QKi
-
UKMT
The United Kingdom
Mathematics Trust
Handbooks
Handbooks
Number Three
A Mathematical
Olympiad Primer
Geoff C Smith
The United Kingdom Mathematics Trust
A Mathematical Olympiad Primer
© 2007 United Kingdom Mathematics Trust
All rights reserved. No part of this publication may be reproduced or
transmitted in any form or by any means, electronic or mechanical,
including photocopy, recording, or any information storage and retrieval
system, without permission in writing from the publisher.
Published by The United Kingdom Mathematics Trust
Maths Challenges Office, School of Mathematics, University of Leeds,
Leeds, LS2 9JT, United Kingdom
http://www.ukmt.org.uk
First published 2008.
ISBN 978-1-906001-03-2
Printed in the UK for the UKMT by Cromwell Press, Trowbridge, Wiltshire.
Typographic design by Andrew Jobbings of Arbelos,
http://www.arbelos.co.uk
Typeset with lATpX.
The books published by the United Kingdom Mathematics Trust are grouped into series.
The EXCURSIONS IN Mathematics series consists of monographs which focus on a
particular topic of interest and investigate it in some detail, using a wide range of ideas and
techniques. They are aimed at high school students, undergraduates and others who are
prepared to pursue a subject in some depth, but do not require specialised knowledge.
1. The Backbone of Pascal's Triangle, Martin Griffiths
The HANDBOOKS series is aimed particularly at students at secondary school who are
interested in acquiring the knowledge and skills which are useful for tackling challenging
problems, such as those posed in the competitions administered by the UKMT and similar
organisations.
1. Plane Euclidean Geometry: Theory and Problems, A D Gardiner and C J Bradley
2. Introductions to Number Theory and Inequalities, C J Bradley
3. A Mathematical Olympiad Primer, Geoff C Smith
The PROBLEMS series consists of collections of high-quality and original problems of
Olympiad standard.
1. New Problems in Euclidean Geometry, David Monk
The YEARBOOKS series documents all the UKMT activities, including details of all the
challenge papers and solutions, lists of high scorers, accounts of the IMO and Olympiad
training camps, and other information about the Trust's work during each year.
Digitized by the Internet Archive
in 2023 with funding from
Kahle/Austin Foundation
https://archive.org/details/mathematicalolymOOOOgeof
For Maximilian Geoffovich.
Contents
9
Series Editor's Foreword
Preface
I
Theory
Introduction
What do you need to know? ........................................................
What is a proof?.................................................................................
The square root of 2 ........................................................................
Numbers..............................................................................................
How did proof start?...........................................................................
Proof and mathematics competitions............................................
2
4
7
in
11
12
Algebra
Inequalities...........................................................................................
Arithmetic-Geometric mean inequality........................................
Other inequalities.................................................................................
Polynomials...........................................................................................
13
13
14
15
16
Combinatorics
The Dirichlet principle........................................................................
Sequences..............................................................................................
Choosing.................................................................................................
Double counting.................................................................................
Colouring arguments........................................................................ ...
21
21
21
23
27
27
•••
Vlll
A Mathematical Olympiad Primer
4
Geometry
Triangle notation.................................................................................
Special triangles.....................................................................................
Similarity and congruence..................................................................
Trigonometry........................................................................................
Triangle area formulas........................................................................
Circle theorems.....................................................................................
29
30
32
32
34
36
37
5
Number Theory
Prime numbers.....................................................................................
Common divisors.................................................................................
Congruences ........................................................................................
Testing for divisibility........................................................................
43
44
46
47
49
II
Problems
III
Solutions
77
6
BMO11996-97 Solutions
79
7
BMO1 1997-98 Solutions
89
8
BMO1 1998-99 Solutions
99
9
BMO11999-2000 Solutions
109
10 BMO1 2000-01 Solutions
115
11 BMO1 2001-02 Solutions
123
12 BMO1 2002-03 Solutions
129
13 BMO1 2003-04 Solutions
137
14 BMO1 2004-05 Solutions
145
15 BMO1 2005-06 Solutions
153
16 BMO1 2006-07 Solutions
163
Bibliography
173
Series Editor's Foreword
This primer is part of a series whose aim is to help young mathematicians
prepare for competitions, such as the British Mathematical Olympiad, at
secondary school level. Like the two previous volumes in the Handbooks
series, on geometry, number theory and inequalities, this book provides
cheap and ready access to directly relevant material. All these books are
characterized by the large number of carefully constructed exercises for
the reader to attempt.
I hope that every secondary school will have these books in its library.
The prices have been set so low that many good students will wish to
purchase their own copies. Schools wishing to give out large numbers of
copies of these books as prizes should note that discounts may be negoti­
ated with the UKMT office.
London, UK
Gerry Leversha
About the author
Geoff Smith is a Senior Lecturer in Mathematics at the University of Bath.
He has been involved in mathematics enrichment since 1991 when he
helped to found the Royal Institution Mathematics Masterclasses in Bath and
Bristol (and later in Swindon). He became UK team leader at the Interna­
tional Mathematical Olympiad in 2002, and, at the time of going to press,
still holds that post. He was appointed chair of the British Mathematical
Olympiad Subtrust in 2006.
He edited the first two of UKMT's series of books, Plane Euclidean Ge­
ometry and Introductions to Number Theory and Inequalities. Together with
X
A Mathematical Olympiad Primer
Ceri Fiddes, he has made a number of educational DVDs for Highperception Ltd which are targetted at strong young mathematicians and their
teachers. He has written two texts for Springer:- Introduction to University
Mathematics: Algebra and Analysis is to help students make the transition
to university mathematics, and Topics in Group Theory, written with Olga
Tabachnikova, is for a more advanced audience.
He has had 10 successful PhD students, and does research in group
theory and lately also classical and projective geometry. He works out of
his main areas from time to time, and so also has research publications in
computer op-code design, DNA sequencing and snail venom.
He is married with two children and lives in Bath.
Preface
In writing this book for the publishing arm of the United Kingdom Math­
ematics Trust, I hope that this inexpensive and accessible text will enable
many more enthusiastic students to enter the world of senior secondary
school mathematics competitions with confidence. If only in a chronolog­
ical sense, this book is a sequel to Gardiner's The Mathematical Olympiad
Handbook published by Oxford University Press [9]. The styles of these
two books are not the same.
I must acknowledge the help of the geometer and illustrator Christo­
pher J. Bradley for producing large numbers of helpful draft diagrams.
Two other draft diagrams were also kindly provided by Bill Richardson.
The many problem setters and the Problem Setting Committee of the British
Mathematical Olympiad also deserve thanks for creating such interest­
ing papers, as do the many BMOS and UKMT volunteers and profession­
als who help in various aspects of BMO organization. I have sometimes
drawn for inspiration upon the Solutions Booklets published by UKMT. The
authors of those booklets have done an excellent job. This text does not re­
place the booklets, which often contain more solutions than are presented
here, but give less commentary.
Bath, UK
Geoff Smith
Part I
Theory
Chapter 1
Introduction
Mathematics competitions for younger secondary students do not usually
involve the notion of proof. Rather students are expected to use mental
agility to count, calculate and spot patterns. Of course there are limits to
the amount of ingenuity that can be packed into questions of this type, and
at some stage the focus must shift to reasoning. In the UK, this definitely
happens in the first round of the British Mathematical Olympiad, widely
known as BMO1. The vast majority of candidates sitting this exam are
in the age range 15-18. However, there is also a multi-stage competition
called IMOK which is aimed at slightly more junior students, and at that
level the setters are already trying to foster increasing mathematical so­
phistication.
The organization which administers the BMO forms part of a larger ed­
ucational charity called the United Kingdom Mathematics Trust. This body
operates outside the public examinations system. In addition to running
many national mathematics competitions at various levels, UKMT en­
gages in other activities aimed at promoting excellent mathematics educa­
tion at school level.
In the hope that this text may have a significant readership outside the
United Kingdom, this work has not been written solely with the needs
of BMO candidates in mind. School level mathematics competitions now
exist in almost all economically developed countries, and the points made
in this text are of wide applicability.
A Mathematical Olympiad Primer
4
What do you need to know?
If you are approaching BMO1 or a similar competition, there is a tempta­
tion to learn some advanced theory as a sturdy crutch on which to lean.
In some countries this may indeed be useful, but in the UK it is not. How­
ever, learning how to write down a correct proof in geometry is essential.
More generally, you have to learn how to write out a logical argument.
You must distinguish between a statement of the form “A implies B" and
its converse statement which is "B implies A”. One can be true without the
other being true. If you are asked to show that “A is true if, and only if, B
is true", then you have two jobs on your hands. You must show that if A
is true, then B is true. Also you must show that if B is true, then A is true.
You must make sure that you know how to perform an argument by
induction. Let us give an example. Let P(n) denote the statement that the
sum of the first n positive integers is n (n + 1) / 2. We are interested in prov­
ing that this is a true statement. However, P(n) is not really a single state­
ment. Rather it is a collection of infinitely many statements P(l), P(2),
P(3) and so on. This point of view allows the following strategy. First
prove that P(l) is a true statement. Since P(l) asserts that 1 = (1 x 2)/2,
we know that it is true. Next show that if P(r) is true, then P(r + 1) is true
(here r is an arbitrary positive integer). In our case, when we suppose that
P(r) is true, it means that we accept that
1+2 + 3 + • • • + r =
Add r + 1 to each side to obtain
2(r + 1)
and so P(r + 1) is true.
The principle of mathematical induction allows us to take the facts that (a)
P(1) is true and (b) if P(r) is true, then P(r + 1) is true (for every positive
integer r), and conclude that P(n) is true for every positive integer n.
If you are intending to engage in a regional competition such as the
Balkan Mathematical Olympiad, the second round of the British Mathe­
matical Olympiad or the International Mathematical Olympiad itself, it is
Chapter 1: Introduction
5
true that there is a body of useful knowledge, an unofficial syllabus, which
it would be wise to assimilate. Indeed, it is wise to learn a little beyond
the syllabus in order to be sure that the papers are inside your "comfort
zone". Engel's text [8] is a wonderful introduction to more advanced prob­
lem solving and contains some marvellous questions (Can three dimensional
space be expressed as a union of non-overlapping circles? is my favourite). Past
IMO papers and shortlists are available on the internet via Art of Prob­
lem Solving and elsewhere, though it is particularly convenient to have so
many past problems (and their solutions) available in Reiman's book [14],
The current UK IMO deputy leader Ceri Fiddes presents discussion on,
and solutions of, the problems of IMOs 2005, 2006 and 2007 [10] on DVDs.
It is a common experience for a candidate to have a wretched time the
first time that he or she sits BMO1, and to do much better second time
around. This is because, in common with all other human activities, one
gets better with practice. The ability to solve BMO1 problems at speed is
not a form of magic; rather it is usually the consequence of having done
dozens (or possibly hundreds) of similar questions before. The internet is
awash with past examination papers of national and regional mathematics
competitions. Search under AoPS (for Art of Problem Solving) to find many
past papers and a lot of discussions concerning their solutions.
