Question

Let P be the set of all points in the plane and L be the set of all lines of the plane. Find, with proof, whether there exists a bijective function $f:P\to L$ such that for any three collinear points A, B, C, the lines $f\left(A\right),f\left(B\right)$ and $f\left(C\right)$ are either parallel or concurrent.

• A. No such funtions exists.
• B. There exists a funtion.
• C. The solution has infinite functions.
• D. The solution is an integer.

Answer 1

The correct option is A No such funtions exists.

Let $A_i,i=1,2,3$ be three distinct points in the plane and $l_i=f\left(A_i\right)$.
We claim that if $l_1,l_2,l_3$ are concurrent or parallel, then $A_1,A_2,A_3$ are collinear. Indeed, suppose that $l_1,l_2,l_3$ intersect at M and that $A_1,A_2,A_3$ is a non-degenerated triangle. Then for any point B in the plane we can find points $B_2{B}_3$ on the lines $A_1,A_2,A_3$, respectively, such that $B_2{B}_3$ are collinear. Because $A_1,A_2,A_3$ are collinear it follows that $f\left(B_2\right)$ is a line passing through M. The same is true for $f\left(B_1\right)$, hence also for $f\left(B\right)$.This contradicts the surjectivity of $f$. A similar argument can be given if $l_1,l_2,l_3$ are parallel. We conclude that the restriction of $f$ to any line $l$ defines a bijection from $l$ to a pencil of lines (passing through a point or parallel). Consider two pencils $P_1$ and $P_2$ of parallel lines. The inverse images of $P_1$ and $P_2$ are two parallel lines $l_1,l_2$ ($P_{1}and{P}_2$ have no common lines, hence $l_1$ and $l_2$ have no common points). Let $P_3$ be a pencil of concurrent lines whose inverse image is a line $l$, clearly not parallel to$l_1,l_2$. Let l' be a line parallel to l. Then f(l') is a pencil of concurrent lines and it follows that there is a line through the points corresponding to l and l' whose inverse image would be a point on both l and l', a contradiction. Hence no such functions exists.