BMO 2017

We start with a lemma. Lemma. If the triangle $ABC$ 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 $ABC$ is acute, $O$ is in the interior. Consider the line that passes through $O$ and is parallel to $BC$. 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 $l_A$ 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 $l_A$ as $A$. Now, consider the perpendicular bisector of $BC$ and assume that $P$, w.l.o.g, is on the same side of this line as $C$. Under these circumstances, $|PB|\ge |OB|$. Thus, the smaller circle centered at $P$ must exclude $B$. In the obtuse case, let $\angle BAC\ge 90^{\circ }$. Then $BC$ is the longest side. The circle with diameter $BC$ contains all three vertices. Therefore, $r(A,B,C)$ is not greater than $\frac{1}{2}|BC|$. 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 $A_1,A_2,...,A_n$ such that the circumcircle of $ABC$ 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 $AB$ 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 $ABC$ contains all of $A_1,A_2,...,A_n$. 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, $$r(A_1,A_2,...,A_n)=\text{circumradius of }{A}_i{A}_j{A}_k=r(A_i,A_j,A_k).$$ If $A_i{A}_j{A}_k$ is an obtuse triangle, let $A_i$ 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 $$r(A_1,A_2,...,A_n)=\frac{1}{2}|A_j{A}_k|=r(A_i,A_j,A_k)$$ 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 $\angle A_{j}X{A}_k$ 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, $\angle A_{j}X{A}_k\ge \angle A_j{A}_i{A}_k\ge 90^{\circ }$. 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.