$. So $M_EM_FM_BM_C$ are cyclic, and so on. Since $X$, $Y$, and $Z$ aren't the same point (otherwise the orthocenter is undefined and the problem is dead anyways), we have that all $6$ midpoints are cyclic, as desired. $\blacksquare$ Now note that since $M_AZ = M_DY$, $YZ$ and $M_AM_D$ share a midpoint, and so $XYZ$ also shares the same circumcenter as the $6$ midpoints. Now, apply complex numbers to find that the centroids of $M_AM_CM_E$, $M_BM_DM_F$, and $XYZ$ are collinear. Taking a $\times 3$ homothety at their shared circumcenter, we find that their orthocenters are collinear as well, and we are done! edit: if the orthocenter is still defined just as $X = Y = Z$ then the problem still follows from homothety things so yay.

using the sketch from post #2:

Let $H$ be the orthocenter of $\triangle XYZ$, $M_1M_2M_3$ be the medial triangle of $EAC$, and $N_1N_2N_3$ be the medial triangle of $FBD$. By parallel sides $X$, $Y$, $M_3$ are collinear. We have $$\text{Pow}(M_3,(XYZ)) = M_3Y \cdot M_3X = \frac 14 AB \cdot DE,$$which is symmetric by the length condition. Thus by symmetry all of $M_1$, $M_2$, $M_3$, $N_1$, $N_2$, $N_3$ have equal power wrt ($XYZ$), so they lie on a circle concentric with $(XYZ)$. Note that $O_1$, the center of $(ACE)$, is the orthocenter of $\triangle M_1M_2M_3$. Similarly the center $O_2$ of $(BDF)$ is the orthocenter of $\triangle N_1N_2N_3$. Use complex numbers with the center of $(XYZ)$ as origin. Compute $$h = x+y+z = \frac 12 (a+d) + \frac 12 (b+e) + \frac 12 (c+f) = \frac 12 (a+b+c+d+e+f)$$$$o_1 = m_1+m_2+m_3 = \frac 12 (a+c) + \frac 12 (c+e) + \frac 12 (e+a) = a+c+e$$$$o_2 = n_1+n_2+n_3 = \frac 12 (b+d) + \frac 12 (d+f) + \frac 12 (f+b) = b+d+f.$$It follows that $h = \frac 12 (o_1+o_2)$, implying the conclusion.

mira74 wrote: The key claim is that $A+C+E-2O$ is the circumcenter of $ACE$. What does $A+C+E-2O$ mean?

sugar_rush wrote: What does $A+C+E-2O$ mean? If you choose any coordinate axes, if $A$ has coordinates $(x_A,y_A)$, and similar for the other points, then $A+C+E-2O$ has coordinates \[(x_A+x_C+x_E-2x_O,y_A+y_C+y_E-2y_O).\]Equivalently, you can choose an arbitrary origin, and treat every point as the vector from the origin to that point. Then, $A+C+E-2O$ is the point you get from adding the vector to $A$, the vector to $C$, and the vector to $E$, and subtracting two times the vector to $O$.

