[vnmath.com]-balkan mo 1994 2009

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11-th Balkan Mathematical Olympiad Novi Sad, Yugoslavia – May 10, 1994 1. An acute angle XAY and a point P inside it are given. Construct (by a ruler and a compass) a line that passes through P and intersects the rays AX and AY at B and C such that the area of the triangle ABC equals AP 2 . (Cyprus) 2. Let m be an integer. Prove that the polynomial x 4 - 1994x 3 + (1993 + m)x 2 - 11x + m has at most one integer zero. (Greece) 3. Let (a 1 ,a 2 ,...,a n ) be a permutation of the numbers 1, 2,...,n, where n 2. Determine the largest possible value of n-1 k=1 |a k+1 - a k |. (Romania) 4. Find the smallest number n> 4 for which there can exist a set of n people, such that any two people who are acquainted have no common acquaintances, and any two people who are not acquainted have exactly two common acquaintances. (Acquaintance is a symmetric relation.) (Bulgaria)

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11-thBalkanMathematicalOlympiadNoviSad,YugoslaviaMay10,19941. An acute angleXAYand a pointPinside it are given. Construct (by aruler and a compass) a line that passes throughPand intersects the raysAXandAY atBandCsuch that the area of the triangleABCequalsAP2. (Cyprus)2. Letm be an integer. Prove that the polynomialx41994x3+ (1993 +m)x211x +mhas at most one integer zero. (Greece)3. Let(a1, a2, . . . , an)beapermutationof thenumbers1, 2, . . . , n, wheren 2. Determine the largest possible value ofn1

k=1|ak+1ak|.(Romania)4. Findthesmallest number n>4for whichtherecanexist aset of npeople, suchthatanytwopeoplewhoareacquaintedhavenocommonacquaintances,and any two people who are not acquainted have exactlytwo common acquaintances. (Acquaintance is a symmetric relation.)(Bulgaria)12-thBalkanMathematicalOlympiadPlovdiv,BulgariaMay9,19951. Denex y =x +y1 +xy. Evaluate (. . . (((2 3) 4) 5) . . . ) 1995.(FYR Macedonia)2. Circlesc1(O1, r1)andc2(O2, r2), r2>r1, intersectatAandBsothatO1AO2 = 90. The lineO1O2 meetsc1 atCandD, andc2 atE andF(in the order C E DF). The line BE meets c1 at K and AC at M,and the lineBD meetsc2atL andAFatN. Prove thatr2r1=KEKM LNLD.(Greece)3. Leta andb be natural numbers witha > b and 2 | a +b. Prove that thesolutions of the equationx2 (a2a + 1)(x b2 1) (b2+ 1)2= 0are natural numbers, none of which is a perfect square. (Albania)4. Letn bea natural number andSbetheset of points(x, y) withx, y {1, 2, . . . , n}. LetTbe the set of all squares with the verticesw in the setS. We denote by ak (k 0) the number of (unordered) pairs of points forwhich there are exactly k squares in Thaving these two points as vertices.Show thata0 = a2 + 2a3. (Yugoslavia)13-thBalkanMathematicalOlympiadBacau,RomaniaApril30,19961. LetObe the circumcenter andG be the centroid of a triangleABC. IfRandrarethecircumcenterandincenterofthetriangle, respectively,prove thatOG

