# 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.