Construct parallelograms \(FABA'\), \(ABCB'\), \(BCDC'\), \(CDED'\), \(DEFE'\), \(EFAF'\). In general, observe that \(\overline{AF'}\parallel\overline{EF}\parallel\overline{BC}\parallel\overline{AB'}\) and (in directed lengths) \(B'F'=CB-EF=E'C'\), so \(\triangle D'B'F'\) and \(\triangle A'E'C'\) are homothetic and directly congruent. This implies they are translations of each other. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=red; pen sec2=lightred; pen tri=purple; pen qua=pink+fuchsia; pen fil=invisible; pen sfil=invisible; pen sfil2=invisible; pen tfil=invisible; pair O1,Dp,Bp,Fp,MD,MB,MF,A,EE,C,F,D,B,Ap,Ep,Cp,O2,MA,ME,MC,X,Y,Z; O1=origin; Dp=dir(120); Bp=dir(210); Fp=dir(330); MD=(Bp+Fp)/2; MB=(Fp+Dp)/2; MF=(Dp+Bp)/2; A=Fp+unit(Fp-Bp)*0.8; EE=intersectionpoint(Fp--(Fp+100*(Dp-Fp)),circle(O1,abs(A-O1))); C=intersectionpoint(Dp--(Dp+100*(Bp-Dp)),circle(O1,abs(A-O1))); F=A+EE-Fp; D=EE+C-Dp; B=C+A-Bp; Ap=B+F-A; Ep=F+D-EE; Cp=D+B-C; O2=circumcenter(Ap,Ep,Cp); MA=(Ep+Cp)/2; ME=(Cp+Ap)/2; MC=(Ap+Ep)/2; X=(A+D)/2; Y=(B+EE)/2; Z=(C+F)/2; filldraw(X--Y--Z--cycle,tfil,tri+dashed); filldraw(MD--MB--MF--cycle,tfil,tri); filldraw(MA--ME--MC--cycle,tfil,tri); filldraw(circumcircle(Dp,Bp,Fp),sfil2,sec2+dashed); filldraw(circumcircle(Ap,Ep,Cp),sfil2,sec2+dashed); draw(Fp--A,sec); draw(Ep--D,sec); filldraw(Dp--Bp--Fp--cycle,sfil,sec); filldraw(Ap--Ep--Cp--cycle,sfil,sec); filldraw(A--B--C--D--EE--F--cycle,fil,pri); dot("\(A\)",A,E); dot("\(B\)",B,SE); dot("\(C\)",C,SW); dot("\(D\)",D,W); dot("\(E\)",EE,NW); dot("\(F\)",F,NE); dot("\(D'\)",Dp,Dp); dot("\(B'\)",Bp,W); dot("\(F'\)",Fp,NE); dot("\(A'\)",Ap,N); dot("\(E'\)",Ep,SW); dot("\(C'\)",Cp,SE); dot("\(X\)",X,E); dot("\(Y\)",Y,SE); dot("\(Z\)",Z,W); [/asy][/asy] Claim: If \(AB\cdot DE=BC\cdot EF=CD\cdot FA\), then the circumcenters of \(\triangle D'B'F'\) and \(\triangle ACE\) coincide. Proof. Observe that \[\operatorname{Pow}(A,(D'B'F'))=AB'\cdot AF'=BC\cdot FE,\]which is fixed, so \(A\), \(C\), \(E\) have equal power with respect to \((D'B'F')\). \(\blacksquare\) Now let \(O_1\) and \(O_2\) be the circumcenters of \(\triangle D'B'F'\) and \(\triangle A'C'E'\). I contend that in general, the midpoint of \(\overline{O_1O_2}\) coincides with the orthocenter of \(\triangle XYZ\). Let \(M_DM_BM_F\) and \(M_AM_EM_C\) be the medial triangles of \(\triangle DBF\) and \(\triangle AEC\), so their orthocenters are \(O_1\) and \(O_2\). It is easy to check that \(X\), \(Y\), \(Z\) are the midpoints of \(\overline{M_DM_A}\), \(\overline{M_BM_E}\), \(\overline{M_FM_C}\). But we know \(\triangle M_DM_BM_F\) and \(\triangle M_AM_EM_C\) are translations of each other, and \(\triangle XYZ\) is their vector average, so we conclude the orthocenter of \(\triangle XYZ\) is the midpoint of \(\overline{O_1O_2}\). This completes the proof.

Let $A_1$ be the midpoint of $FB$, $B_1$ be the midpoint of $AC$, and so on. Once you get that $A_1B_1C_1D_1E_1F_1$ is cyclic (which has been proven above so I won't prove it again) and that $Z = A_1D_1\cap B_1E_1$, etc, you can actually directly finish with complex once you realize that the orthocenter of $XYZ$ is the midpoint of the orthocenters of $A_1C_1E_1$ and $B_1D_1F_1$. Indeed, set up $(A_1B_1C_1D_1E_1F_1)$ to be the unit circle, with values $a, b, c, d, e, f$; then by standard complex formulas you get that $z = \frac{ad(b+e)-be(a+d)}{ad-be}$, and similar for $x$ and $y$. Now, setting $h = \frac{a+b+c+d+e+f}{2}$ (which is the midpoint of the orthocenters of $ACE$ and $BDF$), you just need to check that $ZH\perp XY$. Since $XY$ lies on $C_1F_1$, this is equivalent to $$\frac{z-h}{\overline{z}-\overline{h}} = -\frac{c-f}{\overline{c}-\overline{f}} = cf$$After a not terrible amount of computation, this condition actually nicely reduces to $$\frac{(b+e)-(a+d)}{(\overline{b} + \overline{e}) - (\overline{a} + \overline{d})} = -cf$$which is equivalent to the vectors $(\vec{B} - \vec{A}) + (\vec{E} - \vec{D})$ being parallel to $\vec{F_1} - \vec{C_1}$ after you plug in $\vec{A_1} = \frac{\vec{F} + \vec{B}}{2}$, etc. However, this is obvious from the definitions of $F_1, C_1$. You can similarly check that $XH\perp YZ$ and $YH \perp ZX$, so you're done.

