c o l e c Ń i a
Această lucrare este rezultatul colaborării dintre Editura Paralela 45 şi Societatea Română de ŞtiinŃe Matematice.
Editor: Călin Vlasie Corectură: autorii Tehnoredactare: Carmen Rădulescu Coperta colecŃiei: Andrei Mănescu Machetare & prepress: ART CREATIV
Descrierea CIP a Bibliotecii NaŃionale a României ŞERBĂNESCU, DINU Training Problems for the Junior Balkan Mathematical Olympiads – The Romanian Experience / Dinu Şerbănescu, Mircea Fianu, Ioan Şerdean. - Piteşti : Paralela 45, 2010 ISBN 978-973-47-0957-1 I. Fianu, Mircea II. Şerdean, Ioan 51(075.33)(076)
Copyright Editura Paralela 45, 2010
D INU Ş ERBĂNESCU , F LORICA B ANU , M IRCEA F IANU , Ş TEFAN S M Ă R Ă NDOIU ND OIU , M ARIUS P ERIAN ERIA N U , I OAN Ş ERDEAN , G ABRIEL P OPA
Training Tra ining Problems for the Junior Balkan Matehmatical Olympiads The Romanian Experience
FOREWORD The Junior Balkan Mathematical Olympiad started fourteen years ago as an alternative of the Balkan Olympiads for Junior High students. Since, the competition has become very popular not only in South-Eastern European countries but all over the world. Some teams from Asia and Europe participate as invited countries to the competition. But the most important achievement of this competition is in my opinion the long list of nice problems produced for these kids; usually extremely elementary but containing deep mathematical ideas. Maybe this is why, in the Balkan countries, the junior teams usually graw to become in years the senior teams for the BMO and IMO. The booklet is devoted to the Romanian problems used for the qualification tests of the Romanian team in the last decade. One can remark the great number of authors, some of them being IMO students at the time the problems were given. It is a great job the authors have done here, collecting all those problems and editing nice solutions. Radu Gologan
PROBLE PROBL E M S
Chapter I ALGEBRA ALGEBRA Problem 1 Prove that for any real numbers a, b, c such that 0 < a, b, c < 1, the following inequality holds abc + (1 − a )(1 − b)(1 − c) < 1 . Dinu Şerbănescu, 2002
Problem 2 Let a, b, c be positive real numbers with abc = 1. Prove that 3 6 1+ ≥ . a + b + c ab + bc + ca Mircea Lascu and Vasile Cârtoaje, 2003
Problem 3 Find the positive real numbers a, b, c which satisfy the inequalities 4(ab + bc + ca) – 1 ≥ a2 + b2 + c2 ≥ 3(a3 + b3 + c3). LaurenŃiu Panaitopol, 2004
Problem 4 Let a, b, c be positive real numbers such that a + b + c ≥ a+b+c≥
1 1 1 + + . Prove that a b c
3 . abc
Cezar Lupu, 2005
Problem 5 Let a, b, c be positive real numbers such that (a + b)(b + c)(c + a) = 1. Show that 3 ab + bc + ca ≤ . 4 Cezar Lupu, 2005
Problem 6 Let a, b, c be positive real numbers with a + b + c = 3. Prove that (3 – 2a)(3 – 2b)(3 – 2c) ≤ a2b2c2. Robert Szasz, 2005
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Problem Problem 7 Prove that
a 3 b3 c 3 + + ≥ a + b + c , for all positive real numbers a, b and c. bc ca ba
* * * , 2005
Problem 8 Prove that 2
a b c 3 a+b b+c c+a + + + + ≥ ⋅ a b b c a 2 c for all positive real numbers a, b and c.
Cezar Lupu, 2006
Problem 9 Suppose a, b, c are positive real numbers with the sum equal to 1. Prove that: a2 b2 c2 + + ≥ 3(a 2 + b 2 + c 2 ) . b c a Mircea Lascu, 2006
Problem 10 Let x, y, z be positive real numbers such that 1 1 1 + + =2. 1+ x 1+ y 1+ z Prove that 8xyz ≤ 1. Mircea Lascu, 2006
Problem 11 Let a and b be integer numbers. Show that there exists a unique pair of integers x, y so that (x + 2y – a)2 + (2x – y – b)2 ≤ 1. Adrian Zahariuc, 2007
Problem 12 Suppose a, b, c are positive real numbers satisfying: 1 1 1 + + ≥1 . a + b +1 b + c +1 c + a +1 Show that a + b + c ≥ ab + bc + ca. Andrei Ciupan, 2007
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Problem 13 Prove that x3 + y 3 + z 3 3 ≥ xyz + | ( x − y )( y − z )( z − x) | , 3 4 for any real numbers x, y, z ≥ 0.
Viorel Vâjâitu, 2007
Problem 14 Let a, b, c be positive real numbers with ab + bc + ca = 3. Prove that 1 1 1 1 . + + ≤ 2 2 2 1 + a (b + c) 1 + b (c + a ) 1 + c (a + b) abc Vlad Matei, 2008
Problem 15 Determine the maximum value of the real number k such that 1 1 1 (a + b + c) + + −k≥ k , a+b b+c a+c for all real numbers a, b, c ≥ 0 with a + b + c = ab + bc + ca. Andrei Ciupan, 2008
Problem 16 Let a, b, c > 0 be real numbers with the sum equal to 3. Show that: a+3 b+3 c+3 + + ≥ 3. 3a + bc 3b + ca 3c + ab Dinu Şerbănescu, 2009
Problem 17 1 1 1 + + . Prove that a b c a b c 1 1 1 + + ≥ + + . b c a ab bc ca
Let a, b, c be positive real numbers such that a + b + c ≥
Cezar Lupu, 2009
Problem 18 Let A be a non-empty subset of R with the property that for every real numbers x, y, if x + y ∈ A, then xy ∈ A. Prove that A = R. Eugen Păltănea, 2001
11
Problem 19 For any positive integer n, let f ( n) =
4n + 4n 2 − 1 2 n + 1 + 2n − 1
.
Compute the sum f (1) + f (2) + … + f (40). Titu Andreescu, 2002
Problem 20 Five real numbers of absolute values not greater than 1 and having the sum equal to 1 are written on the circumference of a circle. Prove that one can choose three consecutively disposed numbers a, b, c, such that all the sums a + b, b + c and a + b + c are nonnegative. Dinu Şerbănescu, 2003
Problem 21 A set of 2003 positive integers is given. Show that one can find two elements such that their sum is not a divisor of the sum of the other elements. Valentin Vornicu, 2003
Problem 22 Consider the numbers an = 3n + n 2 − 1
and bn = 2( n 2 + n + n 2 − n ) ,
for all n = 1, 2, …, 49. Prove that there are integers A, B so that a1 − b1 + a2 − b2 + ... + a49 − b49 = A + B 2 . Titu Andreescu, 2004
Problem 23 Let a < b ≤ c < d be positive integers such that ad = bc and
d − a ≤ 1 . Prove that a is a square. Dinu Şerbănescu, 2004
Problem 24 The real numbers a1, a2, …, an satisfy the relation 2 a12 + a22 + ... + a100 + (a1 + a2 + ... + a100 ) 2 = 101 . Prove that |ak| ≤ 10, for all k = 1, 2, …, 100. Dinu Şerbănescu, 2004
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Problem 25 Consider the triangular array 0 1 1 2 0 1 1 2 3 4
3 5… 2 3… 5 8… 7 11…
defined by the conditions: i) on the first two rows, each element starting with the third is the sum of the two preceding elements; ii) on the other rows each element is the sum of the two elements placed above on the same column. a) Prove that all the rows are defined according to condition i). b) Consider 4 consecutive rows and let a, b, c, d be the first element in each of these rows, respectively. Find d in terms of a, b and c. Dinu Şerbănescu, 2004
Problem 26 Let A be a set of positive integers with the properties: i) if a ∈ A, then all positive divisors of a are elements of A; ii) if a, b ∈ A and 1 < a < b, then 1 + abc ∈ A. Prove that if the set A has at least 3 elements, then A = N*. Valentin Vornicu, 2004
Problem 27 Let n > 3 be a positive integer. Consider n sets, each having two elements, such that the intersection of any two of them is a set with one element. Prove that the intersects of all sets is nonempty. Sever Moldoveanu, 2005
Problem 28 Consider two distinct positive integers a and b having integer arithmetic, geometric and harmonic means. Find the minimum value of |a – b|. Mircea Fianu, 2005
Problem 29 Find all real numbers a and b satisfying 2(a2 + 1)(b2 + 1) = (a + 1)(b + 1)(ab + 1). Valentin Vornicu, 2006
Problem 30 Show that the set of real numbers can be partitioned into subsets having two elements. * * *, 2006
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Problem 31 The set of positive integers is partitioned in subsets with infinite elements each. The question (in each of the following cases) is if there exists a subset in the partition such that any positive integer has a multiple in this subset. a) Prove that if the number of subsets in the partition is finite the answer is yes. b) Prove that if the number of subsets in the partition is infinite, then the answer can be no (for a certain partition). * * *, 2006
Problem 32 Find all non-empty subsets A of the set {2, 3, 4, 5, …} so that for any n ∈ A, both n2 + 4 and [ n ] + 1 also belong to A. Lucian łurea, 2007
Problem 33 An irrational number x, 0 < x < 1 is called suitable if its first 4 decimals in the decimal representation are equal. Find the smallest positive integer n such that any real number t, 0 < t < 1 may be written as a sum of n distinct suitable numbers. Lucian łurea, 2007
Problem 34 Let n ∈ N* and let a1, a2, …, an be positive real numbers so that 1 1 1 a1 + a2 + ... + an = 2 + 2 + ... + 2 . a1 a2 an Prove that for any m = 1, 2 …, n, there exist m numbers among the given ones with the sum not less than m. Andrei Ciupan and Flavian Georgescu, 2008
Problem 35 Let A be a finite set of positive real numbers satisfying the property: For any real numbers a > 0, the sets 1 {x ∈ A | X > a} and x ∈ A | x < a have the cardinals of the same parity.
Show that the product of all elements in A is equal to 1. Dinu Şerbănescu, 2009
14
Problem 36 Let a and b be positive integers. Consider the set of all non-negative integers n for which the number n
1 1 a + + b + 2 2 is an integer. Show that the set is finite.
n
* * *, 2009
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Chapter II GEOMETRY Problem 37 Let ABC be an arbitrary triangle. A circle passes through B and C and cuts again the lines AB and AC in D and E, respectively. The projections of the points B and E on CD are denoted by B′ and E′, respectively. The projections of the points D and C on BE are denoted by D′ and C′, respectively. Prove that the points B′, D′, E′ and C′ lie on the same circle. Dan Brânzei, 2001
Problem 38 Let ABCD be a rectangle. We consider the points E ∈ CA, F ∈ AB, G ∈ BC such that DE ⊥ CA, EF ⊥ AB and EG ⊥ BC. Solve in the set of rational numbers the equation ACx = EFx + EGx. Dan Brânzei, 2001
Problem 39 Let ABCD be a quadrilateral inscribed in the circle O. For a point E ∈ O, its projections K, L, M, N on the lines DA, AB, BC, CD, respectively, are considered. Prove that if N is the orthocenter of the triangle KLM for some point E, different from A, B, C, D, then this holds for every point E of the circle O. Dan Brânzei, 2001
Problem 40 Let ABCDEF be a hexagon with AB || DE, BC || EF, CD || FA and in which the diagonals AD, BE and CF are congruent. Prove that the hexagon can be inscribed in a circle. Dan Brânzei, 2001
Problem 41 Find the minimal area of a rectangular box of volume strictly greater than 1000, given that the side lengths are integers. Dinu Şerbănescu, 2001
Problem Problem 42 Let ABCD be a parallelogram of center O. Let M and N be the midpoints of BO and CD, respectively. Prove that if the triangles ABC and AMN are similar, then ABCD is a square. Dinu Şerbănescu, 2002
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Problem 43 Let ABC be an isosceles triangle such that AB = AC and 'A = 20°. Let M be the foot of the altitude 1 from C and let N be a point on the side AC such that CN = BC . Find the measure of the angle 'AMN. 2 Dinu Şerbănescu, 2002
Problem 44 Let ABCD be a unit square. For any interior points M and N such that the line MN does not contain any vertices of the square, denote by s(M, N) the least area of a triangle having vertices in the set {A, B, C, D, M, N}. Find the least number k such that s(M, N) ≤ k, for all such points M, N. Dinu Şerbănescu, 2002
Problem 45 The diagonals AC and BD of a convex quadrilateral meet at O. Let m be the measure of the acute angle formed by these diagonals. For any angle xOy of measure m, the area inside the angle that is in the interior of the quadrilateral is constant. Prove that ABCD is a square. Mircea Fianu, 2002
Problem 46 Let C1(O1) and C2(O2) be two circles such that C1 passes through O2. Point M lies on C1 such that M ∉ O1O2. The tangents from M at C2 meet again C1 at A and B. Prove that the tangents from A and B at C2 – others than MA and MB – meet at a point located on C1. Dinu Şerbănescu, 2002
Problem 47 Five points are given in the plane such that each of 10 triangles defined by them has area greater than 2. Prove that there exists a triangle of area greater than 3. LaurenŃiu Panaitopol, 2002
Problem 48 Consider n > 2 concentric circles and two lines D1, D2 which meet at P, a point inside all the circles. The rays determined by P on the line D1 meet the circles in points A1, A2, …, An and A1′, A2′ , ..., An′ respectively and the rays on D2 meet the circles at points B1, B2, …, Bn and B1′, B2′ , ..., Bn′ (points with the same indices lie on the same circle). Prove that if the arcs A1B1 and A2B2 are equal, then the arcs AiBi and Ai′Bi′ are equal, for all i = 1, 2, …, n. Dinu Şerbănescu, 2002
Problem 49 Let ABC be a triangle and a = BC, b = CA and c = AB be the lengths of its sides. Points D and E lie in the same halfplane determined by BC as A. Suppose that DB = c, CE = b and that the area of DECB is maximal. Let F be the midpoint of DE and let FB = x. Prove that FC = x and 4x3 = (a2 + b2 + c2)x + abc. Dan Brânzei, 2002
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Problem 50 Consider a rhombus ABCD with center O. A point P is given inside the rhombus, but not situated on the diagonals. Let M, N, Q, R be the projection of P on the sides (AB), (BC), (CD), (DA), respectively. The perpendicular bisectors of the segments MN and RQ meet at S and the perpendicular bisectors of the segments NQ and MR meet at T. Prove that P, S, T and O are the vertices of a rectangle. Mircea Fianu, 2003
Problem 51 Two circles C1(O1) and C2(O2) with distinct radii meet at points A and B. The tangent from A to C1 intersects the tangent from B to C2 at point M. Show that both circles are seen from M under the same angle. Dinu Şerbănescu, 2003
Problem 52 Let E be the midpoint of the side CD of a square ABCD. Consider the point M inside the square such that 'MAB = 'MBC = 'BME = x. Find the angle x. LaurenŃiu Panaitopol, 2003
Problem 53 Suppose ABCD and AEFG are rectangles such that the points B, E, D, G are collinear (in this order). Let the lines BC and GF intersect at point T and let the lines DC and EF intersect at point H. Prove that points A, H and T are collinear. Mircea Fianu, 2003
Problem 54 Two unit squares with parallel sides overlap by a rectangle of area 1/8. Find the extreme values of the distance between the centers of the squares. Radu Gologan, 2003
Problem 55 Consider a circle of center O and V a point outside the circle. The tangents from V touch the circle at points T1, T2. Let T be a point on the small arc T1T2 of the circle. The tangent at T intersects the line VT1 in the point A and the lines TT1 and VT2 intersect in the point B. Let M be the intersection point of the lines OM and AB. Prove that lines OM and AB are perpendicular. Mircea Fianu, 2004
Problem 56 Let ABC be an acute triangle and let D be a point on the side BC. Points E and F are the projections of D on the sides AB and AC, respectively. Lines BF and CE meet at point P. Prove that AD is the bisector line of the angle BAC if and only if lines AP and BC are perpendicular. Sever Moldoveanu, 2004
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Problem 57 Consider a triangle ABC with the side lenghts a, b, c so that a is the largest. Prove that the triangle is right-angled if and only if ( a + b + a − b )( a + c + a − c ) = (a + b + c) 2 . Virgil Nicula, 2004
Problem 58 Consider the triangle ABC with AB = AC and a variable point M on the line BC so that B is between M and C. Prove that the sum between the inradius of AMB and the exradius of AMC corresponding to the angle M is constant. Virgil Nicula, 2004
Problem 59 Let ABC be a triangle inscribed in the circle K and consider a point M on the arc BC that does not contain A. The tangents from M to the incircle of ABC intersect the circle K at the points N and P. Prove that if 'BAC = 'NMP, then triangles ABC and MNP are congruent. Valentin Vornicu, 2004
Problem 60 Let M, N, P be the midpoints of the sides BC, CA, AB of the triangle ABC, respectively, and let G be the centroid of the triangle. Prove that if the quadrilateral BMGP is cyclic and 2BN = 3 AB , then the triangle ABC is equilateral. BMO shortlist 2004
Problem 61 Circles C1 and C2 intersect at points A and B. The tangent line from A to C2 meets C1 at point C and the tangent line from A to C1 meets C2 at point D. A ray from A, interior to the angle 'CAD, intersects C1 at M, C2 at N and the circumcircle of the triangle ACD at P. Prove that AM = NP. Mircea Fianu, 2005
Problem 62 Points M and N are given on the sides AD and BC of a rhombus ABCD. Line MC meets the segment BD at T and line MN meets the segment BD at U. Line CU intersects the side AB at Q and – finally! – line QT intersects the side CD at P. Show that the triangles QCP and MCN have the same area. Mircea Fianu, 2005
Problem 63 A point M is given inside an equilateral triangle ABC. Denote by A′, B′, C′ the projections of the point M on the sides BC, CA, AB, respectively. Prove that lines AA′, BB′, CC′ are concurrent if and only if point M lies on an altitude of the triangle. Laurentiu Panaitopol, 2005
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Problem 64 Let ABC be a triangle with BC > CA > AB and let G be the centroid of the triangle. Prove that 'GCA + 'GBC < 'BAC < 'GAC + 'GBA. Dinu Şerbănescu, 2005
Problem 65 Three circles C1(O1), C2(O2), C3(O3) share a common point Q and meet again pairwisely at points A, B, C. Show that if points A, B, C are collinear then points Q, O1, O2, O3 are cocyclic. Simpson, 2005
Problem 66 Let AB and BC be two consecutive sides of a regular polygon with 9 vertices inscribed in a circle of center O. Let M be the midpoint of AB and let N be the midpoint of the radius perpendicular to BC. Find the measure of the angle 'OMN. * * *, 2005
Problem 67 A piece of cardboard has the shape of a pentagon ABCDE in which BCDE is a square and ABE is an isosceles triangle right-angled at A. Prove that the pentagon can be divided in 2 different ways in three parts that can be rearranged in order to recompose a right isosceles triangle. * * *, 2005
Problem 68 Let ABC be a triangle right at C and consider points D, E on the sides BC, CA, respectively such BD AE = = k . Lines BE and AD meet at point O. Show that 'BOD = 60° if and only if that AC CD k= 3. Marcel ChiriŃă, 2006
Problem 69 In a plane 5 points are given such that all triangles having vertices at these points are of area not greater than 1. Show that there exists a trapezoid which contains all points in the interior (or on the sides) and having the area not exceeding 3. Marcel ChiriŃă, 2006
Problem 70 Consider a circle C of center O and let A, B be points on the circle with 'AOB = 90°. Circles C1(O1) and C2(O2) are internally tangent to C at points A, B, respectively, and – moreover – are tangent to themselves. Circle C3(O3), located inside the angle 'AOB, is externally tangent to C1, C2 and internally tangent to C. Prove that points O, O1, O2, O3 are vertices of a rectangle. * * *, 2006
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Problem 71 Suppose ABCD is a cyclic quadrilateral of area 8. Prove that if there exists a point O in the plane of the triangle such that OA + OB + OC + OD = 8, then ABCD is either an isosceles trapezoid or a square. Flavian Georgescu, 2006
Problem 72 Let ABC be a triangle and let A1, B1, C1 be the midpoints of the sides BC, CA, AB, respectively. Show that if M is a point in the plane of the triangle such that MA MB MC = = =2, MA1 MB1 MC1 then M is the centroid of the triangle. Dinu Şerbănescu, 2006
Problem 73 Let ABC be a triangle and D a point inside the triangle, located on the median from A. Show that if 'BDC = 180° – 'BAC, then AB ⋅ CD = AC ⋅ BD. Eduard Băzăvan, 2006
Problem 74 Consider a trapezoid ABCD with the bases AB and CD so that with the diameters AD and BC are secant; denote by M and N their common points. Prove that the midsection point of the diagonals AC and BD belongs to the line MN. Sever Moldoveanu, 2007
Problem 75 Let ABCD be a convex quadrilateral. The incircle ω1 of triangle ABD touches the sides AB, AD at points M, N respectively, while the incircle ω2 of triangle CBD touches the sides CD, CB at points P, Q, respectively. Given that ω1 and ω2 are tangent, show that: a) the quadrilateral ABCD is circumscriptible; b) the quadrilateral MNPQ is cyclic; c) the incircles of triangles ABC and ADC are tangent. Vasile Pop, 2007
Problem 76 Let ABC be an acute-angled triangle with AB = AC. For any point P inside the triangle ABC consider the circle centered at A with radius AP and let M and N be the intersection points of the sides AB and AC with the circle. Determine the position of the point P such that MN + BP + CP is minimum. Francisc Bozgan, 2007
Problem 77 Let ABC be a triangle. Points M, N, P are given on the sides AB, BC, CA respectively, so that CPMN is a parallelogram. Lines AN and MP meet at point R, lines BP and MN meet at point S, while Q is the intersection point of the lines AN and BP. Show that S[MRQS] = S[NQP]. Mircea Lascu, 2007
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Problem 78 Circles ω1 and ω2 meet at points A and B. A third circle ω3, which intersects ω1 at points D and B, is internally tangent to ω2 at point C and tangent to the line AB at point F, and lines DE and AB meet at point G. Let H be the mirror image of F across G. Find the measure of the angle 'HCF. Lucian łurea, 2007
Problem 79 Let ρ be a semicircle of diameter AB. A parallel line to AB intersects semicircle in C and D so that points B and C lie on opposite sides of the line AD. The parallel line from C to AD meets ρ again at point E. Lines BE and CD meet at point F and the parallel line from F to AD intersects AB at point P. Prove that the line PC is tangent to the semicircle ρ. Cosmin PohoaŃă, 2007
Problem 80 Let ABC be a triangle right-angled at A and let D be a point on the side AC. Point E is the mirror image of A across BD and point F is the intersection of the line CE with the perpendicular line from D to CB. Show that the lines AF, DE and CB are concurrent. Dinu Şerbănescu, 2007
Problem 81 Let ABC be an acute-angled triangle. Consider the equilateral triangle A′UV, with A′ ∈ (BC), U ∈ ∈ (AC), V ∈ (AB) such that UV || BC. Similarly, define points B′ ∈ (AC) and C′ ∈ (AB). Show that the lines AA′, BB′ and CC′ are concurrent. Vasile Pop, 2008
Problem 82 Let ABC be a triangle and D the midpoint of BC. On the sides AB and AC there are points M, N, respectively, other than the midpoints of these segments, so that AM 2 + AN 2 = BM 2 + CN 2 and 'MDN = 'BAC. Prove that A = 90°. Francisc Bozgan, 2008
Problem 83 Consider an acute-angled triangle ABC, the height AD and the point E where the diameter from A of the circumcircle intersects the line BC. Let M, N be the mirror images of D across the lines AC and AB. Show that 'EMC = 'BNE. Dinu Şerbănescu, 2008
Problem 84 Let d be a line and let M, N be two points on d. Circles α, β, γ, δ centered at A, B, C, D are tangent to d in such a manner that circles α, β are externally tangent at M, while circles γ, δ are externally tangent at N. Moreover, points A and C lie on the same side of line d. Prove that if there exists a circle tangent to all circles α, β, γ, δ, containing all of them in the interior, then lines AC, BD and d are concurrent or parallel. Flavian Georgescu, 2008
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Problem 85 Let ABCD be a quadrilateral with no two opposite sides parallel. The parallel from A to BD meets the line CD at point F and the parallel from D at AC meets the line AB at point E. Consider the midpoints M, N, P, Q of the segments AC, BD, AF, DE respectively. Show that lines MN, PQ and AD are concurrent. Dinu Şerbănescu, 2008
Problem 86 Consider a rhombus ABCD. Point M and N are given on the line segments AC and BC respectively, such that DM = MN. Lines AC and DN meet at point P and lines AB and DM meet at point R. Prove that RP = PD. Cristinel Mortici, 2009
Problem 87 Let ABCD be a quadrilateral. The diagonals AC and BD are perpendicular at point O. The perpendiculars from O on the sides of the quadrilateral meet AB, BC, CD, DA at M, N, P, Q, respectively, and meet again CD, DA, AB, BC at M′, N′, P′, Q′, respectively. Prove that points M, N, P, Q, M′, N′, P′, Q′ are concyclic. Cosmin PohoaŃă, 2O09
Problem 88 Consider a regular polygon A0A1…An–1, n ≥ 3, and m ∈ {1, 2, …, n – 1}, m ≠ n/2. For any number i ∈ {1, 2, …, n – 1}, r(i) be the remainder of i + m at the division by n. Prove that no three segments AiAr(i) are concurrent. * * *, 2009
Problem 89 Consider K a polygon in plane, such that the distance between any two vertices is not greater than 1. Let X and Y be two points inside K. Show that there exist a point Z, lying on the border of K, such that XZ + YZ ≤ 1. * * *, 2009
Problem 90 Show that in any triangle ABC with A = 90° the following inequality holds: (AB – AC)2(BC 2 + 4AB ⋅ AC)2 ≤ 2BC 6. Lucian Petrescu, 2009
Problem 91 Let ABC be a triangle and A′ the foot of the internal bisector of angle BAC. Consider the perpendicular line from A′ on BC. Define analogously the lines dB and dC. Prove that lines dA, dB and dC are concurrent if and only if triangle ABC is isosceles. * * *, 2009
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Chapter III NUMBER THEORY Problem 92 Find n ∈ Z such that the number
4n − 2 is rational. n+5
Dan Popescu, 2001
Problem 93 Three students write on the blackboard next to each other three two-digit squares. In the end, they observe that the 6-digit number thus obtained is also a square. Find this number! Mircea Becheanu, 2001
Problem 94 Determine all positive integers a < b < c < d with the property that each of them divides the sum of the other three. * * *, 2001
Problem 95 Let n be a non-negative integer. Find the non-negative integers a, b, c, d, such that a2 + b2 + c2 + d 2 = 7 ⋅ 4n. LaurenŃiu Panaitopol, 2001
Problem 96 Let n ≥ 2 be a positive integer. Find the positive integers x such that x + x + ... + x < n ,
for any number of radicals. Ion Dobrotă, 2001
Problem 97 Let k, n, p be positive integers such that p is a prime number, k < 1000 and
k =n p .
