Problem 1309

Let the center of the circle be $O$, the vertices of the cyclic quadrilateral be $A_{1}, A_{2}, A_{3}, A_{4}$, and the midpoints of the sides be $B_{1}$, $B_{2}, B_{3}, B_{4}$ as shown in the diagram. One of the diagonals is the diameter of the circle, for example, $A_{2} A_{4}$; the intersection of the diagonals is $M$, which lies on the diameter $A_{2} A_{4}$. The midpoint of $A_{1} A_{3}$ is $K$. The midpoints of the sides, as is well known, form the vertices of a parallelogram: thus, $B_{1} B_{3}$ and $B_{2} B_{4}$ bisect each other at point $I$.

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We will now prove that $I$ is the midpoint of the distance connecting the midpoints of the two diagonals, $OK$. Indeed, in the triangle $A_{1} A_{2} A_{3}$, the line segment connecting the midpoints of $A_{1} A_{2}$ and $A_{1} A_{3}$, $B_{1} K \parallel A_{2} A_{3}$ and $B_{1} K = \frac{1}{2} A_{2} A_{3}$; furthermore, in the triangle $A_{4} A_{2} A_{3}$, the line segment connecting the midpoints of $A_{4} A_{2}$ and $A_{4} A_{3}$, $O B_{3} \parallel A_{2} A_{3}$ and $O B_{3} = \frac{1}{2} A_{2} A_{3}$. Therefore, $B_{1} K \parallel O B_{3}$, so the quadrilateral $B_{1} K B_{3} O$ is a parallelogram, and its diagonal $B_{1} B_{3}$ bisects the diagonal $OK$, meaning $B_{1} B_{3}$ passes through the midpoint of $OK$.

Similarly, $B_{2} O B_{4} K$ is also a parallelogram, and its diagonal $B_{2} B_{4}$ also passes through the midpoint of $OK$. Thus, $B_{1} B_{3}$ and $B_{2} B_{4}$ intersect at the midpoint $I$ of the distance $OK$.

Suppose that the point $M$ is fixed. Then the geometric locus of the point $K$ is a circle with diameter $OM$. Since $OI = \frac{1}{2} OK$, the geometric locus of the point $I$ is also a circle, whose diameter is $\frac{1}{2} OM = ON$, where $N$ is the midpoint of the distance $OM$. This implies that the point $I$ cannot be chosen arbitrarily for the construction to be feasible: the point $I$ must lie on the circle with diameter $ON$.

If $I$ satisfies the above condition, then we draw a perpendicular from $M$ to $OI$: this perpendicular intersects the circle at two points, which are the vertices of the quadrilateral, while the other two vertices are the endpoints of the diameter determined by $OM$.

Note: If the point $I$ is fixed inside the circle $O$, then the geometric locus of the point $M$ is the chord of the circle that is perpendicular to $OI$ and whose distance from $O$ (in the direction of $OI$) is $2OI = OK$.

For any quadrilateral, it is also true that $I$ is the midpoint of the distance connecting the midpoints of the two diagonals, as can be seen from the proof. In the proof, we cannot use anything that the quadrilateral is cyclic.

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However, in the case of a cyclic quadrilateral, the midpoints $K$ and $L$ of the diagonals always lie on the circle with diameter $OM$. Therefore, $I$ must lie inside this circle. If the circle $O$ is already given, along with $M$, and $I$ is such that it lies inside the circle with diameter $OM$, and the center of this circle is $N$, then we draw a perpendicular to $NI$ at point $I$, which intersects the circle $(N)$ at points $K$ and $L$. The lines $MK$ and $ML$ will be the diagonals of a cyclic quadrilateral. This cyclic quadrilateral satisfies the conditions that $M$ is the intersection of the diagonals and $I$ is the intersection of the lines connecting the midpoints of the opposite sides.