Problem: A convex quadrilateral $ABCD$ satisfies $AB \cdot CD = BC \cdot DA$.
Point $X$ lies inside $ABCD$ so that
\[ \angle XAB=\angle XCD \quad \text{ and } \quad \angle XBC=\angle XDA. \]
Prove that $\angle BXA + \angle DXC=180\dg$.

\end{enumerate}

Solution: We present two solutions by inversion.
The first is the official one.
The second is a solution via inversion, completed by USA5 Michael Ren.

\paragraph{Official solution by inversion.}
In what follows a convex quadrilateral is called
\emph{quasi-harmonic} if $AB \cdot CD = BC \cdot DA$.

\begin{claim*}
  A quasi-harmonic quadrilateral is determined
  up to similarity by its angles.
\end{claim*}
\begin{proof}
  Do some inequalities.
\end{proof}
\begin{remark*}
  This could be expected by degrees of freedom;
  a quadrilateral has four degrees of freedom up to similarity;
  the pseudo-harmonic condition is one
  while the measures of angles $\angle A$, $\angle B$, $\angle C$, $\angle D$
  (summing to $360\dg$) provide three degrees of freedom.
  (Note that the point $X$ plays no role in this comment.)
\end{remark*}

Performing an inversion at $X$, one obtains a
second quasi-harmonic quadrilateral
$A^\ast B^\ast C^\ast D^\ast$ which has the same angles
as the original one, $\angle D^\ast = \angle A$,
$\angle A^\ast = \angle B$, and so on.
Thus by the claim we obtain similarity
\[ D^\ast A^\ast B^\ast C^\ast \sim ABCD. \]
If one then maps $D^\ast A^\ast B^\ast C^\ast$,
onto $ABCD$, the image of $X^\ast$
becomes a point isogonally conjugate to $X$.
In other words, $X$ has an isogonal conjugate in $ABCD$.

It is well-known that this is equivalent to
$\angle BXA + \angle DXC = 180\dg$,
for example by inscribing an ellipse with foci $X$ and $X^\ast$.

\paragraph{Second solution: ``rhombus inversion'', by Michael Ren.}
Since
\[ \frac{AB}{AD} = \frac{CB}{CD} \]
and
\[ \frac{BA}{BC} = \frac{DA}{DC} \]
it follows that $B$ and $D$ lie on an Apollonian circle $\omega_{AC}$
through $A$ and $C$,
while $A$ and $C$ lie on an Apollonian circle $\omega_{BD}$
through $B$ and $D$.
We let these two circles intersect at a point $P$ inside $ABCD$.

The main idea is then to
perform an inversion about $P$ with radius $1$.
We obtain:
%\paragraph{Rhombus construction.}
\begin{lemma*}
  The image of $ABCD$ is a rhombus.
\end{lemma*}
\begin{proof}
  By the inversion distance formula, we have
  \[ \frac{1}{A'B'} = \frac{PA}{AB} \cdot PB = \frac{PC}{BC} \cdot PB = \frac{1}{B'C'} \]
  and so $A'B' = B'C'$.
  In a similar way, we derive $B'C' = C'D' = D'A'$,
  so the image is a rhombus as claimed.
\end{proof}

Let us now translate the angle conditions.
We were given that $\dang XAB = \dang XCD$, but
\begin{align*}
  \dang XAB &= \dang XAP + \dang PAB = \dang PX'A' + \dang A'B'P \\
  \dang XCD &= \dang XCP + \dang PCD = \dang PX'C' + \dang C'D'P
  \intertext{so subtracting these gives}
  \dang A'X'C' &= \dang A'B'P + \dang PD'C' = \dang (A'B', B'P) + \dang (PD', C'D') \\
  &= \dang (A'B', B'P) + \dang (PD', A'B') = \dang D' P B'. \tag{1}
\end{align*}
since $\ol{A'B'} \parallel \ol{C'D'}$.
Similarly, we obtain
\[ \dang B'X'D' = \dang A'PC' \tag{2}. \]
We now translate the desired condition.
Since
\begin{align*}
  \dang AXB &= \dang AXP + \dang PXB = \dang PA'X' + \dang X'B'P \\
  \dang CXD &= \dang CXP + \dang PXD = \dang PC'X' + \dang X'DP'
\end{align*}
we compute
\begin{align*}
  \dang AXB + \dang CXD &= (\dang PA'X' + \dang X'B'P) + (\dang PC'X' + \dang X'D'P) \\
  &= -\left[ \left( \dang A'X'P + \dang X'PA' \right)
    + \left( \dang PX'B' + \dang B'PX' \right) \right] \\
  &\quad- \left[ \left( \dang C'X'P + \dang X'PC' \right)
    + \left( \dang PX'D' + \dang D'PX'  \right) \right] \\
  &= \left[ \dang PX'A' + \dang BX'P + \dang PX'C' + \dang D'X'P  \right] \\
  &\quad+ \left[ \dang A'PX' + \dang X'PB' + \dang C'PX' + \dang X'PD' \right] \\
  &= \dang A'PB' + \dang C'PD' + \dang B'X'C + \dang D'X'A
\end{align*}
and we wish to show this is equal to zero, i.e.\
the desired becomes
\[ \dang A'PB' + \dang C'PD' + \dang B'X'C + \dang D'X'A = 0. \tag{3} \]
In other words, the problem is to show (1) and (2) implies (3).

