Can one number the vertices of a Fano plane (say from 1 to 7) in such a way that the sum of the numbers of collinear vertices are equal in each line?
We can do this, if we "number" the vertices with triples of elements of $\Bbb{F}_2=\{0,1\}$ with componentwise arithmetic done modulo two. This comes from the construction of the Fano plane as a projective plane over $\Bbb{F}_2$. For example, the triple $\{(1,1,1),(0,1,0),(1,0,1)\}$ is a line of the Fano plane, and
$$(1,1,1)+(0,1,0)+(1,0,1)=(2,2,2)\equiv(0,0,0)\pmod2.$$
All the lines of the Fano plane sum up to $(0,0,0)$.
But can we achieve the same using real numbers as labels of points, and instead of requiring them to sum up to zero, to sum up to some constant value independent of the line?
Best Answer
Only in the trivial way: by giving all points the same value.
If the values are $a,b,c,d,e,f,g$, with the values on any line adding to $S$, then summing all $7$ lines gives us $7S$, and it counts every value $3$ times. So we have $$ 3a + 3b + 3c + 3d + 3e + 3f + 3g = 7S \implies a+b+c+d+e+f+g = \frac73 S. $$ The sums using the value $a$ are, without loss of generality, $a+b+c$, $a+d+e$, $a+f+g$. If all three are equal to $S$, then by adding them up we get $$ (a+b+c) + (a+d+e) + (a+f+g) = 3S \implies 2a + (a+b+c+d+e+f+g) = 3S \\ \implies 2a + \frac73 S = 3S, $$ so $a = \frac13S$. This can be done for any of the values, by symmetry. Therefore all values must equal $\frac13S$, which is the trivial solution.
The same argument holds for any finite projective plane: all we use here is that each point is on the same number of lines, each line contains the same number of points, and (for the second step) there is a unique line through any two points, so that the lines through a fixed point see every other point exactly once.