22. FileGocHOMOMoi

ContentsChapter 1.1 1.2 Chapter 2.1 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2. Examples for practice . . . . . . . . . . . . . . . . . . . . . . . Arithmetic problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Junior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Senior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 Undergraduate problems . . . . . . . . . . . . . . . . . . . . 2.1.4 Olympiad problems . . . . . . . . . . . . . . . . . . . . . . . 2.2 Algebraic problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Junior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Senior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Undergraduate problems . . . . . . . . . . . . . . . . . . . . 2.2.4 Olympiad problems . . . . . . . . . . . . . . . . . . . . . . . 2.3 Geometric problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Junior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Senior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Olympiad problems . . . . . . . . . . . . . . . . . . . . . . . 2.4 Analysic problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Junior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Senior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 Undergraduate problems . . . . . . . . . . . . . . . . . . . . 2.4.4 Olympiad problems . . . . . . . . . . . . . . . . . . . . . . . 2.5 Problems of Other Topics . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Junior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Senior problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.3 Undergraduate problems . . . . . . . . . . . . . . . . . . . . 2.5.4 Olympiad problems . . . . . . . . . . . . . . . . . . . . . . . Chapter 2. Exercises for training . . . . . . . . . . . . . . . . . . . . . . . 3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Exercises from Mathematic Reflextion . . . . . . . . . . . . 3.1.2 Exercises from OP from Around the World . . . . . . . . . 3.2 Problems of Hanoi Open Mathematical Olympiad . . . . . . . . . 3.2.1 Hanoi Open Mathematical Olympiad 2006 . . . . . . . . . 3.2.2 Hanoi Open Mathematical Olympiad 2007 . . . . . . . . . 3.2.3 Hanoi Open Mathematical Olympiad 2008 . . . . . . . . . 3.2.4 Hanoi Open Mathematical Olympiad 2009 . . . . . . . . . 3.2.5 Hanoi Open Mathematical Olympiad 2010. Senior Section 3.3 Singapore Open Mathematical Olympiad 2009 . . . . . . . . . . . . 3.3.1 Junior Section . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Senior Section . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 2 14 15 15 36 45 47 60 60 77 86 88 101 101 119 138 153 153 156 159 170 174 174 180 184 185 196 196 196 210 215 215 216 219 221 224 226 226 229 234 Chapter 1 Introduction 1.1 Preface Although mathematical olympiad competitions are carried out by solving problems, the system of Mathematical Olympiads and the related training courses cannot involve only the techniques of solving mathematical problems. Strictly speaking, it is a system of mathematical advancing education. To guide students who are interested in mathematics and have the potential to enter the world of Olympiad mathematics, so that their mathematical ability can be promoted efficiently and comprehensively, it is important to improve their mathematical thinking and technical ability in solving mathematical problems. Technical ability in solving mathematical problems does not only involve producing accurate and skilled computations and proofs, the standard methods available, but also the more unconventional, creative techniques. It is clear that the usual syllabus in mathematical educations cannot satisfy the above requirements, hence the mathematical olympiad training books must be self-contained basically. The book is based on the lecture notes used by the editor in the last 25 years for Olympiad training courses in BAC GIANG SPECIALIZING UPPER SECONDARY SCHOOL. Its scope and depth significantly exceeds that of the usual syllabus, and introduces many concepts and methods of modern mathematics. The core of each lecture are the concepts, theories and methods of solving mathematical problems. Examples are then used to explain and enrich the lectures, and indicate their applications. And from that, a number of questions are included for the reader to try. Detailed solutions are provided in the book. The examples given are not very complicated so that the readers can understand them more easily. However, the practice questions include many from actual competitions which students can use to test themselves. These are taken from a range of countries, e.g. China, Russia, the USA and Singapore. The questions are for students to practise, and test students’ ability to apply their knowledge in solving real competition questions. Each section can be used for training courses with a few hours per week. The test questions are not considered part of the lectures, since students can complete them on their own. Acknowledgments My great thanks to Doctor of Science, Professor Nguyen Van Mau, and Doctor Associate Professor Nguyen Vu Luong for their strong support. I would also like to thank my colleagues, MA Bach Dang Khoa, MA Tran Thi Ha Phuong, MA Nguyen Danh Hao and MA Tran Anh Duc for their careful reading of my manuscript, and their helpful suggestions. This book would be not written today without their efficient assistance. Nguyen Van Tien 1 2 Chapter 1. Introduction 1.2 Glossary Abel summation. For an integer n > 0 and reals a1 , a2 , ..., an and b1 , b2 , ..., bn, ∑ ai bi = bn ∑ ai + ∑ ((bi − bi+1 ) ∑ aj ) n n n−1 i i=1 i=1 i=1 j=1 Angle bisector theorem. If D is the intersection of either angle bisector of angle ABC with line AC, then BA DA = ⋅ BC DC Arithmetic mean-geometric mean (AM-GM) inequality. If a1 , a2 , ..., an are 1 n n nonnegative numbers, then their arithmetic mean is defined as ∑ ak and their n k=1 geometric mean is defined as (a1 a2 ⋯an ) n . The arithmetic mean - geometric mean inequality states that 1 1 1 n ∑ ak ⩾ (a1 a2 ⋯an ) n . n k=1 with equality if and only if a1 = a2 = ⋯ = an . The inequality is a special case of the power mean inequality. Arithmetic mean-harmonic mean (AM-HM) inequality. If a1 , a2 , ..., an are n 1 n positive numbers, then their arithmetic mean is defined as ∑ ak and their harmonic n k=1 1 mean is defined as ⋅ The arithmetic mean - harmonic mean inequality states that 1 n 1 ∑ n k=1 ak 1 n 1 ∑ ak ⩾ n 1 1 n k=1 ∑ n k=1 ak ( ∑ ak ) ( ∑ n or k=1 1 ) ⩾ n2 . k=1 ak n with equality if and only if a1 = a2 = ⋯ = an . Like the arithmetic mean-geometric mean inequality, this inequality is a special case of the power mean inequality. Bernoulli’s inequality. For x > −1 and a > 1, with equality when x = 0. (1 + x)a ⩾ 1 + ax, 3 1.2. Glossary Binomial coefficient. Cnk = n! , k!(n − k)! the coefficient of xk in the expansion of (x + 1)n . Binomial theorem. (x + y)n = ∑ Cnk xn−k y k = ∑ Cnk xk y n−k . n n k=0 k=0 Brianchon’s theorem. If hexagon ABCDEF is circumscribed about a conic in the projective plane such that A ≠ D, B ≠ E, and C ≠ F , then lines AD, BE, and CF concur. (If they lie on a conic in the afine plane, then these lines either concur or are parallel.) This theorem the dual to Pascal’s theorem. Brocard angle. ( See Brocard points.) Brocard points. Given a triangle ABC, there exists a unique point P such that ∠ABP = ∠BCP = ∠CAP and a unique point Q such that ∠BAQ = ∠CBQ = ∠ACQ. The points P and Q are the Brocard points of triangle ABC. Moreover, ∠ABP and ∠BAQ are equal; their value φ is the Brocard angle of triangle ABC. Cauchy-Schwarz inequality. For any real numbers a1 , a2 , ..., an , and b1 , b2 , ..., bn n (∑ a2i ) (∑ b2i ) i=1 i=1 n 2 ⩾ (∑ ai bi ) . n i=1 with equality if and only if ai and bi are proportional, i = 1, 2, ..., n. Centrally symmetric. A geometric figure is centrally symmetric (centrosymmetric) about a point O if, whenever P is in the figure and O is the midpoint of a segment P Q, then Q is also in the figure. Centroid of a triangle. Point of intersection of the medians. Centroid of a tetrahedron. Point of the intersection of the segments connecting the midpoints of the opposite edges, which is the same as the point of intersection of the segments connecting each vertex with the centroid of the opposite face. Ceva’s theorem and its trigonometric form. Let AD, BE, CF be three cevians of triangle ABC. The following are equivalent: (i) AD, BE, CF are concurrent; For integers a. Convex hull... Concave up (down) function.. k i=1 Circumcenter. Center of the circumscribed circle or sphere. . k. Introduction AF BD CE ⋅ ⋅ = 1. Circumcircle. . f (a1 )) and (b1 . Congruence. there exists a unique integer a such that 0 ⩽ a < ∏ ni and a ≡ ai (mod ni ) for i = 1. b. Concave up and down functions are also called convex and concave.. b+eiθ (a−b) corresponds to the image of A under a rotation through θ about B. If we introduce a Cartesian coordinate system in the Euclidean plane... . F are points and a. Given integers a1 . nk . and n with n ≠ 1. f are the corresponding complex numbers. . . b. respectively. A function f (x) is concave up (down) on [a. . f (b1 )) for all a ⩽ a1 < x < b1 ⩽ b.4 Chapter 1. Suppose that A. B.. ˆ given an angle θ. . there exists a convex set T such that every convex set containing S also contains T . (iii) sin ∠EBC sin ∠F CA sin ∠DAB (ii) Cevian. where we identify the line naturally with R. Complex numbers in planar geometry. Circumscribed circle. ˆ the absolute value of a − b equals AB. b + λ(a − b) corresponds to thee image of A under a homothety of ratio λ centered at B. Let k be a positive integer. F B DC EA sin ∠ABE sin ∠BCF sin ∠CAD ⋅ ⋅ = 1. Chinese remainder theorem.. 2. .. Given a nonempty set of points S in Euclidean space. (c − b) equals ∠ABC (directed and modulo 2π). Then: Ð→ ˆ a + (c − b) corresponds to the translation of A under the vector BC. a ≡ b (mod n) (or ”a is congruent to b modulo n”) means that a − b is divisible by n. We call T the .. a2 . ak and pairwise relatively prime positive integers n1 . b] ⊆ R if f (x) lies under (above) the line connecting (a1 . one can translate much of the language of geometry in the Euclidean plane into language about complex numbers. we can assign a complex number to each point in the plane by assigning α + βi to the point (α. ˆ given a real scalar λ. n2 .. A cevian of a triangle is any segment joining a vertex to a point on the opposite side. β) for all reals α and β. A function g(x) is concave up (down) on the Euclidean plane if it is concave up (down) on each line in the plane. ˆ the argument of (a − c) Using these facts.. then F + V − E = 2. V. C. According to a formula of Euler’s. A directed angle contains information about both the angle’s measure and the angle’s orientation (clockwise or counterclockwise). D lies either on the internal angle bisector of angle ABC. an is a permutation (b1 . and E are the number of faces. . De Moivre’s formula. then ∠ABC = ∠ADC if and only if A. n! n! n! n! + − + ⋯ + (−1)n 1! 2! 3! n! Desargues’ theorem.. (in planar geometry). This is a special case of an invariant of topological surfaces called the Euler characteristic. Cyclic polygon. (cos a + i sin a)n = cos na + i sin na. One often takes directed angles modulo π or 2π. C. Euler’s formula. Directed angles. b2 . C. Some important features of directed angles modulo p follow: ˆ If A. . These features show that using directed angles modulo π allows one to deal with multiple possible configurations of a geometry problem at once. Let O and I be the circumcenter and incenter. there are exactly n! − derangements of n items. if ∠ABC = 2∠ADC.2. then ∠ABC = ∠ABD if and only if B. For 2 2 example. Then OI 2 = R2 − 2rR. If two directed angles sum to zero. 1 or on the external angle bisector of angle ABC we cannot write ∠ADC = ∠ABC to 2 determine which line D lies on.. then they have the same angle measure but opposite orientations. D are points such that ∠ABC and ∠ABD are welldefined. C. one cannot divide directed angles by 2. D are points such that ∠ABC and ∠ADC are welldefined.. Euler’s formula. D are concyclic. ˆ If A. π π ˆ Because 2(θ) = 2( + θ). of a triangle with circumradius R and inradius r. Derangement.. . but at the expense of possibly losing important information about a configuration. respectively. Polygon that can be inscribed in a circle. B. and edges of a planar graph. Two triangles have corresponding vertices joined by lines which are concurrent or parallel if and only if the intersections of corresponding sides are collinear. B. bn ) of these items such that bi ≠ ai for all i. (for planar graphs) If F.. D are collinear. Glossary convex hull of S.. B. but θ ≠ + θ. vertices. A derangement of n items a1 .5 1. For any angle a and for any integer n. all lie on the same circle. BC. n Fermat number. . a2 . F1 = 1. Feuerbach’s theorem. If a0 . a1 . The sequence F0 . If f is a function such that f (x) = a0 + a1 x + a2 x2 + ⋯. then ap ≡ a(modp) for all integers a. Introduction Euler line. Excircles or escribed circles. Generating function. One is the inscribed circle. CA. The orthocenter. The excenter opposite A is the center of the excircle opposite A. there are four circles tangent to the lines AB. See excircles. A number of the form 22 for some positive integer n. and Fn+2 = Fn+1 + Fn for all n ⩾ 0. If p is prime. . . Euler’s theorem is a generalization of Fermat’s little theorem. Exradius. aφ(m) ≡ a(modm). then the generating function for the sequence is the infinite series a0 + a1 x + a2 x2 + ⋯.. where φ(m) is the number of positive integers less than or equal to m and relatively prime to m. Fermat’s little theorem. the midpoints of the three sides. Given a triangle ABC. Excenters. The nine-point circle of a triangle is tangent to the incircle and to the three excircles of the triangle. Fibonacci sequence. Let R be the circumradius of the triangle. The feet of the three altitudes of any triangle.. The nine-point circle of the triangle has radius R/2 and is centered at the midpoint of the segment joining the orthocenter and the circumcenter of the triangle. Euler’s theorem. which lies in the interior of the triangle. F1 . and the midpoints of segments from the three vertices to the orthocenter. defined recursively by F0 = 0. One lies on the opposite side of line BC from A. is a sequence of numbers... centroid and circumcenter of any triangle are collinear. The radius of the three excircles of a triangle. and similarly for the other two sides. and is called the excircle (escribed circle) opposite A. The centroid divides the distance from the orthocenter to the circumcenter in the ratio of 2 : 1.6 Chapter 1. the Feuerbach circle or the nine-point circle of the triangle. Feuerbach circle. Given relatively prime integers a and m with m ⩾ 1. The line on which these three points lie is called the Euler line of the triangle. it lies on the internal angle bisector of A and the external angle bisectors of B and C. . The area of a triangle with sides a. ∏ (∑ aij ) n m i=1 j=1 wi i ⩾ ∑ ∏ aw ij m n j=1 i=1 Homothety. If n > d and C1 . Suppose that AB ∥ DE. . Heron’s formula. A graph is called connected if for any two vertices v and w. BE. Graph. P. where the edges are distinct unordered pairs of distinct vertices. Glossary then we also refer to f as the generating function for the sequence.7 1.. If the points C and D divide AB internally and externally in the same ratio. Helly’s theorem. Then lines AD.. vertices with degree 1. . BC ∥ EF. and AB is said to be harmonically divided by the points C and D. The four points A. (i.. where k is called the magnitude of the homothety.. v2 . there exists a path from v to w.e.. and CF concur at a point X. The degree of a vertex is the number of edges which ontain it. Two triangles ABC and DEF are homothetic if they have parallel sides. Harmonic conjugates. then there is a point in common to all the sets. vi+1 ) are distinct edges. D be four points on a line in that order. . as given by a special case of Desargues’ theorem. Let w1 . wn be positive real numbers whose sum is 1. Furthermore... then the points C and D are said to be harmonic conjugates of each other with respect to the points A and B. A graph is a collection of vertices and edges.. A path is a sequence of vertices v1 . C. B.. P ′ are collinear and the ratio OP ∶ OP ′ = k is constant (k can be either positive or negative). Homothetic triangles. .. b. if C and D are harmonic conjugates with respect to A and B. A homothety (central similarity) is a transformation that fixes one point O (its center) and maps each point P to a point P ′ for which O. A connected graph which contains no cycles is called a tree. and every tree contains at least two leaves. A cycle of the graph is an ordered collection of vertices v1 . AC ∶ CB = AD ∶ DB). Let A. We say that the two vertices in one of these unordered pairs are adjacent and connected by that edge. denoted by (ABCD). Harmonic range. C. vn such that vi is adjacent to vi+1 for each i. v2 . c is equal to √ a+b+c s(s − a)(s − b)(s − c). D are referred to as a harmonic range. For any positive real numbers aij .2. B.. and CA ∥ F D. vn such that v1 ≡ vn and such that the (vi . . then A and B are harmonic with respect to C and D. each d+1 of which have nonempty intersection. where s = ⋅ 2 H¨ older’s inequality. Cn are convex subsets of Rd .. If C and D are harmonic with respect to A and B. BC. and analogous equations hold for AB 2 and BC 2 . q p .8 Chapter 1. Law of quadratic reciprocity. and ∠CAP = ∠QAB. and CA.. If f is concave up on an interval [a. Inscribed circle. the circle with center O and radius r. a Kummer’s Theorem. With this definition. it intersects ω and the tangents to the circle and to ω at either intersection point are perpendicular. . we see that P is also the isogonal conjugate of Q. Lines through O invert to themselves (though the individual points on the line are not all fixed). Law of cosines. λ2 . 4. Lattice point. There exists a unique point Q in the plane such that ∠ABP = ∠QBC. Isogonal conjugate .. In a triangle ABC.. not point-by-point) if and only if it is orthogonal to ω. CA2 = AB 2 + BC 2 − 2AB ⋅ BC cos ∠ABC. b].. ∠BCP = ∠QCA. y) for which x and y are both integers. 2. We call Q the isogonal conjugate of P . where the angles in these equations are directed modulo π. the inequality is reversed. Inversion of center O and ratio r. b] and λ1 . Jensen’s inequality. We also refer to this map as inversion through ω. Incenter. . Incircle. x2 . the lattice points are the points (x. Given nonnegative integers a and b and a prime p. Circles not through O invert to other circles not through O. λn are nonnegative numbers with sum equal to 1.. q are distinct odd primes. Center of inscribed circle. then λ1 f (x1 ) + λ2 f (x2 ) + ⋯ + λn f (xn ) ⩾ f (λ1 x1 + λ2 x2 + ⋯ + λn xn ) for any x1 . 3. Let ABCbe a triangle and let P be a point in the plane which does not lie on any of the lines AB. Introduction some homothety centered at X maps triangle ABC onto triangle DEF. then (p−1)(q−1) p q ( ) ( ) = (−1) 4 . If p. pt ∣Ca+b if and only if t is less than or equal to the number of carries in the addition a + b in base p. xn in the interval [a. Given a point O in the plane and a real number r > 0. Lines not through O invert to circles through O and vice versa. In the Cartesian plane. Key properties of inversion are: 1. A circle other than ω inverts to itself (as a whole. the inversion through O with radius r maps every point P ≠ O to the point P ′ on Ð→ the ray OP such that OP ⋅ OP ′ = r 2 . that is.. If the function is concave down. a multiset is a set in which an element may appear more than once. b = bk pk + bk−1 pk−1 + ⋯ + b1 p + b0 . HB F C GA Minkowski’s inequality. The object in the ith row and j th column of matrix A is denoted ai. . then Legendre symm bol ( ) is defined to equal 0 if n ∣ m. In a triangle ABC with circumradius equal to R one has sin A sin B sin C = = = 2R. an.9 1.. k. Given a triangle ABC. bi < p are integers for i = 0. 1. In a square n × n matrix A. 1} are distinct multisets.. . 2} and {2. A matrix A with m rows and n columns is an m × n matrix. 2. AB. If a matrix has the same number of rows as it has columns.. let a and b be two positive integers such that a = ak pk + ak−1 pk−1 + ⋯ + a1 p + a0 . .2. CA. n i=1 n i=1 n i=1 Multiset. If m is an integer and n is a positive prime. 2. 3. and positive reals a1 . Glossary q p where ( ) and ( ) are Legendre symbols.. the main diagonal consists of the elements a1. Suppose that S is a collection of functions on a set T . T can be partitioned into its orbits under S. b2 . Nine point circle. For instance. G.1 . a real number r ⩾ 1. such that S is closed under composition and each f ∈ S has an inverse. then the matrix is called a square matrix. where 0 ⩽ ai . bn .. and −1 if n m is a quadratic nonresidue modulo n. a2 . sets of elements such that a and b are in the same set if and only if f (a) = b for some f ∈ S..2 . AH BF CG ⋅ ⋅ = −1. q p Law of sines.j .. Then F.. {1. . 2. Then ⋯Cab11 Cab00 (mod p) Cabk−2 Cab ≡ Cabkk Cabk−1 k−2 k−1 Matrix.) Orbit.. using directed lengths. Lucas’s theorem.. let F. an and b1 . Given a positive integer n. we have 1 r 1 r 1 r (∑(ai + bi )r ) ⩽ (∑ ari ) + (∑ bri ) .. H are collinear if and only if. Informally. 3. . BC AC AB Legendre symbol.n . Let p be a prime. G. 1 if m is a quadratic residue modulo n. A matrix is a rectangular array of objects.. Menelaus’ theorem. (See Feuerbach circle. a2. respectively. H be points on lines BC. let B denote the number of lattice points on its boundary and let I denote the number of lattice points in its interior. The area of P is given by the formula 1 I + B − 1.. φ(ab) = φ(a)φ(b) for all a. Pole-polar transformation. Given a non self-intersecting polygon P in the coordinate plane whose vertices are at lattice points. or else each pair of opposite sides is parallel. 2. If ABCDEF is a hexagon inscribed in a conic in the projective plane. then we may denote a permutation π of S by {y1. This equation has infinitely many integer solutions in x and y.. For all positive integers n. and lines that do not pass through O to points. Point of intersection of the altitudes. (If the hexagon is inscribed in a conic in the afine plane. If P ≠ O is a point then the polar of P is the . some box contains at least two objects. If n objects are distributed among k < n boxes. such that each pair of opposite sides intersects at most one point. 2 Pigeonhole principle. b relatively prime. . Let S be a set.. . such that g d = 1.. that is. then the Diophantine equation x2 − Dy 2 = 1 in x and y is known as a Pell’s equation.. Pick’s theorem.10 Chapter 1. The pole-polar transformation with respect to C maps points different from O to lines. where yk = π(xk ). Function f (x) is periodic with period T > 0 if f (x + T ) = f (x) for all x. φ(n) is defined to be the number of integers in {1. If S = {x1 . y2 . Introduction Order. A permutation of S is a one-to-one function π ∶ S → S that maps S onto S. If D is a prime congruent to 3 modulo 4. Periodic.. Pell’s equations. Phi function. there exists a smallest positive integer d. This function is multiplicative. Let C be a circle with center O and radius R... then either the above result holds. . xn } is a finite set. Orthocenter of a triangle.) This theorem the dual to Brianchon’s theorem. named the order of g.. then the three intersection points formed in this manner are collinear. x2 . Permutation. n} which are relatively prime to n. yn }. Pascal’s theorem. Given a nonzero element g of a finite field. . then the pole of q is the point Q that has polar q.. called a primitive element of F . y. Power mean inequality. a field F with p elements contains an element g. i. b. B) denote the distance between the objects A and B. Then M−∞ ⩽ Ms ⩽ Mt ⩽ M+∞ for all s ⩽ t. . Mt = (a1 xt1 + a2 xt2 + ⋯ + ak xtk ) t . 0)}. on.. Also.. P X ⋅ P Y = P O2 − r2 no matter which line is drawn. M0 = xa11 xa22 ⋯xann . and the equivalence class containing (a. γ) ∈ K3 ∖ {(0. The elements of K are called points.... there exists an integer k such that g k = h. P )d(O. under equivalence by scalar multiplication (that is. The pole-polar transformation with respect to the circle C is also called reciprocation in the circle C. x2 . Suppose that we are given a fixed point P which lies either outside.11 1. where d(A. c) is often denoted [a.. xk }. Projective plane. a2 . The projective plane over K is the set of equivalence classes of K3 ∖ {(0.2. p′ ) = R2 .. Draw a line through P which intersects the circle at X and Y . the set of solutions [x. with the following property: for any nonzero element h of F . it does not depend on which line was drawn. given (α. c) and (d. The power of the point P with respect to C is defined to be the product of the signed distances P X and P Y. b. M+∞ = max{x1 . Glossary Ð→ line p′ that is perpendicular to ray OP and satisfies d(O. b. z] to αx + βy + γz = 0 . For positive numbers x1 . For each prime p. Polynomial in x of degree n. β. c] or [a ∶ b ∶ c]. 0)}. f ) are equivalent if and only if (a. or inside a fixed circle C with center O and radius r. b. xk }. f κ) for some κ ∈ K). ... Primitive element. . If q is a line that does not pass through O. e. n k=1 Power of a point theorem. eκ. Let K be a field. x2 . x2 . Function of the form f (x) = a0 + a1 x + a2 x2 + ⋯ + an xn ∶= ∑ ak xk with an ≠ 0. More precisely. Let a1 . an be any positive numbers for which a1 + a2 + ⋯ + an = 1. where (a.e.. c) = (dκ. . xn we define M−∞ = min{x1 .. The power of a point theorem states that this quantity is a constant. 1 where t is a non-zero real number. Introduction is called a line in the projective plane over K. lines) are said to ”intersect in” or ”lie on” a unique line (resp. ω3 be three circles whose centers are not collinear. write a = BC. Solid triangle inequality. b = CA. c = AB. Let ω1 . This function is multiplicative. Let ω1 and ω2 be two non-concentric circles. Radical axis. b relatively prime. Stewart’s theorem. point). ω2 . . For positive numbers x1 . that is. Solution to the equation z n − 1 = 0. and d = AD. B. The locus of all points of equal power with respect to these circles is called the radical axis of ω1 and ω2 . AC ⋅ BD = AB ⋅ CD + AD ⋅ BC. For all positive integers n. m = BD. σ(n) is defined to be the sum of all positive integer divisors of n. √ x21 + x22 + ⋯ + x2n x1 + x2 + ⋯ + xn ⩾ ⋅ n n Sigma function. ω2 . σ(ab) = σ(a)σ(b) for all a. the feet of the perpendiculars from P to the sides of line called the Simson line of P with respect to ∆ABC all lie on a line called The Simson line of P with respect to ∆ABC. Root of unity. xn . This point is called the radical center of ω1 .. Radical axis theorem. Ptolemy’s theorem. C.12 Chapter 1. There is exactly one point whose powers with respect to the three circles are all equal. Simson line. Then d2 a + man = c2 n + b2 m. x2 . Root of an equation. we have ∠AP B + ∠BP C > ∠AP C.. For any point P on the circumcircle of ∆ABC. n = DC. P in three-dimensional space which are not coplanar. ω3 . Solution to the equation. In a triangle ABC with cevian AD. Given four points A. Any two distinct points (resp. In a convex cyclic quadrilateral ABCD. . This formula can be used to express the lengths of the altitudes and angle bisectors of a triangle in terms of its side lengths.. Root Mean Square-Arithmetic Mean Inequality. defined by t0 = 0 and the recursive relations t2k = tk . n(n + 1) . The sequence t0 .. 1 + tan 2 a cos 2a = cos2 a − sin2 a = 2 cos2 a − 1 = 1 − 2 sin2 a = tan 2a = 2tan a .13 1. cos 3a = 4 cos3 a − 3 cos a.2. The binary representation of n contains an odd number of 1′ s if and only if tn is odd. 1 − tan 2 a triple-angle formulas: sin 3a = 3 sin a − 4 sin3 a. t2k+1 = 1 − t2k for all k ⩾ 1. tan x + 1 = 2 cos x addition and subtraction formulas: sin(a ± b) = sin a cos b ± cos a sin b. where n is some positive in2 Trigonometric identities. 1 − 3tan 2 a half-angle formulas: sin2 a 1 − cos a = 2 2 1 − tan 2 a 1 + tan 2 a . Glossary Thue-Morse sequence. sin2 x + cos2 x = 1. t1 ... . 2 sin x 1 2 = sec 2 x. 1 + cot 2 x = 1 = csc2 x. cos(a ± b) = cos a cos b ∓ sin a sin b. A number of the form teger. 3tan a − tan 3 a tan 3a = . tan a ± tan b tan (a ± b) = . 1 ∓ tan atan b double-angle formulas: sin 2a = 2 sin a cos a = 2tan a . Triangular number. each nonnegative integer n can be written uniquely in the form ∞ n = ∑ αk Fk k=0 where ak ∈ {0. . F1 . (n − 1)! = −1 (mod n) Zeckendorf representation.14 Chapter 1. be the Fibonacci numbers 1. If n > 1 be a positive integer. This expression for n is called its Zeckendorf representation. cos a cos b difference-to-product formulas: sin a − sin b = 2 cos a−b a+b sin . 1} and (ak . 1) for each k. ak+1 ) ≠ (1. that is. then if and only if n is prime. 2 2 a+b a−b cos a − cos b = −2 sin sin . . 2 2 sum-to-product formulas: sin a + sin b = 2 sin a−b a+b cos . 2. 2 2 tan a − tan b = sin(a − b) ... 2 sin a sin b = − cos(a + b) + cos(a − b). .. cos a cos b product-to-sum formulas: 2 sin a cos b = sin(a + b) + sin(a − b). Each nonnegative integer n can be written uniquely as a sum of nonconsecutive positive Fibonacci numbers. cos a + cos b = 2 cos 2 2 tan a + tan b = sin(a + b) .. 2 2 a−b a+b cos . Wilson’s theorem.. Introduction cos2 a 1 + cos a = . 2 cos a cos b = cos(a + b) + cos(a − b). Let F0 . 1 2. 16 (mod 20) and we have six distinct possible residues.Chapter 2 Examples for practice 2. x23 . Second solution Note that for all integer x we have x2 ≡ 1. Solution 15 . 1 (mod 5). x22 . we have at least two perfect squares. x24 . there are two squares x2i and x2j with the same residue and they satisfy the requirement. out of these four perfect squares. Example 2. we have n2 = −1. Hence their diference is divisible by 20 and our proof is complete. Again by the Pigeonhole Principle. 16 (mod 20). 4) = 1. 4∣ (a2 − b2 ). and we are done. 8. note that for any integer n.1 Arithmetic problems Junior problems Example 2. If we have seven arbitrary perfect squares x21 . x25 . but gcd(5. 8. 4. 4. 2. Third solution Observe that by the Pigeonhole Principle. By the Pigeonhole principle we conclude that among seven perfect squares we must have at least two that have the same residue modulo 20.1. x26 . such that a2 = b2 (mod 5). thus we have 20∣ (a2 − b2 ). 2. Hence.1. This implies that 5∣ (a2 − b2 ). 0. First solution It is easy to check that perfect squares can give one of the following residues: 1. there are at least four perfect squares which all have the same parity. Now. Also. x27 .2. Prove that among seven arbitrary perfect squares there are two whose diference is divisible by 20. Find all nine-digit numbers aaaabbbb that can be written as a sum of fifth powers of two positive integers. 2∣(a − b) and 2∣(a + b) since both a and b have the same parity. by the pigeonhole principle. say a2 and b2 . n a multiple of 11. not in 11. so m5 + n5 formed in this way ends in 61. a digit u must exist such that (10u+9)5 must end in 99999. 7. and 800v must finish respectively in 100 (impossible) or in 600. and m5 + n5 leaves remainders 0. 4.16 Chapter 2. Now. and m5 + n5 ends in the same two digits as 55 + 45 + 55 10u + 44 50v. yielding m = 55. We conclude that the only number of the form aaaabbbbb that is the sum of two fifth . hence v = 2 or v = 7. 2}. Clearly u must be odd. v such that u5 + v 5 = aaaa. and no solution exists in this case. hence either m = 10u + 5 and n = 10v + 5 or m = 10u and n = 10v. Combining the numbers whose fifth powers do not exceed 104. ±7. m = 10u + 5 and n = 10v + 4. the last four digits of m5 + n5 are also the last four digits of 53 1000(u2 + v 2 ) + 55 10(u + v) + 2 ⋅ 55 . 35 = 243 = 1 (mod 11). 1. then m + n is a multiple of 5 with multiplicity 2 = a = 1. 45 = 1024 = 1 (mod 11) and 55 = 3125 = 1 (mod 11). yielding u + v + 1 = 8. 3. 4. 0. (5u+v)5 = v 5 (mod 25) for any integers u. ±8. u + v + 1 must be a multiple of 4 for the last three digits to be 0. or m + n = 125. n cannot finish in the same digit or otherwise m5 + n5 would finish in 8. u = 5 and v = 2 results in m5 + n5 = 513050000.mainders −2. Trying all possible alternatives with u + v = 7. absurd. Note that 55 divides m5 + n5 = (m + n)5 − 5mn(m + n)3 + 5m2 n2 (m + n). In either case. 0. If b = 9. 2 modulus 11. 25 = 32 = −1 (mod 11). ±14. aaaabbbb0 is clearly a multiple of 11. −1. 1. either m or n is a multiple of 5. ±6. 85 ends in 68 and 333 ends in 93. 0. Finally. Since 15 = 1 (mod 11). m = 33 must be a multiple of 11. either m + n = 250 and either m5 or n5 exceeds 1010 . hence so is the other. or n = 24 not a multiple of 11. if b = 0. or any fifth power n5 leaves remainders −7. 0. then b ∈ {0. n is a multiple of 11. and 250u ends in 750 or 250. yielding u = −1 (mod 20). If b = 1. and wlog m5 ends in 0 and n5 ends in 5. Finally. or b = 0 with m + n a multiple of 5. Examples for practice By Newton’s binomial formula. In the first case. Now. or since 95 = 59049. Moreover. we find that this result is only true when 65 + 15 = 7777. Since aaaabbbbb = m5 + n5 for integers m. ±2. 50 ⋅ 94 u = 328050u. 9} for the last two digits of m5 +n5 to be equal. ±1. and n = 10u + 8 for some digit u. then m5 + n5 leaves re. In the first case. either m is a multiple of 10 and n5 finishes in 99999. impossible for a digit. hence the last four digits of 5000(u2 + v 2 − 1) + 1250(u + v + 1). Now. hence wlog m ends in 3 and n ends in 8. n = 3 (mod 5) and exactly one of m. then b ∈ {9. and since no digit may be congruent to −1 modulus 11. 1. Together with the previous result. hence 50u must end in 950. or b = 1 with m. because if m + n is a multiple of 53 . Since 885 > 999999999. and clearly one ends in 0 iff the other one ends in 0. 1. and 855 > 999999999. −1. and 1250(u + v + 1) must end in 0000. n. we find that u = 6 results in m5 > 999999999. v. 2 modulus 5. so no solution exists in this case. clearlym. since u + v + 1 = 16 produces either u or v = 8. 555 + 245 = 511246999. 1. when m = 10u and n = 10v. or m is odd and finishes in 5 and n5 must finish in 4. In the second case. hence u2 + v 2 − 1 is even. No solution is possible in this case. which are also the same three digits as 125 + 24 + 250u + 800v = 800v + 250u + 149. yielding either a = 1 and m2 n2 is a multiple of 54 . when n is respectively congruent to 3. or a = 2 and m2 n2 is a multiple of 52 . 7 modulus 25. But 745 > 999999999. and u = 4 and v = 3 results in 237050000. but (10u + 8)5 − 85 is a multiple of 100. we may conclude that the possible values for b are b = 9 with wlog m = 0 (mod 5) and n = 4 (mod 5) and exactly one of m. the problem is equivalent to finding all digits u. Let n be an integer greater than 2.4.3. 2) if x ∈ Ir . y. from where it follows that w = −1. 1.k + ) n−1 n (⊠) Example 2. The second case implies that w = −1 − (x + y + z) = −1. y. z. r ⩾ 1. Solution Let’s suppose k = x < k + 1 and then nk = nx < nk + n. We write n−1 n−1 r+1 r ) ∶= ⋃ Ir . −1) with all posible permutations among x. Arithmetic problems powers is 777700000 = 605 + 105 . 1. where {a} denotes the fractional part of a. Jr = [k + satisfy the inequality. −2). z. 1. Find all quadruples (x.1. 1). which is a contradiction.17 2. 1. 2) or (1. k + 1) = ⋃ [k + . It follows that (x. which is a contradiction. [k. hence x may take one of the following values: . w 2 − w = −1 − (xy + yz + zx) = 2 and w 3 = xyz + 1 = −1. or x + w + 1 divides −2. w) are (1. −2. −2) or (0. (1. it satisfies {x} = {nx}. k + n n r=1 r=1 r For x ∈ Ir the inequality {x} ⩽ {nx} reads as x − k ⩽ nx − nk − r that is x ⩾ k + n−1 The compatibility conditions are x ∈ Ir ⇔ k + r r+1 r ⩽k+ ⩽k+ n n−1 n and we conclude that: 1) if x ∈ I0 . w) of integers satisfying the system of equations x + y + z + w = xy + yz + zx + w 2 − w = xyz − w 3 = −1. 2). −1. y. It follows that all posible quadruples (x. 1. The first case implies that w = −1 − (x + y + z) = −3 and w 3 = 1 + xyz = 1. y + z. y. This implies that (x + y. −1. (−1. r+1 r . The third case implies that w = −1−(x+y+z) = −2 and w 3 = xyz+1 = 1. Find all real numbers x such that {x} = {nx}. z (⊠) Second solution Note first that (x+w +1)(y +w +1)(z +w +1) = w 3 −1+(w +1)(−w 2 +w −1)+(w +1)2 (−w −1)+(w +1)3 = −2. First solution Note that (x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) − xyz = 2. z) equals some permutation of (0. z + x) equals some permutation of (1. (⊠) Example 2. only those x in the subinterval Jr ⊂ Ir . −2). or w = 1. 2. (⊠) Example 2. hence no solution in integers exist in this case. . leading to w 3 − 1 = xyz = (w + 3)(w 2 − 3w − 5) = w 3 − 14w − 15. ˆ x = −w. 1. for x = 0 and yz = −1. p1 = 2. . First solution Let pn = (2 + √ n √ n 5) + (2 − 5) 2 √ n √ n 5) − (2 − 5) qn = √ 2 5 √ √ n for n = 1. n ∈ N (2.. and (x.1) and (2 + with initial conditions p0 = 1. Then (2 + 5) = pn + qn 5. then y + z = −1 − w − x = −1. and yz = −1 + w − w 2 + (w + 2) = 1 + 2w − w 2 . −2). and yz = −1 + w − w 2 + 2(1 − w) = 1 − w − w 2 . . or y = z = 1 because y + z = 2. Restoring generality.5. . q0 = 0. . with solutions y = 1 and z = −2. It is clear that (pn )n⩾0 and (qn )n⩾0 are sequences of nonnegative integers and since √ √ √ bn 1+ 5 bn √ 5 = pn + qn 5 an + bn = pn + qn 5 ⇔ (an + ) + 2 2 2 . for x = 1 and yz = −2. leading to w 3 − 1 = xyz = w 3 + w. ˆ x = 1 − w. for x = −2 and yz = 1. and yz = −1 + w − w 2 − w = −w 2 − 1. Examples for practice ˆ x = −w − 3. or w = −1. leading to w 3 − 1 = xyz = (w + 2)(w 2 − 2w − 1) = w 3 − 5w − 2. z) a permutation of (1. q1 = 1. all solutions are w = −1. then y + z = −1 − w − x = 2. then y + z = −1 − w − x = 1. and both obtained sequences satisfy the same recurrence xn+1 = 4xn + xn−1 . Note that (y − z)2 = (y + z)2 − 4yz = 8 is not a perfect square. Note that (y − z)2 = (y + z)2 − 4yz = 1 + 8 = 32 for y − z = ±3. ˆ x = −w − 2. 1. and yz = −1 + w − w 2 + 2(w + 3) = −w 2 + 3w + 5. or y = −2 and z = 1. Note that (y − z)2 = (y + z)2 − 4yz = 0.18 Chapter 2. y. no solution in integers exists in this case because no integer w satisfies 5w = −1. 2. n = 0. . Find the sequences of integers (an )n⩾0 and (bn )n⩾0 such that √ √ n 1+ 5 (2 + 5) = an + bn 2 for each n ⩾ 0. or w = −1. then y + z = −1 − w − x = −2. leading to w 3 − 1 = xyz = (w − 1)(w 2 + w − 1) = w 3 − 2w + 1. i. exchanging 5 by − 5 in the first term results in √ √ exchanging 5 by − 5 in the second term. b1 = 2. an = . Prove that for each positive integer k the equation x31 + x32 + ⋯ + x3k + x2k+1 = x4k+2 has infinitely many solutions in positive integers with x1 < x2 < ⋯ < xk+1 . (2 + √ n √ √ n √ n √ √ n ( 5 − 1) (2 + 5) + ( 5 + 1) (2 − 5) 5) − (2 − 5) ..e. Arithmetic problems ⎧ bn ⎪ ⎪ ⎪ ⎪an + 2 ⇔ ⎨b ⎪ n ⎪ ⎪ ⎪ ⎩2 a ⇔{ n bn = pn = qn = pn − qn n ∈ N.19 2. b1 = 2. In explicit form √ √ n √ n √ n √ √ n ( 5 − 1) (2 + 5) + ( 5 + 1) (2 − 5) (2 + 5) − (2 − 5) . (⊠) Example 2. First solution Since 13 + 23 + ⋯ + k 3 = then by substitution k 2 (k + 1)2 = (1 + 2 + ⋯ + k)2 4 x1 = x. with initial conditions a0 = a1 = 1 and b0 = 0. xk+2 = (1 + 2 + ⋯ + k)y in original equation.1) with initial conditions a0 = 1. we obtain (13 + 23 + ⋯ + k 3 ) x3 + (1 + 2 + ⋯ + k)2 y 4 = (1 + 2 + ⋯ + k)4 y 4 ⇔ x3 = ay 4 .. = 2qn we have that (an )n⩾0 and (bn )n⩾0 are sequences of integers and can be defined independently by recurrence (2. b0 = 0. (⊠) √ √ bn = 5 5 Second solution √ √ Using Newton’s binomial formula. yielding (2 + bn = √ n √ n √ n √ √ n 5) + (2 − 5) = 2an + bn and (2 + 5) − (2 − 5) = bn 5. xk = kx. bn are integers for all n may be easily proved considering that they are solutions of the recursive equations an = 4an−1 + an−2 and bn = 4bn−1 + bn−2 ..6. . an = . xk+1 = (1 + 2 + ⋯ + k)y 2 .. √ √ 5 5 (⊠) The fact that an . a1 = 1.1. x2 = 2x. hence we get an 2 (⊠) Example 2. The fundamental solution to this Pell equation is (3. 2). There are infinitely many such integers since the relation tn+k = u is equivalent to the Pell’s equation (2n + 2k + 1)2 − 2u2 = 1. n(n + 1) > n + k. Then (x2 + 1) (y 2 + 1) = x2 y 2 + x2 + y 2 + 1 = (xy − 1)2 + (x + y)2 = p2 + q 2 . z) of integers satisfying the system of equations ⎧ z2 ⎪ ⎪(x2 + 1) (y 2 + 1) + ⎨ 10 ⎪ ⎪ ⎩(x + y)(xy − 1) + 14z = 2010 = 1985. ⋯. It is clear that for s big enough we have ns ⩾ 1 and infinite family of solutions. Examples for practice (k − 1)(k + 2)(k 2 + k + 2) . Solution Note that z = 10k for some integer k because z2 = 2010 − (x2 + 1) (y 2 + 1) 10 is an integer. (⊠) Second solution For any positive integer n we have the well-known identity: that is 13 + 23 + ⋯ + n3 + (n + 1)3 + ⋯ + (n + k)3 = ( (n + k)(n + k + 1) ) . xk+2 = (1 + 2 + ⋯ + k)a5 n3 for any positive integer n is a solution to the original equation and obviously. y. xk = ka7 n4 . Since x = a7 n4 . where √ √ 2ns + 2k + 1 + us 2 = (3 + 2 3)s for s big enough such that ns ⩾ 1.20 Chapter 2.. 2 2 (n + k)(n + k + 1) n(n + 1) ) + (n + 1)3 + ⋯ + (n + k)3 = ( ) ( 2 2 2 2 Consider the positive integers n such that the triangular number tn+k = (n+k)(n+k+1) is 2 2 a perfect squares. then x3 = ay 4 . xk+1 = ns (ns + 1). We can take x1 = ns + 1. xk+2 = us . Find all triples (x. . xk = ns + k.. Let p = x + y and q = xy − 1. hence all these integers are given by the sequence (ns ).7. x1 < x2 < ⋯ < xk+1 . xk+1 = (1 + 2 + ⋯ + k)a1 0n6 . x2 = 2a7 n4 . y = a5 n3 for 4 any positive integer n . Hence where a = (1 + 2 + ⋯ + k)2 − 1 = x1 = a7 n4 .. 21 2.1. Arithmetic problems and the system becomes p2 + q 2 + 10k 2 { pq + 140k = 2010 p2 + q 2 ⇔{ = 1985 pq = 2010 − 10k 2 = 1985 − 140k (iii) Since (p −q)2 = 2010 −10k 2 −2(1985 −140k) = −10(k −14)2 then only k = 14 can provide solvability to (iii). And for k = 14, (iii) becomes { Hence, p2 + q 2 pq x+y { xy = 50 ⇔ p = q = 5. = 25 =5 x =4 x =1 ⇔{ or { =4 y =1 y =4 and triples (4, 1, 140), (1, 4, 140) are all integer solutions of the original system in integers. (⊠) Example 2.8. Let n be a positive integer. Find the least positive integer a such that the system x + x2 + ⋯ + xn = a { 12 x1 + x22 + ⋯ + x2n = a. has no integer solutions. Solution First, we notice that if xi ≠ 0, 1 for an integer component xi then x2i > xi and we have a contradiction a = x21 + x22 + ⋯ + x2n > x1 + x2 + ⋯ + xn = a. Hence any component xi is 0 or 1 and the system has integer solutions for a = 1, ..., n: take x1 = ⋯ = xa = 1 and xa+1 = ⋯ = xn = 0. Therefore the least positive integer a such that the system has no integer solutions is n + 1: x1 + x2 + ⋯ + xn ⩽ x21 + x22 + ⋯ + x2n ⩽ n < a = n + 1. (⊠) Example 2.9. Let n be a positive integer relatively prime with 10. Prove that the hundreds digit of n20 is even. Solution If n is prime with 10 it is prime with 5 and 2. Since ϕ(52 ) = 5⋅4 = 20 and ϕ(23 ) = 22 ⋅1 = 4, where ϕ(n) is Euler’s totient function, then n20 = 1 (mod 25) and n20 = (n4 )5 = 15 = 1 (mod 8), or n20 = 1 (mod 200), and uniqueness of this residue modulus 200 is guaranteed by the Chinese Remainder Theorem, hence the last three digits of n20 are a01, where a is even. The conclusion follows. (⊠) Example 2.10. Find all integers n for which 9n + 16 and 16n + 9 are both perfect squares. Solution If 9n + 16 and 16n + 9 are both perfect squares then n ⩾ 0 and the number pn = (9n + 16)(16n + 9) = (12n)2 + (92 + 162 )n + 122 22 Chapter 2. Examples for practice is also a perfect square. Since (12n + 12)2 ⩽ (12n)2 + (92 + 162 )n + 122 < (12n + 15)2 it follows that if n > 0 then we must have pn = (12n + 13)2 or pn = (12n + 14)2 . The former condition gives n = 1 and the latter, n = 52. Therefore, n = 0, n = 1, and n = 52 are the only integers n for which the expressions 9n+16 and 16n+9 simultaneously return perfect squares. (⊠) Example 2.11. Find all n for which there are n consecutive integers whose sum of squares is a prime. Solution If n = 1 any integer square cannot be a prime, so n is at least 2. In cases n = 2 and n = 3 we can find consecutive numbers 1, 2 and 2, 3, 4 for which sum of squares is 5 and 29, so prime numbers. Suppose now n > 3. If a ∈ Z such that A = a2 + (a + 1)2 + ⋯ + (a + n − 1)2 is prime, then A = na2 + 2(1 + ⋯ + (n − 1))a + 12 + ⋯ + (n − 1)2 = n (a2 + (n − 1)a + (n − 1)(2n − 1) ) = p ⋅ B. 6 Case 1: n is prime. Because n > 3, we have 2 ∤ n, 3 ∤ n, so 6 ∤ n, and because A is prime it follows 6 ∣ (n − 1)(2n − 1), B = 1 and A = n. But A is a sum of n > 3 integer consecutive squares, so A > n, contradiction. Case 2: n is not prime. If p prime, p ∣ n, p > 3, we get m(pm − 1)(2pm − 1) ) = pB = p, 6 because A is prime. But A is a sum of n > 3 integer consecutive squares, so A > n > p, contradiction. We get n = 2a3 b, a = 1, b = 1. If a = 2, n = 4m, A = 2(2ma2 + 2m(4m − 1)a + 3) = 2B = 2, because A is prime. But for the same above reason, A > n > 2, contradiction. If b = 2 we have the same approach. If n = 2 ⋅ 3 = 6, we can find integers −1, 0, 1, 2, 3, 4 for which the sum of squares is 31, prime number. Finally the answer is n ∈ {2, 3, 6}. (⊠) n = pm, A = p (ma2 + m(pm − 1)a + Example 2.12. Find all positive integers n for which (n−2)!+(n+2)! is a perfect square. Solution Note the identity (n − 2)! + (n + 2)! = (n − 2)! ⋅ (n2 + n − 1)2 Thus, if (n − 2)! + (n + 2)! is a perfect square, then (n − 2)! must also be a perfect square. For n = 2, 3 we have that (n − 2)! = 1, so n = 2 and n = 3 are solutions. In the case of n ⩾ 4, we note that the largest prime number smaller than or equal to n − 2, which we will denote by p, can only appear once in the prime factorization of (n − 2)!, implying that (n − 2)! is never a perfect square for n ⩾ 4. If it were to appear at least twice in the factorization, then it is necessary to have n − 2 = 2p. But, by Bertrand’s postulate, we know that there is a prime number in between p and 2p, contradicting p’s maximality. Hence, n = 2 and n = 3 are the only solutions. (⊠) 2.1. Arithmetic problems 23 Example 2.13. Find all primes p and q such that both pq −555p and pq +555q are perfect squares. First solution Since pq − 555p = p(q − 555) is a perfect square, p divides q − 555 and q > 555. Therefore there exists an integer a ⩾ 1 such that q − 555 = ap (α) Likewise q divides p + 555, so there exists an integer b ⩾ 1 such that p + 555 = bq (β) From (α) and (β) it follows that p + 555 = b(555 + ap) ⇒ (1 − ab)p = 555(b − 1) = 0 ⇒ 1 − ab = 0 ⇒ a = 1, b = 1. Therefore q − p = 555, so p = 2 (otherwise q − p would be an even number) and q = 557. (⊠) Second solution We have pq − 555p = p(q − 555) and pq + 555q = q(p + 555), hence p∣(q − 555) and q∣(p + 555), so pq∣(q − 555)(p + 555) = pq + 555q − 555p − 5552 . Thus pq∣ (3 ⋅ 5 ⋅ 37(p + 555 − q)) . But note thatp + 555 = qk hence p∣(3 ⋅ 5 ⋅ 37 ⋅ (k − 1)). If p = 3 ⇒ 3∣(q − 555) ⇒ 3∣q ⇒ q = 3 contradiction since q > 555. Analogously for p = 5 and p = 37. Thus we arrive to p∣(k − 1), hence k = ph + 1. But q − 555 = pr and hence p + 555 = q(ph + 1) ⇒ p = pqh + pr ⇒ 1 = qh + r ⇒ h = 0, r = 1 ⇒ k = 1 Hence q = p + 555, if p ⩾ 3 we obtain that p + 555 is even, contradiction, so p = 2 and q = 557 is the solution to our problem. (⊠) Example 2.14. Find all pairs (x, y) for which x! + y! + 3 is a perfect cube. Solution Let x!+y!+3 = z 3 with x, y ⩾ 7. We have that z 3 = 3 mod 7, so this gives a contradiction since the cubic residues modulo 7 are only 0, 1, 6. Thus, we can suppose without loss of generality that x ⩽ 6. Now let us do the case work: A) x = 1 y! + 4 = z 3 and y ⩾ 7 ⇒ z 3 = 4 mod 7; impossible, so y ⩽ 6, but for any value of y we obtain a cube. B) x = 2 y! + 5 = z 3 and y ⩾ 7 ⇒ z 3 = 5 mod 7; impossible, so y ⩽ 6, and we obtain the solution y = 5. C) x = 3 y! + 9 = z 3 and y ⩾ 7 ⇒ z 3 = 2 mod 7; impossible, so y ⩽ 6, and we obtain the solution y = 6. D) x = 4 y! + 27 = z 3 ; we postpone this for now. E) x = 5 y! + 123 = z 3 and y ⩾ 7 ⇒ z 3 = 4 mod 7; impossible, so y ⩽ 6, and we obtain the solution y = 2. 24 Chapter 2. Examples for practice F) x = 6 y! + 723 = z 3 and y ⩾ 7 ⇒ z 3 = 2 mod 7; impossible, so y ⩽ 6, and we obtain the solution y = 3. Now, let’s turn back to D). First, note that the case y = 1, 2, 3, ..., 8 can be easily checked manually. For y > 8 we have 81∣y!; so z = 3z1 and y!/27 = z13 − 1. Now, 3∣z13 − 1, soz = 1 mod 3. But if z = 3k + 1; then y! = (3k + 1)3 − 1 = 27k 3 + 27k 2 + 9k = 9k(3k 2 + 3k + 1). 27 Thus y! = 243k(3k 2 + 3k + 1). Furthermore, let f (n) to be equal to the power of 3 dividing n and note that f (n) = log3 n, so f (y!) = [ 243k(3k 2 + 3k + 1) 243k(3k 2 + 3k + 1) 243k(3k 2 + 3k + 1) ]+[ ]+[ ] + ⋯, 3 9 27 where [x] denotes as usual the integer part of x. But f (y!) = f (243k(3k 2 + 3k + 1)) = 5 + f (k) = 5 + log3 k, so we can see that for k ⩾ 1, [ 243k(3k 2 + 3k + 1) ] = 81k(3k 2 + k + 1) > 5 + log3 k 3 and hence the equation has no solution. Finally, we get that the solutions to original equation are (x, y) = (2, 5); (x, y) = (5, 2); (x, y) = (3, 6); (x, y) = (6, 3). (⊠) Example 2.15. Prove that there are infinitely many pairs (p, q) of primes such that p6 +q 4 has two positive divisors whose difference is 4pq. Solution Let p = 2 and let q be odd prime number. Then and we have p6 + q 4 = 64 + q 4 = (8 + 4q + q 2 ) (8 − 4q + q 2 ) 8 + 4q + q 2 − 8 − 4q + q 2 = 8q = 4pq. (⊠) Example 2.16. Prove that there is no n for which ∏(k 4 + k 2 + 1) is a perfect square. n k=1 First solution Let g(k) = k 4 + k 2 + 1. If we evaluate g(1), g(2), g(3), etc., we notice an interesting pattern: g(1) = 1 ⋅ 3, g(2) = 3 ⋅ 7, g(3) = 7 ⋅ 13, g(4) = 13 ⋅ 21, and so on. This is because g(k)factors as g(k) = (k 2 − k + 1)(k 2 + k + 1), and g(k + 1) = ((k + 1)2 − (k + 1) + 1)((k + 1)2 + (k + 1) + 1) = (k 2 + k + 1)(k 2 + 3k + 3). So multiplying g(1), g(2), ..., g(n), we see that there are a sequence of squares in the product (32 , 72 , 132 , ...), but with a final factor of n2 + n + 1 left over at the end. It follows that since n2 + n + 1 cannot be a square (since it lies between n2 and (n + 1)2 ), the product there is no n for which ∏(k 4 + k 2 + 1) is a perfect n k=1 square. Thus. 1 Because k=1 k 4 + k 2 + 1 = (k 2 + k + 1)(k 2 − k + 1) and (k + 1)2 − (k + 1) + 1 = k 2 + k + 1. the numbers (2n + 6)2 + (n + 5)2 + (n + 4)2 + (n + 2)2 + (n + 2)2 = 8n2 + 50n + 85. m must be odd with m2 ∣ 2009. n n−1 k=1 k=1 Moreover. because the difference between (n + 1)2 and n2 is 2n + 1. it follows that m = 7 √ and n = 16. (⊠) √√ √ n + n + 2009 is an integer. Because 2009 = 72 ⋅ 41. (⊠) Second solution First. Clearly. for each non-negative integer n. (2n + 5)2 + (n + 6)2 + (n + 4)2 + (n + 3)2 + (n + 2)2 = 8n2 + 50n + 90.25 2. Solution Consider. n2 + n + 1 cannot be a perfect square for any n > 1. Thus n = 16 is the only integer n for which (⊠) . (2n + 7)2 + (n + 5)2 + (n + 3)2 + (n + 2)2 + (n + 1)2 = 8n2 + 50n + 88. note that ∑ (k 4 + k 2 + 1) = 3. which is not a perfect square. Find all positive integers n for which First solution √√ √ n + n + 2009 = m for some integer m.17. (2n + 8)2 + (n + 4)2 + (n + 3)2 + (n + 1)2 + (n + 1)2 = 8n2 + 50n + 91. Prove that there are infinitely many penta-sequences of length 7.sequences of length 7. Now. (2n + 6)2 + (n + 5)2 + (n + 4)2 + (n + 3)2 + (n + 1)2 = 8n2 + 50n + 87. (⊠) Example 2. The conclusion follows. (2n + 7)2 + (n + 4)2 + (n + 4)2 + (n + 2)2 + (n + 1)2 = 8n2 + 50n + 86. √ √ n + n + 2009 is an integer. suppose n > 1. Call penta -sequence a sequence of consecutive positive integers such that each of them can be written as a sum of five nonzero perfect squares. for each non-negative integer n. Then Suppose √ m4 − 2009 n= 2m2 For n to be an integer. these numbers form different penta.18. Arithmetic problems cannot be a square. it follows that ∏(k 4 + k 2 + 1) = (n2 + n + 1) ∏(k 2 + k + 1)2 . (2n + 8)2 + (n + 4)2 + (n + 2)2 + (n + 2)2 + (n + 1)2 = 8n2 + 50n + 89. Example 2.1. Therefore. . a3 + 2a. since n is integer. We have b2 − a2 = 2009 = 72 ⋅ 41 giving the systems b − a = 41 { b + a = 49 or b − a = 2009 { b + a = 1.19. n = 16 is the only solution.. n + 2009. hence 1 (a + 1)3 = a2 + 3a + 3 + ⩽ a2 + 3a + 4. a3 + a. k 2 is a divisor of 2009.. Examples for practice Second solution √ √ Suppose there is a positive integer k such that n + n + 2009 = k 2 . Solution√ √ √ Let a = [ 3 n] . k ⩽ 23 + . So let n = a2 . We get a3 = ab < (a + 1)3 . The desired number is given by 3 ∑ (3a + 4) + 24 = 3 ⋅ 11 a=1 11 ⋅ 12 + 44 + 24 = 198 + 44 + 24 = 266.20. we seek to find positive 2 . a ∈ Z∗ . First system gives the solution n = 16. hence k = 23.e. (⊠) Third √√solution √ √ √ √ √ For n + n + 2009 to be an integer n.e. Note that the largest a with a3 < 2012 is a = 12 and we have 123 + 12k ⩽ 2012 is equivalent 2 to 12k ⩽ 284. contradiction. Thus. b positive and a + b a perfect square. a3 ⩽ n < (a + 1)3 .26 Chapter 2. Then a ⩽ 3 n < a + 1. It is clear that k = 1. a a √ All positive integers n with the property that [ 3 n] divides n are given by a2 ⩽ b < a3 . Find all n for which the number of diagonals of a convex n-gon is a perfect square. the only square numbers that divide 2009 are 12 and 72 . i. (⊠) √ Example 2. Second system implies a = −1004. Because [ 3 n] divides n. a3 + (3a + 3)a. 2 (⊠) Example 2. √ √ 2009 Taking the conjugate we obtain n + 2009 − n = 2 ⋅ k √ 2009 2 Subtracting the above equations we get 2 n = k − 2 ⋅ k √ From this equation we infer that n must be rational and hence integer. n+ n + 2009 must be perfect squares. Solution Since the number of diagonals of a convex n-gon is integer solutions to the Diophantine equation n(n − 3) = m2 2 n(n − 3) . If k = 7 we get n = 16 and it is easily verifed that √ √ √ 16 + 16 + 2009 = 7 is an integer. it follows that n = ab for some positive integer b.. n + 2009 = b2 with a. However. How many positive integers n less than 2012 are divisible by [ 3 n]. i. 27 2. (x2 )2y+1 = x4y+2 = xp−1 = (−1)2y+1 = −1 mod p which is impossible since (noting that p cannot divide x) xp−1 = 1 (by Fermat’s ”little” theorem)... 