a) Prove that if the equation k + 100 x = (n + x) p has a non-zero integer solution, then p is a divisor of 10. b) Find the number of all non-negative solutions of the above equation. Mircea Fianu, 2002
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Problem 98 Find all positive integers a, b, c, d such that a + b + c + d – 3 = ab + cd. Dinu Şerbănescu, 2002
Problem 99 Let n be an even positive integer and let a, b be two coprime positive integers. Find a and b such that a + b is a divisor of an + bn. Dinu Şerbănescu, 2002
Problem 100 100 Let a be an integer. Prove that for any real number x, x3 < 3, both numbers 3 − x 2 and 3 a − x3 cannot be rational. LaurenŃiu Panaitopol, 2002
Problem 101 101 The last four digits of a perfect square are equal. Prove that all of them are zeros. LaurenŃiu Panaitopol, 2002
Problem 102 102 Let m, n > 1 be integer numbers. Solve in positive integers xn + yn =2m. LaurenŃiu Panaitopol, 2002
Problem 103 103 Let p, q be two distinct primes. Prove that there are positive integers a, b such that the arithmetic mean of all positive divisors of the number n = paqb is an integer. LaurenŃiu Panaitopol, 2002
Problem 104 104 4 Consider the prime numbers n1 < n2 < … < n31. Prove that if 30 divides n14 + n24 + ... + n31 , then among these numbers one can find three consecutive primes. Vasile Berghea, 2003
Problem Problem 105 Let n be a positive integer. Prove that there are no positive integers x and y such as n + n + 1 < x + y < 4n + 2 . Dinu Şerbănescu, 2003
Problem 106 Let a be a positive integer such that the number an has an odd number of digits in the decimal representation, for all n > 0. Prove that the number a is an even power of 10. Vasile Zidaru, 2003
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Problem 107 Find all positive integers n for which there exist distinct integers a1, a2, …, an such that 1 2 n a + a + ... + an . + + ... + = 1 2 2 a1 a2 an Dinu Şerbănescu, 2004
Problem 108 Let p, q, r be primes and let n be a positive integer such that pn + qn = r2. Prove that n = 1. LaurenŃiu Panaitopol, 2004
Problem 109 A finite set of positive integers is called isolated if the sum of the elements in any proper subset is a number relatively prime with the sum of the elements of the isolated set. Find all nonprime integers n for which there exist positive integers a, b such that the set A = {(a + b)2, (a + 2b)2, …, (a + nb)2} is isolated. Gabriel Dospinescu, 2004
Problem 110 110 Find the greatest integer n, n > 10 such that the remainder of n when divided by each square n between 2 and is an odd integer. 2 Adrian Stoica, 2005
Problem 111 111 Let k, r ∈ N* and let x ∈ (0, 1) be a rational number given in decimal representation: x = 0,a1a2a3a4… Show that if the decimals ak, ak+r, ak+2r, ak+3r, … are canceled, the new number thus obtained is still rational. Dan Schwarz, 2005
Problem 112 112 Find all positive integers n and p if p is prime and n8 – p5 = n2 + p2. Adrian Stoica, 2005
Problem Problem 113 113 For any positive integer n let s(n) be the sum of its digits in decimal representation. Find all numbers n for which s(n) is the largest proper divisor of n. LaurenŃiu Panaitopol, 2006
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Problem 114 Consider the integers a1, a2, a3, a4, b1, b2, b3, b4 with ak ≠ bk for all k =1, 2, 3, 4. If {a1, b1} + {a2, b2} = {a3, b3} + {a4, b4}, show that the number |(a1 – b1)(a2 – b2)(a3 – b3)(a4 – b4)| is a square. Note. For any sets A and B, we denote A + B = {x + y | x ∈ A, y ∈ B}. Adrian Zahariuc, 2006
Problem 115 115 For a positive integer n denote r(n) the number having the digits of n in reverse order – for example, r(2006) = 6002. Prove that for any positive integers a and b the numbers 4a2 + r(b) and 4b2 + r(a) can not be simultaneously squares. Marius Ghergu, 2006
Problem 116 Find all integers n, n ≥ 4 such that [ n ] + 1 divides n – 1 and [ n ] – 1 divides n + 1. Marian Andronache, 2007
Problem 117 Solve in positive integers the equation (x2 + 2)(y2 + 3)(z2 + 4) = 60xyz. Flavian Georgescu, 2007
Problem 118 Determine all positive integers n which can be represented in the form n = [a, b] + [b, c] + [c, a], where a, b, c are positive integers. Note: [p, q] is the lowest common multiple of the integers p and q. Adrian Zahariuc, 2007
Problem 119 Let p be a prime number, p ≠ 3, and let a, b be integer numbers such that p | a + b and p2 | a3 + b3. Show that p2 | a + b or p3 | a3 + b3. * * *, 2008
Problem 120 120 Prove that for any positive integer n there exists a multiple of n with the sum of its digits equal to n. Mihai Bălună, 2008
Problem 121 121 Let n ∈ N, n ≥ 2. Consider the integers a1, a2, …, an with 0 < ak ≤ k, for all k = 1, 2, …, n. Given that a1 + a2 + … + an is an even number, prove that one can choose the signs ‘+’ and ‘–’ such that a1 ± a2 ± … ± an = 0. * * *, 2008
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Problem 122 122 A sequence of integers a1, a2, …, an is given so that ak is the number of all multiples of k in this sequence, for any k = 1, 2, …, n. Find all possible values for n. Cristian Mangra, 2008
Problem 123 123 Let a, b be real numbers with the property that the integer part of an + b is an even number, for all n ∈ N. Show that a is an even integer. Dinu Şerbănescu, 2002
Problem 124 Find all primes p, q satisfying the equation 2pq – qp = 7. Francisc Bozgan, 2008
Problem 125 Find all pairs of integers (m, n), n, m > 1 so that mn – 1 divides n3 – 1. Francisc Bozgan, 2008
Problem 126 For all positive integers n define an = 233 ... 3 , where digit 3 occurs n times. Show that the number n times
a2009 has infinitely many multiples in the set {an | n ∈ N*}. Cristian Lazăr, 2009
Problem 127 Find all non-negative integers a, b, c, d such that: 7a = 4b + 5c + 6d. Petre Stângescu, 2009
Problem 128 A positive integer is called saturated if any prime factor occurs at a power greater than or equal to 2 in its factorisation. For example, numbers 8 = 23 and 9 = 32 are saturated; moreover, they are consecutive. Prove that there exist infinitely many saturated consecutive numbers. * * *, 2009
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Chapter IV COMBINATORICS Problem 129 Consider a 1 × n rectangle and some tiles of size 1 × 1 of four different colours. The rectangle is tiled in such a way that no two neighboring square tiles have the same colour. a) Find the number of distinct symmetrical tilings. b) Find the number of tilings such that any consecutive square tiles have distinct colours. Dan Brânzei, 2002
Problem 130 130 1 and the central square is colored 3 black. The remaining eight squares are analogously divided into nine squares each, and central squares are colored in black. Prove that after 1000 steps the total area of the black region exceeds 0.999. Cristinel Mortici and Costel Chiteş, 2002
A square of side 1 is decomposed into 9 equal squares of sides
Problem 131 131 An equilateral triangle of side 10 is divided into 100 unit equilateral triangles by lines parallel to the sides of the triangle. Find the number of (not necessarily unit) equilateral triangles in the configuration described above such that the sides of the triangles are parallel to the sides of the initial one. Dinu Şerbănescu, 2002
Problem 132 132 Show that one can color all the points of a plane using only two colors such that no line segment has all points of the same color. Valentin Vornicu, 2003
Problem 133 133 Consider a cube and let M, N be two of its vertices. Assign the number 1 to these vertices and 0 to the other six vertices. We are allowed to select a vertex and to increase with a unit the numbers assigned to the 3 adjacent vertices – call this a movement. Prove that there is a sequence of movements after which all the numbers assigned to the vertices of the cube became equal if and only if MN is not a diagonal of a face of the cube. Marius Ghergu and Dinu Şerbănescu, 2004
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Problem 134 An array 8 × 8 consists of 64 unit squares. Inside each square are written the numbers 1 or –1 so that in any 2 × 2 subarray the sum of the four numbers equals 2 or –2. Prove that there exist two rows of the array which are equal. Marius Ghergu, 2004
Problem 135 In a chess tournament each of the players have played with all the others two games, one time with the white pieces and then with the black pieces. In each game the winners sets one point and both players receive 0.5 points if the game ends with draw. At the end of the tournament, all the players end with the same number of points. a) Prove that there are two players with the same number of draws. b) Prove that there are two players with the same number of losses when playing the white. Marius Ghergu, 2004
Problem 136 A regular polygon with 1000 sides has the vertices colored in red, yellow or blue. A move consists in choosing to adjacent vertices colored differently and coloring them in the third color. Prove that there is a sequence of moves after which all the vertices of the polygon will have the same color. Marius Ghergu, 2004
Problem 137 A country has six cities with airports and two rival flight companies. Any two cities are connected by flights so that on each route between two cities one may travel with exactly one of the two flight companies. Prove that you can visit 4 cities in a cycle flying with the same air company (that is, there exist four cities A, B, C, D and a company which operates on the routes A ↔ B, B ↔ C, C ↔ D and D ↔ A). Dan Schwarz, 2005
Problem 138 A phone company starts a new type of customer service. A new client can choose k phone numbers in this network which are call-free – regardless if is called or if calling. A group of n students decide to take advantage of this promotion. • Show that if n ≥ 2k + 2 then there will exist 2 students which will be charged when speaking. • Show that if n = 2k + 1 then there exists a way of arranging the free calls so that in this group everybody speaks free to anyone else. Valentin Vornicu, 2005
Problem 139 The positive integers from 1 to n2 are placed arbitrarily on squares of an n × n chessboard. Two squares are called adjacent if they have a common side. Show that two opposite corner squares can be joined by a path of 2n – 1 adjacent squares so that the sum of the numbers places on them is at least n3 2 + n − n +1 . 2 Radu Gologan, 2005
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Problem 140 140 An 7 × 7 array is divided in 49 unit squares. Find all integers n ∈ N* for which n checkers can be placed on the unit squares so that each row and each line have an even number of checkers. (0 is an even number, so there may exist empty rows or columns. A square may be occupied by at most 1 checker.) Dinu Şerbănescu, 2006
Problem 141 141 A rectangular cardboard is divided successively into smaller pieces by a straight cut; at each step, only one single piece is divided in two. Find the smallest number of cuts required in order to obtain – among others – 251 polygons with 11 sides. Marian Andronache, 2007
Problem 142 142 Consider a n × n array divided into unit squares which are randomly colored in black or white. Three of the four comer squares are colored in white and the fourth is colored in black. Prove that there exists a 2 × 2 square which contains an odd number of white squares. Lidia Ilie, 2007
Problem 143 143 Consider the numbers from 1 to 16. A solitaire game is played in the following manner: the numbers are paired and each pair is replaced by the greatest prime divisor of the sum of the numbers in that pair – for example, (1, 2); (3, 4); (5, 6); …; (15, 16) produces the sequence 3, 7, 11, 5, 19, 23, 3, 31. The game continues similarly until one single number is left. Find the greatest possible value of the number which ends the game. Adrian Stoica, 2007
Problem 144 Eight persons attend a party, and each participant has at most three others to whom he/she cannot speak. Show that the persons can be grouped in 4 pairs so that each pair can converse. Mihai Bălună, 2007
Problem 145 A set of points is called free if there is no equilateral triangle whose vertices are among the points in the set. Show that any set of n points in the plane contains a free subset of at least n points. Călin Popescu, 2007
Problem 146 A 8 × 8 square board is divided into 64 unit squares. A “skew-diagonal” of the board is a set of 8 unit squares with the property that each row or column of the board contains only one unit square of the set. Checkers are placed in some of the unit squares so that each “skew-diagonal” has exactly 2 squares occupied by checkers. Prove that there exist two rows or two columns which contain all the checkers. Dinu Şerbănescu, 2007
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Problem 147 To obtain a square P of side length 2 cm divided into 4 unit squares it is sufficient to draw 3 squares: P and another 2 unit squares with a common vertex, as shown below:
Find the minimum number of squares sufficient to obtain a square of side length n cm divided into n unit squares (n ≥ 3 is an integer). * * *, 2009 2
Problem 148 148 The plane is divided into a net of equilateral triangles of side length 1, with disjoint interiors. A checker is placed initially inside a triangle. The checker can be moved into another triangle sharing a common vertex (with the triangle hosting the checker) and having the opposite sides (with respect to this vertex) parallel. A path consists in a finite sequence of moves. Prove that there is no path between two triangles sharing a common side. Vasile Pop, 2009
Problem 149 Show that there exist (at least) a rearrangement a0, a1, a2, …, a63 of the numbers 0, 1, 2, …, 63, such that ai – aj ≠ aj – ak, for any i < j < k ∈ {0, 1, 2, …, 63}. * * *, 2009
Problem 150 Let A = {1, 2, …, 2006}. Find the maximal number of subsets of A that can be selected such that the intersection of any 2 distinct subsets has 2004 elements. * * *, 2006
Problem 151 Let 1 ≤ m < n be positive integers, and consider the set M = {(x, y) | x, y ∈ N*, 1 ≤ x, y ≤ n}. Determine the least value v(m, n) with the property that for any subset P ⊆ M with |P| = v(m, n) there exist m + 1 elements Ai = (xi, yi) ∈ P, i = 1, 2, …, m + 1, for which the values xi are all distinct, and yi are also all distinct. Vasile Pop, 2007
Problem 152 Ten numbers are chosen at random from the set 1, 2, 3, …, 37. Show that one can select four distinct numbers from the chosen ones so that the sum of two of them is equal to the sum of the other two. Vasile Pop, 2008
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Problem 153 Let m, n ∈ N* and A = {1, 2, …, n}, B = {1, 2, …, m}. A subset S of the set product A × B has the property that for any pairs (a, b)(x, y) ∈ S, then (a – x)(b – y) ≤ 0. Show that S has at most m + n – 1 elements. Dinu Şerbănescu, 2007
Problem 154 154 In the interior of a circle centered in O a number of 1200 points A1, A2, …, A1200 are considered such that for every i, j with 1 ≤ i ≤ j ≤ 1200, the points O, Ai and Aj are not collinear. Prove that there exist the points M and N on the circle, with m('MON) = 30°, such that in the interior of the angle 'MON lie exactly 100 points. * * *, 2001
Problem 155 Consider a convex polygon with n ≥ 5 sides. Prove that there are at most
n(2n − 5) triangles of 3
area 1 with the vertices among the vertices of the polygon. Andrei NeguŃ, 2004
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34
FORMAL SOLUTIONS
35
36
Chapter I ALGEBRA Solution to Problem 1 Observe that x < 3 x for x ∈ [0, 1]. Thus, we have
abc < 3 abc and
(1 − a )(1 − b)(1 − c) < 3 (1 − a )(1 − b)(1 − c) . By the AM-GM inequality, we get a+b+c abc < 3 abc ≤ 3 and (1 − a ) + (1 − b) + (1 − c) (1 − a )(1 − b)(1 − c) < 3 (1 − a)(1 − b)(1 − c) ≤ . 3 Summing up, we obtain a +1− a + b +1− b + c +1− c =1 , abc + (1 − a)(1 − b)(1 − c) < 3 as desired.
