Mathematical puzzle on the coordinate planes.

Question

Recently, I come across this quite interesting mathematical puzzle:
Consider the ten points $(0,0)$, $(1,2)$, $(3,3)$, $(4,1)$ and $A, B, C, D, E, F$ on the coordinate plane. It is known that if any five points are chosen from these ten points, there must exist four points which are concyclic. Furthermore, the distance between any two points among $A, B, C, D, E, F$ is not less than x. Find the greatest possible value of x.

Answer 1

Let the points be $P_1, P_2..., P_{10}$. The solution will proceed in 3 steps. Before we start, we shall recall the following fact which will be used extensively. 'Any 3 non-collinear points determine a unique circle. In other words, if two circles intersect at 3 distant points, these 2 circles are identical.'

(1) Five out of the ten points are concyclic
Assume on the contrary that no set of 5 points is concyclic. Consider {$P_1, P_2, P_3, P_4, P_5$}. The given information shows that 4 points from the set are concyclic. Suppose T={$P_1, P_2, P_3, P_4$} is a set of concyclic points. Next consider the choice {$P_1, P_2, P_3, P_5, P_6$}. If U= {$P_1, P_2, P_3, P_5$} is a set of concyclic points, then T and U are on the same circle, which means {$P_1, P_2, P_3, P_4, P_5$} is a set of 5 concyclic points. It contradicts the assumption. With similar reasoning, we can prove the {$P_1, P_2, P_3, P_6$} cannot be concyclic as the set intersects T at 3 points. Therefore, the 4 concyclic points in U must be $P_5, P_6$ and 2 points from {$P_1, P_2, P_3$}. By symmetry on points $P_6, P_7, P_8, P_9$, the same conclusion applies when $P_6$ is replaced by any of $P_7, P_8, P_9$ in the above argument. However, there are only 3 choices of 2 points from {$P_1, P_2, P_3$}, namely {$P_1, P_2$}, {$P_1 P_3$} and {$P_2, P_3$}. Therefore, among the 4 choices of points {$P_1, P_2, P_3, P_5, P_n$} where n=6,7,8 or 9, two of them must have chosen the same set of 2 points from {$P_1, P_2, P_3$} for concyclicity. WLOG, we may assume V={$P_1, P_2, P_5, P_6$} and W={$P_1, P_2, P_5, P_7$} are two sets of concyclic points. It implies that {$P_1, P_2, P_5, P_6, P_7$} is a set of 5 concyclic points, a contradiction. Therefore, there must be 5 concyclic points.

(2) Nine out of ten points are concyclic
In the view of (1), we assume $P_1, P_2, P_3, P_4, P_5$ are concyclic, and that they lie on circle X. We want to show that at least 4 points in the set {$P_6, P_7, P_8, P_9, P_{10}$} also lie on X. Suppose on the contrary that two points, say $P_6$ and $P_7$, are not on X. Consider {$P_1, P_2, P_3, P_6, P_7$}. With similar arguments as in (1), $P_6$ and $P_7$ must be concyclic with exactly 2 points from {$P_1, P_2, P_3$}. Let Y={$P_1, P_2, P_6, P_7$} be concyclic. Next consider {$P_1, P_2, P_4, P_6, P_7$}. Considering circles X and Y, we can see that the only possibility for which $P_6$ and $P_7$ do not lie on X is that Z={$P_3, P_4, P_6, P_7$} is a set of concyclic points. Finally, consider {$P_1, P_3, P_5, P_6, P_7$}. Again, both $P_6$ and $P_7$ cannot be excluded from the set of concyclic points within this choice. However, with these two points in the set, they will be on Y or Z of $P_1$ or $P_3$ is included. Either case shows that $P_6$ is on X, a contradiction.

(3) Finding the greatest possible value of X
Note that the 4 points $(0,0)$, $(1,2)$, $(3,3)$, $(4,1)$ are not concyclic. As 9 out of the 10 given points are concyclic, all 6 points $A, B, C, D, E, F$ should be concyclic with 3 of the 4 known points. When 6 points are on the circle, the most even distribution, namely, being vertices of a circumscribed regular hexagon, gives the greatest minimum distance. It is clear that the minimum distance in this distribution equals the radius of the circle. It is not hard to see that $(0,0)$, $(1,2)$, $(3,3)$ have the greatest circumradius among all choices of 3 out of the 4 known points, which is $\frac{5\sqrt{2}}{2}$.
Therefore the answer is $\frac{5\sqrt{2}}{2}$.