olymmath

OPM 1996 Problem 6

  • Final day 2

In a regular polygon with 134134 sides, 6767 diagonals are drawn so that exactly one diagonal emerges from each vertex. We call the length of a diagonal the number of sides of the polygon included between the vertices of the diagonal and which is less than or equal to 6767 . If we order the lengths of the diagonals in ascending order, we obtain a succession of 6767 numbers (d1,d2,...,d67)(d_1,d_2,...,d_{67}) . It will be possible to draw diagonals such that

a) (d1,d2,...,d67)=2...2⏟6,3...3⏟61(d_1,d_2,...,d_{67})=\underbrace{2 ... 2}_{6},\underbrace{3 ... 3}_{61} ?

b) (d1,d2,...,d67)=3...3⏟8,6...6⏟55.8...8⏟4(d_1,d_2,...,d_{67}) =\underbrace{3 ... 3}_{8},\underbrace{6 ... 6}_{55}.\underbrace{8 ... 8}_{4} ?

Dataset text AI-MO AoPS. Quoted with source.

Official text: Portuguese Mathematical Olympiad 1996 official paper, page 1.

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