One special case of Helly's theorem (for $\text{radius}=1$ circles)

Question

There are $n$ points on the plane. Any $3$ of them can be covered with a radius $1$ circle. Prove that there is a radius $1$ circle that covers all the points.
Came to this when tried to prove an easy case of Helly's theorem(for $r = 1$ circles).
So it's obvious that the solution can not use Helly's.
Definitely I am missing something, need a wise hint...

Answer 1

We proceed by induction: Assume we have $x_1,\dots,x_{n-1}$ in a circle $C=B(x_0,1)$ of radius 1 and center $x_0$, and let $x_n$ be such that for all $i,j\in [1,n-1]$, there is a circle of radius 1 containing $x_i,x_j,x_n$. Now set $C_i=B(x_i,1)$ and $$ R:=\bigcap_{i=1}^n C_i $$ the intersection of the radius 1 circles around each $x_1,\dots,x_{n-1}$. Then $R\ne \emptyset$, since $x_0\in R$. There are three cases:

1. $R$ is only one point ($R=x_0$), and there are only two points $x_i,x_j$ at a distance 1 from $R$. Then $x_i,x_j$ are opposite points and necessarily $x_n\in R$, since the only circle that contains $x_i,x_j$ is $C=B(x_0,1)$.

2. $R$ is only one point ($R=x_0$), and there are three points $x_i,x_j,x_k$ at a distance 1 from $R$, and not contained in any semicircle with center $x_0$. Assume $d(x_k,x_n)=\min\{d(x_i,x_n),d(x_j,x_n),d(x_k,x_n)\}$. Then $d(x_n,x_0)=d(x_n, C_i\cap C_j)\le 1$, since $x_i,x_j,x_n$ are contained in a circle of radius 1, and so $x_n\in C$.

3. $R$ has more than one point. In this case the boundary of $R$ consists of arcs of circles of radius 1. Let $x$ be the nearest corner of $R$ to $x_n$, i.e., $d(x_n,x)=min\{d(x_n,c)\ :\ c\text{ a corner of $R$}\}$. Then $x$ is the intersection of two arcs of the circles, say $C_i, C_j$. Then $d(x_n,R)=d(x_n, C_i\cap C_j)\le 1$, since $x_i,x_j,x_n$ are contained in a circle of radius 1. Choose one point $c_n$ in $R$ with $d(x_n,c_n)\le 1$, and then $x_1,\dots,x_n\in B(c_n,1)$.

The equalities $d(x_n,x_0)=d(x_n, C_i\cap C_j)$ in case 2 and $d(x_n,R)=d(x_n, C_i\cap C_j)$ in case 3 are central. In the case 2. the equality can be proven geometrically drawing the perpendicular bisectors to $x_i,x_k$ and $x_j,x_k$ and proving that $x_n$ must be in the segment of the plane determined by these lines. In the case 3., we have to draw the bisectors to $x$ and the next corners of $R$, i.e., the line through $x_i$ and the middle point of the arc of $C_i$ contained in $R$ and the line through $x_j$ and the middle point of the arc of $C_j$ contained in $R$.