MathDB

2009 IMO Shortlist

Part of IMO Shortlist

Subcontests

(8)
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IMO Shortlist 2009 - Problem C3

Let nn be a positive integer. Given a sequence ε1\varepsilon_1, …\dots, εn−1\varepsilon_{n - 1} with εi=0\varepsilon_i = 0 or εi=1\varepsilon_i = 1 for each i=1i = 1, …\dots, n−1n - 1, the sequences a0a_0, …\dots, ana_n and b0b_0, …\dots, bnb_n are constructed by the following rules: a_0 = b_0 = 1,   a_1 = b_1 = 7, \begin{array}{lll} a_{i+1} = \begin{cases} 2a_{i-1} + 3a_i, \\ 3a_{i-1} + a_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_i = 0, \\ \text{if } \varepsilon_i = 1, \end{array} & \text{for each } i = 1, \dots, n - 1, \$$15pt] b_{i+1}= \begin{cases} 2b_{i-1} + 3b_i, \\ 3b_{i-1} + b_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_{n-i} = 0, \\ \text{if } \varepsilon_{n-i} = 1, \end{array} & \text{for each } i = 1, \dots, n - 1. \end{array} Prove that an=bna_n = b_n.
Proposed by Ilya Bogdanov, Russia

IMO Shortlist 2009 - Problem G3

Let ABCABC be a triangle. The incircle of ABCABC touches the sides ABAB and ACAC at the points ZZ and YY, respectively. Let GG be the point where the lines BYBY and CZCZ meet, and let RR and SS be points such that the two quadrilaterals BCYRBCYR and BCSZBCSZ are parallelogram. Prove that GR=GSGR=GS.
Proposed by Hossein Karke Abadi, Iran

IMO Shortlist 2009 - Problem N3

Let ff be a non-constant function from the set of positive integers into the set of positive integer, such that a−ba-b divides f(a)−f(b)f(a)-f(b) for all distinct positive integers aa, bb. Prove that there exist infinitely many primes pp such that pp divides f(c)f(c) for some positive integer cc.
Proposed by Juhan Aru, Estonia
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IMO Shortlist 2009 - Problem C8

For any integer n≥2n\geq 2, we compute the integer h(n)h(n) by applying the following procedure to its decimal representation. Let rr be the rightmost digit of nn. [*]If r=0r=0, then the decimal representation of h(n)h(n) results from the decimal representation of nn by removing this rightmost digit 00. [*]If 1≤r≤91\leq r \leq 9 we split the decimal representation of nn into a maximal right part RR that solely consists of digits not less than rr and into a left part LL that either is empty or ends with a digit strictly smaller than rr. Then the decimal representation of h(n)h(n) consists of the decimal representation of LL, followed by two copies of the decimal representation of R−1R-1. For instance, for the number 17,151,345,54317,151,345,543, we will have L=17,151L=17,151, R=345,543R=345,543 and h(n)=17,151,345,542,345,542h(n)=17,151,345,542,345,542. Prove that, starting with an arbitrary integer n≥2n\geq 2, iterated application of hh produces the integer 11 after finitely many steps.
Proposed by Gerhard Woeginger, Austria

IMO Shortlist 2009 - Problem G8

Let ABCDABCD be a circumscribed quadrilateral. Let gg be a line through AA which meets the segment BCBC in MM and the line CDCD in NN. Denote by I1I_1, I2I_2 and I3I_3 the incenters of △ABM\triangle ABM, △MNC\triangle MNC and △NDA\triangle NDA, respectively. Prove that the orthocenter of △I1I2I3\triangle I_1I_2I_3 lies on gg.
Proposed by Nikolay Beluhov, Bulgaria
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IMO Shortlist 2009 - Problem C6

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

IMO Shortlist 2009 - Problem G6

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

IMO Shortlist 2009 - Problem N6

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
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IMO Shortlist 2009 - Problem A2

Let aa, bb, cc be positive real numbers such that 1a+1b+1c=a+b+c\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c. Prove that: 1(2a+b+c)2+1(a+2b+c)2+1(a+b+2c)2≤316.\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}. Proposed by Juhan Aru, Estonia

IMO Shortlist 2009 - Problem C2

For any integer n≥2n\geq 2, let N(n)N(n) be the maxima number of triples (ai,bi,ci)(a_i, b_i, c_i), i=1,…,N(n)i=1, \ldots, N(n), consisting of nonnegative integers aia_i, bib_i and cic_i such that the following two conditions are satisfied: [*] ai+bi+ci=na_i+b_i+c_i=n for all i=1,…,N(n)i=1, \ldots, N(n), [*] If i≠ji\neq j then ai≠aja_i\neq a_j, bi≠bjb_i\neq b_j and ci≠cjc_i\neq c_j Determine N(n)N(n) for all n≥2n\geq 2.
Proposed by Dan Schwarz, Romania

IMO Shortlist 2009 - Problem N2

A positive integer NN is called balanced, if N=1N=1 or if NN can be written as a product of an even number of not necessarily distinct primes. Given positive integers aa and bb, consider the polynomial PP defined by P(x)=(x+a)(x+b)P(x)=(x+a)(x+b). (a) Prove that there exist distinct positive integers aa and bb such that all the number P(1)P(1), P(2)P(2),…\ldots, P(50)P(50) are balanced. (b) Prove that if P(n)P(n) is balanced for all positive integers nn, then a=ba=b.
Proposed by Jorge Tipe, Peru
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IMO Shortlist 2009 - Problem A7

Find all functions ff from the set of real numbers into the set of real numbers which satisfy for all xx, yy the identity f(xf(x+y))=f(yf(x))+x2 f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2
Proposed by Japan

IMO Shortlist 2009 - Problem G7

Let ABCABC be a triangle with incenter II and let XX, YY and ZZ be the incenters of the triangles BICBIC, CIACIA and AIBAIB, respectively. Let the triangle XYZXYZ be equilateral. Prove that ABCABC is equilateral too.
Proposed by Mirsaleh Bahavarnia, Iran

IMO Shortlist 2009 - Problem N7

Let aa and bb be distinct integers greater than 11. Prove that there exists a positive integer nn such that (an−1)(bn−1)(a^n-1)(b^n-1) is not a perfect square.
Proposed by Mongolia
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IMO Shortlist 2009 - Problem A1

Find the largest possible integer kk, such that the following statement is true: Let 20092009 arbitrary non-degenerated triangles be given. In every triangle the three sides are coloured, such that one is blue, one is red and one is white. Now, for every colour separately, let us sort the lengths of the sides. We obtain b1≤b2≤…≤b2009the lengths of the blue sides r1≤r2≤…≤r2009the lengths of the red sides and w1≤w2≤…≤w2009the lengths of the white sides  \left. \begin{array}{rcl} & b_1 \leq b_2\leq\ldots\leq b_{2009} & \textrm{the lengths of the blue sides }\\ & r_1 \leq r_2\leq\ldots\leq r_{2009} & \textrm{the lengths of the red sides }\\ \textrm{and } & w_1 \leq w_2\leq\ldots\leq w_{2009} & \textrm{the lengths of the white sides }\\ \end{array}\right. Then there exist kk indices jj such that we can form a non-degenerated triangle with side lengths bjb_j, rjr_j, wjw_j.
Proposed by Michal Rolinek, Czech Republic

IMO Shortlist 2009 - Problem C1

Consider 20092009 cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two player, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 5050 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. (a) Does the game necessarily end? (b) Does there exist a winning strategy for the starting player?
Proposed by Michael Albert, Richard Guy, New Zealand