Hong Kong Mathematics Olympiad 1995-1996Heat Event (Individual)

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1995年IMO中国国家队选拔考试试题

1995Day 11Find the smallest prime number p that cannot be represented in the form |3a −2b |,where a and b are non-negative integers.2Given a ﬁxed acute angle θand a pair of internally tangent circles,let the line l which passes through the point of tangency,A ,cut the larger circle again at B (l does not pass through the centers of the circles).Let M be a point on the major arc AB of the larger circle,N the point where AM intersects the smaller circle,and P the point on ray MB such that ∠MP N =θ.Find the locus of P as M moves on major arc AB of the larger circle.Corrected due to the courtesy of[url=http://www.mathlinks.ro/Forum/proﬁle.php?mode=viewproﬁleu=2616]zhaoli.[/url]321people take a test with 15true or false questions.It is known that every 2people have at least 1correct answer in common.What is the minimum number of people that could have correctly answered the question which the most people were correct on?/This ﬁle was downloaded from the AoPS −MathLinks Math Olympiad Resources Page Page 1http://www.mathlinks.ro/1995Day 21Let S ={A =(a 1,...,a s )|a i =0or 1,i =1,...,8}.For any 2elements of S ,A ={a 1,...,a 8}and B ={b 1,...,b 8}.Let d (A,B )= i =18|a i −b i |.Call d (A,B )the distance between A and B .At most how many elements can S have such that the distance between any 2sets is at least 5?2A and B play the following game with a polynomial of degree at least 4:x 2n +x 2n −1+x 2n −2+···+x +1=0A and B take turns to ﬁll in one of the blanks with a real number until all the blanks are ﬁlled up.If the resulting polynomial has no real roots,A wins.Otherwise,B wins.If A begins,which player has a winning strategy?3Prove that the interval [0,1]can be split into black and white intervals for any quadratic polynomial P (x ),such that the sum of weights of the black intervals is equal to the sum of weights of the white intervals.(Deﬁne the weight of the interval [a,b ]as P (b )−P (a ).)Does the same result hold with a degree 3or degree 5polynomial?/This ﬁle was downloaded from the AoPS −MathLinks Math Olympiad Resources Page Page 2http://www.mathlinks.ro/。

香港数学竞赛历届比赛题目

27. Common stocks and uncommon profits and otherwritingsFisher, Philip A.332.63FIS28. Stocks for the long run Siegel, Jeremy J.29. Advanced Accounting Ng, Patrick.657 NG30. A random walk down Wall Street Malkiel, Burton Gordon.31. Security analysis Graham, Benjamin.33. 瑪蒂達Dahl, Roald.34. 就愛馬英九范植明35. 全食物排毒密碼: 吃對食物體內排毒全攻略36. 向偵探挑戰37. 我也是名偵探38. 青銅葵花曹文軒 857.7 550539. 地鐵求生121 Holman, Felice.40. 野地獵歌Rawls, Wilson.41. 走了一位老師之後Peck, Richard.42. 天才投資家森生文乃43. 獲利大師黑谷薰44. 世界投資家列傳(漫畫版) 田中憲45. 儒家與現代人生傅佩榮46. 赤壁之戰與三國計謀47. 中國中學生作文大賽2006-2007855 203548. 三四０年代香港新詩論集831.8 758249. 並放集(第八期)50. 二十一世紀香港中學生文選51. 嶺聲(第五期)52. 棠棣53. 六羽集54. 元朗信義中學校園關愛四十五週年紀念文集(2003-04年度)55. 李中文集56. 2004 青年文學創作營文集855 002257. 灣仔堂基道學校(上午) 作文選集58. 小說創未來59. 並放集(第七期)60. 凌雲健筆61. 揚芬集62. 青年心, 回歸情, 精彩十年63. 聚精會文。

Hong Kong Mathematics Olympiad (2007 2008)

Hong Kong Mathematics Olympiad (2007 − 2008)Heat Event (Group)香港數學競賽 (2007 − 2008)初賽項目(團體)除非特別聲明，答案須用數字表達，並化至最簡。

Unless otherwise stated, all answers should be expressed in numerals in their simplest form.1. 已知 5的小數部分為 A 及 5− 的小數部分為 B 。

設 C A ，求 C的值。

B =+Given that the decimal part of 5+ is A and the decimal part of 5− is B . Let,find the value of C .C A B =+2. 有一批糖共 x 粒，x 為正整數，這批糖能分別為 851 人及 943 人所均分。

求 x 的最小可能值。

A total number of x candies, x is a positive integer, can be evenly distributed to 851 people as well as 943 people. Find the least possible value of x .3. 如圖一，正四面體 ABCD 的邊長為 2 cm 。