R(R 2r). (Greece)2. Letp> 5 be a prime. ConsiderX= {p n2|n N}. Prove that thereare two distinct elementsx, y Xsuch thatx = 1 andx | y. (Albania)3. InaconvexpentagonABCDE, M, N, P, Q, RarethemidpointsofthesidesAB, BC, CD, DE, EA, respectively. If thesegmentsAP, BQ, CR,DMpassthroughasinglepoint, provethatENcontainsthatpointaswell. (Yugoslavia)4. Show that there exists a subsetA of the set {1, 2, . . . , 21996 1} with thefollowing properties:(i) 1 A and 21996 1 A;(ii) Every element of A\ {1} is the sum of two (possibly equal) elementsofA;(iii) A contains at most 2012 elements. (Romania)14-thBalkanMathematicalOlympiadKalabaka,GreeceApril30,19971. Suppose thatO is a point inside a convex quadrilateralABCD such thatOA2+OB2+OC2+OD2= 2SABCD,whereSABCDdenotes the area ofABCD. Prove thatABCD is a squareandO its center. (Yugoslavia)2. Let A = {A1, A2, . . . , Ak}beacollectionofsubsetsofann-elementsetS. Ifforanytwoelementsx, y SthereisasubsetAi Acontainingexactly one of the two elementsx, y, prove that 2k n. (Yugoslavia)3. CirclesC1andC2touch each other externally atD, and touch a circle internally atBandC,respectively. LetA be an intersection point of and the common tangent toC1 andC2 atD. LinesAB andACmeetC1and C2 again at K and L, respectively, and the line BC meets C1 again atMandC2 again atN. Show that the linesAD, KM, LNare concurrent.(Greece)4. Determine all functionsf: R R that satisfyf(xf(x) +f(y)) = f(x)2+y for allx, y. (Bulgaria)15-thBalkanMathematicalOlympiadNicosia,CyprusMay5,19981. Consider the nite sequence

k21998

, k = 1, 2, . . . , 1997. How many distinctterms are there in this sequence? (Greece)2. Let n 2beaninteger, andlet0 0.(Cyprus)4. Letm andn be coprime odd positive integers. A rectangleABCDwithAB = m and AD = n is divided into mn unit squares. Let A1, A2, . . . , Akbe the consecutive points of intersection of the diagonal AC with the sidesof the unit squares (whereA1 = A andAk = C). Prove thatk1

j=1(1)j+1AjAj+1 =m2+n2mn.(Bulgaria)21-stBalkanMathematicalOlympiadPleven,BulgariaMay7,20041. A sequence of real numbersa0, a2, a2, . . . satises the conditionam+n +amnm+n 1 =a2m +a2n2for allm, n N withm n. Ifa1 = 3, determinea2004. (Cyprus)2. Find all solutions in the set of prime numbers of the equationxyyx= xy219.(Albania)3. LetObe an interior point of an acute-angled triangleABC. The circlescenteredatthemidpointsofthesidesofthetriangleABCandpassingthrough pointO, meet in pointsK, L, Mdierent fromO. Prove thatOis the incenter of the triangleKLMif and only ifOis the circumcenterof the triangleABC. (Romania)4. Aplaneisdividedintoregionsbyanitenumberof lines, nothreeofwhichareconcurrent. Wecall tworegionsneighboringiftheircommonboundary is either a segment, a ray, or a line. One should write an integerin each of the regions so as to full the following two conditions:(a) The product of the numbers from two neighboring regions is less thantheir sum;(b) Thesumofallthenumbersinthehalfplanedeterminedbyanyofthe lines is equal to zero.Prove that this can be done if and only if not all the lines are parallel.(Serbia and Montenegro)22-ndBalkanMathematicalOlympiadIasi,RomaniaMay6,20051. The incircle of an acute-angled triangleABCtouchesABatDandACat E. Let the bisectors of the angles ACB and ABC intersect the lineDEatXandY respectively, andletZbethemidpointof BC. Provethat the triangleXY Zis equilateral if and only if A = 60. (Bulgaria)2. Find all primesp such thatp2p + 1 is a perfect cube. (Albania)3. Ifa, b, c are positive real numbers, prove the inequalitya2b+b2c+c2a a +b +c + 4(a b)2a +b +c.When does equality occur? (Serbia and Montenegro)4. Let n 2 be an integer, and let S be a subset of {1, 2, . . . , n} such that Sneither contains two coprime elements, nor does it contain two elements,one of which divides the other. What is the maximum possible number ofelements ofS? (Romania)23-rdBalkanMathematicalOlympiadAgros,CyprusApril29,20061. Ifa, b, c are positive numbers, prove the inequality1a(1 +b) +1b(1 +c) +1c(1 +a) 31 +abc.2. AlinemintersectsthesidesAB, ACandtheextensionof BCbeyondCof the triangleABCat pointsD, F, E, respectively. The lines throughpointsA, B, CwhichareparalleltommeetthecircumcircleoftriangleABCagain at pointsA1, B1, C1,respectively. Show that the linesA1E,B1F,C1D are concurrent.3. Determine all triples (m, n, p) of positive rational numbers such that thenumbersm+1np, n +1pm, p +1mnare integers.4. Given a positive integer m, consider the sequence (an) of positive integersdened by the initial terma0 = a and the recurrent relationan+1 =