Unlike many national bodies, UKMT strongly discourages the posting
on the internet of solutions to their examinations in an organized way. The
purpose of this policy is to give students access to some problems on the
internet where the solutions are not immediately accessible.
This book is for the strong willed. You should never look at a solution
until you have struggled with the problem for at least an hour or two. At
that stage, if you want help, you can go to the part of the book where the
problem is solved, and (keeping the solution covered with hand or paper)
look to see if the discussion preceding the solution gives you enough of a
push to enable you to solve it yourself.
You may find that you solve the problem in a way very different from
the method suggested in this book. This is particularly likely to happen
when doing geometry questions. I would be very interested to learn of
any solutions which are significantly better (shorter, more stylish) than
the ones presented here. When writing this text, I had access to the BMO
booklets which give solutions. Sometimes I was able to find solutions
which were improvements on those already on offer, and sometimes not.
The competition BMO1 is definitely beyond the threshold between cal­
culation and reasoning. It assumes a bare minimum of knowledge about
6
A Mathematical Olympiad Primer
the traditional areas (number theory, geometry, combinatorics and alge­
bra) but the best BMO1 questions require insight and wit to answer them
correctly. It is true that some BMO1 questions ask you to perform a calcu­
lation, but it is never the answer which matters. Rather it is the method
which the examiners want to see. Proof plays a very limited role in sec­
ondary school mathematics in the UK, but it is at the heart of competitions
at olympiad level.
What is a proof?
A mathematical proof is a completely convincing logical argument which
underpins, and is the guarantee of, the truth of a mathematical statement.
If we can find a proof for a mathematical statement, then experience in­
dicates that it actually has lots of proofs, and in terms of establishing its
truth, they are all equally good. However, as far as humans are concerned,
it seems that some proofs are better than others. A proof with appeal will
provoke praise from mathematicians. Terms such as beautiful, elegant, neat,
sweet, cute are used as marks of appreciation, and of course there are in­
formal terms of abuse for unattractive proofs including ugly, disgusting,
boring and revolting.
What are the attributes which makes a proof attractive? Well, brevity
helps, and the use of ideas rather than calculation also has appeal. Ide­
ally a proof should not only establish the truth of something, but also
should set it in context, making links with other mathematical truths. It
also seems to help if the proof contains an element of surprise, striking
out along an unexpected path. A dramatic denouement is always welcome.
This happens when you are reading the argument and it all seems to be
going well but you do not see the end coming, until an interesting trick
brings matters suddenly to a head.
There was a celebrated Hungarian mathematician called Paul Erdos
who used to speak of a proof from The Book. Here the conceit is that the
Almighty keeps a book of mathematical truths and their proofs, and of
course has the "perfect" proof for each result.
Chapter 1: Introduction
7
The square root of 2
In ancient Greece, about 550 BC, a school of mathematicians associated
with Pythagoras flourished. The theorem of Pythagoras asserts that the
square on the hypotenuse of a right-angled triangle is equal to the sum of
the squares on the other two sides. This result has achieved iconic status.
The statement means that the area of the square on the hypotenuse is equal
to the sum of the areas of the squares on the other two sides. In algebraic
terms, a2 + b2 = c2.
Figure 1.1: The theorem of Pythagoras
If you have two line segments of length 1 with a common end and
which meet at right angles, then by this Pythagorean result, the distance
between their other ends is \/2. To see this, suppose that the length is x,
then Pythagoras's theorem tells us that
so x2 = 2. The positive solution to this equation is called \/2. Notice that
1.42 = 1.96 is too small and 1.52 = 2.25 is too big, so \/2 is a bit more than
1.4. Now 1.412 = 1.9881 which is quite close to 2. This means that
141
100
19881
10000
A Mathematical Olympiad Primer
is ( !<> .<• io ’ It is reasonable to ask if you can find positive whole numbers
in .aid ii so th.il
Indeed, I he school ol Pythagoras asked this very question 2,500 years ago,
.ind did not gel the answer they were expecting. They discovered that no
sin h hi and ii exist. At a time when numbers and religion were mixed
together, this was a serious matter. We are not sure which proof was the
lirsl lo he dis( overed, but we will give several arguments, all of which are
valid mathematical proofs.
I’KihH m (. ON I KA DICTION
Suppose that m,n are positive whole num-
hers such that
(1.1)
Il eithei m or ii is even, we may cancel a factor of 2 from top and bottom
id the ratio m/n, and repeat this process until we get stuck. This allows us
lo assume that at least one of tn and n is odd.
Now multiply both sides of Equation (1.1) by n2. We get
m2 = 2n2.
I he square ol an even number is even, and the square of an odd number
is odd Now hi is clearly even, so tn is even. We know that at least one of
nt and n is o«.ld, so ii is odd.
Since in is even we have in = 2k for some positive whole number k.
Pul Hus into Equation (1.2) to obtain 4T = 2n2 and, cancelling a factor of
\\ c gel
ii2 = 2k2.
(1-3)
Now A is an even number, so n is even. However, we already know that
•i is odd. Do not panic. Mathematics has not fallen to bits. All our de­
ductions wen' correct, so our initial assumption must have been incorrect.
Therefore there are no positive integers tn and n such that
I'KiC'i id I i| set \ 1 We will now take the previous argument by contra­
diction, and recast it in another way. We begin as before.
Chapter 1: Introduction
Suppose that m, n are positive whole numbers such that
\n/
This time we do not have to do any initial cancelling. Now
zn2 = 2n2
so n < m and m is even, so m = 2k for a positive integer k. Therefore
4k2 = 2n2 so
n2 = 2k2.
Now n2 = 2k2 so k < n. Equation (1.5) has exactly the same shape as
Equation (1.4). We can exploit this. Let
= m, x3 = n so
/y \2
( - ) = 2
\*2/
and Xi > *2- We have discovered a positive integer *3 = k < x3 so that
We can apply this argument again and again, and obtain a sequence of
positive integers
where, for each z,
Moreover,
xi > x2 > x3
We have obtained an infinite chain of positive integers which get strictly
smaller as you proceed along the chain. This is absurd (and contradicts
the correct assertion that there are only finitely many positive integers less
than any given positive integer). Once again, we can conclude that there
are no positive integers m and n such that
A Mathematical Olympiad Primer
10
I'KOOF IW I'ARllA Suppose that m.n are positive whole numbers such
that
(1.6)
We can write m
2' m where ,v is a whole number and is not negative,
and m' is odd. This is because we can pull out factors of 2 from m until
we get stuck. Similarly n
2hi where y is a whole number and is not
negative and n is an odd positive whole number. Therefore
and so
(1.7)
Now 2 '
2
' since 2,v is even but 2i/4 I is odd. Therefore one of these
powers divides the other giving a result which is even (we do not know
which wax the division happens but that does not matter). Effecting this
division we learn that either in ‘ or n ~ is even. This is absurd because
both tn' and n are odd.
n
We have made a detailed study of \ 2. The reader should try to dex elop similar arguments concerning \ 3 and \ 20. W hat happens when
\ ou trv to use such methods to show that \ 4 is irrational’?
Numbers
We begin a taxonomy (a classification) of numbers. The positive whole
numbers I 2 3 4 .. are called the natural numbers, and the set which con­
sists ot all of them is w ritten \. The whole numbers (positive, negative
and zero) are called integers. and the set of these is written Z. Now 37 is
both a natural number and an integer, so we can write 37 E IN and 37 € Z.
\lso 2 t .’but 2 e IN. Notice that IN C Z, because the natural numbers
form a subset of the integers. Those numbers corresponding to points on
the number line are called real numbers, and the set of all such numbers is
written 1R.
Some real numbers can be written as ratios of integers, but thanks to
Tx* thagoras.
some cannot. Those which can be written as ratios of integers
v
cJ
I
It is not
Chapter 1: Introduction
11
are called rational numbers. The set of all these is Q. We therefore have a
hierarchy
NcZcQcR.
We sometimes become a little sloppy with language, and refer to the el­
ements of N as naturals, elements of Q as rationals, and elements of 1R
as reals. This notation is all perfectly standard, in use throughout planet
Earth, except that, in the face of prevailing orthodoxy, there is a French
heresy that 0 E N.
How did proof start?
Proof seems to have begun in ancient Greece. A book by Euclid called The
Elements attempted to organize the geometry known at the time into a de­
ductive system. From a modern point of view the foundations are sloppy,
and the use of geometric diagrams poses all sorts of difficulties since the
given pictures do not always cover all possible configurations. However,
one should not be churlish. This book was pioneering the technique of
proof, the method of deducing deeper truths from simpler ones. The Ele­
ments also includes a theory of ratios of lengths, all carefully worked out
with proofs and so on, and also some number theory, but somehow the
notion of proof has become tied in the popular imagination to geometry,
and not to the theory of proportion.
For a couple of thousand years, those children in the Western world
lucky enough to receive an education would get their geometry from Eu­
clid's book, and the only proofs they would see were geometric. These
days all of pure mathematics is organized by proofs. The more abstract
mathematics gets, the more you need to use proofs, otherwise you lose
all track of what you are doing. Proof pervades the subject, and is by no
means confined to geometry.
We wrote earlier that a mathematical proof is a "completely convincing
logical argument which underpins, and is the guarantee of, the truth of a
mathematical statement". There is a dirty little secret here. One reader's
convincing logical argument is another person's gibberish. A proof must
be tailored to the reader. If you are writing in a learned journal about
some recondite fragment of mathematics which is of interest only to a
small community of experts, then it is fair to assume that the readers pos­
sess a massive amount of knowledge about the background of the subject,
12
A Mathematical Olympiad Primer
and that they will happily accept a short argument which alludes to var­
ious facts which, to them, are well known and well understood. If, on
the other hand, you are writing a proof of the very same facts for student
readers, you have to give much more detailed arguments, and take spe­
cial care to clarify difficult points. Thus, when you are writing a proof,
you have to judge the level of formality and detail which is appropriate
for the occasion. Our context is mathematics competitions, and we shall
give arguments which are appropriate for that environment.
Proof and mathematics competitions
A good mathematics competition problem will test the student in many
ways. In the case of a hard International Mathematical Olympiad problem,
the candidate may be required to produce a sequence of ingenious ideas
which can be wrapped together to form a proof. In easier problems, the
examiners sometimes leave traps for the unwary student — things which
are not hard, but are easy to overlook. Such problems are testing technical
proficiency rather than extreme ingenuity.
There is a widespread delusion that mathematics is a magical talent,
that some people have a special gift for it, but most people cannot do
mathematics well, no matter how hard they try. This is nonsense, and
is an excuse for idle students and teachers not to engage. It is of course
true that some people find mathematics (music, dancing, skiing etc) easier
than others. Except in the case of people with a physical or intellectual im­
pairment, such easy competence seems to be closely related to the amount
of interest the person has in the activity in question. If you think about
algebra in your every free moment, you will find that you become rather
good at it. This is not to deny that some people have extraordinary tal­
ents. However, the strongest mathematicians I have met all clearly enjoy
thinking.
