11.3 Convex Polygons 185In the sequel, for the sake of brevity we will say that certain points are
ingeneral position, if no three of them are on a line.
Imagine that our five points (in general position) are drawn on the black-
board. Let us hammer nails into the points (only in imagination!) and
stretch a rubber band around them. The rubber band encloses a convex
polygon, the vertices of which are among the original points, and the other
points are inside of this polygon. This polygon is called theconvex hullof
the points (Figure 11.8).
FIGURE 11.8. The convex hull of a set of points in the plane.Using this construction, it is quite easy to solve our problem. What can
we get as the convex hull of our five points? If this is a pentagon, then any
four of these points form a convex quadrilateral, and we are done. If the
convex hull is a quadrilateral, then this quadrilateral itself is the desired
convex quadrilateral. So we may assume that the convex hull is a triangle.
Let’s denote the vertices of this triangle byA, B, andC, and the two other
given points byDandE. The pointsDandEare inside the triangleABC,
so the line throughDandEintersects the circumference of the triangle in
two points. Assume that these two points are on the sidesABandAC(if
this is not the case, then we can relabel the vertices of the triangle). Now
you can easily show that the pointsB, C, D, andEform the vertices of a
convex quadrilateral (see Figure 11.9).
The next fact in the line concerns convex pentagons.
If we have 9 points in the plane in general position, then one
can choose five points among them that are the vertices of a
convex pentagon.
The proof of this statement is too lengthy to give it as an exercise, but
the reader may try to work out a proof.
11.3.1Draw eight points in the plane in such a way that no five of them span
a convex pentagon.
One can give 16 points in general position in the plane such that no 6
of them form a convex hexagon. Nobody has been able to construct 17