## The Mathematical Olympiad Handbook: An Introduction to Problem Solving Based on the First 32 British Mathematical Olympiads 1965-1996Mathematical Olympiad competitions started in Hungary at the end of the nineteenth century, and are now held internationally. They bring together able secondary school pupils who attempt to solve problems which develop their mathematical skills. Olympiad problems are unpredictable and have noobvious starting point, and although they require only the skills learnt in ordinary school problems they can seem much harder. The Mathematical Olympiad Handbook introduces readers to these challenging problems and aims to convince them that Olympiads are not just for a select minority. The bookcontains problems from the first 32 British Mathematical Olympiad (BMO) papers 1965-96 and gives hints and outline solutions to each problem from 1975 onwards. An overview is given of the basic mathematical skills needed, and a list of books for further reading is provided. Working through theexercises provides a valuable source of extension and enrichment for all pupils and adults interested in mathematics. |

### Contents

Problems and problem solving | 1 |

A little useful mathematics | 7 |

Some books for your bookshelf | 41 |

Background to the problem papers | 53 |

32nd BMO 1996 | 89 |

31st BMO 1995 | 95 |

29th BMO 1993 | 103 |

28th BMO 1992 | 109 |

23rd BMO 1987A | 142 |

22nd BMO 1986 | 152 |

21st BMO 1985 | 162 |

20th BMO 1984 | 168 |

19th BMO 1983 | 177 |

17th BMO 1981 | 188 |

15th BMO 1979 | 196 |

14th BMO 1978 | 202 |

27th BMO 1991 | 115 |

26th BMO 1990 | 122 |

25th BMO 1988 | 128 |

24th BMO 1987B | 136 |

12th BMO 1976 | 213 |

11th BMO 1975 | 219 |

### Common terms and phrases

Alternatively angles answer approach base begin British Mathematical Olympiad calculate centre choose circle coefficients common complete Conclude condition cone contains cuts David defined denote diagram digits divides divisible draw easy equal equation exactly example Explain expression fact factor Finally fixed formula four function geometry give given Hence idea inequality involving least length lies look means meet multiple namely natural obtain original pairs parallel perfect square perpendicular plane polynomial positive integers possible problem proof prove quadrilateral question radius rational real numbers relation remainder result roots satisfy sequence side solution solve sphere square suggest Suppose things triangle triangle ABC true u₁ vertices write