5046. Assuming to the contrary that p = 4y + 3. (◻) Considering now the original problem. then for each factor fi .. 2 that is 3. by letting n = 3N and m = 3M . p is of the form 4y + 1 or 4y + 3. Since 4x2 + 1 is odd. Proof. Proof: Since p is odd. let n = 2m. it can only have odd divisors. each of which must be congruent to 1 mod 4 by lemma 1..21. It follows that every factor is congruent to 1 mod 4..1. then the condition that N has n2 divisors greater than 1 is equivalent to τ (N) = (e1 + 1)(e2 + 1).(ek + 1) = 4m2 + 1 where τ is the number of divisors function and the prime factorization of N is N = pe11 pe22 . Since the non negative solutions of the Pell’s equation X 2 − 2Y 2 = 1 are √ k √ Xk + Yk 2 = (3 + 2 2) for k ⩾ 0.. First solution The following results will be of use. (◻) Lemma 2. . we have that (note that Xk is odd) nk = 3Nk = 3(Xk + 1) for k ⩾ 0. Note that n2 + m2 = 0 (mod 3) which implies that both n and m are divisible by 3. showing that p = 3 mod 4 is impossible will establish the claim. Each factor is either 1 or some product of (odd) prime factors of 4x2 + 1. Arithmetic problems that is n2 − 3n − 2m2 = 0 for n ⩾ 3 and m ⩾ 0. 27. 150. (⊠) Example 2.fk . then p ≡ 1 mod 4. 6. 867..pekk ... Prove that N is the fourth power of an integer. fi = 1 mod 4. For an even integer n consider a positive integer N having exactly n2 divisors greater than 1. If an odd prime p divides x2 + 1 for some integer x. If 4x2 + 1 = f1 f2 . the equation becomes N 2 − N − 2M 2 = 0 that is (2N − 1)2 − 2(2M)2 = 1. Lemma 1. So. we have x2 = −1 mod p. (⊠) Second solution Let N be factored (into primes) as N = pe11 ⋯pekk . we rewrite the first term as 2(2k+1)−1 − 1 = 4k − 1. ̂ denotes the (minor) arc between points X and Y. thus ei + 1 ≡ 1 ( mod 4) so ei is divisible by 4 for all i. and DA intersect the inner circle at point A1 . Thus 1 ≡ −1 ( mod p). WLOG let p be prime. It follows that N is a fourth power. then it would contradict the prime property). np−1 ≡ n4k+2 ≡ (n2 ) 2k+1 ≡ (−1)2k + 1 ≡ −1 ( mod p). CD. Examples for practice By lemma 2 each of the factors ei + 1 is of the form 4yi + 1. so ei + 1 must be either 1 mod 4 or 3 mod 4 for 1 ⩽ i ⩽ k. It is well known that N has (e1 + 1) × ⋯ × (ek + 1) = n2 + 1 divisors (including 1). respectively. . Solution Let the tangency points of AB. (⊠) Example 2. First. but since p ⩾ 3. However. Solve in prime numbers the equation: xy + y x = z. we have the desired contradiction.22. suppose that x = 2. and let P be the intersection point of A1 C1 and B1 D1 . First solution We first note that neither x and y equal nor are they both odd. n2 + 1 ≡ 1 ( mod 4). and where XY A1 C1 ⊥ B1 D1 as desired. thus by Fermat’s Little Theorem. (y + 1)(y + 2) is divisible by three since all primes greater than 3 are either one or two less than a multiple of three (if some prime was three less. otherwise it must have some prime factor also of the form 3 mod 4 (since p is 3 mod 4. Without loss of generality. Then. for otherwise z wouldn’t be prime. Thus ∠A1 P B1 = 90o.28 Chapter 2. its prime factors cannot all be 1 mod 4). Proof: Assume for a contradiction that there exists some number p such that p = 4k+3 and n2 + 1 is divisible by p. BC. implying that N is a fourth power. Second. Since n is even. Prove that A1 C1 ⊥ B1 D1 . np−1 ≡ 1 ( mod p).23. and D1 . z = 2y − 2 + y 2 + 2 = 2(2y−1 − 1) + (y + 1)(y + 2) − 3y. 180o = ∠DAB + ∠BCD ̂ ̂ ̂ ̂ ̂ ̂ ̂ ̂ C A 1 D1 + D1 A1 + A1 B1 − B1 C1 1 B1 + B1 C1 + C1 D1 − D1 A1 + = 2 2 ̂ ̂ =A 1 B1 + C1 D1 = 2∠A1 P B1 . C1 . since ABCD is cyclic. Note that n must then be relatively prime to p. B1 . therefore each ei is a multiple of 4. (⊠) Example 2. Let ABCD be a quadrilateral inscribed in a circle and circumscribed about a circle such that the points of tangency form a quadrilateral A1 B1 C1 D1 . Thus n2 ≡ −1 ( mod p). We will show that n2 + 1 has no factors of the form 3 mod 4. or. The equation then reduces to 2y + y 2 = z. since y is an odd prime of the form 2k + 1 (for some speciff c integer k). z = 17 or x = 3. Thus. We see that 23 + 32 = 17 is a solution. If p > 3 is a prime then 2p + p2 = 0 ( mod 3). y even and the other odd (since otherwise z would be even and greater than 2). 3) = 1 ∶ we can apply Fermat’s Theorem. y can’t be both odd primes. it follows that 26n±1 + (6n ± 1)2 is divisible by 3. or (3. Proof. 2 ≠ 0 ( mod 3) ⇒ 2k ≠ 0 ( mod 3) ⇒ (2k − 1) (2k + 1) ≡ 0 ( mod 3) Thus 22k ≡ 1 ( mod 3) ⇒ 22k + 1 = 2 ( mod 3). we can observe that x.1. y = 2. 17). Arithmetic problems Since 4 ≡ 1 mod 3. . say x. It follows from Lemma 1 and Lemma 2. 17) by symmetry between x and y. so p3−1 ≡ p2 ≡ 1 ( mod 3).29 2. or z would be even. We’ve shown that z is divisible by three for y > 3. Lemma 3. It follows that the only solution in prime numbers is x = 2. If p = 3 is a prime then p2 = 1 ( mod 3). and since for p odd. (⊠) Second solution We must have one of x. 2. Proof. We now have the equation 2y + y 2 = z. Lemma 2. y = 3. it is easy to show that 4k ≡ 1 mod 3. The following lemmas hold: Lemma 1 . 3. one of them must be equal to 2. (◻) (◻) (◻) It follows from Lemma 3 that y must be equal to 3: the only one solution is (2. Proof. and 2p = (−1)p = −1 mod 3 (6n ± 1)2 = 36n2 ± 12n + 1 = 1 mod 3. so. If a is an odd number then 2a ≡ 2 ( mod 3). z = 17.24. are there any others? No: all primes greater than 3 are of the form 6n ± 1. (⊠) Third solution First of all. (⊠) Example 2. Let a and b be integers such that ∣b − a∣ is an odd prime. Prove that P (x) = (x − a)(x − b) − p is irreducible in Z[x] for any prime p. If p is such a prime then (p. 4k −1 is divisible by three. Therefore a solution must be of the form 2 p + p2 = z where p is an odd prime and z is prime. Then. since ∣(r − a) − (r − b)∣ = ∣a − b∣ is odd. or (r − a)(r − b) = p. However.. positive integers a. The discriminant of this quadratic is (a + b)2 − 4ab + 4p = (a − b)2 + 4p = q 2 + 4p where q = ∣b − a∣. 2.. Thus. take r = 2 2 if n ≡ r (mod q). Since xy = p > 0. y are both negative. and if b is odd. p. where [Q] is equal to 1 or 0 as the proposition Q is true or false. n ≡ r (mod q) is always false for n = 1. so letting y = q + 2t for some positive integer t we now have 4p = 2t(2q + 2t) ⇒ p = t(q + t). chose r = p. In either case.30 Chapter 2. or r = . xy ⩾ 1 ⋅ (1 + 3) > 2 = p. Then. Since q is an odd prime we cannot have t = 1 (since then p = q + 1 which is imposible for p. Prove that for any odd integer d > 0 there is an integer r such that the numerator of the rational number [n ≡ r (mod q)] nd n=1 p ∑ is divisible by p. Without loss of generality let them both be positive. there must be some integers r and s such that P (x) = (x − r)(x − s). If b is even. p − 1. q prime with q odd). and the rational number has numerator 0 divisible by p.25. y must have the same parity as q. their product. we have that b − 2r ≡ 0 (mod q). Let x = r − a and y = r − b. the desired contradiction. so p = 2. y are both positive or x. which is also impossible since then the prime p would factor. Then since ∣x − y∣ = ∣a − b∣. If p > q. It follows that the discriminant cannot be a square. therefore t > 1. since P (x)is a quadratic. either x. so the problem is equivalent to showing that q 2 + 4p cannot be a square. Examples for practice First solution We want to show that (x − a)(x − b) − p = x2 − (a + b)x + ab − p = 0 has no solutions in integers when ∣b − a∣ is an odd prime and p is prime. b exist such that p = aq + b with b < q. must be even.. Let p and q be odd primes. take q+b b . . (⊠) Example 2. Suppose to the contrary that for some integer y q 2 + 4p = y 2 ⇒ 4p = y 2 − q 2 = (y − q)(y + q) In order for (y − q)(y + q) to be divisible by 4. Solution If p < q. and without loss of generality let x < y. (⊠) Second solution Assume for a contradiction that P (x) = (x − a)(x − b) − p is indeed reducible over Z[x] for some prime p. therefore the polynomial is irreducible over the integers. then p − n ≡ b − r ≡ r (mod q). which is an odd prime. . P (r) = 0. from where it follows that (3a + 2x + y)(x2 + xy + y 2 ) = (3a + 3) ⋅ 3 > 9. We have that (3a + 2x + y)(x2 + xy + y 2) > 3a(x2 + xy + y 2) ⩾ 3. (⊠) . Let us prove also that any nice integer divisible by 3 must be divisible by 9. (⊠) Example 2. 1 nd + (pn)d 1 + = ⋅ nd (p − n)d nd (p − n)d nd + (p − n)d = p (nd−1 − nd−2 (p − n) + nd−3 (p − n)2 − . Since (3a + 2x + y)(x2 + xy + y 2) > 0 we have that any nonzero nice integer is nonnegative. Therefore. .. and the integers greater than 9 of the form 9a. from where it follows that x ≡ y (mod 3). p − 1} such that n = r (mod q) may be grouped up in distinct pairs of the form (n. a common factor p will always appear in the numerator.. Find all integers that can be represented as a3 + b3 + c3 − 3abc for some positive integers a. since n and p − n are both smaller than prime p... which implies the claim. for some nonnegative integers x. From these we conclude that all the nice integers are 0. Taking x = 0. their contribution to the total sum is But for odd d. Let us proceed to find which integers are nice. Assume without loss of generality that b = a + x and c = a + x + y. Adding any number of such fractions. all numbers n ∈ {1. Solution Let us say an integer is nice if it can be represented as a3 + b3 + c3 − 3abc for some positive integers a.26. Arithmetic problems Since n and p − n cannot be equal because p is odd. For each one of these pairs.1. and the numerator of this fraction is divisible by p. Suppose that x. Therefore. y = 0 it follows that any positive integer of the form 3a + 2 is nice. a3 + b3 + c3 − 3abc = (3a + 2x + y)(x2 + xy + y 2 ). Let us prove first that 1 and 2 are not nice. but not its denominator. y = 1 it follows that any positive integer of the form 3a + 1 is nice. 3a + 2x + y ≡ x2 + xy + y 2 ≡ 0 (mod 3). We have that 0 ≡ (3a + 2x + y)(x2 + xy + y 2 ) ≡ (y − x) ⋅ (x − y)2 ≡ (y − x)3 (mod 3). For x = y = 0 it follows that 0 is nice. From the previous result we have that x ≡ y (mod 3).. p − n). Let us prove that 9 is not nice. from where it follows the claim. Taking x = y = 1 it follows that any positive integer of the form 9(a + 1) is nice. y. and we are done. + (p − n)d−1 ) . but never in the denominator. c. any positive integer greater than 3 of the form 3a + 1 or 3a + 2. Taking x = 1. b. y are not both zero.31 2. b. and c. 2. 21. 6.32 Chapter 2. 15. these are the integers divisible by 3 and 5. so the product of primes that must divide n grows faster than n. and 24. and if both divide n.. 12. that is there is always a prime between m and 2m. 9. 45. Then q12 = q22 + q32 + q42 + q52 + 9. Suppose there are 0 ⩽ a ⩽ 5 primes between q2 . Then 1 = 1 + 1 ⋅ b + 4(4 − b) (mod 8) or 3b = 0 (mod 8). After n = 49. Therefore. their quadratic residues modulo 8 are 1 if qi is odd or 4 if qi = 2. (⊠) Solution Every square is 0 or 1 modulo 3 and clearly q1 ≠ 3. In summary. 21.. q5 not equal to 2. 2m + 3 = 7 > n ⩾ 5 = 2m + 1. 24. 18. 9. 5. 8. they are coprime. but n must be divisible by 3. 15. 5 and 7. or 4m + 12m + 9 > n ⩾ 4m2 + 4m + 1. q6 not equal to 3. which is absurd. 45. 2. 2. Assume next that m = 1. and necessarily m ⩽ 3. 6. Therefore. and n < 3. 3. Therefore.. and 25 ⩽ n < 49. which is larger than 4m2 − 1. The integers that we are looking for are then 1 through 9. . Recall Bertrand’s postulate. or n ⩾ 8. q6 such that q12 = q22 + ⋯ + q62 . 21. ˆ Suppose a = 4. n ⩾ 2(4m2 − 1). and 4m2 + 12m + 9 > 8m2 − 2. which are 30 and 45 in this range. 4. 12. or 4m2 − 12m − 11 < 0. 18. 12. 5. then 4m2 − 12m − 11 =√ (m2 − 11) + 3(m − 4)m > 0. and 7. Since qi are primes. Assume now m = 2. 24. Since 2m + 1 and 2m − 1 are positive odd integers with difference 2.. 7. 2m + 3 > n ⩾ 2m + 1.. 5. Then. that is multiples of 105. 18. 12. Then 2m+3 = 9 > n. 15. up to n = 92 = 81. 4. . the result is trivially true. . Now. 2 n] . 8. (⊠) Second solution For n ⩽ 25 such integers can be computed as 1. if m ⩾ 4. 24. these are the integers divisible by 3. that is multiples of 15. 18. 9 ⩽ n < 25 and n must be divisible by 3. 36 and it’s easy to see that there are no solutions. 21. This prime is greater than 4 for n ⩾ 9.27. and n < 81. Then. Using induction we can √ see that √ every time we jump from n to 4n we get at least one more prime in the range [ n. 30 and 45. Then. . or n = 30. which are coprime. 7. 15. n must be divisible by 15. 3. In order to get some intuition. Assume 2m + 1 is the largest odd integer √ now that m ⩾ 1 is√an integer such that 2 not exceeding n. q2 . Example 2. which are coprime. the only √ positive integers n such that a divides n for all odd positive integers a not exceeding n are 1. which has the only solution b = 0. where m is any integer with m > 1. consider what happens for n ⩾ 25: up to n = 72 = 49.28. Find all √ positive integers n such that a divides n for all odd positive integers a not exceeding n. Clearly q1 = 2 and suppose that there are 0 ⩽ b ⩽ 4 primes between q2 . from which a = 1 or a = 4. ˆ Suppose a = 1. but n must be divisible by 3 and 5.. Then 1 = 1 ⋅ a + 0(5 − a) (mod 3). then q1 + q2 can be only 9. 18. then their product 4m2 − 1 must also divide n. First solution √ √ If 1 is the largest odd integer not exceeding n. n must√be divisible by 105. and m ⩾ 2. Examples for practice Example 2. 12.. a contradiction.. 30. or n = 9. Assume that m = 3. Then q12 = q22 + 4 ⋅ 32or(q1 − q2 )(q1 + q2 ) = 36.. Find all primes q1 . Since q1 + q2 > q1 − q2 . Let n be an integer. (30n3 + 5)3 − (30n3 − 5)3 − (30n2 )3 + 1 = 251 (⊠) Second solution For every n ∈ Z.30. First solution First we will prove that for any positive real numbers x.29.3) .3) to positive real numbers a and b with a + b = 2 we obtain a + b m−n ) = an + bn . (⊠) Example 2. 2. it suffices to show that x3 + y 3 + z 3 + w 3 = 251 has infinitely many solutions over the integers. z = 2. Find all integers m such that am + bm ⩾ an + bn for all positive real numbers a and b with a + b = 2.33 2.2) with PM-AM inequality x + y m−n xm−n + y m−n ⩾( ) 2 2 gives inequality xm + y m ⩾ (xn + y n ) ( x + y m−n ) 2 Applying (2. Arithmetic problems Hence the solutions to the problem are (5. Prove that the equation x3 + y 3 + z 3 + w 3 = 2008 has infinitely many solutions in integers. we have that no two of these solutions can be identical: this terminates our proof. 2. (⊠) Example 2. Combination (2. 2. Let us note that and thus we are done. y and any natural m ⩾ n following inequality holds We have (xn + y n )(xm−n + y m−n ) ⩾ xm + y m ⋅ 2 (2. the 4-tuple (x = 10 + 60n3 .2) 2(xm +y m )−(xn +y n )(xm−n +y m−n ) = xm +y m −xn y m−n −xm−n y n = (xn −y n )(xm−n −y m−n ) ⩾ 0. 2. y = 10 − 60n3 . 3) and its permutations with 5 fixed. First solution A Since 2008 = 8 ⋅ 251 = 23 ⋅ 251.1. Since n3 = m3 implies n = m. w = −60n2 ) provides us with a solution to the given equation. a + b ⩾ (a + b ) ( 2 m m n n (2. The last result implies that the given relation is only satisfied for all integers m ⩾ n. Since f (x) = f (2 − x) it suffices consider 1 ⩽ x < 2 fixed.34 Chapter 2. Assume that the statemnt is true for k = l and let us prove. Examples for practice Suppose that there is natural m < n such that for any positive real numbers a and b with a + b = 2 holds inequality am + bm ⩾ an + bn ⇔ am + (2 − a)m = an + (2 − a)n . or an+l+1 + bn+l+1 ⩾ an+1 b + bn+1 a. The case k = 0 is readily checked. To note that g ′′ (x) = m(m − 1)xm−2 and m(m − 1) ⩾ 0. We need to prove that h(x) ⩾ 0. To prove this. but in this case we consider 0 < x ⩽ 1. let us prove that for any nonnegative positive integer k and for any positive integer n. We have g ′ (n) = xn ln (x) + (2 − x)n ln (2 − x). (⊠) Second solution First. To note that t′′ (x) = xn−2 [n(n − 1) ln (x) + 2n − 1] > 0. we need to find all integers m such that am + bm ⩾ 2. This completes the induction and proves that the statemnt is true for all nonnegative integers. Let f (x) = m x + (2 − x)m with 0 < x < 2. If n ⩽ −1 then m ⩽ n. Then we obtain inequality 2m = lim(am + (2 − a)m ) = lim(an + (2 − a)n ) = 2n a→2 a→2 which contradict to inequality 2m < 2n ⇐ m < n. The last inequality is equivalent to (a − b)2 (an+l−1 + an+l−2 b + ⋯ + bn+l−2 a + bn+l−1 ) ⩾ 0 and since (a − b)2 ⩾ 0 and an+l−1 + an+l−2 b + ⋯ + bn+l−2 a + bn+l−1 > 0 we have that inequality holds if and only if a = b or k = 0. Thus. so all integers m satisfy. The idea is the same. Thus an+l+1 + bn+l+1 ⩾ an+l + bn+l ⩾ an + bn or an+l+1 + bn+l+1 ⩾ an + bn . we will prove that g(n) is non decreasing. that it is true too for k = l + 1. If we have that a + b = 2 the inequalities is equal to 2an+l+1 + 2bn+l+1 ⩾ (an+l + bn+l )(a + b). we use mathematical induction on k. that is to say h(x) = xn ln (x) + (2 − x)n ln (2 − x). We have to prove that an+l+1 + bn+l+1 ⩾ an+l + bn+l . answer is {m∣m ∈ N and m ⩾ n}. Let f (x) = xn + (2 − x)n . Since g(x) = xm is convex in this interval we have that f (x) = g(x) + g(2 − x) ⩾ 2g(1) = 2 by Jensen’s inequality. (⊠) Third solution Let n = 0 or n = 1. If n ⩾ 2 then m ⩾ n. b ⩾ 0. (⊠) . an+k + bn+k ⩾ an + bn . Since t(x) = xn ln (x) is convex if 1 ⩽ x < 2 we have that h(x) = t(x) + t(2 − x) ⩾ 2t(1) = 0 by Jensen’s inequality and we are done. Now suppose n ⩾ 2 fixed. Now let g(n) = xn + (2 − x)n . if a + b = 2 and a ⩾ 0. Therefore we want to show that (x−1)⋅ x⋅ (x+1)⋅ (x2 +1) is always divisible by 30 = 2 ⋅ 3 ⋅ 5. Proof. 5 ⋅ n + 2. It follows that x5 − x is always divisible by 30 and hence the conclusion. and in some cases a factor of 5 as well. Because 6∣(x−1)x(x+1) so 6∣ (x5 − x) . ˆ If x ≡ 1 (mod 5) then 5∣(x − 1). 3. First note that x5 − x = (x − 1) ⋅ x ⋅ (x + 1) ⋅ (x2 + 1). But gcd(5. ˆ If x ≡ 0 (mod 5) then 5∣x. x = 5 ⋅ n + 2. The only triples x − 1. (⊠) Third solution We will prove a lemma first: Lemma. We have x5−x = (x−1)x(x+1) (x2 + 1) . 5 ⋅ n + 3. . Find all primes p and q such that 24 does not divide q + 1 and p2 q + 1 is a perfect square. a5 + b5 + c5 + d5 ≡ a + b + c + d ≡ 0 (mod k) where k = 2. Coming back to the problem. a5 + b5 + c5 + d5 ⋮ 30. so x2 + 1 = 25 ⋅ n2 + 20 ⋅ n + 5. 5. which also gives a factor of 5. (⊠) Second solution We will prove that x5 ≡ x (mod 30). then 30∣ (x5 − x). b. 5∣ (x5 − x) . 6) = 1. x. 5 ⋅ n + 4. Let x be an integer. (⊠) Example 2. x2 ≡ x (mod 2) then x5 ≡ x3 ≡ x (mod 3) and x5 ≡ x (mod 2) for any integer x. ˆ If x ≡ 4 (mod 5) then 5∣(x + 1). In the second case.31. it follows that 30∣ (x5 − x) . So a5 + b5 + c5 + d5 is divisible by 30. x3 − x = (x − 1)x(x + 1) divisible by 3 and x5 − x = x (x2 − 1) (x2 + 1) = x (x2 − 1) (x2 − 4 + 5) = (x − 2)(x − 1)x(x + 1)(x + 2) + 5x (x2 − 1) divisible by 5. so x2 + 1 = 25 ⋅ n2 + 30 ⋅ n + 10.1.35 2. x. In the following assume x − 1 > 0. Thus. we have a5 + b5 + c5 + d5 = a5 + b5 + c5 + d5 − (a + b + c + d) = a5 − a + b5 − b + c5 − c + d5 − d ⋮ 30. So for all integer x. x = 5 ⋅ n + 3. ˆ If x ≡ 2 (mod 5) then 5∣ (x2 + 1) . which gives a factor of 5. For any natural m > 1 product of m consequtive integers allways divisible by m. Then there will always be factors of 2 and 3 among x − 1. Therefore. 5 ⋅ n + 3. First Solution Because x5 ≡ x (mod 5). ˆ If x ≡ 3 (mod 5) then 5∣ (x2 + 1) . x3 ≡ x (mod 3). Prove that a5 + b5 + c5 + d5 is divisible by 30. Arithmetic problems Example 2.32. and x + 1. or 5 ⋅ n + 2. from the Lemma. d be integers such that a + b + c + d = 0. x + 1 for which there is not a factor of 5 are those of the form 5 ⋅ n + 1. Let a. c. In the first case. then 2 x − x = (x − 1)x divisible by 2. We have p2 q + 1 = x2 ⇔ p2 q = (x − 1)(x + 1). 2 2 Let k = 4. (1) ∶ x − 1 = q. then p2 ≡ 1 (mod 3) and p2 ≡ 1 (mod 8) which contradicts with the condition that 24 doesn’t divide q + 1. Find the least odd positive integer n such that for each prime p. all the solutions are: (p. 1. 3 it is easy to see that when p = 2 there are less than four prime divisors. (1) ⇒ q + 1 = p2 − 1 ◇ If p = 3.33.2 (⊠) Senior problems Example 2. If p is odd we have (p2 + 2p + 2. then M is divisible by 2. we have solution q = 11. x + 1 = p2 or (2) ∶ x − 1 = p2 . the n2 − 1 + np4 + p8 is divisible by at least four primes.36 Chapter 2. p4 + 5 − p4 − 8p2 − 4 − 4p3 − 4p) . p4 + 5) = (p2 + 2p + 2. 3. (3. 7. (2) ⇒ p2 + 2 = q ◇ For p = 3. Since x2 = p2 q + 1. and 2 is the only even prime. 5. ◇ If p ≠ 3. x + 1) = 1 the other cases are not possible. q > 2 in that case x > 2. (3. If p = 2. we have solution q = 7. x2 = 1 (mod 8) ⇒ p2 q + 1 = 1 (mod 8) ⇒ p2 q = 0 (mod 8) Because both p and q are primes. number M = 4 First solution Let n = 2k + 1 with k nonnegative integer. q) ∈ {(2. 7). 2). x − 1 and x + 1 will be both odd. and they are relatively prime because gcd(x − 1. 11)}. 2) = 1 So. Since x is even. 2. ◇ If p ≠ 3. x is odd. (p2 + 2p + 2. Since x > 2 and the gcd(x − 1. For k = 0. x + 1 = q. ie. then M = (p4 + 4)(p4 + 5) = (p2 + 2p + 2)(p2 − 2p + 2)(p4 + 5). M = p8 + np4 + = (p4 + n2 − 1 n 2 1 = (p4 + ) − 4 2 4 n−1 n+1 ) (p4 + ) = (p4 + k)(p4 + k + 1). x + 1) = gcd(x − 1. So. then p2 + 2 = 0 (mod 3) which cannot be prime. p2 − 2p + 2) = (p2 + 2p + 2. 2. Examples for practice Solution ˆ If one of p or q is even. p = q = 2 ˆ If both p. 4p) = 1.1. 3 the result does not hold for p = 2. d = 1. it is enough to prove that two primes divide (p2 + 2p + 2)(p2 − 2p + 2) and another two divide p4 + 5. Arithmetic problems = (p2 + 2p + 2. To prove that another prime divides p4 + 5 it is enough to prove that 4 ∤ p4 + 5. Then. p4 + 5) = (p2 − 2p + 2.37 2. then 21 is the greatest power of 2 dividing p4 + 5.4) . Let (p2 + 2p + 2. then n2 − 1 + np4 + p8 = k(k + 1) + (2k + 1)p4 + p8 = (p4 + k)(p4 + k + 1). which is a contradiction. p2 − 2p + 2 and p4 + 5 are pairwise coprime. and so n = 9 is the least desired number. Since both p2 + 2p + 2 and p2 − 2p + 2 are odd. For p = 2 the result holds. Since (p2 + 2p + 2)(p2 − 2p + 2) = (p4 + 5) − 1 we have that (p2 + 2p + 2. p4 + 5) = (p2 − 2p + 2. p2 − 2p + 2) = 1. Assume that p is an odd prime. First solution √ (a) Let 2α = 1 + 5. For k = 4 we have (p4 + 4)(p4 + 5) = (p2 + 2p + 2)(p2 − 2p + 2)(p4 + 5). This results follows from the fact that 4∣p4 +3. This implies that k = 4 is the least integer value. 1. p2 − 2p + 2) = d. 1) = 1. As p4 + 5 = 2 (mod 4) for all odd p. 4p3 − 8p2 + 4p + 1) = (p2 − 2p + 2. 4p3 + 8p2 + 4p + 1) = (p2 + 2p + 2. Note that d is odd and that d∣4p. This implies that d∣p. 1) = 1. We prove that k = 4 is the least integer that satisfies the condition. (a) integer an such that Prove that for each positive integer n there is a unique positive (1 + √ n √ √ 5) = an + an + 4n . 2. from where we conclude that n = 9 is the least odd positive integer that satisfies the condition. With this we have 2n α n = √ √ an + an + 4n . prove that an is divisible by 5 ⋅ 4n−1 and find the quotient. Thus p2 + 2p + 2. p4 + 5) = 1. as we wanted to prove. (⊠) Second solution Let n = 2k + 1. there is another prime different from 2 and from all the divisors of p2 + 2p + 2 and p2 − 2p + 2 which divides p4 + 5. (b) When n is even. 4 Note that for k = 0. (2.1. In order to prove that two primes divide (p2 +2p+2)(p2 − 2p + 2) it is enough to prove that (p2 + 2p + 2. If d = p then p∣ (p2 + 2p + 2) .34. This implies that any prime that divides (p2 + 2p + 2)(p2 − 2p + 2) does not divide p4 + 5 and viceversa. and (p2 − 2p + 2. 4p3 + 8p2 + 4p + 1 − 4p3 − 8p2 − 4p) = (p2 + 2p + 2. Note that 2∣ (p4 + 5) . Example 2. Therefore. then a2m = 42m−1 (L4m − 2) = 42m−1 ⋅ 5F2 = 5 ⋅ 4m−1 ⋅ (2m F2m )2 .. n = 1. (2. From this relation it is shown that a2m is divisible by 5 ⋅ 4m−1 and has the quotient value (2m F2m )2 . n = 1.. then we have an = x2n − 4n = 5yn2 . If n is even. n = 1. .5) Subtracting 2an from both sides and squaring the resulting value leads to This is reduced to an = 4 n−1 [4n (α2n − 1) − 2an ]2 = 4an (an + 4n ). say n = 2m.. (2. consider an = x2n − 4n and we have √ √ √ √ √ √ √ √ an + an + 4n = x2n − 4n + x2n = 5yn2 + x2n = yn 5 + xn = (1 + 5)n . (⊠) .6) α2n − 1 2 2 ) = 4n−1 (αn − (−1)n β n ) = 4n−1 (α2n + β 2n − 2) = 4n−1 (L2n − 2) ( n α where Lm is the mth Lucas number.4) Squaring both sides leads to √ 4n (α2n − 1) = 2an + 2 an (an + 4n ). Then hence (1 − √ n √ 5) = xn − yn 5.38 Chapter 2. x2n − 5yn2 = (−4)n .5) (2.. (b) If n is an even value. 2. (⊠) Second solution √ √ n (a) Let (1 + 5) = xn + yn 5. In this case we get an = 5 ⋅ 4n−1 Fn2 . where √ n √ n √ n √ n 2n 1+ 5 1− 5 1 ) −( ) ) = 2n−1 Fn . (b) If n is even. . Hence it has been shown that an is a positive integer and is given by an = 4n−1 (L2n − 2) . where xn . yn = √ ((1 + 5) − (1 − 5) ) = √ (( 2 2 2 5 2 5 where Fn is the nth Fibonacci number. Examples for practice (2.. 2. .. 2. consider an = 5yn2 − 4n and we have √ √ √ √ √ √ √ √ an + an + 4n = 5yn2 − 4n + 5yn2 = x2n + 5yn2 = xn + yn 5 = (1 + 5)n . (i) If n is odd. hence 5 ⋅ 4n−1 ∣an and the quotient is Fn2 . yn are positive integers. n nn (n + 1)n+1 ⎡ ⎤ n times ⎢ ⎥ ⎢ ⎥ ³¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ⎢ ⎥ ⎢ 1 1 ⎢ na + (1 − a) + ⋯ + (1 − a) ⎥⎥ n n a ⋅ (1 − a) = ⋅ (na) ⋅ (1 − a) ⩽ ⎢ ⎥ n n⎢ n+1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ = 1 n n+1 nn ( ) = ⋅ n n+1 (n + 1)n+1 1 The equality holds for: na = 1 − a ⇔ a = ∈ (0. 1).GM) . and k. . a2 (1 − a3 )n . . for a ∈ (0. . a2 . ak .39 2. n+1 Using the proved inequality for a1 .. ... . . ⋅ ak ⋅ (1 − a1 ) ⋅ (1 − a2 ) ⋅ . ⋅ (1 − ak ) > [ (n + 1)n+1 n n But.35. we have a(1 − a)n ⩽ Let’s prove this inequality. . Arithmetic problems Example 2. 1). we get: a1 (1 − a1 )n ⩽ a2 (1 − a2 )n ⩽ nn (n + 1)n+1 nn (n + 1)n+1 ⋯ ⋯ ak (1 − ak )n ⩽ nn ⋅ (n + 1)n+1 k (∗) (by AM .. prove that the following inequality holds min{a1 (1 − a2 )n . If a1 .. . Therefore a1 (1 − a2 )n > a2 (1 − a3 )n > nn ⋅ (n + 1)n+1 nn (n + 1)n+1 nn (n + 1)n+1 ⋯ ⋯ ak (1 − a1 )n > Multiplying these relations up.1. ak ∈ (0. ak (1 − a1 )n } ⩽ Solution ( Method ”Reductio ad absurdam”) Let’s suppose that the inequality doesn’t hold.. n are integers such that k > n ⩾ 1. we get nn ⋅ (n + 1)n+1 nn ] a1 ⋅ a2 ⋅ . a2 . . . 1). being ϕ(9) = 6 we have that p6 + q 6 + 1 ≡ 3 (mod 9) by Euler theorem.36. (729. 2. We proceed similarly in the cases where n + 8 = w 3 and 8n − 27 = u3 are both perfect cubes (hence (2w)3 − u3 = 91 is the difference between two perfect cubes) and n + 8 = w 3 and 27n − 1 = v 3 are both perfect cubes (hence (3w)3 − v 3 = 217 is the difference between two perfect cubes). (2v)3 − (3u)3 = 272 − 8 = 721. the values obtained for u. (⊠) Example 2. 8) and (u. n in these two cases are exactly the values found in the case where we assume that 8n − 27 = u3 and 27n − 1 = v 3 are both perfect cubes. 8). then 3 is coprime with both p and q (because they are prime). 64. 27. (−8. Examples for practice Multiplying these relations up. y) of positive integers such that x2 + y 2 = p6 + q 6 + 1. the pairs of perfect cubes for which this happens are (4096. that assumption is false. 216. yielding respectively 2v = 16. It follows respectively that n = 19 and n = 0. for some primes p and q. we get a1 ⋅ a2 ⋅ . the difference of any two perfect cubes one of which is in absolute value larger than 4096 being clearly larger than 4096 − 3375 = 721. Then. v) = (−3. −9. 2197. It follows that either at most one of the given numbers is a perfect cube. . −1) (the other two cases yield non-integral values for u. contradiction. 8n − 27. u and w. 27n − 1 are perfect cubes? Solution The first positive perfect cubes are 1. respectively. v. 512. −2. (⊠) Example 2. v that make this possible. −16. . and 721 must be the difference between two perfect cubes. or all three are perfect cubes. −729) and (−3375. 1331.40 Chapter 2. ⋅ (1 − ak )n ⩽ [ k nn ] . (n + 1)n+1 This inequality contradicts (*). In fact. In this case. with solutions (u. 729. Solution Suppose that 3 ≠ q and 3 ≠ p. −4096). 4096. 3375). ⋅ ak ⋅ (1 − a1 )n ⋅ (1 − a2 )n ⋅ . Hence x2 + y 2 ≡ 0 (mod 9) and 3 ≡ 0 (mod 9). . 1728. 8. 27n − 1 = v 3 a perfect cube. So 9 is coprime with both p and q. hence no n exists such that exactly two of the given numbers are perfect cubes. so is n + 8. 2744. w. .37. 125. v) = (5. for n + 8 = 33 and n + 8 = 23 . we find by inspection that the only integral values of w. 9. and 8n − 27 = u3 a perfect cube. In both cases. also yield. By inspection. Therefore. Then we should have that without loss of generality p = 3. 3375. 343. Is there an integer n such that exactly two of the numbers: n + 8. which follows from the initial assumption. But then we should have x2 + y 2 ≡ 0 (mod 3) which implies that x ≡ 0 (mod 3) and y ≡ 0 (mod 3). or whenever 8n − 27 and 27n − 1 are simultaneously perfect cubes. v. . 1000. Find all pairs (x. Assume that 8n − 27 = u3 and 27n − 1 = v 3 are both perfect cubes for some integers u. v). −15 and 3u = 15. perfect cubes in both cases. Find the least prime that can be written as 2011 integers a and b. So we have to solve x2 + y 2 = 794 ≡ 2 (mod 397). b ⩾ 1. clearly a + b > 1. 53. and p ⩾ 59. Then q = 2. 13) and(13. Now. we have 493 + 103 118649 = = 59. or a + b = 2011 and a2 − ab + b2 is the lowest prime. Since a. or ab = ≡ 66(mod100). b such that a + b = p is prime and a2 − ab + b2 = 2011. then ab ⩽ (a − b)2 + ab = 1. y + z − x. Let x. then a3 + b3 = (a + b)(a2 − ab + b2 ) is the product of two primes. contradiction. and a3 + b3 = 2 is clearly not a multiple of 2011. Note is always 3 an integer since p2 ≡ 2011 ≡ 1 (mod 3) for primes p ≠ 3. and p2 − 2011 ≡ −2(mod100). while if a2 −ab+b2 = 2011. 3 Then (a−b)2 = p2 −4ab ≡ 9−64 ≡ 45 (mod 100) cannot be a perfect square. z + x − y . then a2 −ab+b2 > 2011⋅500. then p2 − 32011 p2 ≡ 9 (mod 100). (⊠) Example 2.41 2.1. x + y − z. Prove that (x + y + z)3 is a sum of four cubes of nonzero integers. z + x − y ≠ 0. Solution Since 2011 is prime (it is not divisible by any prime up to 43. So if q is odd. 2011 2011 and this is therefore the lowest prime of this form that can be found. then x2 + y 2 ≡ 3 (mod 4). or p ⩾ 47. z be integers such that 3xyz is a perfect cube. and it is less than 2 47 = 2209). Note however that a2 − ab + b2 = 3(a − b)2 + (a + b)2 (a + b)2 ⩾ 4 4 √ or if a+b = 2011. b) a permutation of (10. But we have that 397 ≡ 1 (mod 4) and it is prime.38. This is however not a solution when at least one of xyz. Therefore the unique solutions are (25. where the first three terms in the RHS are clearly nonzero cubes. 49). or p ≠ 47. (⊠) a3 + b3 for some positive Example 2. p2 − 32011 Clearly. then a+b ⩽ 4 ⋅ 2011 and a + b < 90 because 2011 < 452 = 2025. Let us therefore look for values a. and 24xyz = 23(3xyz) is also a nonzero cube. 3ab = (a + b)2 − (a2 − ab + b2 ) = p2 − 2011. or a = b = 1. If a2 − ab + b2 = 1. 25). x + y − z. So we get this factorization in primes of Z[i] ∶ 794 = (1 + i)(1 − i)(6 + 19i)(6 − 19i). Arithmetic problems Now 36 + q 6 + 1 ≡ 2 + q 6 (mod 4).39. First solution If xyz. y. Note however that for (a. note that (x + y + z)3 = (x + y − z)3 + (y + z − x)3 + (z + x − y)3 + 24xyz. if p ≡ ±3 (mod 50). y + z − x. hence either a + b is the lowest prime and a2 − ab + b2 = 2011. z. then (x + y + z)3 = s3 = (7s)3 + (−6s)3 + (−5s)3 + (−s)3 is the sum of four nonzero cubes. (1) In particular. Now subtracting the two equalities in (1) we find that (nd + λ)b′ = (md + λ)a′ . Prove that gcd(a. Note therefore that for any three integers x. or if x + y + z = s ≠ 0. since a′ ∣(nd + λ)b′ and gcd(a′ . (x + y + z)3 = 0 = a3 + b3 + (−a)3 + (−b)3 is the sum of four nonzero cubes. Finally. (⊠) Second solution Let 3xyz = w 3 . +1}. Let a and b be integers such √ that a m − b n = a − b. b) and write a = da′ and b = db′ with gcd(a′ . and md + λ = µb. Now if x + y − z = 0 then (x + y + z)3 = 8z 3 but 1 = (−1)3 + (7)3 + (−5)3 + (−6)3 and hence 8z 3 = (−2z)3 + (14z)3 + (−10z)3 + (−12z)3 .40. (3) Whereas from (2) we have λ = µ(a′ − nε(a′ − b′ )). +1}. analogously if y + z − x = 0 or z + x − y = 0. Examples for practice is zero. λ) = 1. hence µ ∈ {−1. we conclude that d2 = ∣a − b∣. and mda′ − λb′ = 1. and again. b′ ) = 1. Solution Let d = gcd(a. Then for any a. if x + y + z = 0. ndb′ − λa′ = 1. We have the identity (x + y + z)3 = (x + y − z)3 + (y + z − x)3 + (z + x − y)3 + (2w)3 which can be proved expanding the cubes in both sides. multiplying both sides of (3) by d and using the fact that ∣εµ∣ = 1. which is the desired conclusion. y. This implies that a′ ∣(nb − 1). (⊠) . b) = ∣a − b∣. regardless of whether 3xyz is a perfect cube or not. (x+y+z)3 may be written as the sum of four nonzero cubes. From a2 m − b2 n = a − b we get (ma − 1)a′ = (nb − 1)b′ . (2) On the other hand. gcd(d. In that case. Finally. and consequently there exists an integer λ such that nb − 1 = λa′ and ma − 1 = λb′ or. (⊠) 2 2 Example 2. b ≠ 0. µ is a common divisor of λ and d which are coprime. b′ ) = 1. we conclude that a′ ∣(nd + λ) and that there exists an integer µ such that nd + λ = µa. where m and n are consecutive integers. recalling that m and n are consecutive integers we get from (2) that d = εµ(a′ − b′ ). so a′ ∣(nb − 1)b′ and gcd(a′ . So.42 Chapter 2. with ε = n − m ∈ {−1. b′ ) = 1. we can use that 73 = 343 = 216 + 125 + 1 + 1 = 63 + 53 + 13 + 13 . and n is divisible by the sum of its digits (18) since n is even and also divisible by 9 by the decimal integer divisibility test for 9 (i. p). First solution For j ⩾ 1. 3) and (2. and the sum of its digits in even positions is 0. 2. n − 2) = (n + 1. and is clearly divisible by 11 since the sum of its digits in odd positions is 11. from which p = 3. But (n + 1. then n′ = (10m a)4 + (10m b)4 + (10m c)4 + (10b d)4 − 4(10m a)(10m b)(10m b)(10m c) = 104m n has the same sum of digits as n. where n and k are positive integers and p is a prime. Then n3 − n + 1 − (n2 + n + 1) = n(n + 1)(n − 2) > 0. (⊠) Example 2. b. Prove that there are infinitely many positive integers n that can be expressed as a4 +b4 +c4 +d4 −4abcd. k. then n = (10j )4 + (10j )4 + (10j−1)4 + (10j−1 )4 − 4(10j )(10j )(10j−1 )(10j−1 ). 3) can be 3 or 1.41. let n = 196 ⋅ 104j−2 + 2 ⋅ 104j−4 . c. such that n is divisible by the sum of its digits. Suppose n > 2. Then n5 + n4 + 1 = 0 (mod 9).42. since if n = a4 + b4 + c4 + d4 − 4abcd. Arithmetic problems 43 Example 2. and if n is divisible by the sum of its digits. If n + 1 = pr it contradicts the first relation. d are positive integers. First solution It’s easy to see that (1. satisfying the equation n5 + n4 + 1 = pk . Subtracting the first relation from the second we get n(n + 1)(n − 2) = pr (ps−r − 1). (⊠) Second solution Note that it suffices to find one. Clearly r > 0 and p doesn’t divide n.2. Find all triples (n. Hence we have n2 + n + 1 = pr and 3 n − n + 1 = ps where r + s = k and s > r.1. where a. n5 + n4 + 1 = (n2 + 1)2 − n2 + n5 − n2 = (n2 + 1 − n)(n2 + 1 + n) + n2 (n − 1)(n2 + 1 + n) = (n2 + n + 1)(n3 − n + 1). but checking this congruence for every residue modulo 9 it’s easy to see that there are no solutions. Note now that (105 )4 + (104 )4 + (2 ⋅ 103 )4 + 14 − 4 ⋅ 2 ⋅ 105+4+3 = 1020 + 1016 + 8 ⋅ 1012 + 1 has sum of digits 11. (⊠) . The conclusion follows.e. the sum of the digits is divisible by 9). so is n′ . 1. 7) are solutions. We’ll prove that there aren’t any other. and so p divides both n + 1 and n − 2. and Cn−1 ≡ (−1)k (mod n) is equivalent to (n − 1) ⋅ (n − 2) ⋅ . pr ) = ps . k = Cn−1 This is clearly true since n − j ≡ (−j) (mod n) for each j ∈ {1. k − 1}. 1).43. If n is not prime.. it is trivially verified that n = 1 and n = 2 both give solutions: (n. 3). ⋅ (n − k) ≡ (−1)k k! (mod n). then k! is prime with n. (2. Because n3 − n + 1 = (n − 1)(n2 + n + 1) − (n − 2) and n2 + n + 1 = (n − 2)(n − 3) + 7. (n − 1) ⋅ (n − 2) ⋅ . and 0 < < 1 for n ⩾ 3. if and only if n is a prime. (⊠) Third solution For n = 1 we have p = 3 and k = 1. n − 1. since pk = n5 + n4 + 1 = (n2 + n + 1)(n3 − n + 1) there exist two positive integers r and s such that n2 + n + 1 = pr and n3 − n + 1 = ps with r ⩾ s. p.. n3 − n + 1) = gcd(ps . and that s(n) > r(n) for n ⩾ 3.. since (k − 1)! is prime with n. n−k ≡ −1 (mod n). Prove that Cn−1 = (−1)k (mod n) for each k = 1. (⊠) k Example 2. since otherwise we would have s(n) s(n) (n − 2) (n − 2) = p(b−a) .. n − 2) = gcd(n − 2. k n ≡ 0 (mod n). 1. 7) therefore ps = 1 or ps = 7. k (⊠) . Therefore any r(n) r(n) r(n) r(n) solution must satisfy n < 3. Next note that for n ⩾ 3. 2. n − 1}.44 Chapter 2. Then.. we have that gcd(n2 + n + 1. . .. 2. the result k Cn−1 = (−1)k (mod n) is equivalent to or equivalently. . k) ∈ {(1. b. ⋅ (n − k) ⋅ k! k If n is prime. Let n be an integer greater than 2. k. and n − j ≡ (−j) (mod n) for all j ∈ {1. This concludes the proof. Examining these two cases we obtain two solutions (n. but = n−1− . 2. and (n... 7. k. that is there are at most two solutions. for any k < n. Examples for practice Second solution First note that n5 + n4 + 1 = (n2 + n + 1) ⋅ (n3 − n + 1) ∶= r(n) ⋅ s(n). 3. Now suppose that n ⩾ 2. we cannot have r(n) = pa and s(n) = pb for any prime p and positive integers a. 7). p) = (2.. p) = (1. n3 − n + 1) = gcd(n2 + n + 1.... 2. b > a. We have that gcd(n2 + n + 1. let k be the least divider of n which is larger than 1. absurd. Finally. 2)}. Solution Clearly.. 1. Consider the function k 2 f (x) = nx 1 1 ln (1 + ) + 1 − (1 + x) n n n−x and we study it for all n ⩾ 2. c (a + 1) − a Since c ∈ (a. Thus the derivative is positive and therefore f (x) ⩾ 0.3 Undergraduate problems Example 2. Then f (x) = exists c ∈ (a. Arithmetic problems 2. Lemma 1.8) ⇔ n3 (x + 1) ⩾ (nx + n − x)(n2 − x2 ) ⇔ n3 x + n3 ⩾ n3 x + n3 − n2 x − nx3 − nx2 + x3 ⇔ n2 x + nx2 + nx3 ⩾ x3 . a + 1) such that 1 and due to the Mean-Value theorem there x ln (a + 1) − ln a 1 = f ′ (c) = = ln (a + 1) − ln a. n ⩾ 2 prove that √ n n 1 n )+1 1 + ⩽ log (1 + k n k−1 (2. it must be that 1 1 1 < < a+1 c a . 0 < x = . a + 1). it is enough to prove that 1 + nx n2 ⩾ ⋅ (2. Let f (x) = ln x.7) First solution n n Let = x. Hence √ x n 1+x⩽1+ ⋅ n To prove that the derivative of f is positive for any 0 < x < n and for any fixed n ⩾ 2.1. We have √ n 1 n2 1+x 1) lim f (x) = 0 . For all positive a 1 1 < ln (a + 1) − ln a < ⋅ a+1 a Proof. For all integers k. (⊠) Second solution We will use the following simple result. 2) f (x) = ( − ) x→0 n (nx + n − x)(n − x) 1+x ′ Recall Bernoulli’s inequality: (1 + x)a ⩽ 1 + ax for 0 ⩽ a ⩽ 1. The proof is completed. which is clearly true.45 2.44.8) (nx + n − x)(n − x) 1 + x We have (2. z) = x2 + y 2 + z 2 − xy − yz − zx over all triples (x. by the AM-GM inequality we have √ √ k+n k+1k+2 k+n k+n k+1 k+2 n n + +⋯+ ⩾n ⋯ =n . z have the same parity. we have A ⩽ ln k − ln (k − 1) + ln (k + 1) − ln k + . k k+1 k+n−1 k k+1 k+n−1 k We also have k+n 1 1 1 k+1 k+2 + +⋯+ =n+( + +⋯+ ) ∶= n + A. y. hence 52 divides d s 3s2 + d2 = f (x. any perfect square leaves a remainder equal to −1. z). y. y. y. where s = u + v and d = u − v. . z) of positive integers for which 2010 divides f (x. 25 25 But taking s1 = 16. x. y. Solution If wlog z is odd and x. z) is odd. y are odd and z is even. and 20100 divides f (x. Therefore. z). and we y−z x−y . y. Now.45. while if wlog x. s are not both multiples of 5. + ln (k + n − 1) − ln (k + n − 2) n ) = ln (k + n − 1) − ln (k − 1) = ln (1 + k−1 Combining the obtained results. 0. Returning to the problem. (⊠) Example 2. hence if d.46 Chapter 2. z) = 25 ⋅ 804 = 20100. Defining s1 = and d1 = . then 4020 = 22 ⋅ 3 ⋅ 5 ⋅ 67 divides f (x. and if 2010 divides f (x. then z 2 is the only odd term. y. y. y. y. z) is again odd. d1 = 6. we find 3s21 + d21 = 768 + 36 = 804. . ⩽ (n + ln (1 + n k−1 n k−1 and the inequality is proved. then 3s2 + d2 cannot be a multiple of 5. hence not a multiple of 2010. then x2 . z) 20100k = = 804k. y. k k+1 k+n−1 k k+1 k+n−1 Due to the result of the lemma. y. y 2 . z). y are even. . z). f (x. or f (x. xy are the only odd terms. z).v = may define u = 2 2 3s2 + d2 = 4(u2 + v 2 + uv) = (x − y)2 + (y − z)2 + (x − y)(y − z) = f (x. we find that 5 5 3s21 + d21 = f (x. Examples for practice Combining these two results yields the desired inequalities. Find the minimum of f (x. y. f (x. we finally obtain √ n 1 k+1 k+2 k+n n 1+ ⩽ ( + +⋯+ ) k n k k+1 k+n−1 n 1 n 1 )) = 1 + ln (1 + ). 1 modulus 5. z). 1. z) = 20100 for all positive integer z as it is easily checked by direct calculation. . y ∈ N} √ √ √ √ and let s = 3 + 2 2 and for any z = x + y 2 ∈ Z ( 2) denote z = x − y 2 . Note that if z ∈ Sol (that is zz = −1. x. k = 1.. In this notation √ equation x2 − 2y 2 = −1.. n 4 2 1 (2n2 + 2n + 1) Q2 4 where Q = ∏nk=2 ak .. . i. 2.4 Olympiad problems Example 2. Indeed. z + 50.e.e. N ( 2) = {x + y 2 ∣ x. z). √ Sol = {z ∣ z ∈ N ( 2) and zz = −1} √ √ Note that for z0 = 1 + 2 ∈ N ( 2) we have z0 z 0 = −1 and since ss = 1 then for zk = sk z0 we also have zk z k = −1. z ∈ N ( 2) . P is the square of a rational number if and only if 2n2 +2n+1 is the square of a positive integer. making the problem trivial. Therefore 2m2 − (2n + 1)2 = 1 and then our problem is finding the solutions to the equation x2 − 2y 2 = −1 P = a1 an+1 Q2 = in positive integers.. or f (z + 160. n. x.1. Since ak+1 = k 2 + k + and 4 2 2 k4 + then 1 1 2 = (k 2 + ) − k 2 = ak ak+1 . .47 2. First solution 1 1 1 Let P = ∏nk=1 (k 4 + ) and let ak = k 2 − k + .. Denote the set of all such solutions by Sol. since clearly f (x. 2. Arithmetic problems with equality for example for s = 80 and d = 30. ie u = 55 and v = 25. if and only if 2n2 + 2n + 1 = m2 for some positive integer m. k = 1. y ∈ Z} . Let √ √ √ √ Z ( 2) = {x + y 2 ∣ x. x) = 0 is a multiple of 2010 for all positive integer x. Also it √is clear that z0 is smallest element in Sol.46. i. Find all positive integers n for which 1 1 1 P = (14 + ) (24 + ) ⋯ (n4 + ) 4 4 4 is the square of a rational number. z ∈ N ( 2)) and z ≠ z0 then sz ∈ Sol. Therefore. (⊠) Note We have restricted ourselves to positive values of f (x. y. y ∈ N becomes zz = −1. k ∈ N. 2.. Since x = 2n + 1 then all natural n for which 2n2 + 2n + 1 is a square of integer should be elements of set xk − 1 . x0 = 1. n2 = 17. . Since z ≠ z0 ⇒ z > z0 ⇒ x. In particular. n1 = 3. . k = 1. . . 2. Then x zk+1 = szk ⇔ { k+1 yk+1 = 3xk + 4yk = 2xk + 3yk ⇒ xk+2 − 3xk+1 = 2xk + 3(xk+1 − 3xk ) ⇔ xk+2 − 6xk+1 + 7xk = 0. Sol = {zk ∣ k ∈ N}. We will prove that Sol = {zk ∣ k ∈ N}. Let z = x + y 2 then x2 = 2y 2 − 1 and √ √ √ sz = (3 − 2 2) (x + y 2) = 3x − 4y + (−2x + 3y) 2. and moreover. Since z0 < sk z < z1 = sz0 = 7 + 5 2 then sk+1 z < z0 and sk+1 z ∈ Sol. Second solution Multiplying by 4 does not affect the property of our number M . n4 = 360. That is the contradiction because z0 is smallest element in Sol. k ∈ N .that our number can be written in this form: ∏ (4k 4 + 1) = ∏ ((2k 2 + 2k + 1)(2k 2 − 2k + 1)) (⊠) . n0 = 0. √ Let zk = xk + yk 2. all solutions of problem are the terms of the sequence (nk )k⩾1 defined recursively nk+2 = 6nk+1 − 7nk − 1.} 2 By substitution xk = 2nk + 1 in xk+2 − 6xk+1 + 7xk = 0 we obtain {nk ∣ nk = 2nk+2 + 1 − 12nk+1 − 6 + 14nk + 7 = 0 ⇔ nk+2 = 6nk+1 − 7nk − 1 by Thus. Since (zk )k⩾0 is unbounded √ from above then there is k such that zk < z < zk+1 . y ⩾ 2. y ⩾ 3 because x is odd and 2y 2 − 1 isn’t square of integer for y = 2.48 Chapter 2. . . x. n1 = 3. Suppose that exist z ∈ Sol which not belong to the sequence z0 < z1 < z2 < ⋯ < zk <. n3 = 80. x1 = 7. . we have and 3x ⩾ 4y ⇔ 9x2 ⩾ 16y 2 ⇔ 9 (2y 2 − 1) ⩾ 16y 2 ⇔ 2y 2 ⩾ 9 ⇐ y ⩾ 3 3y ⩾ 2x ⇔ 9y 2 ⩾ 4x2 ⇔ 9y 2 ⩾ 4 (2y 2 − 1) ⇐ y 2 + 4 ⩾ 0. Examples for practice √ √ sz ⋅ sz = ss ⋅ zz = 1 ⋅ (−1) = −1. . Thus. we can consider M = ∏ (4k 4 + 1) Observe. It remains to prove sz ∈ N ( 2) . . (3. j ⩾ 1 odd and m ⩾ 2. (⊠) . Then. 2). Arithmetic problems As the factors will be repeated. 1) for all positive integers m. 3) or (n.10) by (2.47.that minimal roots of first equation are x0 = 1. Solution We will show that the pairs are (m. where f Euler’s totient function. Finally assume that j = 1. y0′ = 2 √ √ √ √ √ √ √ (2.we have the equation (2n + 1)2 − 2c2 = −1 Using Pell’s equation.9) Then. Hence km + s − 2 ⩽ k. We first consider the case when j ≠ 1. with k ⩾ 1. except the last factor. then if m = 1 then n = 1. that is k(m − 1) ⩽ 2 − s ⩽ 1 which implies that k = 1. 2n + 1 = xk = (A + B − 2) (A + B) ⇔n= ⋅ 2 4 Substitute the values of A and B and derive √ √ (1 + 2)2k+1 − (1 − 2)2k+1 − 2 n= ⋅ 4 (⊠) Example 2.we can consider two general cases: x2 − 2y 2 = −1 and (2. 2) and (2. by 1). Then f (h) = j and by 1) we have h = j = 1 which is a contradiction. We first note that: 1) since f (x) is odd iff x = 1 or x = 2 then f (f (nm )) = n for n odd iff n = 1 and for all positive integers m. h ⩾ 1 odd and f (f (nm )) = f (f (2km j m )) = f (2km−1 ⋅ 2s h) = 2km+s−2 f (h) = n = 2k j. m = 2. √ √ √ xk 1 − yk 2 = (1 − 2)2k+1 = B is the result of division of (2. (2n + 1)2 + 1 = 2(2n2 + 2n + 1) = 2c2 Consequently. that is (n. m) = (4. m) = (2. y0 = 1 and the minimal roots of the second ecuation are x′0 = 3. we will obtain squares. 4).9) x2 − 2y 2 = 1 Obviously. ∏ (4k 4 + 1) = ∏ ((2k 2 + 2k + 1)(2k 2 − 2k + 1)) = (2n2 + 2n + 1) ⋅ ∏(2k 2 − 2k + 1) Let 2n2 + 2n + 1 be equal to c2 . n) of positive integers such that f (f (nm )) = n.10) xk 1 + yk 2 = (3 1 + 2 2)n ⋅ (1 1 + 2) = (1 + 2)2k+1 = A Consequently. So we can assume that n = 2k j.1. Find all pairs (m.49 2. f (jm) = 2s h with s ⩾ 1. Then. 2) since f (n) < n when n > 1. and s = 1. Then n = 2k and f (f (2km )) = 2km−2 = 2k which implies that k(m − 1) = 2. c = 9. Since xy. c = 7. and again c ≠ ∣xy − zw∣ in either case. This means that x. and a = x2 − y 2 is a multiple of 8. Every square of an odd number 2m+1 is 4m(m+1)+1.50 Chapter 2. z. 16. a = x2 − y 2 cannot leave a remainder of 2 when divided by 4. define integers x′ = . ˆ a = 8. and so do z. 3 (when c and one of a. b. 2012. w have opposite parity. 16 and no solution exists in this case. 8. we take a = 0. y have the same parity. and similarly zw is also even. xy = 3. 40}). c are all even). or a. b is odd. or since exactly one of m. or a2 + b2 + 2c2 = x4 + y 4 + z 4 + w 4 − 4xyzw = n. xy = 3 × 1 = 3. 471 are not perfect squares. w are all odd. b = 43. or 4 (when a. b = 13. Since b = z 2 − w 2 is odd. By the same reason. Therefore. and the other two and c are even). . b are odd and c is even. z. z. or x. hence 2 2 2 2 x′4 + y ′4 + z ′4 + w ′4 − 4x′ y ′z ′ w ′ = 2016 = 126 = 6 (mod 8). finding only the following solutions: ˆ a = 0. w. we find that no such multiples of 8 exist such that 2016 − a2 − b2 is twice a perfect square. Solution Define x2 − y 2 = a. 2 (when either a. y. 16. y ′ = . ˆ a = 8. y z w x In the first case. or a2 and b2 leave remainders of 0 or 1 when divided by 8. 2016. zw must also have the same parity. since n = 2012 = 4 (mod 8). 24. hence x. Then. and a = 0 results in x = y. and xy is even. since n = 2016 is a multiple of 8. When x. zw = 462. b odd. z 2 − w 2 = b and xy − zw = c. c are all even. m+1 is even. Examples for practice Example 2. it follows that zw = 21×22 = 462. c are all even). By sheer trial (we need ”only” to try a ⩽ b ∈ {0. and xy is odd. we would need a. y are both odd. z ′ = and w ′ = . 2012. 2016. a = x2 − y 2 and b = z 2 − w 2 are multiples of 8. Again by the previous arguments. contradiction. Note that 194 + 3 × 184 − 4 × 19 × 183 = 2017. hence zw is even.48. or wlog b. 32. 1 (when one of a. false since 453. either x. b = 37. ˆ a = 40. c = 17. Again by sheer trial. every odd perfect square leaves a remainder of 1 when divided by 8. b are odd. b. We will next prove that no solution exists for n = 2011. Since b = 43 = (z +w)(z −w) is prime. or there is a solution for n = 2017. y have opposite parity. c are odd and a is even. or x2 = 462 ± 9. 32. n = 2011 = 3 (mod 8). or no solutions exist in this case either. By the previous arguments. w are all even. y. 2017. and the other one is even). c = 11. Finally. hence 2c2 leaves a remainder of 0 or 2 when divided by 8 because every even perfect square is a multiple of 4. b. and again c ≠ ∣xy − zw∣. or are all odd. Similarly zw = 19 × 18 = 342. and c ≠ ∣xy − zw∣. Find the least integer n ⩾ 2011 for which the equation x4 + y 4 + z 4 + w 4 − 4xyzw = n is solvable in positive integers. hence c is even. there may be solutions only when a. xy = 99 or xy = 21. the lowest integers n = 2011 such that the proposed equation has a solution could be 2011. 8. Then zw = 42. b are even and c is odd). b = 43. 40 and try to express 2016 − a2 as the sum of an odd perfect square b2 and twice another odd perfect square 2c2 . No solution exists either with n = 2012. It follows that the possible remainders of n when divided by 8 are 0 (when a. 24. i j=0 By using this formula for i = 0. Considering each case. it is easy to see that if 5 divides A. v. No solutions exist either when n = 2011. we must have u = v (mod 3) and so u3 − v 3 = (u − v)A is a multiple of 9. Thus 2010(u2 + v 2 ) is a multiple of 9. v must be both even). we deduce that u2 + uv + v 2 = 67. Cn+1 ) = gcd (Cn+1 . Cn+1 . .51 2. Therefore. Finally. . . . C2n . which is not the case. (⊠) Example 2. .49. . . C2n−1 . u. (⊠) Example 2. Thus A is a divisor of 67 and since u. n−j j n = ∑ Cn+1 Cn+i+1 Ci . . . . we get that any odd common First solution Using the identity Cm+1 − Cm m divisor of the numbers is also a common divisor of n n−1 n−1 n−1 . . Thus u2 + uv + v 2 divides 2010(u2 + v 2 ) and since it is relatively prime to u2 + v 2 . Prove that each odd common divisor of n n n n . The equation becomes d(u3 − v 3 ) = 2010(u2 + v 2 ). Cn+1 . . . Hence there is one solution. we deduce that v = 2. . Next. we get that any odd common divisor of the original set of numbers is also a common divisor of n 2 1 . xy = 99 or xy = 21. Then zw = 90. Let n be a positive integer. k k = C k−1 . We claim that A = u2 + uv + v 2 actually divides 67. then 5 divides both u. . Cn+1 . Find all pairs (x. C2n−2 C2n−1 k Repeating this argument till we get Cn+1 for some k. . Clearly u ⩾ v. . y = 2d with d = 318. x = 7d. (⊠) Second solution By Vandermonde’s identity. a contradiction. b = 19. v ⩾ 1. v relatively prime positive integers. . if 3 divides A. Cn+1 Cn+1 and of their sum. (⊠) . which equals 2n+1 − 2 = 2(2n − 1). . y) of positive integers such that x3 − y 3 = 2010(x2 + y 2 ). Arithmetic problems ˆ a = 40. Cn+1 . . It is immediate that A is odd (if not. .50. . c = 5. We conclude that the minimum such n is n = 2017. First solution Write x = du. y = dv with d ⩾ 1 and u. Cn+2 a ∶= gcd (Cn+1 Therefore. . . . and we’re done. Cn+1 . we find that 1 n−1 n n n n ) . u = 7 and so d = 318. if d is divisor of a then d divides the sum k = 2n+1 − 2 = 2(2n − 1) ∑ Cn+1 n k=1 It follows that if d is odd then d divides 2n − 1.1. we deduce that u2 + uv + v 2 divides 2010 = 2 × 3 × 5 × 67. and c ≠ ∣xy − zw∣. then 67 ⩾ 3v 2. we conclude that any odd common divisor of the original set also divides 2n − 1. n. C2n−1 C2n is a divisor of 2n − 1. thus v = 4. letting x = gcd(a. 1 = i = k. b) gcd(a. k. b) ⎡ ⎤ ϕ(b) ⎢ ⎥ a a a ⎢( ) − 1⎥⎥ ≡ 0 mod ( . Because b is odd pi > 2 for i = 1. Prove that there is an integer x such that 1 < x ⩽ a and both a and b divide xϕ(b)+1 − x. z) of integers such that √ √ √ x3 + y 3 + z 3 − xyz = 2010 max{ 3 x − y. y. If x = x0 (mod P 2 ). Let P = p1 p2 ⋯pk . 3 z − x} 3 (⊠) . that So. applying Euler’s theorem this is also divisible by b. For each i. Let vp (a) be the degree of p in the canonical factorization of a. Let that solution be x0 . b) ⎥ ⎣ ⎦ Therefore the required x is a since this also satisfies 1 < x ⩽ a. If x ≡ x0 (mod P 2 ). i = 1. But x ≡ x0 ≡ pi + 1 (mod p2i ). b) gcd(a. Find all triples (x. Hence vpi (n) = vpi (xn − 1) − 1 ⩾ mai − 1 ⩾ (m − 1)ai (because xn ≡ 1 (mod bm )). 2. Consider the sysmte of conguences x ≡ pi + 1 (mod p2i ). and gcd(pi. Now from Dirichlet’s theorem there are infinitely many primes p such that p ≡ x0 (mod P 2 ). ⎢ gcd(a. 3 y − z. We know that pi is odd. Proof. (m−1)ai ∣n. b) ⎢ gcd(a. then x is solution of system (*). then from condition xn = 1 (mod bm ) it follows that bm−1 ∣n. It is true for all i. a a we get. (⊠) Example 2. . Let a and b be positive integers such that a does not divide b and b does not divide a. . Suppose p ∈ P is a prime and a is a positive integer. b). Let m and n be positive integers. We have x0 = pi + 1 (mod p2i ) for all i. x (xϕ(b) − 1) is divisible by x and when gcd(b. so vpi (x − 1) = 1. b) Example 2. Solution We have. gcd(a. Solution Let b = pa11 pa22 ⋯pakk be the canonical factorization of b. .52. (∗) By the Chinese Remainder Theorem the system (*) has solution. . Claim. x) = 1. k. pi ∣x0 − 1. . . using gcd ( . 