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.