若該正四面體的體積是 cm 3，求 R 的值。

In Figure 1, ABCD is a tetrahedron with side length 2 cm . I f the volume of the tetrahedron is 3,find the value of R .圖一Figure 14. 已知 x 為正整數及 x < 60。

若 x 恰有 10 個正因子，求 x 的值。

Given that x is a positive integer and x < 60 . If x has exactly 10 positive factors, find the value of x .5. 已知 90° < θ < 180° 及sin θ=。

英文的数论题小书3解答

GlossaryArithmetic functionA function deﬁned on the positive integers that is complex valued.Arithmetic-Geometric Means InequalityIf n is a positive integer and a1,a2,...,a n are nonnegative real numbers,then1 nni=1a i≥(a1a2···a n)1/n,with equality if and only if a1=a2=···=a n.This inequality is a special case of the power mean inequality.Base-b representationLet b be an integer greater than1.For any integer n≥1there is a unique system (k,a0,a1,...,a k)of integers such that0≤a i≤b−1,i=0,1,...,k,a k=0 andn=a k b k+a k−1b k−1+···+a1b+a0.Beatty’s theoremLetαandβbe two positive irrational real numbers such that1α+1β=1.The sets{ α , 2α , 3α ,...},{ β , 2β , 3β ,...}form a partition of the set of positive integers.190104Number Theory ProblemsBernoulli’s inequalityFor x>−1and a>1,(1+x)a≥1+ax,with equality when x=0.B´e zout’s identityFor positive integers m and n,there exist integers x and y such that mx+by= gcd(m,n).Binomial coefﬁcientn k=n!k!(n−k)!,the coefﬁcient of x k in the expansion of(x+1)n.Binomial theoremThe expansion(x+y)n=nx n+n1x n−1y+n2x n−2y+···+nn−1xy n−1+nny n.Canonical factorizationAny integer n>1can be written uniquely in the formn=pα11···pαk k,where p1,...,p k are distinct primes andα1,...,αk are positive integers. Carmichael numbersThe composite integers n satisfying a n≡a(mod n)for every integer a. Complete set of residue classes modulo nA set S of integers such that for each0≤i≤n−1there is an element s∈S with i≡s(mod n).Glossary191Congruence relationLet a,b,and m be integers,with m=0.We say that a and b are congruent modulo m if m|(a−b).We denote this by a≡b(mod m).The relation“≡”on the set Z of integers is called the congruence relation.Division algorithmFor any positive integers a and b there exists a unique pair(q,r)of nonnegative integers such that b=aq+r and r<a.Euclidean algorithmRepeated application of the division algorithm:m=nq1+r1,1≤r1<n,n=r1q2+r2,1≤r2<r1,...r k−2=r k−1q k+r k,1≤r k<r k−1,r k−1=r k q k+1+r k+1,r k+1=0This chain of equalities isﬁnite because n>r1>r2>···>r k.Euler’s theoremLet a and m be relatively prime positive integers.Thenaϕ(m)≡1(mod m).Euler’s totient functionThe functionϕ(m)is deﬁned to be the number of integers between1and n that are relatively prime to m.Factorial base expansionEvery positive integer k has a unique expansionk=1!·f1+2!·f2+3!·f3+···+m!·f m,where each f i is an integer,0≤f i≤i,and f m>0.192104Number Theory ProblemsFermat’s little theoremLet a be a positive integer and let p be a prime.Thena p≡a(mod p).Fermat numbersThe integers f n=22n+1,n≥0.Fibonacci sequenceThe sequence deﬁned by F0=1,F1=1,and F n+1=F n+F n−1for every positive integer n.Floor functionFor a real number x there is a unique integer n such that n≤x<n+1.We say that n is the greatest integer less than or equal to x or theﬂoor of x and we write n= x .Fractional partThe difference x− x is called the fractional part of x and is denoted by{x}.Fundamental theorem of arithmeticAny integer n greater than1has a unique representation(up to a permutation)as a product of primes.Hermite’s identityFor any real number x and for any positive integer n,x ++1n++2n+···++n−1n= nx .Legendre’s formulaFor any prime p and any positive integer n,e p(n)=i≥1np i.Glossary 193Legendre’s functionLet p be a prime.For any positive integer n ,let e p (n )be the exponent of p in the prime factorization of n !.Linear Diophantine equationAn equation of the forma 1x 1+···+a n x n =b ,where a 1,a 2,...,a n ,b are ﬁxed integers.Mersenne numbersThe integers M n =2n −1,n ≥1.M¨o bius functionThe arithmetic function µdeﬁned by µ(n )=⎧⎨⎩1if n =1,0if p 2|n for some prime p >1,(−1)k if n =p 1···p k ,where p 1,...,p k are distinct primes .M¨o bius inversion formulaLet f be an arithmetic function and let F be its summation function.Then f (n )= d |nµ(d )F n d .Multiplicative functionAn arithmetic function f =0with the property that for any relatively prime positive integers m and n ,f (mn )=f (m )f (n ).Number of divisorsFor a positive integer n denote by τ(n )the number of its divisors.It is clear that τ(n )= d |n1.194104Number Theory ProblemsOrder modulo mWe say that a has order d modulo m,denoted by ord m(a)=d,if d is the smallest