there is at most one point at distance exactly $1$ from three distinct points in the plane Consider $n\geq 1$ points in the plane. I want to show that there is at most one point at distance exactly $1$ from any three distinct points in the plane. 
I want to prove this as part of a larger proof I am writing for a particular problem. This fact seems obvious, yet it is not that easy for me to prove. I will assume that I am working with the Euclidean distance. 
Is there any way to prove this fact using the triangle or reverse triangle inequalities?
Thank you!

Take any pair of the three points. The set of points that are equidistant from those two points form a line; the perpendicular bisector of the line joining those two points. Now use the other point and one of first two points to form an equidistant line from those two points also. 
If these two lines intersect, they do so at only one point. This is the unique point that is equidistant from all three points.  (It is also the circumcenter of the triangle created by those three points).
If the three points are collinear, the two lines formed will not intersect and there is no equidistnt point.
Note that the two equidistant set lines must be distinct as the original three points are distinct.
The point that is equidistant from the three specified points may or may not be at distance $1$, of course. But there is at most one such point.