Solution to Problem 2 1 1 1 Setting x = , y = , z = , we have xyz = 1. The inequality rewrites as a b c 3 6 1+ . ≥ xy + yz + zx x + y + z Since (x + y + z)2 ≥ 3(xy + yz + zx), it follows that 3 9 1+ . ≥1+ xy + yz + zx ( x + y + z)2 It suffices to observe that 9 6 1+ , ≥ 2 x+ y+z ( x + y + z) which reduces to 2
3 1 − ≥ 0 . x+ y+ z
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Solution to Problem 3 Using the Chebyshev inequality we derive (a + b + c)(a2 + b2 + c2) ≤ 3(a3 + b3 + c3), hence a + b + c ≤ 1. On the other hand, 4(ab + bc + ca) – 1 ≥ a2 + b2 + c2 ≥ ab + bc + ca, therefore ab + bc + ca ≥ 1. As 3(ab + bc + ca) ≤ (a + b + c)2 ≤ 1, we obtain a + b + c ≥ 1, thus a + b + c = 1. Consequently, a + b + c = 1 and 3(ab + bc + ca) = (a + b + c)2, which imply 1 a=b=c= . 3 Solution to Problem 4 The given condition rewrites abc(a + b + c) ≥ ab + bc + ca. From the Cauchy-Schwarz inequality we have ab + bc + ca ≥ 3abc(a + b + c) , so the conclusion follows. Solution to Problem 5 Let s = a + b + c. The given condition rewrites as 1 = (s – a)(s – b)(s – c) = s(ab + bc + ca) – abc, 1 + abc so ab + bc + ca = . From the AM-GM inequality we obtain a+b+c 1 3 3 a + b + c = ((a + b) + (b + c) + (c + a )) ≥ ⋅ (a + b)(b + c)(c + a ) = . 2 2 2 1 On the other side, 1 = (a + b)(b + c)(c + a) ≥ 8 ⋅ ab ⋅ bc ⋅ ca , hence abc ≤ . Consequently 8 1 2 3 ab + bc + ca ≤ 1 + ⋅ = , 8 3 4 as needed. Solution to Problem 6 Wlog, assume that a ≤ b ≤ c. 3 3 3 If a + b ≤ c, then c > , while a < and b < . In this case (3 – 2a)(3 – 2b)(3 – 2c) < 0 ≤ a2b2c2 2 2 2 and we are done. 3 If a + b > c, then a, b, c are the side lengths of a triangle with semiperimeter p = . The inequality 2 rewrites 8(p – a)(p – b)(p – c) ≤ a2b2c2.
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Using the formulas S2 = p(p – a)(p – b)(p – c) and abc = 4RS we obtain 8S2 ≤ 16pR2S2 or 1 ≤ 3R2. 3 This inequality can be proved in many ways, for example from 2p ≤ 3 3 R, since p = . 2 Solution to Problem 7 The inequality rewrites as a4 + b4 + c4 ≥ abc(a + b + c). Using successively the inequality x2 + y2 + z2 ≥ xy + yz + zx we get a4 + b4 + c4 ≥ a2b2 + b2c2 + c2a2 = (ab)2 + (bc)2 + (ca)2 ≥ abc(a + b + c) as desired. Solution to Problem 8 a b c Let = x, = y and = z. The inequality rewrites successively: b c a 3 1 1 1 x2 + y2 + z2 + 2xy + 2yz + 2zx ≥ x + y + z + + + ⇔ 2 x y z 1 1 1 3 1 1 1 ⇔ x2 + y2 + z2 + 2 + + ≥ x + y + z + + + ⇔ 2 x y z x y z 1 1 1 ⇔ 2(x2 + y2 + z2) + + + ≥ 3(x + y + z). x y z From AM-GM inequality we get 1 1 1 2x2 + = x 2 + x 2 + ≥ 3 ⋅ 3 x 2 ⋅ x 2 ⋅ = 3 x . x x x Summing with the analogously relations we get the conclusion.
Solution to Problem 9 First solution. Using Cauchy-Schwarz inequality we get: (a 2 + b 2 + c 2 ) 2 a2 b2 c2 a4 b4 c4 + + = 2+ 2+ 2≥ 2 . b c a ba cb ac a b + b 2c + c 2 a The inequality is reduced to a2 + b2 + c2 ≥ 3(a2b + b2c + c2a) or (a + b + c)(a2 + b2 + c2) ≥ 3(a2b + b2c + c2a). The latter rewrites as a(a − b) 2 ≥ 0 , which is obvious.
∑
Second solution. Since a + b + c = (a + b + c)2 = 1, the inequality gives successively: a2 b2 c2 + + − (a + b + c) ≥ 3(a 2 + b 2 + c 2 ) − (a + b + c) 2 , b c a or a2
∑ b
− 2a + b ≥
∑ ( a − b)
2
,
39
hence
∑
(a − b) 2 ≥ b
∑ ( a − b)
2
.
Since a, b, c ≤ 1, we are done. Solution to Problem 10 Canceling the denominators we get 1 = xy + yz + zx + 2xyz. From AM-GM inequality we get 1≥4⋅
4
2 x3 y 3 z 3 ,
so 1 ≥ (8xyz)3. The conclusion follows. Solution to Problem 11 Solving for a and b the system of equations x + 2 y − a = s 2 x − y − b = t , one has (a + 2b) + ( s + 2t ) x = 5 y = ( 2 a − b) + ( 2 s − t ) . 5 Restating the claim, one has to prove that there exists a unique pair of integers s, t ∈ Z with s2 + t2 ≤ 1 so that both numbers (a + 2b) + (s + 2t), (2a – b) + (2s – t) are divisible by 5. Notice that (a + 2b) + (s + 2t) + 2[(2a – b) + (2s – t)] = 5(a + s), so x ∈ Z ⇔ y ∈ Z. 2 2 Since s + t ≤ 1 ⇔ (s, t) ∈ {(0, 0), (1, 0), (0, 1), (–1, 0), (0, –1)}, it follows that (a + 2b) + (s + 2t) ∈ {a + 2b – 2, a + 2b – 1, a + 2b, a + 2b + 1, a + 2b + 2}, so there is exactly one pair (s, t) with 5 | (a + 2b) + (s + 2t)|, as needed. Solution to Problem 12 The Cauchy-Schwarz inequality gives (a + b + 1)(a + b + c2) ≥ (a + b + c)2, so a + b + c2
1
∑ (a + b + c) ≥ ∑ a + b + 1 ≥ 1 . 2
cyc
Then 2
∑a + ∑a cyc
and the claim follows immediately.
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cyc
cyc
2
≥ (a + b + c) 2 =
∑a cyc
2
+2
∑ ab , cyc
Solution to Problem 13 Let p = |(x – y)(y – z)(z – x)|. Recall the identities: x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) and 1 x2 + y2 + z2 – xy – yz – zx = [(a – y)2 + (y – z)2 + (z – x)2]. 2 Using AM-GM inequality, we have 3 (1) x2 + y2 + z2 – xy – yz – zx ≥ 3 p 2 . 2 On the other hand, since |x – y| ≤ x + y, |y – z| ≤ y + z and |z – x| ≤ z + x, it follows that 2(x + y + z) ≥ |x – y| + |y – z| + |z – x|. Applying again the AM-GM inequality gives 3 (2) 2(x + y + z) ≥ 3 p , 2 and the claim follows from the inequalities (1) and (2). Solution to Problem 14 ab + bc + ca 3 ≥ (abc) 2 . As ab + bc + ca = 3, then 3
Using the AM-GM inequality we derive abc ≤ 1. Now 1 = 2 1 + a (b + c) as claimed.
∑
1
1
1
1
∑1 + a(ab + ac) = ∑1 + a(3 − bc) = ∑ 3a + (1 − abc) ≤ ∑ 3a =
ab + bc + ca 1 , = 3abc abc
Solution to Problem 15 Observe that the numbers a = b = 2, c = 0 fulfill the condition ab + bc + ca = a + b + c. Plugging 1 1 1 into the given inequality, we derive that 4 + + − k ≥ k , hence k ≤ 1. 4 2 2 We claim that the inequality holds for k = 1, proving that the maximum value of k is 1. For this, rewrite the inequality as 1 1 ab + bc + ca 1 (ab + bc + ca) + + − 1 ≥ 1 ⇔ ≥ ab + bc + ca + 1 ⇔ a+b a+b b+c a+c ab ab ⇔ + c ≥ ab + bc + ca + 1 ⇔ ≥1. a+b a+b ab ab Notice that ≥ , since a, b, c ≥ 0. Summing over a cyclic permutation of a, b, c we a+b a+b+c get ab ab ab + bc + ca ≥ = = 1, a+b a+b+c a+b+c as needed.
∑
∑
∑
∑
∑
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Alternative solution. The inequality is equivalent to the following: a+b+c 1 1 1 S= + + ≥k. a + b + c + 1 a + b b + c a + c Using the given condition, we get 1 1 1 a 2 + b 2 + c 2 + 3(ab + bc + ca) + + = = (a + b)(b + c)(c + a) a+b b+c a+c =
a 2 + b 2 + c 2 + 2(ab + bc + ca) + a + b + c (a + b + c)(a + b + c + 1) = , (a + b + c)(ab + bc + ca) − abc (a + b + c) 2 − abc
hence (a + b + c) 2 . (a + b + c) 2 − abc Now it is clear that S ≥ 1, and the equality occurs for abc = 0. Therefore k = 1 is the maximum value.
S=
Solution to Problem 16 Observe that 3a + bc = a2 + ab + bc + ca = (b + a)(c + a), hence a+3 1 = (3 + a)(b + c) = (a + b)(b + c)(c + a ) cyc cyc 3a + bc
∑
=
∑
1 (a + b)(b + c)(c + a )
∑
(9 − a 2 ) =
cyc
(27 − a 2 − b 2 − c 2 ) ≥3 ⇔ (a + b)(b + c)(c + a)
⇔ 27 ≤ ∑ a + 3∏ (3 − a ) = ∑ a 2 + 3(27 − 3(a + b + c) + 3(ab + bc + ca) − abc) ⇔ 2
⇔ 27 ≤ ∑ a 2 + 9∑ ab − 3abc ⇔ 18 + 3abc ≥ 7(ab + bc + ca).
Multiplying by 3 in the last inequality gives: 2(a + b + c)3 + 9abc ≥ 7(ab + bc + ca)(a + b + c) ⇔ ⇔ 2(a3 + b3 + c3 + 6abc + a 2b) + 9abc ≥ 7(3abc + a 2b) ⇔
∑
∑
sym
⇔ 2
∑ a ≤ ∑ a b ⇔ ∑ (a + b)(a − b) 3
2
sym
2
≥ 0.
sym
Solution to Problem 17 The inequality rewrites as a2c + b2a + c2b ≥ a + b + c. Using Cauchy-Schwarz inequality, we have 2 b2 c2 1 1 1 a + + ≥ ( a + b + c) 2 . + + 1 1 1 c a b a b c
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This implies a2c + b2a + c2b ≥
(a + b + c) 2 (a + b + c) ≥ (a + b + c) ≥ 1⋅ (a + b + c) , as claimed. 1 1 1 1 1 1 + + + + a b c a b c
Solution to Problem 18 Let a ∈ A. Then a + 0 ∈ A, hence 0 = 0 ⋅ a ∈ A. For any real number b, b + (–b) = 0 ∈ A, hence 2 –b ∈ A. Thus, A contains all negative numbers. Let c > 0; we have − c − c < 0 , so − c − c ∈ A. It follows that c = (− c )(− c ) ∈ A, hence A = R. Solution to Problem 19 Let a = 2n + 1 and b = 2n − 1 . Then a2 + b2 = 4n, ab = 4n2 – 1 and a2 – b2 = 2. Hence f ( n) =
a 2 + b 2 + ab a 3 − b3 1 = 2 2 = ( 2n + 1)3 − (2n − 1)3 a+b 2 a −b
and thus 1 f (1) + f (2) + ... + f (40) = ( 33 − 13 + 53 − 33 + ... + 813 − 793 ) = 2 1 1 1 = ( 813 − 13 ) = (93 − 1) = (729 − 1) = 364 . 2 2 2
Solution to Problem 20 First, we prove that at most two of the sums a + b, b + c, c + d, d + e and e + a can be negative. b
a
c e d
Indeed, assume that two non-consecutive sums (say a + b and c + d) are less than 0. Then 1 – e = = (a + b) + (c + d) < 0 and so 1 < e, a contradiction. Thus, if three sums are negative, then two of them are not consecutive, which is false. Moreover, if two sums are negative, then these must be consecutive; in other words, at least three consecutive sums are nonnegative. Let a + b, b + c, c + d be greater than or equal to zero. If one of the sums d + e or e + a is negative, then a + b + c = 1 – (d + e) or b + c + d = 1 – (e + a) are at least 1, hence is positive and we are done. Finally, consider the case when all sums a + b, b + c, c + d, d + e, e + a are positive. Suppose that a + b + c < 0; then b > (a + b) + (b + c) > 0. Thus, if a + b + c, b + c + d, c + d + e are negative, then b, c, d are positive and we are done.
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Solution to Problem 21 Let a1 < a2 < … < a2003 be the elements of the set. We prove the claim by contradiction. Numbers a1 + a2003, a2 + a2003, …, a2002 + a2003 divide the sum S = a1 + a2 + … + a2003, since a + b | S – a – b if and only if a + b | S. Hence S = ki(ai + a2003) for all i = 1, 2002 , where ki are integers. Since ai + a2003 < S < 2003a2003 < 2003(ai + a2003), it follows that ki ∈ {2, 3, …, 2002} for all i = = 1, 2002 . By Pigeonhole principle, there is a pair of indices i ≠ j such that ki = kj, a contradiction. Solution to Problem 22 The key idea is to observe that 1 ( n − 1 − 2 n + n + 1) 2 . 2 As 2 n ≥ n − 1 + n − 1 , it follows that the sum is 4 2 − 5 .
an – bn =
Solution to to Problem 23 Consider the integers 0 < m ≤ n < p so that b = a + m, c = a + n and d = a + p. mn is an integer, then Then a(a + p) = (a + m)(a + n) and a + p ≤ a + 1 + 2 a . As p = m + n + a m+n a ≤ mn and p ≥ m + n + 1. On the other hand, 1 + 2 a ≥ n + m + 1 ⇒ a ≥ ≥ mn , hence 2 a ≥ mn. Consequently, a = mn and m = n, so a is a square. Solution to Problem 24 Assume by contradiction that |ak| > 10, for some k. Wlog, let k = 1. Then a12 > 100 and 2 a22 + a32 + ... + a100 + s2 < 1 ,
where s = a1 + a2 + … + a100. On the other hand, the Cauchy-Schwarz inequality yields 2 a12 = (s – a2 – a3 – … – a100)2 ≤ 100( a22 + a32 + ... + a100 + s 2 ) < 100 , a contradiction. Solution to Problem 25 a) In the array below a b b d e f g h i where i is an element of the third row, observe that i = c + f = (a + b) + (d + e) = (a +d) + (b + c) = = g + h. The same argument holds for all the other rows, by induction. b) We prove that d = 2b + 2c – a. Indeed, from the array
44
a
x y z b t u c v d we derive d = u + v = b + t + (u + z) = 2(b + t) +(x + y)= 2b + 2(t + y) + x – y = = 2b + 2c + x – (x + a) = 2b + 2c – a. Solution to Problem 26 It is obvious that 1 ∈ A, since 1 is a divisor of any integer. Consider a, b two elements of A with 1 < a < b. Since at least one of a, b or 1 + ab is even, then 2 is an element of A. We induct on n ≥ 6 to prove that n ∈ A. Assume that k ∈ A for all k = 1, 2, …, n – 1. If n is odd, then n = 2p + 1 with 1 < 2 < p ∈ A, hence n ∈ A. If n is even, then n = 2p. As above, 2p – 1 and 2p + 1 are elements of A and consequently 1 + (2p – 1)(2p + 1) = 4p2 ∈ A. The first property implies n = = 2p ∈ A, as needed. To complete the proof, we need to show that 3, 4, 5 ∈ A. For this, consider a > 2 an element of A. Then 1 + 2 ⋅ a ∈ A, 1 + 2(1 + 2a) = 3 + 4a ∈ A and 1 + (1 + 2a)(3 + 4a) = 4 + 10a + 8a2 ∈ A. If a is even, then 4 | 4 + 10a + 8a2 and so 4 ∈ A. If a is odd, then choose a to be 4 + 10a + 8a2 and again 4 ∈ A. Next, as 1 < 2 < 4 ∈ A we have 1 + 2 ⋅ 4 = 9 ∈ A and so 3 ∈ A. Finally, 7 = 1 + 2 ⋅ 3 ∈ A, 15 = = 1 + 2 ⋅ 7 ∈ A, hence 5 ∈ A and we are done. Solution to Problem 27 Let A = {a, b} and B = {a, c} be two of the given sets. A third set C intersect both of them, so C is {b, c} or {a, d}. In the first case, a fourth set cannot have in common with each of the first three sets exactly one element, so C = {a, d}. Any set from the other n – 3 > 1 sets that do not contain a should contain simultaneously the elements b, c and d, a contradiction. Hence the claim is proved. Solution to Problem 28 a+b 2ab Let x = and y = . Then ab = xy and a2 – a(a + b) + ab = a2 – 2ax + xy = 0, so x(x – y) = 2 a+b = (a – x)2 is a square. Let d be the greatest common divisor of x and y and write x = dx′, y = dy′ with x′, y′ ∈ Z and (x′, y′) = = 1. Then d 2x(x′ – y′) is a square and (x′, x′ – y′) = 1, so both x′ and x′ – y′ are squares; write x′ = m2 and x′ – y′ = n2. Recall that the geometric mean ab = xy is an integer, so xy = d 2x′y′ = d 2m2 (m2 – n2) is a square. Consequently m2 – n2 = p2, where p is an integer. To conclude, to this point we have x = dm2, y = dp2 and since a = x ± x( x − y ) , b = = x ∓ ( x( x − y ) , it follows that a = dm(m ± n) and b = dm(m ∓ n). Therefore |a – b| = 2dmn with the minimum of 30 obtained for d = 1, m = 5, n = 3 – don’t forget the Pythagorean triples!