2. . where ϕ is Euler’s totient function. b) ≡ 0 mod lcm(a. Prove that for each odd positive integer b there are infinitely many primes p such that pn = 1 (mod bm ) implies bm−1 ∣n.51. therefore So pi bm−1 = ∏ pi k i=1 (m−1)ai ∣n.52 Chapter 2. . b) = 1. Examples for practice Example 2.53. pj ) = 1 whenever i ≠ j. therefore by a well knwon lemma vpi (xn − 1) = vpi (x − 1) + vpi (n). but 3y + 1. We then have two possible cases: ˆ If u = 0. v. contradiction. and x + y + z = 6030. and 6 (x + y + z)(u6 + v 6 + w 6 ) = 12060u. and 2010 is odd. r) is a solution for any integer r. or x = y = z. with equality iff u = v = w = 0. hence x. If x. 3y + 1 are not. v. then 0 = u3 + v 3 + w 3 ⩽ 0. yielding respectively 3y + 1 = 6030 or 3y + 2 = 6030. y are both odd. Prove that the equation x3 + y 3 = 2010 x−y is not solvable in positive integers. z) is an integral solution of the proposed equation if x = y = z = r for some integer r. hence x−y . Note also that u ⩾ 0. hence clearly u3 + v 3 + w 3 = 0. yielding (x + y + z)u5 = 6030 = 2 ⋅ 32 ⋅ 5 ⋅ 67. hence u = v = w = 0. since if u ⩽ 0.53 2. where we have assume wlog by cyclic symmetry in the variables that u = max{u. impossible since 6030 is a multiple of 3. z) = (r. y. y have the same parity and x − y is even. u = 1. w}. then u = v = w = 0. and x+y is divisible by 4 but not by 8. Now. (⊠) Example 2. y are both odd. x3 + y 3 x2 − xy + y 2 = is odd. ie (u. then x + y and x − y xy(x + y) = 2010 − x2 + y 2 is a are both divided by the lowest of both multiplicities. Now. y have opposite parity.54. and since one of v. v = 3 y − z. hence 2 divides x + y with multiplicity a + 1. y are both even. while v 3 + w 3 = −u3 . w is zero. where a is x+y the multiplicity with which 2 divides x − y. w is zero. then x3 + y 3 is odd. v. −u).1. w) is some permutation of (u. hence (x. w is necessarily zero. then the other one necessarily equals −u. y. hence a = 1. x−y is divisible by 2 but not by 4. either z = y or z = x. or if x. and w = 3 z − x. (x + y) + (x − y) = 2x is divided by 2 with multiplicity 1. which is known to have integral solutions only if at least one of u. 0. or since the RHS is not divisible by any fifth power. If x. r. since one of v. Arithmetic problems Solution √ √ √ Denote u = 3 x − y. It follows that (x. First solution Assume that x. equality must hold. while x3 + y 3 + z 3 (x + y + z)(x2 + y 2 + z 2 − xy − yz − zx) − xyz = 3 3 = (x + y + z)(u6 + v 6 + w 6 ) . and both are divided by 2 with different multiplicity. Note that any x = y = z results in both sides of the proposed equation being zero. ˆ If u > 0. where x = y + 1. then x2 − xy + y 2 = 67k for some positive integer k. or divisible by at least 23 = 8 if x. we have 2010 = x3 + y 3 x3 + y 3 3 x3 + y 3 3 x+y 3 ⩾ = ( )⩾ ( ) x−y 44 44 3 44 3 hence we obtain x + y ⩽ 96. then x + y must be a multiple of 3. That is 672 = 30(x − y) + 3x(67 − x). x − y are divisible by 2 with multiplicity at least a + 1. If x + y is divisible by 67. (⊠) Second solution Assume that the equation is solvable in positive integers. or 2x = (x + y) + (x − y) is a multiple of 3. y 2 are also multiples of 4. and x. equation with no integer solutions. We can write x3 + y 3 x3 − y 3 > = x2 + xy + y 2 = (x − y)2 + 3xy > (x − y)2 . since x + y ⩽ 96 and 67 is a prime number. It is clear that x > y. If x. 2010 = x−y x−y √ and get x − y < 2010. since 67 is not divisible by 3. since x − y divides xy(x + y). It follows that no positive integral solutions exist for the proposed equation. contradiction. divisible by 2 if x. while if z = 4 then z 3 + 9z 2 + 54z = 64 + 144 + 216 = 424 < 562. y are multiples of 3. absurd. hence x3 + y 3 is a multiple of 27. hence x. The equation is equivalent to k(x + y) = 30(x − y). or since x2 . and x + y. Writing x as a function of y. If x2 − xy + y 2 is divisible by 67. hence (x + y)2 − 3xy = 30(x − y). x. 0. y are both odd. . It follows x − y ⩽ 44. y are divisible by 2 with the same multiplicity a. In this case we get x2 − xy + y 2 = 30(x − y) and x + y = 67. then necessarily x + y = 67. 27 z 3 + 9z 2 + 54z = 562. Examples for practice multiple x − y of 4. We conclude that x − y = 18. hence x − y is divisible by 2 with multiplicity at least 3a ⩾ 3. 1 modulus 9.54 Chapter 2. Therefore. and x − y is a multiple of 3 because 2010 is not a multiple of 9. or since x3 + y 3 is a multiple of 3 because 3 divides 2010. then x − y ⩽ 43 Since 452 = 2025 > 2010 = x−y must be an even multiple of 9. The equation is equivalent to (x + y)(x2 − xy + y 2) = 2 ⋅ 3 ⋅ 5 ⋅ 67 ⋅ (x − y). Note now that the LHS 3 increases strictly with z. the proposed equation becomes y3 + y 2 + 18y = 562. we have x ⩽ 44. and if z = 5 then z 3 + 9z 2 + 54z = 125 + 225 + 270 = 620 > 562. or since y ⩾ 1. y are both odd. where we have defined z = y because y is clearly divisible by 3. then x3 + y 3 is a multiple of 9. and x − y must be a multiple of 9. It is well known that any perfect cube leaves remainder −1. y are both even. y are not multiples of 3. On the other hand. x3 + y 3 > x2 . 2010 is a multiple of 4. . m divides N . We may therefore express. hence relatively prime with all primes less than 2012.. Determine all primes that do not have a multiple in the sequence an = 2n n2 + 1. 2010! Prove that there are infinitely many n such that an is an integer with no prime factors less than 2010. hence the equation is not solvable. n ⩾ 1. for all positive integers k.55 2. For each positive integer n define an = (n + 1)(n + 2)⋯(n + 2010) . Arithmetic problems that is (30 − k)x = (30 + k)y. The conclusion follows.. . First solution Let N be the least common multiple of {1. 2012}. 2. akN P as a product of integers. . let n = ∏ pa .. For any a > 10. and take n = kNP . (⊠) Second solution Let P be the set of primes less than 2012... hence we have k = 3a for some positive integer 1 ⩽ a ⩽ 9.. p∈P Hence the first 11 digits of n+2012 and 2012 in base p coincide. Because x2 can’t be divisible by 67 it follows that 3a2 +100 is divisible by 67.. Then x2 (3a2 + 100) = 67 ⋅ a ⋅ (a + 10)2 . hence 30 + k x2 [(30 + k)2 − (30 − k)(30 + k) + (30 − k)2 ] = 67k(30 + k)2 . 2012}. If p ∈ P then the base p representation of 2012 has at most 11 digits: (2012)p = c10 c9 .. and by Lucas’ Theorem 10 1 0 2012 C1 ⋯C00 = 1 (mod p) C10 ≡ Cs0 ⋯C11 Cn+2012 which means that p does not divide an for any p ∈ P . where k is any positive integer. Replacing a = 1. Note that. and m m m prime with P .55. It is clear that k is divisible by 3. 3.. 2. 2.c1 c0 . (⊠) Example 2.. all relatively prime with all primes less than 2012. 9 is easy to see that 3a2 + 100 has no this property.. It follows y = and we get 30 − k x. P the product of all primes less than 2012.1. Example 2. x2 (3k 2 + 302) = 67k(30 + k)2 . for any N n + m kNP + m = = kP +1 is an integer relatively m ∈ {1. .56. 1837. a2009 √ be distinct √ positive integers not exceeding 10 . for some positive integer k. First. (⊠) Stronger result. Examples for practice Solution We will prove that 2 and the primes congruent to −1 modulo 8 are all the primes that do not have a multiple in the sequence. Furthermore. there is an i ∈ {2006. Solution Let √ √ √ ( x + 1)2 ( x − 1)2 3(x + 1) + 8030 x − = ⋅ f (x) = 2006 2009 2006 ⋅ 2009 6 Then f is an √ increasing √function2 of x > 0... (⊠) 6 Example 2. Consider next that a1 < 1742. Prove that there are indices i. 2009} such that aj ⩾ 174. whereas −1 is not a quadratic residue. take n = (p − 1)(p − q − 1) − 1. This clearly implies that the congruence 2n n2 + 1 = 0 does not have solution. Hence f (a1 ) ⩽ f (106 ) < 808. Assume first that a1 ⩾ 1742 . ] contains at most 808 integers.. .57. In this case. q < p such that p ∣ (q 2 + 2.. √ that ∣ iai − a1 ∣ > 174.) Now. i. and see that Fermat’s little theorem gives 2n n2 + 1 ≡ 2p−2 q 2 + 1 ≡ 0 (mod p).. let p be a prime of the form 8k + 3. that is. 2009} such that ai ∉ I . and therefore the 2 ( a1 − 1) ( a1 + 1) interval I = [ . and this proves the claim in this case. which proves the claim in this second case. −1 and 2 are not quadratic residues modulo p. Hence f (a1 ) ⩽ f (10 ) < 3. j such that ∣ iai − jaj ∣ ⩾ 1. Take n = (p − 1)(p − q). 2009} such that ai ∉ I . check q = 2 Fermat’s little theorem. Finally. a2 . and note that here 2 is a quadratic residue. ] contains at most three integers. which implies that −2 is a quadratic residue modulo p.. . completing the proof. Let a1 . such that p ∣ (q 2 + 1) (for example. Let p be such a prime and recall that in this case there is a positive integer p−1 !) . Thus √ √ √ √ jaj − a1 > 1836 ⋅ 174 − 1742 > 174. Next. consider the primes of the form 4k +1.. By the pigeonhole 2009 2006 √ √ principle.. consider a prime p = 8k − 1 for some positive integer k. .. then there is aj ∈ {1836. By the pi2009 1201 geonhole there is an i ∈ {1201.56 Chapter 2. q < p.. 2n n2 + 1 ≡ n2 + 1 ≡ q 2 + 1 ≡ 0 (mod p). ∣ iai − a1 ∣ > 1. by q. .. Let √ √ ( x + 174)2 ( x − 174)2 − ⋅ f (x) = 1201 2009 Then f is an increasing function of x > 0. which implies √ principle. . note that it is clear that 2 satisfies the statement.e.. thus yielding our conclusion. there exists a positive integer q. 1202. and therefore √ √ ( a1 − 174)2 ( a1 + 174)2 the interval I = [ . Clearly all such pairs (x. Let S(n) be the number of pairs of positive integers (x.. since ∣a1 ∣ = 2 < 1 directly results in a1 = 0. y) = 1.. the result is trivial. On the other hand.58. a2 . It is well known that τ (n) = (a1 + 1)⋯(ak + 1) ⇒ τ (n2 ) = (2a1 + 1)⋯(2ak + 1). Solution Let n = pa11 ... Assume that ∣a1 + 2a2 + . not all zero. .. the expansion will contain a summand 2r corresponding to each divisor of n with exactly r distinct prime factors. .1... 3 the ai being integers. if we substitute each pai i (with ai > 0) in the above expresion by 2. . . Arithmetic problems Example 2.n}.. the result will be ∑d∣n S(d). 3 First solution.59. + 2k−1 ak ∣ > 2k .. ∣a1 + 2a2 + . y) such that xy = n and gcd(x.. an be integer numbers. such that a1 + a2 + ⋯ + an = 0. Prove that ∑ S(d) = τ (n2 ). ∑ S(d) = (1 + 2 + 2 + ⋯ + 2)⋯(1 + 2 + 2 + ⋯ + 2) ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ d∣n ak 2′ s a1 2′ s = (1 + 2a1 ) (1 + 2a2 ) ⋯ (1 + 2ak ) = τ (n2 ). if for each subset M of {pa11 . y) = 1. i. hence S(n) = 2k . Example 2.... d∣n where τ (s) is the number of divisors of s.. + 2k−1 ak ∣ ⩽ 2k for all k ∈ {1. Now. The divisors of n are the summands in the expansion of the product (1 + p1 + p21 + ⋯ + pa11 )(1 + p2 + ⋯ + pa22 )⋯(1 + pk + ⋯ + pakk ). x then xy = n and gcd(x. 2. n}. 2..... We shall prove by induction that a1 = a2 = ⋯ = an = 0.57 2.e. pakk } we put x = ∏m∈M m and y = n . Let a1 . y) may be obtained in this way. For k = 1. Prove that for some k ∈ {1. 3 (⊠) . Therefore.pakk be the prime decomposition of n. Since we have at least m + n − 1 = m + 1 rooks... First solution We show that the least number of rooks is 2n − 1. Examples for practice If the result is true for i = 1. 3 The result follows. The thesis holds trivially when m + n ⩽ 6. Therefore ∣a1 + 2a2 + .. ∣ak ∣ ⩾ 1. Otherwise the total number of rooks is greater than or equal to 2k +2. 3 2 < 1.. By placing the 2k + 1 rooks on the (k + 1) × (k + 1) chessboard. .. Similarly there is at least one column containing one rook or no rooks.. there is at least one row containing one rook or no rooks. . Find the least number of rooks such that no matter how they are placed on an n × n chessboard there are two rooks that do not attack each other.. If the rooks are less than m+n−1. but at the same time they are under attack by third rook. The property does not hold for this displacement. Since ak is an integer.. a21 .. we can place them along the first column and along the first row but not at the top left corner (there are m + n − 2 places). (⊠) Second solution We will show that the least number of rooks such that the property holds in a m × n chessboard is m + n − 1.. Now we consider a m × n chessboard with m + n > 6. which is not true. that is. and 2k−1 ∣ak ∣ = 2k−1 > 2 k−1 2k ⋅2 = ⋅ 3 3 (⊠) Example 2. By the induction assumption.58 Chapter 2. By placing the rooks on a12 . + 2k−1 ak ∣ = 2k−1∣ak ∣. It follows that 2k − 1 rooks are sufficient for the (k + 1) × (k + 1) chessboard. yielding ∣ak ∣ ⩽ (⊠) a1 = a2 = . Let n be an integer greater than 1.. a1n . k − 1. This completes the solution. We can assume without loss of generality that n ⩾ m and therefore n > 3. We combine the undeleted parts of the (k + 1) × (k + 1) chessboard to obtain a k × k chessboard which contains at least 2k − 1 rooks. The standard algebraic notation of the n × n chessboard is used. and again ak = 0. the result is clear.. which is not true. 2. n > 1 and m > 1. an1 . n > 1 and m > 1. they are sufficient. For n = 2. then there is a row with at least two rooks. We now suppose that the result is true for n = k ⩾ 1. Second solution Let k be the smallest integer such that ak is non-zero. Select any such row and any such column and delete them from the (k + 1) × (k + 1) chessboard.. Otherwise we can cancel these two . n0 > 1. We will prove by induction that 2n − 1 rooks are sufficient. a31 . = ak−1 = 0 (if k > 1) and ∣ak ∣ > 0. and we assume that our thesis holds for any m0 × n0 chessboard such that m0 + n0 < m + n and m0 > 1. If there is at least another rook in the corresponding colums then the property holds. we see that 2n − 2 rooks are not sufficient. . a13 . then 2k ⩾ ∣a1 + 2a2 + ⋯ + 2k−1 ak ∣ = 2k−1 ∣ak ∣. All the ai are then zero.60. Select any 2k − 1 rooks. by the inductive hypothesis. then resulting in c = 1 and 2ab > (2a − b − c)(b − c) ⩾ abc + 1. (2a − b − 1)(b − 1) = ab + 1. b = n2 + 4n + 3 and c = n. c) of distinct positive integers such that abc + 1 divides more than one of the numbers (a − b)2 . the property holds in this smaller checkboard and therefore it holds also in the initial one.59 2. If abc + 1 divides (a − b)2 and (a − c)2 . (c − a)2 . however. Taking a = n3 + 6n2 + 10n + 4. Let n be a positive integer. 4. then a > b > b − c. Since m + n > m + (n − 2). (c − a)2 . The conclusion of part b) follows. If abc + 1 divides (b − c)2 . and abc + 1 ⩽ (b − c)2 < ab. b. The conclusion of part a) follows.61. (⊠) Example 2. and since 2a > 2a − b − c > 0 and b > b − c > 0. c) = c = n. m > 1 and n − 2 > 1. clearly min(a. in none of these cases does abc + 1 divide (a − b)2 or (a − c)2 . (b − c)2 . 8. then it divides (a − c)2 − (a − b)2 = (2a − b − c)(b − c) < 2ab. b. 6. (◻) Assume now wlog that a > b > c. then b = 3. the following relation holds: (n3 + 6n2 + 10n + 4)(n2 + 4n + 3)n + 1 = n6 + 10n5 + 37n4 + 62n3 + 46n2 + 12n + 1 = (n3 + 5n2 + 6n + 1)2 = (n3 + 6n2 + 10n + 4) − (n2 + 4n + 3)2 . c) = n and abc + 1 divides one of the numbers (a − b)2 . b.1. Arithmetic problems columns obtaining a m × (n − 2) chessboard with at least m + (n − 2) − 1 rooks. (⊠) . 9. b) there is no triple (a. Prove that a) there are infinitely many triples (a. clearly impossible. Solution For any positive integer n. c) of distinct integers such that min(a. (b − c)2 . yielding respectively a = 9. b. or a = b2 4 = b+2+ ⋅ b−2 b−2 Since b − 2 divides 4. and abc + 1 divides (a − b)2 = abc + 1. 3 Example 2. a2 + d2 3 x2 + 1 3 a3 + d3 ⩽( ) ⇔ x3 + 1 ⩽ 2 ( ) 2 a+d x+1 a = x. Examples for practice 2. d be positive real numbers. and we are done. so we obtain that a3 + d3 a2 + d2 3 ⩽( ) 2 a+d with equality if and only if a = d. b.2 2. b. Prove that Solution ( (⊠) a+b 3 c+d 3 a2 + d2 3 b2 + c2 3 ) +( ) ⩽( ) +( ) 2 2 a+d b+c Using the well known inequality ( x + y 3 x3 + y 3 ) ⩽ . we have that 2 2 c + d 3 a3 + b3 c3 + d3 a+b 3 ) +( ) ⩽ + ( 2 2 2 2 with equality if and only if a = b and c = d. 4 4 3 2t = 1 a contradiction. c.1 Algebraic problems Junior problems Example 2. Now notice that if we let But.63.62.2. Let a. Let a. we have d (2. Analogously b2 + c2 3 b3 + c3 ⩽( ) 2 b+c .13) ⇔ x6 − 3x5 + 3x4 − 2x3 + 3x2 − 3x + 1 ⩾ 0 ⇔ (x − 1)4 (x2 + x + 1) ⩾ 0 is true. Thus b2 + c2 ⩾ .60 Chapter 2. c be real numbers such that ∣a∣3 ⩽ bc. thus 3 1 ⩽ a6 + b6 + c6 ⩽ (bc)2 + b6 + c6 27 = (b2 + c2 )3 + (bc)2 (1 − 3(b2 + c2 )) ⩽ (b2 + c2 )3 + ( b2 + c2 2 ) (1 − 3(b2 + c2 )) 2 (b2 + c2 )2 (4(b2 + c2 ) + 1 − 3(b2 + c2 )) 4 2 ( 13 ) (b2 + c2 )2 1 1 2 2 (1 + (b + c )) < = (1 + ) = .11) (2. 1 1 ⋅ Prove that b2 + c2 ⩾ whenever a6 + b6 + c6 ⩾ 3 27 First solution 1 Assume the contrary that b2 + c2 < ⋅ Then note that a6 ⩽ (bc)2 . (⊠) . Second solution By expanding the terms (x2 + y 2 + z 2 )2 and (xy + yz + zx)2 and equating the terms on both sides leads to 1 1 x4 + y 4 + z 4 + (x2 y 2 + y 2z 2 + z 2 x2 ) − xyz(x + y + z) ⩾ 0. Hence it is enough to prove that x4 + y 4 + z 4 = 0 which is obvious.14) ⩾ 2 (x2 y 2 + y 2 z 2 + z 2 x2 ) + xyz(x + y + z) ⩾ x2 y 2 + y 2z 2 + z 2 x2 + 2xyz(x + y + z) 2 = (xy + yz + zx)2 ⩾ (xy + yz + zx)2 = RHS(2. 3 (⊠) Remark. 3 3 By the AM-GM inequality we have x2 y 2 + y 2z 2 + z 2 x2 ⩾ xyz(x + y + z). (⊠) Example 2.2. Hence the original inequality is true with equality if and only if a = b = c = d. for any real u. 3 First solution The original inequality will follow from the following sharper inequality 2 2 (x2 + y 2 + z 2 ) + xyz(x + y + z) − x2 y 2 + y 2z 2 + z 2 x2 ⩾ (xy + yz + zx)2 . y. Prove that 2 (x2 + y 2 + z 2 )2 + xyz(x + y + z) ⩾ (xy + yz + zx)2 + (x2 y 2 + y 2 z 2 + z 2 x2 ). v. 3 (2. x2 y 2 + y 2z 2 + z 2 x2 ⩾ xyz(x + y + z). (x2 + y 2 + z 2 ) ⩾ 3 (x2 y 2 + y 2 z 2 + z 2 x2 ) 2 and Therefore. w we have Then u2 + v 2 + w 2 ⩾ uv + vw + wu ⇔ (u + v + w)2 ⩾ 3(uv + vw + wu) ⇔ (u − v)2 + (v − w)2 + (w − u)2 ⩾ 0.61 2. Equality in (2. Let x. LHS(2.14). z be real numbers.14) occurs if and only if x = y = z and in original inequality equality occurs if and only if x = y = z = 0.64. Algebraic problems with equality if and only if b = c.12) Indeed. Equality occurs if and onyl if x = y = z = 0. sec a sec b π and sin2 b + cos 2a ⩾ . b1 ) H(an . bn ) S= = an + bn (a1 + ⋯ + an )(b1 + ⋯ + bn ) a1 + b1 +⋯+ ] [ (a1 + ⋯ + an ) + (b1 + ⋯ + bn ) a1 b1 an bn (a1 + ⋯ + an )(b1 + ⋯ + bn ) 1 1 1 1 [ +⋯ + +⋯ ] (a1 + ⋯ + an ) + (b1 + ⋯ + bn ) a1 an b1 bn 1 1 1 1 +⋯ + +⋯ a1 an b1 bn ⩾ n2 = 1 1 + a1 + ⋯ + an b1 + ⋯ + bn The last inequality is true because 1 1 1 + ⋯ ⩾ n2 . b ∈ (0. For n ⩾ 2. (c) 1 = cos x. . Examples for practice Example 2. y) = 2xy be the harmonic mean of the positive real numbers x+y x and y. . find the greatest constant C such that for any positive real numbers a1 . Let H(x. ) such that sin2 a + cos 2b ⩾ 2 2 2 Prove that 1 cos6 a + cos6 b ⩾ ⋅ 2 Solution We will use the following well-known trigonometric identities (a) sin2 x = 1 − cos2 x. b1 + ⋯ + bn ) H(a1 . . (b) cos 2x = 2 cos2 x − 1. Let a.62 Chapter 2. b1 + ⋯ + bn ) +⋯+ H(a1 . Example 2. bn ) H(a1 + ⋯ + an . .66. . . b1 + ⋯ + bn ) H(a1 + ⋯ + an . b1 ) H(an . .13) . Equality occurs if and only if a1 = ⋯ = an and b1 = ⋯ = bn so C = n2 .65. . bn the following inequality holds Solution Let S= Note that 1 1 C ⩽ +⋯+ ⋅ H(a1 + ⋯ + an . sec x The inequalities can be written as cos2 b cos a − cos3 a ⩾ 1 2 (2. b1 . a1 an a1 + ⋯ + an 1 1 1 + ⋯ ⩾ n2 b1 bn b1 + ⋯ + bn by the AM-HM inequality. an . 14) and adding them we get cos6 a + cos6 b ⩾ 1 ⋅ 2 Example 2. . ).14) π The signs of the inequalities are preserved because cos x is positive when x ∈ (0. 2 Now by squaring both sides of (2. y. Prove that √ √ √ x+y y+z z+x √ 1 + + + 4 xyz ⩽ (abc + 5a + 5b + 5c). Let x. Solve in positive real numbers the system of equations Solution We have ⎧ ⎪ x1 + x2 + ⋯ + xn ⎪ ⎪ ⎨1 1 1 1 ⎪ + + ⋯ + + ⎪ ⎪ xn x1 x2 ⋯xn ⎩ x1 x2 (⊠) =1 = n3 + 1. Algebraic problems and cos2 a cos b − cos3 b ⩾ 1 2 (2. 3a ⩾ ax y z + + . 2 3 3 3 (⊠) Example 2. For n = 2 we get ⎧ ⎪ x1 + x2 ⎪ ⎪ ⎨1 1 1 ⎪ + + ⎪ ⎪ ⎩ x1 x2 x1 x2 =1 =9 1 2 2 1 which is (n. . x2 ) ∈ {(2. bc c b . (2. 2 2 2 4 Solution From the given condition. c. )} . b. If n = 1 we get a contradiction.67. 1 1 1 n2 + ⋯+ ⩾ = n2 x1 x2 xn x1 + x2 + ⋯ + xn and Thus.63 2.13) and (2. x1 . z be nonnegative real numbers such that ax + by + cz = 3abc for some positive real numbers a.2. 1 1 = nn ⩾ x1 + x2 + ⋯ + xn n x1 x2 ⋯xn ) ( n n3 + 1 ⩾ nn + n2 which implies that n ⩽ 2.68. ) . a2 b2 + b2 c2 = 2b2 . Now x ≠ y ≠ z ≠ 0 and after multiplying all three equations we get 3(x2 − xy + y 2 )3(y 2 − yz + z 2 )3(z 2 − zx + x2 ) =1 (x2 + xy + y 2 )(y 2 + yz + z 2 )(z 2 + zx + x2 ) which can be written as (1 − 2xy 2yz 2zx 1 ) (1 − ) (1 − )= ⋅ 2 2 2 (x − y) + 3xy (y − z) + 3yz (z − x) + 3zx 27 It is not difficult to see that each factor of the LHS is greater than a contra. c2 a2 + a2 b2 = 2a2 . z ≠ 0. If we assume that y = −z then the first equation becomes 4x4 = 6x4 . (⊠) Example 2. implying a = b = c = 1 and x = y = z = 1. z + x = 2b2 and abc = ax by cz = = bc ca ab This implies b2 c2 = c2 a2 = 2c2 .64 Chapter 2. contradiction.diction. y + z = 2a2 . then x = z or y = −z. If a = b = c = x = y = z = 1. as well as the condition of the problem. Because the given system is symmetric we can assume that x. then it is clear that y = z = 0. hold. Solution Without loss of generality assume that x = 0. Examples for practice 3b ⩾ x by z + + . Assume that x = y. the equality. that is b2 + a2 = c2 + b2 = a2 + c2 = 2. So the only possible solution is x = y = z. Solve in real numbers the system of equations ⎧ (x + y)(y 3 − z 3 ) = 3(z − x)(z 3 + x3 ) ⎪ ⎪ ⎪ ⎪ ⎨(y + z)(z 3 − x3 ) = 3(x − y)(x3 + y 3 ) ⎪ ⎪ ⎪ 3 3 3 3 ⎪ ⎩(z + x)(x − y ) = 3(y − z)(y + z ). 1 which leads to 3 (⊠) . c ca a x y cz 3c ⩾ + + ⋅ b a ab Then 3(a + b + c) ⩾ hence x+y y+z z+x ax by cz + + +( + + ) c a b bc ca ab y+z z+x x+y + 2c) + ( + 2a) + ( + 2b) c a b ax by cz + abc + + + bc ca ab√ √ √ √ ⩾ 2 2(x + y) + 2 2(y + z) + 2 2(z + x) + 4 4 xyz abc + 5(a + b + c) ⩾ ( and the conclusion follows. y. The equality holds if and only if x + y = 2c2 .69. y. z) is 1 1 .72. otherwise the right hand side would be 0 which is a contradiction of (* * *). If x. y. b − c. and thus. respectively and let S = w + x + y + z. occurring for example if x = y = z = ⋅ 2 3 Example 2. we have that. So that. Algebraic problems Example 2. 1 1 1 1 w 3 + x3 + y 3 + z 3 = 0 (∗) and + + + ≠ 0 (∗∗) x y z u Since a.70.71. y. and we’re done. c − d.65 2. z) ⩽ + + x+y y+z z+x = x+y y+z z+x x+y+2 1 + + = = ⋅ 4 4 4 2 2 Therefore the maximum of E(x. find the maximum of E(x. z. b. x. y. z > 0 and x + y + z = 1. z) = xy yz zx + + ⋅ x+y y+z z+x Solution By the AM − GM inequality. d − a by w. c and d are distinct. (∗ ∗ ∗) S 3 = 6(wxy + wxz + wyz + xyz) + 3(w 2 + x2 + y 2 + z 2 )S ⇒ S(S 2 − 3(w 2 + x2 + y 2 + z 2 )) = 6(wxy + wxz + wyz + xyz) Which implies that S ≠ 0. ∣a∣ ⩽ ∣a2 47 52 and ∣b∣ ⩽ 2 ⋅ 2 − 3ab ∣ ∣b − 3a2 ∣ (⊠) . ( x+y 2 y+z 2 z+x 2 ) ) ) ( ( 2 2 2 E(x. using (*). Solution √ √ √ √ 3 3 3 3 Let a − b. d be distinct real numbers such that Prove that 1 1 1 1 √ +√ +√ +√ ≠0 3 3 3 3 a−b b−c c−d d−a √ √ √ √ 3 3 3 3 a − b + b − c + c − d + d − a ≠ 0. c. we can conclude that w. y. (⊠) Example 2. x y z u Furthermore. Let a. x. 1 1 1 1 wxy + wxz + wyz + xyz = wxyz ( + + + ) ≠ 0. z ≠ 0. b. y.2. Let a and b be rational numbers such that Prove that a2 + b2 ⩽ 17. Define x = a2 . 1. Observing that X + iY = (a + ib)3 (easily checked). absurd. since otherwise 3 would be the square of a rational. z) as follows: f n(x. the inequality is equivalent to ∑(a4 c2 b + c4 b3 ) ⩾ ∑ 2a3 b2 c cyc but this follows from the AM-GM: Example 2. b. Let a. Examples for practice First solution Let X = a3 −3ab2 and Y = 3a2 b−b3 . c be positive real numbers.66 Chapter 2. y. hence x(x − 3y)2 ⩽ 472 and y(3x − y)2 ⩽ 522 . we deduce that a + b = ∣a + ib∣ = ((a + ib) ) = ((X + Y ) ) = (X 2 + Y 2 ) 3 = (472 + 522 ) 3 . y. 2 2 2 3 2 3 2 2 1 2 2 3 1 Since 472 + 522 = 4913 = 173 . Adding these two inequalities results in (x + y)3 ⩽ 472 + 522 = 173 . it follows that a2 + b2 ⩽ 17. y = b2 . a+b cyc ∑ Solution Clearing the denominators. 2 (x − y)z n+2 + (y − z)xn+2 + (z − x)y n+2 ⋅ (x − y)(y − z)(x − z) (⊠) Prove that fn (x. y.73. for example when a = ±1 and b = ±4. 1 (⊠) Second solution Note that a = b = 0 clearly satisfies a2 + b2 < 17. Prove that a2 b2 (b − c) ⩾ 0. Note that equality is reached whenever equality holds simultaneously on both inequalities given as condition in the problem statement. we have ∣X∣ ⩽ 47 and ∣Y ∣ ⩽ 52. Let fn (x. z) = xn+2 y n+2 z n+2 + + (x − y)(x − z) (y − x)(y − z) (z − y)(z − x) . (⊠) Example 2. Solution It is easily verified that 1 1 1 + + = 0. (x − y)(x − z) (y − x)(y − z) (z − y)(z − x) This allows one to transform fn (x.74. or x + y = a2 + b2 ⩽ 17. 1) for all positive integers n. z) = cyc (a4 c2 b + c4 b3 ) ⩾ a3 b2 c. y. From the hypothesis. therefore a2 − 3b2 and b2 − 3a2 are nonzero rationals. z) can be written as a sum of monomials of degree n and find fn (1. This answers the first part of the question. Let a. (n + 1)(n + 2) (n + 1) + n + ⋯ + 2 + 1 = ⋅ (⊠).76.75. fn (x. c be nonzero real numbers such that ab + bc + ca ⩾ 0. As for the second part. Algebraic problems = = = = −xn+2 y n+2 z n+2 −xn+2 + + + (y − x)(y − z) (z − y)(z − x) (y − x)(y − z) (z − y)(z − x) z n+2 − xn+2 y n+2 − xn+2 + (y − x)(y − z) (z − y)(z − x) y n+1 + y n x + ⋯ + xn y + xn+1 z n+1 + z n x + ⋯ + xn z + xn+1 + (y − z) (z − y) (y n+1 − z n+1 ) + x(y n − z n ) + ⋯ + xn (y − z) y−z and finally fn (x. that is. 1. 1) is just the total number of monomials in the above sum. fn (1. Thus. (⊠) Example 2. find the minimum possible value of x + y. y. If x and y are positive real numbers such that √ √ (x + x2 + 1) (y + y 2 + 1) = 2012. Solution √ z2 − 1 ⋅ Let z = x + x2 + 1. b. z) is the sum of all monomials xa y bz c over all triples (a.67 2. z) = (y n +y n−1 z+⋯+yz n−1 +z n )+x(y n−1 +y n −2z+⋯+yz n−2 +z n−1 )+⋯+xn−1 (y+z)+xn . c) of nonnegative integers satisfying a + b + c = n. y. We have z > 0 and x = 2z (i) . b.2. 2 Example 2. Prove that a2 ab bc ca 1 + 2 2+ 2 ⩾− ⋅ 2 2 +b b +c c +a 2 Solution We have ab (a + b)2 1 3 3 ab ( ( = + ) − = )− ∑ ∑ 2 2 2 2 2 2 2 2 cyc 2 (a + b ) 2 cyc a + b cyc a + b ∑ ⩾ ∑( cyc =1+ (a + b)2 3 2 (a2 + b2 + c2 ) + 2(ab + bc + ca) 3 ) − = − 2 (a2 + b2 + c2 ) 2 2(a2 + b2 + c2 ) 2 (ab + bc + ca) 3 1 − ⩾− 2 2 2 (a + b + c ) 2 2 where in the last step we have used the fact that ab + bc + ca ⩾ 0. 68 Chapter 2.M inequality we get ∑ (b2 ¿ Á (a2 − 1)2 (b2 − 1)2 (c2 − 1)2 − 1) 3 Á À ⩾ 3 = 3.(2. c be real numbers greater than 1 such that c+a a+b b+c + 2 + 2 ⩾ 1. 2 a −1 b −1 c −1 Prove that bc + 1 2 ca + 1 2 ab + 1 2 10 ( 2 ) +( 2 ) +( 2 ) ⩾ ⋅ a −1 b −1 c −1 3 First solution Observe that ( ( ( Therefore ∑( Now observe that b + c 2 (b2 − 1)(c2 − 1) bc + 1 2 ) − ( ) = .16) and by A.M-G. Examples for practice From hypothesis y + From (i) and (ii).15) (2. a2 − 1 a2 − 1 (a2 − 1)2 ca + 1 2 c + a 2 (c2 − 1)(a2 − 1) ) − ( ) = . Let a. b2 − 1 b2 − 1 (b2 − 1)2 ab + 1 2 a + b 2 (a2 − 1)(b2 − 1) ) − ( ) = ⋅ c2 − 1 c2 − 1 (c2 − 1)2 (b2 − 1)(c2 − 1) b+c 2 bc + 1 2 ) = ( ) + ∑ ∑ a2 − 1 a2 − 1 (a2 − 1)2 ∑( bc + 1 2 ) ⩾ a2 − 1 (∑ bc + 1 2 ) 1 a2 − 1 ⩾ ⋅ 3 3 (2. we get y = ⋅ y2 + 1 = z 2 ⋅ 2012 ⋅ z (ii) √ 2011 2012 2011 2012 z 2 − 1 20122 − z 2 + = (z + )⩾ ⋅ x+y = 2z 2 ⋅ 2012 ⋅ z 2 ⋅ 2012 z 2012 The equality occurs for z = obtain 2012 or equivalently z 2 = 2012.17) (⊠) .77. b.15).16) and (2.17) we obtain ∑ ( 2 a −1 3 3 (2. Then from (i) and (ii) we z 2011 x=y= √ 2 2012 √ 2011 2012 ⋅ So min(x + y) = 2012 (⊠) Example 2. (a2 − 1)2 (a2 − 1)2 (b2 − 1)2 (c2 − 1)2 − 1)(c2 1 10 bc + 1 2 ) ⩾3+ = ⋅ By virtue of (2. √ 2012 20122 − z 2 . Find the minimum of 1 1 1 + + ⋅ a b c First solution Note that a + b + c + 2 = abc implies Then 3 + 2(a + b + c) + (ab + bc + ca) = abc + ab + bc + ca + a + b + c + 1..78. By the Cauchy-Schwarz inequality 1 ∑ (1 + ) ∑ a cyc cyc hence the minimum is 1 3 9 1 ⩾ 9 ⇔ ∑ (1 + ) ⩾ ⇔ ∑ ⩾ ⋅ 1 a 2 2 cyc a cyc 1+ a 1 3 ⋅ 2 (⊠) Second solution 1 1 1 Let x = . we obtain 1 =1⇔∑ cyc a + 1 cyc ∑ 1 1+ 1 a = 2. a2 . Let a1 . z the inequality is x + y + z = 1 and y z x + + y+z x+z x+y 3 That the minimum of the above expression is is the content of Nesbitt’s inequality 2 which is well known. Prove that a2 an a1 + ⋯ + an a1 ⋅ + +⋯+ ⩽ 2 2 2 (1 + a1 ) (1 + a1 + a2 ) (1 + a1 + ⋅ ⋅ ⋅ + an ) 1 + a1 + ⋯ + an . b. x x y y z z In terms of the variable x. c be positive real numbers such that a + b + c + 2 = abc.c = = .z = ⋅ By trivial algebra we observe that a + b + c + 2 = abc 1+a 1+b 1+c is equivalent to x + y + z = 1 and then a= 1−x y+z 1−y x+z 1−z y+x = .y = .69 2. One of the many proof available is x y z (x + y + z)2 1 3 + + ⩾ = ⩾ y + z x + z x + y 2(xy + yz + zx) 2(xy + yz + zx) 2 having employed Cauchy-Schwarz. (⊠) 3 Example 2. Algebraic problems Example 2. Thus we have xy + yz + zx ⩽ 1 which is obvious. an be positive real numbers.. Let a.. (a + 1)(b + 1)(c + 1) = ∑ ((a + 1)(b + 1)) cyc and dividing by the left-hand side.b = = . .79.2. y. Let a ⩾ b ⩾ c > 0. transforms into a2n ⩾ 0. Examples for practice Solution We will prove the result by induction. One may get arbitrarily close to equality when a1 . b. the result is equivalent to a1 a1 . Example 2. the proof is completed. Prove that 1 1 1 (a − b + c) ( − + ) ⩾ 1. ⩽ 2 (1 + a1 ) 1 + a1 clearly true with a strict inequality since 1 + a1 > 1. Let a. ⩾ n ∑k=1 ai k=1 ai n ∑ we have 322 1 1 1 1 1 1 +⋯+ + +⋯+ + +⋯+ ⩾ a a b b c c 10a + 11b + 11c ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ 10 times 11 times 11 times . the inequality being strict for all n.81. a b c Solution We write the inequality as follows 1 1 1 1 − + ⩾ ⋅ a b c a−b+c It is equivalent to (⊠) a+c a+c ⩾ ac b(a − b + c) Therefore it is enough to check that ac ⩽ b(a − b + c) or (b − a)(b − c) ⩽ 0. . 1 n2 . then for the result to be true with a strict inequality for n. and which after performing the products in both sides and simplifying. (⊠) Example 2. which is in turn arbitrarily close to zero. a2 . If the result is true with a strict inequality for n − 1.. both sides being arbitrarily close to a1 +a2 +⋯+an .. c be positive real numbers. For n = 1. The conclusion follows. trivially true.80. an all tend to 0. Prove that 1 1 1 1 1 1 + + ⩽ + + ⋅ 10a + 11b + 11c 11a + 10b + 11c 11a + 11b + 10c 32a 32b 32c Solution Recall the well-known inequality. it suffices to show that an sn sn − an + ⇔ (sn − an )(1 + sn )2 ⩽ (sn + s2n − an )(1 + sn − an ) ⩽ 1 + sn − an (1 + sn )2 1 + sn where we have denoted sn = a1 +a2 +⋯+an .. Hence. The last inequality is true due to the given condition.70 Chapter 2. 2. y. b ∶= . ) (a3 + b3 + c3 ) ⩾ ( 1 1 1 + 2 + 2 ) (a + b + c) 2 a b c (a2 + b2 + c2 ) (a3 + b3 + c3 ) ⩾ (a2 b2 + b2 c2 + c2 a2 ) (a + b + c). z be positive real numbers with x ⩽ 2. c ∶= . y ⩽ 3 and x + y + z = 11. z be positive real numbers. cyc cyc cyc cyc which turns out to be just the immediate ∑ a5 ⩾ ∑ ab2 c2 = abc(ab + bc + ca). Algebraic problems which is equivalent to 11 11 32 10 + + ⩾ ⋅ 32a 32b 32c 10a + 11b + 11c Similarly 10 11 32 11 + + ⩾ ⋅ 32a 32b 32c 10a + 11b + 11c and 11 11 10 32 + + ⩾ ⋅ 32a 32b 32c 10a + 11b + 11c Adding the three inequalities we get the desired result.71 2. (⊠) Example 2. ∑ a5 + ∑ a3 (b2 + c2 ) ⩾ ∑ a3 (b2 + c2 ) + ∑ ab2 c2 . cyc cyc which can be seen for example as a consequence of the AM-GM Inequality. (⊠) Example 2.82. Let x. + ⩽Á y z z cyc cyc y cyc cyc x so it would suffice to prove that (x2 y 2 + y 2 z 2 + z 2 x2 ) ( 1 1 1 1 1 1 + 3 + 3 ) ⩾ (x2 + y 2 + z 2 ) ( + + ) 3 x y z x y z 1 1 1 Now. Prove that √ √ √ √ 1 1 1 1 1 1 1 1 1 + +y + +z + ⋅ 2(x2 y 2 + y 2z 2 + z 2 x2 ) ( 3 + 3 + 3 ) ⩾ x x y z y z z x x y Solution Note that by the Cauchy-Schwarz Inequality we have that √ ∑x cyc ¿ ¿ Á 1 1 Á 1 1 À∑ x2 ∑ ( + ) = Á À2 ∑ x2 ∑ 1 . Let x. let a ∶= . Prove that xyz ⩽ 36.83. y. .e. then the inequality to be proven can be rewritten as x y z ( 1 a2 b2 + 1 b2 c2 + 1 c2 a2 which is equivalent with i. 6 3 6 3 We have equality if and only if x = 2. and the problem is solved. Prove that (a + b)2 c2 + ⩾ 4b. (and by AM-GM inequality) . The lemma can be proved by straight forward calculations. α. β be positive real numbers then a2 b2 (a + b)2 + ⩾ ⋅ α β α+β Proof. b. we have xyz = = 1 3x + 2y + z 3 1 ⋅ (3x) ⋅ (2y) ⋅ z ⩽ ( ) 6 6 3 1 2x + y + 11 3 1 2 ⋅ 2 + 3 + 11 3 ( ) ⩽ ( ) = 36. z = 6. Second solution By the Cauchy-Schwarz inequality. Lemma 2. c a a+c (a + b + c)2 ⩾ 4b.84. y = 3.72 Chapter 2. Let a. b. Examples for practice Solution By AM-GM inequality. a+c The preceding inequality is equivalent to (a − b + c)2 ⩾ 0. (⊠) Example 2. c be positive real numbers. Let a. c a First solution We will use the following lemma. we have ⎛√ c LHS ⋅ (c + a) ⩾ ⎝ √ √ (a + b)2 √ c2 ⎞ + a c a⎠ = (a + b + c)2 = (b + (a + c))2 √ 2 = (2 b ⋅ (a + c)) = 4b ⋅ (a + c) = RHS ⋅ (a + c). Based on the lemma we have that so it suffices to prove that (⊠) (⊠) (a + b)2 c2 (a + b + c)2 + ⩾ . (⊠) Third solution After clearing denominators we need to prove that f (b) = ab2 + (2a2 − 4ac)b + a3 + c3 ⩾ 0.73 2. Considering f as a quadratic function in b we can easily check that the discriminant = −4ac(2a − c)2 ⩽ 0. b. (⊠) (2a2 − 4ac)2 − 4a(a3 + c3 ) Fourth solution The proposed inequality follows from: a3 + 2a2 b + ab2 + c3 − 4abc ab2 + 2(a − 2c)b + a2 + c3 (a + b)2 c2 + − 4b = = c a ac ac = = = = = a [b2 + 2(a − 2c)b + (a − 2c)2 − (a − 2c)2 + a2 ] + c3 ac a [(b + a − 2c)2 + 4ac − 4c2 ] + c3 ac a(a + b − 2c)2 + 4a2 c − 4ac2 + c3 ac a(a + b − 2c)2 + c(4a2 − 4ac + c2 ) ac (a + b − 2c)2 (2a − c)2 + ⩾ 0. Let a. 2a2 + b2 + 3 4a + 2b b b c c ⩾. ⩽ . we have 2b2 a a ⩽ . c be positive real numbers. we conclude that f (b) ⩾ 0 for all b ⩾ 0. Since the leading coefficient of this quadratic function is positive.2. Algebraic problems Dividing both sides by (a + c) we obtain the desired result. 2 2 + b + 3 2b + c + 3 2c + a + 3 cyc 4a + 2b (i) . c a (⊠) Example 2. Prove that 2a2 b c 1 a + 2 2 + 2 ⩽ ⋅ 2 2 + b + 3 2b + c + 3 2c + a + 3 2 Solution We have 2a2 + b2 + 3 = 2(a2 + 1) + (b2 + 1) ⩾ 4a + 2b ⇒ Similarly. 2 2 2 +c +3 4b + c 2c + a + 3 4c + a and we obtain 2a2 a b c a + 2 2 + 2 ⩽∑ .85. b. Let a.GM inequality. (ii) clearly true by the AM . a+b+c a + 2b b + 2c c + 2a Because from the AM-GM inequality (a + b + c)2 ⩾ 3(ab + bc + ca). 2 cyc 2a + b cyc 4a + 2b ∑ Which is equivalent to ab2 + bc2 + ca2 ⩾ 3abc.86. Prove that (⊠) ab bc ca a+b+c + + ⩽ ⋅ 3a + 4b + 2c 3b + 4c + 2a 3c + 4a + 2b 9 Solution The inequality is equivalent to 9ab 9bc 9ca + + ⩽ a + b + c. Examples for practice Now we prove a 1 a ⩽ ⇔∑ ⩽ 1. c be positive real numbers. a + 2b b + 2c c + 2a 3 But from AM-GM inequality (a + 2b)(b + 2a) ⩾ 9ab ⇔ ab b + 2a ⩽ ⋅ a + 2b 9 Adding the other two similar inequalities we get that bc ca b + 2a c + 2b a + 2c a + b + c ab + + ⩽ + + = ⋅ a + 2b b + 2c c + 2a 9 9 9 3 The proof is complete. (⊠) . From (i) and (ii) follows that 2a2 b c 1 a + 2 2 + 2 ⩽ ⋅ 2 2 + b + 3 2b + c + 3 2c + a + 3 2 Example 2.74 Chapter 2. 3a + 4b + 2c 3b + 4c + 2a 3c + 4a + 2b From Cauchy-Schwartz inequality it is known that 9ab ab ab ab ⩽ + + ⋅ 3a + 4b + 2c a + b + c a + b + c a + 2b Adding the other 2 similar inequalities we get that it suffices to show that 2(ab + bc + ca) ab bc ca + + + ⩽ a + b + c. we conclude that 2(ab + bc + ca) 2(a + b + c) ⩽ ⋅ a+b+c 3 So it is enough to prove that ab bc ca a+b+c + + ⩽ . a). n i=1 On the other hand. 1 1 bc − 1 3 ca − 1 3 ab − 1 3 ⩾ (ab ) + (bc ) + (ca ) ab bc ca √ √ √ 3 3 3 = ab − 1 + bc − 1 + ca − 1 = RHS(2. c). Let a1 . c ⩾ 1 be real numbers such that a + b + c = 2abc. we ab bc ca 1 ab − 1 bc − 1 ca − 1 3 + + ) LHS(2. ab bc ca ab bc ca get Applying H¨ older’s inequality to the triples (a. Let a. (b. (⊠) . (2.18).88.. an be positive real numbers such that ∑ 2 i=1 ai Prove that ∑ ai aj ⩽ 1⩽i<j⩽n 1 = n − 1. c.87. Algebraic problems n Example 2. we have n a2i (∑ni=1 ai ) 1 = ⩾ ∑ 2 2 n + ∑ni=1 a2i i=1 ai + 1 i=1 ai + 1 2 1=n−∑ n Thus 2 (∑ ai ) − ∑ a2i ⩽ n.75 2.. b. ). .18) = (a + b + c) ⋅ (b + c + a) ⋅ ( ab bc ca 1 3 1 3 1 and we are done. . Prove that √ √ √ √ 3 3 3 3 (a + b + c)2 = ab − 1 + bc − 1 + ca − 1. b. we have n i=1 2 2 ∑ ai aj = (∑ ai ) − ∑ a2i ⋅ n i=1 1⩽i<j⩽n Therefore n i=1 ∑ ai aj ⩽ 1⩽i<j⩽n (⊠) n ⋅ 2 1 Equality holds when a1 = a2 = ⋯ = an = √ ⋅ n−1 Example 2. +1 n ⋅ 2 Solution From the hypothesis and the Cauchy-Schwarz Inequality..18) First solution We write the given equality as follows 1 1 1 ab − 1 bc − 1 ca − 1 + + =2⇒ + + = 1.2. ( ab − 1 bc − 1 ca − 1 . 76 Chapter 2. a+b+c =∑ cyc Therefore cyc √ √ 3 3 (a + b + c)2 = ∑ ab − 1 and we are done. Let a and b be positive real numbers. 2 + b + 1 b + c + 1 c + a2 + 1 Prove that ab + bc + ca ⩽ 3. Let a. Prove that a6 + b6 a4 + b4 a2 + b2 ⩾ ⋅ ⋅ a4 + b4 a3 + b3 a + b First solution The above inequality is equivalent to (a6 + b6 )(a3 + b3 )(a + b) ⩾ (a4 + b4 )2(a2 + b2 ). (⊠) . Example 2.89.90. we have 4 or Similarily. Thus ab + bc + ca ⩽ 3 and we are done. ⩽ ⋅ b2 + c2 + 1 (a + b + c)2 c2 + a2 + 1 (a + b + c)2 1⩽ or equivalent to 1 2 + c2 ⩽ a2 + b2 + 1 (a + b + c)2 1 1 1 6 + a2 + b2 + c2 + + ⩽ a2 + b2 + 1 b2 + c2 + 1 c2 + a2 + 1 (a + b + c)2 (a + b + c)2 ⩽ 6 + a2 + b2 + c2 a2 + b2 + c2 + 2(ab + bc + ca) ⩽ 6 + a2 + b2 + c2 . Solution By applying the Cauchy-Scharz inequlity. we obtain Therefore 1 2 + a2 1 2 + b2 ⩽ . c be positive real numbers such that a2 1 1 1 + 2 2 + 2 ⩾ 1. (⊠) cyc Example 2. b. Examples for practice Second solution a+b−c thus by the AM-GM inequality We are given that 0 ⩽ (ab − 1) ⩽ 2c √ (a + b − c) + a + b ⩾ ∑ 3 ab(a + b − c) 3 cyc cyc √ √ = ∑ 3 2abc(ab − 1) = ∑ 3 (a + b + c)(ab − 1). Multiplying these inequalities. . a4 + b4 = y . Algebraic problems Using the Cauchy-Scharz inequality. the conclusion follows. ab = z .2.2 (⊠) Senior problems Example 2. xn .91. It follows that x1 + ⋯ + xn = (2x1 − x2 ) + ⋯ + (2xn − x1 ) ⩽ It then suffices to show that 4x31 4x3n + ⋯ + ⋅ (x1 + x2 )2 (xn + x1 )2 1 4 ⩽ (x1 + x2 )2 (a + b)2 (ax1 + bx2 )(ax2 + bx1 ) (2. Our inequality transforms to x x + z2t ⩾ y y + zt or after some algebra x ⩾ yz. Let x1 . b > 0. a2 + b2 = t. . Prove that the following inequality holds x31 x3n x1 + ⋯ + xn +⋯+ ⩾ ax1 + bx2 )(ax2 + bx1 ) (ax1 + bxn )(axn + bx1 ) (a + b)2 First solution We have 4x31 − (x1 + x2 )2 (2x1 − x2 ) = 2x31 − 3x21 x2 + x32 = (x1 − x2 )2 (2x1 + x2 ) ⩾ 0 with equality if and only if x1 = x2 .2. If we back substitute we obtain which is exactly the same as a6 + b6 ⩾ (a4 + b4 )ab (a5 − b5 )(a − b) = 0 and thus we are done. we have (a6 + b6 )(a2 + b2 ) ⩾ (a4 + b4 )2 . (⊠) Second solution After multiplying out the right side the inequality becomes a6 + b6 a6 + a4 b2 + b4 a2 + b6 ⩾ 4 a4 + b4 a + a3 b + b3 a + b4 or a6 + b6 a6 + b6 + a2 b2 (a2 + b2 ) ⩾ 4 4 ⋅ a4 + b4 a + b + ab(a2 + b2 ) To simplify calculations let use make the following substitutions a6 + b6 = x . 2. .19) . (a3 + b3 )(a + b) ⩾ (a2 + b2 )2 .77 2. Equality occurs when a = b. . Equality occurs when a = b. a. .78 Chapter 2. = xn is necessary and sufficient for equality in (2. a2 − 4ab + 4b2 ⩽ 1 1 and 4a2 − 4ab + b2 ⩽ ⋅ a b (⊠) .. (⊠) Second solution Note that (ax1 + bx2 )(ax2 + bx1 ) = (a2 + b2 )x1 x2 + ab(x21 + x22 ) ⩽ Hence (x2 + x2 )(a + b)2 ⋅ 2 x31 2x31 ⩾ (ax1 + bx2 )(ax2 + bx1 ) (a + b)2 (x21 + x22 ) so we need to prove that 2x31 2x32 2x3n ⩾ x1 + ⋯ + xn + + ⋯ + x21 + x22 x22 + x23 x2n + x21 We suppose the contrary. Note that.19). . xn positive real num. there exist x1. Example 2. clearly true with equality if and only if x1 = x2 or a = b. . . .92. The conclusion follows. then equality holds in the proposed inequality if and only if x1 = x2 = ⋯ = xn . since x1 = x2 = .bers such that 2x32 2x31 2x3n < x1 + ⋯ + xn + + ⋯ + x21 + x22 x22 + x23 x2n + x21 by symmetry we have the following inequality 2x33 2x31 2x32 < x1 + ⋯ + xn + + ⋯ + x21 + x22 x22 + x23 x2n + x21 summing up we obtain x31 + x32 x32 + x33 x3n + x31 < x1 + ⋯ + xn + + ⋯ + x21 + x22 x22 + x23 x2n + x21 contradiction since x3 + y 3 x + y ⩾ x2 + y 2 2 So the original inequality is true. Let a and b be positive real numbers such that Prove that a + b ⩽ 2. = xn .. that is to say. with equality if and only if x1 = x2 = . 1 1 ∣a − 2b∣ ⩽ √ and ∣2a − b∣ ⩽ √ ⋅ a b Solution We start by squaring both inequalities. Examples for practice which is equivalent to (x1 − x2 )2 (a − b)2 ⩾ 0. . so we’re done. 2 a+b+c 2 2 6x − 4x(a + b + c) + (a + b + c) = 6 (x − ) ⩾0 3 3 2 and we’re done. 2 ⩾ a + b Which is what we wanted to prove. cyc ∑ a3 (b − c) = (a − b)(b − c)(a − c)(a + b + c). 3 or equivalently. (⊠) Example 2. Therefore the left-hand side equals ∑ 6x2 a2 (b − c) − 4xa3 (b − c) + a4 (b − c). b = c or c = a. 6 cyc Solution Note that the coefficients of x4 and x3 vanish. cyc It results that the inequality is equivalent to 1 6x2 − 4x(a + b + c) + (a2 + b2 + c2 + ab + bc + ca) = (a2 + b2 + c2 − ab − bc − ca). (⊠) . we eliminate the fractions in each inequality. By multiplying by 4 and applying H¨ older’s Inequality.93. cyc ∑ a4 (b − c) = (a − b)(b − c)(a − c)(a2 + b2 + c2 + ab + bc + ca).2. Let a. cyc Note that each of the coefficients of the polynomial on the left-hand side vanish when a = b. 8 ⩾ (1 + 1)(1 + 1)(a3 + b3 ) ⩾ (a + b)3 And finally.79 2. Therefore (a − b)(b − c)(a − c) divides each of them. we have that. a3 + b3 ⩽ 2. Now add the two inequalities together. It is easy to show that ∑ a2 (b − c) = (a − b)(b − c)(a − c). Prove that for each real number x the following inequality holds 1 ∑(x − a)4 (b − c) = (a − b)(b − c)(a − c)[(a − b)2 + (b − c)2 + (c − a)2 ]. a3 − 4a2 b + 4b2 a ⩽ 1 and 4a2 b − 4ab2 + b3 ⩽ 1. c be real numbers such that a > b > c. Algebraic problems Next. b. ie s = 3 and d = 1 for x = 2. or since x ≠ y (if x = y ≠ 0 the LHS of the first equation would be zero. x2 + y 2 = ⋅ 2 2 Therefore. (243d − s)(243s + d) = sd(s2 + d2 )(s2 − d2 )(243s + d) = (s4 − s2 d2 + d4 )(s2 + d2 )(243d − s). ⎪sd(s2 + d2 )(s2 − d2 ) ⎨ 4 2 2 4 2 2 ⎪ ⎪ ⎩(s − s d + d )(s + d ) = 243s + d. it follows that 243s2 d(s2 − d2 ) + sd2 (s2 − d2 ) = 243d(s4 − s2 d2 + d4 ) − s(s4 − s2 d2 + d4 ). 4xy (x4 − y 4) = 121x − 122y and x.94. in contradiction with the problem statement). Since s2 + d2 > 0 (otherwise x = y = 0. and no other solutions exist. we find that d5 = 1. which after simplification yields s5 = 243d5 . while x4 − y 4 = sd(x2 + y 2 ) = (s2 + d2 ) sd(s2 + d2 ) . 4xy = s2 − d2 . (⊠) Second solution Since (x4 + 14x2 y 2 + y 4 ) (x2 + y 2 ) = 122x + 121y. but the RHS would not). y = 1.80 Chapter 2. the system may be rewritten as We can then obtain ⎧ ⎪ = 243d − s. or s = 3d. These values can be clearly shown to satisfy the system by plugging them into the given equations. where we have defined x + y = s and x − y = d. Solve in nonzero real numbers the system of equations First solution Note first that ⎧ ⎪ x4 − y 4 ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ x4 + 14x2 y 2 + y 4 ⎪ ⎪ ⎩ = 121x − 122y 4xy 122x + 121y = x2 + y 2 x4 + 14x2 y 2 + y 4 = 4(x2 + y 2 )2 − 3(x2 − y 2 )2 = s4 − s2 d2 + d4 . Examples for practice Example 2. (α) We have (α) ⇔ (x2 + y 2) (x4 + 14x2 y 2 + y 4) (x − y) − 4xy (x2 − y 2 ) (x + y) = x2 + y 2 . y ≠ 0 then (x4 + 14x2 y 2 + y 4 ) (x2 + y 2 ) (x−y)−4xy (x4 − y 4 ) (x+y) = (122x+121y)(x−y)−(121x−122y)(x+y). Substitution in both equations yields d6 = d. z) are (a. Let t = x + y then x2 − y 2 = t. The last relation is equivalent to (∣x − y∣ − 1)2 + (∣y − z∣ − 1)2 + (z − x)2 ⩽ 0. Algebraic problems ⇔ (x4 + 14x2 y 2 + y 4 ) (x − y) − 4xy (x2 − y 2 ) (x + y) = 1 ⇔ (x − y)5 = 1 ⇔ x − y = 1.96. x2 + y 2 + z 2 + 1 = xy + yz + zx + ∣x − y + z − y∣. The desired triples (x. (x − y)2 + (y − z)2 + (z − x)2 + 2 ⩽ 2∣x − y∣ + 2∣y − z∣. First solution Since all zeros of P (x) are negative real numbers. where a ∈ R. Prove that b + c ≠ 4. y. (x − y)2 + (y − z)2 + (z − x)2 + 2 = 2∣x − y + z − y∣. a + 1. y = .2. Let a. =3 Example 2. (a. x2 + y 2 = and the equation x4 − y 4 = Hence. β. b. (⊠) Example 2. 121x − 122y = 121 − 2 2 2 121x − 122y becomes 4xy t (t4 − 1) t−1 = 121 − ⇔ t(t4 − 1) + t − 1 = 242 ⇔ t5 = 243 ⇔ t = 3. c = αβγ.81 2. We get ∣x − y∣ = 1.95. a). c be real numbers such that a < 3 and all zeros of the polynomial x3 + ax2 + bx + c are negative real numbers. y = 1. γ. z) of real numbers such that First solution We can write hence It follows (⊠) x2 + y 2 + z 2 + 1 = xy + yz + zx+ ∣ x − 2y + z ∣ . Find all triples (x. a − 1. a). 4xy = t2 − 1. t−1 t−1 t2 + 1 . and a = α + β + γ. . b = αβ + βγ + γα. y. ∣y − z∣ = 1 and x = z. it follows that P (x) = x3 + ax2 + bx + c = (x + α)(x + β)(x + γ) for some positive real numbers α. 2 2 ⎧ ⎪ ⎪x − y ⎨ ⎪ ⎪ ⎩x + y =1 ⇔ x = 2. (⊠) Third solution Since all zeros of the polynomial p(x) = x3 + ax2 + bx + c are negative real numbers. Examples for practice By AM-GM inequality 1 b 2 1 a ⩾ ( ) ⩾ c3 3 3 hence. Moreover 1 pq + qr + pr = (p + q + r) 3 < 3. √ ( 3 9 a b c 3 abc a b c But this last inequality is equivalent to 1 1 1 3 √ ⩽( + + ) 3 a b c abc which follows directly from the AM-GM Inequality. q. 2 If b + c = 4 we would have 4 = pq + qr + pr + pqr < 3 + pqr < 4 contradiction.97. + 2 + 2 ⩽ √ 3 + bc 2b + ca 2c + ab 3 abc a b c Thus. (or simply by multiplicity for multiple roots. we have This implies that 1 1 1 = ⩽ √ ⋅ 3 2a2 + bc a2 + a2 + bc 3a abc 1 1 1 1 1 1 1 ( + + ). if −α. so we are done. that is b + c < 4. hence 3b ⩽ a2 < 9. 2a2 + bc 2b2 + ca 2c2 + ab 9 a b c Solution By the AM-GM Inequality.) we see that p′ (x) = 3x2 + 2ax + b. (⊠) . On the other hand. (⊠) Example 2. has also negative real zeros. its discriminant must be positive. Let a. then c = αβγ and a = α + β + γ. or b < 3. a 3 Thus. a < 3 implies that b < 3 and c < 1. c = pqr a < 3 implies by the AGM 3 > p + q + r = 3(pqr) 3 and then pqr < 1. In particular. (⊠) Second solution Let p. then by Roll’s theorem. From c < 1 and b < 3 we 3 conclude that b + c < 4. r the absolute values of the zeroes of the polynomial. b. c be positive real numbers.82 Chapter 2. it is suficient to prove that 2a2 1 1 1 1 2 1 1 1 1 + + ) ⩽ ( + + ) . Viete’s formula yields a = p + q + r. −β and −γ are the negative real zeros of p. by the AM-GM inequality we see that c ⩽ ( ) < 1. b = pq + qr + pr. Prove that 1 1 1 1 1 1 2 1 + + ⩽ ( + + ) . c are 1. (⊠) 2003 + . hence of a2 and of a. Now. Under this constraint. b(c + a). and the other one is . Let a. These values can 4 be easily shown to satisfy the relation given in the problem statement. c(a + b)}. then the maximum is achieved when b = c = 1. a2 s2 = x. x= 2 and finally. and s needs to be minimized so that x = a2 s2 is maximized. It remains thus to find the maximum value of a under these two constraints. Find the greatest possible value of max{a(b + c). let us see that it is otherwise. √ √ √ √ 2011 ⋅ 1995 2001 + 1995 = max{a(b + c). it would seem at first sight that bc would need to be minimized. or equivalently. the values of p = abc and s = b + c are known. wherefrom the largest of both (clearly positive real) roots is √ √ √ 2011 − 2s2 + (2011 − 2s2 )2 − 4s4 2011 − 2s2 + 2011 2011 − 4s2 = ⋅ x= 2 2 It finally follows that. b. and s2 ⩾ 4bc = 4 by the AM-GM inequality. s are known and fixed. We may thus perform substitution bc = 1. the previous relation between a and s may be rewritten as x2 − (2011 − 2s2 )x + s4 = 0.83 2. and the square of this maximum is √ 2003 + 2011 ⋅ 1995 . Note first that (b2 + 1)(c2 + 1) = s2 + b2 c2 − 2bc + 1 = s2 + (bc − 1)2 . and the relation given in the problem statement becomes (a2 + 1)2 s2 = 2011a2 . and a needs to be maximized. is obtained when bc = 1.2. c(a + b)} = 2 2 √ √ 2001 + 1995 with equality if two of a. c be positive real numbers such that (a2 + 1)(b2 + 1)(c2 + 1) ( 1 a2 b2 c2 + 1) = 2011. a2 + 1 = 2011 p2 ⋅ ⋅ s2 + (bc − 1)2 p + 1 Thus the maximum of the RHS. since p. Algebraic problems Example 2. we need to maximize as. b(c + a). b. since p = abc is known.98. Solution Assume that when a(b + c) reaches its maximum possible value (which by symmetry in the variables will also be the maximum that we are looking for). since bc = 1. b. Solving f (x) = x we get only a real solution x = 4. yi = b 3 . (⊠) √ √ √ Example 2. we have f (x) < x. Suppose that f (x) > x. .101. Prove that 8(a2 + b2 + c2 ) ⩾ 3(a + b)(b + c)(c + a). And.21) it follows that (a + b + c)3 ⩽ 9 (a2 + b2 + c2 ) (2. c be positive real numbers such that a + b + c = 1. so the solution of our system is x = y = z = 4.20) 3 1 2 On the other hand. Hence we have f (f (f (x))) = x and the same for y and z.84 Chapter 2. y = f (z). Let a. applying f again. p = . Let a. Then we have f (f (x)) < f (x) and x = f (f (f (x))) > f (f (x)).22) we get the desired inequality. b. c be positive real numbers such that a + b + c = 3. in the H¨ older’s inequality 2 we get yields Since 1 p p q ∑ xi yi ⩽ ( ∑ xi ) ( ∑ yi ) n n k=1 k=1 n 1 q k=1 1 √ √ √ 23 2 1 2 1 2 1 a + b + c = a 3 a 3 + b 3 b 3 + c 3 c 3 ⩽ ( a + b + c) (a2 + b2 + c2 ) 3 .22) From (2. Examples for practice Example 2. The equality holds if a = b = c = 1. Prove that √ √ 3 √ 3 (aa + ba + ca )ab + bb + cb (ac + bc + cc ) ⩾ ( 3 a + b + 3 c) .21) (2. taking xi = a 3 . contradiction. 3 9 (2. √ √ √ 2 (a + b + c)3 ⩽ ( a + b + c) (a2 + b2 + c2 ) √ √ √ a + b + c = 3 from (2.100.20) and (2. (⊠) Example 2. Solution By AM-GM inequality we have 2a + 2b + 2c 3 8 3(a + b)(b + c)(c + a) ⩽ 3 ( ) = (a + b + c)3 . z = f (x) where f (x) = ( + ) is a strictly decreasing x 4 function over the positive real numbers. q = 3. Solve the system of equations √ 1 √ 1 √ 1 7 x− = y− = z− = ⋅ y z x 4 Solution 1 7 2 We have x = f (y).99. and the same for y and z. In a similar way we discard the case f (x) < x and we have f (x) = x. c) majorizes (1/3. Using AM-GM. f (x) = ln (ax + bx + cx ).85 2.S. 1/3.. a + b ⩾ . it follows by the majorization inequality for convex functions that which is our inequality. 1 1 1 f (a) + f (b) + f (c) = f ( ) + f ( ) + f ( ) 3 3 3 f ′′ (x) = (ab)x ( ln a − ln b)2 + (ac)x ( ln a − ln c)2 + (bc)x ( ln b − ln c)2 (ax + bx + cx)−2 > 0.. and R. we can assume a ⩾ b ⩾ c so that (a. (⊠) Second solution By the symmetry of both the L. xn be positive real numbers. . (⊠) Example 2. The proof is completed. . b. 3 3 The inequality is √ √ √ 3 ln (aa + ba + ca ) + ln (ab + bb + cb ) + ln (ac + bc + cc ) = 3 ln ( 3 a + b + 3 c) and if we show that f ′′ (x) > 0.H. Let x1 . a + b + c = 1. x2 ..H. we may suppose x1 x2 ⋯xn = 1 and then prove (x1 + x2 + ⋯ + xn )2n = n2n−1 S where S = x21 + x22 + ⋯ + x2n . 2 (S + n(n − 1))n ⩾ n2n−1 S.S. Prove that √ √ n−1 x21 + x22 + ⋯ + x2n x1 + x2 + ⋯ + xn n ( ) ⩾ ( n x1 x2 ⋯xn ) ⋅ n n Solution By homogeneity. 1/3) : 1 2 a ⩾ .2. (x1 + x2 + ⋯ + xn )2 = S + 2 ∑ xi xj 1⩽i<j⩽n = S +2⋅ hence it suffices to show that 2 n(n − 1) ⋅ ((x1 x2 ⋯xn )n−1 ) n(n−1) = S + n(n − 1). Algebraic problems First solution By H¨ older’s Inequality (aa + ba + ca ) 3 (ab + bb + cb) 3 (ac + bc + cc) 3 = a 1 1 1 a+b+c 3 +b a+b+c 3 +c a+b+c 3 = ∑a cyc and by cubing both sides we get the desired result.102. 2 First solution From weighted AM-GM inequality we conclude that xy y x ⩽ ( xy + yx x+y 2xy x+y x + y x+y ) =( ) ⩽( ) . a−b b−c c−a Solution By expanding. S + n(n − 1) = S + n + n + ⋯ + n ⩾ n (Snn−1 ) n 1 and so (S + n(n − 1))n ⩾ nn Snn−1 = n2n−1 S. ∑( cyc a − b cyc a − b a(b − c)(c − a) + b(c − a)(a − b) + c(a − b)(b − c) a = (a − b)(b − c)(c − a) cyc a − b ∑ = It follows that ab(c − a) + bc(a − b) + ca(b − c) b a + 1 = ∑( )( ) + 1.3 a b c + + = 0.103. (⊠) Example 2. Prove that ( 2 2 2 b c a + 1) + ( + 1) + ( + 1) ⩾ 5. x+y x+y 2 where the last inequality again follows from AM-GM √ 2xy x+y x+y ⩽ ⇔ xy ⩽ ⋅ x+y 2 2 (⊠) . Let a.86 Chapter 2. Let x and y be positive real numbers. a−b b−c c−a (⊠) Undergraduate problems Example 2. by AM-GM again. b. (a − b)(b − c)(c − a) b−c cyc a − b a 2 a a b c 2 ) + 2∑ =2+( + + ) ⩾2 ∑( a−b b−c c−a cyc a − b cyc a − b as desired. Equality occurs if and only if 2. the inequality becomes On the other hand. c be distinct real numbers.104. a 2 a ) + 2∑ ⩾ 2. Examples for practice Now. Prove that xy y x ⩽ ( x + y x+y ) .2. as desired. 1) . Let then h′ (x) = ln x and 1−x h(x) = x ln x + (1 − x) ln (1 − x) 1 1 1 h′ (x) < 0 ⇔ x ∈ (0.87 2. Firstly. we obtain xy y x ⩽ ( x+y x+y x + y x+y ⇔ (xy) 2 ⩽ ( ) . Since xy y x ⩽ (xy) x+y 2 x x−y ⇔ x2y y 2x ⩽ xx+y y x+y ⇔ xy y x ⩽ xx y y ⇔ ( ) ⩾ 1. h′ (x) > 0 ⇔ x ∈ ( . 1) 2 2 which can proved by employing calculus.2. Normalization of the inequality by the condition x + y = 1 yields xy y x ⩽ 2xy ⇔ xy y x ⩽ ( 2xy x+y ) x+y 1 1 ⩽ xx y y ⇔ ⩽ xx (1 − x)1−x . 2 2 (⊠) x y y x 2xy xx + y y x + y ⩽ x +y = x+y x+y x+y then original inequality immediately follows from the obtained stronger inequality and the inequality xy y x ⩽ ( 2xy x+y ) x+y 2xy x+y ⩽ . Thus we obtained the chain of inequalities x+y 2 xy y x ⩽ ( x+y 2xy x+y x + y x+y ) ⩽ (xy) 2 ⩽ ( ) . we will prove the stronger inequality x+y xy y x ⩽ (xy) 2 . y (xy) 2 ⩽ 1 x + y x+y ) . Algebraic problems Second solution Let us take a more indepth look at this inequality. h′ (x) = 0 ⇔ x = ⋅ 2 2 2 (⊠) . Indeed. 2 Third solution By weighted AM-GM. ) . x ∈ (0. x+y 2 Remark. Let a1 . yn−k . ⋯. if n is odd. a + a21 + a22 + ⋯ + a2n ⩾ m(∣a1 ∣ + ∣a2 ∣ + ⋯ + ∣an ∣). a2 . 2.2. x2 .1) 2 1 1 h(x) ⩾ h ( ) ⇔ ⩽ xx (1 − x)1−x .106. we have to prove that ∑ x2i + ∑ zj2 + a ⩾ m ∑ xi + m ∑ zj Denote x = k n−k k n−k i=1 j=1 i=1 j=1 1 n−k 1 k ∑ zi and z = ∑ zj . 2 n n −1 Solution The first step is to dismiss the absolute value sign. . k = 0. n − k}. 2 2 (⊠) π Example 2. 1). ∀ x ∈ (0. . √ √ a an where m = 2 . Clearly. x∈(0.e. and m = 2 . an into a sequence of non-negative real numbers x1 . the indicated product telescopes. y2 . ⋯.4 1 + 2 cos π (−1)n+1 (−1)n−1 = = ⋅ 1 + a0 1 + a0 1 + a0 (⊠) Olympiad problems Example 2. We separate the sequence a1 .105. ⋯..88 Chapter 2. Examples for practice 1 Thus min h(x) = h ( ) . i. Denote zj = −yj . xk and a sequence of negative real numbers y1 . an be real numbers such that a1 + a2 + ⋯ + an = 0. Let n be a given positive integer and let ak = 2 cos n−k . .. a2 . . 2 Prove that n−1 (−1)n−1 ⋅ ∏ (1 − ak ) = 1 + a0 k=0 Solution Note that 1 − ak = (1 − 2 cos = 1 + 2 cos 1 − 2 (1 + cos 1 + 2 cos =− π 1 − 4 cos2 π = ) pi 2n−k 1 + 2 cos π 1 + 2 cos n−k 2n−k 2 π π 2n−k−1 π 2n−k ) 2n−k =− 1 + 2 cos 2n−k π 2n−k−1 π 1 + 2 cos n−k 2 1 + ak+1 ⋅ 1 + ak Thus. and ∏(1 − ak ) = (−1)n n−1 k=0 2. . Prove that for a ⩾ 0. j ∈ {1.. n − 1.. k i=1 n − k j=1 . 1. ⋯. if n is even. Let n be a positive integer. k ∑ zj2 ⩾ (n − k) ( n−k j=1 z1 + z2 + ⋯ + zn−k 2 ) = (n − k)z 2 . The conclusion follows. 4 If n is even. f (x) is a polynomial of degree 2n − 1 with the roots x = cos . as a1 + a2 + ⋯ + an = 0. then x x x xf (x) = g(y) = y n−1 − 22 (n2 − 1) n−3 24 (n2 − 1) (n2 − 22 ) n−5 y + y +⋯ 3! 5! . n−k Using the AM-GM inequality we get √ √ LHS ⩾ 2 ax k (1 + √ 2k anx k )= √ ⋅ n−k k(n − k) n2 .89 2. 2n 1 1 1 Consider f ( ) and substitute y = 2 . the maximum of k(n − k) is (⊠) Example 2. 1 ⩽ k ⩽ 2n − 1. the above inequality becomes kx2 (1 + k ) + a ⩾ 2mkx. Prove that 2 2 1 (n − 1) .2. Algebraic problems k ∑ x2i i=1 x1 + x2 + ⋯ + xk 2 ⩾ k( ) = kx2 . 4 2 n −1 If n is odd.107. n−k After all. = kπ 2 3 k=1 cos 2n n ∑ First solution From De Moivre’s fomula and calculus one can show: sin 2nθ 23 n(n2 − 1) = (−1)n+1 [2n cos θ − cos3 θ + ⋯] sin θ 3! Then we can Define a function f (x) = (−1)n+1 2nx [1 − 22 (n2 − 1) 2 24 (n2 − 1) (n2 − 22 ) 4 x + x + ⋯] 3! 