[asy][asy] size(10cm,0); defaultpen(fontsize(10pt)); pair X = dir(115); pair Y = dir(210); pair Z = dir(330); path w = circle((0,0),1.75); pair A1 = IP(L(Y,Z),w,0); pair B1 = IP(L(X,Z),w,0); pair C1 = IP(L(X,Y),w,0); pair D1 = Y+Z-A1; pair E1 = X+Z-B1; pair F1 = X+Y-C1; pair A = C1+E1-A1; pair C = A1+E1-C1; pair E = A1+C1-E1; pair B = D1+F1-B1; pair D = B1+F1-D1; pair F = B1+D1-F1; draw(A1--D1^^B1--E1^^C1--F1,linewidth(1)); draw(A--B--C--D--E--F--cycle); draw(A--C--E--cycle,blue+linewidth(1)); draw(B--D--F--cycle,deepgreen+linewidth(1)); draw(A--D^^B--E^^C--F,red+linewidth(1)); draw(w); dot("$A$",A,dir(0),blue); dot("$B$",B,dir(-45),darkgreen); dot("$C$",C,dir(225),blue); dot("$D$",D,dir(180),darkgreen); dot("$E$",E,dir(135),blue); dot("$F$",F,dir(45),darkgreen); dot("$A'$",A1,dir(A1),blue); dot("$B'$",B1,dir(B1),darkgreen); dot("$C'$",C1,dir(90),blue); dot("$D'$",D1,dir(D1),darkgreen); dot("$E'$",E1,dir(-60),blue); dot("$F'$",F1,dir(F1),darkgreen); dot("$X$",X,2.5*dir(95),red); dot("$Y$",Y,2*dir(225),red); dot("$Z$",Z,2*dir(30),red); dot("$O$",(0,0),dir(90)); [/asy][/asy] One of the best hard geo I have ever seen! First, we define some notations. Let $A'$, $C'$, $E'$ denote the midpoints of $CE$, $EA$, $AC$, and let $B'$, $D'$, $F'$ denote the midpoints of $DF$, $FB$, $BD$. Let $O$ be the circumcenter of $\triangle XYZ$. First, we collect some immediate observations. $A'$, $D'$, $Y$, $Z$ all lie on the mid-segment of $BC$ and $EF$. Similarly, sets $\{B',E',X,Z\}$ and $\{C',F',X,Y\}$ are colinear. $A'Z=D'Y=0.5EF$, so $OA'=OD'$. Similarly, $OB'=OE'$ and $OC'=OF'$. In fact, we will incorporate the length conditions and prove the following. Claim: $A'$, $B'$, $C'$, $D'$, $E'$, $F'$ are all concyclic with center $O$. Proof: Using the length conditions, we get $$XB'\cdot XE' = \frac{AF}{2}\cdot\frac{CD}{2} = \frac{DE}{2}\cdot\frac{AB}{2} = XC'\cdot XF',$$so $B'$, $E'$, $C'$, $F'$ are concyclic. However, from the observation above, the center must be $O$, so by repeating this with other two sides, we get the claim. $\blacksquare$ It suffices to show that the orthocenters of $\triangle A'C'E'$, $\triangle B'D'F'$, $\triangle XYZ$ are colinear, but these three triangles have the same circumcenter $O$. Thus, it suffices to show that the three centroids are colinear (by Euler line). The easiest way is to use vectors. We have \begin{align*} A'+D' &= Y+Z, \\ B'+E' &= X+Z, \\ C'+F' &= X+Y, \end{align*}so summing up gives $$\frac{A'+C'+E'}{3} + \frac{B'+D'+F'}{3} = 2\cdot\frac{X+Y+Z}{3},$$done.