positive integer such that a d≡1(mod m).Perfect numberAn integer n≥2with the property that the sum of its divisors is equal to2n. Pigeonhole PrincipleIf n objects are distributed among k<n boxes,some box contains at least two objects.Prime number theoremThe relationlim n→∞π(n)n/log n=1,whereπ(n)denotes the number of primes less than or equal to n.Prime number theorem for arithmetic progressionsFor relatively prime integers a and r,letπa,d(n)denote the number of primes in the arithmetic progression a,a+d,a+2d,a+3d,...that are less than or equal to n.Thenlim n→∞πa,d(n)n/log n=1ϕ(d).This result was conjectured by Legendre and Dirichlet and proved by Charles De la Vall´e e Poussin.Sum of divisorsFor a positive integer n denote byσ(n)the sum of its positive divisors including 1and n itself.It is clear thatσ(n)=d|nd.Glossary195Summation functionFor an arithmetic function f the function F deﬁned byF(n)=d|nf(d).Wilson’s theoremFor any prime p,(p−1)!≡−1(mod p).Zeckendorf representationEach nonnegative integer n can be written uniquely in the formn=∞k=0αk F k,whereαk∈{0,1}and(αk,αk+1)=(1,1)for each k.Further Reading1.Andreescu,T.;Feng,Z.,101Problems in Algebra from the Training of theUSA IMO Team,Australian Mathematics Trust,2001.2.Andreescu,T.;Feng,Z.,102Combinatorial Problems from the Training ofthe USA IMO Team,Birkh¨a user,2002.3.Andreescu,T.;Feng,Z.,103Trigonometry Problems from the Training ofthe USA IMO Team,Birkh¨a user,2004.4.Andreescu,T.;Feng,Z.,A Path to Combinatorics for Undergraduate Stu-dents:Counting Strategies,Birkh¨a user,2003.5.Feng,Z.;Rousseau,C.;Wood,M.,USA and International MathematicalOlympiads2005,Mathematical Association of America,2006.6.Andreescu,T.;Feng,Z.;Loh,P.,USA and International MathematicalOlympiads2004,Mathematical Association of America,2005.7.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2003,Mathematical Association of America,2004.8.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2002,Mathematical Association of America,2003.9.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2001,Mathematical Association of America,2002.10.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2000,Mathematical Association of America,2001.11.Andreescu,T.;Feng,Z.;Lee,G.;Loh,P.,Mathematical Olympiads:Prob-lems and Solutions from Around the World,2001–2002,Mathematical As-sociation of America,2004.198104Number Theory Problems12.Andreescu,T.;Feng,Z.;Lee,G.,Mathematical Olympiads:Problems andSolutions from Around the World,2000–2001,Mathematical Association of America,2003.13.Andreescu,T.;Feng,Z.,Mathematical Olympiads:Problems and Solutionsfrom Around the World,1999–2000,Mathematical Association of America, 2002.14.Andreescu,T.;Feng,Z.,Mathematical Olympiads:Problems and Solutionsfrom Around the World,1998–1999,Mathematical Association of America, 2000.15.Andreescu,T.;Kedlaya,K.,Mathematical Contests1997–1998:OlympiadProblems from Around the World,with Solutions,American Mathematics Competitions,1999.16.Andreescu,T.;Kedlaya,K.,Mathematical Contests1996–1997:OlympiadProblems from Around the World,with Solutions,American Mathematics Competitions,1998.17.Andreescu,T.;Kedlaya,K.;Zeitz,P.,Mathematical Contests1995–1996:Olympiad Problems from Around the World,with Solutions,American Mathematics Competitions,1997.18.Andreescu,T.;Enescu,B.,Mathematical Olympiad Treasures,Birkh¨a user,2003.19.Andreescu,T.;Gelca,R.,Mathematical Olympiad Challenges,Birkh¨a user,2000.20.Andreescu,T.,Andrica,D.,An Introduction to Diophantine Equations,GILPublishing House,2002.21.Andreescu,T.;Andrica,D.,360Problems for Mathematical Contests,GILPublishing House,2003.22.Andreescu,T.;Andrica,D.,Complex Numbers from A to Z,Birkh¨a user,2004.23.Beckenbach,E.F.;Bellman,R.,An Introduction to Inequalities,New Math-ematical Library,V ol.3,Mathematical Association of America,1961. 24.Coxeter,H.S.M.;Greitzer,S.L.,Geometry Revisited,New MathematicalLibrary,V ol.19,Mathematical Association of America,1967.25.Coxeter,H.S.M.,Non-Euclidean Geometry,The Mathematical Associa-tion of America,1998.Further Reading199 26.Doob,M.,The Canadian Mathematical Olympiad1969–1993,Universityof Toronto Press,1993.27.Engel,A.,Problem-Solving Strategies,Problem Books in Mathematics,Springer,1998.28.Fomin,D.;Kirichenko,A.,Leningrad Mathematical Olympiads1987–1991,MathPro Press,1994.29.Fomin,D.;Genkin,S.;Itenberg,I.,Mathematical Circles,American Math-ematical Society,1996.30.Graham,R.L.;Knuth, D.E.;Patashnik,O.,Concrete Mathematics,Addison-Wesley,1989.31.Gillman,R.,A Friendly Mathematics