45
Solution to Problem 29 First solution. Consider the given equation as quadratic in a: a2(b2 – b + 2) – a(b + l)2 + 2b2 – b + 1 = 0. The discriminant is ∆ = –(b – 1)2(7b2 – 2b + 7), hence we have solutions only for b = 1. It follows that a = 1. Second solution. Using Cauchy-Schwarz inequality, we get 2(a2 + 1) ≥ (a + 1)2, 2(b2 + 1) ≥ (b + 1)2 and (a2 + 1)(b2 + 1) ≥ (ab + 1)2. Multiplying, we obtain 2(a2 + 1)(b2 + 1) ≥ (a + 1)(b + 1)(ab + 1), hence the equality case occur in all the inequalities, so a = b = 1. Solution to Problem 30 For example, consider R \ Z partitioned in {–x, x} and Z in {2n, 2n + 1}. Another example: split R into disjoint intervals [2n, 2n + 1), with n ∈ Z. Then take pairs (x, x + 1) from each interval [2n, 2n + 1), with x ∈ [2n, 2n + 1). Solution to Problem 31 a) Let Ak be the partition classes, with k = 1, 2, …, r. Assuming that the answer is no, there exist nk, k = 1, 2, …, r, such that no multiples of nk is in Ak. But n1n2…nr lies in one of the sets Ak and is multiple of any nk, false. b) We exhibit a partition for which the answer is no. Let Ak be the set of all numbers written only with the first k primes at any positive power; moreover, put 1 ∈ A1. Fixing k, the number p1p2…pk+1 have no multiples in Ak. Solution to Problem 32 We claim that A = {2, 3, 4, 5, …}. Let m be the smallest element of the set A. Since [ m ] + 1 ∈ A, we have m ≤ [ m ] + 1 ≤ which gives m = 2.
m + 1,
Notice that [ n 2 + 4 ] = n for all n ≥ 2. Indeed, n2 ≤ n2 + 4 < (n + 1)2 = n2 + 2n + 1, for all n ≥ 2. Using both hypothesis, we have n ∈ A ⇒ n2 + 4 ∈ A ⇒ [ n 2 + 4 ] + 1 ∈ A ⇒ n + 1 ∈ A. The conclusion follows by induction. Solution to Problem 33 At first, we look for a lower margin of n. The number 0.1111 can be written as a sum of n distinct suitable numbers, all starting with 0.0000…, therefore lower than 0.0001. Hence if 0.1111 = a1 + a2 + + … + an, with ak < 0.0001, then 0.1111 < n ⋅ 0.0001, or n > 1111. Thus n is at least 1112. We claim that 1112 is the requested number. Let t ∈ (0, 1) be a real number. If t > 0.1111, choose a suitable number of the form y = 0.xxxx… so that y < t ≤ y + 0.1111. The other suitable numbers will have the form 0.0000…, so they are different from y. We have 0 < t – y ≤ 0.1111. Because t − y 0.1111 t−y 0< ≤ < 0.0001, the first four decimals of the number u = are all equal to 0. Choose 1112 1112 1112 an irrational number e, small enough not to change the first four decimals of the numbers u + e,
46
u + 2e, …, u + 1111e, and such that all – plus y + u + e – are left irrationals. Then 1111⋅1112 ⋅ e t – y = (u + e) + (u + 2e) + (u + 3e) + … + (u + 1111e) + u − . 2 As previously stated, number e was selected so that all summands are irrational, suitable numbers with the first four decimals 0 and paiwisely distinct – in other words, different from the last one. Consequently, 1111⋅1112 ⋅ e t = (y + u + e) + (u + 2e) + (u + 3e) + … + (u + 1111e) + u − . 2 The number y + u + e starts with 0.xxxx, while the others with 0.0000, so they are suitable. Since t is now represented as required, the proof is concluded. Solution to Problem 34 It is clear that we need to prove that a1 + a2 + … + an ≥ n. Let us notice that this is enough: let m = = 1, 2, …, n and assume that any selection of m numbers from the given ones has the sum less than 1. Then add the n inequalities a1 + a2 + ... + am < m, a2 + a3 + ... + am+1 < m, ⋮
an + a1 + ... + am−1 < m, to get n(a1 + a2 + … + an) < nm, which is a contradiction. Back to the top, let g be the geometrical mean of the numbers a1, a2, …, an and suppose that a1 + a2 + … + an < n. a + a + ... + an By AM-GM inequality, we have g ≤ 1 2 < 1 , while n 1 1 1 + + ... + 2 a1 + a2 + ... + an a12 a22 1 an 1> = ≥ 2, n n g which gives g > 1, a contradiction.
Solution to Problem 35 The main idea is to show that x ∈ A if and only if
1 ∈ A. Since A is finite, one may pair the x
1 elements of A in x, , with number 1 left alone if belongs to A. Then it is obvious that the product x of all elements is equal to 1. 1 Let b > 1 be an arbitrary real number. Applying the hypotesis for a = b and a = we find that the b 1 sets {x ∈ A | x > b} and x ∈ A | x < have the cardinals of the same parity, and also the sets b
47
1 1 x ∈ A | x > and x ∈ A | x < 1 = b have the cardinals of the same parity. b b 1 1 This implies that the sets L = x ∈ A | x > \ {x ∈ A | x > b} = x ∈ A | x ∈ , b and R = b b 1 1 = {x ∈ A | x < b} \ x ∈ A | x < = x ∈ A | x ∈ , b have the cardinals of the same parity. But the b b set L may have at most one element which is not in R and vice versa, which shows the sets L and R 1 have the same number of elements. Thus both numbers b and are or aren’t elements of A, as b claimed.
Solution to Problem 36 n
n
1 1 The number a + + b + is an integer if and only if 2n divide (2a + 1)n + (2b + 1)n. 2 2 n k For n = 2k, 2 = 4 divides (2a + 1)n + (2b + 1)n = (4a2 + 4a + 1)k + (4b2 + 4b + 1)k ≡ 2 (mod 4), implying k ≤ 1, hence the set of even integers satisfying the claim is finite. For n = 2k + 1, one has (2a + 1)n + (2b + 1)n = (2a + 1 + 2b + 1)[(2a + 1)2k – (2a + 1)2k–1(2b + 1) + + (2a + 1)2k–2(2b + 1)2 – … + (2b + 1)2k]. The second factor is a sum of odd summands, each summand being odd, thus being also an odd integer. The number 2a + 2b + 2 has only a finite number of divisors of the form 2k, therefore the claim is proved.
48
Chapter II GEOMETRY Solution to Problem 37 First solution. Let I be the intersection point of the lines BE and CD. The quadrilaterals BD′B′D and CE′C′E are cyclic, hence 'BDB′ = 'B′D′I and 'CEC′= 'IE′C′. Since BDEC is also cyclic, 'BDB′= 'CEC′. It follows that 'B′D′I = 'IE′C′, so B′D′E′C′ is a cyclic quadrilateral. A D
E
E′ D′ C′ B′ B
C
Second solution. Using the power of a point theorem, one has: IB′ ⋅ ID = ID′⋅ IB IC′⋅ IE = IE′⋅ IC IE ⋅ IB = ID ⋅ IC From these one easily obtains IB′⋅ IE′= ID′⋅ IC′, which proves that the quadrilateral B′D′E′C′ is cyclic, using the reciprocal of the power of a point theorem. Solution to Problem 38 Denote AD = a, AB = b. We have AC2 = a2 + b2, CE = b2/AC, AE = a2/AC and EF/a = AE/AC = a3 b3 = a2/AC. It follows that EF = and, analogously, EG = . Thus, the equation is equivalent to AC 2 AC 2 (a2 + b2) = (a3x + b3x)2. 2 Notice that x = is a solution. Moreover, for a = b the solution is unique. 3 b Suppose a ≠ b; wlog a > b. Set k = ∈ (0, 1) to obtain (1 + k2)3x = (1 + k3)x. a
49
2 , then k3x > k2 ⇒ 1 + k3x > 1 + k2 > 1 ⇒ (1 + k3x)2 > (1 + k2)2 > (1 + k2)3x. 3 2 If x < , we use a similar argument. 3 2 Thus, the only solution is x = . 3
If x <
Solution to Problem 39 Let F and G be the projections of E on the diagonals BD and AC. From Simson’s theorem, it follows that the triplets of points (K, L, F), (M, N, F), (K, G, N) and (M, L, G) are collinear. The point N is the orthocenter of the triangle KLM if and only if KL ⊥ MN and ML ⊥ KN. Let F′ and G′ be the points in which EF and EG intersect the second time the circle. We have KF || AF′ and MG || CF′. Thus KL ⊥ MN is equivalent to AF′⊥ CF′ and then to O ∈ AC. Similarly, ML ⊥ KN is equivalent to O ∈ BD. Thus, ABCD is a rectangle. It is easy to see that in this case, N is the orthocenter of the triangle KLM for any position of the point E. Solution to Problem 40 We first notice that ABDE is an isosceles trapezoid. The segments AB and DE have the same perpendicular bisector. Let O and R the center and radius of the circumcircle of the triangle ABC. One can see that the perpendicular bisectors of DE and CF also pass through O, hence O is the center of the circle circumscribed around DCF, with radius R′. Finally, since ACDF is an isosceles trapezoid, it follows that R = R′. Solution to Problem 41 Let a ≤ b ≤ c the lengths of the parallelepiped’s. We have abc ≥ 1001 and c ≥ 11. By analysing the cases c ∈ {11, …, 21} one finds that a = 8, b = 9 and c = 14 is the solution. Solution to Problem 42 From the similarity of the triangles AMN and ABC, we obtain AM AN = (1) AB AC and (2) 'MAN = 'BAC or 'BAM = 'CAN. The relations (1) and (2) imply the similarity of the triangles BAM and CAN. Hence, we obtain the proportions: AN AB BM = = (3) AN AC CN and 'ABM ≡ 'ACN. The last equality implyies that ABCD is a rectangle. 1 1 1 To conclude the proof, notice that BM = BD = AC and CN = AB. Hence the last equality in 4 4 2
50
(3) becomes
AB AC = , that is 2AB2 = AC 2 = AB2 + BC 2, which proves that ABCD is a square. AC 2 AB
Solution to Problem 43 Let P the midpoint of BC. Since MP is a median in the right-angled triangle MBC, it follows that PB = MP = PC = CN. A
20°
R M 20° B
N P
C
The point R is considered such that PCNR is a parallelogram (in fact a rhombus). Notice that 'RPM = 'RPB – 'MPB = 'ACB – (180° – 2'MBC) = 60°, and BP = MP, therefore MPR is an equilateral triangle. Hence MR = RP = RN and 'MRN = 'MRP + + 'PRN = 60° + 80° = 140°. Then 'RMN = 'RNM = 20°, 'ANM = 20° + 80° = 100° and the required angle 'AMN is equal to 60°. Solution to Problem 44 Let E, F be the midpoints of AD and BC repectively and let M, N be the midpoints of OE and OF. 1 1 1 It is easy to check that S ( M , N ) = , therefore k ≥ . We claim that k = . 8 8 8 B
A
E
D
M
B
A M
N
N
F
C
D
C
1 , for any interior point M. We may assume that N is an 2 interior point of the triangle AMD. Therefore area[AND] + area[ANM] + area[DNM] + area[BMC] = 1 1 = . Consequently, one of the summands exceeds . The conclusion follows. 2 8
Notice that area[BMC] + area[AMD] =
51
Solution to Problem 45 Consider 'AOD = m ≤ 90°. As the angles 'AOD and 'BOC are equal to m, we find area[AOD] = = area[BOC]. It follows 1 1 AO ⋅ DO ⋅ sin m = BO ⋅ CO ⋅ sin m , 2 2 AO BO = that is . Since 'AOB = 'DOC, the triangles AOB and DOC are similar and consequently CO DO AB is parallel to DC. A
B
E m m
O m D
F
C
Draw line EF that contains O, such that 'AOE = 'COF = m and E ∈ (AB), F ∈ (DC). The triangles AOE and COF are similar and have the same area, that is they are congruent. It follows that AO = OC and in the same way BO = OD, implying AD || BC. Moreover area(COF) = area(BOC), and since ABCD is a parallelogram, we find area[BOC] = area[DOC]. Hence D = F and m = 'COF = 'DOF = = 'BOC = 90°. We thus proved that ABCD is a rhombus. To conclude, consider the bisector lines OM and ON of angles 'AOD and 'DOC respectively, where M ∈ (AD), N ∈ (DC). It is easy to check that 'MON = m = 90°, whence area[MON] = 1 = area[AOD]. Thus area[DON] = area[ACM], that is area[AOM] = area[DOM] = area[AOD]. It 2 follows that OM is a median in the triangle AOD, that is AO = OD, which proves that the rhombus ABCD is a square. Solution to Problem 46 Since O2 is at equal distances from the tangents MA and MB, it follows that MO2 is a bisector line of the angle 'AMB or of the exterior angle defined by MA and MB. N
B
O1
M
O2
A
In each case one obtains O2A = O2B. Reflecting the figure with respect to the line O1O2, the circles C1 and C2 remain fixed, A reflects in B and M reflects in N. It is obvious that NA, the reflected of MB is tangent to C2 and the same is valid for NB. Observe that N is on C1, proving thus the claim.
52
Solution to Problem 47 Denote by A, B, C, D, E the five given points. If the pentagon ABCDE is concave, we can suppose that D is located inside the triangle ABC or inside the quadrilateral ABCD. In first case area[ABC] = area[ABD] + area[DBC] + area[DAC] > 6 > 3. In the second case, D is inside the one of triangles BCE, ACE, ABC or ABD. Suppose, without loss of generality, that D is inside to the triangle BCE. Then area[BCE] ≥ area[BDC] + area[CDE] > 4 > 3. Consider now the case when ABCD is a convex pentagon. Let M and N be the intersection points of BE with AC and AD respectively. A N B
E
M
C
D
The following result will be useful. Lemma. Let PQRS be a quadrilateral and T a point on the side PQ. Then area[TRSR] ≥ min(area[PRS], area[QSR]). The proof consists of simply observing that the distance from T to SR is bounded up and below by the distances from P and Q to SR. 1 1 In our case, suppose that BM ≥ BE, which yields BM ≥ ME. Then 3 3 1 area[BDE] = area[BDM] + area[MDE] ≥ area[MDE] + area[MDE] = 2 3 3 3 = area[MDE] ≥ min(area[CDE], area[ADE]) > ⋅ 2 = 3. 2 2 2 1 1 The case when NE ≥ BE is similar. It is left to consider the case when MN ≥ BE. We then 3 3 have: 1 2 area[AMN] ≥ area[ABE] > , 3 3 1 2 area[MND] ≥ area[BED] > , 3 3 and area[MCD] ≥ min[area[BCD], area[ECD] > 2. Summing up, we conclude 2 2 area[ACD] > 2 + + > 3, 3 3 and the proof is complete.
53
Solution to Problem 48 Let O be the common center of the n circles and α = A1B1 = A2B2 (the arcs are directly oriented). Rotate the figure around the center O by angle α such that A1, A2 become B1, B2 respectively. The above rotation R maps lines into lines, that is R(D1) = D2, since D1 = A1A2 and D2 = B1B2. Moreover, any circle Ci is invariant under the rotation. As R(Ai) lies on D2 and on Ci, we get that R(Ai) = Bi, that is AiBi = α. In the same way we get R ( Ai′) = Bi′ and Ai′Bi′ = α. This concludes the proof. Solution to Problem 49 Let DECB be the quadrilateral of maximal area. It is easy to prove that 'DBE = 'DCE = 90°. E
F D
A
C
B
DE = x and that the quadrilateral DBCE is cyclic. By Ptolemey’s theorem 2 we have DC ⋅ BE = BC ⋅ DE + DB ⋅ CE. Squaring, we get (4x2 – b2)(4x2 – c2) = (2ax + bc)2, that is 4x2 – (b2 + c2)x2 + b2c2 = 4a2x2 + 4abcx + b2c2. From this we obtain 4x3 = (a2 + b2 + c2)x + abc, as desired.
It follows FB = FC =
Solution to Problem 50 Firstly, observe that triangles RSN and QSM are congruent (S.S.S.), hence 'PMS = 'PNS and 'PQS = 'PQS. It follows that P, S, M, N are concyclic and P, S, Q, R are concyclic. A M
B
P
R
T
S
D
O
N Q C
On the other hand, as 'BNP + 'BMP = 180°, points B, N, S, M are concyclic, thus P, S, N, B, M are points on the circle C1(O1) of diameter BP. Likewise, points P, S, Q, D, R he on the circle C2(O2) of diameter DP. Since PS is the common chord of the circles C1 and C2, lines PS and O1O2 are perpendicular. As O1 and O2 are the midpoints of the segments BP and DP, lines O1O2 and BD are parallel, so PS ⊥ BD and
54
then PS || AC. Likewise, PT || BD and consequently PS ⊥ PT. Furthermore, because O1O2 is middle line in the triangle PBD one find that S lies on the segment BD. Analogously, T ∈ (AC). Thus, PSOT is a rectangle. Solution to Problem 51 We have to prove that 2'O1MA = 2'O2BM, which is equivalent to O1 A O2 B = . (1) AM BM A O1
O2 B M
1 The lenght of the common chord AB is equal to 2O1A ⋅ sin AB = 2O1A ⋅ sin'BAM, regardless if 2 AB is the small arc or the great arc AB. Similarly, AB = 2O2B ⋅ sin'ABM, hence O1 A O2 B (2) = . sin 'ABM sin 'BAM By the Law of Sines in the triangle ABM we derive that MA MB (3) = . sin 'ABM sin 'BAM OA OB From the relations (2) and (3) we obtain 1 = 2 , as desired. AM BM
Solution to Problem 52 Observe that 'MAB + 'MBA = 'MBC + 'MBA = 90°, hence 'AMB = 90°. Let F be the midpoint of the side AB. Then MF = FA = FB =
1 AB, so 'MBF = 'MBF. It follows 2
that 'EMF = 'EMB + 'BMF = 'MAB + 'MBA = 90°.
55
C
D
M
A
B
1 In the right triangle MEF, the leg MF is equal to EF. hence 'MEF = 30°. We obtain 'MBF = 2 1 1 = 'MFA = MEF = 15° and x = 75°. 2 2
Solution to Problem 53 Lines CD and FG meet at M and lines BC and EF meet at N. As DHEA and FHCT are cyclic quadrilaterals, it follows that 'FTC = 180° – 'FHC = 'DAE and 'DAH = 'DEH. T
C F
H D
G
E
B N
A M
Since 'DMG = 90° – 'FTC = 90° – 'DAE = 'DAG, it follows that the quadrilateral ADGM is cyclic. Hence 'DAM = 'FGE and consequently 'MAH = 'DAM + 'DAH = 'FGE + 'DEH = 90°. Likewise, 'NAH = 90° and therefore points M, A, N are collinear. In the triangle TMN, point H is the orthocenter. Thus A, H, T lie on the altitude of the triangle, as desired. Solution to Problem 54 Let MNPQ be the rectangle at the intersection of the unit squares with centers A and B. Set MN = x and PQ = y, hence 1 xy = , x, y ∈ [0, 1]. 8 The parallel from A to MN intersects the parallel from B to NP at point C. It is easy to observe that AC = 1 – x and BC = 1 – y, so
B Q M A
56
P N
C
AB2 = (1 – x)2 + (1 – y)2 = x2 + y2 – 2(x + y) + 2 = 1 7 = x2 + 2xy + y2 – 2(x + y) – + 2 + (x + y)2 – 2(x + y) + = 4 4 3 = ( x + y − 1)2 + . 4 It follows that the minimal value of the distance between the centers A and B is equal to
3 and it 2
1 2+ 2 2− 2 2− 2 2+ 2 ; i.e. x = ,y= or x = ,y= . 8 4 4 4 4 9 To find the maximal value of AB observe that 0 ≤ (1 – x)(1 – y) = 1 – x – y + xy = – (x + y), so 8 9 1 1 1 x + y ≤ . On the other hand, we have x + y ≥ 2 xy = , therefore – 1 ≤ x + y – 1 ≤ . As 8 2 2 8
is obtained for x + y = 1, xy =
2
2
1 3 9 1 1 1 − 1 , we find that AB2 ≤ − 2 + 1 = − 2 = 2 − 2 . Thus AB ≤ 2 – ≤ 2 4 4 2 8 2 1 equality when x = y = . 2 2 3 1 Consequently, ≤ AB ≤ 2 − . 2 2
1 2
, with
Solution to Problem 55 The approach of the problem is to see no circles in the figure. Instead, recall that a quadrilateral ABCD is orthogonal if and only if (1) AB2 – CD2 = AD2 + BC 2. Using successively the Pytagoras theorem we have BA2 – BT22 = BA2 – (BO2 – OT22 ) = BA2 – (BO2 – OT12 ) = = BA2 – (BA2 – AT12 ) = AT12 = AO2 – OT12 = OA2 – OT22 . so the conclusion follows from the relation (1). Solution to Problem 56 Let a, b, c, x, y be the lengths of the sides BC, CA, AB, BD, DC, respectively and let A′ be the foot of the altitude from A in the triangle ABC. Notice that x + y = a. Due to Ceva theorem the claim is equivalent to BD CF AE AB BD ⋅ ⋅ =1⇔ = . DC FA EB AC DC As CF = y cos C, FA = b – y cos C, BE = x cos B, AE = c – x cos B, BA′ = c cos B and A′C = b cos C, the equivalence rewrites cy(c – x cos B) = bx(b – y cos C) ⇔ xb = cy.
57
Indeed, we have a2 + c2 − b2 a 2 + b2 − c 2 = b 2 x − bxy ⇔ 2ac 2ab ⇔ a(c2y – b2x) = xy(c2 – b2) ⇔ c2y(a – x) = b2x(a – y) ⇔ ⇔ c2y2 = b2x2 ⇔ cy = bx, c 2 y − cxy
as claimed. Solution to Problem 57 Squaring both sides of the equality yields 2(a + a 2 − b 2 )(a + a 2 − c 2 ) = (a + b + c) 2 . It is easy to observe that the equality holds if a2 = b2 + c2. To prove the converse statement, assume
that a2 > b2 + c2. Then
a 2 − b 2 > c and
a 2 − c 2 > b, hence
2(a + a 2 − b 2 )(a + a 2 − c 2 ) > 2(a + b)(a + c) = = 2a2 + 2(ab + bc + ca) > a2 + b2 + c2 + 2(ab + bc + ca) = (a + b + c)2,
false. The case a2 ≤ b2 + c2 leads similarly to a contradiction and we are done. Solution to Problem 58 The idea is to prove that the sum of the radii is equal to the altitude h from A of the triangle ABC; a hint is to assume that line MA is parallel to BC. The Stewart relation gives AM 2 ⋅ BC + AC 2 ⋅ MB = AB2 ⋅ MC + MB ⋅ MC ⋅ BC, so AM 2 ⋅ BC = = AB2(MC – MB) + MB ⋅ MC ⋅ BC, hence AM 2 = AB2 + MB ⋅ MC. Let r be the inradius of triangle AMB and R the exradius of triangle AMC corresponding to the angle 'M. Since 2 ⋅ areaAMB r= , AM + MB + AB and 2 ⋅ areaAMC R= , MA + MB − AB then r MB = , h AM + MB + AB and R MC = . h AM + MB − AB Thus r + R = h ⇔ MB(MA + MB – AB) + MC(MA + MB + AB) = (MA + MB + AB)(MA + MB – AB) ⇔ ⇔ MB(MA + MB – AB) = (MA + MB + AB)(MA – AB) ⇔ ⇔ MB ⋅ MC + MB(MA – AB) = MA2 – AB2 + MB(MA – AB) ⇔ MA2 = AB2 + MB ⋅ MC, so the claim holds.