USAMO 2021/6 wrote: Let $ABCDEF$ be a convex hexagon satisfying $\overline{AB} \parallel \overline{DE}$, $\overline{BC} \parallel \overline{EF}$, $\overline{CD} \parallel \overline{FA}$, and \[ AB \cdot DE = BC \cdot EF = CD \cdot FA. \]Let $X$, $Y$, and $Z$ be the midpoints of $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$. Prove that the circumcenter of $\triangle ACE$, the circumcenter of $\triangle BDF$, and the orthocenter of $\triangle XYZ$ are collinear. What a great and rich problem! A pinnacle of olympiad geometry . Took me ~5 hours to find the following solution, but the amount of satisfaction is immeasurable. Let us first define $M_{XY}$ to be the midpoint of segment $XY$. Claim 01. $M_{AC}, M_{BD}, M_{CE}, M_{DF}, M_{EA}, M_{FB}$ are concyclic, and we call this circle $\Gamma$. Proof. We will divide this proof into two parts: Part 01. $M_{BF}, Z, M_{CE}$ and $M_{DF}, Z, M_{AC}$ are collinear. We will first prove that $M_{BF}, Z, M_{CE}$ are collinear. Note that since $M_{BF}$ and $Z$ are the midpoints of segments $BF$ and $CF$. Then $M_{BF} Z \parallel BC$. Similarly, we get $M_{BF} Z \parallel BC \parallel EF \parallel M_{EC}Z$. This is enough to prove that $M_{BF}, Z, M_{CE}$ are collinear. Similarly, we could prove that $M_{DF}, Z, M_{AC}$ are collinear. Part 02. $M_{BF}, M_{CE}, M_{DF}, M_{AC}$ are concyclic. Note that the length condition translates to \[ \frac{AB}{BC} = \frac{FE}{ED} \]Since $AB \parallel DE, BC \parallel EF$, we have $\measuredangle ABC = \measuredangle DEF$. These two conditions force $\triangle ABC \sim \triangle FED$. Similarly $\triangle BCD \sim \triangle AFE$ and $\triangle CDE \sim \triangle BAF$. Now, taking homothety $\mathcal{H}(F,2)$ maps $\triangle M_{BF} M_{DF} Z \mapsto \triangle BDC$ and $\mathcal{H}(C,2)$ maps $\triangle M_{AC} M_{CE} Z \mapsto \triangle AEF$. Furthermore, from the previous part, we get that $M_{BF} M_{CE} \cap M_{DF} M_{AC} = Z$, therefore \[ \measuredangle M_{BF} M_{DF} M_{AC} \equiv \measuredangle M_{BF} M_{DF} Z = \measuredangle BDC = \measuredangle FEA = \measuredangle ZM_{EC} M_{AC} \equiv \measuredangle M_{BF} M_{EC} M_{AC} \]implying the desired concyclic condition. Claim 02. The circumcircle of $\triangle XYZ$ and $\Gamma$ are concentric. Proof. To do this, we'll prove that $M_{AC} X = M_{DF} Y$. Notice that the homothety $\mathcal{H}(C,2)$ proves $M_{AC} X = \frac{1}{2} FA$ and the homothety $\mathcal{H}(D,2)$ proves $M_{DF} Y = \frac{1}{2} FA$, which implies the desired equality. Now, this implies that the midpoint of $XY$ is the same as midpoint of $M_{AC}M_{DF}$. Furthermore, since $M_{AC} M_{DF}, M_{BD} M_{AE}$ lies on $\Gamma$, then their perpendicular bisector intersects at $O$, center of $\Gamma$, which by definition must be also circumcenter of $\triangle XYZ$ as well. Now, we are ready to simplify the problem. Notice that $W := N_9(\triangle ACE), N_9(\triangle BDF), O(\triangle XYZ)$ are the same point. To prove that $O(\triangle ACE), O(\triangle BDF), H(\triangle XYZ)$ are collinear. We can take homothety $\mathcal{H} \left( W, \frac{1}{3} \right)$ and by a famous theorem on Euler line states that $H(\triangle), N_9(\triangle), G(\triangle), O(\triangle)$ lies on the Euler line of the triangle with the ratio $HN_9 : N_9 G : GO = 3 : 1 : 2$. Therefore, it suffices to prove that \[ G(\triangle ACE), G(\triangle BDF), G(\triangle XYZ) \]are collinear. We'll now finish the problem by proving a more general statement: Claim 03. Let $A,B,C,D,E,F$ be six points in plane, no three of which are collinear. Let $X,Y,Z$ be the midpoints of $AD, BE, CF$. Then the centroid of $\triangle ABC, \triangle DEF$ and $\triangle XYZ$ are collinear. Proof. Note that we have $X = \frac{1}{2} (A + D), Y = \frac{1}{2} (B + E), Z = \frac{1}{2} (C + F)$. Therefore, we get \[ \frac{2}{3} (X + Y + Z) = \frac{1}{3} (A + C + E) + \frac{1}{3} (B + D + F) \]Therefore, $G(\triangle XYZ)$ is the midpoint of segment formed by $G(\triangle ACE)$ and $G(\triangle BDF)$, and we are done.