Competition,The Mathematical As-sociation of America,2003.32.Greitzer,S.L.,International Mathematical Olympiads,1959–1977,NewMathematical Library,V ol.27,Mathematical Association of America, 1978.33.Holton,D.,Let’s Solve Some Math Problems,A Canadian MathematicsCompetition Publication,1993.34.Kazarinoff,N.D.,Geometric Inequalities,New Mathematical Library,V ol.4,Random House,1961.35.Kedlaya,K;Poonen,B.;Vakil,R.,The William Lowell Putnam Mathemat-ical Competition1985–2000,The Mathematical Association of America, 2002.36.Klamkin,M.,International Mathematical Olympiads,1978–1985,NewMathematical Library,V ol.31,Mathematical Association of America, 1986.37.Klamkin,M.,USA Mathematical Olympiads,1972–1986,New Mathemat-ical Library,V ol.33,Mathematical Association of America,1988.38.K¨u rsch´a k,J.,Hungarian Problem Book,volumes I&II,New MathematicalLibrary,V ols.11&12,Mathematical Association of America,1967. 39.Kuczma,M.,144Problems of the Austrian–Polish Mathematics Competi-tion1978–1993,The Academic Distribution Center,1994.40.Kuczma,M.,International Mathematical Olympiads1986–1999,Mathe-matical Association of America,2003.200104Number Theory Problemsrson,L.C.,Problem-Solving Through Problems,Springer-Verlag,1983.usch,H.The Asian Paciﬁc Mathematics Olympiad1989–1993,Aus-tralian Mathematics Trust,1994.43.Liu,A.,Chinese Mathematics Competitions and Olympiads1981–1993,Australian Mathematics Trust,1998.44.Liu,A.,Hungarian Problem Book III,New Mathematical Library,V ol.42,Mathematical Association of America,2001.45.Lozansky,E.;Rousseau,C.Winning Solutions,Springer,1996.46.Mitrinovic,D.S.;Pecaric,J.E.;V olonec,V.Recent Advances in GeometricInequalities,Kluwer Academic Publisher,1989.47.Mordell,L.J.,Diophantine Equations,Academic Press,London and NewYork,1969.48.Niven,I.,Zuckerman,H.S.,Montgomery,H.L.,An Introduction to the The-ory of Numbers,Fifth Edition,John Wiley&Sons,Inc.,New York,Chich-ester,Brisbane,Toronto,Singapore,1991.49.Savchev,S.;Andreescu,T.Mathematical Miniatures,Anneli Lax NewMathematical Library,V ol.43,Mathematical Association of America, 2002.50.Sharygin,I.F.,Problems in Plane Geometry,Mir,Moscow,1988.51.Sharygin,I.F.,Problems in Solid Geometry,Mir,Moscow,1986.52.Shklarsky,D.O;Chentzov,N.N;Yaglom,I.M.,The USSR Olympiad Prob-lem Book,Freeman,1962.53.Slinko,A.,USSR Mathematical Olympiads1989–1992,Australian Mathe-matics Trust,1997.54.Szekely,G.J.,Contests in Higher Mathematics,Springer-Verlag,1996.55.Tattersall,J.J.,Elementary Number Theory in Nine Chapters,CambridgeUniversity Press,1999.56.Taylor,P.J.,Tournament of Towns1980–1984,Australian MathematicsTrust,1993.57.Taylor,P.J.,Tournament of Towns1984–1989,Australian MathematicsTrust,1992.Further Reading201 58.Taylor,P.J.,Tournament of Towns1989–1993,Australian MathematicsTrust,1994.59.Taylor,P.J.;Storozhev,A.,Tournament of Towns1993–1997,AustralianMathematics Trust,1998.60.Yaglom,I.M.,Geometric Transformations,New Mathematical Library,V ol.8,Random House,1962.61.Yaglom,I.M.,Geometric Transformations II,New Mathematical Library,V ol.21,Random House,1968.62.Yaglom,I.M.,Geometric Transformations III,New Mathematical Library,V ol.24,Random House,1973.Indexarithmetic functions,36base-b representation,41 Beatty’s theorem,60 Bernoulli’s inequality,145B´e zout’s identity,13binomial theorem,5canonical factorization,8 Carmichael numbers,32 ceiling,52Chinese remainder theorem,22 complete set of residue classes,24 composite,5congruence relation,19 coprime,11decimal representation,41 Diophantine equations,14 division algorithm,4Euclidean algorithm,12Euler’s theorem,28Euler’s totient function,27 factorial base expansion,45 Fermat numbers,22,70 Fermat’s little theorem,28 Fibonaccinumbers,45sequence,45fractional part,52fully divides,9fundamental theorem ofarithmetic,7geometric progression,9greatest common divisor,11Hermite identity,63inverse of a modulo m,26least common multiple,16of a1,a2,...,a n,16Legendre function,65linear combinations,14linear congruence equation,22linear congruence system,22linear Diophantine equation,38Mersenne numbers,71multiplicative arithmetic functions,18M¨o bius function,36M¨o bius inversion formula,37number of divisors,17order d modulo m,32perfect cube,2perfect numbers,72perfect power,2perfect square,2pigeonhole principle,93prime,5203204Indexprime number,5quotient,4reduced complete set of residueclasses,28 relatively prime,11 remainder,4square free,2sum of positive divisors,18 summation function,36twin primes,6Wilson’s theorem,26 Wolstenholme’s theorem,115 Zeckendorf representation,45。

