Problem: Let $ABCDEF$ be a convex hexagon satisfying
$\ol{AB} \parallel \ol{DE}$,
$\ol{BC} \parallel \ol{EF}$,
$\ol{CD} \parallel \ol{FA}$, and
\[ AB \cdot DE = BC \cdot EF = CD \cdot FA. \]

Let $X$, $Y$, and $Z$ be the midpoints
of $\ol{AD}$, $\ol{BE}$, and $\ol{CF}$.
Prove that
the circumcenter of $\triangle ACE$,
the circumcenter of $\triangle BDF$, and
the orthocenter of $\triangle XYZ$ are collinear.

\end{enumerate}

Solution: We present two solutions.

\paragraph{Parallelogram solution found by contestants.}
Note that the following figure is intentionally
\emph{not} drawn to scale, to aid legibility.
We construct parallelograms $ABCE'$, etc as shown.
Note that this gives two congruent triangles $A'C'E'$ and $B'D'F'$.
(Assuming that triangle $XYZ$ is non-degenerate,
the triangles $A'C'E'$ and $B'D'F'$ will also be non-degenerate.)
\begin{center}
\begin{asy}
pair B = dir(123);
pair C = dir(150);
pair D = dir(250);
pair E = -conj(D);
pair F = B*C/E;
pair A = D*C/F;

pair Ap = C+E-D;
pair Cp = E+A-F;
pair Ep = A+C-B;
pair Bp = D+F-E;
pair Dp = F+B-A;
pair Fp = B+D-C;

filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightcyan, blue);
draw(A--Cp, red);
draw(C--Ep, red);
draw(E--Ap, red);
draw(B--Fp, deepgreen);
draw(D--Bp, deepgreen);
draw(F--Dp, deepgreen);
draw(A--D, gray);
pair X = midpoint(A--D);

pair M = midpoint(Cp--Ep);
pair N = midpoint(Bp--Fp);
draw(M--N, gray);

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("$A$", A, dir(A));
dot("$A'$", Ap, dir(90));
dot("$C'$", Cp, dir(160));
dot("$E'$", Ep, dir(310));
dot("$B'$", Bp, dir(90));
dot("$D'$", Dp, dir(180));
dot("$F'$", Fp, dir(315));
dot("$X$", X, dir(X));
dot("$M$", M, dir(350));
dot("$N$", N, dir(270));

/* TSQ Source:

B = dir 123
C = dir 150
D = dir 250
E = -conj(D)
F = B*C/E
A = D*C/F

A' = C+E-D R90
C' = E+A-F R160
E' = A+C-B R310
B' = D+F-E R90
D' = F+B-A R180
F' = B+D-C R315

A--B--C--D--E--F--cycle 0.1 lightcyan / blue
A--Cp red
C--Ep red
E--Ap red
B--Fp deepgreen
D--Bp deepgreen
F--Dp deepgreen
A--D gray
X = midpoint A--D

M = midpoint Cp--Ep R350
N = midpoint Bp--Fp R270
M--N gray

*/
\end{asy}
\end{center}
\begin{claim*}
  If $AB \cdot DE = BC \cdot EF = CD \cdot FA = k$,
  then the circumcenters of $ACE$ and $A'C'E'$ coincide.
\end{claim*}
\begin{proof}
  The power of $A$ to $(A'C'E')$ is
  $AE' \cdot AC' = BC \cdot EF = k$;
  same for $C$ and $E$.
\end{proof}


\begin{center}
\begin{asy}
size(10cm);
pair B = dir(123);
pair C = dir(150);
pair D = dir(250);
pair E = -conj(D);
pair F = B*C/E;
pair A = D*C/F;

pair Ap = C+E-D;
pair Cp = E+A-F;
pair Ep = A+C-B;
pair Bp = D+F-E;
pair Dp = F+B-A;
pair Fp = B+D-C;

pair Ma = midpoint(Cp--Ep);
pair Mc = midpoint(Ep--Ap);
pair Me = midpoint(Ap--Cp);
pair Mb = midpoint(Dp--Fp);
pair Md = midpoint(Fp--Bp);
pair Mf = midpoint(Bp--Dp);

pair X = midpoint(A--D);
pair Y = midpoint(B--E);
pair Z = midpoint(C--F);

draw(Ap--Cp--Ep--cycle, red);
draw(Bp--Dp--Fp--cycle, deepgreen);
draw(X--Y--Z--cycle, blue);
draw(Ma--Mc--Me--cycle, lightred);
draw(Md--Mf--Mb--cycle, lightgreen);

draw(Mc--Mf, gray);
draw(Me--Mb, gray);
draw(Ma--Md, gray);

dot("$A'$", Ap, dir(180));
dot("$C'$", Cp, dir(275));
dot("$E'$", Ep, dir(0));
dot("$B'$", Bp, dir(0));
dot("$D'$", Dp, dir(180));
dot("$F'$", Fp, dir(275));

dot("$X$", X, dir(340), blue);
dot("$Y$", Y, dir(225), blue);
dot("$Z$", Z, dir(130), blue);
dot(Ma, red);
dot(Mc, red);
dot(Me, red);
dot(Mb, deepgreen);
dot(Md, deepgreen);
dot(Mf, deepgreen);

dot(orthocentercenter(Ma, Mc, Me), red);
dot(orthocentercenter(Mb, Md, Mf), deepgreen);
dot(orthocentercenter(X, Y, Z), blue);
\end{asy}
\end{center}

