Impossible to intresect

Let ${\rm P_1, P_2, P_3}$ and ${\rm Q_1, Q_2, Q_3}$ be points on a sphere. Prove that it is impossible to connect each ${\rm P_i}$ with each ${\rm Q_i}$ with $9$ curves of the sphere that intersect only at the given points. 

Solution

This is a very classic problem, but its solution may not be that known.

We are using Euler's formula which says that if we have $V$ vertices, $E$ edges and $F$ fields (in the case that the graph is connected) then:

$$V-E+F=2 \Leftrightarrow 6  - 9 + F =2 \Leftrightarrow F= 5$$

We cound the pairs $(e, f)$ where $e$ is an edge that belongs in the circumference of a field $f$. Since each edge belongs at two fields at the same time , we have $2e=18$ such pairs. However in the circumference of each field we have at least four edges ⁽¹⁾ therefore we will have also at least $4f=20$ pairs, which is a contradition.



⁽¹⁾ The edges of the circumference of a field are alternating between ${\rm P_i, \; Q_i}$ . Therefore in the circumference of the field we must have an even number of both vertices and edges.