奥数教程英文版

奥数教程英文版Mathematics has long been a subject of fascination and challenge for students around the world. From the fundamental operations of arithmetic to the intricate theories of calculus, the field of mathematics encompasses a vast and ever-evolving landscape of knowledge. One particular branch of mathematics that has gained significant attention in recent years is Olympiad mathematics, a highly competitive and demanding discipline that pushes the boundaries of mathematical understanding.The Olympiad mathematics curriculum is designed to challenge the most talented and dedicated students, offering them the opportunity to engage with complex problems and develop their analytical and problem-solving skills to the highest degree. However, the specialized nature of this curriculum can often present a significant barrier to entry, particularly for students who do not have access to comprehensive educational resources or guidance.This is where the need for an English textbook on Olympiad mathematics becomes increasingly apparent. By providing acomprehensive and accessible resource for students to explore the intricacies of this subject, we can help to democratize the world of Olympiad mathematics and make it more inclusive and accessible to a wider audience.One of the key advantages of an English textbook for Olympiad mathematics is the ability to reach a global audience. In many parts of the world, English has become the lingua franca of education and academic discourse, making it an essential tool for students and educators alike. By offering an Olympiad mathematics textbook in English, we can ensure that students from diverse linguistic backgrounds can engage with the material and develop their skills in this highly specialized field.Moreover, an English textbook can also serve as a valuable resource for educators who are tasked with teaching Olympiad mathematics. By providing a comprehensive and well-structured curriculum, teachers can more effectively plan their lessons, deliver engaging content, and support their students in their pursuit of mathematical excellence.The content of an Olympiad mathematics textbook in English should be carefully curated to ensure that it covers the full breadth and depth of the subject matter. This may include topics such as number theory, combinatorics, geometry, and algebra, as well as moreadvanced concepts such as graph theory, probability, and combinatorial optimization.Each chapter should be designed to build upon the previous one, with a clear progression of difficulty and complexity. The textbook should also incorporate a variety of practice problems and exercises, ranging from basic to highly challenging, to allow students to test their understanding and develop their problem-solving skills.In addition to the core content, the textbook should also include supplementary materials such as historical context, real-world applications, and links to additional resources. This will help to situate the mathematical concepts within a broader context and inspire students to explore the subject matter beyond the confines of the textbook.One of the key challenges in creating an Olympiad mathematics textbook in English is ensuring that the content is both rigorous and accessible. The language used should be clear and concise, with a focus on explaining complex ideas in a way that is easy to understand. The textbook should also include a glossary of mathematical terms and symbols to help students navigate the specialized vocabulary.Another important consideration is the visual presentation of thetextbook. The inclusion of high-quality illustrations, diagrams, and mathematical visualizations can greatly enhance the learning experience and help students to better grasp the underlying concepts.Overall, the development of an English textbook for Olympiad mathematics has the potential to significantly impact the field of mathematics education. By providing a comprehensive and accessible resource for students and educators alike, we can help to cultivate a new generation of mathematical thinkers and problem-solvers, who are equipped with the skills and knowledge necessary to excel in this highly competitive and rewarding discipline.。