\begin{claim*}
  Triangle $XYZ$ is the vector average
  of the (congruent) medial triangles of
  triangles $A'C'E'$ and $B'D'F'$.
\end{claim*}
\begin{proof}
  If $M$ and $N$ are the midpoints
  of $\ol{C'E'}$ and $\ol{B'F'}$,
  then $X$ is the midpoint of $\ol{MN}$ by vector calculation:
  \begin{align*}
    \frac{\vec M + \vec N}{2}
    &= \frac{\frac{\vec C' + \vec E'}{2} + \frac{\vec B' + \vec F'}{2}}{2} \\
    &= \frac{\vec C' + \vec E' + \vec B' + \vec F'}{4} \\
    &= \frac{(\vec A + \vec E - \vec F) + (\vec C + \vec A - \vec B)
      + (\vec D + \vec F - \vec E) + (\vec B + \vec D - \vec C)}{4} \\
    &= \frac{\vec A + \vec D}{2} = \vec X. \qedhere
  \end{align*}
\end{proof}
Hence the orthocenter of $XYZ$
is the midpoint of the orthocenters
of the medial triangles of $A'C'E'$ and $B'D'F'$ --- that is,
their circumcenters.

\paragraph{Author's solution.}
Call $MNP$ and $UVW$ the medial triangles
of $ACE$ and $BDF$.

\begin{center}
\begin{asy}
size(11cm);
pair M = dir(253.96);
pair N = dir(14.7);
pair P = dir(133.84);

pair U = dir(283.18);
pair V = dir(44.37);
pair W = dir(163.96);

pair A = N+P-M;
pair C = P+M-N;
pair E = M+N-P;

pair B = V+W-U;
pair D = W+U-V;
pair F = U+V-W;

draw(A--C--E--cycle, lightred);
draw(M--N--P--cycle, lightred);
draw(B--D--F--cycle, lightgreen);
draw(U--V--W--cycle, lightgreen);

pair X = midpoint(A--D);
pair Y = midpoint(B--E);
pair Z = midpoint(C--F);

draw(A--D, gray);
draw(B--E, gray);
draw(C--F, gray);

draw(unitcircle, dotted);

draw(M--V, deepcyan+1);
draw(B--C, deepcyan+1);
draw(E--F, deepcyan+1);

draw(A--B, blue+1);
draw(N--W, blue+1);
draw(D--E, blue+1);

draw(A--F, purple+1);
draw(U--P, purple+1);
draw(D--C, purple+1);

dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$P$", P, dir(P));
dot("$U$", U, dir(U));
dot("$V$", V, dir(V));
dot("$W$", W, dir(W));
dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));

/* TSQ Source:

!size(11cm);
M = dir 253.96
N = dir 14.7
P = dir 133.84

U = dir 283.18
V = dir 44.37
W = dir 163.96

A = N+P-M
C = P+M-N
E = M+N-P

B = V+W-U
D = W+U-V
F = U+V-W

A--C--E--cycle lightred
M--N--P--cycle lightred
B--D--F--cycle lightgreen
U--V--W--cycle lightgreen

X = midpoint A--D
Y = midpoint B--E
Z = midpoint C--F

A--D gray
B--E gray
C--F gray

unitcircle dotted

M--V deepcyan+1
B--C deepcyan+1
E--F deepcyan+1

A--B blue+1
N--W blue+1
D--E blue+1

A--F purple+1
U--P purple+1
D--C purple+1

*/
\end{asy}
\end{center}

\begin{claim*}
  In trapezoid $ABDE$, the perpendicular
  bisector of $\ol{XY}$ is the same as the
  perpendicular bisector of the midline $\ol{WN}$.
\end{claim*}
\begin{proof}
  This is true for any trapezoid:
  because $WX = \half AB = YN$.
\end{proof}

\begin{claim*}
  The points $V$, $W$, $M$, $N$ are cyclic.
\end{claim*}
\begin{proof}
  By power of a point from $Y$, since
  \[ WY \cdot YN = \half DE \cdot \half AB
    = \half EF \cdot \half BC
    = VY \cdot YM. \qedhere \]
\end{proof}

Applying all the cyclic variations of the above two claims,
it follows that all six points $U$, $V$, $W$, $M$, $N$, $P$
are concyclic, and the center of that circle
coincides with the circumcenter of $\triangle XYZ$.

\begin{remark*}
  It is also possible to implement ideas from the first solution here,
  by showing all six midpoints have equal power to $(XYZ)$.
\end{remark*}

\begin{claim*}
  The orthocenter of $XYZ$ is the midpoint
  of the circumcenters of $\triangle ACE$ and $\triangle BDF$.
\end{claim*}
\begin{proof}
  Apply complex numbers with the unit circle coinciding
  with the circumcircle of $NVPWMU$.
  Then
  \begin{align*}
    \opname{orthocenter}(XYZ) &= x+y+z = \frac{a+b+c+d+e+f}{2} \\
    \opname{circumcenter}(ACE) &= \opname{orthocenter}(MNP) \\
    &= m+n+p = \frac{c+e}{2} + \frac{e+a}{2} + \frac{a+c}{2} = a+c+e \\
    \opname{circumcenter}(BDF) &= \opname{orthocenter}(UVW) \\
    &= u+v+w = \frac{d+f}{2} + \frac{f+b}{2} + \frac{b+d}{2} = b+d+f.
    \qedhere
  \end{align*}
\end{proof}