Home > Analysis, Combinatorics, Geometry, Olympiad > Arcs on a circle

Arcs on a circle


Suppose that we have a finite set of arcs on a circle, with the property that every two of them intersect. Prove that there exists a diameter which intersects all arcs.

Proof: Take a fixed diameter AB and project all arcs on that diameter. Every two such projections must intersect, therefore using Helly’s theorem, the intersection of all such projections, P is not empty. Since each of the projectioins is a line segment, their intersection is also a line segment. If the intersection contains the center of the circle, then the diameter perpendicular to AB solves our problem. If not, we use a continuity argument to reduce the problem to the preceeding case.

The argument is as follows: Imagine that we rotate AB such that AB still contains the center (i.e. it still remains a diameter). Then the projection P will initially ‘move’ towards or outwards the center of the circle, depending on the direction of the rotation. We choose the direction such that the projection moves towards the center of the circle. The projection moves continuously on the rotated diameter AB. Now, when we have made a rotation of angle \pi, the diameter AB will coincide with the initial diameter, with endpoints switched. But now, the two rays composing the diameter have interchanged, i.e. OA \mapsto OB, OB \mapsto OA, so the projection has moved from a ray to another one. The continuity of the movement implies that at some moment, the projection P must contain the center of the circle, and we are done.

Another argument: (See: Vasile Pop, Geometrie Combinatorica, a very good Romanian book with problems concerning Combinatorial Geometry) Denote I_0 the projection, and suppose that I_0 does not contain the center O; wlog suppose that I_0 \subset OA. As we rotate the diameter AB with angle \alpha, denote for each \alpha with I_\alpha the new projection on the rotated diameter. Suppose that at each angle \alpha the center does not belong to any of the projections. By the continuity of the movement, we see that I_\alpha must belong to the same ray OA as we rotate the diameter AB. Then I_\pi is on the same diameter as the initial I_0, which would imply that I_\pi=I_0. But this is a contradiction, since I_\pi cannot be on the same radius as I_0 due to the continuity.

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: