Saudi Arabia booklet 2024

1) Let $T$ be the intersection of the tangent at $A$ of $(O)$ with $BC$. Then, because $ABDC$ is a harmonic quadrilateral, $TD$ is also tangent to $(O)$. It follows that $A,D,G$ both belong to the circle of center $T$, which is also the Apollonius circle at vertex $A$ of triangle $ABC$. On the other hand, $(T)$ and $(O)$ are orthogonal so $OA,OD$ are tangent to $(T)$, then it is easy to check that $GO$ is the symmedian of triangle $ADG$. Let $K$ be the second intersection of $AG$ and $(O)$. Then, $$\angle BEK=\angle BAK=\angle BGK⟹KE=KG.$$ Similarly, $KF=KG$ so $KE=KF$ implies that $K$ is the midpoint of the minor arc $EF$ of $(O)$. It follows that $AG$ is the angle bisector of $\angle EAF$.

Let $M$ be the midpoint of $BC$ and take $A^{\prime }\in (O)$ such that $A{A}^{\prime }\parallel BC$ then since $M{A}^{\prime }=MA=MG$ implies that $G,M,A^{\prime }$ are collinear. On the other hand, since $$A^{\prime }(AD,BC)=-1\text{ and }A{A}^{\prime }\parallel BC,$$ we have $A^{\prime },M,D$ are collinear. From there, it can be deduced that $GD$ passes through $A^{\prime }$. Since $K,O,A^{\prime }$ are also collinear, $A^{\prime }$ is the midpoint of the major arc $EF$ of the circle $(O)$. Then, $K$ is the center of the circumcircle of $GEF$ and $A^{\prime }E\perp KE$, $A^{\prime }F\perp KF$ so $A^{\prime }$ is the intersection point of two tangent lines of $(K)$ at $E,F$. Therefore, $GD$ is the symmedian of triangle $GEF$.

2) We will prove that $D$ is the tangent point of the external Mixlinear circle with respect to vertex $A$ in triangle $AEF$. Indeed, let $D^{\prime },U,V$ be the tangent points of the external Mixtilinear circle $(\omega )$ of triangle $AEF$ with $(O)$, $AE,AF$ respectively. Then, $G$ is the midpoint of $UV$, and $D^{\prime }U,D^{\prime }V$ respectively pass through the midpoints of the arc $AE$ (contains $F$) and $AF$ (contains $F$) of $(O)$. Note that $$\angle ACB=\angle GCB=\angle BCF=\angle BAF$$ so $B$ is the midpoint of the arc $AE$ (contains $F$) of $(O)$ so $B,D^{\prime },V$ are collinear. Similarly, $C,D^{\prime },U$ are collinear. Hence, we have $UV\parallel BC$. Since $(\omega )$ and $(O)$ are tangent, so we have $$\angle F{D}^{\prime }V=\angle FC{D}^{\prime }+\angle D^{\prime }UV=\angle FC{D}^{\prime }+\angle D^{\prime }CB=\angle BCF=\angle FGV$$ implies that $D^{\prime }FVG$ is cyclic. Similarly $D^{\prime }EUG$ is also cyclic. Let $D^{\prime }$ be the opposite ray of $D^{\prime }G$ then $$\angle F{D}^{\prime }x=\angle FVG=\angle AVU=\angle AUV=\angle EDx$$ so $D^{\prime }$ passes through the midpoint of the major arc $EF$ of $(O)$, which is $A^{\prime }$. From here, $D\equiv D^{\prime }$. Finally, let $I$ be the incenter of triangle $AEF$ and consider $\Omega$ to be the inversion of center $A$ and power $AE\cdot AF$, union to the reflection about $AG$. Then, according to the properties of this transformation which preserve the tangency and the isogonal properties, one can get: $$\begin{array}{c}\Omega :(I)\to (\omega )\text{ and }\Omega :BC\leftrightarrow (O)\\ ⟹\Omega :J\leftrightarrow D.\end{array}$$