Selected Problems from the Final Round of National Olympiad

Since $AB$, $BC$, $CD$, and $DA$ are diameters (Fig. 9), $AOB$, $BOC$, $COD$, and $DOA$ are right angles. Hence $AOC$ and $BOD$ are straight angles, i.e., the diagonals of the quadrangle meet at $O$. The circles drawn on the opposite sides $AB$ and $CD$ cannot have common points besides $O$, since otherwise one circle would pass through three collinear points (two vertices of the quadrangle and the point $O$). Consequently, the circles drawn on $AB$ and $CD$ must touch at $O$. Let $P$ and $Q$ be the centers of these circles, respectively; then $O$ lies on the segment $PQ$.

Fig. 9

$|AP|=|PO|$ and $|CQ|=|QO|$, isosceles triangles give $\angle BAC=\angle PAO=\angle POA=\angle QOC=\angle QCO=\angle DCA$. Thus $AB$ and $CD$ are parallel. Similarly, the remaining sides are parallel. Thus $ABCD$ is a rectangle; the diagonals of a rectangle are perpendicular only if the rectangle is a rhombus.

Fig. 9