There are several types of problem which fall loosely into this category.
Problems involving polynomials and polynomial equations are certainly
algebra questions. For our purposes, problems which involve inequalities
are often regarded as algebra. This is less reasonable, since some inequal­
ities are actually translations of geometric statements into algebra.
For example, Euler proved that, given any triangle, IO2 -R2 2Rr
R(R — 2r). Here R is the circumradius (the radius of the circumcircle), r the
inradius (the radius of the incircle or inscribed circle), O the circumcentre
and 1 the incentre (see Figure 2.1 on the following page).
A square is never negative, so R(R — 2r) > 0 and we obtain Euler's
inequality 2r < R.
Inequalities
If you are asked to show that x>y, you can equally well try to show that
x — y > 0. Thus any comparison of real numbers amounts to showing
that some quantity is positive or zero. The ugly word "non-negative" is
sometimes used to mean this.
A square of a real number is never negative. Indeed, if x,y,z are
real numbers, then x2 + y2 4- z2
0. More generally, if x-\,X2, ■ ■ ■ ,xn are
?al numbers, then x2 + x2 + ■ ■ ■ + x2 > 0. We can go even further. If
1, A2,..., A„ are positive real numbers, then
2
A]X 1
A Mathematical Olympiad Primer
14
Figure 2.1: O and R; I and r
If you have an inequality a < b, then you can multiply it by a positive
quantity c and get another inequality ca < cb. If, however, you multiply
by a negative quantity d, then the inequality is reversed. You find that
db < da.
Inequalities may be added provided that they are the same way round.
Thus, if a < b and c < d, then a + c < b + d. However, subtraction of
inequalities will not always give a true inequality. For example 2 < 5 and
1 < 6. If you subtracted these you would get 1 < — 1, which is nonsense.
Arithmetic-Geometric mean inequality
Suppose that x, y are non-negative real numbers, then x = a2 and y = b2
for suitable non-negative real numbers a and b. Now (a — b)2 > 0 so
or put in terms of x and y,
We have established the so-called AM-GM inequality. The left-hand
side is the GM or geometric mean of x and y; the right-hand side is the AM
or arithmetic mean of x and y.
Chapter 2: Algebra
15
The natural generalization to n non-negative variables is also true, and
is also called the AM-GM inequality. It states that
Other useful means you may wish to look up are the quadratic mean
(also called the root mean square) and the harmonic mean.
Other inequalities
Any inequality which appears in BMO1 is likely to submit to either a sum
of squares argument or the AM-GM inequality. However, there are lots
of other famous algebraic and geometric inequalities, and students think­
ing about addressing BMO2 or the IMO will no doubt want to add these
techniques to their repertoire. Christopher J. Bradley's Number Theory and
Inequalities [2] is an excellent place to look.
Perhaps the most widely used inequality which has not been men­
tioned already is the Cauchy-Schwarz inequality. For those who know
about the scalar or dot product of vectors, this has the geometric meaning
that the modulus of the dot product of two vectors is at most the prod­
uct of their lengths. Equality is achieved precisely when one vector is a
multiple of the other.
We now give the algebraic form of the Cauchy-Schwarz inequality. You
need know nothing about vectors to understand the statement of this re­
sult.
Suppose that alz a2, ■ ■ ■, an and blz b2,..., bn are real numbers. Then
a^b^ + a2b2 + ■ ■ ■ T Qnbn
with equality if, and only if, there is a real number A such that either
(, a2,... / ayi) — [Ab], Ab2/..., Ab^)
or
(fq z b2,..., bfi) — (Aai, Aa2,... ? Aa^i
A A (athematical Olympiad Primer
Polynomials
1'olx nomial factorizations are often useful. Perhaps the most famous is the
s.m.oes. Thus we have a polynomial identity
x2 - y2 = (x + y)(x - y).
F\ ex aluating this poh nomial identity using specific integer values for x
and \ ou can tactori.e some numbers quite easily. For example, here is a
challenge: factorize
in vour head. It is actually very easy.
goo
Kipp
(30 + l)(30 - 1) =31 x 29.
In some countries it is or has been standard usage to say that an equalitx between poh nomials is an :.;ez:hfu and to use three horizontal lines
rather than two to indicate this tact. From a mathematical point of view,
this is pointless I lowever. it is also harmless, and perhaps it has pedagog­
ical advantages.
Notice that if i i are integers (and not just pohnomial variables), then
Ki ■ r
^.1 i
2.i is even. Thus the integers x + y and x — y are either
both ex en or both odd Since i and i, have temporarily become integers it
would be inappropriate to use the symbol
For even positive integer *:
2 we have a polynomial identity
= (l-L
X’: 1
It •• is odd x ou get a factorization of x’’ — y': bv replacing v by —y. Howex er such a trick does not work when r. is even. For example
and replacing y by
y we get a polynomial identity
x — y = (x + y)(jr — x“y + xy4 — y)
so \\ e Vget a second factorization of v4 v4. In fact both of these factorizations can be obtained from the single factorization
x4 - y4 = (x - y)(x + y)(x2 + y2).
th allowing
v. i and M. toV ’ take integer \ alues we obtain factorizations of
integers.
Chapter 2: Algebra
17
The degree of a polynomial
P(X)
+ • • • + 4?0
is n provided that an
0. This gives a definition of degree for any non­
zero polynomial. For reasons which will become clear shortly, we define
the degree of the zero polynomial to be -co where this is pronounced
minus infinity. You should not think too deeply about this; it is simply a
formal convenience.
You may regard the following statements as obvious, but in fact you
could w’rite out proofs.
Suppose that /(X) and g(X) are polynomials, and that a, b are con­
stants, then
deg/(X)<g-(X) = deg/(X) + degg(X),
degaf(X) — deg /(X) provided that a
and
0
deg(fl/(X) + hg(X)) < max{deg/(X),degg(X)}.
Notice that if /(X) and g(X) are non-zero polynomials of the same
degree, then a and b may be chosen so that «/(X) + bg(X) has smaller
degree than f(X) and g(X), but af(x) + bg(x)
0.
A careful examination of how these statements apply when /(X) or
#(X) is the zero polynomial reveal that we must endow the symbol — oo
with certain properties.
1. — co < n for any non-negative integer n;
2. — co + n = n —co
co for any non-negative integer n;
3. — co -|----co = — oo.
These properties are in tune with intuition about how an infinitely large
negative number should behave, which inspires the notation —co. How­
ever, it is only notation, and everything would work just the same if we
used the word banana rather than the symbol —co, provided we endowed
the word with the correct properties.
Notice how we have chosen notation to make mathematical statements
easier. If we had (foolishly) decided that one should not assign a degree to
the zero polynomial, or perhaps more likely, argued that the zero polyno­
mial is a constant polynomial, and so should have degree 0 like all other
A Mathematical Olympiad Primer
18
constant polynomials, then we would be in big trouble. A statement such
as
deg/(X)g(X) = deg/(X) + degg(X)
would have to carry a health warning of the form "provided that neither
/(X) nor g(X) is the zero polynomial", and every time that we used it, we
would have to verify that /(X) and g(X) were non-zero. Life is too short
for such nonsense.
There is a very similar situation which arises when deciding whether
or not 1 should be a prime number. If 1 were prime, then 2 x 1 = 2 x 1 x 1,
so you would not be able to say "every natural number can be factorized
into prime numbers in a unique way". We definitely do want to make this
statement, so we are forced to insist that 1 is not a prime number (though
once upon a time, it was). I can feel a quiz question coming on: what do
1, Pluto, Piltdown Man and the Empire State Building have in common?
Polynomial division
Suppose that /(X) and g(X) are polynomials, and that g(X) is not the
zero polynomial. It may or may not be the case that g(X) divides /(X). In
other words, there may or may not be a polynomial t/(X) so that /(X) =
(f(X)g(X). If there is, then we say that
/(X)
,(X) =
g(X)
is the quotient of/(X) by g(X).
Even if g(X) does not divide into /(X) exactly, one can do a sort of
approximate division, with an error term (the remainder) which is "small"
compared with the polynomial by which you are attempting to divide. In
this context, we mean of smaller degree when we say "small". There are
polynomials t?(X) and r(X) with degr(X) < degg(X) such that
/(X) = <;(X)g(X) + r(X).
We can prove it by induction on deg/(X). We have an unusual base
case: if deg/(X) < degg(X), then let r(X) = /(X) and choose q(X)
to be the zero polynomial. If deg/(X) > degg(X), let m = deg/(X) —
degg(X). By choosing an appropriate constant c we may arrange that
/(X) — cX”'g(X) has smaller degree than does / (by subtracting away the
term of highest degree).
Chapter 2: Algebra
19
By induction there are polynomials <?i(X), ri(X) with degri(X) <
degg(X) and
/(X)-cXwg(X)^<71(X)g(X) + r1(X)
and so
/(X) = (^1(X)+cX"I)g(X) + r1(X).
Now put q(X) = t/i(X) + cX'” and r(X) = ri(X) to complete the induc­
tive step.
In common with many inductive arguments, this can be turned into
a procedure. In this case we obtain the process of long division which is
used for dividing one polynomial by another with a small remainder.
An important special case is when the polynomial that you are divid­
ing by has degree 1. By multiplying through by a harmless non-zero con­
stant, we may assume that the dividing polynomial g(X) is X — a for some
constant a. The remainder must have degree less than 1, and so must be a
constant c. Therefore there is a polynomial *?(X) such that
/(X) = <?(X)(X-fl)+c.
Evaluating both sides at a we learn that f(a) = c.
Thus we have proved the remainder theorem:
Remainder Theorem The remainder when X — a is divided into f(X) is f(af
The factor theorem is an important special case of the remainder theo­
rem.
Factor Theorem X — a divides /(X) if, and only if, f(a) = 0.
More is true. A non-zero polynomial of degree n has at most n different
roots. To see this, suppose that a^, a2, ■.., at are different roots of /(X). In
other words /(*?/) = 0 for i = 1,..., t. Then /(X) = (X — )/i (X) and
a2,... ,at are all roots of /i(X). Carrying on in the same way (replacing
f(X) by f-[ (X) and so on) we eventually find that
/(X) = (X-«1)(X-fl2)---(X-«f)/f+i(X).
Therefore t < deg / and we have learned the important result that a non­
zero polynomial /(X) can have at most deg/(X) different roots. This is
an extremely powerful result which is often extremely useful.
20
A Mathematical Olympiad Primer
Note the caution in the previous paragraph. When working with the
integers, 6 and 8 both divide 24, but it does not follow that 6x8 divides
24. Just because each X — a, divides /(X), you cannot immediately de­
duce that their product divides /(X). An argument, such as the one we
supplied, is required.