历年中国参加国际数学奥林匹克竞赛选手详细去向第26届IMO

历年中国参加国际数学奥林匹克竞赛选手详细去向第26届IMO（1985年，芬兰赫尔辛基）吴思皓（男）上海向明中学确规定铜牌上海交通大学王锋（男）北京大学（根据yongcheng先生提供的信息修订）目前作企业软件第27届IMO（1986年，波兰华沙）李平立（男）天津南开中学金牌北京大学方为民（男）河南实验中学金牌北京大学张浩（男）上海大同中学金牌复旦大学荆秦（女）陕西西安八十五中银牌北京大学，现在美国哈佛大学任教林强（男）湖北黄冈中学铜牌中国科技大学第28届IMO（1987年，古巴哈瓦那）刘雄（男）湖南湘阴中学金牌南开大学滕峻（女）北京大学附中金牌北京大学林强（男）湖北黄冈中学银牌中国科技大学潘于刚（男）上海向明中学银牌北京大学何建勋（男）广东华南师范大学附中铜牌中国科技大学高峡（男）北京大学附中铜牌北京大学，现在北大任教第29届IMO（1988年，澳大利亚堪培拉）团体总分第二陈晞（男）上海复旦大学附中金牌复旦大学，美国密苏里大学，美国哈佛大学，现在加拿大Alberta大学数学系任教授韦国恒（男）湖北武汉武钢三中银牌北京大学查宇涵（男）南京十中银牌北京大学，在中科院数学所任副研究员邹钢（男）江苏镇江中学银牌北京大学王健梅（女）天津南开中学银牌北京大学何宏宇（男）以满分成绩获第29届国际数学奥林匹金牌，1993年破格列入美国数学家协会会员，1994年获博士学位，现任亚特兰大乔治大学教授、博士生导师，从事现代数学研究前沿的《李群》《微分几何》等方向的研究，在《李群》的研究上已有重大突破。

第30届IMO（1989年，原德意志联邦共和国布伦瑞克）团体总分第一罗华章（男）重庆水川中学金牌北京大学俞扬（男）吉林东北师范大学附中金牌吉林大学霍晓明（男）江西景德镇景光中学金牌中国科技大学唐若曦（男）四川成都九中银牌中国科技大学颜华菲（女）北京中国人民大学附中银牌北京大学本科，1997年获美国麻省理工博士，现任Texax A&M Uneversity 数学系教授，美国数学会常务理事会成员，Mathematical Reviews评论员。

顺德联谊总会何日东小学课外活动参赛获奖纪录表

2D 徐樂龢 2D 曾珞軒 2E 向洋 2E 梁丞禧 3D 廖培翔 4A 關尹禎 4E 譚丁元 4D 陳昕昱 5C 李卓謙 5D 郭正堯 5C 袁曉匡 6D 廖茜怡
2E 向洋 2D 郭涵銘 4B 盧柏熙 4A 江梓謙 5C 李卓謙
保良局
第四十三屆全港青年學藝比賽－ 2018香港小學數學精英選拔賽
個人
中童組西畫
銅獎
HKT
2016 eye 慈善圖書封面創作比賽
高小組
亞軍
good morning CLASS
J3-J12 全港兒童繪畫比賽 2017-2018
J9 西洋畫組 J10 西洋畫組 J11 西洋畫組
2018亞洲國際數學奧林匹克公開賽 (AIMO Open)晉級賽
小二
小三小四
小五
優數盃組委會
2018「優數盃」國際數學大賽(選拔賽)
小六小學二年級數學組小學四年級數學組小學五年級數學組
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6E 陳柏皓 6E 陳柏皓 6E 關頌軒 6E 陳柏皓
4C 陳冠霖
4D 陳亨 5B 莊瀚榛 5B 陳梓傑 5D 郭正堯 6A 林君濠 6E 關頌軒 6E 陳柏皓 6E 關頌軒 6E 陳嘉偉
6C 王健蘅 6E 陳柏皓 6A 林君濠 5D 鄭焯堯 6C 梁芊悠 5C 李卓謙 6A 陳俊希 6B 林耀升 6C 伍宜婷 6E 陳柏皓
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世界數學邀請賽組委會奧冠教育中心正統心算協會環亞太杯國際數學邀請賽組委會香港數學盃協會中華珠心算數學協會 Math Concept Education

