Saudi Arabian Mathematical Competitions

1) We will use the inversion to solve this problem.
Note that $\triangle AHI \sim \triangle AIB$ and that $\triangle AKI \sim \triangle AIC$, hence
$$AI^{2} = AH \cdot AB = AK \cdot AC =: k.$$
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, $AM^{2} = AN^{2} = k = AI^{2}$; 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 $\triangle AIM \sim \triangle BHL$ and $\triangle AOM \sim \triangle CBL$.

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 $B', C'$ the midpoints of $AB, AC$ and by $Z$ the intersection of $B'C'$ and $AI$.

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

On the other hand, $B'C' \parallel BC$, hence $I$ is the midpoint of $BC$.
Let $T$ be the midpoint of the segment $OI$. Then $XT \parallel B'I \parallel AC$. But $OC' \perp AC$, thus $OY \perp XT$.
Similarly, $OX \perp YT$. Therefore, $T$ is the orthocenter of triangle $XOY$.