# First year french college. Is there a finite set of points of the plane A. Finite set of point A such as...

**Answer to be proved in details**

Such a finite set of points $A$ can't exist.

Denote by $P_1, P_2, P_3$ three points for which the circumscribed circle $\mathcal C$ is achieving the minimum radius $R$ of all circumscribed circles. Let $O$ be its center. By hypothesis $O \in A$.

**Assumption to be proven:** one of the triangles $(O \ P_1 \ P_2)$, $(O \ P_2 \ P_3)$ or $(O \ P_1 \ P_3)$ has a circumscribed radius strictly less than $R$. A contradiction.

Assume such a set $A$ exists.

Let $X,Y$ be two points in $A$ that are the closest of all pairs of points in $A$.

Look at the bisector $l$ of the segment $XY$. For each point $Z\in A\setminus\{X,Y\}$ we can find the circumcentre of $\triangle XYZ$ and it will be on $l$. Thus, some points in $A$ will be on $l$, and so let's choose the point $O\in l$ so that it is the closest to the line $XY$.

- Case 1: $O$ is on the line $XY$, and so $A$ contains three colinear points - a contradiction.
- Case 2: $O$ is not on the line $XY$. Note that $OX=OY\ge XY$ so the angle $\angle XOY$ (opposite to the shortest side $XY$) is acute, and so all three angles of the isosceles triangle $\triangle OXY$ are acute. This means that the circumcentre $O'$ of $\triangle XOY$ lies inside $\triangle XOY$, and therefore $O'\in l$ and is
*closer*to $XY$ than $O$ - a contradiction.