SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) We will use the inversion to solve this problem. Note that $△AHI\sim △AIB$ and that $△AKI\sim △AIC$, hence $$A{I}^2=AH\cdot AB=AK\cdot AC=:k.$$ Let $f$ be the inversion with center $A$ and power $k$. Then $$f(I)=I,f(H)=B,f(B)=H,f(K)=C,f(C)=K.$$ So this inversion sends the line $HK$ to $(O)$, which implies that the intersection of $HK$ and $(O)$ are fixed points under $f$. Therefore, $A{M}^2=A{N}^2=k=A{I}^2$; therefore, $A$ is the center of the circle $(IMN)$.



2) First, we will prove the following lemma: Let $D$ be a point on the segment $BC$ and $H$ be the projection of $D$ on $AC$. Suppose that $I$ is the circumcenter of triangle $ABH$. Denote by $M,N$ the midpoints of $AB,AC$ respectively. Let $MN$ meet $AD$ at $K$. Then $IK\perp AC$.

Denote by $L$ the projection of $B$ on $AC$. It is easy to see that $I,M,O$ are collinear. Further, $$\angle AIM=\angle LHB,\angle AOM=\angle LCB,$$ hence $△AIM\sim △BHL$ and $△AOM\sim △CBL$.



It follows that $$\frac{IM}{IO}=\frac{LH}{LC}=\frac{BD}{BC}=\frac{MK}{MN};$$ which implies that $IK\parallel ON$, i.e. $IK\perp AC$.

Going back to the original problem, we denote by $B^{\prime },C^{\prime }$ the midpoints of $AB,AC$ and by $Z$ the intersection of $B^{\prime }{C}^{\prime }$ and $AI$.



Applying the lemma, we have $XZ\parallel O{C}^{\prime }$ and $YZ\parallel O{B}^{\prime }$; therefore, $OXZY$ is a parallelogram. Hence, $OZ$ passes through the midpoint of $XY$. But $XY\parallel B^{\prime }{C}^{\prime }$, so $Z$ is the midpoint of $B^{\prime }{C}^{\prime }$ and $XY$ is a medial line of triangle $O{B}^{\prime }{C}^{\prime }$.

On the other hand, $B^{\prime }{C}^{\prime }\parallel BC$, hence $I$ is the midpoint of $BC$. Let $T$ be the midpoint of the segment $OI$. Then $XT\parallel B^{\prime }I\parallel AC$. But $O{C}^{\prime }\perp AC$, thus $OY\perp XT$. Similarly, $OX\perp YT$. Therefore, $T$ is the orthocenter of triangle $XOY$.