从timss 透视香港的小学科学学习

从TIMSS透视香港的小学科学学习苏咏梅香港教育学院数社科技学系电邮︰wiso@.hk收稿日期︰二零零七年十一月十七日(于二零零八年七月三日再修定)内容o摘要o 1 香港在SISS 1983-1984的参与o 2 香港在TIMSS 1995的参与o 3 香港在TIMSS 2003的参与o 4 总结：香港小学科学学习的进程o参考文献摘要国际教育成就评价协会（The International Association for the Evaluation of Educational Achievement，简称 IEA）在过去的30年进行了多次的「国际数学及科学趋势研究」，其中在科学方面的调查包括在1970-1971年度进行的 First International Science Study （FISS）、在1983-1984年的Second International Science Study（SISS）、在1995年的Third International Mathematics and Science Study（TIMSS1995）、1999 年的 Third International Mathematics and Science Study–Repeat (TIMSS–R) ，以及在2003年的Trends in Mathematics and Science Study（TIMSS2003），而最新的2007年调查则在协调和策划中。

调查的主要对象为小学四年级学生（约10岁）及中学二年级学生（13岁）。

测验的主题架构是试题发展小组对各国调查该年段已教过和教学上会着重的重要主题，经由考虑各国对各主题内容教学涵盖情形而发展出来的，通过测试及问卷测量学生在数学及科学成绩的状况，从而了解影响学生科学与数学成就的不同因素。

调查的主要目的是提供学生在数学和科学成就的趋势，还有课程、教学、学习环境、家庭背景、以及教师等影响因素的相关资料，以了解各国在其教育改革或课程改革等改进措施的成效。

2004年香港物理奥林匹克竞赛

Hong Kong Physics Olympiad 20042004年香港物理奧林匹克競賽Written Examination筆試Jointly Organized byEducation and Manpower Bureau教育統籌局Hong Kong Physical Society香港物理協會The Hong Kong University of Science and Technology香港科技大學共同舉辦May 30, 20042004年5月30日The following symbols will be used throughout the examination paper unless otherwise specified:g– gravitational acceleration on Earth surface, 9.8 (m/s2)G – gravitation constant, 6.67 x 10-11 (N m2/kg2)e – charge of an electron, -1.6 x 10-19 (A s)ε0– electrostatic constant, 8.85 x 10-12 (A s)/(V m)m e– electron mass, 9.11 x 10-31 kgc – speed of light in vacuum, 3.0 x 108 m/s除非特別說明，本卷將使用下列符號：g–地球表面重力加速度, 9.8 (m/s2)G –重力常數, 6.67 x 10-11 (N m2/kg2)e –電子電荷, –1.6 x 10-19 (A s)ε0–靜電常數, 8.85 x 10-12 (A s)/(V m)m e–電子質量, 9.11 x 10-31 kgc –真空光速, 3.0 x 108 m/sThe following conditions will be applied unless otherwise specified:1)All objects are near Earth surface and the gravity is pointing downwards.2)Neglect air resistance.3)All speeds are much lower than the speed of light.除非特別說明，本卷將使用下列條件：1)所有物體都處於地球表面，重力向下；2)忽略空氣阻力；3)所有速度均遠低於光速.Multiple choice questions (2 points each. Select one answer in each question.)選擇題（每道題二分，每道題選擇一個答案）MC-1Which of the following provides the largest buoyancy to a totally submerged object?(A) shallow sea water (B) shallow fresh water (C) deep fresh water(D) oil (E) oil and fresh water mixture選擇題1下面哪一種情況會使完全浸沒物體受到最大浮力？(A) 淺海水（B）淺淡水（C）深淡水（D）油(E）油與淡水混合物MC-2A sinusoidal wave is traveling along a string. Any point on the string:(A)moves in the same direction as the wave(B)moves periodically with a different frequency from that of the wave(C)moves periodically with the same frequency as the wave(D)moves circularly with a different speed from that of the wave(E)moves circularly with the same speed as the wave選擇題2一正弦波在繩上傳播，繩上任意一點都_____.（A）沿著波傳播的方向運動（B）作與波不同頻率的周期性運動（C）作與波相同頻率的周期性運動（D）作與波不同速率的圓周運動（E）作與波相同速率的圓周運動MC-3The diagram shows four situations in which a source of sound S and a detector D are either moving or stationary. The arrows indicate the direction of motion. The speeds are all the same. Detector 3 is stationary. Rank the situations according to the frequency detected, from lowest to highest.S D S D S D S D1 2 3 4(A) 1, 2, 3, 4 (B) 4, 3, 2, 1 (C) 1, 3, 4, 2(D) 2, 1, 4, 3 (E) None of the above選擇題3下圖給出運動或靜止聲源S與探測器D的四種情況。

