2025 International Mathematical Olympiad China National Team Selection Test

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: $$\begin{array}{rl}4[AXY]& =2[APY]+2[ARY]=[APQ]+[APS]+[ARQ]+[ARS]\\ & =[APS]+[ARQ]=[PSRQ],\end{array}$$ (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 ${\omega }^{\prime }$ be the circle with diameter $OC$. Then $X$ and $Y$ lie on ${\omega }^{\prime }$. Conversely, if a line through $M$ intersects ${\omega }^{\prime }$ 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 $A^{\prime }$, and the lines $PS$ and $QR$ meet at $B^{\prime }$. By polar line properties, $A^{\prime }$ and $B^{\prime }$ 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 $A^{\prime }{B}^{\prime }$, so the midpoint of $A^{\prime }{B}^{\prime }$ is also $M$. By polar properties, $\overrightarrow{OA}\cdot \overrightarrow{OB}=\overrightarrow{O{A}^{\prime }}\cdot \overrightarrow{O{B}^{\prime }}=|OP{|}^2$. Moreover, $$\overrightarrow{OA}\cdot \overrightarrow{OB}=(\overrightarrow{OM}+\overrightarrow{MA})\cdot (\overrightarrow{OM}-\overrightarrow{MA})=|OM{|}^2-|MA{|}^2,$$ and similarly, $\overrightarrow{O{A}^{\prime }}\cdot \overrightarrow{O{B}^{\prime }}=|OM{|}^2-|M{A}^{\prime }{|}^2$. Thus, $|MA|=|M{A}^{\prime }|$, implying that $A^{\prime }$ and $B^{\prime }$ 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 ${\omega }^{\prime }$ 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 ${\omega }^{\prime }$), 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: $$S_{AXY}=\frac{1}{2}XY\cdot d(A,XY)=\frac{1}{2}XY\cdot AM\cdot \sin \angle AMX=\frac{1}{2}AM\cdot X_1{Y}_1.$$ 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 ${\omega }^{\prime }$ onto $m$ has maximal length. We first prove the following claim: If a line $l$ through $M$ intersects ${\omega }^{\prime }$ at $X$ and $Y$, and the segment $XY$ meets $m$ at $N$ satisfying $$\frac{XN}{YN}=\frac{YM}{XM},(\star )$$ then $l$ is the desired line.

Indeed, let the tangents to ${\omega }^{\prime }$ at $X$ and $Y$ meet at $T$. Consider another line $l^{\prime }$ through $M$ intersecting ${\omega }^{\prime }$ at $X^{\prime }$ and $Y^{\prime }$. Suppose $l^{\prime }$ 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 $l^{\prime }$ 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 ${\omega }^{\prime }$ at $K$ and $L$. Then $MK$ and $ML$ are tangent to ${\omega }^{\prime }$. 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. $\square$