5! kπ In fact. the maximum of k(n − k) is . From the the condition kx = (n − k)z. we would like to prove that kx2 + (n − k)z 2 + a ⩾ m(kx + (n − k)z). . xn we have n P ′ (0) 1 = − (2. ∑ ∑ ∑ i i a0 P (x) i=1 i=1 i=1 x − xi n ′ ′ and plugging x = 0 the conclusion follows. 2n Pn (x) = 4n−1 x ∏ (x − cos2 n−1 k=1 kπ ). Examples for practice kπ This is a polynomial of degree n − 1 in y whose roots are y = sec 2 with 1 ⩽ k ⩽ n − 1. 3 3! 2n i=1 and we are done. . a0 ≠ 0 with non-zero roots x1 . 2n (⊠) . n ∈ Z. . then n n P (x) P ′ (x) 1 ′ = ( ln (x − x )) = ( ln (x − x )) = ( ln ) = .. In particular. (⊠) Second solution Note that for any polynomial P (x) = a0 xn +a1 xn−1 +⋯+an−1 x+an . n − 1 and n 2π (n − 1)π π ). . 2n Using Viete’s theorem we find that the sum of coefficients is kπ 22 (n2 − 1) n−1 2 2 (n − 1) = = ∑ sec 2 . ′ Tn+1 (x) sin(n + 1)θ Let Un (x) ∶= = be the Chebishev Polynomial of the n+1 sin θ Second Kind. . sin nθ kπ Because = 0 if and only if θ = . Un−1 (x) = 2n−1 (x − cos ) (x − cos ) ⋯ (x − cos n n n as the coefficient of xn in Un (x) is 2n .. Then Un (x) satisfies to recurrence Un+1 (x) = 2xUn (x) − Un−1 (x). Let P (x) = a0 (x − x1 )(x − x2 ). k = 1. . . we get Un−1 (x) = 0 if and sin θ n kπ only if x = cos . 2.23) ∑ P (0) i=1 xi Proof.90 Chapter 2. then 2 x kπ ). x2 . U2n−1 (x) = 22n−1 ∏ (x − cos 2n−1 k=1 = 22n−1 (x − cos kπ ) 2n kπ n−1 (2n − k)π nπ n−1 ) ∏ (x − cos ) ∏ (x − cos ) 2n k=1 2n k=1 2n = 22n−1 x ∏ (x2 − cos2 n−1 k=1 √ U2n−1 ( x) Let Pn (x) ∶= √ .(x − xn ). n ∈ N and U0 (x) = 1. U1 (x) = 2x. . .2. where α = β = 0. with P0′ (0) = 0.27). Since bn+1 − bn = bn − bn−1 we have bn − bn−1 = −1 and ∑nk=1 (bk − bk−1 ) = −n. because a0 = a1 = 0. (2. then an = 3 2n (n2 − 1) and Thus an = 3 Pn′ (0) n−1 1 an 2 2 (−1)n =− = = (n − 1) ⋅ ∑ kπ Pn (0) bn 3 k=1 cos2 2n (−1)n+1 (2. then polynomial Pn (x) satisfy the recurrence Pn+1 (x) = 2(2x − 1)Pn (x) − Pn−1 (x). b1 = −1. Algebraic problems Note that U2n−1 (x) can be Defined by the recurrence U2n+1 (x) = 2 (2x2 − 1) U2n−1 (x) − U2n−3 (x). . n ∈ N with P0 (0) = 0.108. (2. with P0′ (x) = 0. . Since U2n−1 (x) is divisible by 2x. Pn (0) Let bn ∶= . implying bn = −n. xn such that x1 + x2 + ⋯ + xn = n.26) can be rewritten as (−1) (−1)n+1 an+1 − 2an + an−1 = 4n. n ∈ N with a0 = a1 = 0. x2 . From the other hand. b0 = 0. P1′ (0) = 0. n ∈ N with P0 (x) = 0.24) Thus applying (2.26) (2.27) Since sequence (⊠) Example 2. . then ′ ′ (x).25) bn+1 − 2bn + bn−1 = 0. then (2. x3 x1 n x2 √ +√ +⋯+ √ ⩾√ ⋅ x1 + 2x3 x2 + 2x4 xn + 2x2 3 Prove that P (n) is true for n ⩽ 4 and false for n ⩾ 9.23) to the polynomial Pn (x) we obtain n−1 ∑ 1 =− Pn′ (0) ⋅ Pn (0) kπ 2n In particularly. P1′ (x) = 0. 2n (n2 − 1) is particular solution of nonhomogeneous recurrence 3 2 2n (n − 1) + αn + β. (x) = 2(2x − 1)Pn′ (x) + 4Pn′ (x) − Pn−1 Pn+1 ′ ′ (0) = 4Pn′ (0). from (2. with U−1 (x) = 0.24) follows recurrence k=1 cos2 Pn+1 (0) + 2Pn (0) + Pn−1 (0) = 0. .91 2. P1 (0) = 1. (0) + 2Pn′ (0) + Pn−1 Pn+1 P ′ (0) Pn (0) Let an ∶= n n then = −bn = n and (2. P1 (x) = 1.25) can be rewritten as (−1)n (2. Therefore bn − b0 = −n. Let P (n) be the following statement: for all positive real numbers x1 . U1 (x) = 2x. n ∈ N. the hypothesis gives ∣Z∣ = 4 so that the inequality ∣Re(Z)∣ ⩽ ∣Z∣ writes as that is. . Using the fact that x1 x2 + x2 x3 + ⋯ + xn x1 ⩽ n whenever x1 + x2 + ⋯ + xn = n and n ⩽ 4. . An easy calculation yields: Re(Z) = 1 − (ab + bc + cd + da + ac + bd) + abcd and ∣Z∣2 = (a2 + 1)(b2 + 1)(c2 + 1)(d2 + 1). Now. b. x3 . d be real numbers such that Prove that (a2 + 1)(b2 + 1)(c2 + 1)(d2 + 1) = 16. xn ) be the left hand side of the inequality. ∣(ab + bc + cd + da + ac + bd − abcd) − 1∣ ⩽ 4 −3 ⩽ ab + bc + cd + da + ac + bd − abcd ⩽ 5. Solution Consider the complex number Z = (1 + ia)(1 + ib)(1 + ic)(1 + id). we have x2 (x1 + 2x3 ) + ⋯ + x1 (xn + 2x2 ) = 3(x1 x2 + x2 x3 + ⋯ + xn x1 ). x2 . On the other hand.92 Chapter 2. Examples for practice Solution older’s inequality. Using H¨ we obtain S 2 (x2 (x1 + 2x3 ) + ⋯ + x1 (xn + 2x2 )) ⩾ (x1 + x2 + ⋯ + xn )3 = n3 . Chosing x1 . . one easily obtains that the expression is smaller than √ 3 for n ⩾ 9. x2 . −3 ⩽ ab + bc + cd + da + ac + bd − abcd ⩽ 5. . (⊠) Example 2. c. The conclusion follows. as required. (⊠) . The last fact follows from the fact that ab + bc + ca ⩽ and (a + b + c)2 3 ab + bc + cd + da = (a + c)(b + d) ⩽ (a + b + c + d)2 4 The conclusion follows easily for n ⩽ 4. Let a. Let S(x1 .109. x4 close to n and the other 4 n variables equal and close to 0. b. z. y. Let a.b = .110.93 2. hence 2b + c 3 9 a3 a a a2 2 2b + c 2 ⩾ ( a − ) = −∑ ∑ ∑ 2 9 3 cyc 2b + c cyc 9 cyc (2b + c) cyc (2b + c) 3 ∑ a a+b+c 2 2b + c 2 2 )−∑ = − ∑(2b + c) ∑( a − 3 cyc 3 9 3 9 cyc cyc 9 √ 3 a + b + c 3 abc 1 = ⩾ = ⋅ 9 9 3 ⩾ (⊠) . b. a. Set 1 1 1 a = . . If x. Note that 2 2b + c a2 ⩾ a− .2. b. Algebraic problems Example 2. c > 0 we have x3 y 3 z 3 (x + y + z)3 + + ⩾ ⋅ a2 b2 c2 (a + b + c)2 We will also relax the condition abc = 1 to abc ⩽ 1.c = ⋅ x y z Then. ) in original inequality with (a. c) we obtain the equivalent a b c inequality a 1 ⩾ ∑ 3 cyc (2b + c) with abc = 1. we have xyz ⩾ 1 and the left hand side of the inequality is equal to K = (xyz)2 ∑ x3 ⋅ 2 cyc (2y + z) From the above lemma (and since xyz ⩾ 1)we have √ (x + y + z)3 x + y + z 3 3 xyz 1 K⩾ = ⩾ ⩾ ⋅ 9(x + y + z)2 9 9 3 (⊠) Second solution Since 1 3 ( ) 1 1 a = 2 = 5 2 a (b + 2c) 2 2 2 1 1 a5 b2 c2 ( + ) ( + ) c b c b 1 1 1 then by replacing ( . c be positive real numbers such that abc = 1. Prove that 1 a5 (b + 2c)2 + 1 b5 (c + 2a)2 First solution We will make use of the following lemma + 1 c5 (a + 2b)2 ⩾ 1 ⋅ 3 Lemma 3. Finally all triples (x. S are positive or both are negative. we get P ⋅ S > 0. so z ∈ {−1. y = 1. Let a. We can write: y = z + 1. 0. From first relation we have 3z 3 +3z 2 (1−p)+z(p−1)2 = 0. From first relation we have 3z 3 + 3z 2 (p + 2) + z(p + 2)2 + 2p + 1 = 0. Find all triples (x. Prove that a+b+1 b+c+1 c+a+1 (a + 1)(b + 1)(c + 1) + 1 + + ⩽ ⋅ 2 3 2 3 2 3 a+b +c b+c +a c+a +b a+b+c First solution From the Cauchy-Schwartz inequality and because abc = 1 we obtain that (a + b2 + c3 )(a + 1 + ab) ⩾ (a + b + c)2 ⇔ So we get that 1 + a + ab 1 ⩽ ⋅ 2 3 a+b +c (a + b + c)2 a+b+1 (a + b + 1)(1 + a + ab) ⩽ ⋅ 2 3 a+b +c (a + b + c)2 In a similar way we obtain the following inequalities and (b + c + 1)(1 + b + bc) b+c+1 ⩽ b + c2 + a3 (a + b + c)2 (c + a + 1)(1 + c + ca) c+a+1 ⩽ ⋅ c + a2 + b3 (a + b + c)2 . −4021. (⊠) Example 2. −2010). If P = a ⋅ b. 1. (1) From here. Examples for practice Example 2. We get z = 0 or 3z 2 + 3z(1 − p) + (p − 1)2 = 0. None of these verifies (1) Case 2: a = x − y = −2011 = −p − 1. From z = 0. S = a + b. (1. 1). Case 1: a = x − y = 2010 = p. b = y − z = 1. 1. which has negative discriminant. z) of integers such that x2 y + y 2z + z 2 x = 20102 and xy 2 + yz 2 + zx2 = −2010. z∣(2p + 1) = 4021. If a = x − y and b = y − z. Solution Substract both relations and get (x − y)(y − z)(x − z) = 2010 ⋅ 2011. x = z−p. b = 1 and in second a = −2011. −2010.112.111. y. 4021}. x = z + p + 1.94 Chapter 2. 0). z) are (−2010. so both P. (0. c be positive real numbers such that abc = 1. primes decomposition. We can write: y = z+1. y. In first case a = 2010. we have x = −p = −2010. b. b = 1 and there permutations. prime number. we have a ⋅ b ⋅ (a + b) = 2 ⋅ 3 ⋅ 5 ⋅ 67 ⋅ 2011. b = y − z = 1. y. we have: Similarly. Let c = . Algebraic problems So it is enough only to prove that (a + b + 1)(1 + a + ab) + (b + c + 1)(1 + b + bc) + (c + a + 1)(1 + c + ca) (a + 1)(b + 1)(c + 1) + 1 ⩽ ⋅ (a + b + c)2 a+b+c The last one is equivalent to ∑(a + b + 1)(1 + a + ab) ⩽ (a + b + c)(a + 1)(b + 1)(c + 1) + a + b + c. (⊠) Second solution By Cauchy-Schwarz Inequality.2. Solution The inequality in question immediately follows from the identity T ∶= (x2 + xy + y 2)(y 2 + yz + z 2 )(z 2 + zx + x2 ) = 3(x2 y + y 2 z + z 2 x)(xy 2 + yz 2 + zx2 ) + ((x − y)(y − z)(z − x))2 .113. What remains is to prove this identity. or cyc 3 ∑ a+3+2 ∑ a2 +2 ∑ ab+ ∑ ab(a+b) ⩽ abc⋅ ∑ a+3abc+ ∑ ab(a+b)+ ∑ a2 +2 ∑ ab+2 ∑ a. cyc cyc cyc cyc cyc cyc cyc cyc cyc which is true. We work in C. 2 Then. but here is a more conceptual proof: √ 1+i 3 2 2 2 2 2 2 Denote a = x y + y z + z x and b = xy + yz + zx . we can prove it by expanding.95 2. Prove that for all real numbers x. (⊠) (⊠) Example 2. Of course. z the following inequality holds (x2 + xy + y 2)(y 2 + yz + z 2 )(z 2 + zx + x2 ) ⩾ 3(x2 y + y 2 z + z 2 x)(xy 2 + yz 2 + zx2 ). c3 = −1 and thus ⎞ ⎛ ⎞ ⎛ ⎛ ⎞ 3 2 2 2 ⎟ 2 2 2⎟ 2⎜ ⎟ ⎜ ⎜ (x + cy)(y + cz)(z + cx) = ⎜ c + 1 ⎟ xyz + c ⎜x y + y z + z x⎟ + c ⎜xy + yz + zx ⎟ ⎟ ⎜ ´¹¸¹ ¶ ⎠ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ ⎠ ⎝´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ ⎠ ⎝=−1+1=0 ⎝ =a =b . a+b+c Equality holds when a = b = c = 1. we obtain LHS ⩽ = = (a + b2 + c3 )(a + 1 + ab) ⩾ (a + b + c)2 (a + b + 1)(a + 1 + ab) + (b + c + 1)(b + 1 + bc) + (c + a + 1)(c + 1 + ca) (a + b + c)2 (a + b + c)(ab + bc + ca + a + b + c + 3) (a + b + c)2 (a + 1)(b + 1)(c + 1) + 1 = RHS. the following inequality holds: √ A+B+C Au2 + Bv 2 + Cw 2 ⋅ ⩾ A B C A+B+C + + u v w . b. A+B +C = 3 m. w = √ . y. B = b. we have 1 1 1 T = (y + cz) (y + z) (z + cx) (z + x) (x + cy) (x + y) c c c 1 1 1 = (x + cy)(y + cz)(z + cx) ⋅ (x + y) (y + z) (z + x) c c c 1 1 1 = c(a + cb) ⋅ (a + b) = (a + cb) (a + b) c c c = a2 + ab + b2 (by (i)) (by (i)) = 3ab + (b − a)2 = 3 (x2 y + y 2z + z 2 x) (xy 2 + yz 2 + zx2 ) + ((x − y)(y − z)(z − x))2 . Prove that a b c m + + ⩾ ⋅ x y z n Solution √ √ √ √ A B C 3 Denote A = 3 a. B. because T = (y 2 + yz + z 2 ) (z 2 + zx + x2 ) (x2 + xy + y 2 ). Examples for practice = ca + c2 b = c(a + cb).96 Chapter 2. u. The same computation with c replaced by of c3 = −1) proves 1 1 3 everywhere (and using ( ) = −1 instead c c 1 1 1 1 1 (x + y) (y + z) (z + x) = (a + b) . y x z √ A B C Clearly n = + + . (⊠) Example 2. C = 3 c. C. v = √ . c and x. or the result that we need to prove is equivalent to showing u v w that. Let a. and define u = √ . v. for all positive reals A. z be positive real numbers such that √ √ √ √ √ √ √ √ 3 3 a + b + 3 c = 3 m and x + y + z = n. b = xy 2 + yz 2 + zx2 . w.114. and a quick computation shows that b − a = xy 2 + yz 2 + zx2 − x2 y + y 2 z + z 2 x = (x − y)(y − z)(z − x) and we are done. c c c Hence. Since a = x2 y + y 2z + z 2 x. c c c c c But any two complex numbers u and v satisfy 1 (u + cv) (u + v) = u2 + uv + v 2 c (i) 1 1 1 (since (u + cv) (u + v) = u2 + (c + ) uv + v 2 and c + = 1 as we can easily see). B. ie.28) cyc cyc ab3 ca3 bc3 + + ) − 3 (a2 c + c2 b + b2 a) c b a (⊠) . C. b. Prove that 8abc a+b+c ⋅ + 4⩽ √ 3 (a + b)(b + c)(c + a) abc Solution Defining a = x3 . 27 ⩾ 4 ⇔ S 4 − 4S 3 + 27 ⩾ 0 ⇔ (S − 3)2 (S 2 + 2S + 3) ⩾ 0 S3 Example 2.115. Let a. with equality holding iff u = v = w. v. which is well known to be always true.28) ⇔ S + and we are done.97 2. Algebraic problems This is the inequality between weighted quadratic and harmonic means of u. b. c = z 3 we come to 4⩽ Now (x3 + y 3 )(y 3 + z 3 )(z 3 + x3 ) ⩽ so we prove Defining S = x3 + y 3 + z 3 8(xyz)3 + 3 ⋅ xyz (x + y 3 )(y 3 + z3)(z 3 + x3 ) (2(x3 + y 3 + z 3 ))3 27 x3 + y 3 + z 3 27(xyz)3 + 3 ⩾ 4. b = y 3. we have 3T ∑ ab = 3 (∑ ab) ( cyc cyc = 3 (∑ a3 + cyc a2 b2 c2 + + + a + b + c) b c a ab3 ca3 bc3 + + + a2 c + c2 b + b2 a + ∑ a2 b + 3abc) c b a cyc = 3 ∑ a3 + 3 ( cyc ab3 ca3 bc3 + + ) + 3 (a2 c + c2 b + b2 a) + 3 ∑ a2 b + 9abc c b a cyc Thus 3 3T ∑ ab − 2 (∑ a) = ∑ a3 − 3abc + 3 ( cyc (2. c be positive real numbers.2. Prove that a2 b2 c2 2(a + b + c)3 + + +a+b+c ⩾ ⋅ b c a 3(ab + bc + ca) Solution Denote LHS of the given inequality by T . w and respective weights A. if and only if √ √ √ 3 3 a b 3c √ =√ =√ ⋅ (⊠) y x z Example 2. xyz (x + y 3 + z 3 )3 (x3 + y 3 + z 3 ) we have xyz (2.116. c be positive real numbers. Let a. 98 Chapter 2. Examples for practice Now by the AM-GM inequality we have √ ab3 ca3 ab3 ca3 + ⩾2 × = 2a2 b c b c b etc,. By adding these inequalities we have ab3 ca3 bc3 + + ⩾ a2 c + c2 b + b2 a. c b a And it remains to prove that ∑ a3 ⩾ 3abc which is known to be true. (⊠) cyc Example 2.117. Let a, b, c, d be positive real numbers such that a2 + b2 + c2 + d2 = 1. Prove that √ √ √ √ √ √ √ √ 1 − a + 1 − b + 1 − c + 1 − d ⩾ a + b + c + d. Solution √ √ The inequality takes the form ∑ ( 1 − a − a) ⩾ 0. We have cyc √ √ 1 − 2a 1 1−a−a 1−a− a= √ √ =√ √ ⩾ √ ∑(1 − 2a) 1−a+ a 1−a+ a 2 cyc where in√the last step we used the estimate (x + y)2 ⩽ 2(x2 + y 2 ) applied to the numbers √ 1 − a, a. Hence, it suffices to prove that ∑(1 − 2a) ⩾ 0 ⇔ 4 − 2(a + b + c + d) ⩾ 0 ⇔ 4(a2 + b2 + c2 + d2 ) ⩾ (a + b + c + d)2 , cyc which is clearly true (for instance using Power Mean.) (⊠) Example 2.118. Let a, b, c, x, y, z ⩾ 0. Prove that (a2 + x2 )(b2 + y 2 )(c2 + z 2 ) ⩾ (ayz + bzx + cxy − xyz)2 . First solution The inequality is trivial if any of x, y, z equals 0. Suppose that xyz ≠ 0. Therefore, dividing by (xyz)2 > 0 it follows that the inequality is equivalent to a b c (m2 + 1)(n2 + 1)(p2 + 1) ⩾ (m + n + p − 1)2 , where (m, n, p) = ( , , ) . x y z After expanding and rearranging some terms it follows that this inequality is equivalent to m2 n2 p2 + (m2 n2 + m + n) + (n2 p2 + n + p) + (p2 m2 + p + m) ⩾ 2mn + 2np + 2pm. From AM-GM it follows that m2 n2 + m + n ⩾ 3mn ⩾ 2mn, from where it is easy to conclude the result. (⊠) Second solution 99 2.2. Algebraic problems By expanding, we have LHS = a2 b2 c2 + x2 b2 c2 + y 2 c2 a2 + z 2 a2 b2 + a2 y 2 z 2 + b2 z 2 x2 + c2 x2 y 2 + x2 y 2z 2 , RHS = x2 y 2 z 2 + 2xyz(abz + bcx + cay − ayz − bzx − cxy). The inequality becomes a2 b2 c2 + x2 b2 c2 + y 2c2 a2 + z 2 a2 b2 + 2xyz(ayz + bzx + cxy) ⩾ 2xyz(abz + bcx + cay). By the AM-GM inequality, we have x2 b2 c2 + xyz ⋅ bzx + xyz ⋅ cxy ⩾ 3xyz ⋅ xbc. Adding two similar inequalities, we obtain x2 b2 c2 + y 2 c2 a2 + z 2 a2 b2 + 2xyz(ayz + bzx + cxy) ⩾ 2xyz(abz + bcx + cay), and we are done. Equality holds if and only if a = b = c = x = y = z = 0. Example 2.119. Let a, b, c be positive real numbers such that abc = 1. Prove that √ √ √ √ 3 3 a + b + 3 c ⩽ 3 3(3 + a + b + c + ab + bc + ca). (⊠) First solution √ √ √ 3 3 Set a = x, b = y, 3 c = z. We have then x, y, z > 0 and xyz = 1. What we have to prove assumes the form 3 (3 + x3 + y 3 + z 3 + (xy)3 + (yz)3 + (zx)3 ) = (x + y + z)3 . Now, recall Schur’s inequality A3 + B 3 + C 3 + 5ABC ⩾ (A + B)(B + C)(C + A). Setting A = xy, B = yz, C = zx, we find (xy)3 + (yz)3 + (zx)3 ) + 5 = (x + y)(y + z)(z + x), since xyz = 1. Then it is enough to prove that 3(3 + x3 + y 3 + z 3 + (x + y)(y + z)(z + x) − 5) ⩾ x3 + y 3 + z 3 + 3(x + y)(y + z)(z + x), or equivalently that x3 + y 3 + z 3 ⩾ 3, which is true by the AM-GM inequality. (⊠) Second solution x y z Setting a = , b = , c = . The inequality becomes y z x √ √ √ √ x y z x y z y z x 3 + 3 + 3 ⩽ 3 3 (3 + + + + + + ) y z x y z x x y z By the Holder inequality, we have: √ √ y z y z x 1 1 1 x 3 3 (3 + + + + + + ) = 3 3(x + y + z) ( + + ) y z x x y z x y z √ √ √ x 3 y 3 z 3 ⩾( + + ) y z x Hence we are done. Equality holds if and only if x = y = z or a = b = c = 1. (⊠) 100 Chapter 2. Examples for practice Example 2.120. Let a0 , a1 , ..., a6 be real numbers greater than −1. Prove that a2 + 1 a2 + 1 a2 + 1 √ 0 +√ 1 +⋯+ √ 6 ⩾ 5. a51 + a41 + 1 a52 + a42 + 1 a50 + a40 + 1 whenever a3 + 1 a3 + 1 a3 + 1 +√ 1 +⋯+ √ 6 ⩽ 9. √ 0 a51 + a41 + 1 a52 + a42 + 1 a50 + a40 + 1 Solution It suffices to prove that a3 + a20 + 2 a3 + a21 + 2 a3 + a26 + 2 √0 + √1 + ⋯ + √6 ⩾ 14. a51 + a41 + 1 a52 + a42 + 1 a50 + a40 + 1 We observe that (a7 = a0 ) a3k + a2k + 2 =∑√ √ ∑√ 5 ak+1 + a4k+1 + 1 k=1 a2k+1 + ak+1 + 1 a3k+1 − ak+1 + 1 k=1 a3k + a2k + 2 6 6 which we rewrite as ∑√ 6 k=1 namely 6 a3k − a −k +1 a2k + ak + 1 √ +∑√ √ a2k+1 + ak+1 + 1 a3k+1 − ak+1 + 1 k=1 a2k+1 + ak+1 + 1 a3k+1 − ak+1 + 1 √ √ 6 a3k − a −k +1 a2 + ak + 1 +∑√ k ⩾ 2l.7 = 14 ∑√ 2 ak+1 + ak+1 + 1 k=1 a3k+1 − ak+1 + 1 k=1 6 the last inequality allowed by the AGM and we are done. (⊠) 101 2.3. Geometric problems 2.3 2.3.1 Geometric problems Junior problems Example 2.121. Let ABC be a triangle with circumradius R. Prove that if the length of one of the medians is equal to R, then the triangle is not acute. Characterize all triangles for which the lengths of two medians are equal to R. First solution Let O be the circumcenter and M be the midpoint of the side BC. Without loss of generality we have that a ⩾ b ⩾ c, we have 1√ 2 1√ 2 1√ 2 2b + 2c2 − a2 ; mB = 2a + 2c2 − b2 ; mC = 2b + 2a2 − c2 , 2 2 2 and we deduce that mA ⩽ mB ⩽ mC . On the other hand, if the triangle is acute angled, then its circumcenter lies int the interior of the triangle. Note that mA > R, because ∠AOM is obtuse, and the equality does not occur. Thus triangle ABC is not acute angled. For the second part it is not difficult to see that if two medians in a triangle are equal, then the triangle is isosceles, because mA = mB = mC ⇔ 1√ 2 1√ 2 2a + 2c2 − b2 ; mC = 2b + 2a2 − c2 ⇔ b = c. 2 2 Let the ABC be isosceles triangle with b = c. By the Law of Sines and the Law of Cosines we have b2 + c2 − a2 2b2 − a2 a2 , cos A = = sin A 2bc 2b2 and if mB = mC = R, we have R= 1 m2B = R2 ⇒ (2a2 + 2b2 − b2 ) = R2 ⇒ 2a2 + b2 = 4R2 4 and finally R2 = a2 a2 = = 2 sin2 A 4(1 − cos 2A) Finally, using (2.29) we get 2a2 + b2 = yielding b = c = √ (2.29) a2 4b4 2 . ⇒ 4R = 4b2 − a2 2b2 − c2 2 )] 4 [1 − ( 2b2 4b4 ⇒ 7a2 b2 − 2a4 = 0 ⇒ a2 (7b2 − 2a2 ) = 0, 4b2 − a2 2 a, and we are done. 7 Second solution Without loss of generality, let us assume that the length of the median from A equals R. The square of the length of this median is given by the least number is no larger than n − 2. 1 yielding tan Btan C = − ⋅ 3 Clearly. then b2 + c2 a2 c2 + a2 b2 − = − . or ABC is isosceles at C. The result is true for the case n = 3. . and using the well known identity tan A + tan B + tan C = tan Atan Btan C. and the median from A is a radius of the circumcircle. since P will be in the interior of A1 A2 A3 only. The result is true for n = 3. Example 2. 1 tan A we find 2tan A − =− and 3tan A 3 7 1 tan 2 A = tan 2 B = . say ma = mb . tan 2 C = ⋅ 7 9 √ √ 2 tan α 2 7 Using that sin2 α = . Since there are exactly n − 2 different triangles A2 A3 Ak that contain P in their interior. Consider a convex polygon A1 A2 . C is obtuse. consider triangles A1 A2 A3 and A3 A4 A5 in an n-gon. If the result is true for n − 1 ⩾ 3. or the we find that sin A = sin B = 2 4 4 1 + tan α lengths of two √ medians √ √ in a triangle are equal to R if and only if it is similar to a triangle with sides 2.. One possible solution is that triangle ABC is right triangle at A. n ⩾ 4 (if n = 4. B and C cannot be simultaneously acute. and no other triangle Ai Aj Ak may contain P on its sides or in its interior. Examples for practice b2 + c2 a2 a2 − = + bc cos A = R2 sin2 A + 4R2 sin B sin C cos A. 4 sin B sin C = − cos(B + C) = − cos B cos C + sin B sin C. each triangle Ai Aj Ak will have non void intersection with the interior of A2 A3 Q if and only if one of its sides is A2 A3 . and ABC is either rectangle or obtuse. Find the least number of triangles Ai Aj Ak that contain P on their sides or in their interiors. 2 4 4 2 Equating this result to R and grouping terms in one side of the equality yields cos A(cos A − 4 sin B sin C) = 0.An and a point P in its interior. 2. If n = 4. 7. sin C = . The number cannot be less than n − 2. in which case the midpoint of BC is also the circumcenter. Since A = B. Otherwise. in which case A2 A3 Q is contained in it. then A5 = A1 )..102 Chapter 2. If the lengths of two medians are equal to R.122. Solution We prove that a point P may be found such that it is not contained in the interior or on the sides of more than n − 2 triangles. denote by Q the point where diagonals A1 A3 and A2 A4 intersect. 2 4 2 4 yielding a = b. Clearly. we prove this by induction. P is on the sides or in the interior of at least one of the triangles thus generated. Let la . = 4R sin A sin B sin C = Second solution By Mollweide’s formula we have A−B sin A−B a − b sin A − sin B 2 cos A+B 2 sin 2 2 = = = c sin C 2 sin C2 cos C2 cos C2 and lc = 2ab C cos . B. Assume without loss of generality. lc be the lengths of the angle bisectors of a triangle. Consider now the partition of the n-sided polygon on triangles by drawing all diagonals A2 Ak .30). a+b 2 sin A−B a2 − b2 2 = ⋅ lc 2abc (⊠) . and this triangle is different to the n − 3 previously considered. Geometric problems If n = 4.. we get cos ( A−B 2 ) LHS(2. P cannot be simultaneously on the sides or in the interior of both triangles.An . or the number of triangles that contain P on their sides and in their interior is no less than n − 2. lb .30) = sin B−C sin C−A sin A−B 2 2 2 + + 2R sin A sin B 2R sin B sin C 2R sin C sin A ) ) ) cos ( A−B cos ( B−C cos ( C−A 2 2 2 1 1 1 sin(A − B) sin(B − C) sin(C − A) = 2 + 2 + 2 2R sin A sin B 2R sin B sin C 2R sin C sin A 1 sin(A − B) sin(B − C) sin(C − A) = [ + + ] 4R sin A sin B sin B sin C sin C sin A 1 ∑cyc (sin A cos B − cos A sin B) sin C 4R sin A sin B sin C 0 1 = 0 = RHS(2. Therefore.An . or (n−1)-gon A1 A2 A3 A5 . P is either on the common boundary A1 A3 of these triangles.30) lc la lb where A. Example 2. b.3. and by hypothesis of induction n − 3 triangles Ai Aj Ak may be found that contain P on their sides or in their interiors. or completely outside one of them. If n ⩾ 5. C are the angles of the triangle. j.103 2.123.. Prove the following identity sin B−C sin C−A sin A−B 2 2 2 + + =0 (2. and so this is the least number. c are the sides of the triangle. where i.. Clearly.. where a. k ≠ 2. contains P in their interior. First solution 2R sin A sin B Using the fact that lc = . either (n−1)-gon A1 A3 A4 . hence. since they only have one common vertex A3 which cannot be P . on the other hand. (d) Line l1 is horizontal. In the Cartesian plane call a line ”good” if it contains infinitely many lattice points. n). suppose the slope of L1 is rational but neither ±1.31) holds (similarly we can do for (2. Once again. If the slope of L1 is undefined. the b hypothesis that lines l1 and l2 intersect at an angle of 45o imply that one and only one of above relations holds α2 = α1 + 45o (2. we get tan α2 = tan (α1 + 45o ) = tan α1 + tan 45o 1 − tan α1 tan 45o . Now. (c) Line l1 is vertical. where a. and let α denote the angle of inclination of L1 . Finally. in either situation.104 Chapter 2. (⊠) Second solution Let us suppose that l1 and l2 are two lines that satisfy the conditions stated in the hypothesis. then the slope of L2 is ±1. L2 contains one lattice point and has rational slope so must therefore contain infinitely many lattice points and is good. the slope of L1 is ±1. respectively. Then tan α is rational and tan (α ± 45o) = tan α ± tan 45o tan α ± 1 = 1 ∓ tan αtan 45o 1 ∓ tan α is also rational. b ∈ Z ∖ {0}. then L2 is either a horizontal line or a vertical line. Further. and a − b ≠ 0. and that the coordinates of the lattice point at which those lines meet are (m. assuming that (2. Without loss of generality we may assume that l1 is a ”good” line. we obtain as desired. because L2 is known to contain one lattice point it must therefore contain infinitely many lattice points and is good. lc la lb 2abc (⊠) Example 2. its slope must either be undefined or rational. a + b ≠ 0.31) α2 = α1 − 45o . (b) The slope of line is l1 is -1. The purported result clearly holds in any one of the following cases: (a) The slope of line l1 is 1. As L1 contains two lattice points. Examples for practice With this and the two similar results. suppose that L1 is good. If line l1 falls into neither of those categories below. (2.32) α2 = α1 + 45o where α1 and α2 are the elevation angles of lines l1 and l2 . First solution Let L1 and L2 be lines that intersect at a lattice point at an angle of 45o. sin B−C sin C−A sin A−B (a2 − b2 ) + (b2 − c2 ) + (c2 − a2 ) 2 2 2 + + = = 0. in either situation. then so is the other. If.124. we infer that its slope is a rational a number of the form . Prove that if one of the lines is good. Furthermore.32)). Two lines intersect at a lattice point at an angle of 45o degrees. again. L2 contains one lattice point and has rational slope so must therefore contain infinitely many lattice points and is good. R) are externally tangent. so O1 A is the internal bisector of ∠CAN. Tangent lines from O1 to C2 intersect C2 at A and B. (⊠) Example 2. d′ ) is a lattice point. l1 contains an infinite number of lattice points. b. Circles C1 (O1 . Now. c and d are all integers. Prove that EF ∩O1 O2 = AD ∩BC. B. Hence. line l2 is represented by the equation or equivalently. since O2 A = O2 B. say. This implies if l1 is ”good”. Now.33) possesses an infinite number of solutions in integers.3. as shown in Figure 2. Note that triangle P QR is a right isosceles triangle. consider the points on this plane as complex numbers. Without loss of generality we may assume that line l1 is ”good”. d′ ) on line l2 . r) and C2 (O2 . Geometric problems = a b +1 a+b = ⋅ 1 − ab b − a Hence.e. and we are done. First solution Let M be the midpoint of O1 O2 and observe that A. we have ∠O2 CA = ∠BCO2 .33) Since gcd(b−a. then so is line l2 . with QP = QR and ∠P QR = 90o . if an arbitrary point Q(c. By the same argument.125. y − n = (tan α2 ) (x − m) = a+b (x − m) b−a (b − a)y − (a + b)x = (b − a)n − (a + b)m. Let O1 A∩O2 C = {E} and O1 B ∩O2 D = {F }. Construct a perpendicular on l1 passing through Q such that it intersects l2 at R(c′ . l1 and l2 . i. C.105 2. Recall that a complex number a + ib when multiplied by eiθ rotates it by an angle θ in the counterclockwise direction. d′ ). We show that R itself is a lattice point. Each one of these solutions corresponds with a lattice point in l2 . D lie on the circunference with center M and radius MO1 . Since O1 C = O1 D we have∠CAO1 = ∠O1 AD. Figmr31 Let N = AD ∩ BC and note that N lies on O1 O2 because AD is the reflection of BC across the line O1 O2 . so are c′ and d′ . thus proving that R(c′ . (⊠) Third solution Let the two lines. since a. Therefore we have π (c′ − c + i(d′ − d))ei 2 = a − c + i(b − d). while tangent lines from O2 to C1 intersect C1 at C and D. d) on line l1 is a lattice point. . Ð→ Ð→ We note that QR rotated by 90o in the counterclockwise direction coincides with QP . intersect at a lattice point P (a.1. we obtain c′ = b + c − d and d′ = c + d − a. Solving for c′ and d′ . Let Q(c. so O2 C is the internal bisector of ∠ACN. which implies d − d′ + i(c′ − c) = a − c + i(b − d). (2. then so is R(c′ . −(a+b))∣(b−a)n−(a+b)m. the diophantine equation in (2. b) at an angle of 45o . d) be an arbitrary lattice point on l1 . it follows that AD. AO1 ∩ CO2 = E and BO1 ∩ DO2 = F are collinear. Examples for practice Figure 2. AE = EM implies that ∠EAM = ∠EMA = (180o − 48o ) = 66o . Consequently we have 1 1 ∠CNE = ∠CNA = ∠DNB = ∠F NB 2 2 and this implies E. O2 all lie on the circle of 2 diameter O1 O2 . BC. EF and O1 O2 concur. so that the points A. In the same way 2 π ∠O1 CO2 = ∠O1 DO2 = . (⊠) Second solution We will show that the result holds in the more general case of any two circles such that neither of them is contained in the other. so ∠CNE = ∠ENA. 2 . Thus EF ∩ O1 O2 = AD ∩ BC = {N} and the claim is proved. Since we just proved that EF goes through P . In other words. π Note that ∠O1 AO2 = ∠O1 BO2 = since O1 A is tangent to C2. Similarly F is the incenter of ∆BND. so ∠DNF = ∠F NB. BC and EF concur. F are collinear. Find the angles of triangle AMB. D. Now applying Pascal’s theorem to the cyclic hexagon ADO2 CBO1 it follows that P ∶= AD ∩ BC. Now it remains to observe that by symmetry O1 O2 is the perpendicular bisector of AB and CD. O1 . as we wanted. (⊠) Example 2.106 Chapter 2. AD. B. In the interior of a regular pentagon ABCDE consider the point M such that triangle MDE is equilateral. Morever. Solution We have that ∠MEA = ∠AED − ∠MED = 108o − 60o = 48o . and thus P lies on O1 O2 .126.1: Therefore E is the incentre of ∆CNA. N. C. z2 x2 z2 x2 y2 y2 + + = + + (y + z)2 (z + x)2 (x + y)2 (y + z)2 (z + x)2 (x + y)2 Which. y. on transposition. Similarly we obtain four y+z y+z other relations. (⊠) Example 2.3. and A1 C = . C1 be the intersectionsof lines AP. ∆AP B. Find P such that A1 B 2 + B1 C 2 + C1 A2 = AB12 + BC12 + CA21 . CP with sides BC. Figure 2. ∆AP C as x. BA1 ∆AA1 B ∆BP A1 ∆AA1 B − ∆BP A1 ∆AP B y = = = = = ⋅ A1 C ∆AA1 C ∆CP A1 ∆AA1 C − ∆CP A1 ∆AP C z ay az Let AB = BC = CA = a. ∠AMB = 180o − ∠MAB − ∠ABM = 84o . B1 . Let P be a point situated in the interior of an equilateral triangle ABC and let A1 . is equivalent to y2 − z2 z 2 − x2 x2 − y 2 + + = 0.107 2. z respectively. CA. Solution Let us denote the areas of ∆BP C. (y + z)2 (z + x)2 (x + y)2 .127. Geometric problems By symmetry ∠ABM = ∠MBC = 2∠ABC = 54o . AB.2: Substituting these values in our first relation we obtain. BP. Finally ∠MAB = ∠EAB − ∠DCM = 108o − 66o = 42o . respectively. then BA1 = . we have. P must lie on any of the three medians of ∆ABC. Find the minimum of AP ⋅ AQ.128. Q ∈ AB. (⊠) ̂ = 90o and let d be a line passing trough Example 2. ∆CP A have the same area. S(MNP Q) = Second solution a x + (ha − x) aha 1 a ⋅ x(ha − x) ⩽ ⋅( ) = = S(ABC). So. So.129. therefore. ha ha 2 4 2 2 (⊠) . Then. N be the projections of I on AC. which rewrites into cyc x + y n This relation can be simplified to ∑ (x − y)(z − x)(z − y) = 0.108 Chapter 2. Prove that 1 S(MNP Q) ⩽ S(ABC). 2 First solution Let x = NP and a = BC. if P is such that at least two of the triangles ∆AP B. We have IM = IN = r and from the similarity of the triangles P MI. Let ABC be an acute triangle and let MNP Q be a rectangle inscribed in the triangle such that M. P ∈ AC. The sign of equality holds iff MP = NQ ⇔ AP = AQ ⇔ AP Q = AQP (⊠) Example 2. ∆BP C. (x + y)(y + z)(z + x) Hence the first relation actually holds true if and only if x = y or y = z or z = x. AB respectively. we have AP ⋅ AQ = (AM + MP )(AN + NQ) = AM ⋅ AN + AM ⋅ NQ + MP ⋅ AN + MP ⋅ NQ = 2r 2 + r(NQ + MP ) ⩾ 2r 2 + 2rpMP ⋅ NQ = 2r 2 + 2r 2 = 4r 2 . Examples for practice x−y = 0. ̂ ̂ = 45o. INQ we find P M ⋅ NQ = r 2 . where r is the inradius of triangle ABC. Let ABC be a triangle with A the incenter of the triangle and intersecting sides AB and AC in P and Q. P Q ha − x a(ha − x) ⇔ PQ = ⋅ = CB ha ha Hence. then the first relation holds good. respectively. Solution Let M. Since P Q ∥ BC then ∆QAP ≃ ∆BAC and. N ∈ BC. 3 a+x b+x ab 1 1 + ) ⩾ 0. NP = z. 2 Solution The parallel to AB intersects MN at T . Let MP = y. f (c) − f (a) = (c − a) ( + 3 2a(a + c) (b + a)(b + c) 1 1 2a + b b ab 1 1 5 2a + b − ab ( + )= − − ( + ) = (a − b) ⩾ 0.109 2. b. 3 2a a + b 3 2 4 a b 12 (1) . (⊠) Example 2. c be the side-lenghts of a triangle with the largest side c. 1]. So. Let a. Geometric problems Since ABC is an acute triangle then the rectangle MNP Q is contained into the triangle ABC. Prove that MN 4 4 OP + ( ) = MP 2 ⋅ NP 2 . Consider the function We have and f (a) = f (x) = a+b+x 1 1 − ab ( + ).131. It is clear that AMO and OT P are right isosceles triangles. The desired inequality follows by noting that max 2t(1 − t) = t∈[0. we can denote AM = MO = x and OT = T P = t. So we must prove that u4 + x4 = y 2 ⋅ z 2 .130. we have y 2 ⋅ z 2 = [(x + t)2 + t2 ] ⋅ [(x − t)2 + t2 ] = (x2 + 2t2 + 2xt) ⋅ (x2 + 2t2 − 2xt) = (x2 + 2t2 )2 − 4x2 t2 = x4 + 4t4 . Let ABCD be a square of center O. BM = tBH. Let AH be the height from A to BC. and OP = u for convenience. S(MNP Q) = QM(BC − BM − CN) = t(1 − t)AH ⋅ BC = 2t(1 − t)S(ABC).1] 1 ⋅ 2 (⊠) Example 2. So we have T N = x − t. then Therefore QM = P N = tAH. By Pythagoras. Prove that ab(2c + a + b) a + b + c ⩽ ⋅ (a + c)(b + c) 3 First solution Without loss of generality assume that c ⩾ a ⩾ b. The parallel through O to AD intersects AB and CD at M and N and a parallel to AB intersects diagonal AC at P . CN = tCH for some t ∈ [0.3. From triangle OT P we have 2t2 = u2 then 4t4 = u4 and we are done. Furthermore. Ib . I is the center of the nine-point circle. . hence abc ⩽ c3 . Feuerbach’s Theorem tells us that the nine-point circle is internally tangent to the incircle. has equality if and only if triangle ABC is equilateral and so we’re done. Ic C1 are concurent. ) with pole I and ratio . CA. AB. But Euler’s Triangle Inequality. a+b+c We have c2 + c(a + b) 3abc ⩽ c( ) = c2 . R = 2r. The original inequality can be written as (⊠) (i) 3abc ⩽ c2 + a(c − b) + b(c − a). Moreover. b. IIc are in fact the midpoints of the arcs BC. Let ABC be a triangle with incenter I and let A1 . abc ⩽ ac2 . It is wellknown that the 2 2 midpoints of the segments IIa .133. Let I be the incenter of triangle ABC and let A1 . abc ⩽ bc2 and a(c − b) + b(c − a) ⩾ 0. (ii) (⊠) Example 2. Examples for practice Note that. if we let r be inradius and R be the circumradius of triangle ABC then R the condition that the radius of the nine-point circle. First solution 1 1 Consider the homothety H(I. Since the nine-point circle and the incircle have the same center and are internally tangent we can conclude that these circles have the same radius. the given inequality holds for any positive a. AB not containing the vertices of the triangle. c with the largest c. then prove that triangle ABC is equilateral. AB. prove that lines Ia A1 . is the same as the radius of the 2 incircle tells us that R = r ⇒ R = 2r 2 which is the equality case of Euler’s Triangle Inequality. (⊠) Second solution . therefore. CA. C. the lines Ia A1 . If IA1 = IB1 = IC1 . Ib B1 . IIb .132. Second solution Assume that c = max{a. c}. Ic denote the excenters corresponding to sides BC. C1 be the symmetric points of I with respect to the midpoints of sides BC. C1 be the feet of altitudes from vertices A. (⊠) Example 2. Ic C1 are concurrent since the mediators of the triangle ABC are concurrent. B.110 Chapter 2. Ib B1 . B1 . respectively. Solution We start by noting that IA1 = IB1 = IC1 implies that I is the circumcenter of triangle A1 B1 C1 whose circumcircle is of course the nine-point circle. CA. Thus. If Ia . B1 . b. a+b+c a+b+c Combining (i) and (ii) the inequality is proved. where p denotes the semiperimeter and r the inradius of triangle ABC. .3. respectively. AI 2 = (p − a)2 + r 2 by Pythagoras theorem. Besides. r. knowing the incenter we can draw . Solution We are given the foot of the altitude from A and the midpoint M of BC. the foot of the altitude from A. the point that satisfies (ha − 2r)(β + γ) = (hb − 2r)(γ + α) = (hc − 2r)(α + β) is simultaneously on lines Ia A1 . hb − r. Thus A1 = (−r. CA. Denoting by ra . ra ). and the midpoint of the side BC. and the midpoint of hb hc BC is Ma = (0. Solution We begin with the following preliminary result: Lemma. so by applying the Lemma for this triangle. However. (⊠) Example 2. rc the exradii corresponding to sides BC. rb . ) . i.111 2. clearly Ia = (−ra . b + c = 2a. β.135. Then. the points on line Ia A1 satisfy RR α β γ RRRR RR R RRR α+β α+γ 1 1 RRRR = ∣ 0 = RR−1 ∣ RR hb − 2r hc − 2r. Ic C1 . r being the inradius. Let ABCD be a trapezoid (AB ∥ CD) with acute angles at vertices A and B. B.134. Prove that AC bisects the segment EF if and only if AF + EF = AB. The conclusion follows. Line BC and the tangent lines from A and E to the circle of center D tangent to AB are concurrent at F. r). I = (r. ma be the bisector and median from A. Give a straightedge and compass construction of a triangle ABC starting with its incenter I. hc − r). ra . we obtain 9 [(p − a)2 + 4bc(a + b + c)(b + c − a) (p − a)(p − b)(p − c) ]= . the lines IG and BC are parallel if and only if b + c = 2a. C. hc are the respective lengths of the altitudes from 2 2 A. Geometric problems In exact trilinear coordinates. AB. Proof. so the line determined by the two is precisely the line BC. Let la .e. γ). where ha . Ib B1 . (⊠) Example 2. p (b + c)2 so using 3AG = 2ma . Returning to the original problem we have that AF + EF = 2AE and note that G is the midpoint of EF if and only if C is the centroid of triangle AEF. hb . we arrive to desired conclusion. we have that r2 = (p − a)(p − b)(p − c) bc(a + b + c)(b + c − a) and la2 = . Now. RR RRR−r hb − r hc − r RRR or by cyclic permutations. or in trilinear coordinates (α. p (b + c)2 which after several simple algebraic manipulations becomes (b + c − 2a)(2a + 5b + 5c) = 0. We have that IG is parallel to BC if and only if AI = AG. Let ABC be an arbitrary triangle with incenter I centroid G. Solution We will prove a lemma first: Lemma 4. in order to use this. let us denote x = ∠ADE = ∠ADF . we need to find X. they intersect at the vertex A. The second case gives either D = A or D = B. are therefore the right triangles. because D ∈ AB ∩ Gc = {A. since MD = MX.112 Chapter 2. just draw the lines XD and the altitude from A (which we can draw since we have the foot of the altitude on BC and the line BC). Gb . respectively. Now. So suppose that Ga . For which triangles ABC are Ga . CA. and for D = B we get a right angle at B. this will give us the vertices B and C. just take the tangents from A to the incircle and intersect them with BC . and so D ∈ AB. Then either C ∈ Gc or D ∈ Gc . D. Let D the orthogonal projection of A onto BC and let E and F be points on lines AB and AC respectively such that ∠ADE = ∠ADF. Examples for practice the perpendicular from I to BC and get the incircle of ABC. Gc with diameters BC. By the Lemma we have sin x AE AD = ⋅ EB BD sin(90o − x) . Gb and Gc are concurrent. consider circles Ga . (One easily checks that these triangles satisfy the conditions from the problem. Gb and Gc are concurrent. BF . B}. This is not a problem whatsoever. However.136. so we have the construction of X as the reflection of D in M.137.) (⊠) Example 2. Afterwards. Gb . Given a nondegenerate triangle ABC. and D the antipode of D with respect to the incircle. The only triangles for which Ga . Hence our construction is complete. X are collinear. (⊠) Example 2. In triangles P AB and P AC. the law of sines gives PB AB = sin ∠P AB sin ∠AP B AC AC PC = = ⋅ o sin ∠P AC sin(180 − ∠AP B) sin ∠AP B Dividing the above relations we get the desired result. as shown in figure. Prove that the lines AD. we get that ∠BDA = 90o + 90o = 180o. The first case gives a right triangle with right angle at C. If P is a point on the side BC of a triangleABC we have P B AB sin ∠P AB = ⋅ P C AC sin ∠P AC Proof. and AB. then the points A. where X is the tangency point of the A-excircle with BC. As ∠BDC = ∠CDA = 90o (as D lies on circles with diameter BC and CA). For D = A. Consider a triangle ABC. Gc concurrent? Solution Let C and D be the intersections of Ga and Gb . and CE are concurrent. (⊠) Coming back to the problem. we get a right angle at A. Recall the well-known fact that if D is the tangency point of the incircle with the side BC. and CE are concurrent and we are done. z. We now prove the initial inequality. where S denotes the area of the triangle ABC. dc be the distances from point P to the triangle’s sides. If x. a b c 2 3 and the result follows. Geometric problems A F E x x B C D Figure 2.138. (⊠) Example 2. hb .113 2.b = . Let P be a point inside a triangle ABC and let da .c = . y. Indeed. Solution First need the following Lemma 5. c > 0 we have x3 y 3 z 3 (x + y + z)3 + + ⩾ ⋅ a2 b2 c2 (a + b + c)2 The proof follows from H¨ older’s Inequality. we have 1 x3 y 3 z 3 3 (a + b + c) ( 2 + 2 + 2 ) ⩾ x + y + z. Prove that da ⋅ h2a + db ⋅ h2b + dc ⋅ h2c = (da + db + dc )3 . b.3. hc are the altitudes of the triangle. BF . (⊠) follows upon substituting a = ha hb hc . by Ceva’s theorem the lines AD. db . The result 2S 2S 2S . where ha . Using the above lemma we have d3b d3a d3c da db dc + + = + + a2 b2 c2 (ada )2 (bdb )2 (cdc )2 ⩾ (da + db + dc )3 (da + db + dc )3 = ⋅ 2 (ada + bdb + cdc ) 4S 2 since ada + bdb + cdc = 2S.3: CF DC sin(90o − x) = ⋅ F A AD sin x Therefore AE BD CF AD sin x BD DC sin(90o − x) ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ =1 EB DC F A BD sin(90o − x) DC AD sin x so. a. e . and therefore (λ − ai )(1 − bi ) ⩾ 0. + an = b1 + . (⊠) Example 2. n this implies that ai .. An be a regular n-gon inscribed in a circle of center O and radius R. Let A1 . e .. M respectively.114 Chapter 2. . + an bn ⩾ −nλ + λ(n + 1 − λ) + (n + 1 − λ) = n + 1 − λ2 . Hence it remains to show that (∣x∣n + 1)2 ⩽ (∣x∣2 + 1)n ⇔ 2∣x∣n ⩽ ∑ Cnk ∣x∣2k n−1 k=1 which follows from AM-GM inequality since n ⩾ 3 and ∑ Cnk ∣x∣2k ⩾ n∣x∣2 + n∣x∣2n−2 ⩾ 2n∣x∣n ⩾ 2∣x∣n . (⊠) . we find 0 ⩽ nλ − λ ∑ bi − ∑ ai + ∑ ai bi ... Since e. .. an ∈ [0. . For every i = 1.. An . we have ∏ ∣x − ek ∣ = ∣xn − 1∣ ⩽ ∣x∣n + 1.. we have λ = n + 1 − (a1 + . 1] and λ be real numbers such that a1 + a2 + ⋯ + an = n + 1 − λ.139.. A2 . when M = O. a1 b1 + . + bn = n + 1 − λ.. Let e = exp ( 2π n and let the complex numbers e.... n k=1 by the triangle inequality.e. Let a1 .. Then our inequality is equivalent to ∏ ∣x − ek ∣ ⩽ n k=1 √ n (∣x∣2 + 1) .. bi ⩽ 1 ⩽ λ (where (bi ) is a permutation of (ai )). x correspond to the points A1 . . Solution As a1 .. .. k=1 Solution Let us work in the complex plane with O as the origin and without loss of generality.140... en are the roots of z n − 1 = 0.. Prove that for each point M in the plane of the n-gon the following inequality holds: n n ∏ MAk ⩽ (OM 2 + R2 ) 2 ..... 1]... this yields as desired. . n n n i=1 i=1 i=1 Because a1 + . .. a2 .. an ∈ [0. Examples for practice Example 2. 2 n ) R = 1. n−1 k=1 Equality holds iff ∣x∣ = 0 i. A2 .. e2 .. + an ) ⩾ n + 1 − n = 1.. For any permutation (bi )ni=1 of (ai )ni=1 prove that a1 b1 + a2 b2 + ⋯ + an bn ⩾ n + 1 − λ2 . Summing these inequalities for every i = 1. n.. .. LD = LE. ∠BLA = 60o . (⊠) . AL + BL sin ∠ABL + sin ∠BAL = AB sin ∠ALB 100o + 20o 100o − 20o cos sin 20o + sin 100o 2 2 = = 2 cos 40o = o o sin 60 sin 60 o o sin 100 sin A BC sin 80 = = = ⋅ = o o sin 40 sin 40 sin C AB 2 sin The conclusion follows. On the prolongation of BL take P such that AL = LP. Since ∠CLP = ∠ALB = 60o and LP = LA = LM we see that M and P are symmetrical with respect to AC. hence also ∠MLC = 60o .115 2. (⊠) Third solution First we note that ∠CBA = ∠BCA = 40o . We also have that LC is the bisector of the vertical angle in isosceles triangle DLE. Denote by BL the angle bisector of angle ∠ABC. giving LA = LE. leading to ∠ALB = 60o and ∠CLB = 120o . First solution Let D be a point on BL produced beyond L such that LD = LA and let E be a point on BC such that LE bisects ∠BLC. making BCD isosceles with BC = BD = BL + LD = BL + LA and we are done. therefore BC = BP = BL + LP = BL + AL. hence ∠LBA = 20o and ∠BLA = 60o . Using the Sine Law.3. Prove that AL + BL = BC. Then ∠BLM = ∠BLA = 60o . hence ∠P CB = 2∠ACB = 80o .141. Thus. Geometric problems Example 2. we have ∠ABL = ∠LBE = 20o . we have ∠CP B = 80o and BCP is isosceles. Since ∠CBP = 20o . Let ABC be an isosceles triangle with ∠A = 100o. ∠LCD = ∠ECL = 40o and∠EDC = 90o − ∠LCD = 50o. Therefore. and accordingly it crosses it at right angles at its midpoint. Hence LC is actually the perpendicular bisector of the base DE. Since ∠ABC = ∠BCA = 40o. triangles ABL andEBL are congruent (angle-side-angle). So we have ∠BDC = ∠BDE + ∠EDC = 30o + 50o = 40o + 40o = ∠BCL + ∠LCD = ∠BCD. Let M be the point (on BC) symmetric of A with respect to BL. or ∠ABL = ∠CBL = 20o . ∠B = ∠C = 40o . Consequently. (⊠) Second solution Clearly. ∠BLC = 120o and ∠BLE = ∠ELC = 21∠BLC = 60o = ∠DLC. S = = = = = = b2 p(p − b) + c2 p(p − c) − a2 p(p − b)(p − c) S2 (b2 + c2 − a2 )p2 + p(b + c)(b2 − bc − c2 + a2 (b + c)p − a2 bc) S2 p(b + c − a)(p − b − c) + bc((b + c)p − a) S2 2a2 (b2 + c2 + bc) − a4 − (b4 + c4 − 2b2 c2 ) + 2abc(b + c) − 4a2 bc 4S 2 2(a2 b2 + b2 c2 + c2 a2 ) − a4 − b4 − c4 + 2abc(b + c − a) 4S 2 16S 2 + 4abc(p − a) 4SR(p − a) 4R R = =4+ = 4 (1 + ) = R. and ra = ⋅ S= 4R p−a and the respective cyclic ones. R= ⋅ p−a p−b p−c p 4S The left hand side of the equality may be simplified by using the above facts to obtain L. respectively.142. rc = .116 Chapter 2. 2 2 4S S ra ra and we are done.S. Ω2 . Then we have the well-known identities √ S abc = rp = p(p − a)(p − b)(p − c).143. ( + )− r rb rc rb rc ra First solution We know the follwing facts: ra = S S S abc S .H. Denote by Ω1 . (⊠) Second solution Let S and p be the area and semiperimeter of the triangle. Let ABCD be a quadrilateral whose diagonals are perpendicular. rb = . 2 2 which is true. Prove that the diagonals of Ω1 Ω2 Ω3 Ω4 intersect at the centroid of ABCD. Examples for practice Example 2. Prove that in each triangle a2 R 1 b2 c2 = 4 ( + 1) . r= .H. Ω3 . (⊠) Example 2. Using these the equality becomes b2 p(p − b) + c2 p(p − c) − a2 (p − b)(p − c) − abc(p − a) = 4S2 p p ⇔ (a2 b + ab2 + b2 c + bc2 + c2 a + ca2 − a3 − b3 − c3 − 2abc) = (b + c − a)(c + a − b)(a + b − c). Ω4 the centers of the nine-point circles of triangles ABC. respectively. . CDA. DAB. BCD. are on the same perpendicular bisector of AC. ). MCD . {X. then the centroid of ABCD. Then H = (0. (⊠) Third solution Let Mac and Mbd be the midpoints of the diagonals AC and BD. Y. Let H and O be the orthocenter and the ac circumcenter of triangle ABC. It is well 2 2b a + c b2 − ac . yielding Mac ∈ Od Ob ⊥ AC. ). C. Thus Hd . ). Ω2 = ( c2 − bd b + d a + c d2 − ac a2 − bd b + d . This implies that the line Ω1 Ω3 is the midline of opposite sides of this rectangle: MAB MBC and MCD MDA . Analogoulsly. ) . Since the nine-point circle of the triangle ABC passes through the midpoints of its sides we have Ω1 belongs to the perpedicular bisector of MAB MBC . MAD be the midpoints of the sides AB. B.117 2. C = (c. and we are done. Ω1 Ω3 and Ω2 Ω4 pass through G. 4 4 Now we will find the coordinates of Ω1 . Ω4 = ( . (2) From (1) and (2) it follows that Od Ob ∥ Hd Hb . It follows that Ωd and Ωb are the midpoints of Od Hd and Ob Hb . Let Ox and Hx be the cicrumcenter and the orthocenter of triangle Y ZT. ) . DA. Geometric problems First solution Let MAB . It is known that the centroid G of ABCD coincides with the midpoint of Mac Mbd . Finally. D = (0. − ) and using that BH = 2OF where F is b a + c b2 + ac the feet of the perpendicular to AC from O we obtain that O = ( . The Euler circle’s center in a triangle is the midpoint of the segment determined by the circumcenter and the orthocenter. ) . D}. 0). BC. Similarly. the midpoint of Mac Mbd .3. respectively. Last relation proves that Od Ob Hd Hb is a trapezoid. T } = {A. Z. b). Hb ∈ BD and Hd Hb ⊥ AC. Let A = (a. G=( a+c b+d . known that Ω1 is the midpoint of HO. (1) The quadrilateral ABCD having the diagonals intersecting at a right angle implies that BO and DO are heights in the triangles ABC and ACD. B = (0. Points Od and Oc being cicrumcenters of triangle ABC and ACD. Thus the line Ωd Ωb is the midline of this trapezoid and so passes through G. 0). the intersection of the lines Ω1 Ω3 and Ω2 Ω4 coincides with the intersection of the diagonals of the rectangle MAB MBC MCD MAD which is the centroid of ABCD. hence Ω1 = ( 4 4b Analogously. d). (⊠) . Ω3 belongs to the perpedicular bisector of MCD MDA . Ω3 = ( . 0). Since AC and BD are perpendicular we get that MAB MBC MCD MAD is a rectangle. CD. (⊠) Second solution We proceed by coordinate geometry. MBC . we prove that G ∈ Ωa Ωc . Let the point of intersection of the diagonals be (0. 4c 4 4 4d 4a 4 Clearly. Examples for practice Example 2. Clearly HAIb C. 2 2 Thus I lies on the circumcircle (Oa ) of BIa C. tan βc is the sum of the other two. Now ∠BIC + ∠BIa C = 90o + A A + 90o − = 180o . βb . Ia all lie on a line. Hence Oa . Since a = b cos C + c cos B and tan βa = obtain a 2 − c cos B then. Ib Ob = Ob H and Ic Oc = Oc H.144. First solution If we define βa . tan βb . Oa . Let ABC be a triangle and let Ia . (⊠) Second solution A Let I denote the incenter of triangle ABC. (⊠) Example 2. Therefore Oa is the midpoint of diameter IIa and hence 2[IBOa ] = [IBIa ].145. Prove that the area of triangle Ia Ib Ic is twice the area of hexagon Oa COb AOc B. Ic be its excenters. Therefore A. First solution It is well-known that the angle-bisectors of ABC are the altitudes of Ia Ib Ic with feet A.118 Chapter 2. We have ∠BIa C = 90o − which implies 2 ∠BOa C = 180o − A. Ic AB. we conclude that Oa is the midpoint of minor arc BC of (O). Adding. In a triangle ABC. βc be the angles between medians and altitudes emerging from the same vertex. Again. Denote by H the orthocenter of Ia Ib Ic . Then [Ia Oa C] = [Oa HC] and so on. since BOa = COa . C. cyc cyc (⊠) . βc as oriented angles between medians and altitudes (let it be counterclockwise orientation) then statement of problems becomes: Prove that tan βa + tan βb + tan βc = 0. let βa . βb . Ib . B. we conclude that 2[Oa COb AOc B] = [Ia Ib Ic ]. Oc the circumcenters of triangles Ia BC. I. This leads to [Ia Ib Ic ] = 2[Oa CObAOc B]. Similarly 2[ICObA] = [ICIb A] and 2[IAOc B] = [IAIc B]. Ob . 2[ICOa ] = [ICIa ] so that 2[IBOaC] = [IBIa C]. HAIc B and HBIa C are cyclics and so Ia Oa = Oa H. Oc lie on the circumcircle (O) of ABC. Ob . Ib AC. Denote by Oa . applying the Sine Theorem we c sin B tan βa = b cos C − c cos B 2R sin B cos C − 2R sin C cos B = c sin B 2R sin C sin B sin B cos C − sin C cos B = cot C − cot B = sin C sin B therefore ∑ tan βa = ∑(cot C − cot B) = 0. Prove that one of the numbers tan βa . Similarly. It is obvious that MD ⋅ tan βa = AD Recalling that ∣b2 − c2 ∣ = 2aMD. Geometric problems Second solution Let AM be the median and AD the altitude emerging from vertex A. So we will consider the case it is acute.3.34) When triangle ABC is obtuse.146.2 Senior problems (⊠) Example 2. 2. First solution Using the equality cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1 the initial inequality becomes equivalent to or ∑ cos3 A + 3 ∏ cos A ⩽ ∑ cos2 A 3 ∏ cos A ⩽ ∑ cos2 A(1 − cos A) Now. We find then that tan βb = tan βa + tan βc . we have tan βb = ∣c2 − a2 ∣ ∣a2 − b2 ∣ . We have (2. Let ABC be a triangle. the above inequality is clearly true. we obtain tan βa = ∣b2 − c2 ∣ ∣b2 − c2 ∣ = 2aha 4S where S is the area of triangle ABC.34) ⇔ ∏(cos A)(1 + cos A) ⩽ ∏(1 − cos2 A) ⇔ 8 ∏(cos A) (cos2 A ) ⩽ ∏ sin2 A 2 . it suffices to prove that ∏ cos A ⩽ ∏(1 − cos A) (2.3. by the AM-GM inequality. Prove that cos3 A + cos3 B + cos3 C + 5 cos A cos B cos C ⩽ 1. we have √ 3 2 cos A(1 − cos A) ⩾ 3 ∑ ∏ cos2 A ∏(1 − cos A) Thus. tan βc = ⋅ 4S 4S Without loss of generality assume that a ⩾ b ⩾ c.119 2. cyc A 2B 2C sin sin 2 2 2 cyc Without loss of generality we can assume that ∏ b2 + c2 − a2 > 0.35) cyc Using that cos A = A a2 − (b − c)2 b2 + c2 − a2 and 2 sin2 = .35) ⇔ cos A cos B cos C ⩽ ∏(1 − cos A) = cos A cos B cos C cyc ⇔ cos A cos B cos C ⩽ cos A cos B cos C = 8 sin2 ⇔∏ cyc b2 + c2 − a2 a2 − (b − c)2 ⩽∏ 2bc 2bc cyc ⇔ ∏ (b2 + c2 − a2 ) ⩽ ∏(b + c − a)2 . (2. a2 + b2 > c2 .120 Chapter 2. we have ∑ tan A = ∑ B+C A tan B + tan C ⩾ ∑ tan = ∑ cot . (2. 2 2 2 and the equality holds if and only if triangle ABC is equilateral. (⊠) Second solution Since cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1 we will prove that ∑ cos2 A(1 − cos A) = 3 cos A cos B cos C. Examples for practice ∏ cos2 ⇔ A 2 ⩽ ∏ sin ∏ sin A ∏ cos A A A cos 2 2 B C A ⇔ cot cot cot ⩽ tan Atan Btan C 2 2 2 A B C ⇔ cot + cot + cot ⩽ tan A + tan B + tan C.35. 2 2 2 Indeed. cyc By the AM-GM Inequality we have √ 2 2 3 ∏ cos A(1 − cos A) cos A(1 − cos A) ⩾ 3 ∑ cyc cyc then it suffices to prove √ 3 3 ∏ cos2 A(1 − cos A) ⩾ 3 cos A cos B cos C.1) . cyc Then b2 + c2 > a2 . . we get 2bc 2 2bc (2. c2 + a2 > b2 . Thus. ⩾ 2r ka kb kc Solution Let K be the symmedian point of ∆ABC and let BC = a. Prove that 3R ma mb mc + + ⩾ 3. (2.121 2. r bc ca ab . and R the circumradius of a triangle ABC. and we are done. respectively. AB = c. kc the symmedians. This gives (b2 + c2 )A1 = b2 B + c2 C where A1 is the point of intersection of the lines AK and BC and it follows that Ð→ Ð→ (b2 + c2 )2 k 2 = (b2 + c2 )2 AA21 = (b2 AB + c2 AC)2 = b4 c2 + b2 c4 + b2 c2 (b2 + c2 − a2 ) = b2 c2 (2b2 + 2c2 − a2 ) that is. Geometric problems and. mb . Let ma .3. C with masses a2 .1) ⇔ ∏ (b2 + c2 − a2 ) ⩽ ∏(b + c − a)4 . the required inequalities become 3R b2 + c2 c2 + a2 a2 + b2 ⩾ + + ⩾ 6. kb . mc be the medians. m2a b2 + c2 and similarly.36) cyc cyc ∏ (b2 + c2 − a2 ) = ∏ (b4 − (c2 − a2 ) ) 2 cyc cyc and 2 2 4 ∏ (b + c − a) = ∏ (b2 − (c − a) ) 2 cyc cyc it is enough to prove b4 − (c2 − a2 ) ⩽ (b2 − (c − a)2 ) We have 2 2 (b2 − (c − a)2 ) − b4 + (c2 − a2 ) = b4 − 2b2 (c − a)2 + (c − a)4 − b4 + (c2 − a2 ) 2 2 2 = (c − a)2 ((c + a)2 − 2b2 + (c − a)2 ) = (c − a)2 (2c2 + 2a2 − 2b2 ) = 2(c − a)2 (c2 + a2 − b2 ) ⩾ 0. It is well known that K is the barycentre of A. r the inradius. (b2 + c2 )2 ka2 = 4m2a b2 c2 . 