Maths Olympiad Prep

Library / Auswahlwettbewerb zur Internationalen Mathematik-Olympiade /14 of 17

Auswahlwettbewerb zur Internationalen Mathematik-Olympiade

Combinatorics Difficulty 8.7 Shortlist Prove it Germany

Problem:

The vertices and edges of a regular nn-gon are numbered clockwise from 1 to nn, in such a way that edge no. ii follows vertex no. ii (1≤i≤n1 \leq i \leq n).
Now the vertices are assigned nonnegative integers eie_{i} and the edges are assigned nonnegative integers kik_{i} such that:
(1) The nn-tuple (e1,e2,…,en)(e_{1}, e_{2}, \ldots, e_{n}) is a permutation of the nn-tuple (k1,k2,…,kn)(k_{1}, k_{2}, \ldots, k_{n}).
(2) For every i∈Ni \in \mathbb{N} with 1≤i≤n1 \leq i \leq n we have ki=∣ei+1−ei∣k_{i} = |e_{i+1} - e_{i}|, where en+1=e1e_{n+1} = e_{1}.

a) Prove that for all natural numbers nn, n≥3n \geq 3, such nn-tuples exist that are different from (0,…,0)(0, \ldots, 0).

b) Determine, for each positive natural number mm, the smallest natural number nn with the following property: In the nn-tuples (e1,e2,…,en)(e_{1}, e_{2}, \ldots, e_{n}) and (k1,k2,…,kn)(k_{1}, k_{2}, \ldots, k_{n}), all natural numbers from 00 to mm occur, respectively.

Solution

Solution:

a) A possible example for n≥3n \geq 3 is e1=e2=1e_{1} = e_{2} = 1, ei=0e_{i} = 0 for 3≤i≤n3 \leq i \leq n. This forces k2=kn=1k_{2} = k_{n} = 1 and ki=0k_{i} = 0 otherwise. Clearly all conditions are satisfied.

b) We prove that always n=m+2n = m + 2. The proof proceeds in two steps.

i) We have n≥m+2n \geq m + 2: Since the number 00 occurs in (k1,k2,…,kn)(k_{1}, k_{2}, \ldots, k_{n}), two elements of (e1,e2,…,en)(e_{1}, e_{2}, \ldots, e_{n}) must be equal. Since, moreover, each of the m+1m + 1 natural numbers from 00 to mm must occur at least once, it follows that n≥m+2n \geq m + 2.

ii) We give an example for n=m+2≥3n = m + 2 \geq 3: Let (e1,e2,…,en)=(0,m,1,m−1,…)(e_{1}, e_{2}, \ldots, e_{n}) = (0, m, 1, m-1, \ldots). In this nn-tuple the numbers are continued alternately, decreasing down to the difference 00, so that en−1=ene_{n-1} = e_{n} holds. Thus every natural number from 00 to mm occurs exactly once, and the number ene_{n} occurs exactly twice. The absolute values of the differences clearly occur once each, going downward from mm to 00; the difference ∣en−0∣=en|e_{n} - 0| = e_{n} occurs twice, because en<me_{n} < m. This establishes properties (1) and (2).

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from de; metadata (topic, difficulty) added by this project.