Henceforth drop apostrophes.
Here is the inverted diagram (with apostrophes dropped).
\begin{center}
\begin{asy}
size(12cm);
pair A = (0,3);
pair B = (-8,0);
pair C = (0,-3);
pair D = (8,0);

pair X = (-5.6,7.8);
pair Y = reflect(circumcenter(A,C,X), circumcenter(D,B,X))*X;
pair Q = IP(circumcircle(A, Y, D), circumcircle(B, Y, C));
pair P = conj(Q);

filldraw(A--B--C--D--cycle, opacity(0.1)+lightred, red);
draw(A--C, red);
draw(B--D, red);

draw(B--X--D, lightgreen);
draw(A--P--C, lightgreen);
draw(C--X--A, lightblue);
draw(B--P--D, lightblue);

draw(circumcircle(A, X, C), heavycyan);
draw(circumcircle(B, X, D), heavycyan);
draw(circumcircle(B, Q, C), lightcyan);
draw(circumcircle(A, Q, D), lightcyan);
draw(Q--P, red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Q$", Q, dir(Q));
dot("$P$", P, dir(P));

/* TSQ Source:

A = (0,3)
B = (-8,0)
C = (0,-3)
D = (8,0)

X = (-5.6,7.8)
Y = OP circumcircle A X C circumcircle B X D
Q = IP circumcircle A Y D circumcircle B Y C
P = conj(Q)

A--B--C--D--cycle 0.1 lightred / red
A--C red
B--D red

B--X--D lightgreen
A--P--C lightgreen
C--X--A lightblue
B--P--D lightblue

circumcircle A X C heavycyan
circumcircle B X D heavycyan
circumcircle B Q C lightcyan
circumcircle A Q D lightcyan
Q--P red

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

Let $Q$ denote the reflection of $P$
and let $Y$ denote the second intersection of $(BQC)$ and $(AQD)$.
Then
\begin{align*}
  -\dang AXC &= -\dang DPB = \dang BQD = \dang BQY + \dang YQD = \dang BCY + \dang YAD \\
  &= \dang(BC,CY) + \dang(YA,AD) = \dang YCA = -\dang AYC.
\end{align*}
% again using $\ol{AB} \parallel \ol{CD}$.
Hence $XACY$ is concyclic; similarly $XBDY$ is concyclic.

\begin{claim*}
  $X \neq Y$.
\end{claim*}
\begin{proof}
  To see this: Work pre-inversion assuming $AB < AC$.
  Then $Q$ was the center of $\omega_{BD}$.
  If $T$ was the second intersection of $BA$ with $(QBC)$,
  then $QB = QD = QT = \sqrt{QA \cdot QC}$, by shooting lemma.
  Since $\angle BAD < 180\dg$,
  it follows $(QBCY)$ encloses $ABCD$ (pre-inversion).
  (This part is where the hypothesis that
  $ABCD$ is convex with $X$ inside is used.)
\end{proof}

Finally, we do an angle chase to finish:
\begin{align*}
  \dang DXA &= \dang DXY + \dang YXA = \dang DBY + \dang YCA \\
  &= \dang (DB, YB) + \dang (CY, CA) = \dang CYB + 90\dg \\
  &= \dang CQB + 90\dg = -\dang APB + 90\dg \tag{4}.
\end{align*}
Similarly,
\[ \dang BXC = \dang DPC + 90\dg. \tag{5} \]
%% \dang AXB = \dang DYC + 90\dg, \quad
%% \dang CXD = \dang BYC + 90\dg.
Summing (4) and (5) gives (3).
% $\dang BXC + \dang DXA = (\dang BPA + 90\dg) + (\dang DPC + 90\dg) = -(\dang APB + \dang CPD)$.


\begin{remark*}
  A difficult part of the problem in many solutions
  is that the conclusion is false in the directed sense,
  if the point $X$ is allowed to lie outside the quadrilateral.
  We are saved in the first solution because the equivalence
  of the isogonal conjugation requires $X$ inside the quadrilateral.
  On the other hand, in the second solution,
  the issue appears in the presence of the second point $Y$.
\end{remark*}