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Exercise from Matousek Lectures in Discrete Geometry

Prove that each set $X \subset \mathbb{R}^2$ of diameter at most 1 (i.e., any two points have distance at most 1) is contained in some disc of radius $\frac{1}{\sqrt 3}$

I'm just interested in the proof for 3 points. I wanted to argue that we take one point and then set the next point a unit distance anyway and then see what constraints this produces on the third point. But that doesn't consider cases where the distance between $p_1, p_2 < 1$

Ainsley Pullen
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1 Answers1

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I would go through Helly's theorem. For any $P\in X$, let $\Gamma_P$ a circle with radius $\frac{1}{2}$ and centre at $P$. For any $P,Q\in X$ we have that $\Gamma_P\cap\Gamma_Q\neq\emptyset$. If we enlarge the radii of such disks, we have that for any $P,Q,R\in X$ the disks centered at $P,Q,R$ with radii $\frac{1}{\sqrt{3}}$ intersect somewhere. By Helly's theorem, there is a point at a distance $\leq\frac{1}{\sqrt{3}}$ from every point of $X$.

Jack D'Aurizio
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  • This is approximately what I have for part b where I generalise it to all finite sets. I think I need to elaborate on the step where you go from radius $\frac{1}{2}$ to radius $\frac{1}{\sqrt 3}$ – Ainsley Pullen Mar 22 '16 at 05:17