Darn sniped by IndoMath Solved with some help from DrMath: Let $A_1,B_1,C_1,D_1,E_1,F_1$ be the midpoints of $BF, AC, DB, EC, FD, AE,$ respectively; obviously common midlines imply that $X\in B_1E_1, C_1F_1, Y\in A_1D_1, C_1F_1,$ and $Z\in A_1D_1, B_1E_1.$ From the length and parallelity conditions, we get $\tfrac{FA}{BC}=\tfrac{EF}{CD}$ and $\measuredangle EFA=\measuredangle BCD,$ so by SAS similarity $\triangle AFE\sim\triangle BCD,$ hence $\measuredangle DBC=\measuredangle FAE.$ Since $A_1E_1\parallel BD$ and $A_1D_1\parallel BC,$ this implies that $\measuredangle E_1A_1D_1=\measuredangle DBC,$ and by similar logic we have $\measuredangle E_1B_1D_1=\measuredangle FAE,$ so $A_1B_1D_1E_1$ is cyclic and symmetrically $C_1,F_1\in (A_1B_1D_1E_1)$ as well. We can further observe that by homothety $E_1X=\tfrac{1}{2}AF=B_1Y,$ so $B_1E_1$ and $XY$ share a common perpendicular bisector, which implies that $(XYZ)$ and $(A_1B_1C_1D_1E_1F_1)$ are concentric. Now let $O_1,O_2,O_3$ be the respective orthocenters of $\triangle B_1D_1F_1, \triangle XYZ, \triangle A_1C_1E_1;$ then setting the origin as the common center of $(XYZ)$ and $(A_1B_1C_1D_1E_1F_1),$ we have \[\vec{O_1}=\vec{B_1}+\vec{D_1}+\vec{F_1}=\vec{A}+\vec{C}+\vec{E}\]\[\vec{O_2}=\vec{X}+\vec{Y}+\vec{Z}=\frac{1}{2}\left(\vec{A}+\vec{B}+\vec{C}+\vec{D}+\vec{E}+\vec{F}\right)\]\[\vec{O_3}=\vec{A_1}+\vec{C_1}+\vec{E_1}=\vec{B}+\vec{D}+\vec{F}.\]Clearly $O_2$ must then be the midpoint of $O_1O_3,$ but $O_1,O_3$ are the circumcenters of $\triangle ACE,\triangle BDF,$ respectively, so we're done. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -12.854144877441623, xmax = 13.775611166043257, ymin = -10.116364244097822, ymax = 7.234731157057853; 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Here is an almost synthetic (except for the trig step and the area argument) solution. By the given conditions, it is easy to see that $\triangle ABC\stackrel{-}{\sim} \triangle FED$ and similarly for others. Let $O_1,R_1$ and $O_2,R_2$ be the circumcenter and circumradius of $\triangle ACE$ and $\triangle BDF$. Let $A_1=\overline{BE}\cap \overline{CF}$, and define $B_1$ and $C_1$ similarly. Claim 1: $R_1=R_2$. Proof. By the similarities, we have $\frac{AE}{BD}=\frac{EF}{CD}$ and similarly for others. Cyclicly multiplying all of them shows that \[\frac{AE}{BD}\cdot\frac{EC}{FB}\cdot\frac{CA}{DF}=1\Longrightarrow AC\cdot CE\cdot EA=BD\cdot DF\cdot FB.\]Since the area of a triangle with side lengths $a,b,c$ and circumradius $R$ is $\frac{abc}{4R}$, to show the claim it suffices to show that $\triangle ACE$ and $\triangle BDF$ have the same area. This follows from the parallelisms. \[[ACE]=[B_1AC]+[A_1CE]+[C_1EA]+[A_1B_1C_1]=[B_1DF]+[A_1FB]+[C_1BD]+[A_1B_1C_1]=[BDF]. \blacksquare\][asy][asy] defaultpen(fontsize(10pt)); size(12cm); real s =1.8; pair X = dir(30); pair Y = dir(250); pair Z = dir(150); path w = circle((0,0),s); pair M = IP(L(Y,Z),w,0); pair N = IP(L(X,Y),w,0); pair P = IP(L(X,Z),w,1); pair U = X+Z-P; pair V = Z+Y-M; pair W = Y+X-N; pair A = P+N-M; pair F = V+W-U; pair E = P+M-N; pair D = U+W-V; pair C = M+N-P; pair B = U+V-W; pair O1 = circumcenter(A,C,E); pair O2 = circumcenter(B,D,F); pair H = midpoint(O1--O2); pair A1 = extension(B,E,C,F); pair B1 = extension(A,D,C,F); pair C1 = extension(A,D,B,E); pair M1 = extension(A,D,O1,foot(O1,B,C)); pair N1 = extension(A,D,O2,foot(O2,B,C)); draw(A--C--E--cycle, blue+1); draw(B--D--F--cycle,red+1); draw(circumcircle(A,C,E), blue); draw(circumcircle(B,D,F), red); draw(A--B--C--D--E--F--cycle, black+1); draw(A--D); draw(B--E); draw(C--F); draw(O1--A,blue); draw(O2--D,red); draw(O1--foot(O1,B,C), dotted); draw(O2--foot(O2,B,C),dotted); draw(H--foot(H,B,C),dotted); draw(O1--O2); draw(anglemark(M1,O1,A,5)); draw(anglemark(D,O2,N1,5)); draw(rightanglemark(B,foot(O1,B,C),O1,4)); draw(rightanglemark(B,foot(O2,B,C),O2,4)); draw(X--Y--Z--cycle); dot("$X$",X); dot("$Y$",Y,dir(270)); dot("$Z$",Z,dir(180)); dot("$A_1$",A1,dir(120)); dot("$B_1$",B1,dir(30)); dot("$C_1$",C1); 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("$F$",F,dir(F)); dot("$O_1$",O1,dir(270), blue); dot("$O_2$",O2,dir(180), red); dot("$H$",H,dir(180)); dot("$M$",M1,dir(350)); dot("$N$",N1,dir(60)); [/asy][/asy] Claim 2: $\angle (O_1A,BC)=-\angle (O_2D,BC)$. Proof. This is just angle chasing. \begin{align*} &&\angle(O_1A,BC) &=\angle(O_2D,BC)\\ &\Longleftrightarrow &\measuredangle O_1AB+\measuredangle ABC &=\measuredangle BCD+\measuredangle CDO_2\\ &\Longleftrightarrow &\measuredangle O_1AC+\measuredangle CAB+\measuredangle ABC &=\measuredangle BCD+\measuredangle CDB+\measuredangle BDO_2\\ &\Longleftrightarrow & 90^\circ-\measuredangle CEA+\measuredangle EFD+\measuredangle DEF &= \measuredangle EFA+\measuredangle AEF+90^\circ-\measuredangle DFB\\ &\Longleftrightarrow &\measuredangle EFD+\measuredangle AEF+\measuredangle DEC &= \measuredangle EFD+\measuredangle AEF+\measuredangle BFA \end{align*}which is true, since $\triangle ABF\stackrel{-}{\sim}\triangle DCE$. $\blacksquare$ Let $H$ be the midpoint of segment $O_1O_2$ ; we claim that $H$ is the orthocenter of $\triangle XYZ$. Since $YZ\parallel EF\parallel BC$, it suffices to show that $HX\perp BC$. Let $M$ and $N$ be points on $\overline{AD}$, such that $\overline{O_1M}$ and $\overline{O_2N}$ are perpendicular to $BC$. Then by claim 2, we have \[\measuredangle MO_1A=90^\circ-\angle (O_1A,BC)=90^\circ-\angle(BC,O_2D)=\measuredangle DO_2N.\]Therefore, by the law of sines, \[AM=\frac{O_1A}{\sin\angle O_1MA}\cdot\sin\angle MO_1A=\frac{O_2D}{\sin\angle O_2ND}\cdot\sin\angle NO_2D = DN.\]Therefore, $X$ is also the midpoint of segment $MN$. Consequently, $\overline{HX}$ is parallel to $\overline{O_1M}$ and hence perpendicular to $\overline{BC}$ as desired.