2 Because (2. (⊠) Example 2.147. b2 . CA = b. = ka2 2bc m2b c2 + a2 m2c a2 + b2 = and ⋅ = kb2 2ca kc2 2ab As a result.35. c2 . therefore. ka . B. K1 be the midpoints of AB. and b + c ⩾ c + a ⩾ a + b so that (a2 + b2 + c2 )(2a + 2b + 2c) 3 (2. Thus. Prove that UV W K is a cyclic quadrilateral. W. which is immediately true 2 2 2 8 because of the two well-known facts below: A sin 2 A B C r = sin sin sin and R ⩾ 2r. Let ABCD be a cyclic quadrilateral whose diagonals are perpendicular to each other. BI = r . we readily obtain (2. (⊠) . W. K = D ∩ A. Note that O. The Varignon paral.37) (since abc = 2Rr(a + b + c)). respectively.lelogram U1 V1 W1 K1 of the quadrilateral ABCD is a rectangle (because AC ⊥ BD). hence U1 . For a point P on its circumscribed circle denote by P the line tangent to the circle at P . W1 . it rewrites as a2 (b + c) + b2 (c + a) + c2 (a + b) ⩽ 6R2 (a + b + c) Assuming that a ⩽ b ⩽ c. As for the left inequality. K clearly cannot be collinear.149. from which we deduce OH 2 = 9R2 − (a2 + b2 + c2 ). the inverse of γ is a circle and so UV W K is a cyclic quadrilateral. K all lie on the inverse of the circle γ.37). CI = r ⋅ B C sin sin 2 2 B C 1 A the inequality above is equivalent to sin sin sin ⩽ . W1 . R 2 2 2 where R is the circumradius of triangle ABC. which holds since b2 + c2 ⩾ 2bc. where r is the inradius of triangle ABC. Solution Since AI = r . K1 lie on a circle γ centered at the centre of the rectangle. V. W = C ∩ D.38) by Chebyshev’s inequality.122 Chapter 2. c2 + a2 ⩾ 2ca and a2 + b2 ⩾ 2ab. Similar results hold for V1 . CD. Solution Let U1 . W. V. Now. V. V = B ∩ C. Prove that AI ⋅ BI ⋅ CI = 8r 3 . W1 . U1 . ÐÐ→ Ð→ Ð→ Ð→ We know that OH = OA + OB + OC. (⊠) Example 2. let H and O be the orthocentre and circumcentre of ∆ABC. Examples for practice The right inequality rewrites as a(b2 + c2 ) + b(c2 + a2 ) + c(a2 + b2 ) ⩾ 6abc. V1 . Let U = A ∩ B. V1 . respectively. we have a2 ⩽ b2 ⩽ c2 a2 (b + c) + b2 (c + a) + c2 (a + b) ⩽ (2. Example 2. Let I be the incenter of triangle ABC. so that U. Since U.148. DA.38) into account. W1 . K. BC. a2 + b2 + c2 ⩽ 9R2 and taking (2. K1 in the circle G are U. V1 . K1 and it follows that the inverses of U1 . U are collinear (on the perpendicular bisector of AB) and that AB is the polar of U with respect to G. BA′ ⊥ Ia A′ . Thus.4: Example 2. the line Ia B is the perpendicular bisector of A′ C ′ and intersects A′ C ′ in its midpoint B1 . CIa C ′ have a common point. The second result follows from the fact that the inverse of Ga is on the line through Ia and Ga . B ′ . AB. Since Ia A′ = Ia C ′ an BA′ = BC ′ . triangles AB ′ D ′ and B ′ Ia D ′ are similar. hence B ′ D ′ ⋅ C ′ D ′ = B ′ D ′2 = Ia D ′ ⋅ AD ′ . the circumcircles of ∆Ia AA′ . triangles BE ′ A′ and A′ E ′ Ia are similar.123 2. First solution Let γ be the excircle. Geometric problems Figure 2. or D ′ lies on the radical axis of both circles. Prove that the circumcircles of triangles AIa A′ . and the power of D ′ with respect to the circumcircles of A′ B ′ C ′ and AIa A′ is the same. CA. situated on the line Ga Ia . Since B ′ is invariant under this inversion. Denote by A′ . (⊠) Second solution Let D ′ be the midpoint of B ′ C ′ . ∆Ia CC ′ invert into the medians A′ A1 . BIa B ′ . the circumcircle of ∆Ia BB ′ inverts into the median B ′ B1 of triangle A′ B ′ C ′ . different from Ia . B ′ B1 . C ′ the tangency points of the excircle of center Ia with the sides BC. Since A′ C ′ is the polar of B with respect to γ. the three circumcircles all pass through Ia and through the inverse of Ga (because Ga lies on the three medians A′ A1 . Let E ′ be the midpoint of C ′ A′ . As a result. the inversion in the circle γ exchanges B1 and B. Similarly. which is median A′ D ′ . while BIa ⊥ A′ C ′ where A′ E ′ = C ′ E ′ by symmetry around the external bisector of angle B. respectively. C ′ C1 ). C ′ C1 . hence A′ E ′ ⋅ C ′ E ′ = A′ E ′2 = BE ′ ⋅ Ia E ′ . while AIa ⊥ B ′ C ′ where B ′ D ′ = C ′ D ′ by symmetry around the internal bisector of angle A.150. Now AB ′ ⊥ Ia B ′ . Thus. Let Ia be the excenter corresnponding to the side BC of triangle ABC.3. and median B ′ E ′ is the radical axis of the . where Ga is the centroid of triangle A′ B ′ C ′ . Clearly. where k > 2 of a circle with center the midpoint of AB. Examples for practice Figure 2. There is therefore at most one more solution to the equation. Ga have the same power with respect to both.151. then P also has the same power with respect to both. (⊠) Example 2. and M is on that circle. it is also on the circumcircle of CIa C ′ . Since Ia Ga is the radical axis of the circumcircles of AIa A′ and BIa B ′ because Ia . Clearly X = M is one solution.5: circumcircles of A′ B ′ C ′ and BIa B ′ . Solution By the median theorem. ie the symmetric of M with respect to the foot of the perpendicular from the midpoint of AB onto BC. since it is well known that AX 2 + BX 2 = k.124 Chapter 2. Find the locus of points X on line BC such that AB 2 + AC 2 = 2(AX 2 + BX 2 ). it is also on the circumcircle of BIa B ′ . has the same power with respect to the four circumcircles. but since it is on the circumcircle of AIa A′ . The conclusion follows. Any other solution is on the intersection of side BC and the circle with center the midpoint of AB passing AB 2 . the point Ga where the medians A′ D ′ . Similarly. which is the symmetric of M with respect to the perpendicular to BC through the center of the circle. median C ′ F ′ (F ′ is the midpoint of A′ B ′ ) is the radical axis of the circumcircles of A′ B ′ C ′ and CIa C ′ . Let ABC be at triangle. There is exactly one . 4 2 where M is the midpoint of side BC. Similarly. B ′ E ′ and C ′ F ′ meet. AM 2 + BM 2 = AM 2 + BC 2 AB 2 + AC 2 = . consider now the second point P where Ia Ga meets the circumcircle of AIa A′ . is the equation through M. iff the second term in the RHS equals 1. E. The bisector of angle A intersects lines BC and OM at L and Q. if and only if A1 A2 . CD = . A1 B1 .154. ̂>B ̂ −C ̂ ⩾ C. B1 A2 = ⋅ sin ∠AA2 C1 sin ∠AA2 B1 Now.125 2. C2 on segments B1 C1 . lines AA1 . (⊠) ̂ ⩾ 2C. where points A1 . Consider points A2 . B1 B2 .152. CA. ie by Thales’ ̂ = 90o . the first term in the RHS equals 1. theorem when C Example 2. . Prove that DM ⩾ 2 Solution Clearly CD = b cos C and 2CM = a. respectively. ̂ The conclusion follows. BB2 . BB2 . ̂ iff B (⊠) Example 2. Let ABC be a triangle with B AB ⋅ from A and by M be the midpoint of BC. B2 . In triangle ABC. when the midpoint of BC is also the foot of the perpendicular from the midpoint of AB onto BC. we find: BD = C1 A2 = AC sin ∠CAD AB sin ∠BAD . BB1 . B1 . respectively. C1 A1 . CA. ∠BAD = ∠C1 AA2 and ∠CAD = ∠B1 AA2 . We then conclude that BD CE AF AB1 CA1 BC1 A1 B2 C1 A2 B1 C2 ⋅ ⋅ ) ⋅ ⋅ =( ⋅ ⋅ )( DC EA F B B1 C A1 B C1 A B2 C1 A2 B1 C2 A1 By Ceva’s theorem. ie exactly one point X that satisfies the given condition. The conclusion follows. Prove that AA2 . CC2 are concurrent iff the LHS equals 1. C1 C2 are concurrent. = ⋅ CD AC AC1 B1 A2 and similarly for its cyclic permutations. AB. sin ∠ADB sin ∠ADC AB1 sin ∠B1 AA2 AC1 sin ∠C1 AA2 . AA2 . CC2 are concurrent if and only if A1 A2 . and again by Ceva’s theorem. since 180o = ∠ADB + ∠ADC = ∠AA2 C1 + ∠AA2 B1 . AB. we find BD AB AB1 C1 A2 ⋅ . CC1 are concurrent at P . Solution Let D. C1 are situated on sides BC. Applying the Sine Law to triangles ADB. B1 B2 .153. BB2 . CC2 intersect lines BC.3. or since by the Sine Law: a = 2R sin A = b cos C + c cos B. respectively. C1 C2 are concurrent. respectively. Let ABC be a triangle with incenter I and circumcenter O and let M be the midpoint of BC. ADC. Prove that AI ⋅ LQ = IL ⋅ IQ. F the second points where AA2 . Geometric problems solution. C1 AA2 and B1 AA2 . and equality is reached where we have used that π − C ̂ = 2C. we have 2DM = 2b cos C − a = b cos C − c cos B = 2R sin(B − C) = 2R sin C = AB. ̂ Denote by D the foot of the altitude Example 2. which are therefore symmetric with respect to BC. if P is on the same half plane as A. ie BC is the perpendicular bisector of T U. Solution (readers build their own figure).126 Chapter 2. . C with respect to γ are respectively PB . since the triangles BLQ and ALC are similar we can write BQ AC = ⋅ LQ LC Now repeated use of the angle bisector theorem in the triangles ABC and ABL yields IQ BQ AC AB AI = = = = ⋅ LQ LQ LC LC IL Therefore AI ⋅ LQ = IQ ⋅ IL. Q is the midpoint of the arc BC not containing A. Let now T. U be the second points where P B. On the other hand. Let ABC be a triangle such that ∠ABC > ∠ACB and let P be an exterior point in its plane such that P B AB = P C AC Prove that ∠ACB + ∠AP B + ∠AP C = ∠ABC. as desired. It is a well-known that the bisector of ∠CAB and the perpendicular bisector OM of BC intersect on the circumcircle of ABC. passes through A and through the point D where the internal bisector of angle A intersects BC. leaving B inside γ and C outside γ because ∠ABC > ∠ACB. ∠AP B = ∠AP T = 180o − ∠ADT = 180o − ∠ADB − ∠BDT = 180o − ∠ADB − ∠BDU = 180o − 2∠ADB − ∠ADU = 180o − 2∠ADB − ∠AP U = 180o − 2∠ADB − ∠AP C. Note that the powers of B. Therefore.155. In other words. and similarly BT = BU. (⊠) Example 2. Observe that BQ = IQ. Examples for practice Solution (readers build their own figure). since by angle chasing ∠QBI = ∠QIB = π ∠BCA − 2 2 (in fact Q is the circumcenter of BIC). P C meet γ (the first one being clearly P in both cases). PC . = = CU c ⋅ CU c ⋅ CU ⋅ P B c ⋅ CU ⋅ P C c ⋅ PC or CT = CU. such that PB BD ⋅ BD′ BA2 c2 = = = PC CD ⋅ CD′ CA2 b2 where D ′ is the point diametrally opposite D in γ. P B AB Note that the relation = clearly defines an Apollonius circle γ with center on P C AC line BC. Note therefore that b ⋅ PB b2 ⋅ PB b2 ⋅ PB CT b ⋅ BT = 1. BC1 = B C 2abc(a + b + c) A 4a2 bc cos cos = sin (a + b)(a + c) 2 2 (a + b)(a + c) 2 ac a2 bc and this yields BC1 ⋅ CB1 = . ) . and these imply that a+c 2 2 2 a 2 BB1 ⋅ CC1 = In addition. in order to get(∗). to establish the inequality we need to show that 1 + cos B + cos C ⩽ 3 sin ( A π + ) 3 6 (∗) Now. 4R sin3 ( A π A π + ) = R [3 sin ( + ) − cos A] . observe that it is enough to prove the inequality for acute angles B and C. Prove that in any triangle. we find A ∠AP B + ∠AP C = 180o − 2∠ADB = 180o − 2 (180o − B − ) = 2B + A − 180o = B − C. its cosine is nonpositive. since if one of them is obtuse or right. 3 6 Solution We know that ra = R(1 − cos A + cos B + cos C). note that by Ptolemy’s inequality for quadrilateral BCB1 C1 . 2 The conclusion follows. 3 6 (⊠) .156. In triangle ABC let BB1 and CC1 be the angle bisectors of B Prove that A a 2bc [(a + b + c) sin − ] . (⊠) ̂ and C.3. On the π other hand. In either case. cos x is concave on the interval (0. ra ⩽ 4R sin3 ( A π + ). therefore by just a+b (a + b)(a + c) plugging in these identities in the inequality above. ̂ Example 2. A π 0+B +C ) = sin ( + ) (∗∗) 1 + cos B + cos C ⩽ cos ( 3 3 6 This proves our inequality.127 2. as from (∗∗) it follows that ra ⩽ 4R sin3 ( A π + ). thus by Jensen’s inequality for concave 2 functions. B1 C1 ⩾ BB1 ⋅ CC1 − BC1 ⋅ CB1 ⋅ BC 2ac B B C p A Now. the desired conclusion follows. Also. B1 C1 ⩾ (a + b)(a + c) 2 2 Solution First.157. 3 6 3 6 Therefore. (⊠) Example 2. Geometric problems and similarly we find the same result if P is on the opposite half plane. BB1 = cos and that cos cos = sin . 39) ⩽ (a + b + c) ln a+b−c+b+c−a+c+a−b = (a + b + c) ln 1 = 0 a+b+c (2. either a = b = c. a b c Clearly.s Inequality we get LHS(2. or the triangle is degenerate with two equal sides and one side of length 0. b−a with equality iff x = 0.39) . the inequality being trivially true. ⎟ ⎜ a b c a+b+c 2 ⎠ ⎝ a c b that is equivalent to a a b b c c (s − c)a (s − a)b (s − b)c ⩽ ( ) ( ) ( ) . (⊠) Third solution The inequality is equivalent to (a + b − c)a (b + c − a)b (c + a − b)c ⩽ aa bb cc It suffices to shows that a ln a+b−c b+c−a c+a−b + b ln + c ln ⩽0 a b c Since the function f (x) = ln x is concave. The triangular inequality guarantees that −1 < < 1 for nonc b−c degenerate triangles. the inequality may be rewritten in the following equivalent form: a log (1 + b−c c−a a−b ) + b log (1 + ) + c log (1 + ) ⩽ 0. Examples for practice Example 2. occurring iff x = 0. Note that if 1+x x > 0. c be the side lengths and let s be the semiperimeter of a triangle ABC. while if x < 0. g(x) = x − log(1 + x) is zero for x = 0. by Jensen. Prove that a a b b c c a b c (s − c) (s − a) (s − b) ⩽ ( ) ( ) ( ) . while g ′ (x) = 1 − 1 . b. Let a. while the RHS is nonnegative. Dividing by the RHS and taking logarithm. then g(x) strictly increases. We need to consider thus only non-degenerate triangles. then g(x) strictly decreases. or a log 1 + ⩽ b − c with equality iff b = c. or the maximum of g(x) is 0. 2 2 2 First solution By the weighted AM-GM inequality we have a+b+c s−a s−b s−c ⎞ ⎛ ⋅ a + ⋅ b + ⋅ c s−c s−a s−b 1 a b c ⎟ ( ) ( ) ( ) ⩽⎜ = a+b+c .128 Chapter 2. Adding this inequality a to its cyclic permutations we obtain the proposed inequality. Equality holds if and only if. then the LHS is identically zero. x ⩾ log(1 + x) for all x > −1. 2 2 2 (⊠) Second solution If the triangle is degenerate. As a consequence.158. CA1 CS = . Prove that AX ⋅ DY = BX ⋅ CY. Geometric problems Hence we are done. Consider a point P inside a triangle ABC. Solution ̂ ̂ ̂ ̂ Note first that BT M =A 1 T C iff CT M = A1 T B since ̂ ̂ ̂ ̂ ̂ BT M + CT M =A 1 T B + A1 T C = BT C. Let ABCD be a cyclic quadrilateral and let {U} = AB ∩ CD and {V } = BC ∩ AD. . Prove that if the circumcircle of triangle BT C is tangent to the line ̂ ̂ B1 C1 . The line that passes through V and is perpendicular to the angle bisector of angle ̂ AUD intersects UA and UD at X and Y . Let AA1 . then BT M =A 1 T C. ⩾ (1 + b c Multiplying these three inequalities we get the desired result. respectively. Since ex = 1 + x ⇔ x = 0. Therefore. e ⩾ (1 + ) . or since SA1 = SB + BA1 = SC − CA1 wlog.3.159. BB1 . By Pappus’ harmonic theorem. a b c Employing the well known inequality ex ⩾ 1 + x and setting x = e b−c a Analogously. then A1 B SB SA1 ⋅ SM = SA1 SB + SC = SB ⋅ SC. or we may exchange B and C at will without altering the problem. CC1 be cevians through P . ̂ ̂ ̂ ̂ ̂ ̂ ̂ ̂ BT M = ST M − ST B = SA 1 T − T CB = π − CA1 T − T CA1 = A1 T C. it is clear that the equality holds iff a = b = c. or the power of S with respect to the circumcircle of A1 T M is SA1 ⋅ SM = ST 2 . and ST is the tangent at T to the circumcircle of A1 T M. we have c−a e ⩾1+ b−c we get a b−c b−c a ⇒ eb−c ⩾ (1 + ) a a a−b c c−a b a−b ) .160. The equality holds when a = b = c. (⊠) Example 2. (⊠) Example 2. Call S the point where the tangent to the circumcircle of BT C meets BC. and T is the intersection of AA1 and B1 C1 . 2 But the power of S with respect to the circumcircle of BT C is ST 2 = SB ⋅ SC. The midpoint M of BC different from A1 . The conclusion follows. (⊠) Fourth solution Observe that the inequality we have to prove is equivalent to (1 + b−c a c−a b a−b c ) (1 + ) (1 + ) ⩽ 1.129 2. DV k − b e + f Then the same operation is done for ∆AUD with a secant XV : AV c k ⋅ ⋅ = 1. Consequently. a = k − b and d = k − e. due to 2 triangle inequalites a.40) a b c abc First solution a+b+c Let s = is semiperimeter of a triangle with sidelengths a.we apply the Menelaus theorem for triangle ∆AUD with a secant BV : Moving k to the left side. c. (c − f )(be − cf ) = 0 ⇔ (Y D − AX)(XB ⋅ CY − Y D ⋅ AX) = 0.Triangle ̂ is perpendicular to XY. Prove that in any triangle with sidelenghts a. respectively. as bisector of U Considering this fact. y ∶= s − b. XB. c = x + y. z > 0. c < s and setting x ∶= s − a. b = z + x.130 Chapter 2. CY. (⊠) Example 2. s = x + y + z. a + b = d + e. b. we get: k = AV b + c k − e ⋅ ⋅ = 1. Y D. y. z ∶= s − c we obtain a = y + z. AF as a. we consider the equation: (k − e)(k + f ) = (k − b)(k + c) (bc − ef ) ⋅ (c − b + e − f ) Now. As we have a cyclic quadrilateral. BU. d. Then. c the following inequality holds: b + c c + a a + b (b + c − a)(c + a − b)(a + b − c) + + + ⩾ 7. . (2.161. b. b. let’s denote UC. b. DV k f Equilizing both parts gives us: f b+c k−e ⋅ ⋅ = 1. c k−b e+f Substituting the value of k: f ⋅ c b+c (b−e)(b+f ) c−b+e−f Factorization yields to: ⋅ (b−e)(e+c) c−b+e−f e+f = 1 ⇔ f (b + c)(c + e) = c(b + f )(e + f ). As Y D ≠ AX we get the desired answer: XB ⋅ CY = Y D ⋅ AX. ∆UXY is isosceles. c. where x. Let k be equal to a + b. e and f. Examples for practice Solution First. (⊠) Second solution We make the well-known substitution a = x + y. x(z + x)(x + y) + y(y + z)(x + y) + z(y + z)(z + x) + 4xyz = 2(x + y)(y + z)(z + x) which is equivalent to x3 + y 3 + z 3 − xy 2 − yz 2 − zx2 − x2 y − y 2 z − z 2 x + 3xyz ⩾ 0. . c = x + y. after multiplying both sides by (x + y)(y + z)(z + x).3. (⊠) Third solution The inequality to be proved may be written in the form Let x = b + c − a c + a − b a + b − c (b + c − a)(c + a − b)(a + b − c) + + + ⩾ 4. original inequality becomes 8xyz 2x + y + z + ⩾7 y+z (y + z)(z + x)(x + y) cyc (2.131 2. Geometric problems Thus. y+z z+x x+y (x + y)(y + z)(z + x) After multiplying both sides by (x+y)(y +z)(z +x). Then a = y + z. y.41) b+c−a c+a−b a+b−c . y. c 2 2 2 are sides of a triangle. We substitute these into (2.40) ⇔ ∑ 8xyz s+x + ⩾7 (s − x)(s − y)(s − z) cyc s − x ⇔∑ ⇔ ∑(s + x)(s − y)(s − z) + 8xyz ⩾ 7(s − x)(s − y)(s − z) cyc ⇔ ∑(s + x)(sx + yz) + 8xyz ⩾ 7(xy + yz + zx − xyz) cyc ⇔ 2s3 − 8s(xy + yz + zx) + 18xyz ⩾ 0 ⇔ (x + y + z)3 − 4(x + y + z)(xy + yz + zx) + 9xyz ⩾ 0 ⇔ ∑ x(x − y)(x − z) = 0. note that x. (Schure Inequality) cyc and we are done. b. b = y + z and c = z + x. The inequality becomes equivalent to: 2x + y + z x + 2y + z x + y + 2z 8xyz + + + ⩾ 7. since a.z = . z are positive. a b c abc (2. where x.y = . z > 0. the inequality becomes equivalent with: 2(x3 + y 3 + z 3 ) + 6xyz = 2xyz ⋅ ∑ xy(x + y) ⇔ ∑ x(x − y)(x − z) ⩾ 0 cyc cyc which is just Schur’s inequality. b = z + x.41) and obtain. then angles GF H and CGF are both obtuse. P Q. which implies that ∠F DG = ∠F HG. Let U = AB ∩ XS and V = AC ∩ XR. (⊠) Example 2. then DHGF is cyclic. The circle with center A and radius AX intersects line AB at P and R and line AC at Q and S such that P ∈ AB and Q ∈ AC. Prove that G is the incenter of triangle AHI. as we wanted to prove. If H lies between C and D.132 Chapter 2. only if ABC is equilateral triangle. Equality occurs only if x = y = z. respectively.. In order to prove that G is in the interior of triangle AHI it is enough to show that F lies between H and I. so ∠GF H = ∠GDH = 180o − ∠F DG − ∠C = 180o − ∠ACG − ∠C.162. Therefore. only if a = b = c. as we wanted to prove. (⊠) Example 2. First solution . If H lies between C and D then DHGF is cyclic. Examples for practice This. Let D be an arbitrary point on BC and let E be its reflection in the side AB. ∠F DG = ∠F HG. so ∠AHG = ∠ACG = ∠ACE = ∠CED = ∠EDG = ∠F DG. Denote by F and G the intersections of AB with lines DE and CE. respectively. which implies that ∠F DG = ∠F HG. This result and the fact that G lies in the internal bisector of angle AHF implies that G 1 lies on the internal bisector of angle AHI. In both cases. i. and this proves that G lies on the internal bisector of angle AHF. Note that ∠AGI = 90o +∠ACG = 90o + ∠AHI. Note 2 that G lies on segment AF and that AGHCis cyclic. which implies that rays CE and HF intersect beyond E and F. Prove that lines BC. If D lies between C and H. Let ABC be a triangle and let X be the projection of A onto BC. is equivalent to x(x − y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y) ⩾ 0 which follows by Schur’s inequality.163. where we have used the fact that ∠F DG = ∠ACG. If D lies between C and H. ∠GF H + ∠CGF = (180o − ∠ACG − ∠C) + (90o + ∠ACG) > 180o . as we proved it before. Solution We will use the trivial fact that if a point P in the interior of XY Z is such that P lies 1 on the bisector of angle Y XZ and ∠Y P Z = 90o + ∠Y XZ then P is the incenter. Let ABC be a right triangle with ∠A = 90o . respectively.e. Let H be the projection of G onto BC and let I be the intersection of HF and CE. which implies that rays CE and HF intersect beyond E and F. in turn. then HDGF is cyclic. UV are concurrent. 2 and this completes the proof. so ∠AHG = ∠F HG. R. by the reciprocal of Menelaus’ theorem. (Oad ) intersect again at A1 . B1 . (ii). The pairs of circles (Oab ). (Obc ). Prove that A1 . and (v) we get: BZ SC BX AR BX BX 2 SC BP ⋅ BR ⋅ SC BP BY = ⋅ ⋅ ⋅ = ⋅ = = = . UV.164. Multiplying (iii) by (iv). It would suffice to prove that = ⋅ Y C ZC From Menelaus’ theorem for triangle ABC and secants P Q. and CX = b∣ cos C∣. B. . DA. Q are collinear. Y. SX. BY UA BV BX2 AR CS c cos2 B 1 + sin C = ⋅ = ⋅ ⋅ = ⋅ Y C CU V A CX2 BR AS bcos2C 1 + sin B c − c sin B P B QA P B = = ⋅ . BC. (Oab ). (Oad ) be the symmetric circles to C(O) with respect to AB. AU AS CX AV AR BX = ⋅ ⋅ CV BR CX Because AX ⊥ BC. Menelaus’ theorem ensures that CX BV AS BX CU AR ⋅ ⋅ = 1 and ⋅ ⋅ = 1. (Ocd ). RX we get: BY BP AQ BP = = . and used that AX = AP = AQ = AR = AS = b sin C = c sin B. are triplets of collinear points. (Ocd ). U. respectively. B1 . (⊠) Second solution Denote by Y the intersection of BC and P Q and by Z the intersection of BC and BY BZ UV. V. CD. = b − b sin C CQ CQ AP where we have applied the Sine Law. or BC. (Obc ). S. using (i). ZC AS CX BR CX CX 2 BR CQ ⋅ CS ⋅ BR CQ Y C (i) (ii) (iii) (iv) (v) (⊠) Example 2. XC UA RB XB V A SC Call Y = BC ∩ UV. D1 lie on a circle of radius R. P. BX = c∣ cos B∣. Geometric problems Since X. so: BX 2 = BP ⋅ BR and CX 2 = CQ ⋅ CS. R) and let (Oab ). Y C AP CQ CQ BZ BU AV = ⋅ . Therefore. P Q. BC is tangent to the circle with radius AX. where a.3. Let ABCD be a quadrilateral inscribed into a circle C(O. Therefore. (Obc ). UV meet at Y. and X. D1 . (Oad ). c are obviously the lengths of the sides opposing vertices A. (Ocd ). again by Menelaus’ theorem. ZC AU CV BU SC BX = ⋅ . C. C1 . C1 . b.133 2. Similarly. D with respect to NP. b. so C1 D1 = AB and C1 D1 ∥ AB. thus OA ⊥ B1 C2 and since a ⊥ B1 C2 we get OA ∥ a.134 Chapter 2. respectively. Let M. DA. Y be the midpoints of OP and QP respectively. Ma Mb and let a. 4 Quadrilateral Mc Mb B1 C2 is cyclic. Mc in triangle AMb Mc be C2 and B1 . Denote by P the intersection of MB and CF and by Q the intersection of MC and BE. Ma in triangle BMa Mc be A2 and C1 . BC1 QO is a rhombus. They have the same centroid G and at the same time G is the center of homothety with the ratio 1. C0 be the midpoints of Mb Mc . Working analogously for other sides of the quadrilateral. Prove that EF bisects the segment P Q at the midpoint of P Q. MN. triangles A0 B0 C0 and ABC are homothetic. Example 2. Since triangle A0 B0 C0 is the complementary triangle of triangle Ma Mb Mc and triangle Ma Mb Mc is the complementray triangle of ABC. AB. This implies that A0 ∈ a. . C1 A2 . Indeed. CA.166. Let ABC be a triangle and let Ma . Let ABC be a triangle with altitudes BE and CF and let M be a point on its circumcircle. QM and P Q respectively.165. This means that line a in triangle A0 B0 C0 is homologous to the radius OA of triangle ABC and so O0 ∈ a. AB. Let H be orthocenter of the triangle and D the intersection of the lines P Q and EF. we get AD1 ∥ XY ∥ OQ and AD1 = 2XY = OQ = R. Prove that the perpendicular bisectors of B1 C2 . A1 B2 . Examples for practice Solution We prove that the quadrilateral A1 B1 C1 D1 is congruent to ABCD and the conclusion follows. We are going to show that AD1 QO is a rhombus. B0 . let X. hence BC1 = AD1 = R and BC1 ∥ OQ ∥ AD1 . Since XY is median line in ∆OP Q and ∆ADD1 . Since OA = R. C1 . Thus ÐÐ→ 1 Ð→ (1) GO0 = GO. respectively. and triangle A0 B1 C2 is isosceles. C1 A2 . the circumcenter of triangle A0 B0 C0 . N. Q be the reflections of O with respect to BC. Mb . B1 . Then it is easy to see that A1 . It follows that ABC1 D1 is a parallelogram. Mb in triangle CMa Mb be B2 and A1 . Ma Mc . Mc be the midpoints of sides BC. the midpoint of Mb Mc . respectively. and A1 B2 are concurrent. the feet of the perpendiculars from vertices Mc . B. c be the perpendicular bisectors of B1 C1 . the feet of the perpendiculars from vertices Ma . Then X is the midpoint of AD and Y is the midpoint of P Q. Analoguously it can be proved that b and c pass through O0 . we obatin the conclusion. Thus A0 B1 = A0 C2 . Solution Denote by E1 and F1 intersections of the altitudes BE and CF with the circumcircle of triangle ABC. Solution Let A0 . Let O and O0 be the centers of their circumcircles. AD1 QO is a rhombus. P. Hence AD1 QO is a parallelogram with AD1 = OQ = R. On the other hand Mb Mc ∥ BC. D1 are reflections of A. and CD. C. having center at A0 . Let the feet of the perpendiculars from vertices Mb . (⊠) Example 2. 3.t. We have AC ⋅ BD = . respectively. EE1 EC F Q BF = . Prove that the product of its diagonals is a constant.r. On the other side we have S = abcd and thus √ √ 2 abcd b a AC ⋅ BD = =2 ⋅ ⋅ cd sin α sin α sin α √ d c =2 ⋅ ⋅ AP ⋅ CP sin ∠ABP sin ∠DBC √ = 4R AP ⋅ CP .6: It is well known that E1 is reflection of H w. Consider a quadrilateral that is incribed in a circle and circumscribed about a circle. F H EP DQ PD (⊠) Example 2.135 2.167. So ∆EE1 C ≡ ∆EHC ∼ ∆F HB then = . . so HE = EE1 and F H BF ∠E1 HC = ∠HE1 C. Then we have P ∈ OI and OP = 2R2 ⋅ OI ⋅ R 2 + d2 2S These two relations imply that the point P is fixed. Geometric problems Figure 2. point E. so ∆F BQ ∼ ∆ECP ⇒ EP EC From (i) and (ii) we obtain that (i) (ii) FQ FH FH F Q EP = = ⇔ = ⋅ EP EE1 EH F H EH Then apply Menelaus Theorem to the triangle QHP and transversal F DE: F Q HE P D DQ ⋅ ⋅ =1⇔ = 1. First solution Let ABCD be a bicentric quadrilateral and let O and I be the circumcenter and incenter. But ∠ABM = ∠ACM. where sin α √ α is the angle between diagonals. Q. and hence Q varies continuously from E to C.7). respectively. BC = DA. Thus the circumcenter of ∆AQU is the midpoint O1 of AU. Points Q and CQ F R = ⋅ Determine the locus R lie on segments CE and BF. such that QE QE of the circumcenter of triangle AQR when Q and R vary. If Q is a point of CE.136 Chapter 2.168. the point R can be costructed in the following way: ˆ through point Q draw a line parallel to BE to intersect BC at point U . From Thales’ theorem we have: CQ CU CU F R = . Let ABC be an acute triangle with altitudes BE and CF. = QE UB UB RB so the point R satisfies the relation: CQ F R = ⋅ QE RB The circle C with diameter AU contains Ha . AB ⋅ CD + BC ⋅ DA = AC ⋅ BD AB 2 + BC 2 = AC ⋅ BD (⊠) Example 2. and the locus of the circumcenter of AQR is clearly the segment joining the midpoints of AB and AC. then P varies continuously from B to C. (⊠) Second solution (See Figure 2. Examples for practice √ But the value 4R AP ⋅ CP is constant because both 4R and AP ⋅CP is constant(power of the point P which is fixed). P R ∥ CF ⊥ AB. where U is a variable point . (⊠) Second solution Let ABCD be a quadrilateral that is inscribed in a circle and circumscribed about a cirle. R because ∠AQU = ∠ARU = ∠AHa U = 90o . As R varies continuously from B to F. This implies that the required locus is the set of mid-points of the cevians AU. (I) According to the Ptolemey theorem. or AP is a diameter of the circumcircle of AQR. ˆ through point U draw a line parallel to CF to intersect AB at point R. First solution BP BR EQ = = ⋅ Clearly. we have and using (I) we get that which is constant. and P Q ∥ BE ⊥ AC. by Thales’ Denote by P the point in segment BC such that P C RF QC theorem. then we have AB = CD. (⊠) Third solution (See Figure 2. P ′ ∈ BC.137 2. Geometric problems Figure 2.7: of BC. Figure 2. such that P R ⊥ AB and P ′ Q ⊥ AC.3. AC. BP ′ EQ BP BR = and ′ = ⋅ P C RF P C QC . In other words the locus is the segment joining the midpoints M.8: so Let P. N of the sides AB.8). Then P ′ Q ∥ BE and P R ∥ CF. Hb . Analogously. so the circumcenter of the triangle ARQ is midpoint of the segment AP. P moves on BC and ints midpoint always lies on the midline of the triangle ABC. Examples for practice But therefore from the given relation BC BC BP BP ′ = ′ ⇔ = ⋅ PC P C P C P ′C So P ′ C = P C and P ≡ P ′ .169. Hd the orthocenters of triangles AKN. hence Ha KP N is a parallelogram. triangles Ha HbK and Hd Hc M are congruent and with all of its correspondent sides parallel. and by Ha . Hb K = Hc M and Hb K∣∣Hc M. NHd Hc L is a parallelogram. so P N ∥ KHa . CLM. Oc . Hb . Analogously. Therefore. (⊠) Third solution Let us prove that Ha Hb ∥ NL. we prove that Ha Hd is parallel to Hb Hc and this completes the proof. Similarly. BKL. Ha Hb is parallel to Hd Hc . Using following properties: ˆ G is between H and O and OH = 3OG. Thus means NHa Hb L is a parallelogram. Hc . Gd the gravity gravity of these triangles. DA. It implies that NHa is parallel and equal to LHb (both are parallel and equal to P K). Similarly. and DMN.3. Denote by Oa . and by Ga . When Q and R vary. respectively. Ob . respectively. we deduce that the quadrilateral P KHa N is a parallelogram. Gb . it follows that Hd M = P N and Hd M∣∣P N. Thus Ha K = Hd M and Ha K∣∣HdM. Hence. (⊠) 2. Let ABCD be a quadrilateral and let P be a point in its interior. Hd are the vertices of a parallelogram. Quadrilateral ARP Q is cyclic (because ∠P RA = ∠P QA = 90o ). Od the circumcircles of triangles AKN. This implies that Ha K = P N and Ha K∣∣P N. Hence. KHa ⊥ AN. Hb . respectively. BC. Hc .3 Olympiad problems Example 2. ˆ Oa is midpoint of AP and Ob is midpoint of BP we get ÐÐÐ→ ÐÐÐ→ ÐÐ→ Ð→ ÐÐ→ ÐÐÐ→ Ha Hb = Ha Oa + Oa A + AB + BOb + ObHb ÐÐÐ→ ÐÐ→ Ð→ ÐÐ→ ÐÐÐ→ = 3Ga Oa + Oa A + AB + BOb + 3Ob Gb . (⊠) Second solution Note that Ha N∣∣KP and Ha K∣∣NP . CD. Prove that Ha . Denote by K. We also have P N–AD. BKL. M. so P K ∥ NHa . Hc . Gc . DMN. N the orthogonal projections of P onto lines AB.138 Chapter 2. First solution We have P K ⊥ AB and NHa ⊥ AK. Hd are the vertices of a parallelogram. Ha . as desired. Similarly. CLM. L. Therefore we conclude that Ha Hb and Hc Hd are parallel and equal. In particular. the quadrilateral P LHb K is also a parallelogram. B ′ . B2 . C2 .3. and let A′ . respectively. B1 . C2 are concyclic. CA. First solution Figure 2.9: Considering the power of A1 with respect to the circumcircle C(O. A′ H) with side BC. C ′ be the midpoints of sides BC. Prove that points A1 . A2 . R) of ∆ABC. Let H be the orthocenter of an acute triangle ABC. (⊠) Example 2. Denote by A1 and A2 the intersections of circle C(A′ . we obtain ÐÐ→ ÐÐ→ BC 2 (2. In the same way we define points B1 . C1 . Hc Hd ∥ NL ∥ Ha Hb and Ha Hc ∥ KM ∥ Hb Hd .139 2. 2 2 Analogously. B2 and C1 . hence Ha Hb Hc Hd is a parallelogram.170. AB.42) OA2 − R2 = A1 C ⋅ A1 B = A1 A′2 − 4 . Geometric problems ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ Ob B + Ob K + Ob L AOa + KOa + NOa ÐÐ→ Ð→ ÐÐ→ + Oa A + AB + BOb + 3 =3 3 3 ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ Ð→ ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ = AOa + KOa + NOa + Oa A + AB + BOb + Ob B + Ob K + Ob L ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ Ð→ Ð→ AOa + Oa A KP + KA NP + NA AB + AB = + + + 2 2 2 2 ÐÐ→ ÐÐ→ ÐÐ→ ÐÐ→ Ð→ Ð→ BOb + Ob B P K + BK P L + BL + + + 2 2 2 ÐÐ→ ÐÐ→ ÐÐ→ Ð→ ÐÐ→ ÐÐ→ Ð→ ÐÐ→ Ð→ Ð→ KP + P K KA + AB + BK NP + P L NA + AB + BL = + + + 2 2 2 2 Ð→ Ð→ NL NL Ð→ = + = NL. Observing that OC ′ = R cos C (since C is acute).43) 4 where we have used the well-known CH = 2OC ′ . B2 . The power of a point theorem implies AB1 ⋅ AB2 = AH ⋅ AK = AC1 ⋅ AC2 . B ′ H. B1 . cos C − sin A sin B = − cos(A + B) − sin A sin B = − cos A cos B so that OA21 = R2 (1 − 4 cos A cos B cos C). If K is the orthogonal projection of C onto AB. B.42) and (2. C ′ H respectively. Examples for practice Let A. C2 are all on the same circle (with centre O).10: Let Ca . Denote by K the second intersection point of Cb and Cc . Thus AH is the radical axis of Cb and Cc . Now. Since B ′ C ′ ∥ BC and AH ⊥ BC we have AH ⊥ B ′ C ′ . (2. C ′ and radii A′ H. A2 .43) readily give OA21 = R2 + 4R2 cos C(cos C − sin A sin B). so K ∈ AH. C denote the angles of the triangle. Cc be the circles with centers A′ . besides H. BC = 2R sin A and A′ H = A1 A′ . B ′ . (⊠) Second solution Figure 2. Owing to the symmetry of the result. C1 . Cb .140 Chapter 2. we see that OA1 = OA2 = OB1 = OB2 = OC1 = OC2 and A1 . we clearly have ∠BCK = 90o − B. hence the law of cosines in triangle CHA′ yields A′ H 2 = A′ C 2 + CH 2 − 2A′ C ⋅ CH ⋅ cos(90o − B) = BC 2 + 4OC ′2 − 2BC ⋅ OC ′ sin B (2. b. Geometric problems so the points B1 . (⊠) Example 2.141 2. Therefore A1 . C1 . C2 are concyclic. A2 lies on the circle with center O and radius OB1 . C2 lies on the circle with center O and radius OB1. C2 are concyclic and the proof is complete.GM since done.GM (x + y) √ whence xy ⩽ 2 1 4 (x + y)(y + z)(z + x) ∑ ( + ) ⩾ 8(x + y + z)2 x+y cyc 2 By clearing the denominators we get ∑ x3 y 2 ⩾ ∑ x3 yz cym cym which follows also by the AM . B2 .171. Similarly. A1 . Let a.3. Prove that √ √ √ abc abc abc + + ⩾ a + b + c. B2 . it follows that B1 . Since the axes of B1 B2 and C1 C2 intersect at the circumcenter O of ∆ABC. and cyclic and we are 2 (⊠) Second solution Applying H¨ older’s inequality it follows that √ 3 ⎛ a2 (−a + b + c) abc ⎞ (∑ ) ⩾ (∑ a) . z = b + c. ∑ bc ⎝ cyc −a + b + c ⎠ cyc cyc It is enough to prove that a2 (−a + b + c) ⋅ bc cyc ∑a ⩾ ∑ cyc . B2 . A2 . 2 2 2 or y = a + b. c = (z + x − y). x = a + c. This sets the inequality as √ √ 1 1 1 (x + y)(y + z)(z + x) ( √ + √ + √ ) ⩾ 2 2(x + y + z) y x z Squaring we get 2 1 (x + y)(y + z)(z + x) ∑ ( + √ ) ⩾ 8(x + y + z)2 xy cyc 2 By AM . c be the sidelengths of a triangle. b = (y + z − x). we can prove that B1 . B2 . B1 . −a + b + c a−b+c a+b−c First solution We change variables 1 1 1 a = (x + y − z). C1 . (x3 y 2 + x3 z 2 ) ⩾ x3 yz. C1 . while out of P X. We may thus associate exactly one intersection point in the interior of ABC to each such pair of pairs of points. On side BC of triangle ABC consider m points. who are on its border. Each one of these two lines passes through one point on BC. and Q is outside P XY . otherwise. another taken from N ∪ S. we will assume that no three lines intersect at a point inside the triangle. for each one of the lines P Q (with P ∈ M and Q ∈ N ∪ S) that concurs at a given point with two or more other lines. To achieve this we first count the intersections between the sides BC and CA. and any pair of points X. S the set of points on sides BC. P is outside QXY . then between BC and AB and finally between 2 points. and another point either on CA or AB. Since the problem statement says nothing about the locations of points on these sides. Join the points from the sides AB and AC with the points on side BC. since it is a point on the border of ABC. otherwise either the lines would be the same. we can move ever so slightly P on BC until it passes through no point where two other lines concur. Therefore. we notice that the maximum number of points of intersection can be obtained as soon as the intersections occur between no more than two segments.172. neither can the points on CA. two will be sides of the quadrilateral and will thus meet outside ABC. thus increasing the number of intersection points (line P Q will now meet each one of the lines with which it concurred at different intersection points. this is also the maximum Since the number of pairs of pairs of points is Cm n+s number of intersection points in the interior of ABC. and between AB and CA there are snC 2 points. Indeed. Determine the maximum number of the points of intersection situated in the interior of triangle ABC. QX. N. First solution Denote respectively by M. Q. except for segment P Q and vertexX. instead of at one single intersection point). of ABC. on CAn points. one pair of points taken from M. consider any pair of points P. Y are the vertices of a convex quadrilateral. Y is outside P QX. X is outside P QY. So we just count all the possible intersections between the segments. and two will be their diagonals and will meet in the interior of ABC . Y ∈ N ∪S. and distinct from X. not on BC. X. Between BC and CA there are Cn2 Cm 2 points. cyc and we’re done. Note that P QX is a triangle contained entirely inside ABC. Note now that we can establish a bijection between the pairs of pairs of points. Note that P QnXY is clearly on line BC. or their intersection point would be on the border. Examples for practice It’s easy to verify that the last inequality is equivalent to Schur’s inequality ∑ a2 (a − b)(a − c) ⩾ 0. or P. Q ∈ M.142 Chapter 2. We may thus associate exactly one pair of pairs of points to each intersection point. (⊠) Example 2. AB. exactly two lines pass through any intersection point inside the triangle. AB. Reciprocally. Similarly. are Cs2 Cm m . (⊠) Second solution First. CA. between BC and AB there AB and CA. no other intersection points of any two lines defined by these pairs of points may be in the interior of ABC. Note also that the two points on BC cannot coincide. not in the interior. 2 C 2 . and on ABs points. QY and P Y. and the number of intersection points in the interior of ABC. 2 P 2 ) = (x − sy)2 + (s(x − t) + y)2 = x2 + y 2 + 1 − t2 = 2 − t2 . without loss of generality. Find the locus of the midpoint of the segment AB. where O is the center of the given circle. hence we get 4MN 2 = 2R2 − OP 2. Hence 1 = ∣w∣2 = (t − sy)2 + s2 (x − t)2 The midpoint of the segment AB is given by M = In fact. that is NM = 1√ 2 2R − OP 2. Cm (⊠) Example 2. respectively. it follows that the desired locus is the circle of center N 1√ 2 2R − OP 2.44) (A + B) . First solution We can assume. (⊠) Hence the required locus is a circle with center 2 2 In the general setting. Now we verify that 2 √ 2 − ∣P ∣2 . Two variable perpendicular lines through P intersect the circle at A and B. Let M be the midpoint of the segment AB and let N be the midpoint of the segment OP . Applying the relation above in the quadrilateral ABP O we obtain AP 2 + R2 + BP 2 + R2 = AB 2 + OP 2 + 4MN 2 . that P = t ∈ [0. P ∣M − ∣ = 2 (2. It is clear that AP 2 + BP 2 = AB 2 . 1] and the circle C = {∣z∣ = 1}. 2 √ 2 − ∣P ∣2 P and radius . (⊠) and radius 2 .173. Let P be point situated in the interior of a circle.143 2.44).3. by (2. if the circle C has √ center at P0 and radius R then the locus is 2R2 − ∣P − P0 ∣ (P0 + P ) and radius ⋅ a circle with center 2 2 (2∣M − Second solution Let ABCD be a quadrilateral and let M and N be the midpoints of sides AB and CD. 2 Since the point N is fixed. Let A = z = x + iy ∈ C then B = w = si(z − P ) + P ∈ C with some s > 0. Geometric problems Hence the maximum number of the points of intersection we are looking for is given by the sum of these three numbers: 2 (Cn2 + Cs2 + sn) . Using the Median Theorem it is easy to prove that the following relation holds : AC 2 + BD 2 + BC 2 + DA2 = AB 2 + CD2 + 4MN 2 . l must be the perpendicular bisector of the line segment A1 A2 in the general case when A1 ≠ A2 (and IA1 if A1 = A2 ). hence it must intersect the interior of segments AC. (b) Prove that A1 .175.144 Chapter 2.. Dissect (any) quadrilateral ABCD in two triangles. C1 . that is A1 = A3 . then dissect one of them into three cyclic quadrilaterals. . 4. we have IB1 = IA1 = IA2 = IC1 . or ABDE is cyclic. E. the isometry R = RBI ○ RAI ○ RCI is opposite as well and since R(I) = I. 2) and one triangle. BF ID. If v ≠ 0. B1 . A3 = Id (A1 ) = A1 . In triangle ABC. BC at E. Thus. CA. dissect the other triangle into one cyclic quadrilateral and one triangle. 2}. (⊠) Example 2. Clearly. Let ABC be a triangle and let A1 be a point on the side BC. consider the circumcenter O. B1 . Thus at the first step we construct an isosceles triangle A1 CB1 with point B1 lying on AC. The process is done in one direction: either clockwise or counterclockwise. 5 respectively for v = 0. where u ⩾ 1 is an integer and v ∈ {0. and where two of the sides of each triangle are on the straight line containing two of the sides of the other. We have thus dissected the original quadrilateral into 3 + v + 3u = n cyclic quadrilaterals. wlog acute at C. Starting with A1 construct reflections in one of the angle bisectors of triangle such that the next point lies on the other side of the triangle. Dissect now this triangle into u triangles (for example dividing one of its sides in u equal parts and joining each point of division with the opposite vertex). 1.lateral is already formed by two triangles joined at one vertex. (b) Since l is the the perpendicular bisector of A1 A2 . A2 . C1 . let I be the incenter and D. As a result. Examples for practice Example 2. any crossed quadri. The circle with center O ′ through A. Prove that any convex quadrilateral can be dissected into n ⩾ 6 cyclic quadrilaterals. CDIE. Since R(A1 ) = A2 . ABC my be dissected into three cyclic quadrilaterals AEIF. Solution Any convex quadrilateral is dissected into two triangles by either of its diagonals. F the points where the incircle touches respectively sides BC. AB. R = Rl and RBI ○ RAI ○ RCI ○ RBI ○ RAI ○ RCI = Rl ○ Rl = Id where Id denotes the identity of the plane. In triangle ABC. C2 lie on the same circle.. (a) As the product of three opposite isometries (reversing the orientation). (a) Prove that the process terminates in six steps. B2 . dissect again this latter triangle into one cyclic quadrilateral and one triangle. We may then proceed as follows: write n = 3 + 3u + v. R must be a reflection in a line l. In fact we get a sequence of points A1 . and take a point O ′ on the perpendicular bisector of AB that is closer to AB than O.174. and if v = 2.. Solution Let RM N denote the reflection in the line MN and let I be the incentre of ∆ABC. and dissect now each one of these u triangles into three cyclic quadrilaterals. After this procedure. 1. A2 . B leaves C outside. any concave quadrilateral is dissected into two triangles by exactly one of its diagonals. we have dissected the original quadrilateral into 3 + v cyclic quadrilaterals (3. At the second step we construct an isosceles triangle B1 AC1 with point C1 on AB. D. . 11: Similarly. we have Rl′ = RCI ○RBI ○RAI and so IC2 = IB2 = IB1 . C1 . A2 .176. Let R and r be the circumradius and the inradius of a triangle ABC with the lengths of sides a. lb . IA1 = IA2 = IB1 = IB2 = IC1 = IC2 and the six points A1 . In conclusion. lc be angle bisectors of a triangle ABC.3. Prove that 2 − 2∑( cyc Solution Note that 2 − 2∑( cyc a 2 r ) ⩽ ⋅ b+c R a 2 r a 2 r ) ⩽ ⇔ 6 − 2∑( ) ⩽4+ b+c R R cyc b + c ⇔ 2 (3 − ∑ ( cyc (b + c)2 − a2 r ⩽4+ ⋅ 2 (b + c) R cyc ⇔ 2∑ Since cos A + cos B + cos C = 1 + a 2 r ) )⩽4+ b+c R 1 1 (b + c)2 − a2 r and ⩽ and = 1 + cos A then R (b + c)2 4bc 2bc (b + c)2 − a2 (b + c)2 − a2 r ⩽ 2 = 2 ∑(1 + cos A) = 4 + ⋅ ∑ 2 (b + c) 2bc R cyc cyc cyc 2∑ Remark. Geometric problems Figure 2. if l′ denote the perpendicular bisector of B1 B2 . (⊠) Example 2. and C2 all lie on a circle with centre I. Noting that (b + c)2 − a2 ala2 = (b + c)2 abc (⊠) . b. c.145 2. Let la . B1 . B2 . P BC.. = 4 cos2 4 4 with equality iff A+B = 180o and simultaneously A = B. we obtain the proposed inequality with both sides multiplied by 4. Adding the cyclic permutations of both sides of this resulting inequality. and that O3 is determined in exactly the same way.. Consider a triangle ABC and a point P in its interior. equality holds iff A = B = 90o . . AB at A1 . 3. C1 .146 Chapter 2..if you take the smallest of the two angles formed at the vertex.An . centered at A2 and A3 ... prove that the polygon A1 A2 . We see that O2 is the intersection of two arcs of circle of radius R. P CA have the same area. The circle C(O. Given a convex polygon A1 A2 . (⊠) Example 2. denote by Ri the radius of the circumcircle of triangle Ai−1 Ai Ai+1 .. P C intersect BC.. and O3 A2 = O3 A3 = O3 A4 = R. 2∑ R R a+b+c cyc 4Rrp cyc abc Example 2. Solution That the polygon A1 A2 .. all circumcenters Oi lie in a zone Z that is exterior to all triangles Ai−1 Ai Ai+1 (where i = 2. 3.An is convex means that ∆Ai−1 Ai Ai+1 is obtuse ..An is cyclic.. n and An+1 is the vertex A1 ) and interior to the polygon A1 A2 . cos A−B B−C C −D D−A 1 + cos + cos + cos ⩾ 2 + (sin A + sin B + sin C + sin D). Examples for practice we can rewrite original inequality in such form ala2 + blb2 + clc2 ala2 r r ala2 ⩽ 4 + ⇔ 2∑ ⩽4+ ⇔ ⩽ r(4R + r). the latter is cyclic. 4 4 4 4 2 Solution We can write 2 + sin A + sin B = 2 + 2 sin A+B A−B A−B cos ⩽ 2 + 2 cos 2 2 2 A−B A−B ⩽ 4 cos . where i = 2. B1 . therefore.178. equality holds in the proposed inequality iff ABCD is a rectangle. n and An+1 is the vertex A1 . we find that all circumcenters must coincide in a unique circumcenter O common to all vertices. Repeating this reasoning.. the circumcenter Oi of ∆Ai−1 Ai Ai+1 is exterior to it. CA. In fact.177. (⊠) Example 2. The conclusion follows. Therefore. ie. n ⩾ 4.. .. since O2 and O3 lie on the same side of A2 A3 . Given that R2 = R3 = ⋯ = Rn . Prove that in any convex quadrilateral ABCD. P B. We have then O2 A1 = O2 A2 = O2 A3 = R. Prove that BA1 CB1 AC1 3 + + = BC CA AB 2 if and only if at least two of the triangles P AB. Lines P A.. Therefore.179. O2 and O3 must coincide. respectively. R) is the circumcircle of the polygon. etc. .An . A non-isosceles acute triangle ABC is given. Therefore. Prove that ̂ > 135o OIH Solution It is relatively well known that OI 2 = R(R − 2r). We may also write after some algebra a2 + b2 + c2 = 8R2 + 8R2 cos A cos B cos C. P BC have the same area. P AC. (⊠) Example 2. on simple calculations. b. The first two relations may be found through the respective powers of I and H with respect to the circumcircle. ∆P BC = y. H be the circumcenter.e when two of triangles P AB. OH 2 = 9R2 − (a2 + b2 + c2 ). or OH 2 − IH 2 − OI 2 Rr − r 2 − d √ ⋅ =√ √ 2OI ⋅ IH 2 r 2 − d R(R − 2r) .147 2. a2 + b2 + c2 = 8R2 + 4d for some d > 0. is equivalent to with (x − y)(z − y)(z − x) =0 (x + y)(y + z)(z + x) Which is true if and only if x = y or y = z or z = x. or OIH will be obtuse at I. ∆P CA = z we have. approaching equality when ABC approaches an equilateral triangle. The third one may be deduced therefrom. r denote the circumradius and inradius of ABC with sidelengths a. IH 2 = 4R2 + 2r 2 − a2 + b2 + c2 ⋅ 2 where as usual R. Geometric problems Solution Denote the area of triangle XY Z by ∆XY Z. the incenter. from the given condition that x y z 3 + + = ⋅ z+x x+y y+z 2 This also implies and so we obtain x y 3 z + + = ⋅ z+x x+y y+z 2 ∑( cyc x x − )=0 x+y z+x Which. since the nine-point center N is the midpoint of OH and IN = R − r because the incircle and nine-point circle are tangent at the Feuerbach point. I. or since ABC is acute. c. respectively.3. and the orthocenter of the triangle ABC. the second one requiring some algebra using the Cosine Law and expressions of the area of the triangle. applying the median theorem to triange OIH.180. Let O. Note that OH 2 − IH 2 − OI 2 = IH 2 + 2r(R − 2r) > 0 because R > 2r and IH 2 > 0. Note that we have ∆P AB BA1 ∆ABA1 ∆P BA1 ∆ABA1 − ∆P BA1 = = = = BC ∆ABC ∆P BC ∆ABC − ∆P BC ∆P AB + ∆P AC Denoting ∆P AB = x. OH 2 = R2 − 4d and IH 2 = 2r 2 − 2d. i. note that cos 20o cos 40o cos 80o = α = 4 cos 20o cos 40o = 1 + 2 cos 20o.that is α3 − 3α2 + 1 = 0. 2 (⊠) ̂ = 20o and side-lengths a. sin 80o sin 160o 1 = ⋅ Let α = .e. Prove that 1√ 2 2 2 7(a + b + c ) + 2(ab + bc + ca).148 Chapter 2. cyc cyc From this inferred ma + mb + mc ⩽ cyc cyc cyc cyc 1√ 2 2 2 7(a + b + c ) + 2(ab + bc + ca). Let ABC be a triangle with sidelengths a. we assume without loss of generality that b ⩾ c. Examples for practice 1 Assume that this quantity is smaller than or equal to √ . α2 (α − 3) = −1. since that the previous identity yields α = 4 cos 20o cos 40o = 2 cos 80o Solution First. thus. mb . and OIH ̂ > 135o . It follows that α − 3 = −2(1 − cos 20o ) = −4 sin 210o. b.181. At this point. note 8 sin 20o 8 sin 20o 1 . Prove that an acute triangle with A fying √ 3 a3 + b3 + c3 − 3abc = min{b. i. c and medians ma . Then. Hence cos OIH 2 (⊠) Example 2. we have cyc 4T 2 = 3 ∑ a2 + 8 ∑ mb mc ⩽ 3 ∑ a2 + 2 ∑ (2a2 + bc) = 7 ∑ a2 + 2 ∑ bc. . ̂ < − √1 = cos 135o . c} is isosceles. b. c satisExample 2. mc . 2 d2 + (R2 + 2r 2 )d + r 4 < 0.182. clearly impossible. ma + mb + mc ⩽ 2 Solution Since √ (2(a2 + c2 ) − b2 ) (2(a2 + b2 ) − c2 ) √ = 4a4 + 2a2 (b2 + c2 ) + 5b2 c2 − 2b4 − 2c4 √ = 4a4 + 2a2 (b2 + c2 ) + b2 c2 − 2(b2 − c2 )2 √ = (2a2 + bc)2 − 2 ((b2 − c2 )2 − a2 (b − c)2 ) √ = (2a2 + bc)2 − 2(b − c)2 (b + c − a)(b + c + a) √ ⩽ (2a2 + bc)2 = 2a2 + bc 4mb mc = and 4 (m2a + m2b + m2c ) = 3 ∑ a2 then denote ma + mb + mc by T. Prove that quadrilateral BCMX is cyclic if and only if quadrilateral ADNY is cyclic. ̂ so 90o ⩾ B ̂ ⩾ 80o ⩾ C ̂ (as the angles B ̂ and Now. OB be b . Hence.184. This proves that ABC is isosceles. Prove that AD. the equality from the hypothesis can be rewritten as a3 − 3abc + b3 = 0. The conclusion follows. Let M and N be the midpoints of AD and BC. Similarly. +∞). since b ⩾ c we have x ⩾ y and B o ̂ both add up to 160o . ACD and ACG. and so it follows that x = y = α. yielding GX ⋅ GM = GA ⋅ GD.3. BC. Solution It is well known that the intersection N of MH with the circumcircle is the same as the intersection of AO with the circumcircle. C sin 20o 3 2 α > 2 and x → x − 3x + 1 is strictly increasing in (2. On the other hand. Therefore H is the orthocenter of triangle AMQ. 2GM = GA + GD. or GX = . Ð → Ð→ Ð→ Ð→ → → If we prove that OH ⊥ AM we are done. and the Euler line of triangle ABC are concurrent. Ð → Ð → Ð → 2 2 2 2 It is clear that a = b = c = r . Let P Q be the intersection of AD with BC. therefore 0 = x3 − 3xy + 1 ⩾ x3 − 3x2 + 1 ⩾ α3 − 3α2 + 1 = 0. Let M be the midpoint of BC and let D be the intersection of MH with the circumcircle of triangle ABC such that H lies between M and D. Let OA be Ð a . respectively. by the Law of Sines. Solution Applying Menelaus’ theorem to triangles CDG. Let H and O be the orthocenter and the circumcenter of triangle ABC. respectively. so QH ⊥ AM. as desired. ⋅ ⋅ = 1. AB ∩CD = {F }. AC ∩BD = {E}.2. (⊠) Example 2. and EF intersects the sides AD and BC at X and Y. where x = and y = ⋅ a a ̂ ⩾ C. GD DX GD − GX GA + GD and since M is the midpoint of AD. we find CF DA GB AX DF CE AE CB GD ⋅ ⋅ =1. (⊠) Example 2. Consider a non-isosceles acute triangle ABC such that AB 2 + AC 2 = 2BC 2 . Hence AN is a diameter of the circumcircle and AD ⊥ MH. Geometric problems 149 Accordingly. we find 2GA ⋅ GD GA AX GX − GA = = . Note therefore that BCMX is cyclic iff GX ⋅ GM = GB ⋅ GC iff GB ⋅ GC = GA ⋅ GD iff ABCD is cyclic. ⋅ ⋅ =1. OC be Ð c. Then: Ð → → Ð → Ð Ð → ÐÐ→ ÐÐ→ a + b +Ð c ⎛Ð b +→ c⎞ → 6OH ⋅ AM = 6 a − 3 2 ⎠ ⎝ Ð → Ð → Ð →→ Ð Ð → Ð →→ → → → → → → → = 2Ð a2− b 2−Ð c 2+Ð aÐ c +Ð a b −2 b Ð c =→ aÐ c +Ð a b −2 b Ð c . In a convex quadrilateral ABCD. we find that ADNY is cyclic iff ABCD is cylcic.183. c b which after dividing both sides by a3 becomes x3 − 3xy + 1 = 0. We have that AH ⊥ MQ and MH ⊥ AQ. x ⩾ sin 80 = α. F D AG BC XD F C EA EC BG DA Multiplying these three equalities. the circles degenerate to straight lines that meet only at one point. and this line is outside ABCD. Assume furthermore wlog that BC < DA (if BC = DA then ABCD would be an isosceles trapezium and AB ∥ CD. P A meet the circumcircle of ABCD. or line P E cannot be defined. R) and let E be the intersection of its diagonals. or AB ∥ CD. Call now A0 . C0 . and trivially AEB and DEC are similar. and call F = AB ∩ CD. Call first P the second point where the circumcircle of ABE and line EF meet. satisfy = . but not on segments CA or DB respectively. which we are assuming not to be true). assuming wlog that AB < CD. and P = E. Note finally that. and similarly AC0 = CA0 . since AP B and CP D are similar. CDP E is also cyclic. and P is also on the circumcircle of CDE. Obviously. or P ∈ EF. P B. B0 .185. Suppose P is the point inside ABCD such that triangle ABP is directly similar to triangle CDP . P C. Trivially. If AB = CD. then the centers of both circles are on the rays CA. or AB0 = BA0 .150 Chapter 2. meet. The power of F with respect to the circumcircle of ABE (which is also the power of F with respect to the circumcircle of ABCD is then F E ⋅ F P = F A ⋅ F B = F C ⋅ F D. and AA0 ∥ BB0 ∥ CC0 ∥ DD0 . Let ABCD be a cyclic quadrilateral inscribed in the circle C(O. If AB ∥ CD. D0 the second points where P D. line EF contains all points Q such that the distances from Q to lines AB and CD. Examples for practice Ð → →2 Ð → Ð →→ → → → → = b 2+Ð c − 2Ð a2+Ð aÐ c +Ð a b −2 b Ð c Ð → → 2 Ð Ð → → → = 2( b − Ð c ) − (→ a −Ð c )2 − (Ð a − b )2 Ð→ Ð→ Ð→ = 2BC 2 − AC 2 − AB 2 = 2BC 2 − AC 2 − AB 2 = 0 (⊠) Example 2. AB) and d(Q. then ∠ABE = ∠BEF = π − ∠DEF = ∠DCE = ∠ABE. DB from C. then ∠P AB = ∠BEF = ∠P ED = ∠P CD. or AA0 BB0 . then ABCD is an isosceles trapezoid. We will now show that P is the point the circumcircles of ABE and CDE. AB) AB respectively d(Q. point P is the intersection of two distinct Apollonius’ circles P C P D CD constructed by taking the ratios of the distances to A and C for one. the diagonals AD0 and DA0 of AA0 D0 D . Assume henceforth then that AB and CD are not parallel. Prove that OP ⊥ P E. and line EF. since ABEP and CDP E are cyclic. the altitudes from D to AB and CD are also proportional to AB and CD. CD) CD passes through the intersection of both lines. d(Q. since it contains E and d(Q. and similarly ∠ABP = ∠AEP = ∠CEF = ∠CDP. or the altitudes from E to AB and CD are proportional to the lengths of the sides AB and CD. Solution P A P B AB Since = = . CD). Therefore. if the circumcircles of ABE and CDE where tangent. B. AA0 CC0 and AA0 D0 D are isosceles trapezii. B and D for the other. AD0 = DA0 . and ABE and CDE are isosceles and similar. or indeed P AB and P CD are similar. Otherwise. Trivially. Since both points where the circles meet are symmetric with respect to the line joining the centers of both circles. ∠ACB0 = ∠ECP = ∠EDP = ∠BDA0 . and P is therefore unique. Therefore. then both points cannot be in ABCD simultaneously. Now. . such that AO is a symmedian. ie. the second point where the circumcircles of ABE and CDE meet. Solution In addition to the usual trig formulas.151 2. Therefore. Example 2. and ABCD is an isosceles trapezium with AB ∥ CD. OP is the internal bisector of angles ∠AP A0 . ∠BP B0 . which is then in the common perpendicular bisector of AA0 . Now. but not a median. we will make use of the following two known results: T ∶= tan A + tan B + tan C = tan Atan Btan C and sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C. in triangle ABC. (⊠) Example 2. BB0 . with AB = AC.e. DD0 . Geometric problems meet at P.3. then as shown P = E. Let ABC be an acute triangle. (⊠) Remark.187. sin A sin B sin C 1 sin 2A + sin 2B + sin 2C = ⋅ cos A cos B cos C 4 cos A cos B cos C 1 sin A sin B sin C = ( + + ) 2 cos B cos C cos C cos A cos A cos B T= = sin B sin C sin A + + cos(B − C) − cos A cos(C − A) − cos B cos(A − B) − cos C ⩾ sin A sin B sin C + + 1 − cos A 1 − cos B 1 − cos C B C p−a p−b p−c p A + + = = cot + cot + cot = 2 2 2 r r r r and the result follows. If both circles are tangent. i. which trivially passes also through O. or P E is the external bisector of angles ∠BP B0 and ∠CP C0 . and P E is the internal bisector of angle ∠BP C = π − ∠BP B0 . and hence perpendicular to their internal bisectors. and A0 B0 C0 D0 is the result of taking the reflection of ABCD with respect to OP. The proof is completed. Prove that tan A + tan B + tan C ⩾ p ⋅ r where p and r are semiperimeter and inradius of triangle ABC. to OP.. respectively. ∠CP C0 and DP D0. Choose points B and C on C. Now. Prove that the circumcircle of triangle ABC passes through a second fixed point.186. ∠BP E = ∠BAE = ∠CDE = ∠CP E. CC0 . Let C be a circle with center O and let A be a fixed point outside C. Note that this solution includes also the way to construct point P. let M be the midpoint of BC. Prove that the equality A = 3B implies the inequality (a2 − b2 )(a − b) = bc2 . AC respectively intersect C.152 Chapter 2. where R is the circumradius of ABC. Claim: P Q is a diameter of C. and since it is also a median.g. Since BC is a chord both in C and C0 . hence P Q is a diameter. N. b = 2R sin B. AC are the respectively symmetric lines of AQ. (a2 − b2 )(a − b) = 8R3 (sin2 A − sin2 B)(sin A − sin B) = 8R3 (sin2 3B − sin2 B)(sin 3B − sin B) = 8R3 (sin 3B − sin B)2 (sin 3B + sin B) = 8R3 (8 cos2 2B sin2 B sin2 B cos B) = 8R3 (sin2 (180o − 4B))(sin B) = 8R3 (sin2 C)(sin B) = bc2 . a = 2R sin A. then AO is the perpendicular bisector of P Q. Q be the second points where AB. Proof : Triangles ABC and AQP are clearly similar. Let ABC be a triangle with sides a. The converse is false in general. and so is ABC. and determine whether the converse also holds. c and corresponding angles A. e. The point N is also clearly the midpoint of arc BC. or it is also on line OO0. c = 2R sin C. Since AB. Examples for practice Solution Let P. C. C = 40o. let C0 be the circumcircle of ABC with center O0 .188. We reach a contradiction. and AO is a symmedian in triangle BAC. AP Q is isosceles in A. Assume now that P Q is a chord of C that is not a diameter. O0 are collinear. B. M. Hence. and let N the second point where the internal bisector of angle ∠BAC intersects the circumcircle of ABC. then it is a median in triangle AP Q. Since AO passes through its midpoint and through the center of C. (⊠) . O. Solution: By the extended law of sines. AP with respect to this internal bisector. (⊠) Example 2. B = 125o . b. for A = 15o . its midpoint M clearly lies on line OO0. Thus. we can also have A = 3B − 360o. hence the internal bisector of angles ∠BAC and ∠QAP is the same. 153 2.4. Analysic problems 2.4 2.4.1 Analysic problems Junior problems Example 2.189. Let a0 = a1 = 1 and an+1 = 1 + a21 a22 a23 a2 + + +⋯+ n a0 a1 a2 an−1 for n ⩾ 1. Find an in closed form. Solution Let us prove by induction that an = n!. The result is not difficult to verify for n = 0, 1. Assume that ak = k! for k ⩽ n − 1. Now ak+1 = (1 + By the induction hypothesis, a2 a2 a21 a22 a23 + + + ⋯ + k−1 ) + k a0 a1 a2 ak−2 ak−1 ak+1 = ak + which is equivalent to ak+1 = a2k ak = (ak−1 + ak ) ak−1 ak−1 k! ((k − 1)! + k!) = k(k − 1)!