IMO2024 Shortlisted Problems

Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively and let the perpendicular bisector of $AB$ intersect the line through $P$ parallel to $AB$ at $R$.

Lemma. Triangles $QAB$ and $RNM$ are similar.

Proof. Let $O$ be the circumcentre of triangle $ABC$, and let $S$ be the midpoint of $CX$. Since $N,S$, and $R$ are the respective perpendicular feet from $O$ to $BC,CX$, and $PR$, we have that quadrilaterals $PRNO$ and $CNSO$ are cyclic. Furthermore, $P,S$, and $O$ are collinear as $PC=PX$. Since $ABCX$ is also cyclic, we have that $$\angle QAB=\angle XCB=\angle PON=180^{\circ }-\angle NRP=\angle MNR.$$ Analogously, we have that $\angle ABQ=\angle RMN$, so triangles $QAB$ and $RNM$ are similar. $\square$



Let $d(Z,\ell )$ denote the perpendicular distance from the point $Z$ to the line $\ell$. Using that $PR\parallel AB$ along with the similarities $QAB\sim RNM$ and $PAB\sim PMN$, we have that $$\frac{d(Q,AB)}{AB}=\frac{d(R,MN)}{MN}=\frac{d(P,MN)}{MN}=\frac{d(P,AB)}{AB},$$ which implies that $PQ\parallel AB$.

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Alternative solution.

Let $BD$ and $AC$ intersect at $T$ and let the line through $P$ parallel to $AB$ intersect $BD$ at $V$. Next, let $Q^{\prime }$ be the foot of the perpendicular from $T$ to $PV$. Finally, let $Q^{\prime }A$ intersect circle $ABC$ again at $X^{\prime }$ and $Q^{\prime }B$ intersect circle $ABD$ again at $Y^{\prime }$.



Claim. $P{Q}^{\prime }$ bisects $\angle B{Q}^{\prime }D$ externally.

Proof. Let $PT$ intersect $CD$ at $L$. Let ${\infty }_{CD}$ be the point at infinity on line $CD$. From the standard Ceva-Menelaus configuration we have $(D,C;L,{\infty }_{CD})$ is harmonic. Hence projecting through $P$ we have $$-1=(D,C;L,{\infty }_{CD})=(D,B;T,V).$$ As $(D,B;T,V)$ is harmonic, and also $\angle V{Q}^{\prime }T=90^{\circ }$ (by construction), the claim follows. $\square$

Now as $$\angle Q^{\prime }PD=\angle BAD=180^{\circ }-\angle D{Y}^{\prime }B=180^{\circ }-\angle D{Y}^{\prime }{Q}^{\prime }$$ we have $Q^{\prime }PD{Y}^{\prime }$ cyclic. By the claim, we have that $P$ is the midpoint of $\mathrm{arc}\widetilde{D{Q}^{\prime }{Y}^{\prime }}$, so $PD=P{Y}^{\prime }$.

Since $Y$ is the unique point not equal to $D$ on circle $ABD$ satisfying $PD=PY$, we have $Y^{\prime }=Y$. Likewise $X^{\prime }=X$ so $Q^{\prime }=Q$ and we are done.

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Alternative solution.

Let $AX$ intersect circle $PCX$ for the second time at $Q^{\prime }$. Then $$\angle A{Q}^{\prime }P=\angle X{Q}^{\prime }P=\angle XCP=\angle XCB=180^{\circ }-\angle BAX=\angle Q^{\prime }AB$$ so $P{Q}^{\prime }$ is parallel to $AB$. Hence, it suffices to show that $Q^{\prime }$ is equal to $Q$. To do so, we aim to show the common chord of circles $PCX$ and $PDY$ is parallel to $AB$, since then by symmetry $Q^{\prime }$ is also the second intersection of $BY$ and circle $PDY$.



Let the centres of circles $PCX$ and $PDY$ be $O_X$ and $O_Y$, respectively. Let the centres of circles $ABC$ and $ABD$ be $O_C$ and $O_D$, respectively.

Note $P,O_X$, and $O_C$ are collinear since they all lie on the perpendicular bisector of $CX$. Likewise $P,O_Y$, and $O_D$ are collinear on the perpendicular bisector of $DY$. By considering the projections of $O_X$ and $O_C$ onto $BC$, and $O_Y$ and $O_D$ onto $AD$, we have $$\frac{P{O}_X}{P{O}_C}=\frac{\frac{PC}{2}}{\frac{PB+PC}{2}}=\frac{\frac{PD}{2}}{\frac{PA+PD}{2}}=\frac{P{O}_Y}{P{O}_D}.$$ Hence $O_X{O}_Y$ is parallel to $O_C{O}_D$, which is perpendicular to $AB$ as desired.