mathcool2009 wrote:

sir quick question are you franzliszt

I do not control the account franzliszt, nor am I the historical virtuoso pianist Franz Liszt

mathcool2009 wrote: I do not control the account franzliszt, nor am I the historical virtuoso pianist Franz Liszt your post would like to disagree

Billybillybobjoejr. wrote: mathcool2009 wrote:

sir quick question are you franzliszt If it were franzlist using bary, your computer would be dead pressing expand.

Danggggg respect bro I don't think I would've been able to do that

An extremely beautiful problem！Thank you to the proposer of this wonderful problem.

here's my funny in-test solution way overdue

8charlimit

As the author of this problem, I am literally asking for thanks (though some people use the term "upvotes"). The above image suggests I give "a few thoughts" about how I made the problem, so here they are.

Solved with rama1728. Very nice! Define point $A'$ so that $ABA'F$ is a parallelogram. Define $B', C', D', E', F'$ similarly. Note that $A'C' \parallel FA \parallel D'F',$ and $A'C' = FA-DC = D'F'.$ Similarly, $C'E' \parallel F'B',$ $E'A' \parallel B'D',$ $C'E' = F'B',$ $E'A'=B'D'.$ By the given length condition, $BA' \cdot BC' = DC' \cdot DE' = FE' \cdot FA',$ thus $B,D,F$ all have equal power wrt $(A'C'E').$ So the center of $(BDF)$ is the center of $(A'C'E'),$ call this $O_1.$ Similarly, the center of $(ACE)$ is the center of $(B'D'F'),$ call this $O_2.$ The orthocenter of the medial triangle of a triangle is its circumcenter. So $O_1, O_2$ are the orthocenters of the medial triangles of $A'C'E'$ and $B'D'F'.$ It's not hard to see $AF'DE',$ $AB'DC'$ are parallelograms, so $X$ is the midpoint of both $B'C'$ and $F'E'.$ So, we can see that the midpoint of the midpoints of $B'F'$ and $E'C'$ is $X$ (since $B'E'C'F'$ is a parallelogram). Using the same argument on $Y,Z$ as well yields that the midpoint of the orthocenters of $A'C'E'$ and $B'D'F',$ which is the midpoint of $O_1O_2,$ is the orthocenter of $XYZ$ as desired. $\blacksquare$

Let $M_1$, $M_2$, and $M_3$ be the midpoints of $CE$, $AE$, $AC$ and $N_1$, $N_2$, and $N_3$ be the midpoints of $DF$, $BF$, and $BD$. Also, let $H$ be the orthocenter of $XYZ$. Note that we can use parallel sides to see that $X$, $Z$, and $M_3$ are collinear. Thus we have \[ \text{Pow}(M_3,(XYZ)) = M_3Z \cdot M_3X = \frac 14 AB \cdot DE \]by midlines. Applying this argument cyclically, and noting the condition $AB \cdot DE = BC \cdot EF = CD \cdot FA$, $M_1$, $M_2$, $M_3$, $N_1$, $N_2$, $N_3$ all lie on a circle concentric with $(XYZ)$. Next, realize that basic orthocenter properties imply that the circumcenter $O_1$ of $(ACE)$ is the orthocenter of $\triangle M_1M_2M_3$, and likewise the circumcenter $O_2$ of $(BDF)$ is the orthocenter of $\triangle N_1N_2N_3$. The rest is just complex numbers; toss on the complex plane so that the circumcenter of $\triangle XYZ$ is the origin. Then we have \[ o_1 = m_1+m_2+m_3 = (c+e)/2+(a+e)/2+(a+c)/2=a+c+e \]\[ o_2 = n_1+n_2+n_3 = (b+d)/2+(d+f)/2+(b+f)/2=b+d+f \]\[ h = x+y+z = (a+d)/2+(b+e)/2+(c+f)/2=(a+b+c+d+e+f)/2.\]Note that from the above we have $h=\frac{o_1+o_2}{2}$, so $H$ is the midpoint of segment $O_1O_2$. In particular, $H$, $O_1$, and $O_2$ are collinear, as required.

