National Olympiad Final Round

We show that $d_6\le 1\le d_7$. For the first inequality, assume arbitrary six points $A_1,A_2,A_3,A_4,A_5,A_6$ being marked in the circle. Let the centre of the circle be $O$. If $A_i=O$ for some $i$, the distance between $A_i$ and any other marked points is at most $1$. Assume in the rest that $A_i=O$ for no $i$. Let $\alpha$ be the smallest angle that arises between some two rays $O{A}_i$ and $O{A}_j$, where $i,j=1,2,3,4,5,6$ (Fig. 26). Fig. 26 is $360^{\circ }$. If $A_i{A}_j>1$ then $A_i{A}_j$ is the largest side of the triangle $O{A}_i{A}_j$, as the lengths of $O{A}_i$ and $O{A}_j$ do not exceed $1$. The angle opposite to the longest side is the largest, whence $\alpha$ should be larger than $60^{\circ }$, contradiction. Thus there exist two marked points at distance at most $1$ from each other. As the choice of the points was arbitrary, this establishes $d_6\le 1$.

On the other hand, when marking the vertices of a regular hexagon inscribed into the circle together with the centre of the circle, the distance between any two consecutive marked points on the circumference is equal to $1$ and their distance from the remaining point is also $1$. Hence $d_7\ge 1$.