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Solution to Problem 59 Let Q be the intersection point of the line segments AB and MP. The tangents from A and M to the A incircle are equal (to r ⋅ cot ). Moreover, the tangents from Q to the incircle are equal, so AQ = MQ. 2 This implies 'QMA = 'QAM, so the arcs AP and BM are equal. In the trapezoid APBM, the diagonals AB and MP are equal, and likewise AC = MN. This concludes the proof. Solution to Problem 60 By the Power of a point theorem we have AG ⋅ AM = AP ⋅ AB, so 4MA2 = 3AB2 and thus AM = 3 = AB = BN. Then AG = GB, so the median GP is also an altitude in the triangle AGB. This implies 2 'BPG = 90°, and since BMGP is cyclic, 'GMA = 90°. It follows that BC = CA and AB = AC, so the triangle is equilateral. Solution to Problem 61 Consider the case when the ray is interior to the angle 'BAD. Then 'CMP = 'MCA + 'CAM = 'MAD + 'CAM = 'CAB. On the other side, 'CPM = 'CDA, since both subtend the chord AC in the circumcircle ACD. From the similarity of the triangles ACD and MCP we derive that MC MP = . AC AD Furthermore, 'ACM = 'NAD and 'CAM = 'ADN, so 'ACM ~ 'DAN. Hence AN MC = , AC AC thus MP = AN and AM = NP, as claimed. Solution to Problem 62 The distance from M to the line BC is equal to the distance from Q to the line DC, since ABCD is a rhombus. Hence it suffices to prove that CP = CN or DP = BN. BU QU NU Using Thales theorem we obtain = = , hence NQ || MC and likewise MP || CQ. UD UC UM BN BQ DP MD DT Triangles MDP and BQC are similar, so . Analogously, = . It follows that = = BQ BC BT MD DC DP BN , therefore DP = BN. = MD MD
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Solution to Problem 63 Denote AB′ = x1, BC′ = y1, CA′ = z1, A′B = x2, B′C = y2, C′A = z2. From the Ceva theorem we have x1y1z1 = x2y2z2. As AB′2 = AP2 – PB′2 (and the analogous relations) we obtain by summation x12 + y12 + z12 = x22 + y22 + z22 . Since x1 + y2 = x2 + z1 = y1 + z2 = AB and x1 + x2 + y1 + y2 + z1 + z2 = 3 ⋅ AB, it follows that x1 + y1 + z1 = x2 + y2 + z2. Thus x1 + y1 + z1 = = x2 + y2 + z2; x1 y1 + x1 z1 + y1 z1 = x2 y2 + x2 z2 + y2 z2 ; x1y1z1 = x2y2z2, so {x1, y1, z1} = {x2, y2, z2}. An easy check shows that the claim holds. Solution to Problem 64 It suffices to prove that 'GAC < 'GBA and 'GBC < 'GAC, given that CA > AB and BC > CA. Since the two claims are equivalent, we will prove the first inequality. First approach The reflection B′ of the point B across AG lies on the parallel line through C at AG. Observe that point C is further than B′ from the perpendicular bisector d of the line segment AG, since d intersects line BC at a point located at the left of point B (with respect to point C) and we are done. Second approach As area[GBA] = area[GAC], the inequality 'GAC < 'GBA (< 90°!) is equivalent to c ⋅ mb > b ⋅ mc. Squaring both sides yields b2 ⋅ (2a2 + 2c2 – b2) > c2 ⋅ (2a2 + 2b2 – c2), hence (b2 – c2)(b2 + c2 – 2a2) > 0, which is implied by a > b > c. Solution to Problem 65 Set Q1, Q2, Q3 the mirror images of Q across O1, O2, O3, in other words consider the diametrically opposite points of Q in the circles C1(O1), C2(O2), C3(O3). It is easy to check that the claim is equivalent to the fact that points Q, Q1, Q2, Q3 lie on the same circle. An inversion of pole Q maps the line A – B – C to a circle A′ – B′ – C′ passing through Q, while points Q1, Q2, Q3 map to points Q1′, Q2′ , Q3′ , which are the projection of Q to lines A′B′, B′C′, C′A′. By Simpson theorem, points Q1′, Q2′ , Q3′ are collinear, thus points Q, Q1, Q2, Q3 lie on the same circle, as needed. Solution to Problem 66 Let P be the intersection point of the circle with the radius ON. Triangle AOP is isosceles at A and 'AOP = 60°, so 'ANO = 90°. On the other side 'AMO = 90°, so AMNO is cyclic. Hence 'OMN = 'OAN = 30°. Solution to Problem 67 The first decomposition is obtain as follows: Consider the midpoint M of the side DE and C′ the intersection point of the lines BE and CM. The right isosceles triangle obtain by reassembling is CAC′, since ∆ABC = ∆AEC′ and ∆CMD = ∆MEC′.
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For the second decomposition consider N the midpoint of DE and let A′ be the intersection point of the lines BD and AM. The right isosceles triangle obtain by reassembling is CAA′, as ∆ANE = ∆NDA′ and ∆ABC = ∆CDA′. Solution to Problem 68 Consider the rectangle ACDP. The hypothesis rewrites as
BD AE = = k , so 'APE = 'BPD and DP AP
AP PD = , hence ∆PAD ~ ∆PEB. PE PB It follows that 'DAP = 'PEB, so APOE is cyclic and hence 'BOD = 'AOE = 'APE. The claim is proved by the following chain of equivalences: AE = 3 ⇔k= 3. 'BOD = 60° ⇔ tan'BOD = 3 ⇔ tan'APE = AP
'APD = 'EPB. Moreover,
Solution to Problem 69 Denote A, B, C, D, E the given points and suppose ABC is the triangle having the maximal area. The distance from D to BC is not greater than the distance from A to BC, hence D – and similarly E – are located between the parallel line from A to BC and its mirror image across BC. Applying the same reasoning for AB and AC, one obtains a triangular (bounded) region A1B1C1 – with ABC as median triangle – in which all points must lie. Points D and E are located in at most 2 of the triangles A1BC, AB1C, ABC1, hence one of the trapezoids APA1B1, BCB1C1 or ABA1B1 contains all points. Since the area of the trapezoids is 3 times the area of ABC, hence not greater than 3, the conclusion is proved. Solution to Problem 70 Let R, r1, r2 be the radii of the circles C, C1, C2 and let r = R – r1 – r2. Consider the point P so that OO1PO2 is a rectangle. From the tangency conditions we get OO1 = R – r1, OO2 = R – r2 and O1O2 = = r1 + r2 = R – r. It is sufficient to prove that the circle centered at P with radius r is C3. To prove this, notice that O1P = OO2 = R – r2 = r + r1, O2P = OO1 = R – r1 = r + r2, and OP = = O1O2 = R – r, so the 3 tangency conditions are fulfilled. Solution to Problem 71 Let α be the measure of the angle determined by the diagonals. Since 8 = OA + OB + OC + OD ≥ AC + BD ≥ 2 ⋅ AC ⋅ BD ≥ 2 ⋅ AC ⋅ BD ⋅ sin α = 2 2S = 8, we get AC = BD = 1 and α = 90°. The claim follows from a simple arch subtraction. Solution to to Problem 72 Let A2, B2, C2 be the mirror images of the point M with respect to points A1, B1, C1. The given condition shows that MA = MA2, MB = MB2, MC = MC2. From the parallelograms AMBC2, BMCA2, AMCB2 we derive that MA = MA2 = BC2 = B2C, MB = MB2 = AC2 = CA, and MC = MC2 = PA2 = AP2. It follows that MA2BC2, MA2CB2 and MB2AC2 are also parallelograms, therefore A, M and A2 are collinear. The conclusion is now obvious.
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Solution to Problem 73 Let E be the mirror image of D across the midpoint of the side BC. We notice that DBEC is a parallelogram and ABEC is cyclic. The equality of the areas of triangles ABE and ACE implies AB ⋅ BE = AC ⋅ CE. We are left only to notice that CE = BD and BE = CD. Solution to Problem 74 Let Q be the intersection point of the diagonals, T the second point of intersection of the line AC with the circle C1 of diameter AD and S the second point of intersection of the line BD with the circle C2 of diameter BC. Then 'ATD = 'BSC = 90°, so DT and SC meet in the orthocenter H of the triangle DQC. Denote by C3 the circle DCTS. The radical axis of the circles C1, C2 is MN, the radical axis of the circles C1, C3 is DT, while the pair of circles C2, C3 has SC as radical axis, hence the radical center of the three circles is H. The line segment MN is the common chord of the circles C1 and C2, thus perpendicular to the line passing through the centers, which is in fact the middle line of the trapezoid. As H ∈ MN, then MH || DC, and since QH || DC the conclusion follows. Solution to Problem 75 Let T ∈ BD be the tangency point of incircles of the triangles ABD and CBD. Notice that BM = = BT = BN and DN = DT = DM. a) We have AB + CD = AM + MB + CP + PD = AN + BQ + CQ + DN = AD + BC, so the quadrilateral ABCD is circumscriptible. b) Triangles AMN, DNP, CQP, BQM are isosceles, so 'QMN + 'NPQ = 360° – ('AMN + 'BMQ + 'QPC + 'NPC) = 1 = 360° – (4 ⋅ 180° – A – B – C – D) = 180°, 2 hence the quadrilateral MNPQ is cyclic. c) Let U be the point where the side AC touches the incircle of triangle ABC. Since AB – BC = AB + AC − BC AD + AC − DC = , so U is also the tangency point of the side = AD – DC, then AU = 2 2 AC with the incircle of triangle ADC, as needed. Solution to Problem 76 For a fixed point P inside the given triangle consider the point Q on bisector line of BC so that AQ = AP. The parallel line d from Q to BC separates the arc MN and the side BC, so d meets the line segment [BP] at a point, say S. The triangle’s inequality gives SP – PC ≥ SC, so BP + PC ≥ BS + SC. On the other hand, with an argument frequently refer to as Heron’s problem we have BS + SC ≥ BQ + + QC, so BP + PC is minimum if P = Q. Let T be the midpoint of the segment MS. Notice that triangle AMQ is isosceles and MT is an altitude in this triangle, hence MT = QZ, where Z is the foot of the altitude from Q onto AC. Then MN + BQ + QC = 2(MT + CQ) = 2(CQ + QZ) is minimum when CZ ⊥ AC. Consequently, the required point is the orthocenter of the triangle ABC, which belongs to the interior of the triangle, since it is an acute-angled one.
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Solution to Problem 77 AM AP MS Let k = . Using Thales Theorem, MP || BC yields = k ,while MN || AC implies =k . AB AC MN CN PR On the other hand, = =k. BC PM Setting S = area[MNP], we have area[ MSP] MS RP area[ NPR] = = = , S MN PM S hence area[MSP] = area[NPR]. Subtracting area[RPQ] from both sides of the latter equality we get the conclusion. Solution to Problem 78 Line AB is the radical axis of the circles ω1 and ω2, and line DE is the radical axis of the circles ω1 and ω3, hence point G is the radical center of the three circles. Since the radical axis of the circles ω3 and ω2 is the tangent line at C to these circles, it follows that the tangents from G to ω – 3 are GF and GC. Then GF = GC and GH = GF, so the triangle HCF is right-angled at C. Therefore 'HCF = 90°. Solution to Problem 79 A “special” position occurs when 'CAB = 60°, when C = E = F. In this case the claim is obvious. Consider the case 'CAB > 60°, where E belongs to the small arc CD and F lie on the segment CD. Notice that 'PFC = 'ADC = 'BCD, hence the trapezoid PBFC is isosceles. On the other hand, as CD || AB and CE || AD, it follows that the arcs AC, BD, DE are equal. Then 'EFC = CE + BD = CE + + DE = CD = 'P'CD, where P' ∈ PC, C ∈ (PP'). The last equality proves the conclusion. Slight changes in notations are required for the case 'CAB < 60°. Solution to Problem 80 Line BC meet DF, AE at points T, G respectively. Using Ceva’s theorem, it suffices to prove that CF EG AD ⋅ ⋅ =1 , FE GA DC since only one or all points D, F, G lies on the sides of the triangle AEC. Observe that 'BAD = 'BED = 'BTD = 90°, so points A, C, E, T, D lies on the circle of diameter not
not
BD. Then 'FDE = 'TBE = α and 'TDC = 'ABC = β. Moreover, DE = DA and AB = BE. The law of sinuses gives FE DE DC FC = , = . sin α sin 'EFD sin 'CFD sin β and since sin'EFD = sin'CFD we have CF DC sin β (1) = ⋅ . FE DA sin α On the other hand,
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EG BE BA AG . = , = sin α sin 'EGB sin 'AGB sin β
hence EG sin α = . GA sin β Multiplying the inequalities (1) and (2) concludes the proof. Alternative solution. Let H be the intersection point of the lines DF and AE. The claim is equivalent AG AH to = , in other words the pairs of points A, E and G, H are harmonical conjugates. GE HE Since I is the midpoint of the segment AE, the claim reduces to IG ⋅ IH = IA2. The segment AI is an altitude in the right-angled triangle ABD, so AI 2 = ID ⋅ IB. Angles 'HTB and 'HIB are right, so the points H, T, I, B are cocyclic. It follows that 'DHI = 'IBG ID IG and further, ∆DHI ~ ∆IBG. Hence = , so ID ⋅ IB = IH ⋅ IG, the conclusion is now obvious. IH IB
(2)
Solution to Problem 81 Consider the equilateral triangle BCA1, constructed in the exterior of triangle ABC. Then points A, A′, A1 are collinear through the homothety of center A which map points U, V in B, C, respectively Since AA1, BB1, CC1 concur in the Fermat-Torricelli point of the triangle ABC, the claim is proved. Solution to Problem 82 Let E and F be the midpoint of the sides AC and AB and let P be the mirror image of D across E. 2
2
2
2
c b c b The relation AM 2 + AN 2 = BM 2 + CN 2 gives + FM + − NE = − FM + + NE , 2 2 2 2 FM NE FM NE hence c ⋅ FM = b ⋅ NE. Then = = , so . Since 'MFD = 'NEP, we get ∆MFD ~ AC AB FD EP ~ ∆NEP, which implies 'MDF = 'NPE. On the other hand 'MDN = 'BAC = 'FDE, so 'MDF = = 'NDE. Now the triangle NPD is isosceles and NE is a median in this triangle, so NE ⊥ DP, in other words A = 90°.
Solution to Problem 83 Observe that AD = AN = AM and 'AND = 'AMC = 90°, due to the reflections across AB and AC. It is known that AD and AE are isogonal cevians, that is 'BAD = 'EAC. Then 'NAE = 'NAB + 'BAE = = 'BAD + 'BAE = 'EAC + 'DAC = 'EAC + 'CAM = 'EAM and consequently ∆NAE = ∆EAM. It follows that 'ENA = 'EMA, so 'BNE = 90° – 'ENA = 90° – 'EMA = 'EMC, as desired. Solution to Problem 84 Let a, b, c, d be the radii of the circles α, β, γ, δ. It suffices to prove that
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a c = , in other words the b d
a is constant while point M varies on line d. b Let R and S be the midpoints of the arcs determined by d on the fifth circle K, the one tangent simultaneously to α, β, γ, δ, and let N be on the same side of d as A. Denote by A1 and B1 the tangency point of α and β to K, respectively. Observe that points A1, M, R are collinear – via the homothety which maps circle a onto circle K – and similarly points B1, M, S are collinear. Since RS is a diameter of K, angles 'RA1S and 'SB1R are right. If lines B1R and A1S meet as point V, then M is the orthocenter of the triangle VRS. Notice that d ⊥ RS. hence V ∈ d; denote by O the intersection point of d and RS. Lines A1S and B1R intersect the circles α and β at points U and Z respectively. Since 'RA1S = a UM RO = 'SB1R = 90°, the segments UM and ZM are diameters in circles α, β, so = . The latter = b ZM SO ratio is constant, as claimed.
ratio
Solution to Problem 85 Let O be the midpoint of AD, R be the intersection point of lines AC and BD and S be the intersection point of lines AF and DE. Since N and Q are the midpoints of the sides DB and DE of the triangle DBE, we have O ∈ NQ and similarly O ∈ MP. Moreover, as DRAS is a parallelogram, the diagonal RS passes through the midpoint O of the other diagonal, AD. Now apply Desargues Theorem for triangles NRM and SPQ, given that O lies simultaneously on lines NQ, MP, RS and we are done.
Solution to Problem 86 The diagonal AC is the internal bisector for the angles 'BAD and 'BCD, so 'BAC = 'CAD = = 'BCA = 'ACD. Observe that the triangles CMN and CMD have 'MCN = 'MCD and MN = MD, hence their circumradius are equal and consequently CNMD is cyclic. It follows that 'MDP = 'MCN = = 'RAP, implying that ARPD is cyclic. Then 'DRP = 'PAD = 'RAP = 'MDP, hence RP = PD, as needed. Solution to Problem 87 A well-known result states that the quadrilateral MNPQ is cyclic. For this, notice that the quadrilaterals BMON, CNOP, DPOQ and AQOP have a pair of opposite right angles, hence all of them are cyclic. It follows that 'OMN = 'OBN, 'OMQ = 'OAQ, 'OPQ = 'ODQ and 'OPN = 'OCN. Summing up yields 'QMN = 'QPN, so MNPQ is cyclic, as claimed. Next we’ll show that the points M′, N′, P′, Q′ lie on the circumcircle K of MNPQ. As an external angle, we get 'NQ′O = 'Q′CO + 'COQ′. Since CNOP is cyclic, then 'Q′CO = 'NPO. On the other hand we have 'Q′OP = 'ODQ, because the sides are perpendicular, and 'ODQ = 'OPQ, as DPOQ is cyclic. Therefore 'NQ′O = 'NPO + 'OPQ = 'NPQ, implying that Q′ belongs to the circumcircle of MNPQ. In the same manner we find that N′, P′, Q′ he on K, which concludes the proof. Solution to Problem 88 n Since m ≠ , no segment AiAr(i) contain the center O of the polygon A0A1…An–1. 2
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The chords AiAr(i) are equal, since all subtend m from the n equal arcs in which the vertices A0, A1, …, An–1 divide the circumcircle of the polygon. Consequently, the distances from O to each of the segments AiAr(i) are equal, hence these segments are all tangent to a non-degenerate circle centered at O. Since through a point one cannot draw more than two tangents to a circle, it follows that there are no three concurrent segments of the form AiAr(i). Solution to Problem 89 Since the points X and Y are inside the polygon K, the line XY with the border of K contains two points Z and Z′, such that X, Y ∈ (ZZ′). The diameter of the polygon K is not greater than 1, hence |ZZ′| ≤ 1. To prove this, let AB the side of K containing Z and let CD the side containing Z′, where A, B, C, D are vertices of the polygon, not necessarily distinct. Notice that AC, AD, BC, BD ≤ 1. In a triangle MNP, a cevian MS, S ∈ (NP) satisfies the inequality MS < max(MN, MP). Therefore ZZ′ < max(ZC, ZD) < max(max(CA, CB), max(DA, DB)) ≤ 1, as claimed. To conclude, observe that (XZ + YZ) + (XZ′ + YZ′) = (XZ + XZ′) + (YZ + YZ′) = 2ZZ′ ≤ 2, hence at least one of the sums XZ + YZ, XZ′+YZ′ is less than or equal to 1. Solution to Problem 90 AB AC Let a = and b = . BC BC Then a2 + b2 = 1 and the inequality rewrites as (a – b)2(1 + 4ab)2 ≤ 2 and furthermore (1 – 2ab)(1 + 4ab)2 ≤ 2. 2 Put x = 2ab to obtain (1 – x)(1 + 2x) ≤ 2 or 4x3 – 3x + 1 ≥ 0, which is equivalent to (2x – 1)2(x + 1) > 0. The latter is obvious, so we are done. Solution to Problem 91 If the triangle ABC is isosceles, the claim is obvious. To prove the converse, observe that A′B2 + B′C 2 + C′A2 = A′C 2 + B′A2 + C′B2. Using the standard notation one has ac ab A′B = , A′C = b+c b+c and the analogous formulas. The above equality rewrites as a 2c 2 a 2b 2 , = (b + c)2 (b + c) 2 hence a 2 (c − b ) =0. b+c Clearing the denominators one obtains (a – b)(b – c)(c – a)(a + b + c)2 = 0, implying that ABC has at least two congruent sides.
