MathDB

6

Part of 2009 IMO Shortlist

Problems(3)

IMO Shortlist 2009 - Problem C6

Source:

7/5/2010
On a 999×999999\times 999 board a limp rook can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A non-intersecting route of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called cyclic, if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over. How many squares does the longest possible cyclic, non-intersecting route of a limp rook visit?
Proposed by Nikolay Beluhov, Bulgaria
modular arithmeticcombinatoricsIMO ShortlistChessboardChess rook
IMO Shortlist 2009 - Problem G6

Source:

7/5/2010
Let the sides ADAD and BCBC of the quadrilateral ABCDABCD (such that ABAB is not parallel to CDCD) intersect at point PP. Points O1O_1 and O2O_2 are circumcenters and points H1H_1 and H2H_2 are orthocenters of triangles ABPABP and CDPCDP, respectively. Denote the midpoints of segments O1H1O_1H_1 and O2H2O_2H_2 by E1E_1 and E2E_2, respectively. Prove that the perpendicular from E1E_1 on CDCD, the perpendicular from E2E_2 on ABAB and the lines H1H2H_1H_2 are concurrent.
Proposed by Eugene Bilopitov, Ukraine
geometrycircumcircletrigonometrysymmetryIMO Shortlist
IMO Shortlist 2009 - Problem N6

Source:

7/5/2010
Let kk be a positive integer. Show that if there exists a sequence a0,a1,…a_0,a_1,\ldots of integers satisfying the condition an=an−1+nkn for all n≥1,a_n=\frac{a_{n-1}+n^k}{n}\text{ for all } n\geq 1, then k−2k-2 is divisible by 33.
Proposed by Okan Tekman, Turkey
algebrapolynomialSequencenumber theoryDivisibilityIMO Shortlist