Well I was hinted (specifically, the condition doesn't mean similar triangles! instead, it means PoP). If you try the similar triangles approach, you quickly run into a roadblock, since the problem is not really projective-based. However you can still use Pascal's and DDIT a few times to get something which still feels extremely clean. If $AF\cap BC=U$, $EF\cap DC=P$, $AE\cap BD=K$, then there exists an involution swapping $(KF,KC),(KA,KB),(KU,KP)$. If we define points $L$ and $M$ similar to $K$, then the involution also swaps $(KM,KL)$. I think for the most part this question is about not giving up and just continuing to push the boundaries of the problem and try new claims. I think I've been avoiding this too much, and I'm generally afraid to even construct new points just for the sake of experimentation. I had actually drawn the six points on the common nine-point circle and the thought of concyclic points had occurred to me, but my foolish self decided to read the solution instead of trying it for real (and it's a 5-second proof anyway). Here we go, though: Let the midpoint of $EC$ be $A_1$ and define other points similarly. Let's also define $O_{\triangle \bullet},G_{\triangle \bullet}, H_{\triangle \bullet}$ in the usual fashion. There are immediate benefits: we can get rid of "circumcenter of $\triangle ACE$" and replace with "orthocenter of $\triangle A_1C_1E_1$", something which is much easier to prove in foresight due to nice homotheties. If you don't find this step, you might end up proving something about radical axes. It's not clean. Notice that $A_1D_1$ coincides with $YZ$ and is the midline of trapezoid $BCEF$ and cyclically. Now we can begin. I claim that $\bullet_1$ lie on the common nine-point circle of $\triangle ACE$ and $\triangle BDF$. Notice by homothety that \[YA_1\cdot YD_1=\frac{1}{2}BC\cdot \frac{1}{2}EF\]which is fixed, which should mean that $A_1,D_1,C_1,F_1$ are cyclic. We get similar results for two other cyclic quadrilaterals. But if the circles are different, then we get a contradiction since the radical axes $A_1D_1$ and cyclically do not concur. Now let's realize that \[YA_1=\frac{1}{2}BC=D_1Z\]which means that $YZ$ and $A_1D_1$ have a shared perpendicular bisector. As a result, the circumcenter of $\bullet_1$ coincides with the circumcenter of $\triangle XYZ$. We should also realize, though, that \[G_{\triangle A_1C_1E_1}=\frac{\overrightarrow{A}+\overrightarrow{C}+\overrightarrow{E}}{3}\]\[G_{\triangle B_1D_1F_1}=\frac{\overrightarrow{B}+\overrightarrow{D}+\overrightarrow{F}}{3}\]\[G_{\triangle XYZ}=\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D}+\overrightarrow{E}+\overrightarrow{F}}{6}\]hence the midpoint of $G_{\triangle A_1C_1E_1}$ and $G_{\triangle B_1D_1F_1}$ is just $G_{\triangle XYZ}$. Since the circumcenters of the three triangles are the same and the centroids are collinear, the orthocenters are collinear by homothety (around the common circumcenter with scale factor $3$).

