The South African Mathematical Olympiad Third Round

Label the sixteen points $A$, $B$, $C$, $\dots$, $P$, as shown below in Figure 1:



A selection of four points satisfying the condition that the distance between every two of the four points is greater than $2$, will be called a valid selection.

Let us first consider a valid selection which includes one of the points of the inner square $FGKJ$. Clearly, only one of these points ($F$, $G$, $K$, $J$) can be used; say we select $F$. The only points from the outer square $ABCDHLPONMIE$ at distance more than $2$ from $F$, are $D$, $L$, $P$, $O$ and $M$. If we choose either $L$ or $O$, then there are not enough points left among the remaining ones on the outer square to form a valid selection. We are therefore forced to select $D$, $P$ and $M$, together with $F$, to obtain a valid selection. Similarly, by symmetry, there are three further valid selections that contain points from the inner square: $\{G,P,M,A\}$, $\{K,M,A,D\}$ and $\{J,A,D,P\}$.

All that remains, is to consider valid selections using only points from the outer square. We note that the only way to choose two points in a valid selection from the same side of the outer square, is to choose two corner points, such as $A$ and $D$. But then the only option for the other two points in the valid selection would be to choose the other two corner points of the outer square, $P$ and $M$, giving the valid selection $\{A,D,P,M\}$. Moving away from corner points, leaves us with the two remaining valid selections (where only one point from each of the sides of the outer square is selected), namely $\{B,H,O,I\}$ and $\{C,L,N,E\}$.

Hence, there are seven possible valid selections.