View source for 1959–1966 IMO Longlist Problems/Czechoslovakia 1

== Problem ==

Given <math>\displaystyle n > 3 </math> points in the plane such that no three of the points are collinear.  Does there exist a circle passing through (at least) 3 of the given points and not containing any other of the <math>\displaystyle n </math> points in its interior?

== Solution ==

The answer is yes.

Since any finite set of at least three coplanar points is contained by a [[convex hull]] with vertices in the set of points, we can select adjacent points <math>\displaystyle A </math> and <math>\displaystyle B </math> on this convex hull.  Clearly all of the other <math>\displaystyle {n-2} </math> points will lie on the same side of the line <math>\displaystyle AB </math>.  Of these other points, we select the point <math>\displaystyle {C} </math> such that the angle <math> \displaystyle ACB </math> is maximized.  Then <math>\displaystyle A,B,C </math> satisfy the conditions of the problem, because if there were some point <math>\displaystyle D </math> inside the circle, since it would be on the same side of line <math>\displaystyle AB </math> as <math>\displaystyle {C} </math>, the angle <math>\displaystyle ADB </math> would be greater than the angle <math>\displaystyle ACB </math>, which is a contradiction.  Q.E.D.

== Resources ==
* [[1959&ndash;1966 IMO Longlist Problems]]


[[Category:Olympiad Combinatorics Problems]]