alternative title: Google Drive Logo Hexagon Since opposite sides are parallel, we may construct points $D^{\prime}$, $F^{\prime}$, and $B^{\prime}$ inside the hexagon such that $ABCD^{\prime}$, $CDEF^{\prime}$, and $EFAB^{\prime}$ are all parallelograms, making the hexagon look like the Google Drive logo. The length condition is equivalent to \[CD^{\prime}\cdot CF^{\prime} = AD^{\prime}\cdot AB^{\prime} = EF^{\prime}\cdot EB^{\prime}\]in other words, $\text{Pow}(C,(D^{\prime}F^{\prime}B^{\prime}))=\text{Pow}(E,(D^{\prime}F^{\prime}B^{\prime}))=\text{Pow}(A,(D^{\prime}F^{\prime}B^{\prime}))$. Thus, the circumcenter of $\triangle ACE$ coincides with that of $\triangle B^{\prime}D^{\prime}F^{\prime}$. Repeating this logic, construct points $E^{\prime}$, $A^{\prime}$, and $C^{\prime}$ inside the hexagon such that $BCDE^{\prime}$, $DEFA^{\prime}$, and $FABC^{\prime}$ are all parallelograms. Then rewrite the length condition as \[FC^{\prime}\cdot FA^{\prime}=DE^{\prime}\cdot DA^{\prime}=BE^{\prime}\cdot BC^{\prime}\]so $\text{Pow}(F,(C^{\prime}E^{\prime}A^{\prime}))=\text{Pow}(D,(C^{\prime}E^{\prime}A^{\prime}))=\text{Pow}(B,(C^{\prime}E^{\prime}A^{\prime}))$, hence the circumcenters of $\triangle BDF$ and $\triangle A^{\prime}C^{\prime}E^{\prime}$ are the same point. It only remains to prove that the circumcenter of $\triangle B^{\prime}D^{\prime}F^{\prime}$, the circumcenter of $\triangle A^{\prime}C^{\prime}E^{\prime}$, and the orthocenter of $\triangle XYZ$ are collinear. We start with the following claim: Claim. $\triangle A^{\prime}C^{\prime}E^{\prime}$ and $\triangle B^{\prime}D^{\prime}F^{\prime}$ are translations of each other. Proof. Use complex numbers. By parallelograms, we have \begin{align*}a^{\prime}=d+f-e \\ b^{\prime}=e+a-f \\ c^{\prime}=f+b-a \\ d^{\prime}=a+c-b \\ e^{\prime}=b+d-c \\ f^{\prime}=c+e-d\end{align*}The triangle formed by $a^{\prime}, c^{\prime}, e^{\prime}$ has sides that are parallel to the vectors \[\left[(d+f-e)-(f+b-a), (f+b-a)-(b+d-c), (b+d-c)-(d+f-e)\right]=\left[d-e-b+a, f-a-d+c, b-c-f+e\right]\]while the triangle formed by $b^{\prime}, d^{\prime}, f^{\prime}$ has sides that are parallel to the vectors \[\left[(e+a-f)-(a+c-b), (a+c-b)-(c+e-d), (c+e-d)-(e+a-f)\right]=\left[e-f-c+b, a-b-e+d, c-d-a+f\right],\]which are the same vectors as above, but in a different order. The claim is proven. $\blacksquare$ We may also verify that \[a-a^{\prime}=a-(d+f-e)=a-d-f+e=(a-f+e)-d=b^{\prime}-d,\]so $AA^{\prime}DB^{\prime}$ is a parallelogram and by a similar logic so are $CC^{\prime}FD^{\prime}$ and $EE^{\prime}BF^{\prime}$. Then, the midpoints of $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ become the midpoints $X$, $Y$, and $Z$ of $\overline{A^{\prime}B^{\prime}}$, $\overline{C^{\prime}D^{\prime}}$, and $\overline{E^{\prime}F^{\prime}}$. We see that $\triangle XYZ$ is the vector average for the medial triangles of $\triangle A^{\prime}C^{\prime}E^{\prime}$, $\triangle B^{\prime}D^{\prime}F^{\prime}$. But the orthocenter of a medial triangle is the circumcenter of the original triangle (use the Euler line and a homothety of ratio $-2$ at centroid), so the orthocenter of $\triangle XYZ$ lies halfway between the circumcenters of $\triangle A^{\prime}C^{\prime}E^{\prime}$, $\triangle B^{\prime}D^{\prime}F^{\prime}$. Q.E.D.

Let $A'$, $B'$, $\dots$ be the midpoints of $CE$, $DF$, $\dots$. Then $D'Y \cdot A'Y = \frac{1}{4}EF \cdot BC = YF' \cdot YC'$ by midlines on $\triangle BEF$ and $\triangle BCE$, so $A'B'C'D'$ is cyclic, and similarly $C'D'E'F'$ and $E'F'A'B'$. It is easy to check by radical axis that $A'B'C'D'E'F'$ is cyclic. Also we get that $X$, $Y$ and $Z$ have equal power wrt $(A'B'C'D'E'F')$ so the circumcenter of $(XYZ)$ lies on the nine-point center of $(AEC)$(since $(A'E'C')$ is the nine-point circle). Also note that the orthocenter of $A'E'C'$ is the circumcenter of $AEC$ since the altitude of $A'$ to $E'C'$ is the perpendicular bisector of $EC$. So it suffices to show that the orthocenters of $A'E'C'$, $B'D'F'$ and $X'Y'Z'$ are collinear. However since these three triangles share a circumcenter it suffices to show that the three centroids of the triangles are collinear by homothety at the circumcenter along the Euler line of the three triangles. However this is obviously true as $\frac{a' + e' + c'}{3} = \frac{a + e + c}{3}$, $\frac{b' + d' + f'}{3} = \frac{b + d + f}{3}$ and $\frac{x + y + z}{3} = \frac{a + b + c + d + e + f}{6}$ which is the midpoint of the previous two centroids, done.