Chapter 3
Combinatorics
Combinatorics, at least at the level of BMO1, is the study of counting and
choosing. One can often get away with using scrupulous accounting and
the careful management of great piles of data to answer some BMO1 ques­
tions correctly. It is admirable to be able to do this, but if there is an easier
way, it should be used.
The Dirichlet principle
This is also called the pigeon-hole principle, or the shoe-box principle. The
idea is that if you have N pigeon-holes, and you place N + 1 letters (or,
for the less squeamish, pigeons) in the pigeon-holes, then one pigeon-hole
must contain at least 2 letters (or pigeons). An obvious extension of the
principle is that if you instead put kN + 1 letters in the N pigeon-holes,
then some pigeon-hole must contain at least k + 1 letters. Like many won­
derful ideas, this sounds deceptively simple. In fact one can sometimes
solve seemingly difficult problems by a cunning construction of letters and
pigeon-holes.
Sequences
The theory of sequences with integer entries hovers somewhere between
number theory and algebra, but we will address it here. A finite sequence
is a just a list of finite length where the members of the list are (usually)
\ Mathematical (Ah niptad /'rimer
numbers. An infinite sequeiu e starts but does not finish and its terms can
be indexed bx the positive integers (and sometimes bx the non negative
integers) A sequence \\ here the n th term is
is sometimes written p;
Here are some examples.
I. 1 2.3 4.5 ... is possible the sequence pi.A where u-.
n for every n.
We cannot be sure tliat this rule determines all tei ms ot the sequence
because we only know the first five terms.
2. 1 I 1.1.1 ... is possible the sequence (J’.A where b..
11
11
tor ex erx< n.
is possibh (lit' sequence (c.A where c.:
4. 1 1 2.3 .5.8. 13 21 ... is possibh- the rtl\)ii<hvt sct;u<-’:ce \J’.A where
l\
/A
I and tor ft ■ we detine l\> bx I',.
। I I.. >
W hen thinking about a sequence (id there ate x arious attributes it
might hax e which are worth considering It could be an .••ic’e.isifiy scquc’.’ce.
which means that i,. ,t ■ i.. tor ex erx m It could be a aw.-x.'i, :r:crms.-H\'
sequence which means that
( ' v. tor ex erv it khere are similar
notions ot a Jo rc-iisiny oi ..•<-.;x.’i, J<\ ■c./snry sequence It might be .A';.
in the sense that there is a number e such that
< c tor ex erx% r:.
Similarh it might be h»int./fi/ ••e.'e.c It it is bounded abox e and below
then it is a botlihled sequence It could be an altc
-:g sequence so that
consecutix e terms hax e opposite signs It could be a - a.: c sequence so
that there is a positive integer ;• w ith the propertx that v.
forexerx
n. It might be a constant sequence where ex erx term is the same
W hen a sequence is gix en bx a simple iterative formula such as the
Fibonacci sequence then it is reasonable to hope that there max be a pleas
ant formula tor the a th term ot the sequence Indeed tor the Fibonacci
sequence
where a and are the distinct roots ot the polynomial \
i
1 phat is
the distinct solutions ot the equation v
\
1
0).
There is a fairly standard olx mpiad problem format which runs like
this. A sequence yi.A is detmed using some initial data phe x alue ot
or
possibh the values ot .q and .; 3 and then some iteratixe (mductix e) deti
nition of a>t in terms ot earlier terms ot the sequence Aon are then asked
Chapter 3: Combinatorics
23
to prove something about the sequence. A very effective approach is to
consider the first few terms of the sequence in order to make an intelligent
guess about the overall pattern. It may be that you need to find a formula
for the n-th term, but maybe not. If you are asked about some number
theoretic property of <22020 (substitute the current year for 2020 probably),
it may be that you can work with the sequence modulo N for a suitable
(possibly prime) number N. This will have the advantage that the terms of
the sequence can be deemed to range over a finite set of remainders. If the
recurrence defines the next term by means of the previous r values, even­
tually some set of r consecutive values will repeat (by the Dirichlet prin­
ciple), and so the sequence will eventually become periodic and therefore
easy to understand.
In case this is not clear to you, we will spell it out. We take our original
sequence (a,) of integers, and replace it by the corresponding sequence
(■ ) of remainders modulo N. Each term bj of the sequence must take one
of N values. The number of different sequences of length r consisting of
remainders modulo N is Nr. These label the pigeon-holes, and there are
Nr of them. Now consider the first r terms of our actual sequence (modulo
N). This must correspond to one of the labels for the pigeon-holes. We
(mentally) pop the number 1 (the index of the first term of our consecutive
string) in that pigeon-hole. Next consider the sequence of consecutive
terms of our sequence which starts with the second term and finishes with
the r + 1-st term. We put 2 into the pigeon-hole corresponding to that
string of consecutive terms, and we carry on in similar fashion. After we
have done this Nr + 1 times, then by the Dirichlet principle, at least one
pigeon-hole contains at least two positive integers. Therefore there are
two occurrences of the same consecutive string of r terms (modulo N). Say
that the first string begins at m and the second at n. Therefore bm = bn,
bm+i = bn+i,... and bm+r_i = bn+r^. Now, providing that the recurrence
is defined nicely, there is hope that, by induction on t, we havebm—n+t — bf
for all integers t > n. We have arranged that the result is true when t =
n,n + 1,.. .,n + r - 1 and after that the recurrence formula may provide
the induction step.
Choosing
Suppose that you have a collection of 30 students in a class, and you want
to form a team of 4 to send to the school principal to complain that the
24
A Mathematical Olympiad Primer
school mathematics lessons are insufficiently stimulating. In how many
ways can you do this?
Before we attempt to give an answer, we must first try to understand
the question. It matters that the people are different. This may seem ob­
vious, but if instead you were dealing with a collection of 30 cards, half
with the letter A written on, and the other with the letter B written on, but
otherwise identical, then the question is different and the answer is much
easier. There are five ways to form a collection of four cards, because you
must pick 0,1, 2, 3 or 4 cards with an A written on, and the cards with a B
written on must form the rest of your collection of four.
Next note that we have said "form a team of four". Actually, in En­
glish, we usually say "pick" or "select" a team of four, and such a choice
of words leads to further complications. If you are picking a team of four,
perhaps the obvious thing to do is to point a finger at four different people
in turn, listing their names. Now, is it a different selection if you pick the
same four people, but you do the choosing in a different order? Since the
method of selection was not specified, then there is no reason to suppose
that it has to be done by a sequence of single choices, and so the order
of choice cannot be important. However, there are circumstances under
which this matters, so be careful.
Here is another thing to worry about. Can the same person be selected
more than once to be in the team? You might think that this is silly, but
consider the following variation of the problem. How many ways can you
write down four numbers, each of which is an integer in the range 1 to 30?
Leave aside the problem as to whether or not it matters in which order
you write the numbers down, and focus on the fact that writing 3, 3, 3,
and 4 is a way of writing four numbers, each in the range 1 to 30.
Of course the context of the question often tells you what you are sup­
posed to count. The current dismal state of person replication technology
is such that we can safely assume that each person can only be se'ected
once for the team to go to see the principal.
So, perhaps we understand the question. The people are distinguish­
able and each person can be in the team at most once. Moreover the prob­
lem does not specify the method of selection so it would not be appropri­
ate for us to assume that this is relevant. It is only the chosen team of 4
which matters.
Now, at last, we can try to answer the question. We will do this from
first principles. Strangely enough, it is best to assume that we do pick
the team one at a time and that the order of selection matters. Then we
Chapter 3: Combinatorics
25
will have to compensate for the fact that we have answered the wrong
question.
There are 30 ways to pick the first member of the team. That person is
no longer available for selection, so there are 29 ways to choose the second
member of the team, and so on. We come up with the answer 30 x 29 x
28 x 27. However, each possible team of four has been counted many
times using our method. Given any possible team of 4, it is possible to
arrange them as first choice, second choice, third choice and fourth choice
in 4 x 3 x 2 x 1 = 24 ways. Each team has been counted 24 times, so the
correct answer to the original problem is
30 x 29 x 28 x 27
= 27,405.
4 x 3 x 2 x 1
These products of consecutive integers are best expressed using facto­
rial notation. Define 0! = 1 and for a positive integer n define n! to be
n x (n - 1)!. Thus 3! = 3 x 2 x 1 =6. The notation 3! is pronounced 3
factorial. There are isolated pockets of heretics who say factorial 3 instead.
In this notation, the answer to our problem is
30!
26!4!'
Indeed, if you have n people and you wish to form a team of r from this
squad, similar reasoning shows that the number of ways to do this is
(n — r)!r!
This notation is so useful and common that it has a shorthand form and a
way to pronounce it. We write
nl
(n — r)!r!
and say n choose r.
These quantities arise in Pascal's triangle (Table 3.1 on the next page).
The top (zero-th) row of Pascal's triangle consists of (q) = 1. The first row
consists of (J) and (}). The entries of the triangle can be calculated using
the formula for ("), or by the rule that each entry is the sum of the two
entries diagonally above it (where missing entries are deemed to be 0).
Using the (”) notation this becomes Table 3.2.
A Mathematical Olympiad Primer
1
1
1
3
13
1
6
4
1
5
1
2
1
10
1
4
1
5
10
1
Table 3.1 : Pascal's triangle
(1)
(?)
Table 3.2 : Pascal's triangle in the (”) notation
Of course these Pascal triangles can be extended indefinitely, with each
row defined inductively in terms of its predecessor. While this notation
has become fairly standard, there are plenty of people who write C” or
"Cr instead of
Watch out for this in books, or flowing from teachers'
pens.
These numbers also arise (for very good reasons) when doing a bino­
mial expansion. Indeed, some people refer to (") as a binomial coefficient.
You can learn interesting things about the rows of Pascal's triangle by
setting x = l,y = 1 or x = l,y = —1. In the first case it becomes clear
why the rows of Pascal's triangle sum to a power of 2. In the second case,
the alternating sum of the entries of Pascal's triangle is 0, except for the
zeroth row.
Chapter 3: Combinatorics
27
Double counting
This is an excellent mathematical method for extracting sunlight from cu­
cumbers. The idea is that you have a finite set, and you count the number
of its members in two ways. You then observe that the two answers must
be the same.
Consider a rectangle of dots, with m rows and n columns. Each row
contains n dots and there are m rows so the total number of dots is m x n.
On the other hand each column contains m dots and there are n columns
so the total number of dots is n x m. Since the rectangle of dots could
have arbitrary dimensions, it follows that m x n = n x m for all positive
integers m and n. This is a proof of the commutativity of multiplication of
positive integers based on double counting.
Here is another less obvious application of this technique.
The Handshaking Lemma At a party, some people shake hands. Each pair of
people shake hands at most once. It follows that the number of people who shake
hands an odd number of times is even.
