IMO2024 Shortlisted Problems

Let $T$ be the midpoint of $\mathrm{arc}\text{overparen}BAC$ and let lines $BA$ and $CD$ intersect $EF$ at $K$ and $L$, respectively. Note that $T$ lies on the perpendicular bisector of segment $BC$.



Since $ABCD$ is cyclic, $\frac{BD}{\sin \angle BAD}=\frac{AC}{\sin \angle ADC}$. From parallel lines we have $\angle DAF=\angle ADC$ and $\angle BAD=\angle EDA$. Hence, $$AF\cdot \sin \angle DAF=BD\cdot \sin \angle ADC=AC\cdot \sin \angle BAD=DE\cdot \sin \angle EDA.$$ So $F$ and $E$ are equidistant from the line $AD$, meaning that $EF$ is parallel to $AD$.

We have that $KADE$ and $FADL$ are parallelograms, hence we get $KA=DE=AC$ and $DL=AF=BD$. Also, $KE=AD=FL$ so it suffices to prove the perpendicular bisector of $KL$ passes through $T$.

Triangle $AKC$ is isosceles so $\angle BTC=\angle BAC=2\angle BKC$. Likewise, $\angle BTC=2\angle BLC$. Since $T,K$, and $L$ all lie on the same side of $BC$ and $T$ lies on the perpendicular bisector of $BC$, $T$ is the centre of circle $BKLC$. The result follows.

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Alternative solution.

Let $AF$ and $DE$ meet $\omega$ at $X$ and $Y$, respectively, and let $T$ be as in Solution 1.

As $BD<AD$, $DY\parallel AB$ and $\angle BAY=\angle DBA<90^{\circ }$, we have $DY<AB$ and $Y$ lies on the opposite side of line $AD$ to $C$. Also from $BD<AD$, we have $B,C$, and $D$ all lie on the same side of the perpendicular bisector of $AB$ which shows $AC>AB$. Combining these, we get $DY<AB<AC=DE$ and, as $Y$ and $E$ both lie on the same side of line $AD$, $Y$ lies in the interior of segment $DE$. Similarly, $X$ lies in the interior of segment $DF$.

Since $AB$ is parallel to $DY$, we have $YA=BD=FA$. Likewise $XD=AC=ED$.



Claim 1. $T$ is the midpoint of $\mathrm{arc}\text{overparen}XY$.

Proof. From $AX\parallel CD$ and $AB\parallel DY$ we have $$\angle CAX=\angle AXD=\angle AYD=\angle YDB.$$ Since $T$ is the midpoint of $\mathrm{arc}\text{overparen}BAC$, we have $\angle BAT=\angle TDC$, so $$\angle TAX=\angle CAX+\angle BAC-\angle BAT=\angle YDB+\angle BDC-\angle TDC=\angle YDT.$$ Recall from above we have $AB<AC$ and analogously, $DC<DB$, which shows that $X,Y$ and $T$ all lie on the same side of line $AD$. In particular, $T$ and $A$ lie on opposite sides of $XY$ so $T$ lies on the internal angle bisector of $\angle XAY$. Since $AF=AY$, we have $△ATF\cong △ATY$, giving $TF=TY$.

Likewise, $TE=TX$, so $TE=TF$, meaning that $T$ lies on the perpendicular bisector of segment $EF$ as required.

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Alternative solution.

From $AF=DB$, $AC=DE$ and $$\angle (AC,AF)=\angle (AC,CD)=\angle (AB,BD)=\angle (DE,DB),$$ triangles $ACF$ and $DEB$ are congruent, so $CF=BE$.

Let $P=BE\cap CF$. Since $$\angle (CP,BP)=\angle (CF,BE)=\angle (AF,DB)=\angle (DC,DB),$$ we have that $P$ lies on circle $ABCD$.



Finally, let $T$ be the Miquel point of the quadrilateral $BCFE$ so $T$ lies on circles $EFP$ and $ABCD$. Note that $T$ is the centre of spiral similarity taking segments $BE$ to $CF$ and since $BE=CF$, this is in fact just a rotation, so $TB=TC$ and $TE=TF$; that is, the perpendicular bisectors of $BC$ and $EF$ meet at $T$, on circle $ABCD$.