Problem: In hexagon $ABCDEF$, which is nonconvex but not self-intersecting,
no pair of opposite sides are parallel.
The internal angles satisfy
$\angle A=3\angle D$, $\angle C=3\angle F$, and $\angle E=3\angle B$.
Furthermore $AB=DE$, $BC=EF$, and $CD=FA$.
Prove that diagonals $\ol{AD}$, $\ol{BE}$, and $\ol{CF}$
are concurrent.

Solution: We present the official solution.
We say a hexagon is \emph{satisfying}
if it obeys the six conditions;
note that intuitively we expect three degrees of freedom
for satisfying hexagons.

Main idea:
\begin{claim*}
In a satisfying hexagon,
$B$, $D$, $F$ are reflections of $A$, $C$, $E$ across
the sides of $\triangle ACE$.
\end{claim*}
(This claim looks plausible because every excellent hexagon is satisfying,
and both configuration spaces are three-dimensional.)
Call a hexagon of this shape ``excellent'';
in a excellent hexagon the diagonals clearly concur (at the orthocenter).

Set $\beta = \angle B$, $\delta = \angle D$, $\varphi = \angle F$.

Now given a satisfying hexagon $ABCDEF$,
construct a ``phantom hexagon'' $A'B'C'D'E'F'$
with the same angles which is excellent (see figure).
This is possible since $\beta + \delta + \varphi = 180\dg$.

\begin{center}
\begin{asy}
size(12cm);

void MA(string s, pair A, pair O, pair B) {
  markangle(11.0, A,O,B);
  label("$"+s+"$", O, 4*dir((pair)incenter(A,O,B)-O));
}

pair A = dir(110);
pair C = dir(205);
pair E = dir(335);
pair B = -E+2*foot(E, A, C);
pair D = -A+2*foot(A, C, E);
pair F = -C+2*foot(C, E, A);

filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightgreen, heavygreen);
draw(A--C--E--cycle, green);
draw(A--D, dashed+green);
draw(B--E, dashed+green);
draw(C--F, dashed+green);

MA("\beta", C,B,A);
MA("\delta", E,D,C);
MA("\varphi", A,F,E);

transform t = shift(-4,0);

add(t*CC());


// Again

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

filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightcyan, heavycyan);
draw(A--C--E--cycle, heavycyan);
draw(A--D, dashed+heavycyan);
draw(B--E, dashed+heavycyan);
draw(C--F, dashed+heavycyan);

MA("\beta", C,B,A);
MA("\delta", E,D,C);
MA("\varphi", A,F,E);

MA("\beta", A,E,C);
MA("\varphi", E,C,A);
MA("\delta", C,A,E);


pair A = t*A;
pair B = t*B;
pair C = t*C;
pair D = t*D;
pair E = t*E;
pair F = t*F;

dot("$A'$", Ap, dir(Ap));
dot("$C'$", Cp, dir(Cp));
dot("$E'$", Ep, dir(Ep));
dot("$B'$", Bp, dir(Bp));
dot("$D'$", Dp, dir(Dp));
dot("$F'$", Fp, dir(Fp));
dot("$A$", A, dir(120));
dot("$B$", B, dir(135));
dot("$C$", C, dir(225));
dot("$D$", D, dir(270));
dot("$E$", E, dir(350));
dot("$F$", F, dir(45));

/* TSQ Source:

!size(12cm);

A := dir 110
C := dir 205
E := dir 335
B := -E+2*foot E A C
D := -A+2*foot A C E
F := -C+2*foot C E A

A--B--C--D--E--F--cycle 0.1 lightgreen / heavygreen
A--C--E--cycle green
A--D dashed green
B--E dashed green
C--F dashed green

!MA("\beta", C,B,A, 0.2);
!MA("\delta", E,D,C, 0.2);
!MA("\varphi", A,F,E, 0.2);

! transform t = shift(-4,0);

!add(t*CC());


// Again

A' = A
C' = C
E' = E
B' = B
D' = D
F' = F

A--B--C--D--E--F--cycle 0.1 lightcyan / heavycyan
A--C--E--cycle heavycyan
A--D dashed heavycyan
B--E dashed heavycyan
C--F dashed heavycyan
! anglepen = heavycyan;

!MA("\beta", C,B,A, 0.2);
!MA("\delta", E,D,C, 0.2);
!MA("\varphi", A,F,E, 0.2);

!MA("\beta", A,E,C, 0.2);
!MA("\varphi", E,C,A, 0.2);
!MA("\delta", C,A,E, 0.2);


A = t*A R120
B = t*B R135
C = t*C R225
D = t*D R270
E = t*E R350
F = t*F R45

*/

\end{asy}
\end{center}


Then it would suffice to prove that:
\begin{lemma*}
  A satisfying hexagon is uniquely determined by its angles
  up to similarity.
  That is, at most one hexagon
  (up to similarity) has angles $\beta$, $\delta$, $\gamma$ as above.
\end{lemma*}
\begin{proof}
  Consider any two satisfying hexagons $ABCDEF$ and $A'B'C'D'E'F'$
  (not necessarily as constructed above!) with the same angles.
  We show they are similar.