PROOF Suppose that person x shakes hands with v(x) people at the party
(the notation is supposed to remind you of the notion of valency in chem­
istry). We count the number of pairs (h,s') where h is a hand that partic­
ipated in a shake s. We count in two ways. First it is the sum of all v(x)
as x ranges over the set R of revellers and second it is twice the number of
shakes. Equating these two we learn that the sum of the quantities v(x) as
x ranges over R is even. Therefore the number of party-goers who shake
hands an odd number of times is even. Put another way, if there were an
odd number of x such that v(x) was odd, then the sum of all v(x) would
be odd.
□
We often use double counting without noticing the fact. For example,
consider how we showed that the sum of the entries in a row of Pascal's
triangle is a power of 2.
Colouring arguments
One colouring argument is very well known and will be familiar to many
readers. Suppose that you remove two opposite corner squares from a
chessboard. Can you cover what is left with dominoes of size 2x1?
28
A Mathematical Olympiad Primer
The answer is no. Colour the small squares black and white in chess­
board fashion. Anywhere a domino is placed, it covers one black square
and one white. The two removed squares are of the same colour, so you
can only put down at most 30 dominoes on the remaining board.
The same argument shows that if any two small squares of the same
colour are removed, then the remaining board cannot be covered with
dominoes.
Chapter 4
Geometry
It is very important to understand congruence and similarity of triangles.
In a sense, that is all you need to understand, because everything else
can be reduced to this. However, it would be foolish to adopt this totally
minimalist policy. A candidate for BMO1 should certainly understand the
elementary circle theorems, a few facts about parallelograms, and a little
light trigonometry.
It is also important to understand what is meant by a general config­
uration. Not all quadrilaterals are parallelograms and not all parallelo­
grams are squares. Make sure that any diagram you draw is not a special
case of what is being described.
The second round of the British Mathematical Olympiad, BMO2, oc­
cupies a space between BMO1 and the IMO, and the setters feel free to
assume a more mature understanding of trigonometry and geometry. I
heartily recommend the text Plane Euclidean Geometry [4] for students ap­
proaching this or a similar more advanced examination. From the point
of view of olympiad competitions, the first two chapters are perhaps of
peripheral interest, but Chapters 3 to 7 contain the results you need to
know and excellent exercises in industrial quantities. Indeed, these five
chapters contain all the facts and theory that you are likely to need to ad­
dress IMO geometry problems. Of course one can never have too much
of a good thing, and once you start to savour geometry, it may be hard
to resist exploring deeper. If you like the style of the later chapters of [4],
then you may find yourself drawn to read some of Bradley's other books,
including Challenges in Geometry [1] and The Algebra of Geometry [3]. You
30
A Mathematical Olympiad Primer
may wish to look at Coxeter's famous text [5], or Coxeter and Greitzer's
accessible Geometry Revisited [6]. We must mention Honsberger's elegant
Episodes in 19th and 20th century Euclidean geometry [11]. For readers happy
with electronic media, there are advanced geometry DVDs presented by
Ceri Fiddes, currently UK IMO deputy leader [6]. Enthusiastic geometry
students who read French will enjoy Lalesco [12]; indeed, there are stu­
dents who may find that reading La Geometric du Triangle is a good way to
improve their French.
We do not attempt an organized development of geometry from the
ground up. We will assume that the reader is familiar with the language
and concepts of elementary geometry, so we are gambling that you know
what parallel means, what vertically opposite angles are, and that you have
seen the theorem of Pythagoras in action, and so on. Rather you should
regard this section as nothing more then a refresher course.
Triangle notation
Unless there is a good reason to vary our policy, we will name triangles by
listing their vertices in anticlockwise order. If a triangle is ABC, then we
may sometimes take the liberty of calling AC AB by the even shorter name
A A (and similarly for other vertices).
The length of the side opposite to AA is called a. We will be careful to
distinguish between a triangle side BC which is the line segment joining B
and C, and the side line BC which is the straight line through B and C, but
extends indefinitely in both directions. Since the same notation BC is used
for both, you have to be careful here. There may be times when you want
to talk about the half-line or ray BC which starts at B and extends through
C indefinitely.
The perpendicular bisectors of the sides meet at the circumcentre of the
triangle (often called) O. This is the centre of the circumcircle, the circle
which passes through the vertices.
The angle bisectors meet at a point called the incentre of the triangle. This
is the centre of the incircle, the unique circle with centre inside the triangle
which is tangent to all three sides.
The medians of a triangle are the three lines which join each vertex to
the midpoint of the opposite side. These three lines meet at a point called
the centroid of the triangle, often called G.
The altitudes of a triangle are the three lines which pass through a trian-
Chapter 4: Geometry
31
Figure 4.1: The circumcentre O and incentre I
Figure 4.2: The centroid G and orthocentre H
gle vertex and are perpendicular to the opposite side. The altitudes meet
at the orthocentre of the triangle, often called H.
It is unlikely to come up in BMO1, but you should know that Euler
proved that OGH are collinear and spaced in the proportion OG : GH =
1 : 2.
The radius of the circumcircle is called the circumradius, and is often
written R. The radius of the incircle is called the inradius, and is often
written r.
When training the UK IMO squad, I sometimes ask the question "If
you are addressing a triangle problem, when should you consider draw­
ing the circumcircle?" The correct answer is "always".
Naturally all readers will know that the angles of a triangle sum to
180°, so that an exterior angle of a triangle is equal to the sum of the interior
opposite angles.
32
A Mathematical Olympiad Primer
Special triangles
A triangle with all sides equal (equally well all angles equal) is said to
be equilateral. Its three angles are all 60°. It is useful to keep in mind a
reference copy, one where the sides all have length 2. By Pythagoras's
theorem, an altitude (viewed as a line segment rather than a line) will
have length y3.
A triangle with two sides equal is said to be isosceles, and it is some­
times handy to describe the vertex where they meet as the apex of the tri­
angle. The angles of the isosceles triangle other than the one at the apex
are sometimes called the base angles. The base angles of an isosceles tri­
angle are equal. It is tempting to say that the converse is true, that if two
angles of a triangle are equal, then the triangle is isosceles, but one must be
a little careful. If the triangle is degenerate there are problems. Suppose
that A lies in the interior of the line interval BC but not at its midpoint.
Notice that AABC = ABCA = 0°. You cannot escape this problem by
saying "I do not consider degenerate triangles as triangles", because you
may be studying three points A, B and C in the plane without knowing
whether or not they are collinear. Look out for broken converses like this.
This particular example of a broken converse was crucial in Problem 4 of
IMO 2003 in Tokyo.
Another special triangle to keep in mind is the isosceles right-angled
triangle with sides 1,1 and y2. The angles of this triangle are 45°,45°
and 90°. There is also the famous 3,4,5 right-angled triangle. The other
angles of this triangle are not particularly interesting, but the inradius of
this triangle (the radius of its incircle) is significant. It is 1.
Similarity and congruence
We assume that the reader is familiar with the fundamental notions of
similar and congruent triangles. You can show that triangles ABC and DEF
are similar by showing that two pairs (and consequently all three pairs)
of corresponding angles are the same. A less obvious method is to show
that one pair of angles are the same, and that their adjacent sides are in
the same ratio. For example, you might show that A A = AD, and that
AB : DE = AC : DE or equivalently AB : AC = DE : DE. Once you
know that two triangles are similar, then you are free to use the facts that
corresponding angles are equal and that the ratios of all three pairs of
Chapter 4: Geometry
33
Figure 4.3 : Similar triangles
corresponding sides are the same.
Sometimes it is important whether a similarity is direct or indirect. Di­
rect similarity is when the letters of corresponding vertices run the same
way round the triangles. This has the precise meaning that one can get
from one triangle to the other by an enlargement (or shrinking) followed
by a rotation and then a translation. In the case of an indirect similarity,
then you have to "turn over" a triangle as well.
Congruence is a special type of similarity where the ratios of corre­
sponding sides is 1. In other words, corresponding sides have the same
length. In a mathematics competition, when you assert that two triangles
are congruent (or similar) you must always give the reason that this is
so. There three letter acronymns available so that you can do this quickly.
Here are legitimate ways to justify congruence.
1. SSS. This indicates that you have a proof that corresponding sides
are equal.
2. ASA. This indicates that you have a proof that there are two pairs
of angles where the two angles in each pair are equal, and that the
sides between these angles are equal.
3. AAS. Two pairs of equal angles and two equal sides in correspond­
ing positions. Indeed, since the angles of a triangle sum to 180°, you
can always turn an AAS justification into an ASA justification.
4. SAS. Two pairs of sides where the lengths in each pair are the same,
and the two angles between these sides are the same.
A Mathematical Olympiad Primer
34
5. RHS. Each triangle has a right angle, the hypotenuses are of equal
length, and some other side in the first triangle has the same length
as another side in the second triangle.
Note that ASS is not legitimate justification for congruence, and anyone
who says that it is has donkey-like attributes.
Trigonometry
We shall assume that you know the definitions of sine, cosine and tangent
for angles in the range 0° to 90°. One can extend these definitions to all
angles as follows. Let X be an arbitrary point on the positive real axis of a
Cartesian co-ordinate plane with origin O. For any angle 9 as shown in the
diagram, let P denote the point distance 1 from O such that XPOX = 9.
Figure 4.4 : P = (cos 9, sin 0)
We define the co-ordinates of P to be (cos 9, sin 9). Then we define tan 9 =
sin#/cos# except where cos# = 0. The theorem of Pythagoras tells us
that for all # we have
cos2 # + sin2 # = 1.
The position of the 2s in the exponent may look curious, but this is the
standard way of writing
(cos#)2 + (sin#)2 = 1.
Chapter 4: Geometry
35
Thus for angles which are obtuse but not reflex (so they are greater
than 90° but less than 180°) we find that cosine is negative, sine is positive
and tangent is negative.
sin(A + B) = sin A cos B + sin B cos A
sin(A — B) = sin A cos B — sin B cos A
cos(A + B) = cos A cos B — sin A sin B
cos(A — B) = cos A cos B + sin A sin B
tan (A + B)
tan(A — B)
tan A + tan B
1 — tan A tan B
tan A — tan B
1 + tan A tan B
sin (—A) = — sin A
cos( — A) = cos A
tan( — A) = — tan A
cos A cos B =
cos(A + B) + cos(A — B)
cos(A — B) — cos(A + B)
sin A sin B =
2
sin(A + B) + sin(A — B)
sin A cos B =
2
cos A + cos B
cos A — cos B = 2 sin (^4-^) sin ()
sin A — sin B
These are standard formulas which you should commit to memory.
The trigonometric functions are defined in terms of right-angled trian­
gles, but in fact they can be used to study arbitrary triangles. The devices
which are used to do this are the sine rule and the cosine rule.
A Mathematical Olympiad Primer
36
Sine rule In standard notation,
sin A
sin B
sin
where R is the radius of the circumcircle of triangle ABC.
