IMO2024 Shortlisted Problems

Solution 1. Let $\Gamma$ be circle $ABC$ and $\omega$ be circle $AYZ$. Let $O,M$, and $S$ be the centres of $\Gamma ,\Omega$, and $\omega$, respectively. Let $AK$ intersect $\Gamma$ again at $P$, and let the angle bisector of $\angle ZAY$ intersect $\omega$ again at $N$.



By power of a point from $K$ to $\Gamma$ and $\Omega$, we have that $KA\cdot KP=KB\cdot KC=KY\cdot KZ$, so $P$ also lies on $\omega$. The pairwise common chords of $\Gamma ,\Omega$, and $\omega$ are then $AP\perp OS,BC\perp OM$, and $YZ\perp MS$, so we have that $\angle OMS=\angle CKY=\angle YKA=\angle MSO$. As $M$ lies on $\Gamma$ and $OM=OS$, $S$ also lies on $\Gamma$. Note that $N$ lies on $MS$ as $NY=NZ$, so $$\angle PAN=\frac{1}{2}\angle PSN=\frac{1}{2}\angle PSM=\frac{1}{2}\angle PAM.$$ Thus, $AN$ bisects $\angle PAM$ in addition to $\angle ZAY$, which means that $\angle ZAK=\angle IAY$ as $K$ lies on $AP$ and $I$ lies on $AM$.

Solution 2. Define $M$ and $P$ as in Solution 1, and recall that $AYPZ$ is cyclic. Let $Q$ be the second intersection of the line parallel to $BC$ through $P$ with circle $ABC$ and let $J$ be the incentre of triangle $APQ$.



Since $PQ$ is parallel to $BC$ and $\angle BAP<\angle PAC$, the angle bisector of $\angle APQ$ is parallel to the angle bisector of $\angle AKC$. Hence, $PJ$ is parallel to $YZ$. As $M$ is the midpoint of $\text{overparen}PQ$ on circle $APQ$, we have that $MP=MJ$. Then since segments $YZ$ and $PJ$ are parallel and have a common point $M$ on their perpendicular bisectors, $PJYZ$ is cyclic with $JY=PZ$. It follows that $J$ also lies on circle $AYPZ$ and that $\angle ZAP=\angle JAY=\angle IAY$.

Solution 3. As in the previous solutions, let $M$ be the centre of $\Omega$. Let $L$ be the intersection of $AM$ and $BC$, and let $L^{\prime }$ be the reflection of $L$ over $YZ$. Let the circle $MYZ$ intersect $AM$ again at $T$.



Note that as $M$ is the midpoint of $\text{overparen}BC$ on circle $ABC$ and $L$ is the foot of the bisector of $\angle BAC$, we have that $MA\cdot ML=M{I}^2=M{Y}^2$. It follows by power of a point that $MY$ is tangent to circle $ALY$, so $\angle LAY=\angle LYM$. Using directed angles, we then have that $$\text{Varangle}AYT=\text{Varangle}MTY-\text{Varangle}MAY=\text{Varangle}MZY-\text{Varangle}LYM=\text{Varangle}ZYM-\text{Varangle}LYM=\text{Varangle}ZYL=\text{Varangle}{L}^{\prime }YZ,$$ where we use the fact that $MY=MZ$ and that $L$ and $L^{\prime }$ are symmetric about $YZ$. Thus, $YT$ and $Y{L}^{\prime }$ are isogonal in $\angle AYZ$. Analogously, we have that $ZT$ and $Z{L}^{\prime }$ are isogonal in $\angle YZA$. This means that $T$ and $L^{\prime }$ are isogonal conjugates in triangle $AYZ$, which allows us to conclude that $\angle ZAK=\angle IAY$ since $L^{\prime }$ lies on $AK$ and $T$ lies on $AI$.