an2ifanis even,an +m ifanis odd.Findallvaluesof aforwhichthissequenceisperiodic(i.e. thereexistsd > 0 such thatan+d = anfor alln).24-thBalkanMathematicalOlympiadRhodes,GreeceApril28,20071. Inaconvexquadrilateral ABCDwithAB=BC=CD, thediagonalsACandBDare of dierent length and intersect at pointE. Prove thatAE = DEif and only if BAD +ADC = 120. (Albania)2. Find all functionsf: R R such that for all realx, y,f(f(x) +y) = f(f(x) y) + 4f(x)y.(Bulgaria)3. Determine all natural numbersn for which there exists a permutationof numbers 1, 2, . . . , n such that the number

(1) +

(2) +

+

(n)is rational. (Serbia)4. Let n 3beaninteger. Let C1, C2, C3bethecircumferencesof threeconvexn-gonsinaplanesuchthattheintersectionofanytwoofthemis a nite set of points. Find the maximum possible number of points inC1 C2 C3. (Turkey)25-thBalkanMathematicalOlympiadOhrid,FYRMacedoniaMay6,2008Anacute-angledscalenetriangleABCwithAC>BCisgiven. LetObeits 1.circumcenter, Hitsorthocenter, andFthefootof thealtitudefromC. LetPbethepoint(otherthanA)onthelineABsuchthat AF=PF, andMbethemidpointofAC. WedenotetheintersectionofPHandBCbyX,theintersectionof OMandFXbyY , andtheintersectionof OFandACbyZ.Prove thatthepointsF, M, Y andZareconcyclic.Does there exist a sequence a1, a2, . . . of positive numbers satisfying both of the 2.followingconditions:(i)

ni=1ai n2foreverypositiveintegern;(ii)

ni=11ai 2008foreverypositiveintegern?Letnbeapositiveinteger. TherectangleABCDwithsidelengths90n + 1 3.and90n + 5ispartitionedintounitsquareswithsidesparalleltothesidesofABCD. LetSbethesetofallpointswhichareverticesoftheseunitsquares.ProvethatthenumberoflineswhichpassthroughatleasttwopointsfromSisdivisibleby4.Letcbeapositiveinteger. Thesequencea1, a2, . . . isdenedbya1=cand 4.an+1=a2n +an +c3foreverypositiveintegern. Findallvaluesofcforwhichthereexistsomeintegersk 1andm 2suchthata2k +c3isthem-thpowerofsomepositiveinteger.Timeallowed: 4.5hours.Each problemisworth 10points.26-thBalkanMathematicalOlympiadKragujevac,SerbiaApril30,20091. Findallintegersolutionsoftheequation3x5y= z2. (Greece)2. InatriangleABC,pointsMandNonthesidesABandACrespectivelyaresuch that MNBC. Let BNand CMintersect at point P. The circumcirclesof trianglesBMPandCNPintersectattwodistinctpointsPandQ. Provethat BAQ = CAP. (Moldova)3. A 912 rectangle is divided into unit squares. The centers of all the unit squares,except the four corner squares and the eight squares adjacent (by side) to them,arecoloredred. IsitpossibletonumeratetheredcentersbyC1, C2, . . . , C96sothatthefollowingtwoconditionsarefullled:1AllsegmentsC1C2, C2C3, . . . C95C96, C96C1havethelength13;2ThepolygonallineC1C2. . . C96C1iscentrallysymmetric? (Bulgaria)4. Determineallfunctionsf: N Nsatisfyingf

f(m)2+ 2f(n)2

= m2+ 2n2forallm, n N. (Bulgaria)Eachproblemisworth10points.Timeallowed: 412hours.