Thus the ratios of two sides of a triangle is the same as the ratio of the
sines of the angles which are opposite to these sides.
Cosine rule In standard notation,
a2 = b2 + c2 — 2bc cos A.
Notice that if A is a right angle, then cos A = 0 so the cosine rule special­
izes to give the theorem of Pythagoras.
The nature of these formulas is such that the sine rule is often the easier
one to use, if you have a choice. The sine rule is a trigonometric way to
avoid arguments concerning similar triangles.
A word of warning is appropriate. The sine of an obtuse angle is posi­
tive. In fact if a is an acute angle, then sin a = sin(180° — a). The sine rule
alone therefore cannot distinguish between an acute angle and its sup­
plement. Another way of viewing this ambiguity is that if X and Y are
distinct points on a circle, then the angle subtended by X and Y is not well
defined. If you specify which arc XY is doing the subtending, then the
answer is well-defined. However, there are two arcs XY of the circle, and
they subtend supplementary angles.
Triangle area formulas
There are very many formulas for the area of a triangle in terms of its
natural attributes (angles, sides, inradius, circumradius etc.). At BMO1
level you should certainly be prepared to use "half base times height" and
-1
2 ab sin C. Both of these formulas for the area of a triangle are in fact three
formulas. In one recent BMO1 question, it might have been useful to know
Heron's formula for the area of a triangle. We put s = (a + b + c) / 2, the
semiperimeter of the triangle, and then Heron's formula for the area of the
triangle is
\/s(s — a)(s — b)(s — c).
Chapter 4: Geometry
37
This is an example of a common phenomenon in mathematics compe­
titions. The BMO1 setting committee would not dream of assuming that
candidates know Heron's formula. Nonetheless, a student who knows
Heron's formula is welcome to use it. The setting committee was care­
ful to ensure that there was a solution available which was entirely free
of Heron's formula. Moreover, the method which did use Heron's for­
mula contained an elephant trap into which an over-confident (and per­
haps over-educated) candidate might fall.
Two simple triangle area formulas which you are unlikely to need for
BMO1 are rs and abc/4R. Here r is the radius of the incircle and R is
the radius of the circumcircle. Triangle area formulas are explored in a
Highperception DVD [10].
Circle theorems
Angles in the same segment
Suppose that A and B are different points on a circle. If C, D are points on
the same arc BA (there are two arcs), then ABC A = ABDA.
Figure 4.5: Angles in the same segment
There is a converse. Suppose that A and B are different points. If C and
D are points on the same side of the line AB such that ABCA = ABDA,
then the points A, B, C and D lie on a circle.
A Mathematical Olympiad Primer
38
Angles subtended by arcs
Suppose that A, B, C and D lie on a circle. We focus on the anticlockwise
arcs AB (from A to B) and CD (from C to D). Suppose that X and Y are
points on the circle with X not on the specified arc AB, and Y not on the
specified arc CD.
Figure 4.6: Angles subtended by arcs
It follows that ZBXA = ZDYC if and only if the specified arcs have
equal lengths.
One must be careful when using this result. The subtended angle determines
the arc length, but the arc length only determines the subtended angle if that angle
is located on the correct arc.
Angles in opposite segments
Suppose that A and B are different points on a circle. If C, D are points
on different arcs BA, then ZBCA + AADB — 180°. A cyclic quadrilateral
ABCD is a quadrilateral with vertices which lie on a circle in that order.
Thus, "opposite angles of a cyclic quadrilateral are supplementary" (that is,
sum to 180°). In turn this means that an exterior angle to a cyclic quadrilaterial is equal to the opposite interior angle.
The converse holds. If ABCD is a quadrilateral and two opposite in­
terior angles sum to 180°, then ABCD is a cyclic quadrilateral. This is an
extremely useful result.
Chapter 4: Geometry
39
Figure 4.7: Angles in opposite segments
Angle at the centre is twice that at the circumference
Suppose that A and B are different points on a circle with centre O. Sup­
pose that C is a point on the circle such that O and C are on the same side
of the line AB, then ABO A = 2ABCA. (The same theorem holds if O and
Figure 4.8: Angle at the centre
C are on different sides of the line AB, provided one interprets ABO A as
being a reflex angle, that is, greater than 180°.)
There is no simple converse to this result.
Angle in the alternate segment
Suppose that a circle and a line are tangent at a point A. Let B and C points
on the circle different from A, and let D be a point on the line such that C
and D are on opposite sides of the line AB, then ABAD = ABC A.
Chapter 4: Geometry
41
Equal tangents
Suppose that P is a point outside a circle, and that A and B are different
points on the circle. If the lines PA and PB are both tangent to the circle,
then the line segments PA and PB have equal length.
Figure 4.11: Equal tangents
Intersecting chords theorem
Suppose that A, B, C and D are four points on a circle, and that the lines
AC and BD meet at a point X. Then triangles ABX and DCX are similar,
and(AX)(XC) = (BX) (XD). (Note that this holds whether X is inside or
outside the circle; we use brackets to indicate that we are taking products
of lengths.)
A Mathematical Olympiad Primer
42
The tangent-secant theorem
Suppose that .A. B, C are on a circle, that X is outside the circle and on the
line .AC. and that XB is tangent to the circle, then (AX)(XC) = (BX)2.
C his is a special case of the intersecting chords theorem, when B = D.)
Figure 4.13: The tangent-secant theorem
There are certainly more circle theorems of wide interest, but this batch
will see you through competitions of the BMO1 type. More advanced re­
sults which you might wish to look up are "the eyeball theorem" and (an
IMO favourite) "Ptolemy's
•r theorem”,
Chapter 5
Number Theory
Problems concerning properties of integers (whole numbers) are often de­
scribed as Number Theory. We will not give a full treatment of elementary
number theory, and so we must point you to places where you can find
such a thing, Harold Davenport's The Higher Arithmetic, Cambridge Uni­
versity Press [7]. This is a singularly well written book, and goes aston­
ishingly far at breakneck speed, while giving the illusion that it is a gentle
introduction for the educated lay reader. A text which is more focussed on
competitive mathematics is Christopher Bradley's Introductions to Number
Theory and Inequalities [2] in this UKMT series. For the reader happy with
electronic media, this author presents a Number Theory DVD for Highperception [10].
We say that the integer b divides the integer a if (and only if) there is an
integer c such that a = be. Therefore 2 divides 6, 6 does not divide 2, 4
divides —8 and —9 divides —18. Some numbers are very rich in divisors,
witness 24,120 and 720. Other integers have very few divisors, for exam­
ple 19 and —47. There is notation to express the statement that a divides
b; we write a b. As usual you use a slash to indicate that a statement is
false. Therefore it is correct to write 3/5.
Fact If m and n are integers such that both m divides n and n divides m, then
m — ±n. (This says that m is plus or minus n.)
You might think that this is so obvious that a proof is not necessary.
Indeed, proofs of elementary facts about whole numbers seem to be little
more than taking one fairly obvious statement and showing that it fol­
A Mathematical Olympiad Primer
44
lows from another arguably more obvious statement. We are not in the
business of giving a logically perfect development of the theory of num­
bers. Such a thing is often done in the first year of university mathematics
courses. Rather we will simply state obvious properties of numbers (or
other objects of mathematical thought) as facts. When it comes to more
complicated and less obvious true statements, we will give proofs show­
ing why they follow from obvious facts. Thus we are building on shoddy
foundations. There is nothing wrong with this, as long as we are honest
about what we are doing.
Fact If the integer m divides the positive integer n, then m < n.
This fact is false if you translate it into French. In France, zero is a positive
integer. Now 1729 divides 0, but it is not the case that 1729 < 0.
Fact If m, n are integers, and n is positive, then there are integers q, r such that
m = qn + r and 0 < r < n.
This says that if you try to divide m by n, you may not succeed, but you
can do so if you allow a small remainder r. For example, if you try to
divide 17 by 5, you can nearly do it;
17 = 3 x 5 + 2.
The remainder 2 is at least 0 and also less than 5, as promised. In fact q
and r are uniquely determined by m and n. They are the only numbers
which will do the job. Note that small remainder has the specific meaning
that 0 < r < n.
Prime numbers
A positive integer with exactly two positive divisors is said to be prime.
This is carefully worded. If m is a positive integer, then m and 1 are both
divisors of m, but if m = 1, these numbers are the same. Therefore 1 is not
a prime number, but
2,3,5,7,11,13,17,19,23,29,31,37,41,43,...
are prime numbers.
Every positive integer bigger than 1 has a prime factor. It is worth
proving this statement. Suppose, for contradiction, that this statement is
Chapter 5: Number Theory
45
false. Therefore there is at least one positive integer n which is a counter­
example to this statement. We may suppose that n is the smallest such
counter-example. Now n is not prime, because n divides n. Therefore n
has a third positive factor u which is neither 1 nor n and so u < n. How­
ever, n is supposed the smallest counter-example so u has a prime factor
P- Now p divides u which divides n so p divides n. This contradicts the
status of n as a counter-example. Therefore there are no counter-examples
to the statement we are trying to prove, which therefore is correct.
It is not an accident that this argument does not have a constructive
flavour. You are simply guaranteed the existence of a prime factor, but
you get no information about where to look to find it. We do not know
of any really good way to find prime factors of large numbers. This is
the basis of many contemporary protocols used to provide encryption of
electronic data (for example, money and health records).
Turning our back on these gratuitous references to the real world, we
look at the mathematical consequences of this result. It follows that every
positive integer bigger than 1 is either prime or a product of primes, since
we may keep on pulling out prime factors until a factorization into prime
numbers is found.
Next we give Euclid's argument that there are infinitely many prime
numbers. For educational reasons we will give it in two flavours. The first
proof has the drama; we show that, if there were only finitely many prime
numbers, then we could produce a contradiction. In the second proof we
show that there is no limit to the size of a finite collection of primes.
PROOF 1 Suppose, hoping for a contradiction, that there are only finitely
many prime numbers 2, 3, .... Multiply them all together and add 1 to
produce a number N. Choose p a prime factor of N. Then p divides both
N and N — 1, so p divides their difference, that is to say p divides 1. No
prime number divides 1, so we have the required contradiction.
□
PROOF 2 Suppose that p\,p2,...,pk is a collection of k different prime
numbers. Multiply them all together and add 1 to produce a number
N. Choose Pk+i a prime factor of N. Now Pk+t cannot divide both N
and N — 1 so the prime number Pk+i must be different from each of the
primes pv p2,---, Pk- Therefore there is a collection of k + 1 different prime
numbers. From any finite set of prime numbers we can make a larger one,
and so there must be infinitely many primes.
□
A Mathematical Olympiad Primer
46
Although prime numbers are defined in terms of their divisors, it turns
out that they have a beautiful property in terms of how they divide into
other numbers. We give it as a fact.
Fact Suppose that p is a prime number. Ifm,n are integers such that p divides
mn, then either p divides m or p divides n (and possibly p divides both).