(k + 1) = (k + 1)!. (k − 1)! Thus an = n! for n ⩾ 1. (⊠) Example 2.190. Prove that if the polynomial P ∈ R[x] has n distinct real zeros, then for any α ∈ R the polynomial Q(x) = αxP (x) + P ′ (x) has at least n − 1 distinct real zeros. Solution Let be a1 < a2 < ⋯ < an be the real roots of P and P (x) = T (x) ∏(x − aj ) n j=1 where T ∈ R[x] has no real roots. Therefore Q(ai ) = αai P (ai ) + P ′ (ai ) = P ′ (ai ) = T (ai ) ∏(ai − aj ) j≠i and it’s clear that in the last product has n−i−1 negative factors. Since T (x) has constant sign, it follows that Q(ai )Q(ai+1 ) < 0 for i = 1, ..., n−1, and by Intermediate Value Theorem there exists at least a zero of Q in (ai , ai+1 ). This means that Q has at least n − 1 distinct real zeros. (⊠) Example 2.191. Let f ∶ R → R be a function such that f (x) + f (x + y) is a rational number for all real numbers x and all y > 0. Prove that f (x) is a rational number for all real numbers x. 154 Chapter 2. Examples for practice Solution For each real number x, consider a real number y > 0 and define u, v, w in the following way u = f (x) + f (x + y), v = f (x − y) + f (x), w = f (x − y) + f (x + y) Since x = (x − y) + y and x + y = (x − y) + 2y, the numbers u, v, w are rational by hypothesis. Therefore 1 f (x) = (u + v − w) 2 is a rational number and we are done. (⊠) Example 2.192. Let m ⩾ 1 and f ∶ [m, +∞) → [1, +∞), f (x) = x2 − 2mx + m2 + 1. (a) Prove that f is bijective; (b) Solve the equation f (x) = f −1 (x); √ (c) Solve the equation x2 − 2mx + m2 + 1 = m + x − 1. Solution (a) The parabola f (x) = (x − m)2 + 1 is symmetric with respect to the axis x = m thus it is bijective on its domain [m, +∞). A more detailed proof about injectivity is the following f (x1 ) > f (x) ⇔ (x1 − m)2 + 1 > (x − m)2 + 1 ⇔ ∣x1 − m∣ > ∣x − m∣ ⇔ x1 − m > x − m ⇔ x1 > x From the calculation above we deduce that the function is increasing as well thus its image is a subset of the interval [f (m), +∞) = [1, +∞). As for the surjectivity we must solve for any y = 1 the equation √ √ (x − m)2 + 1 = y ⇔ ∣x − m∣ = y − 1 ⇔ x = y − 1 + m. (b) The equation f (x) = f −1 (x) may occur only on the bisector of the first and third quadrant y = x thus we must solve √ 1 + 2m ± 4m − 3 2 2 2 (x − m) + 1 = x ⇔ x − (2m + 1)x + m + 1 = 0 ⇔ x = . 2 √ 1 + 2m + 4m − 3 The root x = satisfies x > 1 for any m ⩾ 1. 2 √ (c) x2 − 2mx + m2 + 1 = m + x − 1 is the equation f (x) = f −1 (x) whose solution is precisely that found in (b). √ Indeed the inversion of y = (x − m)2 + 1 yields x = m + x − 1. (⊠) Example 2.193. Let P ∈ R[x] be a nonconstant polynomial and let f ∶ R → R be a function with the intermediate value property such that P ○ f is continuous. Prove that f is continuous. Solution Let x be a real number and consider a sequence xn that converges to x. Then the continuity of P ○ f implies that P (f (xn )) converges to P (f (x)) and so f (xn ) is bounded (as a nonconstant polynomial is a proper map). We claim that f (xn ) converges. If not, there are two real numbers a < b and two subsequences yn and zn of xn such that f (yn ) 155 2.4. Analysic problems converges to a and f (zn ) converges to b. Now, take any c ∈ (a, b). Then for large enough n we have f (yn ) < c < f (zn ) and, since f has the intermediate value property, there exists tn between yn and zn such that c = f (tn ). But since yn and zn converge to x, so does tn . Moreover, P (c) = P (f (tn )) must converge to P (f (x)) by continuity of P ○ f . Therefore we find that P (c) = P (f (x)) for all c ∈ (a, b), which implies that P is constant, a contradiction. So, the sequence f (xn ) converges for all choices of a convergent sequence xn . This immediately implies that if xn tends to x, then f (xn ) tends to f (x), since one can consider the sequence x1 , x, x2 , x, ..., which still converges to x (and for which the sequence obtained after applying f has infinitely many terms equal to f (x)). This obviously implies the continuity of f . (⊠) Remark. All we used about P is that it is a proper map and it is non constant on any nontrivial interval. So, by replacing P with any map g satisfying these two properties, the conclusion still holds. An interesting question is to find all maps g with the following property: if f has the intermediate value property and g ○ f is continuous, then f is continuous. Example 2.194. A sequence (an )n⩾2 of real numbers greater than 1 satisfies the relation ¿ Á (n + 1)! an = Á Á1 + 1 1 Á À ) 2 (a2 − ) ⋯ (an−1 − a2 an−1 for all n > 2. Prove that if ak = k for some k ⩾ 2, then an = n for all n ⩾ 2. Solution 1 1 ) then for n > 2 Letbn ∶= (a2 − ) ⋯ (an−1 − a2 an−1 √ (n + 1)! (n + 1)! an = 1 + ⇔ a2n − 1 = , 2bn−1 2bn−1 and, since we have Letting bn−1 = 1 a1n − 1 bn = an − = , bn−1 an an (i) (n + 1)! (n + 1)! bn = ⇔ bn = ⋅ bn−1 2bn−1 an 2an n! in (i) for n > 2 give us a2n − 1 = (n + 1)an−1 . Thus, 2an−1 a2n+1 = 1 + (n + 2)an , n ⩾ 2. Let ak = k for some k ⩾ 2. Then an = n for any n ⩾ k. Indeed, since ak = k and in supposition an = n, n ⩾ k we obtain an+1 = 1 + (n + 2)an = 1 + (n + 2)n = (n + 1)2 then by induction an = n for any n ⩾ k If k > 2 then for any 2 < n = k from supposition an = n follows an−1 = a2n − 1 n2 − 1 = = n − 1. n+1 n+1 Thus, by induction an = n for any 2 ⩽ n ⩽ k. Finally, we obtain an = n, ∀ n ⩾ 2. (⊠) 156 2.4.2 Chapter 2. Examples for practice Senior problems Example 2.195. Let k be a nonzero real number. Find all functions f ∶ R → R such that f (xy) + f (yz) + f (zx) − k[f (x)f (yz) + f (y)f (zx) + f (z)f (xy)] ⩾ for all x, y, z ∈ R. 3 , 4k Solution 1 Taking x = y = z = 0, the condition becomes (2kf (0) − 1)2 ⩽ 0, or f (0) = . 2k 1 Take x = y = z = 1, the condition becomes again (2kf (1) − 1)2 ⩽ 0, or f (1) = . 2k 1 Take now y = z = 0, the condition becomes f (x) ⩽ . 2k Take finally x = y and z = 1, the condition becomes 2kf (x2 ) + 8kf (x) ⩾ 3 + 8k 2 f 2 (x). But 2kf (x2 ) ⩽ 1, or (2kf (x) − 1)2 ⩽ 0, yielding f (x) = 1 for all real x. 2k (⊠) 1 Example 2.196. Find all continuous functions f on [0, 1] such that f (x) = c if x ∈ [0, ] 2 1 and f (x) = f (2x − 1) if x ∈ ( , 1] , where c is a given constant. 2 Solution We will show that for every positive integer n f (x) = c, x ∈ [0, 1 − 1 ]. 2n We will induct on n. For n = 1 it is the condition of the problem and therefore is 1 1 true. Now suppose it is true for n. Let us prove it for n + 1. If x ∈ [ , 1 − n+1 ], then 2 2 1 2x − 1 ∈ [0, 1 − n ] ⇒ (by induction hypothesis) f (2x − 1) = c ⇒ f (x) = c. Thus f (x) = c 2 if x ∈ [0, 1). Because f is continuous we have f (1) = f ( lim (1 − x→∞ Then f (x) = c for x ∈ [0, 1]. 1 1 )) = lim f (1 − n ) = lim c = c. n x→∞ x→∞ 2 2 (⊠) Example 2.197. Prove that there are sequences (xk )k⩾1 and (yk )k⩾1 of positive rational numbers such that for all positive integers n and k, √ √ n 1+ 5 (xk + yk 5) = Fkn−1 + Fkn , 2 where (Fm )m⩾1 is the Fibonacci sequence. 157 2.4. Analysic problems Solution 1 1 Take xk = Lk and yk = Fk where Ln is the n-th Lucas number. Since 2 2 √ k √ k √ k √ k 1+ 5 1 ⎛ 1+ 5 1− 5 ⎞ 1− 5 Fk = √ ( ) −( ) and Lk = ( ) +( ) 2 2 2 2 ⎠ 5⎝ it follows that √ kn √ √ n 1+ 5 1+ 5 (xk + yk 5) = ( ) = Fkn−1 + Fkn . 2 2 (⊠) Example 2.198. Let f ∶ N → [0, +∞) be a function satisfying the following conditions: (a) f (100) = 10; 1 1 1 (b) + +⋯+ = f (n + 1), for all nonnegative f (0) + f (1) f (1) + f (2) f (n) + f (n + 1) integers n. Find f (n) in closed form. Solution By the condition (b) we have that that is f (n + 1) − f (n) = 1 , ∀ n ∈ N, f (n) + f (n + 1) (f (n + 1)) = 1 + (f (n)) = 2 + (f (n − 1)) = ⋯ = n + 1 + (f (0)) . 2 Letting n = 99, by (a) 2 2 100 = (f (100)) = 100 + (f (0)) 2 2 2 that is f (0) = 0 and finally (note that f (n) ⩾ 0 by hypothesis) f (n) = Example 2.199. Consider the polynomial P (x) = ∑ n 1 xk , n + k + 1 k=0 with n ⩾ 1. Prove that the equation P (x2 ) = (P (x))2 has no real roots. Solution Suppose there exist a real root t to the equation. Since P (t2 ) ⩾ 1 > 0, n+1 it follows that P (t2 ) = (P (t))2 > 0. From Cauchy-Schwarz we get n n 1 1 1 (∑ ) (∑ t2k ) ⩾ ( ∑ tk ) n + k + 1 n + k + 1 n + k + 1 k=0 k=0 k=0 n 2 √ n. (⊠) 158 Chapter 2. Examples for practice which implies that 1 ⩾ 1. k=0 n + k + 1 n ∑ However, we have 1 1 < (n + 1) = 1. n+1 k=0 n + k + 1 n ∑ a contradiction. It follows that the equation P (x2 ) = (P (x))2 has no real roots. Example 2.200. The sequence (xn )n⩾1 is defined by Prove that lim nxn = −2. (⊠) x1 < 0, xn+1 = exn − 1, n ⩾ 1. n→∞ Solution By induction it can be proved that xn < 0 for all n ⩾ 1. On the other hand, since ex − 1 ⩾ x for all x ∈ R, we get that xn+1 = exn − 1 ⩾ xn , and hence, the sequence increases. It follows that x1 < xn < 0 for all n ⩾ 1, and hence, the sequence converges. If l = lim xn , then passing to the limit as n → ∞ in the recurrence relation we obtain that l = el − 1 from which it follows that l = 0. We calculate lim nxn by using Cesaro-Stolz lemma. We have, since ex − 1 x2 lim = 1 and lim x = 2, x→0 x→0 e − 1 − x x that xn+1 ⋅ xn n 1 = − lim lim nxn = − lim −1 = − lim 1 1 n→∞ xn+1 − xn n→∞ n→∞ n→∞ xn xn − xn+1 x2 exn − 1 = (−1) ⋅ 1 ⋅ 2 = −2. lim xn n n→∞ xn n→∞ e − 1 − xn = − lim and the problem is solved. (⊠) Example 2.201. Find all functions f ∶ [0, 2] → (0, 1] that are differentiable at the origin and satisfy f (2x) = 2f 2 (x) − 1, for all x ∈ [0, 1]. Solution Let g(x) = arccos f (x) for all x ∈ [0, 2]. Then for x ∈ [0, 1], cos g(2x) = f (2x) = 2 cos2 g(x) − 1 = cos(2g(x)), thus g(2x) = 2g(x). Hence, for any x ∈ [0, 2], x x x g(x) = 2g ( ) = 4g ( ) = ⋯ = 2n g ( n ) for all n ⩾ 1. 2 4 2 x g( n) 2 Since g is differentiable at 0, we have lim x = k for some constant k. n→∞ 2n Therefore, g(x) = kx for all x ∈ [0, 2]. Considering the range of f, we conclude that π π f (x) = cos(kx) with − < k < . 4 4 Finally, it is easy to verify that all such functions f do satisfy the conditions. (⊠) 159 2.4. Analysic problems 2.4.3 Undergraduate problems Example 2.202. A polynomial p ∈ R[X] is called a ”mirror” if ∣p(x)∣ = ∣p(−x)∣. Let f ∈ R[X] and consider polynomials p, q ∈ R[X] such that p(x) − p′ (x) = f (x), and q(x) + q ′ (x) = f (x). Prove that p+q is a mirror polynomial if and only if f is a mirror polynomial. Solution It is well known that a polynomial p ∈ R[X] has nonzero coefficients only for terms with even degree of x if and only if p(x) = p(−x) for all x; we call such a polynomial an ”even” polynomial (or polynomial with even symmetry). Similarly, a polynomial p ∈ R[X] has nonzero coefficients only for terms with odd degree of x if and only if p(x) = −p(−x) for all x; we call such a polynomial an ”odd” polynomial (or polynomial with odd symmetry). Lemma 6. p is a mirror polynomial if and only if it is either odd or even. Proof. if p is either odd or even, it is clearly a mirror. If p is a mirror, then either p(x) = p(−x) for an infinitude of values of x, or p(x) = −p(−x) for an infinite of values of x. In either case, either finite-degree polynomial p(x) − p(−x) or finite-degree polynomial p(x) + p(−x) has an infinite number of real roots, and needs to be thus identically zero, ie, either p(x) − p(−x) = 0 for all x (and p is even), or p(x) + p(−x) = 0 for all x (and p is odd), or both (and p is identically zero). We prove our statement using induction on the degree n of f , which is by definition equal to the degree of p and q, since the degree of p′ and q ′ is less than the degree of p and q, unless p and q, and therefore also f , are constant. By the previous argument, it is also clearly true that the highest degree of x has the same coefficient in f, p, q. When p(x) + q(x) are constant, thus even, thus mirrors. When n = 1, n = 0, f (x) = p(x) = q(x) = 2 write without loss of generality f (x) = a1 x + a0 with a1 ≠ 0. Then p(x) = a1 x + a0 + a1 and q(x) = a1 x + a0 − a1 , and p(x) + q(x) = 2f (x) = 2a1 x + 2a0 . Since neither f nor p + q may be even, and f is odd if and only if a0 = 0 or p + q is odd, then f is a mirror polynomial if and only if p + q is a mirror polynomial. Assume now that the proposed result is true for all polynomials of degree less than n ⩾ 2, and write without loss of generality f (x) = ∑ ak xk where an ≠ 0. n k=0 Define now polynomials r, s, ∆f, ∆p, ∆q as follows: where ∆f (x) = f (x) − an xn , ∆p(x) = p(x) − r(x), ∆q(x) = q(x) − s(x), Note that r(x) = an ∑ Cnk k!xn−k ; s(x) = an ∑ Cnk (−1)k k!xn−k . n n k=0 k=0 r ′ (x) = an ∑ Cnk (n − k)k!xn−k−1 = an ∑ Cnl l!xn−l = r(x) − an xn ; n n k=0 k=1 s′ (x) = −an ∑ Cnl (−1)l l!xn−l = an xn − s(x), n k=1 q. Then. ∆p(x) + ∆q(x) = p(x) + q(x) − an ∑ Cnk (1 − (−1)k ) k!xn−k . all monomials appearing in f ′′ . and analogously q = f −f ′ +f ′′ −. this can happen if and only if all monomials appearing if f are either of even degree. if and only if p+q is a mirror.. ∆p.. as the higher-order derivatives of a polynomials are all eventually zero.) = f. Thus f is a mirror polynomial if and only if f (x) = g(x2 ) or f (x) = xg(x2 ) for some polynomial g. ∆q(x) + ∆q(x) = q(x) + q(x) − s(x) − s(x) = f (x) − an xn = ∆f (x). if and only if ∆p + ∆q is even or odd (by hypothesis of induction). observe that 2f = (p + q) − (p + q)′′ . thus p + q is a mirror polynomial..) − (f ′ + f ′′ + . or odd degree. ∆q satisfy the conditions given in the problem.. then all monomials appearing in f have either even degree. Examples for practice where we have performed the substitution k = l − 1. For example. so either p(x) = p(−x). if and only if ∆f is even or odd (we obtain ∆f by making 0 one coefficient in f which has the same parity of all other nonzero coefficients in f ). Therefore.e. is a mirror polynomial. because if p1 − p′1 = p2 − p′2 . n k=0 From the First two of the last relations. . and so the sum of all these polynomials has all monomials of even degree. f is a mirror if and only if f is even or odd. This can happen only when one of the two factors is identically 0.. p must be unique. ∆p.. and hence so is their diference 2f . Thus we have found p. then (p1 − p2 ) = (p1 − p2 )′ and a polynomial equals its derivative if and only if it is identicallyzero.160 Chapter 2. while ∆p(x) − ∆p(x) = p(x) − p(x) − r(x) + r(x) = f (x) − an xn = ∆f (x). or p(x) = −p(−x). We are left to prove that f is a mirror polynomial if and only if f + f ′′ + . are also all even or odd (according to whether f is even or odd). or of odd degree. and from the third that the only terms that are different in ∆p + ∆q with respect to p + q are those whose degree has the same parity as n. and hence p+q = 2(f +f ′′ +. By comparing the coefficients of the two polynomials. p − p′ = (f + f ′ + .. ∆q is less than the degree of f. we find that ∆f.. .. The next idea is that p and q can be exhibited in a rather explicit form. (p(x)−p(−x))(p(x)+ p(−x)) = 0. and we are done.. If p + q is mirror polynomial. For the converse. is finite. Moreover. clearly the sum p = f (x) + f ′ (x) + f ′′ (x) + .. p.. then (p + q)′′ is a mirror polynomial of the same type. (⊠) Second solution The condition ∣p(x)∣ = ∣p(−x)∣ is equivalent to p2 (x) = p2 (−x) i.. Since differentiating twice preserves the parity of the degree. or odd. With this definitions.. If f is a mirror polynomial.) (the sum of all derivatives of even order). it is clear that the degree of ∆f. if and only if p + q is even or odd (because ∆p + ∆q is obtained by modifying only coefficients of p+q that have the same parity as n). Second solution Let n n(n − 1) n(n − 1)(2n − 1) − n2 − 1 2 6 = ⋅ 2n 12 (⊠) fn (x) = (x2 1 n(xn + 1) − ⋅ n − 1)(x − 1) (x + 1)2 Let y = −1 − x then..203.).4. x → −1 ⇔ y → 0.161 2. by differentiating the initial relations and manipulating. j (x + 1) ∑n−1 j=0 x ⎡ n−1 ( j−1 i ) ( n−j−1 k ) ⎤ ⎢ ∑j=1 ∑i=0 x ∑k=0 x ⎥ ∑n−1 j(n − j) ⎥ = j=1 lim T = lim ⎢⎢ ⎥ j x→1 x→1 ⎢ 2n (x + 1) ∑n−1 ⎥ j=0 x ⎣ ⎦ = and we are done. we can conclude that p + q = 2(f + f ′′ + . Analysic problems Remark. fn (x) = = = n ((1 + y)n + 1) 1 − 2 2 n ((1 + y) − 1) ((1 + y) − 1) y n (2 + y ∑nk=1 Cnk y k−1) 1 − 2 n k−1 k 2 y (y + 2) (∑k=1 Cn y ) y n (2 + y ∑nk=1 Cnk y k−1) − (y + 2) (∑nk=1 Cnk y k−1) y 2 (y + 2) (∑nk=1 Cnk y k−1 ) . (⊠) Example 2. Let n be an even integer. Evaluate lim [ x→−1 First solution Since n is even we have that lim [ x→−1 1 n (xn + 1) − ] 2 n (x − 1) (x − 1) (x + 1)2 n (xn + 1) 1 n (xn + 1) 1 − ] = lim [ − ] ∶= lim T. and since n is even. 2 n 2 2 n x→1 (x − 1) (x − 1) x→1 (x − 1) (x − 1) (x + 1) (x − 1)2 Note that for x = 1 we have that T= = = j n (xn + 1) − (x + 1) ∑n−1 j=0 x j (x − 1)2 (x + 1) ∑n−1 j=0 x n−1 ∑j=0 (xn + 1 − xj − xn−j ) = j (x − 1)2 (x + 1) ∑n−1 j=0 x n−1 n−1 ∑j=0 (xj − 1) (xn−j − 1) ∑j=1 (xj − 1) (xn−j − 1) = j j (x − 1)2 (x + 1) ∑n−1 (x − 1)2 (x + 1) ∑n−1 j=0 x j=0 x j−1 i n−j−1 k (x − 1)2 ∑n−1 x ) j=1 (∑i=0 x ) (∑k=0 It follows that j (x − 1)2 (x + 1) ∑n−1 j=0 x j−1 n−j−1 n−1 ∑j=1 (∑i=0 xi ) (∑k=0 xk ) = . and by the method exposed above of computing p and q.. It can be proved directly that 2f = (p + q) − (p + q)′′ . x x ) = f ( lim ) = f (0) = 0 n→∞ 2n 2n (⊠) Example 2. and xn+1 = x2n + 1. y1 = 1. 2 2 4 4 2 8 8 4 2 and after n steps we have f (x) ⩽ f ( The limit n → +∞yields f (x) ⩽ lim (f ( n→+∞ n x x ) + ∑ n k 2 k=1 2 n n x x x x ) + ) = lim f ( ) + lim =0+x=x ∑ ∑ n k n k n→∞ n→∞ 2 2 k=1 2 k=1 2 where we have used the continuity of f (x) and f (0) = 0 for writing lim f ( n→∞ and we are done. yn+1 = xn yn for all n. Examples for practice = = = Therefore 2n + ((n − 1)y − 2) (n + Cn2 y + Cn3 y 2 + o(y 2 )) 2ny 2 + o(y 2 ) 2n + n(n − 1)y + Cn2 (n − 1)y 2 − 2n − n(n − 1)y − 2Cn3 y 2 + o(y 2 ) 2ny 2 + o(y 2 ) Cn2 (n − 1)y 2 − 2Cn3 y 2 + o(y 2 ) .162 Chapter 2. . Prove that f (x) ⩽ x for all x ∈ [0.205. Sequences (xn )n⩾1 and (yn )n⩾1 are defined by x1 = 2. +∞) → R be a continuous function such that f (0) = 0 and f (2x) ⩽ f (x) + x for all x ⩾ 0.204. . Prove that for all n ⩾ 1 xn 651 < ⋅ yn 250 . +∞). Solution Note that x x x x x x x x x f (x) ⩽ f ( ) + ⩽ f ( ) + + ⩽ f ( ) + + + . 2n 12 (⊠) Example 2. Let f ∶ [0. 2ny 2 + o(y 2 ) lim fn (x) = lim [ x→−1 y→0 = Cn2 (n − 1)y 2 − 2Cn3 y 2 + o(y 2 ) ] 2ny 2 + o(y 2 ) (n − 1)Cn2 − 2Cn3 n2 − 1 = . and x4 /y4 = 677/260. Solution If x0 is a fixed point then f (x0 ) = f −1 (x0 ) = x0 and f (x0 ) + f −1 (x0 ) = 2x0 = ex0 − 1 which has two solutions: 0 and a some c > 0. ∞) → [0. Therefore c is the only possible fixed point. for n ⩾ 5 xn x2n−1 + 1 xn−1 1 x4 n 1 677 ∞ 1 = = + = +∑ < +∑ ⋅ yn xn−1 yn−1 yn−1 yn y4 k=5 yk 260 k=5 yk Hence it suffices to show that ∞ 1 651 677 1 ⩽ − = ⋅ 250 260 6500 k=5 yk ∑ We have that for n ⩾ 1 xn ⩾ x2n−1 ⩾ x4n−2 ⩾ x8n−3 ⩾ ⋯ ⩾ 22 n−1 . In both cases ex − 1 = f (x) + f −1 (x) < x + 0 = x which is a contradiction because for x < 0 we have that x < ex − 1.4. x→∞ . So. Let f ∶ R ∈ R be a strictly increasing invertible function such that for all x ∈ R. f (x) + f −1 (x) = ex − 1 for all x ∈ R. Since f is a strictly increasing invertible function then f −1 is strictly increasing too. if f has more than one fixed point then it has just these two fixed points: 0 and c. Find all functions f ∶ [0. and different from 0. Take x < 0 then f (x) < f (0) = 0. Moreover f (x) < x otherwise f (x) > x and x > f −1 (x). x3 /y3 = 13/5. Prove that f has at most one fixed point.163 2. and ∞ 1 1 1 1 1 ⩽∑ k = = < ⋅ 4 7⋅8 28672 6500 k=5 8 k=5 yk ∞ ∑ yn−1 = 22 n−1 −1 (⊠) Example 2. yn = xn−1 yn−1 ⩾ 22 n−2 This means that yn ⩾ 22 yn−1 ⩾ 22 n−2 n−1 −1 +2n−3 yn−1 ⩾ ⋯ ⩾ 22 n−2 +2n−3 +⋯+1 ⩾ 23n = 8n for n ⩾ 5. Analysic problems Solution The inequality holds for 1 ⩽ n ⩽ 4 because x1 /y1 = 2. ∞) such that (a) f is multiplicative (b) lim f (x) exists. and. (⊠) Example 2. f −1 (x) < f −1 (0) = 0. x2 /y2 = 5/2. Moreover.206.207. is finite. x2 ⩾ 0. Let f ∶ [0. .209. Because f (x) = xn+8 − 10xn+6 + 2xn+4 − 10xn+2 + xn + x3 − 10x + 1 = (x4 − 10x2 + 1) (xn+4 + xn ) + x3 − 10x + 1 1 1 = (x4 − 10x2 + 1) (xn+4 + xn + ) + 1 − x x √ √ √ √ 1 then f ( 2 + 3) = 1 − √ √ = 2 − 3 + 1. Let n be a positive integer and let f (x) = xn+8 − 10xn+6 + 2xn+4 − 10xn+2 + xn + x3 − 10x + 1. +∞). Then x2 = 5 + 2 6 and more manipulation gives x4 − 10x2 + 1 = 0. Prove that f (x1 ) + f (x2 ) + ⋯ + f (xn ) ⩾ nf (x1 + x2 + ⋯ + xn ) for all x1 . Hence f is identically equal to 1 in (0. √ √ Evaluate f ( 2 + 3) . The first term vanishes and we evaluate the remaining trinomial. Finally √ √ √ √ √ √ √ √ √ √ 3 ( 2 + 3) − 10 ( 2 + 3) + 1 = 11 2 + 9 3 − 10 ( 2 + 3) + 1 = 2 − 3 + 1. Examples for practice Solution Let lim f (t) = c ≠ 0 and let a > 0.. x2 . 2+ 3 (⊠) Example 2.208. xn ⩾ 0.. whereas f (0) can assume any nonnegative real number. then t→∞ c = lim f (at) = lim f (a)f (t) = f (a) lim f (t) = f (a)c t→∞ t→∞ t→∞ which implies that f (a) = 1. The polynomial to be evaluated can be written f (x) = (xn + xn+4 )(x4 − 10x2 + 1) + x3 − 10x + 1. .164 Chapter 2. (⊠) Second solution √ √ Note that 2 + 3 is a root of the polynomial √ √ √ √ √ √ √ √ x4 − 10x2 + 1 = (x − 2 − 3) (x + 2 − 3) (x + 2 + 3) (x − 2 + 3) . Using the binomial √ √ √ √ 3 theorem ( 2 + 3) = 11 2 + 9 3.. +∞) → R be a function such that f (x1 ) + f (x2 ) ⩾ 2f (x1 + x2 ) for all x1 . It’s trivial to check that such functions verify the assumptions. (⊠) Example 2. First solution √ √ √ Let x = 2 + 3. . then m!P (x) is a polynomial with integral coefficients. This holds trivially for m = 1.210. assume now that this holds for m = n − 1 (with n ⩾ 3). (⊠) Example 2. where unless otherwise indicated all summation indices range from 1 to n. (A) f (xj ) + f (∑ xi ) ⩾ 2f (∑ xi ) i≠j i Adding these n inequalities gives ∑ f (xj ) + ∑ f (∑ xi ) ⩾ 2nf (∑ xi ) . and P (n).i≠j Adding these n inequalities gives n ∑ f (xi ) ⩾ ∑ f (xj ) + (n − 1) ∑ f ( ∑ xi ) n n n n i=1 j=1 j=1 i=1. P (n+1). Note by the way that only f (x1 )+f (x2 ) = 2f (x1 +x2 ) for all xi in the domain of f was used above. the following. and we are given that this holds for m = 2. again for 1 ⩽ j ⩽ n.i≠j Subtracting out the common sum and then dividing by (n − 1) gives ∑ f (xi ) ⩾ ∑ f ( ∑ xi ) n n n i=1 j=1 i=1. Then for 1 ⩽ j ⩽ n we have.4. If P (x) is a polynomial of degree m ⩾ 1. i i i≠j j j i≠j Cancelling the common term on both sides and dividing by 2 we have ∑ f (xj ) ⩾ nf (∑ xi ) . that is the result also holds for a wider class of functions. . . therefore by induction the result holds for all m ⩾ 1 as was to be proved.i≠j We also have. P (n+m) are integers for some integer n. j j i≠j (B) i Adding inequalities (A) and (B) gives 2 ∑ f (xj ) + ∑ f (∑ xi ) ⩾ 2nf (∑ xi ) + ∑ f (∑ xi ) . the following.. ∑ f (xi ) ⩾ f (xj ) + (n − 1)f ( ∑ xi ) n n i=1 i=1. Analysic problems Solution We want to show that ∑ f (xi ) ⩾ mf (∑ xi ) m m i=1 i=1 for m ⩾ 1..165 2. j i which establishes the result for m = n. .166 Chapter 2.(x − n − m) is a polynomial of mth degree. hence Imf must be countable too. x = a Let n ⩾ 2. (m + 1)!a is an integer. P (n + 1) are integers. n + 1. since if P (x) = ax + b.. Q(n + m). 2.. the coefficients of m!R(x) are integers. such that P (x) = R(x) are integers for x = n.. P (n + 1). where a0 ≠ 0. such that P (x) ∈ Q for all x ∈ R ∖ Q. The set {0. then P (n + 1) − P (n) = a must be an integer if P (n).. Clearly t = ti for some 1 ⩽ i ⩽ k ⩽ n. n} × Q is countable. P ( ) ∈ Q then 2 2 x a = P (x + 1) − P (x) ∈ Q and b = 2P ( ) − P (x) ∈ Q. holds P (x) ∈ Q for all x ∈ R ∖ Q... First solution If there were such q polynomial than we could build an injection f ∶ R∖Q → {0... Examples for practice Solution The result is clearly true for m = 1. so g −1 exists. define Q(x) = P (x + 1) − P (x). hence R ∖ Q is countable. 1... (⊠) . where a.. y). g(x) = f (x) is a bijection. (⊠) Example 2. . 2 P (x) − b ∈ Q and that contradicts that x ∈ R ∖ Q. . P1 (x) ∈ Q for any x ∈ R ∖ Q. The conclusion follows. Thus we get a contradiction with earlier asumption of the induction.. hence the coefficients of (m + 1)!P (x) = (m + 1)!R(x) − (m + 1)!a(x − n)(x − n − 1).. and for any polynomial P (x) of degree m + 1 with coefficient a ≠ 0 for xm+1 . The equation P (x) = y has k ⩽ n solutions. hence b = P (n) − an must be an integer too. Let P (t) = y ∈ Q. impossible... Then g ∶ R ∖ Q → Imf. or by hypothesis of induction. Define f (t) = (i. n − 1} we should to prove that there is no polynomial P ∈ R[x] of degree n such that P (x) ∈ Q for all x ∈ R ∖ Q.. n}× Q in the following way: take some t ∈ R ∖ Q. Suppose that the statement of problem holds for polynomials of degree m ∈ {1. Suppose the opposite P (x) = a0 xn + a1 xn−1 + ⋯ + an .. Let them be t1 < t2 < . 2. Suppose that P (x) = ax + b. or applying the hypothesis of induction.. Since x + 1 ∈ R ∖ Q then P (x + 1) ∈ Q and for P1 (x) ∶= P (x + 1) − P (x) holds 1 ⩽ degP1 (x) < n. . Assume that the result is true for a given m. . Note that if P (n). 2. Hence.(x − n − m) are integers too. b ∈ R and a ≠ 0. P (n + m + 1) are integers..211. so are Q(n)..... < tk . (⊠) Second solution We will prove the statement of problem using induction on the degree n = 1. 1.. Q(n + 1). Clearly Q(x) is a polynomial with degree m and coefficient (m + 1)a for xm . Since x x x + 1. ∈ R ∖ Q and P (x + 1). Prove that there is no polynomial P ∈ R[x] of degree n ⩾ 1 such that P (x) ∈ Q for all x ∈ R ∖ Q. It is clear why this function is injective. Note therefore that R(x) = P (x) − a(x − n)(x − n − 1). . and so we are done. which is not constant.212. Proof of the above statement follows: Lemma 1: If P (x) and Q(x) satisfy the given condition we must have P (x) = Q ( x(x − 1) x(x + 1) ) − Q( ). √ f (P ) ⩾ AC + BD ⩾ 2 a2 + b2 . passing through P. Therefore. or the range of f is [2 a + b . As P moves continuously from √ the center of ABCD to one of its vertices. with equality iff P is in segment AC. f varies √ 2 2 2 continously. P A + P B = AB. Assume that the maximum of f occurs for some point P in the interior of ABCD. or QC + QD > P C + P D. and the minimum occurs at its center. for all n ⩾ 1. Restoring generality. is on AB or on AD. attained when P is one of the vertices of ABCD. has at most n ⩾ 1 real solutions. while QA+QB = P A+P B because P and Q belong to E1 . the maximum of f is a + b + a2 + b2 . This contradicts the fact that the polynomial P. 2 2 .167 2. this happens when P = A or P = B. Let ∆ be the plane domain consisting of all interior and boundary points of a rectangle ABCD. Since Q ∈ ∆. Clearly. and similarly P B + P D ⩾ BD. Q is outside E2 . then f (Q) > f (P ). Solution Clearly P A + P C ⩾ AC. where as stated above the maximum occurs at the vertices of ABCD. and consider the ellipses.4. a + b + a + b2 ]. Consider one of the points Q where E1 intersects the perimeter of ABCD. and E2 with foci C and D. (⊠) Example 2.213. Both ellipses intersect at P inside ABCD. Define f ∶ ∆ → R. f (P ) = P A + P B + P C + P D. P such that f (P ) is maximum. Wlog. In the first case. such that P (1) + P (2) + ⋯ + P (n) = Q(1 + 2 + 3 + ⋯ + n). E1 with foci A and B. Find all monic polynomials P and Q. (⊠) Example 2. with equality iff P is in segment BD. for P C + P D = AD + AC. whose sides have lengths a and b. Find the range of f. The result is the same in the √ second case by analogous reasoning. and the maximum of f cannot occur in the interior of ABCD. Clearly. with real coefficients. while P C + P D is maximum for the case of the largest ellipse with foci C and D that may be constructed with some intersection point in segment AB. Solution We have the following choices for P (x) and Q(x) (a) P (x) = Q(x) = x (b) P (x) = x3 + bx and Q(x) = x2 + bx for some real b. Analysic problems Third solution Since R ∖ Q is uncountable and Q is countable there is a rational number q ∈ Q such that P (x) = q for an infinite number of x ∈ R ∖ Q. with equality iff P is the center of rectangle ABCD. R(x) = Q ( x(x + 1) (x + 1)(x + 2) ) − Q( ) − P (x + 1). (◻) . P (1) = Q(1) − Q(0). Examples for practice Proof: Define. so the the above is possible iff R(x) is (x + 1)(x + 2) x(x + 1) identically 0. Now R(x) is clearly a polynomial. 2 This condition is clearly true for n ∈ {1. x(x + 1) x(x − 1) Proof: We have P (x) = Q ( ) − Q( ) . 2 n we must have n−1 = 1. 2 2 For any positive integer n we have R(n) = Q ( (n + 1)(n + 2) n(n + 1) ) − Q( ) − P (n + 1) 2 2 = Q(1 + 2 + ⋯ + (n + 1)) − Q(1 + 2 + ⋯ + n) − P (n + 1) = (P (1) + P (2) + ⋯ + P (n + 1)) − (P (1) + P (2) + ⋯ + P (n)) − P (n + 1) = 0.168 Chapter 2. (◻) Lemma 3: If P (x) and Q(x) satisfy the given condition the degree of Q is either 1 or 2. 2} and note for n ⩾ 3 we have 2n−1 > n. since P (x) is monic. Then P (x) = Q ( = [( x(x + 1) x(x − 1) ) − Q( ) 2 2 n n n−1 x(x − 1) x(x + 1) x(x + 1) ) −( ) ] + an−1 [( ) 2 2 2 + ⋯ + a1 x −( n−1 x(x − 1) ) 2 ]+ nx2n−1 The leading term of the P (x) above is easily seen to be n−1 . 2 2 but from the given condition P (1) = Q(1). hence we must have Q(0) = 0. So. Proof: Let Q(x) = xn + an−1 xn−1 + ⋯ + a0 . hence the only possible values for n are 1 and 2. Hence. and. 2 2 (◻) Lemma 2: If P (x) and Q(x) satisfy the given condition we must have Q(0) = 0. P (x + 1) = Q ( ) − Q( ) or equivalently 2 2 P (x) = Q ( x(x − 1) x(x + 1) ) − Q( ). Hence. R(x) vanishes at all integers. and hence can have only finitely many roots if it is not constant. x→0 x→0 x→0 (⊠) . Analysic problems Since. +∞) → [0.214. Let f ∶ (0. hence we must either have Q(x) = x or Q(x) = x2 + bx. lim ⎢f (x) − x→0 x→0 ⎢ 2 2 ⎥ ⎦ ⎣ then lim f (x) = 0. using the well known result that the sum of cubes of first n positive integers is the square of their sum. it is trivial that P (x) = Q(x) = x satisfy the conditions in the problem. lim 2α(x) = 0. Prove that if √ ⎤ ⎡ ⎢ x ⎥⎥ 1 ⎢ f ( )⎥ = 0 and lim (f (x) − 2f (2x)2 ) = 0. Q has to be monic. lim f (x) = 0. x→0 First solution 1√ f (x) then by condition lim α(x) = 0 Let α(x) ∶= f (x)−2f 2 (2x) and β(x) ∶= f (2x)− x→0 2 and √ ⎤ ⎡ ⎢ 1 t ⎥⎥ 1√ ⎢ f (x)) = lim ⎢f (t) − f ( )⎥ = 0 lim β(x) = lim (f (2x) − x→0 x→0 t→0 ⎢ 2 2 2 ⎥ ⎦ ⎣ where t ∶= 2x. We now prove sufficiency. Since f (x) is bounded and lim β(x) = lim α(x) = 0 then x→0 x→0 lim 8f (2x)β(x) = 0. x→0 and. Hence proved. Since 1√ f (2x)− f (x) = β(x) ⇒ f (2x)−2pf (x) = β(x) ⇒ f (x) = 4f 2 (2x)−8f (2x)β(x)+4β 2 (x) 2 and 2f (x) = 4f 2 (2x) + 2α(x) then f (x) = 4f 2 (2x) + 2α(x) − (4f 2 (2x) − 8f (2x)β(x) + 4β 2(x)) = 8f (2x)β(x) − 4β 2 (x) + 2α(x). (⊠) Example 2. lim (−4β 2 (x)) = 0. and the corresponding values of P from Lemma 1 have to be P (x) = x and 2 2 x(x + 1) x(x + 1) x(x − 1) x(x − 1) P (x) = ( ) + b( )−( ) − b( ) = x3 + bx 2 2 2 2 respectively. we get P (1) + ⋯ + P (n) = 13 + ⋯ + n3 + b(1 + ⋯ + n) = (1 + ⋯ + n)2 + b(1 + ⋯ + n) = Q(1 + ⋯ + n). In case Q(x) = x2 + bx and P (x) = x3 + bx. +∞) be a bounded function.4. Q(0) = 0 and degree of Q is 1 or 2. therefore.169 2. .. Solution If n = 2.. Hence a2 − r = p or a2 − r = −p. . not divisible by any odd prime p. since both sequences are clearly strictly increasing. absurd.(x − an ) − p. an+3 = an+2 + 2an+1 + an > bn+3 = bn+2 + 2bn+1 + bn > an+1 + 2an + an−1 = an+2 . an be distinct integers such that q∣(ai − aj ) for all pairs (i. Let p and q be odd primes such that q ∤ (p − 1) and let a1 . no bn for n ⩾ 4 may appear in (an ). and b4 = 12. 5. In the first case. Clearly a2 − r ≠ 1 because a1 and a2 are distinct. If a2 − r = −1.. Then. n = 0.. Let (an )n⩾0 and (bn )n⩾0 be sequences defined by an+3 = an+2 + 2an+1 + an . . (⊠) Example 2. j). yielding p = 1. Prove that P (x) = (x − a1 )(x − a2 ). hence P (x) is irreducible for n = 2.215. and we may exchange x by −x without altering the problem.. a2 .. a2 = 3 and bn+3 = bn+2 + 2bn+1 + bn . a1 − a2 = 1 − p.. we obtain lim [f (2x) − x→0 1√ f (x)] = 0 2 √ Since 2f (2x)+ f (x) is bounded by hypothesis. 1. a5 = 35. P (a1 ) = (a1 − r)(a1 − s) = −p and P (a2 ) = (a2 − r)(a2 − s) = −p. Therefore.. while the only n’s for which an = bn are n = 1 and n = 3 with a1 = b1 = 2 and a3 = b3 = 8. Substitution yields r = a1 − 1 = a2 − p and s = a1 − p = a2 − 1. or by trivial induction.4. 1.4 Olympiad problems Example 2. 3. multiplying the limit by this function we obtain 1 lim (2f (2x)2 − f (x)) = 0 x→0 2 Adding the second condition yields the proposed result. . b2 = 1. while a4 = 16. a0 = 1.. n = 0. is irreducible in Z[x] for n ⩾ 2. b1 = 2. or a2 − r = −p and a2 − s = 1. b5 = 29. and P (x) is not irreducible in Z[x]. and the only values that appear in both sequences are {1.. 2. 6. hence q∣(a2 − a1 ) = p − 1. How many integers do the sequences have in common? Solution Clearly a3 = b3 = 8. a1 = 2. Note that for n = 4.170 Chapter 2. (⊠) 2.216. for any n ⩾ 3. and since P (x) is monic. then P (x) = (x − r)(x − s) for integers r. then a1 − a2 = 2. We will assume in the rest of the problem that n ⩾ 2. b0 = 3. 8}. Note that r − s = p − 1 = 1 − p. where wlog a1 − r = 1 and a1 − s = −p since we may exchange r and s. b6 = 61. a6 = 75. s. absurd.. an > bn > an−1 . Examples for practice Second solution Substituting x by 2x in the first condition. ⩾ ∣∑ ∣x1 ∣∣x2 ∣⋯∣xn−m ∣ x1 x2 ⋯xn−m an If ε = min {∣xi ∣} we have ∑ 1⩽i⩽n 1 εn−m Cnn−m ⩾ ∑ 1 > Cnm = Cnn−m > 0 ∣x1 ∣∣x2 ∣⋯∣xn−m ∣ and then ε < 1. We conclude that distinct i.. ie. −p. n} exist such that. . Then. p} when i n takes all possible values between 1 and n. absurd. 1.. which has degree at most . 1 ⩽ deg(Q(x)) ⩽ 2 for i = 1.. = (−1)n nx1 x2 ⋯xn . x2 . Therefore. 2 2 or it is identically zero. where Q(ai ) and R(ai ) are integers. q∣(ai − aj )∣(Q(ai ) − Q(aj )) = p − 1.. n. Let P (x) = a0 xn + a1 xn−1 + ⋯ + an . . −p. k = 1. 2.. an ≠ 0. Q(ai ) takes at least three diff erent values when i takes all possible values between 1 and n.. or Q(ai ) = −1 and Q(aj ) = −p. j ∈ {1. n. Assume now that Q(ai ) takes only two values k1 . contradiction. xn are the roots of P (x) = 0. by the Viete’s formulae an am = (−1)m ∑ x1 x2 ⋯xm . either Q(ai ) = p and Q(aj ) = 1.. R(x) ∈ Z[x] exist such that Q(x)R(x) = P (x). Q(ai )R(ai ) = P (ai ) = −p 1 ⩽ deg(Q(x)) ⩽ deg(R(x)).. has at least roots. Therefore. Clearly.. where wlog n . 2. 1. and Q(x) = k1 . . an Prove that the polynomial P has at least a zero with the absolute value less than 1. By Vieta’s formula wk ∣ n 1 am ∣= ∑ ∏ an I∈Jn−m k∈I ∣wk ∣ (⊠) . The proof is completed. The result follows. polynomials Q(x). Analysic problems Assume that P (x)is not irreducible in Z[x]. First solution If x1 .. k2 ∈ {−1.. (⊠) Example 2. n} (note that wk ≠ 0 because an ≠ 0). Q(x) − k1 ..217. Therefore.4.. or Q(ai ) ∈ {−1.171 2. There are thus at least values of i for which 2 n n wlog Q(ai ) = k1 . a0 a0 hence and am (−1)m ∑ x1 x2 ⋯xm (−1)n−m = = ∑ an (−1)n nx1 x2 ⋯xn x1 x2 ⋯xn−m ∣(−1)n−m ∣ am (−1)n−m ∣ = ∣ ∣ > Cnm . be a polynomial with real coefficients and have n roots such that there is an m with ∣ am ∣ > Cnm .. .. p} for all i = 1. Second solution The zeros of the polynomial are { Q(x) = an xn + an−1 xn−1 + ⋯ + a0 1 . 2. n} such that ∣Jn−m ∣ = n − m.. Clearly. then we are done by Eisenstein’s Criterion. a contradiction.. Note that the coefficients of xp−1 ... Examples for practice where Jn−m is the set of all subsets of {1. (⊠) Example 2. x0 are divisible by q. and our proof is complete. . . If f (f (x)) > x.. there is no multiple of q among (q + 1)..(p − 1) is not divisible by q. (p − 1)...172 Chapter 2. Hence. Solution Let q be the greatest prime less than p.220. p is one such prime. Let p be a prime.. ⋅ (p − 1).. the left side of this equation is negative.. If q 2 does not divide (p−1)!. For which a does there exist a nonconstant function f ∶ R → R such that f (a(x + y)) = f (x) + f (y)? . 1 ⋅ 2 ⋅ .n − 1 am ∣ ∣ ⩽ ∑ 1 = Cnn−m = Cnm an I∈Jn−m and this contradicts the hypothesis..218. y ∈ R+ . Consider the factors of (p − 1)! = 1 ⋅ 2 ⋅ . xp−2 . Replacing x with f (x) gives f (2f (x)) + f (2f (f (x))) = f (2f (f (x) + f (f (x)))). (⊠) Example 2.. It remains to prove that (q + 1)(q + 2). A similar contradiction occurs if f (f (x)) < x. so f (f (x) + f (f (x)) > f (x + f (x)) and f (x) + f (f (x)) < x + f (x).. ⋅ (q − 1) is not divisible by q.. (q + 2). If all zeros 1 ⩽ 1 and for any integer of P has the absolute value greater or equal than 1 then ∣wk ∣ m ∈ 0. Suppose f ∶ R+ → R+ is a decreasing function such that for all x.. Therefore q 2 does not divide (p − 1)!. Thus f (f (x)) = x as desired. (⊠) Example 2. . Prove that p(x) = xp + (p − 1)! is irreducible in Z[x]. 2. there is at least one prime between q and 2q. ⋅ q ⋅ . Solution: Putting y = x gives f (2x) + f (2f (x)) = f (2f (x + f (x))).219... By Bertrand’s Postulate. f (x + y) + f (f (x) + f (y)) = f (f (x + f (y))) + f (y + f (x)). Subtracting these two equations gives f (2f (f (x))) − f (2x) = f (2f (f (x) + f (f (x)))) − f (2f (x + f (x))). Prove that f (f (x)) = x.. . we may set f (x) = x. which is not allowed. (⊠) Example 2. Show that for any real numbers x and y.2. putting y = 1−a f (y) = f (x) + f (y).221. we have the inequality (P (xy)) = P (x2 )P (y 2). Analysic problems 173 Solution: ax gives For a = 1. (⊠) (a0 x2n + ⋯ + a0 )(a0 y 2n + ⋯ + a0 ) = (a0 xn y n + ⋯ + a0 )2 (⊠) 2 Solution: This actually holds for any polynomial with nonnegative coefficients. If P (x) = a0 xn + ⋯ + an x0 . then by the Cauchy-Schwarz inequality. So only a = 1 works.4. For any other a. Let P (x) be a quadratic polynomial with nonnegative coefficients. so f (x) = 0 for all x. .. hence considering the isosceles triangle with equal sides of length 1 + r and sidelength 2r for the other side. we have r = (1 + r) sin . . How many of the first 2012 terms of these sequences are equal? (⊠) . and by cyclic symmetry. p − 1. with sidelength 2r and circumradius 1 + r. Examples for practice Problems of Other Topics 2. 4. . 1. 2...174 2. 1.. and at most three of them would have the same radius... With this assumption. where the 180o 360o . 1. Consider the sequences are given (a) (an )n∈N∗ ∶ 1..5. 4. where each three consecutive signs + are followed by two signs −. . and each circle of radius r is tangent to another two circles of radius r. 2. 2.1 Junior problems Example 2. Find r.. 3. clearly all centers of the circles with radii r are at a distance 1 + r from the center of the circle with radius 1.. 2. or equivalently different angle is n n 180o n r= ⋅ 180o 1 − sin n sin Example 2. since either one of the four circles contains the other three inside them..5 Chapter 2. Consider n. 1. 3. and at a distance 2r from their closest neighbours of radius r. (b) (bn )n∈N∗ ∶ 1. 2. 2. p..223. p.222. since it is well known that at most four circles can be tangent to each other.224. Solution Notice that Hence (5k − 4)2 + (5k − 3)2 + (5k − 2)2 − (5k − 1)2 − (5k)2 = 25k 2 − 80k + 28. Evaluate T ∶= 12 + 22 + 32 − 42 − 52 + 62 + 72 + 82 − 92 − 102 + ⋯ − 20102 + 20112 . T = 20112 + 25 ∑ k 2 − 80 ∑ k + 28 ∑ 1 = 20112 + 25 402 402 402 k=1 k=1 k=1 402 ⋅ 403 ⋅ 805 − 40 ⋅ 402 ⋅ 403 + 28 ⋅ 402 = 57736735. (n ⩾ 6) circles of radius r < 1 that are pairwise tangent and all tangent to a circle of radius 1. . . Solution We will assume that the problem statement means that n circles or radius r are all tangent to a given circle of radius 1. ... 6 Example 2. 1.. or one of them is in the gap formed by the other three. p − 1. these n centers are at the vertices of a regular n-gon. 3. 3. 3. 1. p − 2. 1.. 2. 2. 1. Let an = ⎨ 2 ⎪ ⎪ ⎩n − n . and thus k = 2m2 + 2m + 1. 2⋅312 +2⋅31+1 = 1985.. Using the above formulae about ak and bk . + a2012 . 2 2 for all k ∈ N∗ . there exists a unique nk ∈ N∗ such that 1 1 nk (nk − 1) < k ⩽ nk (nk + 1). 2⋅22 +2⋅2+1 = 13.. 2⋅322 +2⋅32+1 = 2113 > 2012. But n2 − n + (n + 1) − (n + 1)2 = −2n and so S = −2⋅1+2⋅3+⋯+2⋅2011 = 2((−1+3)+(−5+7)+⋯+(−2005+2007)+(−2009+2011)) = 2012. n must be odd. if 4 divides n2 − n Example 2.. otherwise.175 2. (⊠) First solution It is well known that 4 divides n2 − n if and only if n = 0(mod4) or n = 1(mod4). (⊠) Second solution S = ∑ an = ∑ (a4k+1 + a4k+2 + a4k+3 + a4k+4 ) 2012 502 k=1 k=0 = ∑ [(4k + 1)2 − (4k + 1) + 4k + 2 − (4k + 2)2 + 4k + 3 − (4k + 3)2 + 502 k=0 +(4k + 4)2 − (4k + 4)] = ∑ [16k 2 + 4k − 16k 2 − 12k − 2 − 16k 2 − 20k − 6 + 16k 2 + 28k + 12] 502 k=0 = ∑ 4 = 4 ⋅ 503 = 2012. 502 k=0 (⊠) . Equal terms of the sequences satisfy ak = bk for a k ∈ N∗ . Problems of Other Topics Solution For every k ∈ N∗ . this is equivalent with 2k = n2k + 1. 2 2 It’s not hard to prove that 1 1 ak = k − nk (nk − 1) and bk = nk (nk + 1) + 1 − k.5. Hence the sum is S = ((12 − 1) + (2 − 22 )) + ((3 − 32 ) + (42 − 4)) + ⋯ + ((2007 − 20072) + (20082 − 2008)). The total number of equal terms is therefore 31.225. We conclude that ak = bk if and only if there exists a positive integer n such that 2k = n2 + 1. .. Clearly. and so there should exist a m such that 2k = (2m + 1)2 + 1.. ⎧ ⎪ ⎪n2 − n. Evaluate S ∶= a1 + a2 + . 25. We find that the only equal terms are those with index 2⋅12 +2⋅1+1 = 5. First solution Two cyclic n-gons P and P1 with the same sides. if x < y. Since among all closed curves of the same lenght.. Examples for practice Example 2. . 2 2 2 (⊠) Third solution Note that the triangle flipping operation illustrated by the figure is area pre. Assume by contradiction that the radius R1 of the circle of P1 is larger than the radius R of the circle of P .. a2 By the Law of Cosines we have that a2 = 2R2 − 2R2 cos βi . Thus cos βi = 1 − i 2 2R a2i a2i and we obtain βi = arccos (1 − ) . Because with this operation we can construct all possible cyclic n-gons. then keeping the same circle. the solution is x = R. Let fi (x) = arccos (1 − 2 ) ⇒ βi = f (R) and let 2R2 2x g(x) = f1 (x) + f2 (x) + ⋯ + fn (x). We consider an arbitrary cyclic n-gon of radius R constructed with these matches. and the that part above. On the sides of P1 we construct the arcs of the circle in which the P is inscribed (the arcs do not overlap because P1 is convex). Example 2. Let βi be the central angle that corresponds to the ai . The following equation has an unique solution: (◻) g(x) = 2π. Prove that all of them have the same area. still has the same area after being flipped). Mister Tien has a box with n not necessarily equal matches.227. since g(x) is continuos and strictly decreasing. a2i a2i a2i a2i < 1 − ⇒ arccos (1 − ) > arccos (1 − ) ⇒ fi (x) > fi (y). are inscribed in circles with the same radius and therefore they have the same area (because it is equal to the sum of the areas of the n isosceles triangles whose bases are the sides and the two other sides are equal to the radius).226. an be the n matches. We get a closed curve of lenght 2πR which contains an area larger than πR2 . but not necessarly in the same order. Mister Tien then constructs other cyclic n-gons with these matches. we will prove that g(x) is strictly decreasing. It follows that all cyclic n-gons constructed by Mister Tien have the same radius and the area of them is 1 1 1 S = R2 sin β1 + R2 sin β2 + ⋯ + R2 sin βn . Characterize triangles with sidelengths in arithmetical progression and lengths of medians also in arithmetical progression. (⊠) Second solution Let a1 . He is able to construct with them a cyclic n-gon. Indeed. namely the triangle ABC. 1− 2 2x 2y 2 2x2 2y 2 It suffices to add over all of the i. we get that all of them have the same area.. he can swap any two adjacent sides and therefore he can obtain any permutation of the sides. Note that when Mister Tien constructs a cyclic n-gon.176 Chapter 2. the circle is the one with the largest area. a2 . we have a contradiction.serving (since that part of the n-gon below AC is unchanged. . Without loss of generality a ⩽ b ⩽ c. By the properties of arithmetic progression we have that a + c = 2b . 4 4 From this. [2] By the relations [1] and [2] we have √ √ √ 2b2 + 2c2 − a2 + 2a2 + 2b2 − c2 = 2 2a2 + 2c2 − b2 . [1] and ma + mc = 2mb . Therefore. Assume that a ⩾ b ⩾ c.5. 2mb = ma + mc . 2 and the analoguous ones. b. we have a = b = c and triangle is equilateral. Since the sides and the medians form arithmetic progressions we have 2b = a + c. we obtain the following chain of inequalities 9 2 2 2 (a + b + c ) ⩾ (3mb )2 ⇔ a2 + b2 + c2 ⩾ 4m2 ⇔ a2 + b2 + c2 ⩾ 2a2 + 2c2 − b2 . mc . 2b2 ⩾ a2 + c2 ⇔ (a + c)2 ⩾ 2a2 + 2c2 ⇔ (a − c)2 ⩽ 0. We can then deduce that mc ⩾ mb ⩾ ma . By the well known Apollonius-formulas (application of Stewart’s theorem) for the medians of a triangle we have 1√ 2 ma = 2b + 2c2 − a2 . This means that a = c. c and the corresponding medians ma . we have 1 3 2 2 2 (a + b + c ) = m2a + m2b + m2c ⩾ (ma + mb + mc )2 . 4 Hence.12: First solution We prove that only equilateral triangles have the desired property. using (i) and the relation 4m2b = 2a2 + 2c2 − b2 . Then we have ma ⩽ mb ⩽ mc . and we also know that for reals 4 x. Let ABC be a triangle with sides a.177 2. mb . Hence. y. z we have the inequality 3(x2 + y 2 + z 2 ) ⩾ (x + y + z)2 . Problems of Other Topics Figure 2. (i) 3 It is a known fact that m2a + m2b + m2c = (a2 + b2 + c2 ). (⊠) Second solution We use the habitual notation in a triangle. Examples for practice which equivalent √ a2 + 4b2 + 2 (2b2 + 2c2 − a2 )(2a2 + 2b2 − c2 ) = 4(2a2 + 2c2 − b2 ). The game ends when all the cards were played or no more card can be placed on the table. namely 3 and 14. (⊠) Second solution We claim that Khoa has a winning strategy. The sequence of moves are the following: 22. Note that the perfect squares in the play are: 4. 25. . then Khoa have another winning chain (2 . 9. At each step Tien have no choice other than choosing the card shown in the chain. If Tien chosses 14. both of which are impossible. Tien cannot play since the equation 18+m = n2 is not solvable. 16. On his subsequent moves. 18. 9. 12. because 7 was chosen before. Tien chooses 3. 20. If Tien chosses 7.178 Chapter 2.5 . Doing all the subsequent calculations will yeild c 4 c 3 c 2 c 5 ( ) − 8 ( ) + 6 ( ) − 8 ( ) + 5 = 0. it is clear that equilateral triangle satisfies the problem and we are done.228. It is clear that Tien has two choices for her first move. Tien chooses 14.18). Finally. 13. Case 2. The winner is the one who played the last card. such that the sum of his number and the previous number is a perfect square. 3. 4. a a a a Then the polynomial P (x) = 5x4 − 8x3 + 6x2 − 8x + 5 = (x − 1)2 (5x2 + 2x + 5) and the quadratic c equation 5x2 + 2x + 5 has no real roots. 14.14 . Let us consider these two cases separately. Khoa and Tien play the following game: there are 22 cards labeled 1 through 22. Does Khoa have a winning strategy? First solution The winning strategy for Khoa is to choose the card labeled 2 in the first step.11 .20 -16 9 . Then the sequence of moves continues as follows: 5. 7. Khoa chooses one of them and places it on a table. I. Once again we reach a state when Tien can make no further move. 7. 18. Thus Tien has only one choice in each of her subsequent moves if Khoa sticks to his strategy. Khoa then places one of the remaining cards such that the sum of the numbers on the last two cards played is a perfect square. 20. So Khoa has a winning strategy and we are done. 16. Case 2. 5. On his first move. let Khoa choose 22. (⊠) Example 2. then Khoa chosses 18. let Khoa choose the maximum number available to him. Then = 1 which means a = b = c and the triangle is a equilateral. Case 1. 9.7 .18). and so on. Tien then places one of the remaining cards at the right of the one placed by Khoa such that the sum of the two numbers on the cards is a perfect square. Tien plays 5. 36. Once Khoa places 18. 11. 16. or 18. Tien can now either place 5 or 21. Tien has to either place 7. Thus Khoa is winner (2 .7 . The sequence of moves runs 22. We consider the equation 2+m = n2 to obtain that Tien just can choose 7 or 14 in the second step. Evaluate 2011 2010 6 5 4 3 . 13. − C2012 − ⋯ + C2012 + C2012 − C2012 + C2012 −C2012 Using (1) we get that 2012 2011 3 2 1 0 ) − C2012 − ⋯ + C2012 + C2012 − C2012 + C2012 L = L + 2 (−C2012 2011 3 2 2012 1 0 ).2 Tien plays 10. We obtain that (1 + 2 + ⋯ + 16) + (1 + 5 + 13) + x = 4c. as 2010 2009 3 2 1 ) + C2011 + ⋯ − C2011 − C2011 + C2011 L = −2 + 2012 ⋅ 2 − 2 + 2012 (−C2011 (1) (2) (3) .230. 8. Problems of Other Topics Case 2.1 Tien plays 1. The sequence continues thus: 1.229. 8. (⊠) 1 5 13 Example 2. or 15. 4s − x − 1 − 5 − 13 = 1 + 2 + ⋯ + 16 = 56 is a multiple of 4. (⊠) Second solution Let x the number in the darkened square. 5. we can rewrite (2). First solution Let x the number in the darkened square. Tien plays 21. Then the sequence of moves continues as follows: 21. none of which is possible as they have been played earlier. 5. The squares in the figure on the left side are labeled 1 through 16 such that the sum of the numbers in each row and each column is the same. or x must have a remainder of 1 when divided by 4. Thus Khoa can always force a win by sticking to this strategy. (⊠) Example 2.179 2. That is to say 155 + x = 4c. both of which are not possible as they have been played earlier. 17.The positions of 1. 19. From this we deduce that x = 9 because 1. So x = 9 is the only possible value. ) + (−2C2012 − ⋯ + 2011C2012 + 3C2012 − 2C2012 + 2C2012 = (−2C2012 k−1 Since kCnk = nCn−1 . II. and s the sum in each row and each column. and c the sum in each row and each column. 6. and 13 are given. but 155 ≡ 3 (mod 4) so x ≡ 1 (mod 4) and x could be 1. 15.5. Clearly. The sequence continues as follows: 10. Now Tien can play either 15 or 6. 13 are given. Case 2. and Tien can now play either 1 or 10. 9. 19.II. 17. Case 2. + 2009C2012 + ⋯ − 2008C2012 − 4C2012 + 3C2012 − 2C2012 L ∶= C2012 First solution From the Binomial Theorem we can write that 2012 2011 3 2 1 0 = (−1 + 1)2012 = 0. 5.II. 1. Prove that there is only one possibility for the number in the darkened square and find this number. 10. But all integers not exceeding 16 with remainder 1 modulus 4 have been used except for 9. 6. 8. Now Tien has to play either 3. because one will contain an even number of odd weights and the . and evaluating the resulting expression at x = −1. A square of side 1 weighs 1 gram. we need a minimum of 18 steps.180 Chapter 2. (a) Prove that for each integer g Lan can place the laminated squares on the scales such that after a certain number of steps the difference between the aggregate weights on the two scales is g grams. since there are 9 odd perfect squares among the first 18 perfect squares. or the difference of weight between both scales will be odd. ie. we have n2 − (n + 1)2 − (n + 2)2 + (n + 3)2 = 4.231.5. by placing on one scale the squares with sides n. (3) can be rewrite as L = −4 + 4024 + 2008 ⋅ 0 = 4020. we find (since f ′ (−1) = 0) that the sum given in the problem statement is equal to 2012 + 2010 − 2 = 4020. Lan has a pair of scales that display the weight in grams. (⊠) (1 + x)2012 ⋅ Then expanding this out. by placing the squares of sides 1 and 2 in opposite scales (with respective weights 1 and 4). Finally. and on the other scale the squares with sides n+1. Difference 1 is easily obtained after 1 step. 2. Solution (a) Clearly the weight added to one scale at step n is n2 . 