∑
∑
∑
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Chapter III NUMBER THEORY Solution to Problem 92 4n − 2 a 2 Suppose = , where a and b are coprime integers. One obtains n + 5 b2 2b 2 + 5a 2 22b 2 = − + . n= 5 4b 2 − a 2 4b 2 − a 2 Since gcd(b2, 4b2 – a2) = 1, it follows that 4b2 – a2 divides 22. Observe that 4b2 – a2 ≡ 0 or 4b2 – a2 ≡ 3 (mod 4), hence we have either 4b2 – a2 = –1, or 4b2 – a2 = = 1. The first case leads to b = 0, which is impossible. In the second case, we obtain 2b – a = 1 and 2b + a = 11, hence a = 5, b = 3 and n = 13. Solution to Problem 93 Suppose that the number obtained is n2 = abcdef , where ab, cd , ef ∈ {16, 25, 36, 49, 64, 81}. Since 161616 ≤ n2 ≤ 818181, it follows that 402 ≤ n2 ≤ 904. Thus, n = 100x + 10y + z and x ≥ 4, z ≥ 1. By the squaring algorithm we obtain x2 = ab . Also: (l00x + 10y + z)2 = 104 ab + 103c + 102d + 10e + f, hence 2 ⋅ 103xy + 2 ⋅ 102xz + 102y2 + 2 ⋅ 10yz + z2 = 103c + 102d + 10e + f. Two cases arise: a) x = 4, y ∈ {0, l}; For y = 1, ab = 16, cd = 81, by using the above equality we get a contradiction. For y = 0, ab = 16 and 8z ⋅ 102 + z2 = 102 cd + ef . Since the representation of a number is unique we get cd = 8z and since ef is a two digit number, it follows that 8z ∈ {16, 64}. Therefore, z = 8 and n = 408. b) x > 4 and y = 0. We obtain (200x + z)z = 102 cd + ef . As before we obtain cd = 2xz, ef = z2, z ≥ 4. It follows that x = 8, z = 4, hence n = 804. In conclusion, the students can obtain the numbers 4082 or 8042. Solution to Problem Problem 94 Since a + b + c < 3d and d | a + b + c, it follows that a + b + c = d or a + b + c = 2d. Suppose that a + b + c = d. Since a | b + c + d = 2d – a, it follows that a | 2d and, similarly, b | 2d, c | 2d. Let
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2d = ax = by = cz, where x > y > z > 2. We obtain
1 1 1 1 + + = . x y z 2
1 1 1 1 1 1 1 + + < + + = , so there are no solutions. x y z 6 6 6 2 1 1 3 If z = 5, then + = , and we obtain y = 3, again not possible. x y 5 1 1 1 If z = 4, then + = , and we obtain the solutions (k, 4k, 5k, 10k) and (k, 2k, 3k, 6k), with k ∈ N. x y 4 1 1 1 If z = 3, then + = , and we obtain the solutions (k, 6k, 14k, 21k), (k, 3k, 8k, 12k), (k, 2k, 6k, 9k) x y 6 and (2k, 3k, 10k, 15k), with k ∈ N. Now, suppose that a + b + c = 2d. Analogously, we obtain that a, b, c | 3d, hence 3d = ax = by = cz 1 1 1 2 1 1 1 1 1 1 37 2 with x > y > z > 3 and + + = . Then z ≥ 4, y ≥ 5, x ≥ 6, thus + + ≤ + + = < , so x y z 3 x y z 6 5 4 60 3 there are no solutions in this case.
If z ≥ 6, then
Solution to Problem 95 For n = 0, we have 22 + 12 + 12 + 12 = 7, hence (a, b, c, d) = {2, 1, 1, 1) and all permutations. If n ≥ 1, then a2 + b2 + c2 + d 2 = 0 (mod 4), hence the numbers have the same parity. We analyse two cases. a) The numbers a, b, c, d are odd. We write a = 2a′+ 1 etc. We obtain: 4a′(a′ + 1) + 4b′(b′+ 1) + 4c′(c′ + 1) + 4d′(d′+ 1) = 4(7 ⋅ 4n–1 – 1). The left hand side of the equality is divisible by 8, hence 7 ⋅ 4n–1 must be even. This happens only for n = 1. We obtain a2 + b2 + c2 + d 2 = 28, with the solutions (3, 3, 3, 1) and (1, 1, 1, 5). b) The numbers a, b, c, d are even. Write a = 2a′ etc. We obtain a′ 2 + b′2 + c′2 + d′2 = 7 ⋅ 4n–1 so we proceed recursively. Finally, we obtain the solutions (2n+1, 2n, 2n, 2n), (3 ⋅ 2n, 3 ⋅ 2n, 3 ⋅ 2n, 2n), (2n, 2n, 2n, 5 ⋅ 2n), n ∈ N, and all permutations. Solution to Problem 96 Clearly x ≤ n2, so let x = n2 – p, with p > 0. If the number of radicals is 2, we obtain that x ≤ n2 – n. It is easy to check using induction that all x ≤ n2 – n verify the inequality regardless the number of radicals. Solution to Problem Problem 97 a) By squaring the members of the equation we get k + 100x = n2p + 2nxp + x2p, or 100 = p(2np + x). The conclusion follows from the fact that p is a prime number. k k b) If p = 2, then 50 = 2n + x and 0 ≤ n2 ≤ 25. Since n2 = = < 500, it follows that n ≤ 22 and we p 2
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have 23 solutions. If p = 5, then 20 = 2n + x and 0 ≤ n ≤ 10. Notice that n2 =
k k = < 200 for any n ≤ 10, therefore we p 5
have other 11 solutions. We have 34 solutions in all. Solution to Problem 98 We have ab + ad = 2(a + b + c + d) – 6, so (a – 2)(b – 2) + (c – 2)(d – 2) = 2. Assuming that a is the smallest number among a, b, c, d, we get –1 ≤ a – 2 ≤ 1. If a – 2 = 1 then b – 2 = c – 2 = d – 2 = 1 and a = b = c = d = 3. If a – 2 = 0, then c – 2 = 1 and d – 2 = 2 (or c – 2 = 2 and d – 2 = 1). It follows that c ⋅ d = 12, a = 2, that is b = 6. If a – 2 = –1, then a = 1 and b + c + d – 2 = b = cd. Hence c + d = 2, implying c = d = 1 and b = 1. The solutions are (a, b, c, d) = (3, 3, 3, 3); (1, 1, 1, 1); (2, 6, 3, 4); (6, 2, 3, 4); (2, 6, 4, 3); (6, 2, 4, 3); (3, 4, 2, 6); (3, 4, 6, 2); (4, 3, 2, 6); (4, 3, 6, 2). Solution to Problem 99 As n is even, we have an – bn = (a2 – b2)(an–2 – an–4b2 + … + bn–2). Since a + b is a divisor of a2 – b2, it follows that a + b is a divisor of an – bn. In turn, a + b divides n 2a = (an + bn) + (an – bn), and 2bn = (an + bn) – (an – bn). But a and b are coprime numbers, and so g.c.d.(2an, 2bn) = 2. Therefore a + b is a divisor of 2, hence a = b = 1. Solution to Problem 100 100 Suppose, by way of contradiction, that u =
3 − x 2 and v =
3
a − x3 are rational numbers. It
follows that x2 = 3 – u2 and x3 = a – v3, that is a – v3 = ±(3 – u2) 3 − u 2 . It follows that 3 − u 2 = q has to be rational, and 3 = u2 + q2, both u and q being rationals. m n Let m, n, p be integers with g.c.d.(m, n, p) = 1, such that u = and v = . Then 3p2 = m2 + n2, p p 2 2 that is 3 is a divisor of m + n . It is easy to see that 3 has to be a divisor of both m and n. Furthermore 9 is a divisor of 3p2, implying that 3 divides p. Since g.c.d.(m, n, p) = 1 we get a contradiction. Solution to Problem 101 101 Denote by n2 the perfect square and by a the digit that appears in the last four positions. It easily follows that a is one of the numbers 0, 1, 4, 5, 6 or 9. It follows n2 ≡ a ⋅ 1111(mod 104) and consequently n2 = a ⋅ 1111(mod 16). When a = 0 we are done. Suppose that a is 1, 5 or 9. Since n2 ≡ 0 or 1 or 4 (mod 8) and 1111 ≡ ≡ 7 (mod 8), we obtain 1 ⋅ 1111 ≡ 7 (mod 8), 5 ⋅ 1111 ≡ 3 (mod 8) and 9 ⋅ 1111 = 7 (mod 8). Thus the congruence n2 ≡ a ⋅ 1111 (mod 16) cannot hold. Suppose a is 4 or 6. As 1111 ≡ 7 (mod 16), 4 ⋅ 1111 ≡ 12 (mod 16) and 6 ⋅ 1111 ≡ 10 (mod 16). We conclude that in this case the congruence n2 ≡ a ⋅ 1111 (mod 16) cannot hold either.
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Solution to Problem 102 Let d = g.c.d.(x, y) and x = da, y = db, where (a, b) = 1. It is easy to see that a and b are both odd numbers and an + bn = 2k for some integer k. Suppose that n is even. As a2 ≡ b2 ≡ 1 modulo 8 we have also an ≡ bn ≡ 1 modulo 8. As 2k = an + bn ≡ 2 (mod 8), we conclude k = 1 and a = b = 1, thus x = y = d. The equation becomes xn = 2m–1. It has an integer solution if and only if n is a divisor of m – 1 and m−1
x=y= 2 n . Consider the case when n is odd. From the decomposition an + bn = (a + b)(an–1 – an–2b + … + bn–1), we easily get a + b = 2k = an + bn. In this case a = b = 1, and the proof goes on the lines of the previous case. m −1 is an integer and in that case To conclude, the given equation has solutions if and only if n x = y = 2p. Solution to Problem 103 103 The sum of all divisors of n is given by the formula (1 + p + p2 + … + pa)(1 + q + q2 + … +qb). The number n has (a + 1)(b + 1) positive divisors and their arithmetic mean is (1 + p + p 2 + ... + p a )(1 + q + q 2 + ... + q b ) . m= (a + 1)(b + 1) If p and q are both odd numbers, we can take a = p and b = q and it is easy to see that m is an integer. If p = 2 and q is odd, one can choose again b = q and a + 1 = 1 + q2 + … + qq–1. Then m = 1 + 2 + + 22 + … + 2a, and it is an integer. For p odd and q = 2, we choose a = p and b = p2 + … + pp–1, concluding the proof. Solution to Problem 104 104 4 4 4 Denote S = n1 + n2 + ... + n31 and A = {n1, n2, …, n31}. Firstly, observe that 2 ∈ A, otherwise all numbers ni, i = 1,31 are odd and consequently S must be odd, a contradiction. Then, 3 ∈ A, else ni ≡ –1(mod 3) and ni4 ≡ 1(mod 3) for all i = 1,31 . It follows that S ≡ 31 ≡ 1 (mod 3), a contradiction. Finally, we prove that 5 ∈ A. Indeed, if else, then ni ≡ ±1 (mod 5) or ni ∈ ±2 (mod 5) for all i = 1,31 . Consequently, ni2 ≡ ±1 (mod 5) and ni4 ≡ 1 (mod 5) for all i = 1,31 . Thus, S ≡ 31 ≡ 1 (mod 5), a contradiction. Solution to Problem 105 Assume that such numbers exist. By squaring, (1)
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2n + 1 + 2 n 2 + n < x + y + 2 xy < 4n + 2 .
1 Since 4n + 1 < x + y + 2 xy ≤ 2(x + y), we obtain x + y > 2n + . Numbers x and y are integers, 2
so x + y ≥ 2n + 1. Set a = x – y – (2n + 1) ≥ 0, where a is an integer. The second inequality from (1) gives 2 xy < < 2n + 1 – a, hence 4xy < (2n + 1 – a)2. Numbers 4xy and 2n + 1 – a are also integers, therefore 4xy ≤ ≤ (2n + 1 – a)2 – 1 and then 2 xy ≤ (2n + 1 − a ) 2 − 1 . From (1) we have 2 n 2 + n < a + 2 xy ≤ a + (2n + 1 − a) 2 − 1 ,
hence (2n + 1) 2 −1 − a ≤ (2n + 1 − a) 2 −1 .
(2)
As x + y < 4n + 2, then a = x + y – (2n + 1) ≤ 2n and so a < of the relation (2) we obtain
(2n + 1) 2 − 1 . By squaring both sides
(2n + 1)2 – 1 + a2 – 2a (2n + 1) 2 − 1 < (2n + 1 – a)2 – 1 ⇔ ⇔ – 2a (2n + 1) 2 − 1 < –2a(2n + 1), a contradiction. Solution to Problem 106 Number a has an odd number of digits, hence 102k ≤ a < 102k+1 for some integer k > 0. It suffices to prove that a = 102k. Firstly, observe that 104k ≤ a2 < 104k+2. Number a2 has also an odd number of digits, hence 102k ≤ ≤ a < 10
2k +
102k ≤ a <
1 2
. Next, 108k ≤ a4 < 108k+2 and consequently 102k ≤ a < 10
1 2k + 10 2 n
1 4
. Inducting on n we obtain
for all n > 0.
Assume by contradiction that a ≥ 102k + 1. Then 10 1 10 2 n
2k +
2k +
1 2n
1 > 102k + 1 ⇔ 102 k + 10 2 n − 1 > 1 ⇔
2n
1 1 ⇔ > 1 + 2 k ⇔ 10 > 1 + 2 k , for all n > 0. 10 10 On the other hand, using Bernoulli inequality we find that 2n
1 2n 1 + ≥ 1 + for all n > 0. 2k 102 k 10 2n For sufficiently large n we have 1 + 2 k > 10, a contradiction. 10
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Solution to Problem 107 Rearrange the numbers a1, a2, …, an in ascending order: b1 < b2 < … < bn. Obviously, k ≤ bk, and substituting bk with k, the left-hand side term increases. Furthermore, by the Rearrangements inequality we infer that the maximum value of the left-hand side term is 1 2 n + + ... + . n n −1 1 On the other side, the right-hand side term is greater than or equal to 1 + 2 + ... + n n(n + 1) = . 2 4 We have 1 2 n n n − k +1 = + + ... + = n n −1 1 k =1 k
∑
n n+1 1 1 1 . − n = 1 + (n + 1) = (n + 1) k =1 k k =2 k k =2 k For n > 6 we prove by induction on n that n n+1 1 , ≥ 4 k =2 k n
= (n + 1)
∑
∑
∑
∑
which implies that the given equality cannot hold. Indeed, for n = 7 we have +
7 1 1 = 1.75 ≥ + + … + 4 2 3
1 = 1.51… 8
1 1 ≥ . 4 n +1 We are left with the cases when n = 2, 3, 4, 5, 6. Clearly, the case case n = 2 is impossible. For n = 3 we have the numbers a1 = 1, a2 = 2 and a3 = 3, so n = 3 is a solution. If n = 4, then 1 2 3 4 1 2 3 4 a1 + a2 + a3 + a4 = 2 + + + ≤ 2 + + + < 13 , a1 a2 a3 a4 4 3 2 1 so a1 + a2 + a3 + a4 ≤ 12. By inspection, all the cases: {a1, a2, a3, a4} = {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5} and {1, 2, 3, 6} fail to satisfy the required relation. If n = 5, then 1 2 3 4 5 1 2 3 4 5 a1 + a2 + a3 + a4 + a5 = 2 + + + + ≤ 2 + + + + < 17.4 , a1 a2 a3 a4 a5 5 4 3 2 1 so a1 + a2 + a3 + a4 + a5 ≤ 17. We study the cases {a1, a2, a3, a4, a5} = {1, 2, 3, 4, 5}, {1, 2, 3, 4, 6} and {1, 2, 3, 5, 6} with no success (for an easy argument, observe that 5 must be a5 and so on). Finally, for n = 6 we obtain similarly a1 + a2 + … + a6 ≤ 22, thus {a1, a2, a3, a4, a5, a6} can be {1, 2, 3, 4, 5, 6} or {1, 2, 3, 4, 5, 7}. The last case fails immediately because of 7, and the same outcame is for the first one. Therefore n = 3.
If the inequality holds for n > 7 then it is true for n + 1, as
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Solution to Problem 108 Clearly one of the primes p, q or r is equal to 2. If r = 2 then pn + qn = 4, false, so assume that p > q = 2. Consider the case when n > 1 is odd; we have ( p + 2)( p n−1 − 2 p n−2 + 22 p n−3 − ... + 2n−1 ) = r 2 . Notice that p n−1 − 2 p n−2 + 22 p n−3 − ... + 2n−1 = 2n−1 + ( p − 2)( p n−2 + p n−4 + ...) > 1 and p + 2 > 1 hence both factors are equal to r. This rewrites as pn + 2n = (p + 2)2 = p2 + 4p + 4, which is false for n ≥ 3. Consider the case when n > 1 is even and let n = 2m. It follows that pm = a2 – b2, 2m = 2ab and r = 2 a + b2, for some integers a, b with (a, b) = 1. Therefore a and b are powers of 2, so b = 1 and a = 2m–1. This implies pm = 4m–1 – 1 < 4m, so p must be equal to 3. The equality 3m = 4m–1 – 1 fails for m = 1 and also for m ≥ 2, as 4m–1 > 3m + 1, by induction. Consequently n = 1 – take for example p = 23, q = 2 and r = 5. Solution to Problem 109 n(n + 1)( 2n + 1) 2 b . Assume that n has 6 a prime divisor p > 3. Then p | S and p | (a + b)2 + (a + 2b)2 + … + (a + pb)2 = pa2 + p(p + 1)b + p ( p + 1)(2 p + 1) 2 + b , a contradiction. It remains n = 2k3l, for some integers k, l. Suppose that k > 1. 6 Then 2 | S and so all elements of A must be odd. Taking the subset given by any pair we reach the contradiction. Finally, suppose that l > 1, so 3 | S. If one of the numbers a + b, a + 2b, a + 3b is divisible by 3, then we have a contradiction; if not, then 3 | b. Then 3 | (a + b)2 + (a + 2b)2 + (a + 3b)2 = 3a2 + + 12ab + 14b2, again a contradiction. We are left with n = 6, which satisfies the claim: the set A = {4, 9, 16, 25, 36, 49} is isolated, because the sum of its elements is a prime number (139).
The sum of the elements of the set A is S = na2 – n(n + 1)b +
Solution to Problem 110 110 The required number is 505. At start, note that the remainder of n when divided by 4 is odd, hence n is odd. n is greater than or 2 equal to 2. On the other side, the quotient at a division by an odd square cannot equal 3, as the n remainder would be even. Consequently, there are no positive integers k so that 3 ≤ < 4, in (2k − 1)2 other words there is no k ∈ N with n n < (2k − 1)2 ≤ . 4 3 n n n n Let m ∈ N* so that (2m – 1)2 ≤ < < (2m + 1)2. Then (2m + 1)2 – (2m – 1)2 > − , hence 4 3 3 4 n 8m > . It follows that 12
Furthermore, observe that the quotient of n when divided to a square less then
73
(2m – 1)2 ≤
n < 24 ⋅ m, 4
n < 288 are 9, 25 …, 225. 2 Recall that the quotients at the division by 9, 25, …, 225 are even, so the quotients at the division by 225 and 169 are both 2 (else 4 ⋅ 169 > 576). Thus n = 450 + a = 338 + b with 0 < a < 225, 0 < b ≤ 137 and a, b are odd, so n ≤ 338 + 137 = 505. For n = 505 one can easy check the claim.
so m ∈ {1, 2, …, 6}. Since n < 96 ⋅ m ≤ 576, then the odd squares less then
Solution to Problem 111 111 At first, note that if r = 1, then the new number is obviously rational. Furthermore, if x is a simple rational number (that is with no period), then again the new number is rational. Now assume that x has a period and let p ≥ 1 the number of digits of the period. Let m > 0 be the rank of the last decimal which do not belong to the period, thus x = 0,a1a2…am(am+1…am+p). Let l = k + ir > m, i ∈ N. Then x = 0,a1a2…al(al+1…al+r…al+pr). The number obtained by canceling the decimals ak, ak+r, ak+2r, ak+3r, … will have the period al…al+r–1al+r+1… al+pr–1, therefore it will be a rational number. Solution Solution to Problem 112 112 Suppose that n ≥ p. Then n8 – n2 = n2(n3 – 1)(n3 + 1) ≥ n2n2(n3 + 1) = n7 + n4 ≥ p7 + p5 > p5 + p2, a contradiction. Therefore n < p. As p is prime and p2 | n2(n3 – 1)(n3 + 1), it follows that p2 | n3 – 1 or p2 | n3 + 1, so p2 ≤ n3 + 1. Next we have n2(n3 – 1)(n3 + 1) = p5 + p2 ≤
(n3 + 1)5 + (n3 + 1), hence
n 2 (n3 − 1) ≤ (n3 + 1) n3 + 1 + 1 ,
(1) On the other hand n3 + 1 <
(n 2 − 1)2 , since this rewrites as n2 ≥ n + 2. The relation (1) yields
n2(n3 – 1) ≤ (n3 + 1)(n3 – 1) + 1 = n5 – n3 + n2, then n3 ≤ 2n2. Consequently n = 2 and p2 ≤ n3 + 1 = 9, thus p = 3. Solution to Problem 113 113 The numbers are 18 and 27. Let k be the number of digits of n in decimal representation. Notice that: (1) n = p ⋅ s(n), where p is prime so that any prime divisor of s(n) is greater than or equal to p; (2) s(n)2 ≥ n, so 10k–1 ≤ n ≤ s(n)2 ≤ (9k)2, hence k ≤ 4. We study the following cases: a) If k = 4, then n = abcd , n ≤ s(n)2 ≤ 362 = 1296, so a = 1. Then s(n) ≤ 28, thus n ≤ 282 < 1000, false. b) If k ≤ 3, then abc , so 9(11 ⋅ a + b) = (p – 1)(a + b + c).
