SELECTION EXAMINATION

Let $c_1,c_2,c_3,c_4$ be the circumcircles of the triangles $KLS,LMT,MNS$ and $NKT$, respectively. It is easy to prove that the quadrilaterals $KLMN,TLSN$ and $KSMT$ are parallelograms and so $KM,NL,TS$ will pass through the same point $G$ and therefore the points $K,S,L$ are symmetric of the points $M,T,N$, respectively, with respect to a symmetry with center point $G$. Let $c_1^{\prime }$ be the circumcircle of the triangle $MTN$. Then $c_1^{\prime }$ passes through the center $O$ of the circle $c(O,R)$ and it has diameter $OD=R$ (since $OMD=OND=90^{\circ }$). However the circles $c_1$ and $c_1^{\prime }$ are symmetric with respect to $G$ (since the triangles defined the two circles are point to point symmetric with respect to $G$). Hence the circle $c_1$ has radius $\frac{R}{2}$ and is passing through $O^{\prime }$ (the symmetric point of $O$ with respect to $G$).

Figure 6

Similarly we prove that the circles $c_2,c_3,c_4$ have radius $\frac{R}{2}$ and are passing through $O^{\prime }$. Therefore the centers of the circles $c_1,c_2,c_3,c_4$ belong to the circle with center $O^{\prime }$ and radius $\frac{R}{2}$. The centers of the circles $c_1^{\prime },c_2^{\prime },c_3^{\prime }$ and $c_4^{\prime }$ (which are the midpoints of the segments $OA,OB,OC$ and $OD$, respectively) define a quadrilateral with sides parallel to the corresponding sides of $ABCD$. Hence the centers of the circles $c_1,c_2,c_3,c_4$ (symmetric of the centers of the circles $c_1^{\prime },c_2^{\prime },c_3^{\prime },c_4^{\prime }$) define a quadrilateral similar to $ABCD$.

---

Alternative solution.

We will use the characteristic properties of the Euler circle of a triangle.

Figure 7

The center $O^{\prime }$ of the Euler circle of a triangle belongs to Euler's line of the triangle. It is the midpoint of the segment $OH$ (where $O$ is the circumcentre and $H$ is the orthocenter of the triangle). The radius of the Euler's circle is the half of the radius of the circumcircle of the triangle. From the data of the problem we conclude that the circumcircles of the triangles $KLS$, $LMT$, $MNS$ and $NKT$ ($c_1,c_2,c_3,c_4$, respectively), are the Euler's circles of the triangles $ABC$, $BCD$, $CDA$ and $ABD$. The triangles $ABC$, $BCD$, $CDA$ and $ABD$ are inscribed in the same circle $c(O,R)$. Hence the circles $c_1,c_2,c_3,c_4$ are equal with radius $\frac{R}{2}$. We consider the triangles $ABC$ and $ABD$. In the triangle $ABC$ we consider the centroid $G_1$ (point of intersection of the medians $BS$ and $CK$) and the center $O_1$ of the circle $c_1$ (circumcircle of the triangle $KLS$ and Euler's circle of the triangle $ABC$). In the triangle $ABD$ we consider the centroid $G_4$ and the center $O_4$ of the circle $c_4$ (circumcircle of the triangle $KNT$ and Euler's circle of the triangle $ABD$). Since $G_1$ and $G_4$ are centroids we have: $$\frac{G_{1}K}{G_{1}C}=\frac{G_{4}K}{G_{4}D}=\frac{1}{2}\Rightarrow G_1{G}_4\parallel CD.$$ Since the points $O_1$ and $O_4$ are the centers of the Euler's circles we have: $$\frac{O{G}_1}{O{O}_1}=\frac{O{G}_4}{O{O}_4}=2\Rightarrow G_1{G}_4\parallel O_1{O}_4.$$ Therefore we have: $CD\parallel O_1{O}_4$. Similarly we prove that the other sides of the quadrilateral $O_1{O}_2{O}_3{O}_4$ are parallel to the corresponding sides of the quadrilateral $ABCD$.