A combinatorics problem from from Baltic way math competition (1995) This is a question from Baltic way math competition (1995).  


A polygon with $2n + 1$ vertices is given. Show that it is possible to label the vertices and midpoints of the
    sides of the polygon, using all the numbers $1, 2, \dots , 4n + 2$, so that the sums of the three numbers assigned
    to each side are all equal.   


The official solution is   
First, label the midpoints of the sides of the polygon with the numbers $1, 2, \dots , 2n + 1$, in
clockwise order. Then, beginning with the vertex between the sides labelled by $1$ and $2$, label every second
vertex in clockwise order with the numbers $4n + 2, 4n + 1, \dots, 2n + 2$.
My Question
Is there any way by which this solution can be discovered in exam setup? I want to understand how this kind of problems can be systematically attacked.Thanks in advance.
 A: "I want to understand how this kind of problem can be systematically attacked" - the classic work on systematic problem solving in mathematics is George Polya's How To Solve It - here is a summary.
A: I have done a lot of math contests in the past (some with success and others not so much) and when I first read this questions, the first thing that comes to mind is symmetry. Since each edge is isomorphic to any other edge, the edges should be treated in a "symmetric" way. So the thought would be to take either midpoints or vertices and treat them in a similar way (label with largest numbers, smallest numbers, or whatever else you can think of) and then try to treat vertices in a different but also symmetric way. Rotational assignments like this aren't uncommon with geometry/combinatorics problems. 
This approach applies to most types of problems you see in math competitions, or at least it is a good place to start. Asking yourself "are any aspects of this problem symmetric?" often gives you a better understanding of the problem and sometimes gives you critical information.