  To do this, consider the unit complex numbers
  in the directions $\overrightarrow{BA}$ and $\overrightarrow{DE}$
  respectively and let $\vec x$ denote their sum.
  Define $\vec y$, $\vec z$ similarly.
  Note that the condition $\ol{AB} \nparallel \ol{DE}$
  implies $\vec x \neq 0$, and similarly.
  Then we have the identities
  \[ AB \cdot \vec x + CD \cdot \vec y + EF \cdot \vec z
    = A'B' \cdot \vec x + C'D' \cdot \vec y + E'F' \cdot \vec z
    = 0. \]

  So we would obtain $AB : CD : EF = A'B' : C'D' : E'F'$
  if only we could show that $\vec x$, $\vec y$, $\vec z$
  are not multiples of each other
  (linear dependency reasons).
  This is a tiresome computation with arguments,
  but here it is.

  First note that none of $\beta$, $\delta$, $\varphi$
  can be $90\dg$,
  since otherwise we get a pair of parallel sides.
  Now work in the complex plane,
  fix a reference such that $\vec A - \vec B$ has argument $0$,
  and assume $ABCDEF$ are labelled counterclockwise.
  Then
  \begin{itemize}
    \ii $\vec B - \vec C$ has argument $\pi-\beta$
    \ii $\vec C - \vec D$ has argument $-(\beta+3\varphi)$
    \ii $\vec D - \vec E$ has argument $\pi-(\beta+3\varphi+\delta)$
    \ii $\vec E - \vec F$ has argument $-(4\beta+3\varphi+\delta)$
    % \ii $\vec F - \vec A$ has argument $\pi-(4\beta+4\varphi+\delta)$
  \end{itemize}
  So the argument of $\vec x$ has argument
  $\frac{\pi-(\beta+3\varphi+\delta)}{2} \pmod \pi$.
  The argument of $\vec y$ has argument
  $\frac{\pi-(5\beta+3\varphi+\delta)}{2} \pmod \pi$.
  Their difference is $2\beta \pmod \pi$, and since $\beta \neq 90\dg$,
  it follows that $\vec x$ and $\vec y$ are not
  multiples of each other; the other cases are similar.
\end{proof}

Then the lemma implies $ABCDEF \sim A'B'C'D'E'F$ and we're done.

\begin{remark*}
  This problem turned out to be known already.
  It appears in this reference:
  \begin{quote}
  Nikolai Beluhov, \emph{Matematika}, 2008, issue 6, problem 3.
  \end{quote}
  It was reprinted as Kvant, 2009, issue 2, problem M2130;
  the reprint is available at
  \url{http://kvant.ras.ru/pdf/2009/2009-02.pdf}.
\end{remark*}

\begin{remark*}
  The vector perspective also shows the condition about parallel
  sides cannot be dropped.
  Here is a counterexample from Ryan Kim
  in the event that it is.
  \begin{center}
  \begin{asy}
  pair A = dir(140);
  pair C_0 = conj(A);
  pair E = -A;
  pair B_0 = 2*C_0-E;
  pair D_0 = 2*C_0-A;
  pair F = extension(A, E*E/C_0, E, A*A/C_0);


  pair C = C_0+2*C_0;
  pair B = B_0+2*C_0;
  pair D = D_0+2*C_0;

  filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightcyan, heavycyan);

  draw(A--D, heavygreen);
  draw(B--E, heavygreen);
  draw(C--F, heavygreen);
  draw(A--C_0--E--cycle, blue);

  draw(B_0--C--D_0--C_0--cycle, orange+dashed);

  dot("$A$", A, dir(A));
  dot("$C_0$", C_0, dir(45));
  dot("$E$", E, dir(E));
  dot("$B_0$", B_0, dir(135));
  dot("$D_0$", D_0, dir(315));
  dot("$F$", F, dir(F));
  dot("$C$", C, dir(C));
  dot("$B$", B, dir(B));
  dot("$D$", D, dir(D));

  /* TSQ Source:

  A = dir 140
  C_0 = conj(A) R45
  E = -A
  B_0 = 2*C_0-E R135
  D_0 = 2*C_0-A R315
  F = extension A E*E/C_0 E A*A/C_0


  C = C_0+2*C_0
  B = B_0+2*C_0
  D = D_0+2*C_0

  A--B--C--D--E--F--cycle 0.1 lightcyan / heavycyan

  A--D heavygreen
  B--E heavygreen
  C--F heavygreen
  A--C_0--E--cycle blue

  B_0--C--D_0--C_0--cycle orange dashed

  */
  \end{asy}
  \end{center}
  By adjusting the figure above so that
  the triangles are right isosceles (instead of just right),
  one also finds an example of a hexagon which is satisfying
  and whose diagonals are concurrent,
  but which is \emph{not} excellent.
\end{remark*}