We have already shown that every positive integer bigger than 1 is
either prime of a product of primes. This is an important part of the Fun­
damental Theorem of Arithmetic as the following fact is called.
Fact Every positive integer bigger than 1 can be factorized into prime numbers.
Provided that you do not worry about the order of the factors, this factorization is
unique.
Common divisors
We remind you of the vertical line notation used to indicate divides. We
write a | b to indicate that the integer a divides the integer b. Suppose that
a and b are integers. We let
A(a, b) = {x E
: x | a,x | b}.
The colon here is read as "such that". This notation A(a, b) is home made;
it is far from universally used. Thus A(a,b) is precisely the set of inte­
gers which divide both a and b. Put another way, it is the set of common
divisors of a and b. Thus
A(9,6) = {-3,-1,1,3}.
We might treat ourselves to an adornment and write A+ (a, b) for the set of
positive common divisors of a and b. Thus
(6,9) = 1,3.
Definition Suppose that a, b are integers, and that it is not the case that they are
both 0. The greatest common divisor or gcd of a and b is
gcd(a, b) = maxA(a,b) = maxA+(a, b).
Chapter 5: Number Theory
47
We put the prefix max in front of a finite set of numbers to denote its
maximum element.
Among research mathematicians, the notation gcd has become stan­
dard. However, in secondary schools one often finds the alternative nota­
tion hcf which stands for highest common factor.
Note that A(0,0) = Z = {..., —2, —1,0,1,2,...} and this set has no
greatest element. Given that a, b are not both 0, then g = gcd(a,b) is a
positive integer.
Fact Given that a, b are integers and not both 0, then the set of divisors of g
gcd (a, b) is the set A(a,b) of common divisors of a and b.
Fact Ifa,b are integers and not both 0, and g = gcd(a, b), then there are integers
x and y such that ax + by = g.
There is a procedure for calculating g from a and b which is called
Euclid's algorithm. Unpacking the algorithm gives of a way of finding a
pair of integers x, y which are as stated in the previous fact. Of course,
you are free to just spot them. For example gcd(24,78) = 6 and 6 =
(-3)(24) + (l)(78).
Definition Integers a, b are coprime if (and only if) gcd(a, b) = 1.
We know that if a, b are coprime, then there are integers x, y such that
ax + by = 1. On the other hand, suppose that a, b, x and y are integers such
that ax + by = 1. If z is an integer dividing both a and b, then z divides
ax + by = 1 so z = 1 or -1. Therefore gcd(fl, b) = 1; the integers a and b
are therefore coprime.
Congruences
Consider two integers, one which leaves remainder 3 on division by 5,
and another which leaves remainder 4 on division by 5. The first is 5m + 3
and the second 5n + 4. Their sum is5(m + n + l)+2 and their product
is 5(5mn + 4m + 3n + 2) + 2. Thus the remainder when the sum or the
product is divided by 5 is 2, and this is completely independent of m and
n. Similar facts apply when the remainders take other values. We can
develop an arithmetic of remainders on division by 5.
The addition and multiplication tables are as follows.
48
A Mathematical Olympiad Primer
The laws of algebra of the integers (for example x(y + z) = (xy) ±
(xz) for all x, y and z) are inherited when we work with this arithmetic of
remainders.
We introduce some notation. We say that 5 is the modulus of our arith­
metic of remainders. If two integers a and b leave the same remainder on
division by 5, or what amounts to the same thing, their difference a — b is
divisible by 5, we say that a and b are congruent modulo 5. This is written
a = b mod 5 or sometimes a = b (5). The fact that = looks very much like
an equals sign is no accident. As far as remainders on division by 5 are
concerned, 3 and 13 have the same status; they both leave remainder 3 on
division by 5 (and their difference ±10 is divisible by 5). Thus when we
write 3 = 13 mod 5 we mean that, in this context, 3 and 13 are the same.
Provided that the modulus is the same, congruences may be added,
subtracted or multiplied. Division in the context of modular arithmetic,
on the other hand, is a subtle business, and must not be undertaken lightly.
For example, working modulo 4 we have
2 x 2 = 0 x 2 mod 4.
If we were allowed to divide both sides by 2, we would deduce that
2 = 0 mod 4
which is nonsense.
All is not lost. When you are working with ordinary equations be­
tween integers, you can divide by any number provided that it is not zero.
If you are working modulo 4, then dividing by 1 is fine (rather unsurpris­
ingly). You must not divide by 0 or 2, but in fact you can divide by 3. The
reason is that 3x3 = 1 mod 4. Therefore there is a multiplicative inverse
for 3 (which happens to be 3). Multiplying by the multiplicative inverse
of 3 is the same thing as dividing by 3 in this context.
Chapter 5: Number Theory'
49
Let us look at a larger set-up and work modulo 10. The numbers 0, 2,
4, 6, 8 and 5 do not have multiplicative inverses, but 3x7 = 1 mod 10,
1x1 = 1 mod 10, and 9x9 = 1 mod 10. Therefore modulo 10 you can
divide by 3 by multiplying by 7 (and vice versa), and you can divide by
9 = — 1 mod 10 by multiplying by 9. That information tells a story. The
remainders without multiplicative inverses modulo 10 are 0, 2, 4, 5, 6 and
8. Those with multiplicative inverses are 1, 3, 7 and 9. It does not take a
great detective to spot the pattern. Can you see it?
The only positive common factor of a number from {1,3,7,9} and 10 is
1. On the other hand, each element of {0,2,4,5,6,8} has a common factor
with 10 which is greater than 1. This is not an accident, and tells the correct
story for any modulus. For example, working modulo 15, the invertible
remainders are 1, 2, 4, 7, 8, 11, 13 and 14, whereas 0, 3, 5, 6, 9, 10 and 12
do not have multiplicative inverses. If you want to find out why this is so,
then consult a book on elementary number theory, for example Bradley's
text [2].
This has particular significance when working modulo a prime num­
ber p, because of course none of the numbers 1,2,3,..., p — 1 has a com­
mon factor with p which is greater than 1. This means that all these num­
bers have multiplicative inverses modulo p, and you can divide by any
integer which is not congruent to 0 modulo p.
The first theorem of number theory which is not really obvious is called
Fermat's Little Theorem. We give three versions of the result.
FLT version 1 Suppose that p is a prime number and x is an integer which is
not divisible by p, then p divides xp~^ — 1.
FLT version 2 Suppose that p is a prime number and x is an integer such that
x
0 mod p, then
= 1 mod p.
FLT version 3 Suppose that p is a prime number and x is an integer. Then p
divides xp — x.
Testing for divisibility
Positive integers are usually written using base 10 notation. Thus the re­
sult of adding twelve copies of 1 together is written 12. Of course there are
other ways of representing integers, using a number base different from
10, and even more exotic representations such as Roman numerals. Other
X V,-: V
.7 C v
' 'V:
'.X' v ? v sV •■:?. • tests
c a >■.?■ va e-.'.e c?. ' ?.;'va
»'. •• s
■■ .• ■ '
,w
-e / > v <x. c
-Atexv x
X
X
X
X-
X
Chapter 5: Number Theory
51
10 = — 1 mod 11, so modulo 11 we have
47^10^ + fln-il0?? 1 + • • • +
77 1 + ' * * +
( — 1+ Q().
Part II
Problems
BMO1 1996-97
1. N is a four-digit number which does not end in a zero, and R(N)
is the four-digit integer obtained by reversing the digits of N. For
example R (3275) = 5723.
Determine all such integers N for which R(N) = 4N + 3.
2. For positive integers n, the sequence a\, ai, a$,... is defined by ai = 1
and
( Cl 2
j
77 2
'
• • •
H ^7/ 7
x)
for n > 1. Determine the value of #1997.
3. The Dwarves in the Land-under-the-Mountain have just adopted a
completely decimal currency system based on the Pippin, with gold
coins to the value of 1 Pippin, 10 Pippins, 100 Pippins and 1000 Pip­
pins.
In how many ways is it possible for a Dwarf to pay, in exact coinage,
a bill of 1997 Pippins?
4. Let ABCD be a convex quadrilateral. The midpoints of AB, BC,CD
and DA are P, Q, R and S, respectively. Given that the quadrilateral
PQRS has area 1, prove that the area of ABCD is 2.
5. Let x,y and z be positive real numbers.
(i) If x + y + z > 3, is it necessarily true that
/continued...
56
A Mathematical Olympiad Primer
(ii) lfx + y + z<3, is it necessarily true that
BMO11997-98
1. A 5 x 5 square is divided into 25 unit squares. One of the numbers
1, 2, 3, 4, 5 is inserted into each of the unit squares in such a way that
each row, each column and each of the two diagonals contains each
of the numbers once and only once. The sum of the four numbers
immediately below from top left to botton right is called the score.
Show that it is impossible for the score to be 20. What is the highest
possible score?
2. Let a-[ = 19, fl2 = 98. For n > 1 define an+2 to be the remainder of
an + an+i when divided by 100. What is the remainder when
2,2,
,2
fl2 + fl2 + • • • + ^1998
is divided by 8?
3. ABP is an isosceles triangle with AB = AP and APAB acute. PC is
the line through P perpendicular to BP, and C is a point on this line
on the same side of BP as A. (You may assume that C is not on the
line AB.) D completes the parallelogram ABCD. PC meets DA at
M. Prove that M is the midpoint of DA.
4. Show that there is a unique sequence of positive integers (an) satis­
fying the following conditions:
5. In triangle ABC, D is the midpoint of AB and E is the point of trisec­
tion of BC nearer to C. Given that AADC = ABAE, find ABAC.
BMO1 1998-99
1. I have four children. The age of each child in years is a positive
integer between 2 and 16 inclusive and all four ages are distinct. A
year ago the square of the age of the oldest child was equal to the
sum of the squares of the ages of the other three. In one year's time,
the sum of the squares of the oldest and the youngest will be equal
to the sum of the squares of the other two children.
Decide whether this information is enough to determine their ages
completely, and find all possibilities for their ages.
2. A circle has diameter AB and X is a fixed point on AB lying between
A and B. A point P, distinct from A and B, lies on the circumference
of the circle. Prove that, for all possible positions of P, the quantity
tan AAPX
tan A PAX
remains constant.
3. Determine a positive constant c such that the equation
xy2 -y2 -x + y = c
has precisely three solutions (x, y) in positive integers.
4. Any positive integer m can be written uniquely in base 3 form as a
string of Os, Is and 2s (not beginning with a zero). For example
98 = (1 x 81) + (0 x 27) + (1 x 9) + (2 x 3) + (2 x 1)
= (10122)3-
/continued...
60
A Mathematical Olympiad Primer
Let c(m) denote the sum of the cubes of the digits of the base 3 form
of m; thus, for instance
c(98) = I3 + 03 + l3 + 23 + 23 = 18.
