[ Regular polygons ] [ Continuity considerations ]

A regular 100-gon is placed on a table, with the numbers $1,2, \ldots, 100$ written at its vertices. Then these numbers are rewritten in the order of their distance from the front edge of the table. If two vertices are at the same distance from the edge, the left number is written first, followed by the right number. All possible sets of numbers corresponding to different positions of the 100-gon are written out. Calculate the sum of the numbers that stand in the 13th position from the left in these sets.
Due to the symmetry of the regular 100-gon, each number appears in the sets at the 13th position the same number of times, which means the number of sets is a multiple of 100. If the regular 100-gon is continuously rotated counterclockwise around its center, then, first, all sets will appear, and second, the change of sets, if it occurs, will only happen when one of the diagonals or sides of the 100-gon is parallel to the edge of the table. There are 50 different directions of the sides of a regular 100-gon. Each diagonal is parallel to either a side or a diagonal connecting vertices of the polygon, skipping one vertex, - there are 50 more such different directions. Over a full rotation of the 100-gon, the change of sets occurs no more than twice for each direction, which means no more than 200 times in total. Now, note that the number 1 appears in the first position in at least two sets (when 2 is in the second position and when 100 is in the second position), and the number of sets is a multiple of 100, so there are exactly 200 different sets. Therefore, the sum of the numbers standing in these sets at the 13th position from the left is $2 \cdot(1+2+\ldots+100)=10100$.

## Answer

10100.