How to show that all points are inside of unit circle? There are $n$ points on the plane. Any $3$ of them are inside of a unit circle. How to show that all points are inside of unit circle?
It is needed to prove that if there is a unit circle for each $3$
 points which contains them, there is also a unit circle which contains all $n$ points.
 A: Hint: Flip the question around. Think of unit circles about each point.
Hint: Helly's Theorem, with $d=2$ - In $ \mathbb{R}^2 $, with a collection of convex sets. If every $3$ sets have a non-empty intersection, then all the sets have a non-empty intersection.
Hence Done.
A: How about a reductio ad adsurdum argument.  Assume one point is not inside a unit circle, show a contradiction with the assumption that any 3 points are inside a unit circle.
A: This is clearly false for $n=2$, so I will assume you forgot to say $n\geq3$.
It is intuitively clear that there is some circle of minimal size such that all points are in its convex hull (either on the circle or in its interior). Consider the points that are on such a circle; clearly there are at least two of them by minimality. In fact none of the arcs of the circle separated by the points on it can be more than a semi-circle, or else the circle with diameter the end points of the arc would be smaller and contain all points. So either there are two cases:


*

*There are just two points on the circle, which must be diametrically opposite, and at distance at most $2$ (by considering the condition with a random third point), and our circle is either a unit circle or smaller than one.

*There are three points on the circle with each of the arcs between them at most a semi-circle; these form a non-obtuse triangle, and the circle is both their circumcircle and the smallest circle continuing all three in its convex hull; again it must either be a unit circle or smaller.


It remains to show formally the existence of a smallest circle containing all the points, but this is easily done by induction on the number of points.
