1959–1966 IMO Longlist Problems/Czechoslovakia 1

Problem

Given $\displaystyle n > 3$ 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 $\displaystyle n$ 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 $\displaystyle A$ and $\displaystyle B$ on this convex hull. Clearly all of the other $\displaystyle {n-2}$ points will lie on the same side of the line $\displaystyle AB$ . Of these other points, we select the point $\displaystyle {C}$ such that the angle $\displaystyle ACB$ is maximized. Then $\displaystyle A,B,C$ satisfy the conditions of the problem, because if there were some point $\displaystyle D$ inside the circle, since it would be on the same side of line $\displaystyle AB$ as $\displaystyle {C}$ , the angle $\displaystyle ADB$ would be greater than the angle $\displaystyle ACB$ , which is a contradiction. Q.E.D.

Resources

Page

Toolbox

Search

1959–1966 IMO Longlist Problems/Czechoslovakia 1

Problem

Solution

Resources