If P is the total of points on the plane. Find all functions $f:P\rightarrow P$ such that for every two points $A, B$ of $P$, the points $A, B, f(A), f(B)$ are either collinear or con-cyclic.
I attempted to do it in the following way:
I state $S=(A:f(a)\ne A)$. Then for every three different points $X,Y,Z$ we will write $C(X,Y,Z)$ for the one radius circle or the line which passes through these points.
Afterwards I started taking cases like:
If $S$ is empty, then $f$ is an identity function.
If $|S|=1$, then $f$ is an identity function, in all places apart from one point.
I did not know how to continue it from here on, but I believe that this method is extremely tedious. Could you please explain to me an intuitive solution?
Let $A,B$ be points and $T$ a set of points with $A\notin T$. Define $$f(x)=\begin{cases}B&x=A\\A&x\in T\\x&\text{otherwise}\end{cases}$$ Then $f$ has the desired property because four pints $x,y,f(x),f(y)$ are actually just three points.
Or pick any circle-or-line (or CoL for short) and define $f$ as any map of this CoL to itself while being the identity outside that CoL. Then such $f$ also has this property.
Assume there is $f$ not matching the above. Then we can find two non-fixpoints $A,B$ points such that $A,B,f(A),f(B)$ are four distinct points and are on a unique CoL. We also know that there is another non- fixpoint $C$ not on That COL. Then the CoL through $A,C,f(A)$ and the CoL through $B,C,f(B)$ intersect exactly in $C$ and $f(C)$.
This is not complete yet, but I suspect that one cannot have many non- fuxpoints in such constellations.