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If 9 divides p – 1, since p < a + b + c = 27 we get p = 19. Next 9a = b + 2c, hence a ≤ 3. As a + b + + c ≥ 23 – see (1) – we have no solution. If 9 do not divide p – 1, from 3 | a + b + c and (1) we get p = 2 or p = 3. For p = 3 we have n = 3(a + b + c), so a = 0 and 10 ⋅ b + c = 3(b + c). Consequently, 7b = 2c and n = 27. For p = 2 we get n = 2(a + b + c), so a = 0 and 8b = c, hence n = 18. Solution to Problem 114 Without loss of generality, assume ak > bk, k = 1, 4 . Then a1 + a2 = a3 + a4 and b1 + b2 = b3 + b4. We analyze 2 cases: i) a1 + a2 = a3 + a4 and a2 + b1 = a4 + b3. Subtracting we get |a2 – b2| = |a4 – b4|, |a1 – b1| = |a3 – b3| and the claim is obvious. ii) a1 + b2 = a4 + b3 and a2 + b1 = a3 + b4. By subtraction we obtain |a2 – b2| = |a3 – b3|, |a1 – b1| = = |a4 – b4|, as needed. Solution to Problem 115 Assume by contradiction that the claim holds and let b ≤ a. The number r(b) has at most as many digits as b, so r(b) < 10b ≤ 10a. It follows that (2a)2 < 4a2 + 10a < (2a+ 3)2, hence 4a2 + r(b) = (2a + 1)2 or (2a + 2)2, thus r(b) = 4a + 1 or 8a + 4. Notice that r(b) > a ≥ b, implying that a and b have the same number of digits. Then, as above, we get r(a) ∈ {4b + 1, 8b + 4}. We analyze 3 cases: 1. r(a) = 4b + 1 and r(b) = 4a + 1. Subtracting we get (r(a) – a) + (r(b) – b) = 3(b – a) + 2, which is false since 9 divides r(n) – n for any positive integer n. 2. r(a) = 8b + 4 and r(b) = 4a + 1 (the same reasoning is to be applied for r(b) = 8a + 4 and r(a) = = 4b + 1). Subtracting we obtain (r(a) – a) + (r(b) – b) = 7b + 3a + 3, so 3 divides b. Then 3 divides also r(b) = 4a + 1, so a and r(a) have the remainder 2 when divided at 3. This leads to a contradiction with r(a) = (8b + 3) + 1. 3. r(a) = 8b + 4 and r(b) = 8a + 4. Both r(a) and r(b) have the last digit even, so at least 2. Then a and b have the first digit greater than or equal to 2, so 8a + 4 and 8b + 4 have more digits than a and b. It follows that r(a) < 8b + 4 and r(b) < 8a + 4, a contradiction. Solution to Problem 116 Let m = [ n ] . Since n ≥ 4, then m ≥ 2 is an integer. We have m2 ≤ n < (m + 1)2, so m2 ≤ n ≤ m2 + 2m. 2 Set n = m + k, k = 0, 1, 2, …, 2m. From m – 1 | m2 + k – 1 we get m – 1 | k. On the other hand k < 2(m + 1), thus k = 0 or k = m + 1. If k = 0, from m – 1 | m2 + 1 follows m – 1 | 2, so m = 2 or m = 3, hence n = 4 or n = 9. If k = m + 1, then m – 1 | m2 + m + 2 = m2 – 1 + m – 1 + 4, so m – 1 | 4. We obtain m = 2, 3 or 5, hence n = 7, 13 or 31. Therefore, n ∈ {4, 7, 9, 13, 31}.
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Solution to Problem 117 At first, notice that (x – 1)(x – 2) ≥ 0 for all x ∈ N, (y – 1)(y – 3) ≥ 0 if y ∈ N \ {2} and (z – 1)(z – 4) ≥ ≥ 0 when z ∈ N \ {2, 3}. In other words, if y ≠ 2 and z ∉ {2, 3}, then x2 + 2 ≥ 2x, y2 + 3 ≥ 3y and z2 + 4 ≥ 4z. Multiplying the above inequalities yields (x2 + 2)(y2 + 3)(z2 + 4) ≥ 60xyz, so in all three inequalities the equality must occur. Until now we have the solutions: (x, y, z) = (1, 1, 1), (1, 1, 4), (1, 3, 1), (2, 1, 1), (2, 3, 1), (2, 1, 4), (1, 3, 4), (2, 3, 4). We claim the there are no more solutions. For this, we will show that if z = 2 or z = 3 or y = 2, there are no integers satisfying the given equation. The quadratic residues modulo 5 are 0, 1, 4, so 5 do not divide neither x2 + 2 nor y2 + 3. Since 5 divides 60xyz, it follows that 5 divides z2 + 4, hence z ∈ {5k ± 1 | k ∈ Z}. As a consequence, z ≠ 2 and z ≠ 3. If y = 2, the equation rewrites as 120xz = 7(x2 + 2)(z2 + 4), from which we may notice that 8 divides 2 (x + 2)(z2 + 4). If x, z are even integers, then x2 + 2 is even and z2 + 4 is divisible by 4, but 4 P x2 + 2 and 16 P z2 + 4, so the power of 2 in the right-hand side is at most 4, while in the left hand-side is at least 5, a contradiction. If only one of the numbers x and z is even, the contradiction is reached similarly. Hence y ≠ 2 and the only solutions of the equation are the ones previously obtained. Solution to Problem 118 Any integer which can be represented as described in the problem will be called good. Setting b = c = 1 yields [a, b] + [b, c] + [c, a] = a + 1 + a = 2a + 1, hence any odd integer is good. Notice that [2x, 2y] = 2 ⋅ [x, y]. Therefore, if n can be represented as [a, b] + [b, c] + [c, a], then 2n writes as [2a, 2b] + [2b, 2c] + [2c, 2a] = 2([a, b] + [b, c] + [c, a]), thus all integers with are not powers of 2 are good. We claim that all numbers of the form 2k, k ∈ N are not good. For k = 0 and k = 1 this is obvious, as [a, b] + [b, c] + [c, a] ≥ 1 + 1 + 1 = 3. If k ≥ 2, suppose by contradiction that there exist a, b, c as needed. Let a = 2A ⋅ a1, b = 2B ⋅ b1, c = 2C ⋅ c1, where a1, b1, c1 are odd. Without loss of generality, assume that A ≥ B ≥ C. Then 2k = [a, b] + [b, c] + [c, a] = 2A([al, b1] + [a1, c1]) + 2B ⋅ [b1, c1]. Dividing by 2B, k > B yields 2k–B = 2A–B ⋅ ([al, b1] + [a1, c1]) + [b1, c1]. But [al, b1] + [a1, c1] is even and [b1, c1] is odd, contradiction. Solution to Problem 119 Suppose that p2 P a + b. It suffices to prove that p3 | a3 + b3. Indeed, if p2 | (a + b)3 – 3ab(a + b), we have p | 3ab. As p ≠ 3 is prime, it follows that p | a or p | b. Since p | a + b, we get p | a and p | b. As a consequence, p3 | a3 and p3 | b3, implying p3 | a3 + b3. Solution to Problem 120 120 k Let n ≥ 1 and let 10 , k ∈ N. Consider all the remainders of the numbers 10k at the division by n. Since there are only finitely many residues, there exists a = 0, 1, …, n – 1 so that 10m ≡ a (mod n) for infinitely many values of m ∈ N. Among them select only n, namely 10m1 , 10m2 , ..., 10mn , with m1 > > m2 > … > mn. The number A = 10m1 + 10m2 + ... + 10mn has n digits equal to 1 and the rest equal to 0, has the sum of all digits is n. Moreover, A ≡ na ≡ 0 (mod n), so n | A, which concludes the proof.
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Solution to Problem 121 121 Consider An–1 = an – an–1. Since an ≤ n and an–1 ≥ 1, we have An–1 ≤ n – 1. If An–1 = 0, that is an–1 = an, then a1 + a2 + … + an–2 is even and the claim reduces to the case of n – 2 numbers. If An–1 > 0, then a1 + a2 + … + an–2 is even and the claim reduces to the case of n – 1 numbers. Solution to Problem 122 122 Notice that a1 = n, as 1 divides any aj, and aj ≤ n, for any j = 1, 2, …, n. Consider an array with rows corresponding to 1, 2…., n and columns corresponding to the numbers a1, a2, …, an. Define the entries as follows, on the position (i, aj) put 1 if i divides aj and 0 otherwise. Now observe that the sum of the numbers placed of the ith row is equal to ai, as we see the entries 1 for each multiple of i in the sequence a1, a2, …, an. Hence the sum of all the entries in the array is aj .
∑
On the other hand, on the
a thj th
column we get and entry 1 as long as we count the divisors of aj, so
the sum of the numbers on the j columns is the number of divisors of aj. This implies that the sum of all the entries in the array is the sum of all divisors of the numbers aj. Using this double-counting, together with the obvious fact that the number of divisor of a is less than a – unless a is 1 or 2, show that n = 1 or n = 2. Alternative solution. Recall that a1 = n and ai ≤ n, for all i = 1, 2, …, n. Assume that n > 3. As an–1 ≥ 1, there exists a multiple of n – 1, where n – 1 ≥ 1, in the given sequence; let ak, k > 1 be such a multiple. The condition ai ≤ n shows that ak = n – 1, in other words there are n – 1 multiples of k in the sequence. As n and n – 1 are coprime, k does not divide a1 = n, so k divides a2, …, an. But k ≥ 2 and k | an, therefore an > 1. Thus n must occur at least twice in the sequence, so, beside a1 we have aj = n, j > 1. Hence k | n, a contradiction. As before, n = 1 or n = 2 are the only possible values. Solution to Problem 123 123 Let [an + b] = 2xn, for all integers n > 0. Then (1) 2xn ≤ an + b < 2xn + 1, (2) 2xn+1 ≤ a(n + 1) + b < 2xn+1 + 1. Subtracting (1) from (2) we get 2(xn+1 – xn) – 1 < a < 2(xn+1 – xn) + 1, for all n > 0. As 2(xn+1 – xn) – 1 is an odd integer, it follows that all numbers 2(xn+1 – xn) – 1 must be equal, as otherwise a belongs to two open intervals of lengths 2 having the left margins at a difference of at least 2, which is impossible. Hence 2(xn+1 – xn) – 1 = 2s – 1, s ∈ Z, so xn+1 – xn = s and then xn = ns + p, p ∈ Z, ∀n > 0. Plugging in (1) we get 2p – b ≤ (a – 2s)n < 2p – b + 1, ∀n > 0, so a = 2s, since otherwise the set of the positive integers will have an upper margin. Observe that s is an integer, so a is an even integer, as needed. Solution to Problem 124 It is easy to observe that p is odd and p ≠ q. in other words p ≥ 3 and (p, q) = 1. If q = 2, then 2p+1 = 7 + p2. The only solution is p = 3, as 2n+1 > 7 + n2, for all n ≥ 4. For q ≥ 3, by Little Fermat’s Theorem we get p | 2q – 7 and q | p + 7. Set p + 7 = kq, k ∈ N*. If 2q – 7 ≤ 0, we have q = 3 and p | –1, false.
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If 2q – 7 > 0, then 2q – 7 ≥ p, so 2q ≥ p + 7 ≥ kq, therefore k = 1 or k = 2. For k = 1 we obtain p + 7 = q, so p | 2p + 7. This implies p = 7 and then q = 14, false. Hence k = 2 and p + 7 = 2q. Suppose p > q; as p, q ≥ 3 we get qp ≥ qp and then 7 = 2qp – pq ≥ qp ≥ 33 = 27, a contradiction. Thus q > p and then p + 7 = 2q > 2p, which yields p = 3 or p = 5. For p = 3 we have q = 5, while p = 5 gives q | 12, with no solution. To conclude, the solutions are (p, q) = (3, 2), (3, 5). Solution to Problem 125 The solutions are (k, k2) and (k2, k), with k > 1. We have mn – 1 | (n3 – 1)m – n2(mn – 1) = n2 – m. On the other hand, mn – 1 | m(n2 – m) – (mn – 1)n = n – m2. If n > m2, then mn – 1 ≤ n – m2 ≤ n – 1, so mn ≤ n, false. If n = m2, then obviously m3 – 1 | m6 – 1, so all pairs (m, m2), m > 1 are solutions. If n < m2, from mn – 1 ≤ n3 – 1 we derive that n < m ≤ n 2 . Then mn – 1 ≤ m2 – n < m2 – 1, so n < m. If n2 – m > 0, we obtain mn – 1 ≤ n2 – m < n2 – 1, so m < n, a contradiction. Hence n = m2, which holds, since m3 – 1 | m3 – 1, so all pairs (n2, n), n > 1 are also solutions. Solution to Problem 126 Since is not divisible by 2 nor by 5, the number a2009 has a multiple M with all digits equal to 1. m Let m > 0 be the number of digits of M, that is: M = 11 ... 1 . As a2009+m = a2009 ⋅ 10 + 3M, we get m times
a2009 | a2009+m, as needed. Solution to Problem 127 Considering the remainders modulo 3 we get d = 0, otherwise 1 = 1 + (–1)c, false. The equation rewrites as 7a = 4b + 5c + 1. Assuming b ≠ 0, the left-hand side term is odd while the right hand-side is even, so b = 0 and then 7a = 5c + 2. It is obvious that c ≥ 1. Taking the remainders modulo 5 we obtain 2a = 2, hence a = 1 + 4k and 7(49)2k = 5c + 2. If k ≥ 1 then c ≥ 2. Modulo 25 we get 7(–1)2k = 0 + 2, false. It follows that k = 0, then a = 1 and c = 1. Therefore a = 1, b = 0, c = 1 and d = 0. Solution to Problem 128 Suppose n and n + 1 are saturated. Then the numbers 4n(n + 1) and 4n(n + 1) + 1 = (2n + 1)2 are also saturated, q.e.d.
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Chapter IV COMBINATORICS Solution to Problem 129 a) If n = 2k, there are no such symmetrical tilings (otherwise the k and k + 1 tiles must have the same colour). ⋅ 3 ⋅ ... ⋅3 = If n = 2k + 1, the problem is to count the possible tilings for k + 1 squares. There are 4 ⋅ 3 = 4 ⋅ 3k such tilings. b) There are 4 ⋅ 3 ⋅ 2 ⋅ 2 ⋅ ... ⋅ 2 = 4 ⋅ 3 ⋅ 2n–2 tilings. Solution to Problem 130 130 2
1 1 The first step give rise to one black square of area = . After the second step we obtain eight 3 9 1 8 more squares of side , the black region increasing by 2 . In the same manner, the third step 9 9 1 , that is at this stage the black area increases the black area by 82 = 64 black squares, each of area 27 becomes 1 8 82 + + . 9 9 2 93 We conclude that after 1000 steps, the area of the black region is 2 999 1 8 82 8999 1 8 8 8 + + + ... + 1000 = 1 + + + ... + = 9 9 2 93 9 9 9 9 9 1000
8 1− 1000 1 9 8 = ⋅ =1− . 8 9 9 1− 9 It is left to prove that the last number is greater than 0.001. This follows using a binomial expansion evaluation, namely 1000 1000 1000 1 1000 ⋅ 999 9 1 ⋅ 2 = = 1 + > > 1000 . 2 ⋅ 64 8 8 2 8 1000
8 Consequently, 1 − 9
> 1−
1 = 0,9999 . 1000
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Solution to Problem 131 131 Let us consider the general case, that is to consider the number an of equilateral triangles formed by division in n segments. We shall find a recurrence relation. Consider an equilateral triangle with sides partitioned into n + 1 equal segments and draw the n parallels to each side of the given triangle. We will count all triangles with at least one vertex on BC; the remaining ones are triangles counted in an. Consider first the triangles that have two vertices on BC. When choosing two division points on BC, say M and N with M ∈ (BN), one counts exactly one triangle, namely that one obtained by (n + 2)(n + 1) drawing parallels from M, N to AB, AC respectively. Hence we add new triangles. 2 Consider the triangles with only one vertex on (BC). For each of the n division points we count one unit triangle. Except for the two points closer to B and C respectively, we count n – 2 triangles of side 2, and so on. Therefore, ( n + 2)(n + 1) an+1 = an + + n + ( n − 2) + ( n − 4) + … 2 Changing n in n + 1, we get (n + 3)(n + 2) an+2 = an+1 + + (n + 1) + (n − 1) + (n − 3) = … 2 Summing up, we obtain ( n + 2)(n + 1) (n + 3)(n + 2) (n + 1)(n + 2) (n + 2)(3n + 5) an + 2 = an + + + = an + . 2 2 2 5 It follows that 10(3 ⋅ 8 + 5) a10 = a8 + = a8 + 145 = a6 + 237 = ... = a0 + 315 = 315 . 2 Therefore, the number of triangles is 315. Solution to Problem 132 132 Choose an arbitrary point A in the plane. Points located in the plane at a rational distance from A are colored in red, while the others are colored in blue. Consider an arbitrary segment PQ. We may assume that AP < AQ; if not, take instead P another point of the line segment (PQ). Recall that between two real numbers one can find a rational number q and an irrational number r. The circles centered at A and having the radii q and r intersect the segment PQ at the points M and N respectively. It is obvious that M is colored in red and N in blue, so the claim is proved. Solution to Problem 133 133 Color the 8 vertices of the cube in black or white such that the 4 vertices of the 2 regular tetrahedrons have the same color; notice that the 3 vertices adjacent to a vertex have its opposite color. Therefore, each movement increases the sum of the numbers assigned to the vertices sharing the same color by 3. Consider the cases: 1) MN is a diagonal of a face of the cube. Then M and N have the same color, say black. Assume by contradiction that there is a sequence of movements after which the same number n is assigned to all vertices. Let k1 and k2 be the number of white, respectively black vertices that were selected to
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perform the movements. Then 4n = 3k1 + 2 = 3k2, a contradiction. 2) MN is a diagonal of the cube. Selecting the vertices M, then N, and performing these 2 movements, the number 1 will be assigned to all the vertices, as needed. 3) MN is a side of the cube. The same outcome as in the previous case will occur after 2 movements when selecting the diagonally opposite vertices of M and N. This provides us with a full solution. Solution to Problem 134 The main idea is to observe that two consecutive rows have exactly 4 equal elements, namely those lying on the columns 1, 3, 5, 7 or 2, 4, 6, 8. Moreover, on the other 4 columns the elements are different. Wlog, assume that rows 1 and 2 are equal with respect to the columns 1, 3, 5, 7 and different on the column 2, 4, 6, 8; we call these rows odd equal. If rows 2 and 3 are also odd equal, then rows 1 and 3 are equal, as needed. If not, then rows 2 and 3 are even equal. Now consider the rows 3 and 4; we are done if the rows are even equal, so assume that they are odd equal. Finally, if rows 4 and 5 are odd equal, then rows 3 and 5 are equal, and if rows 4 and 5 are even equal, then rows 1 and 5 are equal. This concludes the proof. Solution to Problem 135 Let n be the number of players in the tournament. The total numbers of matches is n(n – 1), hence each player end up with n – 1 points. a) Assume by contradiction that each player has a different number of draws. As a draw gives 0.5 points, it follows that each player has an odd number of draws. Since the possible cases are: 0, 2, 4, …, 2(n – 1), we infer that each of these numbers is assigned to each of the players. Consider A the player with 0 draws and B the player with 2(n – 1) draws. Each player has played 2(n – 1) matches, hence B obtained a draw in each match played. The match A – B thus ended with a draw, a contradiction, since A has no draws. b) Suppose the contrary. Then each of the n players has 0, 1, …, n – 1 losses when playing the white. Let X and Y be the players with 0 and n – 1 losses, respectively. The player Y has no points when playing the white and n – 1 points, so he won all the matches with the black pieces. This implies that the match X – Y is won by Y, so is lost by X, a contradiction, since X has 0 losses with the white pieces. Solution to Problem 136 Let A1, A2, …, A1000 be the vertices of the polygon. We start with two lemmas. Lemma 1. Three of four consecutive vertices have the same color. Then after a sequence of moves all vertices will have the color of the fourth vertex. Proof. Let the colors be 0, 1 and 2. We have two cases: a) 1110 → 1122 → 1002 → 2202 → 2112 → 0000. b) 1011 → 1221 → 0000. Lemma 2. Any 4 consecutive vertices will turn after several moves in the same color. Proof. Form two pairs of consecutive vertices and change them in the same color – if they do not already have it. Then follow the sequence 1122 → 1002 → 2202 → 2112 → 0000.
