SAUDI ARABIAN IMO Booklet 2023

Let $Z$, $M$ be the midpoints of $BR$, $BC$ respectively, then $ZY$ is the median of the triangle $BCR$ so $ZY=\frac{BC}{2}$ and $ZY\parallel BC$.



It is easy to see that $AH=2OM=BC$ since triangle $OBC$ is right-angled at $O$. Hence $AK=\frac{BC}{2}=ZY$. Let $L$ be the intersection of $BS$, $CR$. Obviously the triangles $ACR$, $ABS$ are right, so $L$ is the orthocenter of the triangle $ARS$. So $L\in (O)$ and $AL$ is the diameter of $(O)$.

Considering the triangle $ARS$ has $\angle RAS=45^{\circ }$ so similar to above, we have $AO=\frac{AL}{2}=\frac{RS}{2}=ZX$ (because $ZX$ is the median of the triangle $BRS$). We also have $ZX\perp AO$ and $ZY\perp AK$ so $\angle XZY=\angle OAK$ leads to $△XYZ\cong △OKA$, and these two triangles have corresponding sides perpendicular, so $XY\perp OK$.

Now let $N$ be the midpoint of $RS$, we will prove that $I$ is the midpoint of $HN$. Since $N$ is the center of the circle passing through $B,C,R,S$, then $NX\perp XL$, $NY\perp YL$, so the points $X,Y,N,L$ belong to the circle of diameter $LN$. Let $D_F$ be the center symmetry $F$ where $F$ is the midpoint of the line segment $XY$. Thus $D_F:X\leftrightarrow Y$. But we have $MX=YN=\frac{CS}{2}$ and $MX\parallel YN\parallel CS$ so $MXNY$ is a parallelogram and $D_F:N\leftrightarrow M$.

Assuming ${\mathcal{D}}_F(L)=G$ then the points $M,G,X,Y$ belong to the circle of diameter $MG$. Otherwise, $MF$ is the median of the triangle $LHG$, followed by $HG\parallel FM$, which $FM\perp BC$ should be $G\in HD$. Infer $\angle GDM=90^{\circ }$ so $D$ also belongs to the circle of diameter $MG$.

Therefore, the center $I$ of $(DXY)$ is the midpoint of $MG$. Also, since $HG=2MF=MN$ and $HG\parallel MN$, $HGNM$ is a parallelogram, so $I$ is also the midpoint of $HN$. Thus $IK\parallel AN$ and we need to prove that $TH\perp AN$. By the four-point theorem, we need to show that $$T{A}^2-T{N}^2=H{A}^2-H{N}^2.$$ Let $R$ be the radius of the circle $(O)$, we have $A{H}^2=(2OM{)}^2=2R^2$. It is easy to see that $OBNC$ is a square, so $ON=2OM=AH$, so $AHNO$ is a parallelogram and $HN=AO$. Hence $$H{A}^2-H{N}^2=H{A}^2-A{O}^2=R^2.$$ We also have $$\begin{array}{rl}T{A}^2-T{N}^2& =(T{O}^2+O{A}^2)-(T{M}^2+M{N}^2)\\ & =(T{M}^2+O{M}^2+O{A}^2)-(T{M}^2+M{N}^2)\\ & =O{A}^2=R^2.\end{array}$$ From this it follows that $T{A}^2-T{N}^2=H{A}^2-H{N}^2$ or $TH\perp AN$, which leads to $TH\perp IK$. This finishes the proof.