MathDB

4

Part of 2000 IMO Shortlist

Problems(4)

Number of integers with F(4n) = F(3n) is F(2^{m+1})

Source: IMO Shortlist 2000, A4

8/10/2008
The function F F is defined on the set of nonnegative integers and takes nonnegative integer values satisfying the following conditions: for every n≥0, n \geq 0, (i) F(4n) \equal{} F(2n) \plus{} F(n), (ii) F(4n \plus{} 2) \equal{} F(4n) \plus{} 1, (iii) F(2n \plus{} 1) \equal{} F(2n) \plus{} 1. Prove that for each positive integer m, m, the number of integers n n with 0≤n<2m 0 \leq n < 2^m and F(4n) \equal{} F(3n) is F(2^{m \plus{} 1}).
functionalgebrafunctional equationIMO Shortlist
Conex polygon is cyclic iff one can assign of real numbers

Source: IMO Shortlist 2000, G4

8/10/2008
Let A1A2…An A_1A_2 \ldots A_n be a convex polygon, n≥4. n \geq 4. Prove that A1A2…An A_1A_2 \ldots A_n is cyclic if and only if to each vertex Aj A_j one can assign a pair (bj,cj) (b_j, c_j) of real numbers, j=1,2,…,n, j = 1, 2, \ldots, n, so that AiAj=bjci−bicj A_iA_j = b_jc_i - b_ic_j for all i,j i, j with 1≤i<j≤n. 1 \leq i < j \leq n.
trigonometrygeometryCyclicpolygonIMO Shortlist
Column/Row contains a block of k adjacent unoccupied squares

Source: IMO Shortlist 2000, C4

8/10/2008
Let n n and k k be positive integers such that 12n<k≤23n. \frac{1}{2} n < k \leq \frac{2}{3} n. Find the least number m m for which it is possible to place m m pawns on m m squares of an n×n n \times n chessboard so that no column or row contains a block of k k adjacent unoccupied squares.
combinatoricsExtremal combinatoricsIMO Shortlistgraph theoryChessboard
Imo Shortlist Problem

Source: IMO Shortlist 2000, Problem N4

2/27/2005
Find all triplets of positive integers (a,m,n) (a,m,n) such that a^m \plus{} 1 \mid (a \plus{} 1)^n.
algebrapolynomialnumber theoryDivisibilityIMO ShortlistZsigmondy