2. (⊠) (b) It is well known (or easily checked by induction) that 12 + 22 + ⋯ + n2 = Since n(n + 1)(2n + 1) ⋅ 6 12 + 22 + ⋯ + 172 = 1785 < 2010 < 2109 = 12 + 22 + ⋯ + 182 . It therefore suffices to show that we may obtain differences 1.2 Senior problems Example 2. difference 2 is obtained by placing in opposite scales the squares with sides 1. and the square with side 4 (weight 16 grams). 3 (total weight 14 grams). At step n she cuts a square of side n from a very large laminated sheet and places it on one of the two scales. The conclusion to part (a) follows. Note first that for any integer n. 4 steps to add 4. by step n+3 we can obtain differences g + 4 and g − 4. (⊠) Second Solution Let f (x) = 2. (b) Find the least number of steps necessary to reach a difference of 2010 grams. Examples for practice But as we have that Cnk = Cnn−k . taking the derivax2 tive. and difference 3 after 2 steps. choosing which squares go on which scale according to whether we want the difference to increase or decrease. n + 3. if by step n−1 we have managed to obtain a difference of g. 3 (difference 0 is present at the initial condition and may be clearly obtained again after 8 steps. and 4 steps to substract 4). n+2. In 18 steps the task is impossible. ρb = ρd . Given that ac = ac. d are collinear. ρd and angles β. c. a and c. On each vertex of the regular hexagon A1 A2 A3 A4 A5 A6 we place a rod. Clearly 0 is the midpoint of diagonals AC and BD. d be the complex numbers corresponding to the vertices A. 2. In either case.5. b. it follows that ABCD is a parallelogram. b and d. It follows that at least one of a + c. and express similarly b. ρc . where ai corresponds to the vertex Ai . we have the following possibilities for three adjacent vertices: A1 A2 A3 . (⊠) Example 2. c. except for the square with side 15. Similarly. (⊠) Example 2. δ = β + π. Solution Denote a = ρa (cos α + i sin α). and similarly for b + d. the maximum number of chains is M ∶= a1 a2 a3 + a2 a3 a4 + a3 a4 a5 + a4 a5 a6 + a5 a6 a1 + a6 a1 a2 . What is the maximum number of such chains that we can create? Solution If by adjacent rods. A3 A4 A5 . b. a + c is either 0 or collinear with 0. a + c = (ρa + ρc ) cos α + i(ρa + ρc ) sin α. A4 A5 A6 . absurd. δ = β or δ = β + π. Note that. hence a. Let a. The minimum is thus 19 steps. γ = α + π. c. 6 ie a difference of 2010 grams may be obtained by placing on one of the scales all squares with sides 3 to 19 inclusive. they are both collinear with 0 since they add up to 0. δ.233. A difference of 2010 may be however obtained after 19 steps. A6 A1 A2 . since 12 + 122 + ⋯ + 192 − 2(152 + 22 + 12 ) = 19 ⋅ 20 ⋅ 39 − 2(225 + 4 + 1) = 2470 − 460 = 2010. we understand consecutive rods. ρa = ∣a∣ being a nonnegative real. b + d. Since ac = ac. is zero. while b + d is also either 0 or collinear with 0. we conclude that ac is real.232. bd = bd and a + b + c + d = 0. γ. or ρa = ρc . whereas if γ = α + π. if α = γ. which is placed on the other scale together with the squares with sides 1. Hence. Problems of Other Topics other an odd number of odd weights. a + c = (ρa − ρc ) cos α + i(ρa − ρc ) sin α.181 2. If a + c and b + d are both nonzero. B. yielding either γ = α or γ = α + π. and clearly a + c = b + d = 0. (⊠) . Taking a ring from any three adjacent rods we can create chains of three rings. D of a convex quadrilateral ABCD. A5 A6 A1 . A2 A3 A4 . prove that ABCD is a parallelogram. On each rod we have ai rings. C. d with respective radii ρb . (⊠) Example 2. and denote xi the number of times that a vertex has been picked before arriving to a given configuration (ie. subtracting from 16n − 2 the same sums of the remaining 4n − 3 elements with the same combinations of signs.182 Chapter 2. up to replacing (4n + 1) by −(4n + 1) which subtracts (8n + 2) from S. we obtain all odd numbers between 16n−1 and 8n2 +6n+1 = (4n+1)(2n+1). Clearly this process can be repeated. if we replace 2 by −2 this subtracts 4 from S. for the second iteration the (4n + 1) term starts out negative. Solution Number the vertices from 1 to n counterclockwise. leaving the rightmost terms with the signs changed alone on the next iteration (that is for the first iteration all terms are initially positive. yields all odd positive integers less than or equal to (2n + 1)(4n + 1). and so on. adds 2 to the number written at that vertex. Prove that. Then. 13 = 5 + 4 + 3 + 2 − 1. since we may subtract min{xi } from each xi . to a given distribution of values in the vertices of the n-gon).235. Consider now what happens if on the left hand side we change the sign of exactly one summand: if we replace 1 by −1 this subtracts 2 from S. and this sequence. Prove that for different choices of signs + and − the expression ±1 ± 2 ± 3 ± ⋯ ± (4n + 1). 11 = 5 + 4 + 3 − 2 + 1. Clearly. 9 = 5 + 4 − 3 + 2 − 1. so all odd numbers between 1 and (2n + 1)(4n + 1) have thus been generated. Zeroes are written at every vertex of a regular n-gon. we obtain all odd integers between 16n − 3 and −8n2 + 10n − 3 = −(2n − 1)(4n − 3). for the third iteration the 4n and (4n + 1) terms start out negative. Examples for practice Example 2. Assume now that the result is true for n−1. (⊠) Second solution Clearly. 7 = 5 − 4 + 3 + 2 + 1. and use cyclic notation such that xn+i = xi . First solution If we take all signs positive we have 1 + 2 + 3 + ⋯ + 4n + (4n + 1) = (2n + 1)(4n + 1) = S. this last integer is negative for any positive integer n.234. Mister Tien picks a vertex. Note that we may thus assume wlog that min{xi } = 0. he will never be able to achieve a configuration in which a 1 is written at one vertex. Every minute. This gives a decreasing sequence of consecutive odd positive integers starting with S and ending with −S. the amount assigned to vertex i is 2xi − xi−1 − xi+1 . since (4n + 1) + 4n + (4n − 1) + (4n − 2) = 16n − 2. 3 = 5−4+3−2+1 and 1 = 5−4−3+2+1. Moreover. and subtracts 1 from the numbers written at the two adjacent vertices. and a zero is written everywhere else. the result is true for n = 1 since 15 = 5 + 4 + 3 + 2 + 1. resulting in the same configuration since each 2xi − xi−1 − xi+1 does not . a −1 is written at another. 5 = 5 − 4 + 3 + 2 − 1. no matter how long Tien plays. The conclusion follows. Clearly. and so on). adding the sum of the remaining 4n − 3 elements with all combinations of signs that produce all positive integers between 1 and (2n−1)(4n−3). then since x − y. we conclude that xi = 0 for the vertex that has 1 assigned. Now. Substituting back into the equation (y + z)(x − y)(x − z) = 1 ∶ (ωx + ω 2 x)(x − ωx)(x − ω 2 x) = 1 ⇔ ωx3 (ω + ω 2 )(1 − ω)2 (1 + ω) = 1.183 2. and x2 + xn = 1. Let x. since if x − y = 0. contradiction. ω−1 Thus we have y = xω. we conclude that xi = 0 for the vertex that has 1 assigned. x y z but since ω ≠ 1 (if ω = 1. we obtain = . Thus we can divide both sides of the equation (y + z)(x − y)(x − z) = (z + x)(y − z)(y − x) by x−y ∶ (y + z)(x − z) + (z + x)(y − z) = 0 ⇔ xy + xz − zy − z 2 + yz + yx − z 2 − zx = 0 ⇔ xy = z 2 . z are non-zero.5. If x = 0. absurd since the value that has 1 assigned would have a non-positive value. and since 0 is assigned to all vertices from the nth down to the vertex that has 1 assigned. then 1 = (y + z)(x − y)(x − z) = 0. Note therefore that xn−1 = 0 or the nth vertex would have a negative value assigned. . Determine all possible values of L ∶= (y + z)(z + x)(x + y). ie xk−1 = xk+1 = 0. so since yz = 0. we obtain ω + ω 2 = −1. Since y z x x z xy = z 2 . y. Problems of Other Topics change. x − z ≠ 0. 1 + ω = −ω 2 . or its value is non-positive. traveling backward from the nth vertex. (1 − ω)2 = −3ω. then x − y = 0). z are non-zero. x2 = 1 and xn = 0. By trivial backwards induction. Then yz(y − z) = (z + x)(y − z)(y − x) = (x + y)(z − x)(z − y) = yz(z − y). Using the identity ω 2 + ω + 1 = 0. Assume that in the ending configuration. Note that we may travel from the k th vertex to the vertex that has 1 assigned. Thus x. x1 = −1. y. The conclusion follows. Solution First.236. z = yω = xω 2 . without going through the vertex that has −1 assigned. y. and z be complex numbers such that (y + z)(x − y)(x − z) = (z + x)(y − z)(y − x) = (x + y)(z − x)(z − y) = 1. Similarly. since 2xk − xk−1 − xk+1 is equal to either 0 or 1. we note that x − y = 0. and 0 is assigned to the k th vertex. Clearly z y x y z y z x ω 3 = ( ) ( ) ( ) = 1. which is a contradiction. and that xk = 0 for some k ≠ 1. (⊠) Example 2. we must encounter the vertex that has 1 assigned. Let ω = = = . it follows that ω2 + ω + 1 = ω3 − 1 = 0. clockwise or counterclockwise. and z = y. Clearly. or wlog by symmetry since we may number the vertices clockwise instead of counterclockwise. zx = y 2 . Hence x1 = 0. then y − z = 0. xk−1 and xk+1 are non-positive. By trivial induction forward or backward. (⊠) . L= 1 u15 (u9 + 1 u15 (u5 + 1) + 1) √ 2 4 = 15 = 8. a20 = u15 + u6 = u6 (u9 + 1). c circles. It is easy to find a configuration of l lines. so the contribution is 2l(c + e). 1 2 1 1 1 1 1 1 Let u = 2 60 and observe that 1 a5 = u15 + u24 = u15 (u9 + 1) . a12 = u15 + u10 = u10 (u5 + 1) . a5 a6 a12 a20 (⊠) Solution We observe that a5 = 2 4 + 2 5 . and e ellipses with such a number of intersection points. so the contribution is 2Cc2 . 4. the only possible value of (x + y)(y + z)(z + x) is ⋅ 3 √ √ n 4 Example 2. Examples for practice so: x3 (−1)(−3ω)(−ω 2 ) = 1 ⇒ x3 (−3) = 1 ⇒ x3 = Now. 3. a6 = 2 4 + 2 3 . ˆ the intersection of a line with a circle or an ellipse yields 2 points. as desired. Now. n = 2. a12 = 2 4 + 2 3 .5. Therefore the final formula is 2Cc2 + 2l(c + e) + 4Ce2 + 4ec + Cl2 . c circles. . ˆ the intersection of two incindent lines yields 1 point so the contribution is Cl2 .238..184 Chapter 2. and e ellipses? Solution ˆ the intersection of two circles yields 2 points. we can compute L = (x + y)(y + z)(z + x) ∶ −1 3 L = (x + ωx)(ωx + ω 2x)(x + ω 2 x) = x3 (1 + ω)(ω)(1 + ω)(1 + ω 2 ) 1 1 −1 (−ω 2 )(ω)(−ω 2 )(−ω) = − (−ω 6 ) = ⋅ = 3 3 3 1 Thus. a6 = u15 + u20 = u15 (u5 + 1) . Prove that √ 1 1 1 1 4 L ∶= + + + = 8. a20 = 2 4 + 2 10 .237. Let an = 2 + 4.bution is 4Ce2 + 4ec. u 2. What is the maximum number of points of intersection that can appear after drawing in a plane l lines..3 + 1 u10 (u5 + 1) + 1 u6 (u9 + 1) (⊠) Undergraduate problems Example 2. ˆ the intersection of an ellipse with a circle or another ellipse yields 4 points. so the contri. contained in (a. Let c1 . . Hence f (r0 ) < ε1 < f (b1 ) < ε0 < f (r0 ) contradiction! We are done. b) red (this point exist because in other case the interval (a.5. y have distinct color then min{f (x). Now we consider Wb1 = (b1 − ε1 . b) such that no contains any monochromatic open interval. (⊠) 2. b) is blue. .. (cn − cn ) because if we have a coloring with two pairs (ci − ci ). r0 + ε) with ε > 0 exists infinitely many blue points. We color the sectors in n − 1 colors using each of the colors at least once. Solution We suppose the contrary: there exist an open interval (a. .. r0 + ε0 ) with 0 < ε0 < f (r0 ). r0 + 2b ). A circle is divided into n equal sectors.4 Olympiad problems Example 2.. because in other case we will have b1 . (cj − cj ) then the n − 4 sectors remaining can not be colored using n − 3 colors. Points on the real axis are colored red and blue.239. f (b1 )} = f (b1 ) < ε0 . bn the only blue points. In the neighborhood Vr0 = (r0 − ε. . since all colors are used. (⊠) 2 . . (c1 − c1 ).5. so we consider b = min{∣r0 − b1 ∣. By analogy we obtain that f (r0 ) < ε1 . Now we consider Vr0 = (r0 − ε0 . Evaluate ∞ ∑ 3n2 − 1 3 2 n=2 (n − n) Solution Since 1 1 1 1 3k 2 − 1 =− 2 + − 2+ 3 2 2 (k − k) 2k 2(k − 1) 2k 2(k + 1)2 we have that thus the 3k 2 − 1 1 1 1 1 = − 2+ − 3 2 2 2 2n 2(n + 1) 8 k=2 (k − k) n Sn ∶= ∑ 1 1 1 1 3 3n2 − 1 = lim Sn = lim ( − 2 + − )= ⋅ 3 2 2 n→∞ n→∞ 2 2n 2(n + 1) 8 8 n=2 (n − n) ∞ ∑ (⊠) Example 2. we can enumerate all colorings counting these that involve each pair. ∣r0 − bn ∣} and set Wr0 = (r0 − 2b .. f (b1 )} = ∣r0 − b1 ∣ < ε0 and hence min{f (r0 ). By each pair we obtain Cn2 (n − 2)! colorings.241. the latter is an monochromatic (red) open interval.. and this color appear exactly two times. We know there exists a function f ∶ R → R+ such that if x. contradiction). contradiction. . let b1 a fixed blue point in this interval. Let r0 ∈ (a. b1 + ε1 ) such that 0 < ε1 < f (b1 ) and r0 ∈ Wb1 .185 2. because in other case we have n − k sectors remaining and n − 2 colors. f (y)} ⩽ ∣x − y∣. b). Problems of Other Topics Example 2. . Prove that every open interval contains a monochromatic open interval.240. cn the n colors. so in total we have n!(n − 1) colorings. impossible. . so we have min{f (r0 ). . .. How many such colorings are there? Solution There exists at least an color repeated by pigeonhole principle. Clearly. Examples for practice Example 2. Assume that n5 = 5. Consider two vertices V1 and V2 . It follows that the n5 subgraphs cover no more than 33 squares. n4 = 7 tetraminoes and n3 = 2 trominoes (leaving n2 = 0). then 2n2 + 3n3 + 4n4 = 19 and n2 + 2n3 + 3n4 = 14.186 Chapter 2. hence n4 ⩾ 8. joined by an edge. or since n2 . yielding n4 ⩾ 6 (two of the n4 subgraphs must be placed on or close to each corner). (⊠) Example 2. ie. and since n2 . Assume that n5 = 4 − d where d ⩾ 0. This square must therefore be covered. hence n5 ⩾ 5. Therefore. n3 ⩾ 0. and n4 = 4. 5) we have 2n2 + 3n3 + 4n4 + 5n5 = 49 and n2 + 2n3 + 3n4 + 4n5 = 38. then 2n2 + 3n3 + 4n4 = 24 and n2 + 2n3 + 3n4 = 18.243. otherwise by Dirichlet’s principle. limiting on one side with the side of the board. and on two sides with sides of this T -shaped tetramino. where two vertices are joined by and edge iff their corresponding squares are covered by the same domino. a graph with k = 5 would represent a cross-shaped pentamino with one square joined to its four neighbours. Mister Tien jumps on the real axis. n4 = 6. Note now that 0 = 3 ⋅ 24 − 4 ⋅ 18 = 2n2 + n3 . The conclusion follows. As in the case of n5 = 5. Assume finally that n5 = 6. we have n2 = n3 = 0. the graph may be decomposed in disjoint subgraphs.. Assume that the board can be covered by 38 dominoes so that no matter which domino is removed. and each n4 subgraph is a T -shaped tetramino. contradiction. such that both of them are ends of other edges. or the board can be covered by 37 dominoes. V3 . Now. Clearly. It is possible however to cover the board with n5 = 3 pentaminoes. ie n5 ⩽ 6. next to each corner square. Prove that if a 7 × 7 square board is covered by 38 dominoes such that each domino covers exactly two squares of the board. it follows that n2 = 0. 4. hence each one of the four corners of the board must be covered by one of the two squares forming the horizontal bar of the T . leaving one ”prisoner” square. these must also have no other edges connecting them to further vertices in the graph. no edge in the graph may join two vertices. If V1 has other vertices joined to it through edges. such that k − 1 of them are joined by an edge to the remaining vertex. then it is possible to remove one domino after which the remaining 37 cover the board. one of them (wlog V2 ) cannot have any other edges. contradiction again. representing each vertex one of the squares in the board.. it follows that one of the four corners of the board must be covered by the n3 subgraph. each one of them with k vertices. whereas n2 + 2n3 + 3n4 = 22 + 4d > 3n2 + 3n3 + 3n4 . n3 = 1. 3. while the other three must be covered by n4 subgraphs. hence n2 + n3 + n4 = 7 + d. on the board. no other vertices or edges are present in each subgraph. can be covered by a domino involved in each one of these n5 subgraphs. Therefore n5 = 6. by another one of the n4 subgraphs. denoting nk the number of subgraphs with k vertices (where clearly k = 2. Note however that each corned must be covered by one of the n4 subgraphs because it cannot be covered by one of the n5 subgraphs. no other vertices involved. or all points that may be reached from V1 through edges of the graph form a tree with root V1 and leaves V2 . from the origin towards point (1. hence no more than 2 squares on each side of the board. at least one square becomes uncovered.. no vertex can be joined to more than 4 other vertices. . or 1 = 3 ⋅ 19 −4 ⋅ 14 = 2n2 +n3 . and none of its corners. two dominoes would cover the same two squares. n3 ⩾ 0. absurd. such that no domino can be removed without leaving uncovered squares. .242. Therefore. Solution Consider a graph with 49 vertices. and one of them could be removed leaving all 49 squares covered. the distance from (1. Can Mister Tien reach point (1. The 10 edges of K5 are colored in two colors. p3 = 5. Mister Tien never reaches point (1. 0). ⋅ 3 3 3 Can a number greater than 70 appear on the board? Solution . (⊠) Example 2. After a few diagrams it is easy to see that also in this case K5 contains a monochromatic C3 or a monochromatic C5 . 0) after the nth jump. b. . Numbers 1 through 24 are written on a board. numbers a. 8 Solution 3 Assume that the number of edges ∣E∣ is greater than n2 . Hence any vertex has exactly two edges of one color and two edges of the other one.5.. Let G be a graph with n ⩾ 5 vertices. pn pk k=1 xn = ∏ (1 − n k=1 1 1 )→ = 0. c may be replaced by 2b + 2c − a 2c + 2a − b 2a + 2b − c . If a vertex has at least three monochromatic edges then there is a monochromatic C3 and we have a contradiction. At any time.. (⊠) Example 2. after a finite number of jups. Prove that there are no 3 more than n2 edges in the graph. . (⊠) Second solution pn − 1 times the distance After the n−th jump.244.245.187 2. hence always positive. xn is a strictly decreasing sequence such that xn = xn−1 − It is well known that n 1 xn−1 = ∏ (1 − ) . 0) but it will never reach it. 8 n2 1 3 (1 − ) ∣E∣ > n2 = 8 2 5−1 implies that there is a K5 in G. Mister Tien can arrive arbitrarily close to (1. Hence x0 = 1 and for n > 0. pk ζ(1) so.). By Turan’s Theorem. 0). 0) will be at least pn before the n − th jump. Problems of Other Topics 1 times its distance to the point (1. 0)? 0) such that the length of the nth jump is pn is the nth First solution Let xn be the distance to the point (1. and the degree of each vertex is 4. where pn prime (p1 = 2. The edges of G are colored in two colors such that there are no monochromatic cycles C3 and C5. p2 = 3. Define its perimeter by ∣(Si ∩ Sj ) ∪ (Sj ∩ Sk ) ∪ (Sk ∩ Si )∣ . Sk of a set with n elements is called a ”triangle”. (1. 1. Note however that. 2. one corresponds to the entire matrix being 1’s. and columns 2 and 3 are equal. since they would result in two columns being equal. 1). . Provided that these two conditions are met. (0. 1). 3. and 2) each row contains at least 2 1’s.. The total number of triangles with perimeter n is then 4n − 3 ⋅ 2n + 2 2((2n−1 ) − 3 ⋅ 2n−1 + 1 (2 ⋅ 2n−1 − 1) (2n−1 − 1) = = ⋅ 6 3 3 2 (⊠) . l) equals 1 if sm is in Sl . c ∈ Sn } is an invariant. Finally. 1). and will appear thus in the three calculations. some are not allowed. A triple of different subsets Si . For any three real numbers a. 1). The conditions of the problem require that: 1) no two columns are equal. or the total number of combinations that are not allowed out of the 4n is 3 (2n − 1) + 1 = 3 ⋅ (2n − 2) . 1. i. each row can take 4 different values. Examples for practice Let Sn = {x1 . 1. . for a total of 4n possible combinations. m = 1. Therefore. (1. In fact. note that in counting these combinations. S2 and S3 are chosen. and that the cardinal of the union equals the cardinal n of S. c we have (n) (n) (n) 2b + 2c − a 2 2c + 2a − b 2 2a + 2b − c 2 ) +( ) +( ) = a2 + b2 + c2 3 3 3 Therefore the function ( I(n) = max{a2 + b2 + c2 ∣ a. b. I(n) does not change during the whole process. 0) and (1.188 Chapter 2. Sj . b. . 1. 0) and (1. and the cardinal of the union would be less than n. (⊠) Example 2. 0 otherwise. Since I(1) = 222 + 232 + 242 = 1589 a number greater than 70 can not appear on the board (after n steps) because we would have I(n) ⩾ 702 = 4900 > 1589 This is a contradiction and the result follows. it is obvious that the sets are distinct.. columns 1 and 2 are equal if and only if the only values taken by the rows are (1. We may calculate in the same way the number of permutations that we need to discard in order to avoid the combinations such that columns 1 and 3 are equal. 0. x2 . for l = 1.e. Prove that the number of triangles with perimeter n is 1 n−1 (2 − 1) (2n − 1) . However. 3 Solution Let us call Sm . otherwise an element of S would not be in any of the intersections. otherwise the corresponding subsets would be equal. 2. out of the 2n combinations such that two columns are equal. x24 } be the numbers written on the board after n iterations. Each triangle may be represented by a n × 3 matrix. out of which there are a total of 2n possible combinations. . we have counted each triangle six times.. where element in position (m. 1. n the elements of the n-element set S out of which subsets S1 . since a permutation of the three columns leaves the triangle unchanged but produces a different matrix. for a total of (4n − 3) ⋅ (2n + 2) allowed combinations.246. . 12. ∠IAB. 2. 11. and likewise for IBC and ICA. Solution Let I be the incenter of ABC. but not in a circle. The sides of the acute triangle ABC are diagonals of the squares K1 . Thus the last two numbers will add up to 1001. 1. K3 . thus x = 19 and the third number. 16 + y ⩽ 16 + 15. Which player has a winning strategy? Solution The second player has a winning strategy: if the first player erases x. 7. Solution If the numbers were in a circle with 16 next to x and y. which is 37 × 19. 8. forcing 16 + x = 16 + y = 25.248.249. Prove that the area of ABC is covered by the three squares. Prove that among any ten points located in a circle of diameter 5. 5. Problems of Other Topics Example 2.252.247. If the first number is 37 and the second number is 1. and the intersection of the remainder with each of eight equal sectors. Solution Divide the circle into nine pieces: a circle of radius 1 concentric with the given circle. 13. ∠IBA < 45o . A regular 1997-gon is divided by nonintersecting diagonals into triangles. The game ends when two numbers remain: the first player wins if the sum of these numbers is divisible by 3. They may be arranged in a line as follows: 16. being a divisor of 38 other than 1 or 19. what is the third number? Solution The last number x must divide the sum of all of the numbers. Show that the numbers from 1 to 16 can be written in a line. 14. the second erases 1001 − x. 4. 3. (⊠) Example 2.251. (⊠) Example 2. so that the sum of any two adjacent numbers is a perfect square. 10. which then must be acute. On a chalkboard are written the numbers from 1 to 1000. Then one checks that two points within one piece have distance at most 2. there exist two at distance less than 2 from each other. since the center of the circle does not lie on any of the diagonals. Two players take turns erasing a number from the board. Prove that at least one of the triangles is acute. the second player wins otherwise. K2 . The numbers from 1 to 37 are written in a line so that each number divides the sum of the previous numbers. (⊠) Example 2. (⊠) Example 2. then 16 + 1 ⩽ 16 + x. (⊠) Example 2. (⊠) . Since the triangle is acute.189 2.5.250. Solution The circumcircle of the 1997-gon is also the circumcircle of each triangle. must be 2. it must lie inside one of the triangles. a contradiction. 6. 15. 9. so the triangle IAB is covered by the square with diagonal AB. Show that for any two factions A and B. A3 to A4 .257. and the circle with radius R contains M. then A1 A2 A3 A4 is a R square with center O contained entirely in M. Solution Suppose the square has vertices as (0. one containing M and the other contained in M. logc logc a > logb logc a. There must be at least one vertex of 2 the triangle on the same side as (0. Thus the left side of the given inequality exceeds Example 2. y) with x. Note that no pair weighs more than twice any other. show that there exist two √ circles. Thus the circle with radius √ is contained 2 in the square and thus in M. then 2 3 3 (⊠) e + h ⩽ 3e ⩽ (f + g). then a + d ⩽ 4a ⩽ 2b + 2c. 2 2 Example 2. the ratio of whose radii is 2. A2 to A3 . y ⩽ . Now pairing the heaviest and lightest pairs gives foursomes. logb (loga b logb c logc a) = 0. and pair off the heaviest and lightest apples. The vertices of triangle ABC lie inside a square K. then loga (loga b) + logb (logb c) + logc (logc a) > 0. b + c ⩽ 3a + d ⩽ 2a + 2d. d and b. loga loga b > logb loga b. Show that if 1 < a < b < c. the complement of A ∪ B is a faction. 1). and so on. (⊠) . then the next heaviest and next lightest. Solution By putting A = B. (⊠) Given a convex polygon M invariant under a 90o rotation. the complement of A ∪ B is also a faction. Examples for practice Example 2. Thus for any factions A and B. If A1 goes to A2 under the rotation. 2y). 0). so A ∪ B is also. A ∪ B is also a faction. 0) and (0. Solution: Since loga b > 1. 0) of the line through (2x. (⊠) Example 2. we see the complement of any faction is a faction. (⊠) Example 2.255. if a.190 Chapter 2.256. more than 2 Solution Sort the apples into increasing order by weight. the rotation of this vertex remains inside the square. Show that the apples may be divided into groups of 4 such that no group weighs 11 times any other group. (0. 1). 0). at least one vertex remains inside the square. if e ⩽ f ⩽ g ⩽ h are pairs. Show that if the triangle is rotated 180o about its centroid. (1. and A4 to A1 . 300 apples are given. Solution Let O be the center of the rotation and A1 a vertex at maximum distance R from O.253. no one of which weighs more than 3 times any other. and without loss of gen1 erality suppose the centroid is at (x. The members of Congress form various overlapping factions such that given any two (not necessarily distinct) factions A and B. c are two groups with a ⩽ b ⩽ c ⩽ d. Since logc a < 1. f + g ⩽ 2e + h ⩽ (e + h). none weighing more than 3 times any other.254. (1. q are integers and 1 ⩽ p. Problems of Other Topics Example 2. (a) Express ∠Ak+1 Bk+1 Ck+1 in terms of ∠Ak Bk Ck .g.. Ak Bk . (⊠) Example 2. and by assumption there are groups of size 1. a contradiction.. Ck Ak . From an initial triangle A0 B0 C0 a sequence A1 B1 C1 . Assuming 1 ⩽ a. Each student is asked how many other students in the class have his first name. There are at most 10 groups by last name. ∠Ck Bk+1 Ak+1 = 90o − . and ∆ = n2 − 4mn = n(n − 4m) < 0. so two students in the group of 11 must also have the same last name. A2 B2 C2 . we must have −2012 ⩽ m. Of the quadratic trinomials x2 + px + q where p.5. (b) Deduce that as k → ∞.258. ∠Ak Bk Ck → 60o .. Solution (a) We have Ak Bk+1 = Ak Ck+1 by equal tangents so triangle Ak Bk+1 Ck+1 is isosceles Ak Ck with ∠Ak Bk+1 Ck+1 = 90o − . . and how many have his last name. It turns out that each number from 0 to 10 occurs among the answers. x2 − 3x + 5). Each student belongs to two groups. b ⩽ 2012.261. we have Bk+1 = Bk (Ak + Ck ) = 90o − . But now the polynomial x2 − nx + mn also has integer coefficients between 1 and 2012. n < 0. but these numbers add up to 66 = 2 × 33. respectively. (⊠) Example 2. there are strictly more polynomials with no real roots. Suppose they exist. Ak+1 .260. 2 2 . Then b + 1 and c + 1 are even integers. 11.191 2. so there is one group of each size from 1 to 11 and no other groups. Similarly. Solution Consider groups of students with the same first name. so b and c are odd and 2 b − 4c ≡ 5 (mod 8) is not a square. which are there more of: those having integer roots or those not having real roots? Solution There are more not having real roots. Ck+1 are the points where the incircle of Ak Bk Ck touches the sides Bk Ck . 2 2 Adding up angles at Bk+1 . (⊠) Example 2. Suppose the group of 11 is a group of students with the same first name. If m ⩽ n are integer roots of x2 + ax + b = 0.259.. Do there exist real numbers b and c such that each of the equations x2 + bx + c = 0 and 2x2 + (b + 1)x + c + 1 = 0 have two integer roots? Solution No. . is formed such that at each stage. q ⩽ 2012. Since one can also write down a polynomial having no real roots not of this form (e. Bk+1 . A class consists of 33 students.. so it has no real roots. Show that there are two students in the class with the same first and last name. and groups of students with the same last name. then m + n = −a and mn = b. 1). Show that all of the circles pass through a single point.. All such circles pass through the point (0. Say the colors are red and black. columns equals the sum of all of the black squares plus twice the sum of some of the red squares. (r2 . . It is given that the sum of the numbers in each row and the sum of the numbers in each column is even.264. etc. Thus the final number N must satisfy and so N = 1. the sum of the numbers in the black squares is even. and so the reflection of (0. then r1 + r2 = −p. etc. it suffices to show the sum of the numbers in the red squares is even. we then have p a = − .263.... After 96 steps. 0). 2N − 1 = (2 ⋅ 49 97 96 1 49 − 1) ⋯ (2 ⋅ − 1) = ( ) ⋅ ( )⋯( ) = 1 1 97 1 2 97 (⊠) Example 2. (⊠) 49 Example 2. and each cell contains an integer. The only such number is 1. a single number remains on the blackboard. In a community of more than six people. k→∞ Bk (Bk − 60o ) (B0 − 60o) − 60o = so Bk − 60o = . Solution. Solution. Examples for practice (b) We have Bk+1 − 60o = 90o − obviously lim Bk = 60o . q) across the horizontal diameter is (0. On a blackboard are written the numbers with k = 1. rows and the first (from the left). . For each parabola y = x2 + px + q meeting the coordinate axis in three distinct points. (r1 . At k each step. Note that 2(2ab−a−b+1)−1 = (2a−1)(2b−1). each member exchanges letters with precisely three other members of the community. (Since the sum of all of the numbers is even. 97. third.262. Since this sum is even. b are erased and 2ab − a − b + 1 is written in their place. Prove that the community can be divided into two nonempty groups so that each member exchanges letters with at least two members of the group he belongs to. Determine all possible such numbers. If (0. and 2 p 2 1 1 2 p + (q − b)2 = (r1 − ) + b2 = (r1 − r2 )2 + b2 or q 2 − 2qb = −q 4 2 4 which gives b = (⊠) q+1 . 2. If (x − a)2 + (y − b)2 = r 2 is the circle..192 Chapter 2. so the product of 2a − 1 over the numbers a on the board never changes. 1). q).265. 2 Example 2. 0) are the three points. and 2 −2 (−2)k (⊠) Example 2. Prove that the sum of all numbers in black cells is even. third. A rectangular grid is colored in checkerboard fashion. with the top left square being red. Solution. two numbers a.) The sum of the first (from the top). a circle through these points is drawn. 1. at most 26 spheres can at least 3 touch S. Problems of Other Topics Solution. Put the people in the cycle into group A and the others into group B. 2. f. Since the volume of X is 27 times this volume.. one-third of the other spheres. 2. d. As for 2013 zeroes. there were at most x edges between A and B. (⊠) Example 2. (⊠) Example 2. To see that 9 is a maximum. a cycle must exist. x + y = z. Any sphere with radius r and touching S 4πr 3 is enclosed within X. z) of elements of A such that x < y and x + y = z. Find a cycle of minimal length and let it have x people. Solution. and the proof is complete. (n + 1) If x ⩽ . c. 2.193 2. This is possible with 2012 zeroes. 3. However. and each of them touches at least one-third of the others? Solution. For any set A of positive integers. 7}. Therefore. it is not possible to have such an arrangement. So B ends up with at least n − 2x > 0 members. transfer him into A. 2 contradicting our choice of the shortest cycle. Any sphere 3 with a radius larger than r and touching S completely encloses a sphere with radius r touching S at the same point. Let n be the number of people. Is it possible to change all of the numbers to 1? And what if we started with 2013 zeroes? Solution. giving n(A) ⩽ 9. It is clear that this gives what we need provided that B does not end up empty. from 0 to 1 and vice versa. e. note that the parity of the sum of the numbers does not change under the operation. y. Find the maximum value of nA given that A contains seven distinct elements. thus. The only permitted operation is to choose a number and change its two neighbors. then A and B satisfy the condition of the problem. (⊠) . Note that each person in A corresponds with at least 2 other people in A. For 2012 zeroes. group them into 503 groups of 4.268. and each transfer reduces this number by at least 1. Around a circle are written 2012 zeroes and one 1. respectively. The maximum of 9 is achieved by A = {1. (⊠) Example 2. i. 3. Is it possible to place 100 solid balls in space so that no two of them have a common interior point. with edges between people who exchange letters. which is less than 33.266. If x ⩾ 5. so it cannot go from even (the initial position) to odd (the desired final position). If a member of B exchanged x letters with two people in A. so the intersection of X and any such sphere is . z ∈ A are there such that x < y < z.5. and form the sphere X with the same center but with three times the radius.e. b. let nA denote the number of triples (x. consider any 7 numbers a<b<c<d<e<f <g how many times each can serve as the middle term. which in particular holds for x ⩽ 4. for each y ∈ A how many x.267. Consider a graph whose vertices correspond to the people. 4. Find a sphere S with minimum radius r. g. he and at most + 1 people in A would form a cycle. Since each vertex has degree greater than 1. in the original groups. 1. 5. The answer is 0. the intersection of X and any sphere touching S is 4πr 3 . 0 times for a. but not with 2013 zeroes. the following algorithm produces 2 satisfactory groups: as long as there exists a person in B corresponding with at least two people in A. 6. then select the second and third zeroes in each group. b are of opposite sign. the first player then writes qn. (⊠) Example 2. (⊠) Example 2. Two players play the following game. Consider the problem of drawing a triangle of maximum area in a semicircle of radius 1. Prove that the six endpoints of the projections are concyclic.272. Does the first player or the second player have a winning strategy? Solution. Adding the second and third rows and columns gives 2x + z = 4. If the second player writes pn.273. One may not write N. Prove that ∣ ∣ ⩾ 1. so its base is at most 2 and its height at most 1. Prove that for x. b are of the same sign. and so have a common point. (⊠) a + b ab Example 2. All that is known is that the sum of the numbers in each row. In turn. We conclude that two triangles of area 1 are drawn in a circle of radius 1.271. they each contain the center of the circle. y the sum of the numbers in the four corners. The first player has a winning strategy: first write pq for p. y. adding the first and fourth rows and columns gives 2y + z = 4. then rotate around that point until a second vertex hits the diameter). A 4 × 4 square is divided into 1 × 1 squares. there may never appear two coprime numbers or two numbers. . Since the diagonals contain the four central squares and the four corners. so again the result is greater than 1.269. and the sum of the numbers in the four central squares? And if so. and each of the diagonals equals 1. (⊠) Example 2. We may assume this triangle has its base along the diameter of the semicircle (translate it towards the diameter until one vertex hits the diameter. Let x be the sum of the numbers in the four central squares. and z the sum of the eight remaining squares. Every subsequent number must be of the form pn or qn for some n > 1 relatively prime to pq.270. where r is the inradius. Each endpoint has distance r 2 from the incenter. (⊠) Example 2. (y 3 + x)(z 3 + y)(x3 + z) ⩾ 125xyz. Show that one cannot draw two triangles of area 1 inside a circle of radius 1 so that the triangles have no common point. and vice versa. The incircle of a triangle is projected onto each of the sides. Solution. we have x + y = 2. what are these sums? Solution. then the fraction has absolute value less than 1 but the exponent is negative.194 Chapter 2. √ Solution. Is it possible to determine from this information the sum of the numbers in the four corners. each column. they write composite divisors of N on a blackboard. Examples for practice Example 2. The first player unable to move loses. then the fraction has absolute value greater than 1 and the exponent is positive. one of which divides the other. The number N is the product of k diferent primes (k ⩾ 3). z = 2. In other words. a−b Solution. Also. If a. If a. the triangle has area at most 1. Let a ≠ ±b be integers. A secret number is written into each small square. q distinct primes. so the result is greater than 1. So they are concyclic. with equality only for the isosceles right triangle with base along the diameter.274. Hence x = y = 1. 276. Thus the maximum is 48. Given a convex 50-gon with vertices at lattice points. one number written on the blackboard is factored into two factors and erased.5. No. and the resulting two numbers are written. (⊠) Example 2. but the first and last vertices encountered cannot belong to diagonals).275. this and the two analogous inequalities imply the claim. The left side is at least (4y + x)(4z + y)(4x + y). By weighted AM-GM. 1 4 (⊠) 4y + x ⩾ 5y 5 x 5 . and changing that by 2 in either direction gives a number congruent to 3 modulo 4. there always is a number congruent to 3 modulo 4: factoring such a number gives one factor congruent to 1 modulo 4. each factor is (independently) increased or diminished by 2. Problems of Other Topics 195 Solution.2. diagonals correspond to pairs passed over at the same time. Is it possible that at some point all of the numbers on the blackboard equal 9? Solution. then laying rectangles on top of it and subdividing the grid as needed. The number 99⋯99 (with 1997 nines) is written on a blackboard. what is the maximum number of diagonals which can lie on grid lines? Solution. which is easily obtained by making a hexagon with 4 diagonals along grid lines. Each minute. There can be at most 24 diagonals parallel to any given line (as we move the line to pass over the vertices. (⊠) . Example 2. Let C(n) be the greatest coefficient of a polynomial with integer coefficients that belongs to Sn . 0. (±1. Y ∈ A2 A3 . Mc be the midpoints of BC. 0. (±2.p52) For n ⩾ 2. Xc lie on a line that is tangent to the incircle of triangle ABC. (O69MR1 . Xb . The incircle (I) of triangle ABC touches the sides BC. 2 Answer. where c is any integer. If n = 2. n2007 < C(n) < 2n . −1). AB. c) = (±1. then we get a = 2c.p45) In triangle ABC let Ma . b. (O70MR1 . respec. (±2. let Sn be the set of divisors of all polynomials of degree n with coefficients in {−1. b. −1).2008. A4 X. then the solutions are (a.2.4. 0. Mb . Exercise 3. 1}. (J146MR1 . (⊠) Exercise 3. Prove that there is a positive integer k such that for all n > k. CA. c for which there is a positive integer n such that √ n √ a + bi 3 ( ) = c + i 3.p41) Find all integers a. The equation is not solvable for n ⩾ 3.1. and Define Xb and Xc analogously. AB at points A.1 3. The line r1 is the reflection of line BC in AI.1 Exercises Exercises from Mathematic Reflextion Exercise 3.2010. 1). Denote by Xa the intersection of r1 and r2 . ±1.p3) Let A1 A2 A3 A4 A5 be a convex pentagon and let X ∈ A1 A2 . Prove that A1 X A2 Y A3 Z A4 U A5 V =1 A2 X A3 Y A4 Z A5 U A1 V 196 .tively.1. A5 Y intersect at P . and line r2 is the perpendicular from A to IMa . Exercise 3. 1). A3 V. If n = 1. b = 2. Z ∈ A3 A4 . ±1. AC. V ∈ A5 A1 be points such that A1 Z.2008.Chapter 3 Exercises for training 3.2008.3. C. B. (O72MR1 . Prove that Xa . A2 U. U ∈ A4 A5 . a2n−1 Exercise 3. (O148MR1 . 2 R 2R R Exercise 3.2010. A3 . c be real numbers such that a2 b ⩾ c2 . Prove that AC 2 < 2 (BC 2 + CD2 ) . Exercise 3. (S149MR1 .p6) Let ABC be a right triangle with AC = 3 and BC = 4 and let the median AA1 and the angle bisector BB1 intersect at O.. .1.p38) Let ABC be a triangle and let A1 . R) and circum. . r). Prove 2 that for all n ⩾ 1.2010. .p10) Let a0 ∈ (0.7. C1 . A4 .11. 1) and an = an−1 − for n ⩾ 1. A2 . en−1 be the nth roots of unity. . yn for which x21 + ⋯ + x2n + b (y12 + + yn2 ) = c.2010. y1 .p15) Prove that in any acute triangle ABC.9.p17) Let A1 A2 A3 A4 be a quadrilateral inscribed in a circle C(O. √ 1 n+1+ n n 1 − < < 2 an a0 2 . Prove that the incenters and circumcenters of triangles ABC and A3 B3 C3 are collinear. A2 be the intersections of the trisectors of angle A with the circumcircle of ABC. e0 . (⊠) Exercise 3. b be complex numbers. (J186MR1-2011. Exercise 3.p14) Let n be a positive integer and let a. b. .10. Exercise 3. Hint: Using to Ptolemy’s inequality and the Cosine Law.scribed about a circle ω(I. Denote by Ri the radius of the circle tangent to Ai Ai+1 and tangent to the extension of the sides Ai−1 Ai and Ai+1 Ai+2 . Exercise 3.5. Find all real numbers x1 . Prove that the sum R1 +R2 +R3 +R4 does not depend on the position of points A1 . .. . Define analogously B3 and C3 . C2 .2010.8.p41) Let n be a positive integer.. . (S148MR1 . Prove that MB NC 4 ⋅ ⩽ ⋅ MA NA 9 Hint: We can to prove the inequality by using analytic geometry. . (S183MR1-2011. (O150MR1 . Exercises Exercise 3. and a.12. Define analogously points B1 . xn .2010.197 3.2010. A line through O intersects hypotenuse AB at M and AC at N . 1 r 2 r r (1 + ) ⩽ cos A cos B cos C ⩽ (1 − ) .p6) Let ABCD be a quadrilateral with ∠A ⩾ 60o .6. (J149MR1 . (S150MR1 . Evaluate the product P = ∏ (a + be2k ) n−1 k=0 n n 2 ⎧ ⎪ ⎪(a 2 − (−b) 2 ) Answer: P = ⎨ n n ⎪ ⎪ ⎩a + b if n is even if n is odd. B2 . Let A3 be the intersection of lines B1 B2 and C1 C2 . (0.p31) Evaluate T = ∑ tan 4 2010 k=1 Answer. (2. and let B1 = l2 ∩ l3 .p11) Let Hn = eHn > 1 1 1 + + ⋯ + . D lie on a line in this order. 2). Prove that each positive integer 2 p 2p 2p n ⩾ p.p12) Let A1 . 0). 2). (2. 0). (0. 2. l2 . (U176MR6-2010. 2 (⊠) Exercise 3. A3 . 2). 2). (O180MR6-2010. Exercise 3. 1). 1). Answer: All the possible pairs of positive integers are of the form (F4n−1 . Using a straight edge and a compas construct parallel lines a and b through A and B. B3 = l1 ∩ l2 where l1 . (2.19. Find the maximum number of non-collinear points contained in the set.p21) In the space. 0). 0.17. A3 be non-collinear points on parabola x2 = 4py. 1. 0. (0.16. 0. (⊠) 3 Exercise 3. B2 = l3 ∩ l1 . consider the set of points (a. and its permutations. p > 0. c ∈ {0.p16) Find all positive integers a and b for which ab − 1 is a positive integer. Prove that 2 3 n √ n n! ⩾ 2Hn .p33) Let p be a prime. such that their points of intersection are vertices of a rhombus. (1. b. L2n+2 ) (for all integers n) and (dn . B. (a2 + 1)2 Exercise 3. 2. (S179MR6-2010. Answer. 2. n ⩾ 2. (O184MR1-2011. taking for example the following ones: (0. 0.p33) Points A. Numbers dn by defining as: dn = √ 1 √ (( 2 + 1)2n + ( 2 − 1)2n ) . [B1 B2 B3 ] (⊠) kπ ⋅ 2011 20113 − 4 ⋅ 2011 + 3 . Exercise 3. Exercises for training Exercise 3. (1.13. c) where a. Exercise 3. A2 . Prove that is a constant and find its [B1 B2 B3 ] value. 2). 1). 0). (1. It is possible to have 16 non-collinear points. 0). 1. 1. Answer. (2. 2}. and parallel lines c and d through C and D. l3 are tangents to the [A1 A2 A3 ] parabola at points A1 . (2. F4n+3 ). respectively. (O183MR1-2011. C. 2. 2. dn+1 ) . (0.14.18. (S185MR1-2011. 1). 1. 0. (0. T = 2011 [A1 A2 A3 ] = 2. (1.198 Chapter 3. 2. A2 . (2. p2 divides (Cn+p ) − Cn+2p − Cn+p . 0). 0. 1. using the de Moivre formula. b. . 2). (L2n . (S184MR1-2011.15. (U170MR5-2010. Let M be the midpoint of side AC and let N be the midpoint of A1 A3 . (J174MR5-2010. Prove that [AS] = 2[SM] if and only if [BP ] = [P Q]. Let Q ∈ (BC).23.2. (a) What is the maximal possible period of a linear recurrence of degree n in Fp ? (b) How many distinct linear recurrences of degree n have this maximal period? Answer. a+b Exercise 3. E.p19) Sequences (xn )n⩾1 . (yn )n⩾1 are defined as follows: x1 = a. Exercises Exercise 3. (U174MR5-2010. c1 . c. (b) (pn − 1)ϕ(pn − 1) . sin ( A B C D + 60o) + sin ( + 60o ) + sin ( + 60o) + sin ( + 60o) 3 3 3 3 1 ⩾ (8 + sin A + sin B + sin C + sin D). (O166MR4-2010.p27) Prove that in any convex quadrilateral ABCD. F .24.p24) Let p be a prime.1.199 3.22. d. Corollary 3. n (⊠) Exercise 3. CA. Prove that K is the foot of the altitude from A if and only if DLMN is a square.21. Prove that lim xn = lim yn = ∞.25. where c0 .p11) Let k > 1 be an odd integer such that ak +bk = ck +dk for some positive integers a.20. xn + yn }.p29) The incircle s of triangle ABC with incenter I is tangent to sides BC and AC at points A1 and B1 ..p7) The incircle of triangle ABC touches sides BC. ak + bk Prove that is not a prime. Line MI meets BB1 in T and line AT meets BC in P . If M is the second intersection point of AK with the circle γ circumscribed to DKNL. with ∣a∣ ≠ ∣b∣ ≠ 0. Denote by L and N the incenters of triangles ABK and ACK. . Points A2 and B2 are diametrically opposite to A1 and B1 in s. y1 = b. prove two following lemmas: Corollary 3. then DM ⊥ LN and AM = AE = AF. AB at D. respectively. (a) pn − 1 . and xn+1 = max{xn − yn . n→∞ n→∞ Exercise 3. (O171MR5-2010. respectively. (S169MR5-2010. Let A3 and B3 be the intersection points of AA2 with BC and BB2 with AC.1. for all n ⩾ 1. K lie on the circle with diameter LN . cn−1 ∈ Fp and c0 ≠ 0. . Exercise 3. The points D. Hint: First... R be the intersection of lines AB and QB1 and NR ∩ AC = {S}. respectively. A linear recurrence of degree n in Fp is a sequence {ak }k⩾0 in Fp satisfying a relation of the form ai+n = cn−1 ai+n−1 + ⋯ + c1 ai+1 + c0 ai for all i ⩾ 0. b. yn+1 = min{xn − yn . xn + yn }. respectively. 3 Exercise 3. Let K be a point on side BC and let M be the point on the line segment AK such that AM = AE = AF . . (J153MR2-2010p3). Then an an+1 − an = L ⇒ lim = L. and −57122. Exercise 3.28. Let a. −2144. −21218. increasing and unbounded. 167334. 1.31. Determine the function f if it is differentiable at x = 1. −34680. Prove that the following inequality holds a+b b+c c+a 2(ab + bc + ca) 13 + + + ⩾ ⋅ a + b + 2c b + c + 2a c + a + 2b 3(a2 + b2 + c2 ) 6 Exercise 3.. a2 .. such that bn is positive. (O160MR32010p25). b.26. Denote the midpoints of sides 1̂ AB. 504008. 19208.. −2680. b2 . Find all integers n such that n2 + 2010n is a perfect square. (S161MR32010p11). +∞). 1608. −3362. such that for each prime p there are infinitely many terms in the sequence that are divisible by p. . dB . The values of n are then −506018. Answer. where O 2 is the point of intersection of segments A1 D1 and B1 E1 . .27. respectively.1. . n→∞ an (b) Find L = lim √ ⋅ n→∞ n for n = 0. −2010. f (x) = x √ 2 or f (x) = x− √ 2 for all positive real x. 10240. Let (an ) and (bn ) be two real sequences. 0. 55112. AB ∥ DE.200 Chapter 3. E1 ..an . 32670. −8576. BC. n. ⋯ be a sequence of positive integers. +∞) be a function satisfying f (f (x)) = x2 for all x ∈ (0. DE. c > 0. D1 . 134. Exercises for training Exercise 3. 1 an Hint: As for (b) we employ the Cesaro-Stolz theorem: Theorem 3. Let f ∶ (0. (⊠) Exercise 3.bn are integers such that 0 ⩽ bi ⩽ ai − 1. If dA . +∞) → (0. Let ABC be a triangle inscribed in a circle of center O and radius R. 670. In a convex hexagon ABCDEF. BC ∥ EF. Prove that every positive rational number less than 1 can be represented as b1 b2 bn + +⋯+ . (J152MR2-2010p2). B1 .. Prove that D̂ 1 OE1 = DEF . . (U158MR32010p15) Let (an )n⩾0 be a sequence with a0 > 0 and an+1 = an + (a) Prove that lim an = +∞. −3618.. 100000.. prove that R3 − (d2A + d2B + d2C ) R − 2dA dB dC = 0. CD ∥ F A and AB + DE = BC + EF = CD + F A. (U161MR32010p20).32. dC are the distances from O to the sides of the triangle. 5766.30... EF by A1 . 1352. Exercise 3. Let a1 . −169344. i = 1. lim n→∞ bn n→∞ bn+1 − bn Exercise 3. 6566. −102010.29. −7776. a1 a1 a2 a1 a2 ⋯an where b1 . . Exercise 3. (O162MR32010p28). Answer. −12250. R. y1 . 2 Exercise 3.35. Exercise 3. (⊠) Exercise 3. Prove that the orthocenter of the triangle AP Q lies on the segment MN.1. Answer. q3 . (S153MR2-2010p10). BB1 . CD. (S152MR2-2010p9). respectively. MA1 MB1 MC1 Evaluate T= BM CM AM + + ⋅ MA1 MB1 MC1 Answer... Prove that the equation m m m xm 1 + x2 + ⋯ + xk = xk+1 has infinitely many solutions in distinct positive integers.38. n ⩾ 2 be relatively prime integers. Q. respectively. N lie on the sides BC and CD.201 3.36. q4 .. CA.. Exercise 3.. xk . AB of triangle ABC such that lines AA1 . C1 are chosen on sides BC. . BC. . Lines AM and AN meet the circumcircle of ABCD again at points P and Q. (S154MR2-2010p11). S the orthogonal projections of X onto AB.37. Prove that there are no rational numbers x1 . Prove that if and only if P A ⋅ AB + RC ⋅ CD = 1 (AD 2 + BC 2 ) 2 QB ⋅ BC + SD ⋅ DA = 1 (AB 2 + CD2 ) .. Let k ⩾ 2 be an integer and let n1 . Exercise 3. respectively so that MN = BM + DN.34. (O155MR2-2010p31). q5 such that q14 + q24 + q34 + q44 + q54 is the product of two consecutive even integers. DA.. (O156MR2-2010p33). (⊠) . Exercise 3. yk such that √ √ √ (x1 + y1 2)2n1 + ⋯ + (xk + yk 2)2nk = 5 + 4 2.. (J189MR2-2011p3) Find all primes q1 . Prove that the equation x2 + y 3 = 4z 6 is not solvable in integers. Denote by P. . Exercises Exercise 3.33.39. B1 . CC1 are concurrent at M and AM BM CM ⋅ ⋅ = 2012. Let k ⩾ 2 be an integer and let m. (J190MR2-2011p4) Points A1 .. q2 . In a cyclic quadrilateral ABCD with AB = AD points M. Let X be a point interior to a convex quadrilateral ABCD. nk be positive integers. q1 = q2 = q3 = q4 = q5 = 2. T = 2010. I. (S192MR2-2011p13). Let (Fn )n⩾0 be the Fibonacci sequence. k = 3.41. Exercise 3. Let O.202 Chapter 3. incenter. and P a point not lying on its sides. R. Let ABC be an isosceles triangle with AB = AC. C Exercise 3. Exercise 3. O. AB BD BC find ∠A. Prove that for any positive integer k the sequence (τ (k + n2 ))n⩾1 is unbounded. y z 3 x +√ +√ ⩽√ ⋅ Prove that √ x+y+z x5 + y 3 + z y5 + z3 + x z 5 + x3 + y Exercise 3. Call XY Z the cevian triangle of P with respect to ABC and consider the points Ya .43. (S197MR3-2011p11). D. (S194MR3-2011p8 ). If 1 1 1 + = . Let ABC be a triangle. respectively.45. Y Za .46. Exercises for training Exercise 3. Answer. Let p be a prime of the form 4k + 3 and let n be a positive integer. and exradii of a triangle ABC. (S198MR3-2011p12). Prove that for any prime p ⩾ 3. Find all positive integers k and n such that k n − 1 and n are divisible by precisely the same primes. y. Prove that √ √ √ 2 √ p ⩽ ra + rb + rc ⩽ √ ⋅ p R r Exercise 3.48. B. Prove that AX. rc be the semiperimeter.44. Let x. H are concyclic if and only if ̂ = 60o . (J201 MR4-2011p3). z be positive real numbers such that (x − 2)(y − 2)(z − 2) ⩾ xyz − 2. Ya Z concur in a point Q that satisfies the cross-ratio (AXP Q) = AP ⋅ AX Exercise 3. (S191MR2-2011p12). Let p. r and ra . inradius. Answer. Point D lies on side AC such that ∠CBD = 3∠ABD. H be the circumcenter.40. (S204MR4-2011p12).42. (⊠) . Za of intersection BC with the parallel lines to AX through Y and Z. where τ (m) denotes the number of divisors of m. and let D be an interior point to triangle ABC such that BC ⋅ DA = CA ⋅ DB = AB ⋅ DC. Prove that for each integer m there are integers a and b such that n n a2 + b2 ≡ m (mod p). Prove that A. p divides F2p − Fp . k = 2 and n = 2. circumradius. (S203MR4-2011p11). ∠A = π ⋅ (⊠) 9 Exercise 3. (O195MR3-2011p23). n = 1.47. and orthocenter of a triangle ABC . Exercise 3. rb . I. respectively intersect at a point L lying on the altitude AD. Answer. Prove that the perpendiculars to OB and OC at F and E. CA intersect the sidelines CA. Find all n such that τ (n) = 6 and 3ϕ(n) = 7!. n = 281 ⋅ 32 = 2529 and n = 41 ⋅ 72 = 2009. √ √ B+C C +A 3 3 p A+B + cot + cot ⩽ + ⋅ 3 3 ⩽ cot 4 4 4 2 2r where p and r denote the semiperimeter and the inradius of triangle ABC. c) = 1 and ab + bc + ca = 0. Prove that for each real number a that is not a zero of P or P there is a real number b such that b2 P (a) + 4bP ′ (a) + 5P ′′(a) = 0.51. b. (⊠) Exercise 3. Prove that in any triangle ABC.53. c.. We ∣a + b + c∣ = u2 + uv + v 2 = (−u − v)2 + (−u − v)v + v 2 = (−u − v)2 + (−u − v)u + u2 . Let A1 A2 .52. (O203MR4-2011p26). and let perpendiculars at B. respectively.55. Let a.54. (U199MR4-2011p14).n π π − ⋅ 2 n Exercise 3. Exercise 3. Let p be a prime. d are positive integers such that ab = cd. with real coefficients.49. have Answer. Exercises Exercise 3. and X2 . c be integers such that gcd(a. b. Prove that min ∠P Ai Ai+1 = i∈1. Let ABC be a triangle with circumcenter O. then a = u(u + v). Exercise 3. AB at E. the Amixtrilinear incircle names the circle tanget to AB.. Exercise 3. Exercise 3. satisfying DL = LA sin2 A.. Prove that ∣a + b + c∣ can be expressed in the form x2 + xy + y 2 . where a. C to BC. (J110MR2-2009 p3).56. respectively. Exercise 3. Let P be a polynomial of degree 5. b. Find all solutions to the equation a + b − c − d = p. With a = du and b = dv. where x and y are integers. F.203 3. Prove that the second intersection of the circumcircle of triangle MX1 X2 with the circumcircle of ABC(different from M ) coincides with the tangency point of the circumcircle with mixtilinear incircle in angle A (As usual. (U203MR4-2011p18).50. (J112MR2-2009 p7). all whose zeros are real. (J114MR2-2009 p10). (O201MR4-2011p24). b = v(u + v) and c = −uv. respectively. (U204MR4-2011p21).1. . Let τ (n) and ϕ(n) denote the number of divisors of n and the number of positive integers less than or equal to n that are relatively prime to n.An be a convex polygon and let P be a point in its interior. AC and to the circumcircle of ABC internally). Let M be an arbitrary point on the circumcircle of triangle ABC and let the tangents from this point to the incircle of the triangle meet the sideline BC at X1 . Let V be the projection from U to BC. (S110MR2-2009 p15).58. P Q. β.60. b. b = t(s − 1). b = t(s + 1). 