, 2025

Let $O$ be the center of $\omega$. For any good quadrilateral $PQRS$ (where $PQ$ and $RS$ meet at $A$, and $PS$ and $QR$ meet at $B$), let the diagonals $PR$ and $QS$ intersect at $C$. By Brocard's theorem, $A, B, C, O$ form an orthocentric system, so $C$ is a fixed point. The existence of a good quadrilateral implies that $A, B, C, O$ are pairwise distinct. By polar line properties, the polar line of $C$ with respect to $\omega$ is the line $AB$. Since $C$ lies inside $\omega$, $AB$ is disjoint from $\omega$.
Let $X$ and $Y$ be the midpoints of diagonals $PR$ and $QS$, respectively. We have:
$$\begin{aligned} 4[AXY] &= 2[APY] + 2[ARY] = [APQ] + [APS] + [ARQ] + [ARS] \\ &= [APS] + [ARQ] = [PSRQ], \end{aligned}$$
(where $[AXY]$ denotes the signed area of triangle $AXY$, and similarly for others) which shows that the area of the good quadrilateral equals four times the area of triangle $AXY$. Thus, the problem reduces to proving that the maximum area of triangle $AXY$ exists and is achieved uniquely.
Let $M$ be the midpoint of $AB$. By properties of the Newton line, $M, X, Y$ are collinear. Let $\omega'$ be the circle with diameter $OC$. Then $X$ and $Y$ lie on $\omega'$. Conversely, if a line through $M$ intersects $\omega'$ at $X$ and $Y$, and the lines $CX$ and $CY$ (if one of $X, Y$ coincides with $C$, replace the corresponding line with the line through $C$ perpendicular to $OC$) intersect $\omega$ at $P, R$ and $Q, S$ respectively, then the lines $PQ$ and $RS$ meet at $A'$, and the lines $PS$ and $QR$ meet at $B'$. By polar line properties, $A'$ and $B'$ lie on $AB$, the polar of $C$ with respect to $\omega$. Since $OX \perp XC$ and $OY \perp YC$, $X$ and $Y$ are the midpoints of $PR$ and $QS$ respectively (even if $X$ or $Y$ coincides with $O$, the conclusion holds). By the Newton line property, $XY$ passes through the midpoint of $A'B'$, so the midpoint of $A'B'$ is also $M$.
By polar properties, $\overrightarrow{OA} \cdot \overrightarrow{OB} = \overrightarrow{OA'} \cdot \overrightarrow{OB'} = |OP|^2$. Moreover,
$$\overrightarrow{OA} \cdot \overrightarrow{OB} = (\overrightarrow{OM} + \overrightarrow{MA}) \cdot (\overrightarrow{OM} - \overrightarrow{MA}) = |OM|^2 - |MA|^2,$$
and similarly, $\overrightarrow{OA'} \cdot \overrightarrow{OB'} = |OM|^2 - |MA'|^2$. Thus, $|MA| = |MA'|$, implying that $A'$ and $B'$ coincide with $A$ and $B$ (possibly in reverse order). Hence, the quadrilateral $PQRS$ constructed this way is necessarily a good quadrilateral, and distinct unordered pairs $(X, Y)$ correspond to distinct good quadrilaterals.
The problem now reduces to proving: Among all lines $l$ through $M$ intersecting $\omega'$ at $X$ and $Y$, there exists a unique line $l$ that maximizes the area of triangle $AXY$.
Let $F$ be the midpoint of $OC$ (i.e., the center of $\omega'$), and let $m$ be the line through $F$ perpendicular to $MA$. Let $X_1$ and $Y_1$ be the projections of $X$ and $Y$ onto $m$. Then:
$$S_{AXY} = \frac{1}{2}XY \cdot d(A, XY) = \frac{1}{2}XY \cdot AM \cdot \sin \angle AMX = \frac{1}{2}AM \cdot X_1Y_1.$$
Thus, the problem further reduces to proving that there exists a unique line $l$ through $M$ such that the projection of its chord $XY$ with $\omega'$ onto $m$ has maximal length.
We first prove the following claim: If a line $l$ through $M$ intersects $\omega'$ at $X$ and $Y$, and the segment $XY$ meets $m$ at $N$ satisfying
$$\frac{XN}{YN} = \frac{YM}{XM}, \qquad (\star)$$
then $l$ is the desired line.

Indeed, let the tangents to $\omega'$ at $X$ and $Y$ meet at $T$. Consider another line $l'$ through $M$ intersecting $\omega'$ at $X'$ and $Y'$. Suppose $l'$ intersects segments $XT$ and $YT$ at $U$ and $V$, respectively. It suffices to show that the projection of $UV$ onto $m$ is shorter than that of $XY$, i.e., $XU \sin \angle XFN > YV \sin \angle YFN$.
Note that $\frac{\sin \angle XFN}{\sin \angle YFN} = \frac{XN}{YN} = \frac{MY}{MX}$. By Menelaus' theorem, $\frac{MY}{MX} \cdot \frac{XU}{YV} = \frac{TU}{TV} > 1$, so $XU \sin \angle XFN > YV \sin \angle YFN$. The case where $l'$ intersects the extensions of $TX$ and $TY$ is similar.
Finally, we prove that a line $l$ satisfying $(\star)$ exists and is unique. Existence suffices, as uniqueness would otherwise lead to a contradiction.

Let $\Omega$ be the circle with diameter $MF$, intersecting $\omega'$ at $K$ and $L$. Then $MK$ and $ML$ are tangent to $\omega'$. Let $KL$ meet $m$ at $J$. Then $MJ$ harmonically divides $XY$, i.e., $\frac{JX}{JY} = \frac{MX}{MY}$. Thus, $\frac{XN}{YN} = \frac{YM}{XM}$ is equivalent to $J$ and $N$ being symmetric about the midpoint $I$ of $XY$. By the perpendicular bisector theorem, $I$ also lies on $\Omega$. The problem now reduces to showing there exists a unique line $l$ through $M$ intersecting $KL, \Omega$, and $m$ at $J, I, N$ respectively, with $JI = IN$.
When $l = ML, JI = 0 < IN$; when $l = MF, IN = 0 < JI$. By continuity, such an $l$ exists. This completes the proof. $\square$