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By the second lemma, after several moves the vertices A1, A2, A3, A4 will have the same color, say red. Likewise, A5, A6, A7, A8 will have the same color. Consider now the vertices A4, A5, A6, A7; the first is red and the other three have the same color. By the first lemma they all will turn red – of course, we do nothing if they were already red. We move on with this procedure until A1, A2, …, A997 turn red (note that 997 = 4 + 3 ⋅ 332, so this requires 332 steps). Now consider the vertices A998, A999, A1000, A1; by the second lemma they all will share the same color. If this is red, we are done. If not, say that they are blue, and taking the vertices A997, A998, A999, A1000 we obtain – using the first lemma – all vertices to be red, except for A1, which is blue. Now A1, A2, A3, A4 turn blue, then A5, A6, A7, A8 and so on. This time, after 333 steps, all the 1000 vertices (1000 = 1 + 3 ⋅ 333) will be colored in blue. Comment. Substituting colors with digits, notice that all moves: 01 → 22, 02 → 11 and 12 → 00 preserve the sum (mod 3). This means that the final color is unique and, of course, is given by the sum of the digits assigned to the vertices of the initial configuration. Solution to Problem 137 137 6 The number of routes between 2 cities is = 15. By the pigeonhole principle one can find that a 2 company – say M – operates at least 8 routes. 6 4 There are = 15 subsets of 4 cities and each pair of cities occurs in = 6 such subsets. Using 4 2 again the pigeonhole principle follows that there exists a subset of 4 cities among which at least 4 routes are operated by the company M (3 ⋅ 15 < 6 ⋅ 8). If those routes form a cycle, we are done. If else, then one can easy observe that among these 4 cities X, Y, Z, T, the routes X ↔ Y, Y ↔ Z, Z ↔ X and X ↔ T are operated by the company M (set the notation accordingly). The other 2 cities, say P and Q, are connected by at least 4 routes operated by M. Even if P ↔ Q is one of them, from P or Q – say P – at least 2 routes to X, Y, Z, T are operated by M. If both destinations are from X, Y, Z, a cycle is obtained, so assume that one of the routes is P ↔ T. In this case we are done if the second route is P ↔ Y or P ↔ Z, hence we are left with the case P ↔ X. From those above, the existence of a third route from P to X, Y, Z or T will provide a cycle. If not, from Q exist 2 routes to X, Y, Z, T. The same line of reasoning shows that we have to consider that on the route Q ↔ T is operated by M. Any of the routes Q ↔ X, Q ↔ Y, Q ↔ Z will close a cycle. To conclude, if the route P ↔ Q is operated by M, then any route from Q to X, Y, Z or T will provide us with the desired cycle.
Solution to Problem 138 1) A person can speak free of charge with at most 2k persons – k that he chooses and other k (at most) that select his numbers among their free calls. Since n ≥ 2k + 2, each person will be charged when speaking to (at least) another one. 2) Assume that all 2k + 1 persons are arranged in a circle. Each person will choose to speak free of charge when calling any of the k persons located – consecutively – at his right. Then any person will speak free of charge with the k persons located at his left, as all of them will choose him among their “favorite numbers”.
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Solution to Problem 139 n 2 (n 2 + 1) , so there is a row r having the sum of the 2 n 2 (n 2 + 1) . Consider the 2n – 2 numbers written on numbers assigned to its squares at least equal to 2 the first and last column – except for the two numbers which belong to the row r – and observe that their sum is at least 1 + 2 + … + (2n – 2) = (n – 1)(2n – 1). Now we can select two “complementary parts” of these columns – in order to complete the row r to (n − 1)(2n − 1) . Since a path – so that the sum of the numbers placed on these n – 1 squares is at least 2 n3 n(n 2 + 1) (n − 1)(2n − 1) n3 1 + = + n 2 − n + , to conclude we only have to notice that + 1 = 2 2 2 2 2
The sum of all integers is 1 + 2 + … + n2 =
n3 1 + – if n odd – and that 2 2 when n is even.
=
n3 n3 1 2 + n – n +1 is the smallest integer greater than + n2 − n + – 2 2 2
Solution to Problem 140 140 One can place 4, 6, …, 40, 42 checkers under the given conditions. We start by noticing that n is the sum of 7 even numbers, hence n is also even. On a row one can place at most 6 checkers, hence n ≤ 6 ⋅ 7 = 42. The key step is to use 2k × 2k squares filled completely with checkers and 2k + 1 × 2k + 1 squares having checkers on each unit square except for one diagonal. Notice that these types of squares satisfies the conditions, and moreover, we can glue together several such squares within the problem conditions. Below we describe the disposal of n checkers for any even n between 4 and 42. For 4, 8, 12, 16, 20, 24, 28, 32 or 36 checkers 1, 2, 3, 4, 5, 6, 7, 8 or 9 2 × 2 squares; notice that all fit inside the 7 × 7 array! For 6 checkers consider a 3 × 3 square – except for one diagonal; then adding 2 × 2 squares we get the disposal of 10, 14, 18, 22, …, 38 checkers. For 40 checkers we use a 5 × 5 and five 2 × 2 squares. Solution to Problem 141 141 Let n be the required number. We claim that n = 2007. With 7 cuts, from the given rectangular piece one can obtain an 11-sided polygon and some triangles. From a triangle, with 8 cuts one can get an 11-sided polygon and some extra pieces, sufficiently enough to continue the same procedure. Hence, using 7 + 8 ⋅ 250 = 2007 cuts one can obtain the 251 requested 11-sided polygons. Denote by k the number of pieces left at the end which are not 11-sided polygons and notice that each has at least 3 sides. Now, observe that with each cut the number of pieces increases by 1 and total the number of vertices increases with at most 4 – actually, with 2, 3 or 4, according to the number of existent vertices through which the cutting line passes.
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Then n = k + 250 and 4n + 4 ≥ v ≥ 11 ⋅ 251 + 3k, where v is the total number of vertices of all polygons at the end. Hence 4n + 4 ≥ 11 ⋅ 251 + 3(n – 250) = 2011 + 3n, so n ≥ 2007, as claimed. Solution to Problem 142 142 Assign the number 0 to each white square and the number 1 to each black square. The claim is achieved if we prove the existence of a 2 × 2 square with an odd sum of the 4 numbers inside. Assume the contrary, so each sum of the 4 numbers inside a 2 × 2 square is even. Summing over all squares we get an even number S. Notice that each square not sharing a common side with the given array occurs 4 times in 5, the squares with only a common side occurs twice, while the 4 squares in the corners only once. But in the four corners there are three 0’s and one 1, so the sum S is even, a contradiction. Remark. The given array may have a rectangular form, and the above solution requires no alteration. However, this remark can easily lead to alternative solutions using induction. Here is a sketch: choose a row with 0 and 1 at endpoints and call it the first row. Suppose that the number below 0 is also 0; arguing by contradiction, we notice that all “doubletons” formed vertically from the first two rows have equal numbers inside, so the second row – which starts with 0 – ends with 1. Deleting the first row of the given rectangular array, the claim is reached by induction. The same line of reasoning is applied to the case when below 0 the number is 1. Solution to Problem 143 143 Let a ♥ b be the greatest prime divisor of a + b. At first, notice that from the initial 16 numbers we obtain 8 primes. The largest prime that can be obtained is 31 = 15 ♥ 16; if this number occurs, the second largest can be 23 = 11 ♥ 12. Otherwise, 29 may occurs twice, from 16 ♥ 13 and 15 ♥ 14, followed by 19 – or lower. From the stage when we are left with 8 primes, and after pairing them we get 4 primes. If a prime is a+b obtained from two odd primes a and b, then a ♥ b ≤ . 2 If else, at least one is 2 and let p be the other. The number p + 2 is prime only if p ∈ {3, 11, 17, 29}. Therefore, if p and q are prime with p ≤ q, then p ♥ g ≤ q + 2. We will prove that the largest number which can end the game is 19. One example to obtain it is exhibit below: (1, 8); (2, 7); (9, 16); (10, 15); (3, 14); (4, 13); (5, 12); (6, 11) → 3, 3, 5, 5, 17, 17, 17, 17 (3, 3); (5, 5); (17, 17); (17, 17) → 3, 5, 17, 17 (3, 5); (17, 17) → 2, 17 (2, 17) → 19. Now, we have to show that the game cannot end with a number strictly greater than 19. Since from the second stage the number cannot increase with more than 2, and since 31 ♥ 2 = 11, we derive that the game will end with a prime p ≤ 31. Suppose by contradiction that p ∈ {23, 29, 31}. If p = 29, as 29 is not sum of two primes, then p is obtained from two of 29. In the previous stage four 29’s are needed, then in the second stage eight 29’s are required, in contradiction with an initial observation. Moreover, we have obtained a stronger result: 29 cannot end the game and cannot occur even in the last pair, since after 2 steps at most one 29 may occur. Suppose that p = 31. Two cases are possible: 31 = 2 ♥ 29 or 31 = 31 ♥ 31. The latter result forbids
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the first case, while the second case requires that the last four numbers are 31, 31, 31, 31 or 31, 31, 29, 2. But among the 8 primes obtained after the first step we have at most two 29’s or one 31, not enough to produce three 31’s or two 31’s and one 29. Assume that p = 23. Again two cases are possible: 23 = 29 ♥ 17 or 23 = 23 ♥ 23. The first case is impossible as shown above, while the second case is allowed if the last four primes are 23, 23, 23, 23 or 29, 17, 23, 23. If all primes are 23, the previous step has eight numbers with the average of 23, which is a contradiction with 8 ⋅ 23 < 1 + 2 + 3 + … + 16 = 8 ⋅ 17. The second case lead similarly to contradiction, since 29 requires two 29’s and the pair of 23’s are given by four numbers with the sum 4 ⋅ 23: 2 ⋅ 29 + 4 ⋅ 23 = 150 < 1 + 2 + … + 16 = 136. The solution is now complete. Solution to Problem 144 Consider an arbitrary grouping in pairs. A pair in which the persons cannot speak will be called “bad”. If there are bad pairs, we prove that some changes can be made to decrease to number of bad pairs. Applying this at most 4 times exhibit a grouping with no bad pairs, and we are done. Label the pairs A, B, C, D and the persons in pair X by X1 and X2. Two persons that cannot converse are called “enemies”, otherwise “friends”. Assume that A is a bad pair. Beside A1, the person A2 has at most two other enemies. Two cases arise: a) If the other enemies of A2 belong to the same pair – call it B, then A1 has at least a friend among C1, C2, D1, D2. Choose C1 as a friend of A1 and swap A1 with C2. The new pairs A and C are good, and the claim is satisfied. b) If not, in at least one of the pairs B, C, D there are only friends of A1. Wlog, say that this pair is B. One of the persons in this pair must be a friend of A2; call this person B1. Now swap A1 with B1 and the new pairs A and B are good, as desired. Remark. Consider the graph with vertices in the eight persons and edges corresponding to each pair of friends. The degree of each vertex is at least 4, so, according to Dirac’s theorem there exists a hamiltonian cycle. Taking 4 edges with ho common endpoint from this cycle, we get 4 good pairs, as needed. Solution to Problem 145 Given a set X of n points in the plane, consider a maximal free subset Y made of m elements, hence such that any point in X \ Y completes an equilateral triangle with (at least) a pair of points from Y. (Any X contains free subsets, since any subset with 1 or 2 elements is obviously free.) But for any pair of points from Y there exist only two points in the plane which complete an equilateral triangle, so m n – m ≤ 2 , that is n ≤ m2, or m ≥ n . 2 One checks the validity of this result for small values (1, 2, 3) of n, too (while the coplanarity restriction is obvious).
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Solution to Problem 146 Label the rows from 1 to 8 and the columns from 1 to 8. The unit square which lies on the row i and the column j will be referred as (i, j). On the skew-diagonal {(i, i) | i = 1, 2, …, 8} there are exactly 2 squares in which checkers were placed; wlog, assume that the squares are (1, 1), (2, 2). Looking at the 6 × 6 sub-array Q determined by the rows 3-8 and the columns 3-8, we see that any “skew-diagonal” of Q, togheter with (1, 1), (2, 2), is a skew-diagonal of the initial array. In view of the given conditions, no checkers are placed in the squares of Q. Now take any skew-diagonal of Q with the squares (2, 1), (1, 2); this is a skew-diagonal of the initial array, and the two checkers are placed inside (2, 1), (1, 2). Up to this point, we know that checkers are placed in the squares on the rows 1-2 or on the columns 1-2. Suppose by way of contradiction that there exist a square located on the first two rows – say (i, m), i = 1, 2, m ≥ 3 – and a square on the first two columns – say (s, j), j = 1, 2, s ≥ 3 – that hold checkers. Then squares (i, m), (s, j), (3 – i, 3 – j) belongs to a skew-diagonal, contradiction. Solution to Problem 147 For n ≥ 3, it suffices to draw 2n – 2 squares, as below: • if n is odd, from each vertex V of P draw squares of sides
n −1 n +1 squares n – 1, n – 2, …, , 2 2
with V as vertex in each square. n n – 1 squares of sides n – 1, n – 2, + 1, with V as 2 2 n vertex in each square and add 2 more squares of side with vertices in 2 opposite vertices of P. 2 To show that 2n – 2 square are needed, divide all four sides in unit segments with 4n – 4 points – other than the vertices of the square – and consider for each point the unit segments perpendicular to the border of the square. This 4n – 4 segments can be covered by no less than 2n – 2 squares, since a square cannot cover 3 such segments, so we are done.
• if n is even, from each vertex V of P draw
Solution to Problem 148 148 Two types of triangles arise after such a partition: “<” and “=”. Suppose that the initial triangle ABC is of type < and notice that all three adjacent triangles – the ones sharing a common side – are of type =. Call A-stripe the portion of plane containing the triangle ABC and bounded by the line BC and the parallel from A to BC. (It is easy to observe that the plane is divided in stripes parallel to this A-stripe.) Two =-triangles adjacent to ABC lie in this A-stripe. We claim that no path from ABC to one of these triangles exists. Indeed, observe that a move changes the type of the triangle hosting the checker, implying that if a path must exists, then it has an odd number of moves. On the other hand, a move changes the position of the checker from a stripe to an adjacent stripe located upwards or downwards. To reach in the end the same A-stripe, such a path must consists in an even number of moves, a contradiction. To conclude the proof, consider the B-stripe and the two =-triangles adjacent to ABC. Repeating the argument, we are done. Alternative Solution. Use four colors – one for ABC and the three adjacent triangles – to indicate the triangles that can be visited starting from any of these triangles. Justify.
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Solution to Problem 149 Suppose A = a0, a1, …, a2 n −1 is a rearrangement of the numbers 0, 1, 2, …, 2n – 1 which satisfies the required condition. Then (2A), (2A + 1) is a rearrangement of the numbers 0, 1, 2, …, 2n+1 – 1, where (2A) = 2a0, 2a1, …, 2 a2 n −1 and (2A + 1) = 2a0 + 1, 2a1 +1, …, 2 a2 n −1 + 1. It is obvious that the rearrangement (2A), (2A + 1) satisfies the claim. Indeed, no triples bi, bj, bk, with bi – bj = bj – bk, b bj b b − 1 b j − 1 bk − 1 i < j < k may occur in (2A) nor in (2A + 1), since either i , , k or i , , belongs to A, 2 2 2 2 2 2 contrary to the fact that A is “free” of triples in arithmetic progressions. Starting with the arrangement 2, 0, 3, 1 for the numbers 0, 1, 2, 3 and applying the above procedure, in 5 steps one has the required rearrangement. Solution to Problem 150 The required number is 2006, the number of the subsets having 2005 elements. For start, notice that each subset must have at least 2004 elements. If there exist a set with exactly 2004 elements, then this is unique and moreover, only 2 other subsets may be chosen. If no set has 2004 elements, then we can choose among the 2006 subsets, with 2005 elements and the set A with 2006 elements. But if A is among the chosen subsets, then any intersection will have more than 2004 elements, false. The claim is now justified. Solution to Problem 151 We claim v(m, n) = mn + 1. Partition M into n sets Pk = {(x, y); n | x + y – k}, k = 1, 2, …, n. The pigeonhole principle now forces (at least) m + 1 elements from P, be them Ai = (xi, yi), to belong to a same Pk. Now, if we assume xi = xj, then from xi + yi – k ≡ xi + yi – k(mod n) it follows n | yi – yj, but as yi, yj ∈ {1, 2, …, n}, it follows yi = yj, i.e. Ai = Aj. Conversely, mn + 1 is the least cardinality of P to warrant the claimed result; for |P| = mn, one can pick P = {{x, y); 1 ≤ x ≤ m, 1 ≤ y ≤ n}; then any m + 1 elements from P, be them Ai = (xi, yi), will share at least one xi = xj (pigeonhole principle again). Solution to Problem 152 Consider all positive differences a – b among all 10 numbers. Since there are C102 = 45 positive differences and all belong to the set 1, 2, 3, …, 36, at least two of them are equal. Let them be a – b and c – d, with a > c. If a, b, c, d are all distinct, we are done; if not, then b = c, so b = c is one of the 8 numbers which are neither the lowest nor the greatest number from the initial ones. Now observe that we have 45 positive differences and 36 possible values for them, so either 3 positive differences are equal or there are 9 pairs of equal positive differences. The first case, gives a – b = c – d = e – f, with a > c > e. Since we cannot have b = c, b = e and d = e, we are done. The second case gives at least one pair of positive differences in which case b = c is excluded, as only 8 candidates for b = c exist, so we are done.
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Solution to Problem 153 153 Consider a set S which satisfies all requirements. For each i ∈ A = {1, 2, …, n}, define Bi ⊂ B the set of all elements j ∈ B for which the pair belongs to the set S – notice that some subsets Bi can be empty. Counting all pairs in S over all second element in each pair, we have |S| = |B1| + |B2| + … + |Bn|. The main idea is to observe the chain of ‘inequalities’ B1 ≤ B2 ≤ … ≤ Bn, where by X ≤ Y we mean that x ≤ y, for any x ∈ X and y ∈ Y, X, Y being sets of integers. (This definition allows the empty set to occupy any position in this chain.) Since B1 ∩ B2 ∩ … ∩ Bn = B = {1, 2, …, m} and any two consecutive subsets Bi share in common at most one element, we get – by sieve theorem – that |S| ≤ m + n – 1, as claimed. Solution to Problem 154 154 Divide the interior of the circle into 12 congruent sectors such that each marked point lies in the interior of some sector. If one of them contains exactly 100 marked points, we are done. If not, we can find a sector A containing less than 100 points and a sector B containing more than 100 points (observe that it is not possible that all sectors contain less than 100 points or more than 100 points). Rotate sector A towards sector B. At each moment at most one marked point gets in or out sector A. It follows that there exists a moment in which the rotating sector contains exactly 100 marked points. Solution to Problem 155 Let A1, A2, …, An be the vertices of the polygon. We start with the following Lemma: Each segment AiAj belongs to at most 2 triangle of area 1 located on the same side of the line AiAj. Proof of the lemma. Indeed, suppose that on the same side of the line AiAj exist the vertices Am, An, Ap, so that the triangles AiAjAm, AiAjAn and AiAjAp have the area 1. Then the points Am, An, Ap will be at the same distance to the line AiAj, hence colinear. This is a contradiction, since the polygon is convex. Consider first the n sides of the polygon. Each of them can form at most 2 triangles of area 1, as all the vertices lie on the same side, hence we have by now at most 2n such triangles. Consider now the n diagonals AiAi+2 – with the cyclic notations: An+j = Aj. Each of them can form at most 3 triangles of area 1, one with Ai+1 and two with the vertices lying on the other side. Thus we have at most 5n = 2n + 3n triangles. n(n − 5) Finally, consider the other diagonals of the polygon. They are , and each of them can form 2 n(n − 5) at most 4 triangles. The final counting is 5n + 4 = n(2n – 5), except that we have counted each 2 n(2n − 5) triangle three times, one time for each side. Therefore, there are at most triangles, as 2 claimed.
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Table of Contents
FOREWORD ............................................................................................................................. 5 PROBLEMS CHAPTER I. ALGEBRA ..........................................................................................................................9 CHAPTER II. GEOMETRY ....................................................................................................................16 CHAPTER III. NUMBER THEORY ........................................................................................................24 CHAPTER IV. COMBINATORICS .........................................................................................................29 FORMAL SOLUTIONS35 CHAPTER I. ALGEBRA .......................................................................................................................37 CHAPTER II. GEOMETRY ..................................................................................................................49 CHAPTER III. NUMBER THEORY ......................................................................................................67 CHAPTER IV. COMBINATORICS .............................................................................................. 79
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