Austrian Mathematical Olympiad

a. We denote the reflections of $A$, $B$, $C$ and $D$ with $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ resp. $D^{\prime }$ and we denote the intersection of the diagonals with $S$. Since the points $A$ and $C$ are reflected in the same line $BD$ and the point $S$ remains invariant under this reflection, the whole line $ASC$ becomes $A^{\prime }S{C}^{\prime }$ after reflection in $BD$. Analogously, the line $BSD$ becomes $B^{\prime }S{D}^{\prime }$ after reflection in $AC$. If we denote the smaller angle between the two diagonals by $\phi$, these two actions on the lines correspond to a rotation of the line $AC$ with center $S$ in direction $BD$ with rotation angle $2\phi$ and a rotation of the line $BD$ with center $S$ in direction $AC$ with rotation angle $2\phi$. Therefore, the angle between the lines $A^{\prime }S{C}^{\prime }$ and $B^{\prime }S{D}^{\prime }$ is the angle $3\phi$. This has to be a multiple of $180^{\circ }$, so that the original angle has to be $0^{\circ }$ or $60^{\circ }$. The first case is not possible since the points of the inscribed quadrilateral cannot lie on a line. We obtain that the four new points lie on a line if and only if the diagonals of the given inscribed quadrilateral make an angle of $60^{\circ }$.

Figure 3: Problem 17

b. Since the reflections do not only preserve the collinearity of $ASC$ and $BSD$, but also the position of $S$ between the two points and the distances to the two points, we want to use the power of $S$ with respect to the circle $ABCD$.

Because of the reflections, we have $$SA=S{A}^{\prime },SB=S{B}^{\prime },SC=S{C}^{\prime },SD=S{D}^{\prime }$$

and since $ABCD$ is an inscribed quadrilateral, we have $$SA\cdot SC=SB\cdot SD.$$ Therefore, we obtain $$S{A}^{\prime }\cdot S{C}^{\prime }=SA\cdot SC=SB\cdot SD=S{B}^{\prime }\cdot S{D}^{\prime }.$$ Since the two lines $A^{\prime }S{C}^{\prime }$ and $B^{\prime }S{D}^{\prime }$ do not coincide in this case, we can apply the properties of the power of a point in reverse, and we get that $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ and $D^{\prime }$ lie on a circle.