Problem 4.
Let ABC be an acute triangle with circumcenter O. Let D be the foot  of the altitude from A to BC and let M be the midpoint of OD. The points Ob and Oc are the circumcenters of triangles AOC and AOB, respectively. If AO=AD, then prove that the points A, Ob, M and Oc are concyclic.

 

Solution
Let $X,Y$ be the reflections of $A$ across $O_c,O_b$ respectively, and $S$ be the midpoint of arc $BC$ in circle $(ABC)$. Note that, $AD=AO=OS$ and $AD \parallel OS,$ hence quadrilateral $ADSO$ is a parallelogram, therefore points $A,M,S$ are collinear and so $M$ is the midpoint of $AS$. We have the following Claim.

Claim: Quadrilateral $AXSY$ is cyclic.
Proof: Note that $\angle AOX=\angle AOY=90^\circ$, and so $O \in XY$. Moreover, if $XY$ intersects $BC$ at point $T$, then $\angle AOT=\angle ADT=90^\circ$, and since $AD=AO$ we obtain that $AT$ bisects $\angle DAO$. Since $AD,AO$ are isogonal, this means $T \in AS$.

To finish, note that

$\angle BXY=\angle BXO=180^\circ-\angle BAO=90^\circ+\angle C=\angle C+\angle ACY=\angle BCY,$

hence $BXCY$ is cyclic. Thus,

$TA \cdot TS=TB \cdot TC=TX \cdot TY,$

as desired $\blacksquare$

Back to the problem, since $AXSY$ is cyclic and $O_c,M,O_b$ are the midpoints of $AX,AS,AY$ respectively, we obtain that $AO_cMO_b$ is cyclic, too, as desired.