We start with a lemma.
Lemma. If the triangle $A B C$ is acute, $r(A, B, C)$ is its circumradius and if it is obtuse, $r(A, B, C)$ is half the length of its longest side.

## Proof.

Let us do the acute case first. The circumcircle contains the vertices, so $r(A, B, C)$ is not greater than the circumradius. Now, let us prove that no smaller circle contains all three vertices. If there is a smaller circle, let its center be $P$. Further, let the circumcenter be $O$.

Since $A B C$ is acute, $O$ is in the interior. Consider the line that passes through $O$ and is parallel to $B C$. Let us call it $l_{A}$ and define $l_{B}$ and $l_{C}$ similarly. Now, consider the set of points that are on the opposite side of $I_{A}$ with respect to $A$. Call this set $S_{A}$ and define $S_{B}$ and $S_{C}$ similarly. It is easily seen (by geometry) that $S_{A} \cap S_{B} \cap S_{C} = \varnothing$. As such, assume $P \notin S_{A}$ without loss of generality. That is to say, $P$ is on the same side of $I_{A}$ as $A$. Now, consider the perpendicular bisector of $B C$ and assume that $P$, w.lo.g, is on the same side of this line as $C$. Under these circumstances, $|P B| \geqslant |O B|$. Thus, the smaller circle centered at $P$ must exclude $B$.
In the obtuse case, let $\measuredangle B A C \geq 90^{\circ}$. Then $B C$ is the longest side. The circle with diameter $B C$ contains all three vertices. Therefore, $r(A, B, C)$ is not greater than $\frac{1}{2}|B C|$. But any smaller circle will clearly exclude at least one of $B$ and $C$.
Now, let us return to the original problem. Note that there must be points $A, B, C$ among $A_{1}, A_{2}, \ldots, A_{n}$ such that the circumcircle of $A B C$ contains all $n$ points. One can see this as follows: First start with a large circle that contains all $n$ points. Then shrink it while keeping the center fixed, until one of the $n$ points is on the circle and call this point $A$. Then shrink it keeping the point $A$ in place and moving the center closer to $A$, until another point $B$ is on the circle. Then keep the line $A B$ fixed while moving the center toward it or away from it so that another $C$ among the $n$ points appears on the circle. It is easy to see that this procedure is doable.
Consider all such triples $A, B, C$ such that the circumcircle of $A B C$ contains all of $A_{1}, A_{2}, \ldots, A_{n}$. Now choose the one among them with the smallest circumradius and let it be $A_{i}, A_{j}, A_{k}$. If $A_{i} A_{j} A_{k}$ is an acute triangle, any smaller circle will exclude one of $A_{i}, A_{j}, A_{k}$ by the lemma above. Therefore,

$$r\left(A_{1}, A_{2}, \ldots, A_{n}\right) = \text{ circumradius of } A_{i} A_{j} A_{k} = r\left(A_{i}, A_{j}, A_{k}\right)$$

If $A_{i} A_{j} A_{k}$ is an obtuse triangle, let $A_{l}$ be its obtuse angle. We wish to prove that the circle with diameter $A_{j} A_{k}$ contains all $n$ points. This will mean that

$$r\left(A_{1}, A_{2}, \ldots, A_{n}\right) = \frac{1}{2} \left|A_{j} A_{k}\right| = r\left(A_{i}, A_{j}, A_{k}\right)$$

and we will be done. If there are no points on the opposite side of $A_{j} A_{k}$ w.r.t. $A_{i}$, then this assertion is clear. If there are some points on that side, choose the one $X$ such that $\measuredangle A_{j} X A_{k}$ is smallest possible. Then the circumcircle of $A_{j} X A_{k}$ contains all $n$ points. However, by the choice of $A_{i}$, the circumradius of $A_{j} X A_{k}$ cannot be less than that of $A_{i} A_{j} A_{k}$. Thus, $\measuredangle A_{j} X A_{k} \geq \measuredangle A_{j} A_{i} A_{k} \geq 90^{\circ}$. As such, the circle with diameter $A_{j} A_{k}$ contains all $n$ points.
!

Figure 1: The circumcircles of $A_{i} A_{j} A_{k}$ and $A_{j} X A_{k}$ as well as the circle with diameter $A_{j} A_{k}$ are shown.

Remark. The problem selection committee recommended formulation of the task to improve.

## C 5

We have $n$ students sitting at a round table. Initially each student is given one candy. At each step each student having candies either picks one of its candies and gives it to one of its neighbouring students, or distributes all of its candies to its neighbouring students in any way he wishes. A distribution of candies is called legal if it can be reached from the initial distribution via a sequence of steps.
Determine the number of legal distributions. (All the candies are identical.)