IMO2024 Shortlisted Problems

Claim. Quadrilateral $WXYZ$ is cyclic, and its circumcentre is $J$. Proof. As $N$ is the midpoint of $\widehat{BAC}$, $W$ and $Z$ lie on $BC$, and $X$ and $Y$ are the second intersections of $NW$ and $NZ$ with $\Gamma$, we have that $WXYZ$ is cyclic. Let the parallel to $BC$ through $N$ intersect $TU$ and $TV$ at $U^{\prime }$ and $V^{\prime }$, respectively. Then $U^{\prime }$ is the intersection of the tangents to $\Gamma$ at $N$ and $X$, so $U^{\prime }N=U^{\prime }X$. As $N{U}^{\prime }\parallel BC$, $U^{\prime }NX$ is similar to $UWX$, so $UW=UX$ as well. Hence, the perpendicular bisector of $WX$ is the internal bisector of $\angle XUW$, which is the external bisector of $\angle VUT$. Analogously, the perpendicular bisector of $YZ$ is the external bisector of $\angle TVU$. This means that the circumcentre of $WXYZ$ is the intersection of the external bisectors of $\angle VUT$ and $\angle TVU$, which is $J$. $\square$



Let $AN$ intersect $BC$ at $L$, so $XY$ passes through $L$ as well. By power of a point from $L$ to $\Gamma$ and circle $WXYZ$, we have that $LA\cdot LN=LX\cdot LY=LW\cdot LZ$, so $WANZ$ is also cyclic. Thus, $A$ is the Miquel point of quadrilateral $WXYZ$. As $WXYZ$ is cyclic with circumcentre $J$ and its opposite sides $WX$ and $YZ$ intersect at $N$, we have that $AN\perp AJ$. Since $AN$ is the external bisector of $\angle BAC$, this implies that $AJ$ is the internal bisector of $\angle BAC$.

Solution 2:

Let the internal and external angle bisectors of $\angle BAC$ intersect $BC$ at $K$ and $L$, respectively. Let $AK$ intersect circle $ABC$ again at $M$, and let $D$ be the intersection of the tangents to $\Gamma$ at $B$ and $C$. Let $\Omega$ be the $T$-excircle of $TUV$, and let $\omega$ be the incircle of $DBC$.

Claim. The points $T$, $K$, and $D$ are collinear. Proof. Note that $BC$ and $XY$ are the polars of $T$ and $D$ with respect to $\Gamma$. By La Hire's Theorem, $TD$ is the polar of $L$ with respect to $\Gamma$. As $(B,C;K,L)=-1$, $K$ also lies on the polar of $L$, thus proving the collinearity. $\square$

Claim. The incentre of $DBC$ is $M$. Proof. We have that $\angle MBC=\angle MAC=\frac{1}{2}\angle BAC=\frac{1}{2}\angle DBC$, so $BM$ bisects $\angle DBC$. Similarly, $CM$ bisects $\angle BCD$, so $M$ is the incentre of $DBC$. $\square$



Claim. The intersection of the common external tangents of $\Omega$ and $\omega$ is $K$. Proof. Let $K^{\prime }$ be the intersection of the common external tangents of $\Omega$ and $\omega$. As $\Omega$ and $\omega$ are both tangent to $BC$ and lie on the same side of $BC$ opposite to $A$, $K^{\prime }$ lies on $BC$. As $T$ is the intersection of the common external tangents of $\Gamma$ and $\Omega$ and $D$ is the intersection of the common external tangents of $\Gamma$ and $\omega$, by Monge's theorem $K^{\prime }$ lies on $TD$. As $K^{\prime }$ lies on both $BC$ and $TD$, it is the same point as $K$. $\square$

Hence, $K$ is collinear with the centres of $\Omega$ and $\omega$, which are $M$ and $J$, respectively. As $K$ and $M$ both lie on the bisector of $\angle BAC$, so does $J$.