2009n can be written as a sum of six nonzero perfect squares. Prove that a3 + abc b3 + abc c3 + abc 3 a3 + b3 + c3 + + ⩾ ⋅ ⋅ (b + c)2 (c + a)2 (a + b)2 2 a2 + b2 + c2 Exercise 3. Let W be the intersection of the angle bisectors of ∠BC1 V and ∠CB1 V . Prove that cos3 β γ α β −γ γ−α α−β sin + cos3 sin + cos3 sin = 0.204 Chapter 3. (S115MR3-2009 p10). Exercises for training Answer. r). Prove that for each integer n ⩾ 3 there are n pairwise distinct positive integers such that each of them divides the sum of the remaining n − 1. Exercise 3. (⊠) Exercise 3. Prove that DE is parallel to BC and 1 1 1 = + ⋅ DE BX CX Exercise 3.62. (O112MR2-2009 p50). γ be angles of a triangle.61.57. (J119MR3-2009 p7). d = (s − p)(t + 1). (S116MR3-2009 p11). Let α. Prove that if 1 1 1 1 1 1 + + = + + A1 A2 A3 A4 A5 A6 A2 A3 A4 A5 A6 A1 then one of its diagonals coincides with OI . The parallel through X to AB meets CA at V and the parallel through X to AC meets AB at W.63.64. 4) a = s(t + p). 3) a = s(t + 1). Points P and Q lie on segment BC with P between B and Q. Prove that there is a point A in the plane such that AP and AQ are the trisectors of angle BAC if and only if P Q is less then BP and CQ. Exercise 3. Let a. (J118MR3-2009 p6). Denote by U the intersection of AA1 and B1 C1 . BB1 . 2) a = s(t − p). Prove that A. c = st. cbe real positive numbers. Hexagon A1 A2 A3 A4 A5 A6 is inscribed in a circle C(O. and QC form a geometric progression in some order. CC1 . Suppose that BP. Let D = BV ∩ XW and E = CW ∩ XV . Prove that for each positive integer n. Exercise 3. Consider triangle ABC with angle bisectors AA1 . V. W are collinear.59. We have four types of solutions: 1) a = s(t − 1). 2 2 2 2 2 2 Exercise 3. (O110MR2-2009 p41). Let X be a point on the side BC of a triangle ABC. c = st. d = (s − 1)(t + p). R) and at the same time circumscribed about a circle ω(I. c = st. c = st. (S114MR2-2009 p22). . d = (t − 1)(s + p). b = t(s − p). where s and t are any non-zero integers. b = t(s + p). Exercise 3. d = (t − p)(s + 1). Exercise 3. Let n be an integer greater than 1 and let x1 . respectively.67. Prove that 4 ⋅ AP ⋅ BP ⋅ CP BP CP AP + + ⋅ ⩾ (da + db )(db + dc )(dc + da ) db + dc da + dc da + db Exercise 3. . n2 1 xk ⩽ 2 − n + 1 − nxk + (n − 1)xk n − 1 Answer. (U115MR3-2009 p18). Let x. x2 .68.1. db .70. Prove that the centers of the nine-point circles of triangles ABC. +√ z 2 + x2 + 7 y2 + z2 + 7 x2 + y 2 + 7 x3 Exercise 3. (J136MR5-2009 p10). Prove that √ (⊠) y3 z3 +√ ⩾ 1. . b. Consider a regular hexagon A1 A2 A3 A4 A5 A6 with center O. respectively. Exercise 3. (J132MR4-2009 p8). and AMN are collinear. hb . lc the angle bisectors of a triangle ABC. dc be the distances from P to the sides of the triangle. z be positive real numbers such that x2 + y 2 + z 2 ⩾ 3. ma .. (S120MR3-2009 p16).66. Let a. Exercises Exercise 3. n = 1.65. Let P be a point interior to a triangle ABC and let da . The total number of inequivalent colorings is 1 6 (n + n3 + 2n2 + 2n) . b. Let a.. c be positive real numbers. Prove that 1 1 3abc 5 1 + + + ⩾ ⋅ a + b b + c c + a 2(ab + bc + ca)2 a + b + c Exercise 3. and la . Equality occurs if and only if x = 1. 6 Exercise 3. Prove that the diameter of the circumcircle of triangle ABC is equal to √ m2a − h2a la2 ⋅ ha la2 − h2a . (O129MR4-2009 p26). hc the altitudes. ha . Let M and N be the midpoints of BP and CQ. (⊠) Exercise 3. Let an = 2 − 1 .205 3. Let ABC be a triangle and let points P and Q lie on sides AB and AC. c be the sides. AP Q. Prove that √ n2 + n4 + 14 √ √ √ a1 + a2 + ⋯ + a119 is an integer. mb . (S127MR4-2009 p8). (S117MR3-2009 p12). y.71. xn be positive real numbers such that x1 + x2 + ⋯ + xn = n.. lb . In how many different ways up to rotation can one color regions Ai OAi+1 (take i mod 6) in n colors? Answer.69.72... Prove that n ∑ k=1 and find all equality cases. mc the medians. (U117MR3-2009 p21). 2. z) = (L2n+2 . L2n . Y. Exercise 3. (O121MR3-20092p36). c be positive real numbers. Prove that if a is an integer that is not divisible by p. Let a.80. c be positive real numbers. (S134MR5-2009 p19). CZ are concurrent at some point P. Mc Ma . Ma Mb . Let a.73. AB at points A1 . (J137MR5-2009 p12). . B. Exercises for training Exercise 3. Mc and let X. Z be the points of tangency of the incircle of triangle Ma Mb Mc with Mb Mc . }. z) of integers satisfying the system of equations ⎧ ⎪ ⎪x + y = 5z ⎨ ⎪ = 5z 2 + 1. (S125MR3-20092p17). y. Exercise 3. (J138MR5-2009 p15). Prove that √ √ √ a2 (b2 + c2 ) b2 (c2 + a2 ) c2 (a2 + b2 ) + + ⩽ a + b + c. respectively. . Find all triples (x. b. AA1 BB1 CC1 AA1 BB1 CC1 Exercise 3. Prove that 1 1 1 1 1 1 + + = 2 max { . then the perimeter of triangle A1 B1 C1 is greater than or equal to the semiperimeter of triangle ABC. C intersect BC. respectively. Prove that 3(a2 b2 + b2 c2 + c2 a2 )(a2 + b2 + c2 ) ⩾ (a2 + ab + b2 )(b2 + bc + c2 )(c2 + ca + a2 ). y.77. BB1. AC. F2n+1 ) or (x. (S124MR3-20092p16). Let p be a prime and let n be an integer. y. F2n+1 ) where Lj and Fj denote the j th Lucas number and the j th Fibonacci number. b. z) = (L2n .79. (S126MR3-20092p21). B1 . Let ABC be a triangle with midpoints Ma . BY.206 Chapter 3.78. ⎪ ⎩xy Answer. Let ABC be a triangle and let tangents to the circumcircle at A. Let a. Find all pairs (p. a) Prove that the lines AX. (J126MR2-20092p10).75. c be positive real numbers. (O134MR5-2009 p36).81. Exercise 3. n > 4. then the polynomial axn −px2 +px+p2 is irreducible in Z[x].74. b. Prove that b2 a b c a+b+c + 2 2+ 2 2⩾ ⋅ 2 +c a +c a +b 2 Exercise 3. a2 + bc b2 + ca c2 + ab Exercise 3. b.76. c be positive real numbers. (⊠) Exercise 3. L2n+2 . Prove that √ √ √ √ √ ab(a + b) + bc(b + c) + ca(c + a) ⩾ (a + b)(b + c)(c + a) + 2abc. b) If AA1 . Mb . respectively. q) of positive integers that satisfy p 1 1 ∣ −√ ∣< 2⋅ q q 2 Exercise 3. All integer solutions are (x. Let a. CC1 are cevians through P . C1 . n→∞ Answer. Hint: Prove Lemma. Qbc . BB2 . Pcd . Similarly. Pcd Qcd . Let ABC be a triangle and let A1 .86. Pda Qda are collinear. Qcd . respectively. √ (U93MR4-2008p29). C2 . A2 . Denote by A1 the tangency point of Ca with G and let A2 be the diametrically opposed point of A1 with respect to Ca .88. They intersect the sides AB. ∠BP C. (S93MR4-2008p17). xn be nonnegative real numbers whose sum is 2. Pbc .. Prove that lines AA2 . (⊠) 3 3 Exercise 3. AC and internally tangent to the circumcircle G of triangle ABC). Exercise 3. p Prove that pq divides Cp+q − Cqp − 1. hb . Determine the maximum.1. Medians A1 M. (S73MR2-2008p17). The maximum value is √ 2 when x1 = 2 and x2 = x3 = ..87.207 3. ∠DP A. BC. (O126MR3-20092p39). Exercise 3.85. define B2 and C2 .. as a function of n. Let ABC be a scalene triangle and let Ca be the A-mixtilinear incircle (the circle tangent to sides AB. (J74MR2-2008 p4). Prove that AA2 . B2 . Cpp−1 are divisible by p. The zeros of the polynomial P (x) = x3 + x2 + ax + b are all real and negative. BB2 . Let p and q be primes. Pda . C1 P in triangle A1 B1 C1 intersect ω at A2 . Exercise 3. Prove that 3 ha − 2r hb − 2r hc − 2r 3 ⩽ + + < ⋅ 5 ha + 2r hb + 2r hc + 2r 2 Exercise 3. ∠CP D.84. If p is prime. Let ABCD be a cyclic quadrilateral and let P be the intersection of its diagonals. Let x0 ∈ (0. A triangle has altitudes ha . lim n→∞ Exercise 3. hc and inradius r. Qda . Cp2 . x2 . √ 1 nxn = √ ⋅ (⊠) 2 . of x21 x22 x2n + + ⋯ + ⋅ 1 + x21 1 + x22 1 + x2n Answer..82. q ⩾ p.. Let n be √ an integer greater than 1 and let x1 . (⊠) Exercise 3. then Cp1 .83. respectively and the extensions of the same sides at Qab .89. . Consider the angle bisectors of the angles ∠AP B. ⋯ . (O96MR4-2008p48). CD. respectively. Hint: Use Viete’s formulae and Schur’s inequality. A3 be the points of tangency of its incircle ω with the triangle’s sides. B1 N. 1] and xn+1 = xn − arcsin(sin xn ). (O123MR3-20092p39). Pbc Qbc . Cp3 . = xn = 0. DA at Pab . CC2 are concurrent at the isogonal conjugate of the Gergonne point. Evaluate lim nxn . O84MR3-2008p45). Prove that 4a − 9b ⩽ 1. n ⩾ 0. CC2 are concurrent. Exercises Exercise 3. Prove that the midpoints of Pab Qab . 2. Exercise 3.. (S102MR5-2008 p24). (J104MR6-2008p2). c be positive real numbers such that abc = 1. (4.97.. Exercises for training Exercise 3. rb . Let E and F be the points of tangency of the incircle with AC and AB. 1 1 1 + +⋯+ x1 x2 xn Exercise 3.93. 6.98.An with ω form another polygon B1 B2 . 5. Given a triangle ABC with orthocenter H. 2. Ib . (J107MR6-2008p7). such that 1 ab + 1 ⩾ √ ∣a − b∣. say a and b. 8. BP.. 1). Ic } are nice if and only if triangle ABC is equilateral. Exercise 3.92.An be a polygon that is inscribed in a circle C(O.. xn be positive real numbers. a b c d Answer. (S103MR6-2008p10). . Ia . Prove that among any four positive real numbers there are two. 1. (9. Ic .. Let x1 ..Bn ) r where P (S) stands for the perimeter of figure S.. 1).208 Chapter 3. 3.Bn .95. Prove that √ max{AP. 1. Prove that EF. r). Find all quadruples (a. CP } ⩾ d2a + d2b + db dc + d2c . OI are concurrent if and only if r 2 = rb rc . (J105MR6-2008p4). respectively. A set of four points in the plane is said to be ”nice” if one can draw four circles centered at these points such that each circle is externally tangent to the other three. Let A1 A2 . and excenters Ia . d) of positive integers such that 1 1 1 1 (1 + ) (1 + ) (1 + ) (1 + ) = 5. . Exercise 3. 3 Exercise 3. C. BC. Ib . prove that {A. .. 9are randomly arranged on a circle. (J103MR6-2008p1). Exercise 3. 1. Let a... incenter I... 1). B. (J106MR6-2008p6). 2.. where ra .. (S105MR6-2008p14). Consider triangle ABC with circumcenter O and incenter I. rc are the radii of the excircles. 2. b.96. The solutions are (24. Exercise 3.94. Let P be a point in the interior of a triangle ABC and let da ⩾ db ⩾ dc be distances from P to the triangle’s sides. (14. 1). The points of tangency of A1 A2 . Prove that there are adjacent numbers whose sum is at least 16.91. Prove that P (A1 A2 . The numbers 1. R) and at the same time circumscribed about a circle ω(I. (S104MR6-2008p12). H} and {I. x2 ..An ) R ⩽ ⋅ P (B1 B2 . (9.. b. 1). Prove that √ n x1 + x2 + ⋯ + xn + ⩾ (n + 1) n x1 x2 ⋯xn . Prove that a2 + b2 b2 + c2 c2 + a2 a+b b+c c+a + + ⩾ + + ⋅ 2 2 2 2 2 a + b + 1 b + c + 1 c + a + 1 a + b + 1 b + c + 1 c + a2 + 1 Exercise 3.90. c.. M = BQ ∩ CP. Prove that √ √ 3 (1 + a)(1 + b)(1 + c) ⩾ 4 4(1 + a + b + c). BC. and N = RQ ∩ P S. Let a. Denote by P and Q the feet of the perpendiculars from D onto AB and AC. Prove that a1a1 a2a2 ⋯anan ⩽ 1. p3 } for which f (1)f (2)⋯f (2n) is a perfect square.99. t(t2 − 1)). N. E. Prove that there is a polynomial Q(x.102. (U106MR6-2008p28). where H is the orthocenter of triangle ABC. y). p3 be distinct primes and let n be a positive integer.101. DA. then AR ⋅ RC = BS ⋅ SD. and abc = 1. L.. Exercise 3. Let R = BE ∩ DP. Line KM meets diagonals AC and BD at P and Q. respectively. (O105MR6-2008p40).209 3.. Let x be a positive real number. and H are collinear. . Let a1 . In a convex quadrilateral ABCD let K. b. 32n + 3 ⋅ (⊠) 4 . (U103MR6-2008p24). 2. and line LN meets diagonals AC and BD at R and S.105. C. Exercises Exercise 3. respectively. Let p1 . Exercise 3.103. p2 . In triangle ABC let D. S = CF ∩ DQ. c be positive real numbers. B. Prove that xx − 1 = ex−1 (x − 1).106. 1 1 1 Exercise 3.. a) Is 3x7 + 3x good? b) Is 3x2008 + 3x7 + 3x good? Answer. Let P (t) be a polynomial with integer coefficients such that P (1) = P (−1). respectively. Prove that if AP ⋅ P C = BQ ⋅ QD.. N be the midpoints of sides AB. . 2n} → {p1 .. A polynomial with integer coefficients is called ”good” if it can be represented as a sum of cubes of several polynomials in x with integer coefficients. respectively. (O106MR6-2008p42). an be positive real numbers such that a1 + a2 + ⋯ + an ⩽ n. (O103MR6-2008p34). Prove that M. p2 . (O107MR6-2008p44). 9x3 − 3x2 + 3x + 7 = (x − 1)3 + (2x)3 + 23 is good. M. with integer coefficients such that P (t) = Q(t2 − 1. a) 3x7 + 3x is good! b) 3x2008 + 3x7 + 3x is not good! Exercise 3.100. F be the feet of the altitudes from vertices A.1. Exercise 3. Exercise 3. Exercise 3. (S108MR6-2008p22). For example. CD. (O104MR6-2008p38).. Find the number of functions f ∶ {1. Answer.104. a2 . (2. 2). (x. 3). and (3. The sequence an is defined by a1 = 1. Prove that the lines EF and O1 O2 are perpendicular.112. m = 3.110. BCF. Let E be the intersection of AC and BD. O2 be the circumcenters of ADF. (OPfAtW9798-p9). Answer. The minimum value is 3 ⋅ (⊠) 4 .2 Chapter 3. y real: ⎧ ⎪ ⎪(x − 1)(y 2 + 6) = y(x2 + 1) ⎨ 2 2 ⎪ ⎪ ⎩(y − 1)(x + 6) = x(y + 1). Find an in closed form.109.111. For any real number b. (⊠) Exercise 3. 3). 1].107. 2). Let k be a positive integer. Find all values of a 4 such that ∣f (x)∣ ⩽ 1 for all x ∈ [0. and an is the n−th positive integer greater than an−1 which is congruent to n modulo k.1. (OPfAtW9798-p7).210 3.113. y) are (2. Solve the system for x. let f (b) denote the maximum of the function 2 + b∣ ∣sin x + 3 + sin x over all x ∈ R. (⊠) 3 Exercise 3. Let ABCD be a trapezoid (AB ∥ CD) and choose F on the segment AB such that DF = CF. (OPfAtW9798-p). an = n(2 + (n − 1)k) ⋅ 2 (⊠) Exercise 3. Exercise 3. Consider the sequence of positive integers which satisfies an = a2n−1 + a2n−2 + a2n−3 for all n ⩾ 3. Exercise 3.108. Find the minimum of f (b) over all b ∈ R. Answer. and let O1 . √ 1 2 Answer. (3. Answer. Let f (x) = x2 − 2ax − a2 − . Find all real numbers m such that the equation (x2 − 2mx − 4(m2 + 1))(x2 − 4x − 2m(m2 + 1)) = 0 has exactly three different roots. Answer. (OPfAtW9798-p3). (OPfAtW9798-p12). Exercises for training Exercises from OP from Around the World Exercise 3. − ⩽ a ⩽ ⋅ (⊠) 2 4 Exercise 3. Prove that if ak = 2011 then k ⩽ 3. (OPfAtW9798-p15). (OPfAtW9798-p3). n) = (98. 1 f (x) = f (x2 + ) . Prove that H. Answer.. α = −1 (mod 3). Let H and O denote the orthocenter and circumcenter of the triangle ABC. (OPfAtW9798-p19).115. Answer.118. Find all values of n for which there exists m such that a1 + a2 + ⋯ + an = 12. (⊠) Exercise 3. (OPfAtW9798-p17). Exercises Exercise 3. (a..211 3. Find the number of nonempty subsets of {1. (⊠) . and Fn is the nth Fibonacci number. n. Prove that the equation x2 + y 2 + z 2 + 3(x + y + z) + 5 = 0 has no solutions in rational numbers.. where ai and bi are natural numbers and ai is squarefree. 2). (OPfAtW9798-p19).116.. (⊠) Exercise 3. (OPfAtW9798-p15). Find all continuous functions f ∶ R → R such that for all x ∈ R. (OPfAtW9798-p19). c) = (1. consider the polynomial Pn (x) = Cn2 + Cn5 x + Cn8 x2 + ⋯ + Cn3k+2 xk n−2 where k = [ ]. (a) k Hint: Use Cnk−1 + Cnk = Cn+1 . are natural numbers.114. 3 (a) Prove that Pn+3 (x) = 3Pn+2 (x) − 3Pn+1 (x) + (x + 1)Pn (x). (⊠) Exercise 3. . (b) Answer. n be natural numbers and m + i = ai b2i for i = 1. F1 = 1. Let ABCD be a convex quadrilateral such that ∠DAB = ∠ABC = ∠BCD.. D are collinear. c such that the roots of the equations x2 − 2ax + b = 0 x2 − 2bx + c = 0 x2 − 2cx + a = 0. Find all natural numbers a.. F2 = 2. Exercise 3.117. (3. .1.120. (m. 2. b. n} which do not contain two consecutive numbers. 4 Answer. Let m. b. 3). 1. (OPfAtW9798-p21). f (x) = c ∶= const ∈ R. O.119. n−1 (b) Find all integers a such that 3[ 2 ] divides Pn (a3 ) for all n ⩾ 3. (⊠) Exercise 3. Answer. Exercise 3. For any natural number n ⩾ 2. 1). (OPfAtW9798-p18). 2. c be positive numbers such that abc = 1.127. Prove that 1 1 1 = + ⋅ BD AD CD Exercise 3. (OPfAtW9798-p40). 5). (⊠) x = y = z = t. (x.124. Answer. 2 2 . Prove that 1 1 3 5 2010 1 < ⋅ ⋅ ⋯ < ⋅ 2012 2 4 6 2011 44 Exercise 3. For each natural number n = 2. respectively. (OPfAtW9798-p22). Prove that ∠OBC = ∠ODC. xn are arbitrary real numbers. max Vn = π n when x1 = ⋯ = xn = ⋅ (⊠) 2 4 Exercise 3.212 Chapter 3. b.122. (OPfAtW9798-p22).123. (OPfAtW9798-p25). Find all real solutions of the system of equations ⎧ x3 ⎪ ⎪ ⎪ ⎪ 3 ⎨y ⎪ ⎪ ⎪ 3 ⎪ ⎩z = 2y − 1 = 2z − 1 = 2x − 1. C. . y) = (2.. Answer. Determine all primes p for which the system has a solution in integers x. Let a. (OPfAtW9798-p24).125. Let ABC be a triangle and M. y. }. (⊠) Exercise 3. determine the largest possible value of the expression Vn = sin x1 cos x2 + sin x2 cos x3 + ⋯ + sin xn cos x1 .. ⎧ ⎪ ⎪p + 1 = 2x2 ⎨ 2 2 ⎪ ⎪ ⎩p + 1 = 2y . Exercises for training Exercise 3. Prove that 1 1 1 1 1 1 + + ⩽ + + ⋅ 1+a+b 1+b+c 1+c+a 2+a 2+b 2+c Exercise 3. t ∈ {1.126. Answer. Let O be a point inside a parallelogram ABCD such that ∠AOB + ∠COD = π. Exercise 3. where x1 . Let D be the intersection of the ray MN with the circumcircle of ABC. (OPfAtW9798-p43).. x2 . The solutions are √ √ −1 + 5 −1 − 5 . p = 7. N the feet of the angle bisectors of B. (OPfAtW9798-p36).121. then ∣P (n)∣ ⩾ 7(6!)2 . (OPfAtW9798-p47). The solutions are (−2. x √ 1− 5 Answer. Answer. Let R be the circumradius of triangle ABC. g(x) = x5 + 5x4 + 3x3 − 5x2 − 1. b. Find all solutions in integers of the equation x3 + (x + 1)3 + (x + 2)3 + ⋯ + (x + 7)3 = y 3 . when x = 0 and P (x) = (x + 1)(x − 1)(x + 2)(x − 2).213 3. Show that if n ∈ Z is not a root of P. (b) f (20 + x) = −f (20 − x). If p = 5. and da . mb .130.132. (OPfAtW9798-p45). Prove that a2 + b2 + c2 ma da + mb db + mc dc = ⋅ 2 Exercise 3. mc the lengths of the altitudes.. x 1 (c) f (x)f (f (x) + ) = 1 for all x > 0. p = 5.128.1. 4). 6). and for each such p. f (1) = ⋅ (⊠) 2 Exercise 3. for example. (⊠) Exercise 3. 17. Equality is satisfied. . Let f be a real-valued function such that for any real x. Let f ∶ (0. Answer. Exercise 3. Answer.(x + 7). −4). Show that AF 2 + BF 2 + CF 2 = 3R2 . Prove that f is odd (f (−x) = −f (x)) and periodic ( ∃ T > 0 such that f (x+T ) = f (x)). x = 13. db . (−4. Let P be a polynomial with integer coefficients having at least 13 distinct integer roots. dc the distances from the vertices to the orthocenter in an acute triangle. Let a. (OPfAtW9798-p59). (OPfAtW9798-p49). ma .129. and let G and H be its centroid and orthocenter. (−3.134. such that both f (x) and g(x) are divisible by p. Define the functions f (x) = x5 + 5x4 + 5x3 + 5x2 + 1 . If p = 17. (⊠) Exercise 3. find all such x. Find f (1). Find all prime numbers p for which there exists a natural number 0 ⩽ x < p.133. (−5. (OPfAtW9798-p48). Let F be the midpoint of GH. +∞) → R be a function such that (a) f is strictly increasing. x = 4. and give an example where equality is achieved. Exercises Exercise 3. c be the sides. (OPfAtW9798-p43). (⊠) Exercise 3. respectively. −6).131.. (OPfAtW9798-p). (a) f (10 + x) = f (10 − x). (b) f (x) > − 1 for all x > 0. (⊠) 3 3 3 3 3 3 Exercise 3. Show 2 BF BC that ∠F CE = ∠F DE and ∠F EC = ∠BDC..139. y. z ∶ ⎧ ⎪ =1 ⎪3(x2 + y 2 + z 2 ) ⎨ 2 2 2 2 2 2 3 ⎪ ⎪ ⎩x y + y z + z x = xyz(x + y + z) . (x. − ) . . Show that f (n) = 1 for all n ∈ N. (OPfAtW9798-p76). z) = (− . y.. y. The positive integers x1 . (b) For some n0 ∈ N. Let a. z satisfying x2 + y 2 + z 2 − 2xyz = 0. Exercise 3. n = 1.. Answer. Solve the following system of equations in real numbers x.137. 2. (OPfAtW9798-p66). 3.136. Find all integers x. Exercise 3. . c be positive real numbers. Exercises for training Exercise 3. x6 = 144. 1 1 1 1 1 1 Answer. y.135. Answer. (OPfAtW9798-p62). (⊠) Exercise 3. xn+3 = xn+2 (xn+1 + xn ). (OPfAtW9798-p73). f (n + f (n)) = f (n). Prove the inequality (c + a − b)2 (a + b − c)2 3 (b + c − a)2 + + ⩾ . f (n0 ) = 1. (OPfAtW9798-p65). (b + c)2 + a2 (c + a)2 + b2 (a + b)2 + c2 5 and determine when equality holds.140. let F be the point on segment AB such that = . Let f ∶ N → N be a function satisfying (a) For every n ∈ N.214 Chapter 3. (OPfAtW9798-p74). x7 satisfy the conditions Compute x7 . 4.138. − . z) = ( . Given a convex pentagon ABCDE with DC = DE π AF AE and ∠BCD = ∠DEA = . b. x7 = 3456. (⊠) Exercise 3. . x = y = z = 0. ) and (x. 9 April 2006 Question 1. What is the last three digits of the sum 11! + 12! + 13! + ⋯ + 2006! . Find the last two digits of the sum 200511 + 200512 + ⋯ + 20052006 . 2. The lines P C and QB intersect at G. E ∈ AB and F ∈ AC with P Q = a and EF = b. The altitude OH of the triangle OAB intersect (O) at C . 3n + 1} with their sum equal to 3n + 1. What is AC if AB = 16cm? Question 8. B.1 215 Problems of Hanoi Open Mathematical Olympiad Hanoi Open Mathematical Olympiad 2006 Junior Section.. Find all points M belonging to the hexagon such that Area of triangle MAC = Area of triangle MCD.2. P Q//BC where P and Q are points on AB and AC respectively. Question 5.3. Question 9. Question 7. What is the last two digits of the number (11 + 12 + 13 + ⋯ + 2006)2 ? Question 2. . 3. On the circle (O) of radius 15cm are given 2 points A. Question 3.2 3.. Find value of BC. Find the value of ∣x3 − y 3∣.. Suppose x and y are two real numbers such that x + y − xy = 155 and x2 + y 2 = 325. What is the smallest possible value of x2 + y 2 − x − y − xy? Senior Section. 9 April 2006 Question 1. The figure ABCDEF is a regular hexagon. Question 4. It is also given EF //BC. Sunday. y.2. z) satisfying the equations x2 + y − z = 100 and x + y 2 − z = 124. Find the number of different positive integer triples (x. Sunday. Problems of Hanoi Open Mathematical Olympiad 3. In ∆ABC. Suppose n is a positive integer and 3 arbitrary numbers are choosen from the set {1. where G ∈ EF. What is the largest possible product of those 3 numbers? Question 6. Find all points M belonging to the hexagon such that Area of triangle MAC = Area of triangle MCD. E ∈ AB and F ∈ AC with P Q = a and EF = b. What is largest positive integer n satisfying the following inequality: n2006 < 72007 ? (A) 7. In ∆ABC. What is the perpendicular distance from C to the circle? Question 7. The figure ABCDEF is a regular hexagon. (D) 10. Which is larger 2 √ 2 . B with AB = 16cm and C is a midpoint of AB. (C) 23. . (D) 37. Question 9. (B) 11. (B) 8.216 Chapter 3. (E) 11. P Q//BC where P and Q are points on AB and AC respectively. Find the last three digits of the sum 200511 + 200512 + ⋯ + 20052006 .2 Hanoi Open Mathematical Olympiad 2007 Junior Section. Find the largest possible value of ∣x3 + y 3 + z 3 − xyz∣? 3. x x ∀ x ≠ 0. Find all polynomials P (x) such that 1 1 P (x) + P ( ) = x + . (C) 9. y. (E) None of the above. Question 8. Exercises for training Question 2. What is the last two digits of the number (3 + 7 + 11 + ⋯ + 2007)2 ? (A) 01. It is also given EF //BC. The lines P C and QB intersect at G.2 1+ √1 2 and 3? Question5. where G ∈ EF. Question 6. Question 3. Let x. Sunday.2. Suppose that alogb c + blogc a = m. Question 2. On the circle of radius 30cm are given 2 points A. Find value of BC. Find the value of clogb a + alogc b ? Question 4. 15 April 2007 Question 1. z be real numbers such that x2 + y 2 + z 2 = 1. (C) 1004. Which of the following is a possible number of diagonals of a convex polygon? (A) 02. Question 6. In a Heron triangle. Question 12. c be positive integers. (D) 6. are located inside an equilateral triangle of side 4. (E) 63. b. c satisfy the equation b = a(a − c). ∠BAC = 600 . (E) 2008. bc ca ab Prove a b c + + ⩾ 1. (D) 2007. b. β) with β − α = . Let be given triangle ABC . 1 Question 5. Let m and n denote the number of digits in 22007 and 52007 when expressed in base 10. (B) 1003. Nine points. (C) 5. Let a. b. Prove√that some three of these points are vertices of a triangle whose area is not greater than 3. (B) 2005. (C) 2006. . b. β) with 1 ⩽ b ⩽ 2007? b (A) 1002. no three of which lie on the same straight line. (D) 54. Then DB is (A) 3. Find all points M such that area of ∆MAB= area of ∆MAC. Prove that (c + a − b)2 (a + b − c)2 3 (b + c − a)2 + + ⩾ ⋅ (b + c)2 + a2 (c + a)2 + b2 (a + b)2 + c2 5 Question 9. Problems of Hanoi Open Mathematical Olympiad 217 Question 3. bc ca ab Question 11. How many possible values are there for the sum a + b + c + d if a. Determine the 2007 a a maximum number of irreducible fractions in (α. If AD bisects ∠BAC and CD = 3cm. (B) 4. Question 8.3. (C) 32. What is the sum m + n? (A) 2004. Question 7. c be positive real numbers such that 1 1 1 + + ⩾ 1. In triangle ABC. ∠ACB = 900 and D is on BC . (E) 7. (B) 21. c. (D) 1005. the sides a. (E) 1006. d are positive integers and abcd = 2007. Question 10. Question 4. Prove that the triangle is isosceles. Let a. Let be given an open interval (α. Calculate the sum 5 5 5 + +⋯+ ⋅ 2 ⋅ 7 7 ⋅ 12 2002 ⋅ 2007 Question 13. A triangle is said to be the Heron triangle if it has integer sides and integer area.2. Let P (x) = x3 + ax2 + bx + 1 and∣P (x)∣ ⩽ 1 for all x such that ∣x∣ ⩽ 1. Suppose that A. Let ABC be an equilateral triangle. AB is the diameter. (C) 3. . such that ij divides xi + xj for any two distinct positive integers i and j. 5 5 and 6 6 in order from greatest to least. (B) 21. Question 8.. let D.. Question 4. (E) None of the above. F be the feet of the perpendiculars from M onto BC. . Which is largest positive integer n satisfying the following inequal. y. 4 4. . B.218 Chapter 3. AB and CD are integers and AE − EB = 3. D are points on a circle. List the numbers √ √ √ √ √ 2. Question 9. respectively. Find the locus of all such points M for which ∆F DE is a right angle. (C) 31. Question 3. (D) 4. x2 . a2 . xn . How many ordered pairs of integers (x. (A) 1. a2007 be real numbers such that . 15 April 2007 Question 1. C. Exercises for training Question 14. Question 5. CD is perpendicular to AB and meets AB at E. AB. z) satsfying the equations x + y − z = 1 and x2 + y 2 − z 2 = 1. (B) 2. Find AE? Question 6. CA. (B) 2. (A) 1. Question 7. 3 3. y) satisfy the equation 2x2 + y 2 + xy = 2(x + y)? Question 15. Senior Section. Find all sequences of integers x1 . E. Prove that the equation ax2 + bx + c = 0 has no rational roots. Question 2...ity: n2007 > (2007)n .. (D) 41. (E) None of the above. Prove that ∣a∣ + ∣b∣ ⩽ 5. Let p = abc be the 3-digit prime number.. What is the last two digits of the number (112 + 152 + 192 + ⋯ + 20072 ) 2 (A) 01. Let a1 . (D) 4. For a point M inside ∆ABC .. Sunday.. (E) None of the above. (C) 3. Find the number of different positive integer triples (x. Question 12. Let p = abcd be the 4-digit prime number. 2007}. Let O. H and F be the circumcenter. How many integers belong to (a. . How many integers from 1 to 2008 have the sum of their digits divisible by 5 ? Question 2. . Sunday. 30 March 2008 Question 1. 2008a).. Let ABC be an acute-angle triangle with BC > CA. Question 14. Calculate the sum 1 1 1 + +⋯+ ⋅ 2 ⋅ 7 ⋅ 12 7 ⋅ 12 ⋅ 17 1997 ⋅ 2002 ⋅ 2007 Question 13. respectively. 3. What is the smallest possible value of x2 + 2y 2 − x − 2y − xy? Question 11. How many ordered pairs of integers (x.2. y) satisfy the equation x2 + y 2 + xy = 4(x + y)? Question 15.. Suppose that the perpendicular to OF at F meet the side CA at P . Prove that ak ∈ [2006. Question 4. Question 10. Prove ∠F HP = ∠BAC. Prove that the equation ax3 + bx2 + cx + d = 0 has no rational roots. where a (a > 0) is given. ∀ x.2.219 3.3 Hanoi Open Mathematical Olympiad 2008 Junior Section. Problems of Hanoi Open Mathematical Olympiad a1 + a2 + ⋯ + a2007 ⩾ (2007)2 and a21 + a22 + ⋯ + a22007 ⩽ (2007)3 − 1. 2.. Find all polynomials P (x) satisfying the equation (2x − 1)P (x) = (x − 1)P (2x). n) of positive integers such that m2 + n2 = 3(m + n). 2008] for all k ∈ {1. Find the coefficient of x in the expansion of (1 + x)(1 − 2x)(1 + 3x)(1 − 4x)⋯(1 − 2008x). orthocentre and the foot of its altitude CH . Question 3. Find all pairs (m. b. What is the smallest possible value of a2 + b2 + c2 ? Senior Section. Sunday. 3] and satisfy the following conditions max{a. 2008b]. n) of positive integers such that the equation x3 − nx + mn = 0 . Suppose x. c} ⩾ 2. Question 6. Let be given a right-angled triangle ABC with ∠A = 900 . n) of positive integers such that m2 + 2n2 = 3(m + 2n). z. Prove that there exists an infinite number of relatively prime pairs (m. b. The sides of a rhombus have length a and the area is S. Question 3. t are real numbers such that ⎧ ∣ − x + y + z + t∣ ⎪ ⎪ ⎪ ⎪ ⎪∣x − y + z + t∣ ⎪ ⎪ ⎨ ⎪ ∣x + y − z + t∣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩∣x + y + z − t∣ Prove that x2 + y 2 + z 2 + t2 ⩽ 1. y. a + b + c = 5. Question 4. Determine EP +EF +P Q? Question 10. Exercises for training Question 5. y and z.220 Chapter 3. The figure ABCDE is a convex pentagon. AC = b. Question 2. where b (b > 0) is given. Find the sum ∠DAC + ∠EBD + ∠ACE + ∠BDA + ∠CEB? Question 8. Let a. Denote by P ∈ BC and Q ∈ BC such that EP ⊥ BC and F Q ⊥ BC. Let P (x) be a polynomial such that Find P (x2 + 1)? P (x2 − 1) = x4 − 3x2 + 3. Question 7. What is the length of the shorter diagonal? Question 9. How many integers are there in (b. 30 March 2008 Question 1. c ∈ [1. ⩽1 ⩽1 ⩽1 ⩽ 1. Let E ∈ AC and F ∈ AB such that ∠AEF = ∠ABC and ∠AF E = ∠ACB. Find all pairs (m. Show that the equation x2 + 8z = 3 + 2y 2 has no solutions of positive integers x. AB = c. a + b + c = 5. Problems of Hanoi Open Mathematical Olympiad 221 has three distint integer roots. a⩽x⩽b Question 6. c} ⩾ 2. respectively. c) of consecutive odd positive integers such that a < b < c and a2 + b2 + c2 is a four digit number with all digits equal. Find all triples (a. Let O be the intersection of BN and CP .3. 29 March 2009 Question 1. b. Let be given a right-angled triangle ABC with ∠A = 900 . (D) 6. b ∈ R a⩽x⩽b where a < b. Let E ∈ AC and F ∈ AB such that ∠AEF = ∠ABC and ∠AF E = ∠ACB. Q be orthocenter of BOC and DOA. ∀ a. M. Determine EP + EF + F Q? 3. Prove that MN ⊥ P Q. AC = b.2. b. c ∈ [1. (D) 54. (E) None of the above. Question 2. Question 5. Consider a convex quadrilateral ABCD. Which is largest positive integer n satisfying the inequality 1 1 1 1 6 + + +⋯+ < ⋅ 1⋅2 2⋅3 3⋅4 n(n + 1) 7 (A) 3. (C) 36.2. Sunday. Consider a triangle ABC . b. N be the centroid of AOB and COD and P. . (B) 41. (C) 5. (E) None of the above. (B) 4. Let a. What is the smallest possible value of a2 + b2 + c2 ? Question 7. What is the last two digits of the number 1000 ⋅ 1001 + 1001 ⋅ 1002 + 1002 ⋅ 1003 + ⋯ + 2008 ⋅ 2009? (A) 25. Find M ∈ BC such that ∠P MO = ∠OMN. For every point M ∈ BC we define N ∈ CA and P ∈ AB such that AP MN is a parallelogram. Question 8. Let O be the intersection of AC and BD. Find all polynomials P (x) of degree 1 such that max P (x) − min P (x) = b − a. Question 10.4 Hanoi Open Mathematical Olympiad 2009 Junior Section. Denote by P ∈ BC and Q ∈ BC such that EP ⊥ BC and F Q ⊥ BC. 3] and satisfy the following conditions max{a. AB = c. Question 9. Suppose that 4 real numbers a. (D) 18. b. Show that there is a natural number n such that the number a = n! ends exacly in 2009 zeros. (A) 15. Prove that m7 − m is divisible by 42 for any positive integer m. c. where b = 210n+1 . b be positive integers such that a + b = 99. 6} and a < b < c such that the number abc + (7 − a)(7 − b)(7 − c) is divisible by 7. Question 14. (E) None of the above. b. Question 8. Question 10. Find all the pairs of the positive integers such that the product of the numbers of any pair plus the half of one of the numbers plus one third of the other number is three times less than 15. 4. (A) 15. Find the smallest and the greatest values of the following product P = ab. S lie in BC such that BR = RS = SC and P. 3. Question 6. d satisfy the conditions a2 + b2 = c2 + d2 = 4 and ac + bd = 2 Find the set of all possible values the number M = ab + cd can take. (E) None of the above. c be positive integers with no common factor and satisfy the conditions 1 1 1 + = a b c Prove that a + b is a square. Question 9. How many positive integer roots of the inequality −1 < there in (−10. Question 7. Exercises for training Question 3. x−1 < 2 are x+1 Question 4. Let ABCbe an acute-angled triangle with AB = 4 and CD be the altitude through C with CD = 3. Let a. How many triples (a. Question 5. (B) 17. Let be given ∆ABC with area (∆ABC) = 60cm2. respectively. Evaluate area (∆P QT ). 2. Let R. 5. c ∈ {1. (C) 19. 10). Prove that a is divisible by 23 for any positive integer n. Suppose that a = 2b + 19. c) where a. Find the distance between the midpoints of AD and BC. Q be midpoints of AB and AC.222 Chapter 3. (D) 21. (C) 17. b. Question 12. b. Find all integers x. Suppose that P S intersects QR at T. Let a. y such that x2 + y 2 = (2xy + 1)2 . Question 13. (B) 16. . Question 11. Given a triangle ABC with BC = 5. Question 5. How many integral roots of the inequality −1 < (−10. c) where a. How many triples (a. b. (D) 18. 2. Problems of Hanoi Open Mathematical Olympiad Senior Section. (B) 16. (C) 19. v) for which 5u2 + 6uv + 7v 2 = 2009. Let be given three positive numbers p. Question 6. CA = 4. F. (A) 15. c ∈ {1. c. (C) 36.223 3. Determine all positive integral pairs (u. Question 8. Question 3. (B) 17. Prove that a is divisible by 23 for any positive integer n. Question 9. (D) 21. y. Which is largest positive integer n satisfying the inequality 1 1 1 6 1 + + +⋯+ < ⋅ 1⋅2 2⋅3 3⋅4 n(n + 1) 7 (A) 3. Question 10. d satisfy the conditions . G lie on the sides BC. Question 11. What is the last two digits of the number 1000 ⋅ 1001 + 1001 ⋅ 1002 + 1002 ⋅ 1003 + ⋯ + 2008 ⋅ 2009? (A) 25. (C) 5. 29 March 2009 Question 1. (C) 17. b. z satisfying the system ⎧ ⎪ ⎪x + y + z ⎨ 3 3 3 ⎪ ⎪ ⎩x + y + z =8 = 8. 5. Find all integers x. Question 7. where b = 210n+1 . 10). Prove that for every positive integer n there exists a positive integer m such that the last n digists in decimal representation of m3 are equal to 8. (D) 54. 6} and a < b < c such that the number abc + (7 − a)(7 − b)(7 − c) is divisible by 7. (B) 4. Question 2. Suppose that 4 real numbers a. (E) None of the above. x−1 < 2 are there in x+1 Question 4. (D) 6. so that EF is parallel to AB and area (∆EF G) = 1. CA. AB = 3 and the points E. (A) 15. (E) None of the above. 4.2. q and r. Give an example of a triangle whose all sides and altitudes are positive integers. Sunday. respectively. b. AB. 3. (E) None of the above. (E) None of the above. Suppose that a = 2b + 19. (B) 41. Find the minimum value of the perimeter of triangle EF G. d be positive integers such that a + b + c + d = 99. (B) : 45625. C. enter only the letters (A. Exercises for training ⎧ a2 + b2 = p ⎪ ⎪ ⎪ ⎪ 2 ⎨c + d2 = q ⎪ ⎪ ⎪ ⎪ ⎩ac + bd = r. B. (D) : 15625. Find all the pairs of the positive integers such that the product of the numbers of any pair plus the half of one of the numbers plus one third of the other number is 7 times less than 2009. let points A. (D) : 9. How many real numbers a ∈ (1. C be located as follows: A is the point where altitude from A on BC meets the outwards facing semicirle drawn on BC as diameter.Given an acute-angled triangle ABC with area S. . Question 14. (C) : 2. b. (B): 1. (D) : 3. (C) : 25625. Points B. The last 5 digits of the number 52010 are (A) : 65625. The number of integers n ∈ [2000. a (A) : 0. Find the smallest and the greatest values of the following product P = abcd. (E) : None of the above.5 Hanoi Open Mathematical Olympiad 2010. Question 3. C are located similarly.224 Chapter 3. (E) : None of the above. B.2. (B) : 1. Let a. 28 March 2010 08h45-11h45 Important: ˆ Answer all 10 questions. Find the set of all possible values the number M = ab + cd can take. 3. Senior Section Sunday. 9) such that the corresponding number 1 a − is an integer. ˆ For the multiple choice questions. ˆ Enter your answers on the answer sheet provided. D or E) corresponding to the correct answers in the answer sheet. (E) : None of the above. Question 1 . c. is (A) : 0. . Question 13. 2010] such that 22n + 2n + 5 is divisible by 7. (C) : 8. Question 2. Question 12. ˆ No calculators are allowed. Evaluate the sum T = (area ∆BCA)2 + (area ∆CAB)2 + (area ∆ABC)2 . Let x. d be the roots of the equation x2 −rx+s = 0. Find the maximum value of T= y z x + + . both roots are prime numbers.2. b be the roots of the equation x2 − px + q = 0 and let c. Question 7. Let P be the common point of 3 internal bisectors of a given ABC. b. y be the positive integers such that 3x2 + x = 4y 2 + y. Suppose that 2(abc + bcd + cda + dab) M= p2 + q 2 + r 2 + s2 is an integer. z > 0. Let a. Determine a. If AP = 3cm. Question 10. Prove that x − y is a perfect integer. find n? Question 9. 2x + y 2y + z 2z + x x. AM respectively. r.225 3. y. Problems of Hanoi Open Mathematical Olympiad Question 4. Determine all positive integer a such that the equation 2x2 − 210x + a = 0 has two prime roots.e. where p. compute the value of ? BN Question 8. The line passing through P and perpendicular to CP intersects AC and BC at M and N. Each box in a 2 × 2 table can be colored black or white. How many different colorings of the table are there? Question 5. c. Question 6. If n and n3 + 2n2 + 2n + 4 are both perfect squares. q. d. s are some positive real numbers. i. BP = 4cm. . ˆ For the multiple choice questions. (C) 34 . (D) 8 . B and C in a row. (C) 6 . Let C1 and C2 be distinct circles of radius 7 cm that are in the same plane and tangenr to each other. (D) 16p . (E) 45359 . (C) 44934 . (B) 42925 . (B) 32 . Given that ∠COD = 30o . (D) 10 . (D) 36 . 2 June 2009 0930-1200 hrs Important: ˆ Answer ALL 35 questions. (E) 2 3 Question 4. (A) 12p . (C) 2 + 3 . find the area of the shades region ABCD. The sequence an satisfy an = an−1 + n2 and a0 = 2009. the radius of quadrant ODA is 4 and the radius of quadrant OBC is 8. (B) -9 . Find a50 . (E) non of the above Question 2. (E) 38 Question 5. D or E) corresponding to the correct answer ˆ For the other short questions.1 Chapter 3. find the maximum value of y. In the diagram below. (A) 30 . Find the number of circles of radius 26 cm in this plane that are tangent to both C1 and C2 . Exercises for training Singapore Open Mathematical Olympiad 2009 Junior Section Tuesday. (B) 4 . Part A. (E) non of the above Question 3. Multiple Choice Questions Question 1. enter your answer on the answer sheet by shading the bubble containing the letter (A. (D) -5 . C. (A) √ √ √ √ √ 5 . Given that x and y are both negative integers satisfying the equation 10x .3 3. (C) 15p . If the line W Z is tangent to the third circle. (A) 42434 . find the langth of XY. (E) non of the above Question 6. (A) 2 . y= 10 − x (A) -10 . write your answer in the answer sheet ˆ No steps are needed to justify your answer ˆ Each question carries 1 mark ˆ No calculators are allowed. B. (D) 45029 . Find the maximum value of k such that the √ following inequality holds: x − 2 + 7 − x ⩾ k.3. Let k be√a real number. (C) -6 .226 3. (B) 3 . (B) 13p . ˆ Enter your answer sheet provided. Three circles of radius 20 are arranged with their respective centres A. Coins of the same size are arranged on a very large table (the infinite plane) such that each coin touches six other coins. find the minimum value of y. If the ratio of the area of the standard model to that of the widescreen model is A ∶ 300. (C) 2 + 3 . (B) 2 − 3 . (B) 7 . The ratio of length to the height of the standard mode is 4 : 3. Find the percetage of the plane that is covered by the coins. (E) − 900 Question 10. (E) 3 + 2 Question 9. Find the minimum number of answer scripts required to guarantee two scripts with at least nine identical answers. while that of the widescreen model is 16 : 9. (A) 1 + √ √ √ √ √ 3 . Question 12. d) satisfying 1 1 1 1 + + + =1 a b c d with the condition that a < b < c < d is (A) 6 . The diagram below shows a pentagon (made up of region A and region B) and a rectangle (made up of region B and region C ) that overlap. (D) 9 . Question 14. Singapore Open Mathematical Olympiad 2009 227 Question 7. (E) 10 Part B. c. (A) − 896 . (E) 18 3π% 3 3 Question 8. √ √ √ 50 20 (A) √ π% . (B) √ π% . (B) − 897 . find the value of ∣x + y + 1∣. The number of positive integral solutions (a. If the ratio of region A of the 16 9 m pentagon to region C of the rectangle is in its lowest term. One is a ”20 inch” standard model while the other is a ”20 inch” widescreen model. Given that x and y are real numbers satisfying the following equations: √ x + xy + y = 2 + 3 2 and x2 + y 2 = 6. There are two models of LCD television on sale. (C) 16 3π% . (C) 8 . find the value of m + n.3. n Question 13. (D) 17 3π% . b. (C) − 898 . Given that y = (x − 16)(x − 14)(x + 14)(x + 16). so both models have the same diagonal length of 20 inches. (D) − 899 . . The number of ways to arrange 5 boys and 6 giels in a row such that girls can be adjacent to other girls but boys cannot be adjacent to other boys is 6! × k.3. The overlapped 3 2 region B is of the pentagon and of the rectangle. find the value of A. (D) 3 − 2 . Television screens are measured by the length of their diagonals. Short Questions Question 11. 2009 students are taking a test which comprises ten true or fales questions. 2. Find the number of integers in the set {1. BCEF. 2p. DEIJ and F GH are 22cm2 . .. Find the area of HIK in cm2 . Given the rectangle ABLJ. Given that a + a+1 b−1 ab − a + b. q. 3n−19. If ∣x∣ + x + 5y = 2 and ∣y∣ − y + x = 7. 7n−45} represents {p. Find the value of m + n. . The sum of 1 1 1 1 1 + + +⋯+ + 2×3×4 3×4×5 4×5×6 13 × 14 × 15 14 × 15 × 16 m in its lowest terms. Question 21. Let p and q represent two consecutive prime number. find the value of x + y + 2009. Exercises for training Find the value of k. Find the area of triangle DBG in cm2 .228 Chapter 3. 2009} whose sum of the digits is 11. 2. r = 0. y) that satisfy the equation √ √ √ √ √ x y + y x + 2009xy − 2009x − 2009y − 2009 = 0. n. n 1 1 = b+ − 2 and a − b + 2 ≠ 0. where the area of ACD.. Question 25.. 2q}. Question 16.. A square is constructed on the side AB and BC as shown. Question 15. The area of the square ABDE is 8cm2 and the area of the square BCF G is 26cm2 . Find the value of n. √ √ √ √ 3 3 Question 23. the set {n−1. Question 19. Find the number of ordered pairs of positive intergers (x. 1. Question 24. 500cm2 and 22cm2 respectively. 38−5n. Evaluate 77 − 20 13 + 77 + 20 13. For some fixed integer n. 3. ABC is a right-angled triangle with ∠BAC = 900 .. Question 20. Find the value of n such that a0 + a2 + a3 + a4 + ⋯ + an−2 + an−1 = 60 − n(n + 1) ⋅ 2 . find the value of Question 17. Given that x + (1 + x)2 + (1 + x)3 + ⋯ + (1 + x)n = a0 + a1 x + a2 x2 + ⋯ + an xn .. is Question 18. Find the intergers part of 1 ⋅ 1 1 1 1 1 1 1 + + + + + + 2003 2004 2005 2006 2007 2008 2009 Question 22. but not necessarily in that order. where each ar is an integer. Singapore Open Mathematical Olympiad 2009 Question 26. 3. x3 . Find the total number of such possible number that are divisible by 7 or 101 but not both. Question 28. Find the value of the smallest positive integer m such that the equation has only integer solutions. Find the largest possible integer calue for the length of third altitude CF. n) such that their produc mn is divisible by 33. Find the number of pairs of (m. x2 + 2(m + 5)x + (100m + 9) = 0 Question 31. 2 June 2009 0930-1200 hrs . Find the value of k. 1. find the number of 13-digit sequences that can be written so that the diffenrence between any two consecutive digits is 1. ∠AOB = ∠AOC = ∠BOC = 900. Question 30. Question 33. and 4. If x1 x2 = 49. OB = 2 and OC = 6. Find the least positive integer n for which 9n + 11 ducible fraction. n − 10 is a non-zero reQuestion 29. Find the value of m + n. A four-digit number consists of two distinct pairs of repeated digit (for example 2211. OAC and OBC of tetrahedron OABC are rightangled triangles. 2626 and 7007). 3.3. x4 denote the four roots of the equation x4 + kx2 + 90x − 2009 = 0.3. find the value of (Area of ∆OAB)2 +(Area of ∆OAC)2 +(Area of ∆OBC)2 +(Area of ∆ABC)2 .229 3. Let x1 . Question 27. Given that OA = 7. In the diagram.e. Question 35. m and n are two positive integer satisfying 1 ⩽ m ⩽ n ⩽ 40. P and Q are 2 points on AB such that 26AP = 22P Q = 11QB. Using the digits 0. m and n are two positive integers of reverse order (for example 123 and 321) such that mn = 1446921630.2 Senior Section Tuesday. Question 34. Question 32. OAB is a triangle with ∠AOB = 900 and OB = 13cm. x2 . the length of the altitudes AD and BE are 4 and 12 respectively. Three sides OAB. i. 2. In a triangle ABC. find the area of the triangle OP Q in cm2 . If the vertical height of P Q = 4cm. Exercises for training Important: ˆ Answer ALL 35 questions. (D) . (A) 1 5 3 1 1 . D or E) corresponding to the correct answer ˆ For the other short questions. find the value of x + y. (D) − . (D) − 1293 . (C) . Part A. In a triangle ABC. 1). (A) 11 . If x and y are real numbers for which ∣x∣ + x + 5y = 2 and ∣y∣ − y + x = 7. (A) − 3 . (E) 15 Question 7. If AD = 3cm. (B) . B. If the distance between A and B is 33 cm. BC and CA. (C) 1 . where x is a real number. (E) 8 16 16 4 3 Question 4. The area of a triangle ABC is 40cm2 . (C) . (A) 1 . DB = 5cm. (B) . (B) 2 . (D) 14 . ˆ For the multiple choice questions. Points D. (B) − 1295 . (E) or − 65 65 65 65 65 65 65 Question 6. (C) 3 . (A) − 1296 . (B) − 1 . Find the value of 3 4 22 + +⋯+ ⋅ 1! + 2! + 3! 2! + 3! + 4! 20! + 21! + 22! . Find the smallest posible value of y. (E) − 1292 Question 3. ˆ Enter your answer sheet provided. (C) 13 . (E) 5 Question 5. Question 2. (D) 4 . find the probability that the equation x2 − ax + b = 0 has real roots. (D) 3 . find the area of triangle AEC in cm2 .230 Chapter 3. Let y = (17 − x)(19 − x)(19 + x)(17 + x). (E) infinitely many. respectively. 5 13 16 56 16 56 56 16 56 or . E and F are on sides AB. (B) 12 . respectively. Suppose that P is a plane and A and B are two points on the plane P. write your answer in the answer sheet ˆ No steps are needed to justify your answer ˆ Each question carries 1 mark ˆ No calculators are allowed. (C) − 1294 . enter your answer on the answer sheet by shading the bubble containing the letter (A. If two real numbers a and b are randomly chosen from the interval (0. sin A = (A) 3 5 and cos B = ⋅ Find the value of cos C. as shown in the figure below. C. and the area of triangle ABE is equal to the area of quadrilateral DBEF. how many lines are there in the plane such that the distance between each line and A is 7 cm and the distance between each line and B is 26 cm. Multiple Choice Questions Question 1. and point F is on side AC such that DF AE BD CF is the angle bisector of ∠ADC. . (E) 6 Part B. Let ABCD be a quadrilateral inscribed in a circle with diameter AC. √ √ √ √ (A) 3 2 . (B) 4 . (A) 1 − (A) 35 . Singapore Open Mathematical Olympiad 2009 231 1 1 1 1 1 1 1 1 . all angles are less than 90o ) in which all angles (in degrees) are positive integers and the largest angle is three times the smallest angle. (C) 2 7 . 100} such that ab + a + b is a multiple of 7. Determine the number of acute-angled triangles (i. Point E is on side AB such that DE is the angle bisector of ∠ADB.. find the length of DE in cm. Find the number of ways in which 4 identical red buttons and 4 identical blue buttons can be put in the envolopes such that each envolope contains exactly one button. (D) 4 2 .e. 99. (E) − 24! 2 23! 2 22! 22! 2 24! Question 8. Find the value of (cot 25o − 1)(cot 24o − 1)(cot 23o − 1)(cot 22o − 1)(cot 21o − 1)(cot 20o − 1). Question 18. ABC is a triangle and D is a point on side BC. Find the number of 2-element subset {a. Find the remainder when is divided by 2009. Find the number of positive divisors of (20083 +(3 ×2008 ×2009)+1))2. . b and c are real numbers greater than 1.3. 3. Question 12. (B) 34 . Find the value of (25 + 10 5) 3 + (25 − 10 5) 3 √ 1 + 2009 ⋅ Find the value of (a3 − 503a − 500)10 . (E) 62 Question 9. (D) 31 . If AD = DC and the area of quadrilateral ABCD is 24cm2 . and let E be the foot of perpendicular from D onto AB. (C) 32 .. (C) 5 .3. (B) 2 6 .. (A) 3 . (D) 1 − . as shown in the figure below. Short Questions Question 11. b} of {1. (C) − . (1! × 1) + (2! × 2) + (3! × 3) + ⋯ + (286! × 286) √ 1 √ 1 Question 14. Suppose that a. Find the value of ⋅ EB DC F A Question 17. There are eight envolopes numbered 1 to 8. Let a = 2 Question 16. 2. 1 1 1 + + ⋅ Find the value of c a 1 + loga2 b ( a ) 1 + logb2 c ( b ) 1 + logc2 a ( cb ) Question 13. and the sum of the nimbers on the envolopes containing the blue buttons. (B) − . Question 15. (D) 6 .. (E) 7 Question 10. 3... find the earea of rectangle ABCD in cm2 . Let x be real number such that x2 −15x+1 = 0. Question 25. Question 29. 1 − 4 sin 250 cos 250 cos 500 Question 27. For example. find x. where n = 1. 65}. Question 21. 300388. z) such that x.. 1 + nan−1 an 1 ⋅ a199 a200 Question 23. Question 26. Let S = {1. Let n be the positive integer such that 1 1 1 1 √ √ + √ √ +⋯+ √ √ = ⋅ 9 11 + 11 9 11 13 + 13 11 n n + 2 + (n + 2) n 9 Find the value of n. find the value of k.232 Chapter 3. . ABCD is a rectangle. Question 22. x < z and y < z. BC = 13cm and AC = 10cm. y. If cos 1000 = tanx. Find the number of positive integers x. such that log x 9 9 x2 < 6 + log3 . find the length of CD in cm. 555333. 3 x Question 28. Given that BE is an angle bisector of ∠ABC and that AD = 13. Find the value of x4 + 1 ⋅ x4 Question 20. Determine the number of ordered triples (x. Question 30. If x. .. 110010100101 is such a sequence but 101011000101 and 110101100001 are not. Find the number of 0 − 1 binary sequences formed by six 0′ s and 1′ s such that no three 0′ s are together. z ∈ S. ..5cm. and a0 = a1 = 1. Points D and E lie on side AC and point F on side BC such that EF is parallel to AB and DF is parallel to EB. 818811. In each of the following 6-digit positive integers: 555555. ABC is a triangle with AB = 10cm. find the largest possible value of z. 3. If the area of triangle BDF is 12cm2 . y are real numbers such that x + y + z = 9 and xy + yz + zx = 24. BC = 40cm. 2. 2. E is the midpoint of AD and F is the midpoint of CE. Points area of ∆AP Q 1 P and Q lie on sides AB and AC respectively such that = ⋅ area of ∆ABC 4 Given that the least posible length of P Q is kcm. 64.. y. Given that an+1 = find the value of an−1 . ABC is a triangle with AB = 5cm. Question 24. Exercises for training Question 19. where x ≠ 9. . Find the largest posible value of xy... Question 33. 87654321.. Let x be a positive integer. Question 34. a2 = 12345687. 2.. Determine the coefficient of x29 in the expansion (1 + x5 + x7 + x9 )16 . Find a2009 − a2008 . 5. we can form 8! (= 40320) 8-digit numbers in which the eight digits are all distinct. . Question 32. Singapore Open Mathematical Olympiad 2009 233 every digit in the number appears at least twice... and let dn denote the greatest common divisor of an and an+1 . a40320 = 87654321.3. let ak denote the k th number if these numbers are arranged in increasing order: 12345678. Find the number of such 6-digit positive integers. 3. 4. Question 31. For 1 ⩽ k ⩽ 40320. 12345768..3. . Let x and y be positive integers such that 27x + 35y ⩽ 945. Question 35. that is. x Here [c] denotes the greatest integer less than or equal to c. 8. let an = n2 + 100. 3. Find the largest possible value of 2a2 − 3b2 . and write a = [log10 x] and b = [log10 . 12345687. Using the digits 1. 100 ]. 7.. a1 = 12345678. 6. For n = 1.. 2. Find the maximun value of dn as n ranges over all positive integers. uk http://www. [4] Titu Andreescu and Zuming Feng. 1998 [14] Websites of mathemathics: ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ http://diendantoanhoc.org/ http://www. Mathematical Olympiads 1998 .com/ http://lib. Engel. 1959-2004.. Bikhauser.bymath.math.vn/hms/ http://sms. J. Springer Verlag.org.Solving Through Problems.Bibliography [1] Nguyen Van Mau. Problems and Solutions: 1965.: 1966. R. K. 2006. The IMO Compendium: A Collection of Problems Suggested for The International Mathematical Olympiads.net www.edu. [6] Titu Andreescu and Zuming Feng.mccme. M.: 2002.ac.mexmat. 1983. [13] A.New York.2000 Problems and Solutions From Around the World.org http://www.nus. 2002. A. Hanoi Open Mathematical Olympiad. 2004. Problems. 1998.org/ . Razvan Gelca. Hanoi. Springer. [12] Kedlaya. Solutions and Commentary. Bikhauser.vn/ http://math. Springer.maths. A path to combinatorics for undergraduates counting strategies. [8] Loren C. HMS. The William Lowell Putnam Mathematical Competition. Problems and Solutions From Around the World. 2000. [2] Dusan Djukic. L. MMA 2002. [5] Titu Andreescu.ams. Mathematical Olympiad Challenges. and Kelly. R.ru/ http://hms.: 1980. [9] Walter Mientka et al. 2009. [3] Titu Andreescu and Zuming Feng. Bikhauser. and Vakil. MMA . The William Lowell Putnam Mathematical Competition 1985-2000. MMA 1997. M. E.co.. MMA . Lectures on Functional Equations and Their Applications.sg/ http://thesaurus.com/ http://www. Greenwood. 1998. Mathematical Olympiads 1996 .Vladimir Jankovic.math.. Poonen.math. Problems and Solutions. Problem . [11] Gleason. [10] Acz’el. Problem .net http://mathnfriend.Solving Strategies.1984. Problem-Solving Strategies. [7] Arthur Engel.. B. 102 Combinatorial Problems from the Training of the USA IMO Team. S. Birkhauser.kalva. Springer.ru/ 234 http://mathforum.Ivan Matic-Nikola Petrovic. Larson.1998.ca/crux/ http://kvant. Documents Similar To 22. FileGocHOMOMoiSkip carouselcarousel previouscarousel nextEdwards 409 Lesson31From Newton’s Theorem to a Theorem of the Inscribable Octagonde thu va loi giai HOMC 2006 - 2016Shifman M. 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