2018年香港数学竞赛决赛项目(个人)

FOR OFFICIAL USEScore foraccuracy⨯ Mult. factor for speed=Team No.+Bonus scoreTime1. 已知 x 2 = y 2 – 4y ，其中 x 及 y 為整數。

求 A = x + y 的最大值。

Given that x 2 = y2– 4y , wherex and y are integers. Determine the largest value of A = x + y . A = 2.已知 y=，且 B 是 y 的最小值，求 B 的值。

Given that y =, and B is the least value of y , determine the value of B .B =3.設 C 為正整數。

已知 144 + (B + 1)C 為平方數，求 C 的值。

Let C be a positive integer. Given that 144 + (B + 1)C is a perfect square, determine the value of C .C =4.已知 1x C x+=，求 D =331x x + 的值。

Given that 1x C x +=, determine the value of D =331x x +.D =FOR OFFICIAL USEScore foraccuracy⨯ Mult. factor for speed=Team No.+Bonus scoreTime1. 77782 – 22232 之值的所有數字之和是 a ，求 a 的值。

Determine the value of a , where a is the sum of all digits of 77782 – 22232 .a = 2.若 b 是乘積 a ⨯(a – 1)⨯(a – 2)⨯ ⨯2⨯1 的尾隨零的數量。

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7. Number of 3-digit numbers using 1 figure 3 times = 6C1 = 6 Number of 3-digit numbers using 1 figure 2 times = 6 × 6 C2 = 90 Number of 3-digit numbers using all different figures = 6P3 = 120 Total number of 3-digit numbers = 216 8.
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9. x2 - y2 + 5 = (x + y )(x - y ) + 5 As x - y = 5, and minimum value of x + y = 6 + 1 = 7 Least value of x2 - y2 + 5 = 7 × 5 + 5 = 40 10. By Cyclic Function property, the polynomial must have factors in cyclic. ie. either (x - y)(y - z)(z - x) together or only (x3 + y3 + z3 ). Assume x2(y - z ) + y2(z - x ) + z2 (x - y) = k (x - y)(y - z)(z - x) Equating coefficients of x2y, k = -1. ie. x2(y - z ) + y2(z - x ) + z2 (x - y) = -(x - y )(y - z )(z - x )
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3. As trailing zero will be formed when a "5" is multiplied to an even number. so, in 100, there are 20 multiples of 5 and 4 multiples of 25. There is 24 trailing zeros of 100! 4. n 200 < 5300 n 2 < 53 n 2 < 125 The largest integral value n that satisfies the inequality is 11. For
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HKMO 1996 Heat Event (Individual)
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HKMO 1996 Heat Event (Individual)
Hong Kong Mathematics Olympiad 1995-1996 Heat Event (individual) 1. If 4x - 3 = 8 x - 2 2 2(x - 3) = 2 3(x - 2) 2x - 6 = 3x - 6 x=0 1+x x2 + 1 1 f( )= 2 + x x x x2 + x + 1 = x2 1+x 2 1 = ( ) x x 1+x 2 1+x = ( ) +1 x x 3 6 3 f(x ) = x - x + 1
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¯àº¡¨¬¤£µ¥¦¡ n 200 < 5300 ªº³Ì¤j¾ãÅé n ¬O¦h¤Ö¡S What is the largest integral value n that satisfies the inequality n 200 < 5300 ? ¤TºØ-±-È $0.1¡A$3 ©M $5 ªº¶l²¼¦@¦³110 -Ó¡A³o²Õ¶l²¼¦@-È¿ú 100¡C ¨D $3 ¶l²¼ªº¼Æ¥Ø¡C A set of 110 stamps of the denominations of $0.1, $3, $5 worths $100 in total. Find the number of $3 stamps in the set of stamps.
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5. Let there be x $0.1-stamps, y $3-stamps and z $5-stamps. x + y + z = 110 0.1x + 3 y + 5 z = 100 x, y, z ≥ 0 From the Indefinite Simultaneous Equations, 29y + 49z = 890 y = 29 and z = 1 Substituting back, x = 80. ie. There are 80 $0.1-stamps, 29 $3-stamps and 1 $5-stamp. 6. Using the Family of Straight Lines, y = mx + 2 m + 2 m(x + 2) - (y - 2) = 0 represents a family of straight lines passing through the intersecting point of