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 $\triangle XYZ \cong \triangle OKA$, 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
$$TA^2 - TN^2 = HA^2 - HN^2.$$
Let $R$ be the radius of the circle $(O)$, we have $AH^2 = (2OM)^2 = 2R^2$. It is easy to see that $OBNC$ is a square, so $ON = 2OM = AH$, so $AHNO$ is a parallelogram and $HN = AO$. Hence
$$HA^2 - HN^2 = HA^2 - AO^2 = R^2.$$
We also have
$$\begin{align*} TA^2 - TN^2 &= (TO^2 + OA^2) - (TM^2 + MN^2) \\ &= (TM^2 + OM^2 + OA^2) - (TM^2 + MN^2) \\ &= OA^2 = R^2. \end{align*}$$
From this it follows that $TA^2 - TN^2 = HA^2 - HN^2$ or $TH \perp AN$, which leads to $TH \perp IK$. This finishes the proof.