Let n be any fixed positive integer. Define the sequence (ur) by
=
n and ur = c(ur-i) for r > 2. Show that there is a positive integer r
for which ur = 1,2 or 17.
5. Consider all functions f from the positive integers to the positive
integers such that
(i) for each positive integer m, there is a unique positive integer n
such that f(n) = m;
(ii) for each positive integer n, we have f(n + 1) is either 4f(ri) — 1
orf(n) - 1.
Find the set of positive integers p such that /(1999) = p for some
function f with the properties (i) and (ii).
BMO1 1999-2000
1. Two intersecting circles Q and C2 have a common tangent which
touches Ci at P and C2 and Q. The two circles intersect at M and N,
where N is nearer to PQ than M is. The line PN meets the circle C2
again at R. Prove that MQ bisects APMR.
2. Show that, for every positive integer n,
121" - 25" + 1900" -
is divisible by 2000.
3. Triangle ABC has a right-angle at A. Among all points P on the
perimeter of the triangle, find the position of P such that AP + BP +
CP is minimized.
4. For each positive integer k > 1, define the sequence (an) by
«o = 1/
an =kn-\- (—l)"fl„_i
for each n > 1. Determine all values of k for which 2000 is a term of
the sequence.
5. The seven dwarfs decide to form four team to compete in the Millen­
nium Quiz. Of course, the sizes of the teams will not all be equal. For
instance, one team might consist of Doc alone, one of Dopey alone,
one of Sleepy, Happy and Grumpy, and one made up of Bashful and
Sneezy. In how many ways can the four teams be made up? (The
order of the teams or the order of the dwarfs within the team does
not matter, but each dwarf must be in exactly one of the teams.)
Suppose that Snow White agreed to take part as well. In how many
ways could the teams then be formed?
BMO1 2000-01
1. Find all two-digit integers N for which the sum of the digits of 10N —
N is divisible by 170.
2. Circle S lies inside circle T and touches it at A. From a point P (dis­
tinct from A) on T, chords PQ and PR of T are drawn touching S at
X and Y respectively. Show that AQAR = 2AXAY.
3. A tetromino is a figure made up of four unit squares connected by
common edges.
(a) If we do not distinguish between the possible rotations of a
tetromino within its plane, prove that there are seven distinct
tetrominos.
(b) Prove or disprove the statement: it is possible to pack all seven
distinct tetrominoes is a 4 x 7 rectangle without overlapping.
4. Define the sequence (an) by
an = n + {y/n}
where n is a positive integer and {%} denotes the nearest integer to
x, where halves are rounded up if necessary. Determine the smallest
integer k for which the terms
ak>
• • • / ^+2000
form a sequence of 2001 consecutive integers.
5. A triangle has sides of length a, b and c and its circumcircle has ra­
dius R. Prove that the triangle is right-angled if and only if a2 + b2 +
c2 = 8R2.
BMO1 2001-02
1. Find all positive integers m, n, where n is odd, that satisfy
2. The quadrilateral ABCD is inscribed in a circle. The diagonals AC,
BD meet at Q. The sides DA, extended beyond A, and CB, extended
beyond B, meet at P.
Given that CD = CP = DQ, prove that ACAD = 60°.
3. Find all positive real solutions to the equation
where |_tj denotes the largest integer less than or equal to the real
number t.
4. Twelve people are seated around a circular table. In how many ways
can six pairs of people engage in handshakes so that no arms cross?
(Nobody is allowed to shake hands with more than one person at
once.)
5. f is a function from Z+ to Z+, where Z+ is the set of non-negative
integers, which has the following properties:(a) /(n + 1) > f(n) for each n C Z ,
(b) f(n + f(m')') = f(n) + m + 1 for all m, n 6 Z+.
Find all possible values of /(2001).
1. Given that
34! = 295 232 799 cd9 604 140 847 618 609 643 5ab 000 000,
determine the digits a, b, c and d.
2. The triangle ABC, where AB < AC, has circumcircle S. The perpen­
dicular from A to BC meets S again at P. The point X lies on the line
segment AC, and BX meets S again at Q. Show that BX = CX if and
only if PQ is a diameter of S.
3. Let x,y, z be positive real numbers such that x2 4- y2 4- z2 = 1. Prove
that
4. Let m and n be integers greater than 1. Consider an m x n rectangu­
lar grid of points in the plane. Some k of these points are coloured
red in such a way that no three red points are vertices of a rightangled triangle two of whose sides are parallel to the sides of the
grid. Determine the greatest possible value of k.
5. Find all solutions in positive integers a, b, c to the equation
a\b\ = a! + b! + c!.
BMO1 2003-04
1. Solve the simultaneous equations
ab + c + d = 3,
be + d + a = 5,
cd + a + b = 2,
da + b + c = 6
where a, b, c, d are real numbers.
2. ABCD is a rectangle, P is the midpoint of AB, and Q is the point on
PD such that CQ is perpendicular to PD. Prove that triangle BQC is
isosceles.
3. Alice and Barbara play a game with a pack of 2n cards, on each of
which is written a positive integer. The pack is shuffled and the cards
laid out in a row, with the numbers facing upwards. Alice starts, and
the girls take turns to remove one card from either end of the row,
until Barbara picks up the final card. Each girl's score is the sum of
the numbers on the chosen cards at the end of the game.
Prove that Alice can always obtain a score at least as great as Bar­
bara's.
4. A set of positive integers is defined to be wicked if it contains no three
consecutive integers. We count the empty set, which contains no
elements at all, as a wicked set. Find the number of wicked subsets
of the set
{l,2,3,4,...,10}.
5. Let p, q and r be prime numbers. It is given that p divides qr - 1,
q divides rp - 1 and r divides pq - 1. Determine all possible values
of pqr.
BMO1 2004-05
1. Each of Paul and Jenny has a whole number of pounds. He says
to her: "If you give me £3, I will have n times as much as you".
She says to him: "If you give me £ n, then I will have 3 times as
much as you". Given that all these statements are true and that n is
a positive integer, what are the possible values for n?
2. Let ABC be an acute-angled triangle, and let D, E be the feet of the
perpendiculars from A, B to BC, CA respectively. Let P be the point
where the line AD meets the semicircle constructed outwardly on
BC, and Q be the point where the line BE meets the semicircle con­
structed outwardly on AC. Prove that CP = CQ.
3. Determine the least natural number n for which the following re­
sult holds: No matter how the elements of the set {1,2,.. .,n} are
coloured red or blue, there are integers x, y, z, w in the set (not neces­
sarily distinct) of the same colour such that x + y + z = w.
4. Determine the least possible value of the largest term in an arith­
metic progression of seven distinct primes.
5. Let S be a set of rational numbers with the following properties:
(a) j e S;
(b) If x E S, then both
E S and
G S.
Prove that S contains all rational numbers in the interval 0 < x < 1.
BMO1 2005-06
1. Let n be an integer greater than 6. Prove that if n - 1 and n + 1 are
both prime, then m2(«2 4-16) is divisible by 720. Is the converse true?
2. Adrian teaches a class of six pairs of twins. He wishes to set up
teams for a quiz, but wants to avoid putting any pair of twins into
the same team. Subject to this condition:
(i) In how many ways can he split them into two teams of six?
(ii) In how many ways can he split them into three teams of four?
3. In the cyclic quadrilateral ABCD, the diagonal AC bisects the angle
DAB. The side AD is extended beyond D to a point E. Show that
CE = CA if and only if DE = AB.
4. The equilateral triangle ABC has sides of integer length N. The tri­
angle is completely divided (by drawing lines parallel to the sides of
the triangle) into equilateral triangular cells of side length 1.
A continuous route is chosen, starting inside the cell with vertex A
and always crossing from one cell to another through an edge shared
by the two cells. No cell is visited more than once. Find, with proof,
the greatest number of cells which can be visited.
5. Let G be a convex quadrilateral. Show that there is a point X in
the plane of G with the property that every straight line through X
divides G into two regions of equal area if and only if G is a parallel­
ogram.
6. Let T be a set of 2005 coplanar points with no three collinear. Show
that, for any of the 2005 points, the number of triangles it lies strictly
within, whose vertices are points in T, is even.
BMO1 2006-07
1. Find four prime numbers less than 100 which are factors of 332 — 232.
2. In the convex quadrilateral ABCD, points M, N lie on the side AB
such that AM = MN = NB, and points P, Q lie on the side CD such
that CP = PQ = QD. Prove that
Area of AMCP = Area of MNPQ
= 1 Area of ABCD.
3. The number 916 238 457 is an example of a nine-digit number which
contains each of the digits 1 to 9 exactly once. It also has the property
that the digits 1 to 5 occur in their natural order, while the digits 1 to
6 do not. How many such numbers are there?
4. Two touching circles S and T share a common tangent which meets
S at A and T at B. Let AP be a diameter of S and let the tangent from
P to T touch it at Q. Show that AP = PQ.
5. For positive real numbers a, b, c, prove that
c)(b + c — a)(c + a — b).
6. Let n be an integer. Show that, if 2 + 2\/l + 12n2 is an integer, then
it is a perfect square.
Part III
Solutions
BMO11996-97 Solutions
Problem 1
N is a four-digit number which does not end in a zero, and R(N) is the
four-digit integer obtained by reversing the digits of N. For example
R(3275) = 5723.
Determine all such integers N for which R(N) = 4N + 3.
Discussion
This problem is about four-digit numbers. The exam was actually sat in
1997 and it should come as no surprise that 7991 = 4(1997) + 3. Always
be on the look out for references to the relevant year. The matter now
hinges on whether or not there are any other four digit numbers with this
property.
There seems to be only one way forward. Express N in its base 10
representation as
N = 1000a + 100b + 10c + d
and go to work. Here a, b, c and d are digits, so they are integers in the
range 0 to 9. Moreover the statement of the problem ensures that neither
a nor d is 0. One then has to write down a formula for 4N + 3 in terms
of a, b, c and d, and then equate this with R(N). There are two techniques
which spring to mind. Congruences modulo 2 and modulo 5 are likely
to be handy, because a lot of terms will disappear (parity arguments are
much the same thing as using congruence modulo 2). A second thing you
A Mathematical Olympiad Primer
80
might do is to eliminate digits because of their size. For example, if a = 9,
then N is more than 9000 so 4N + 3 will certainly be larger than 9999 and
so can not be a 4-digit number. Of course, that argument is very crude,
and you can surely do better.
Solution
Suppose that N = 1000a + 100b + 10c + d satisfies the conditions of the
problem. Since R(N) is odd, it follows that a is odd. Also, since 4N + 3
has only 4 digits, a < 3. Therefore a = 1.
Now R(N) = 4N + 3so
lOOOd + 100c + 10b + 1 = 4000 + 400b + 40c + 4d + 3.
Working modulo 5 we see that 4d = 3 mod 5 so d = 2 mod 5. Therefore
d = 2 or d = 7.
Tidying up we find that
996d + 60c = 390b + 4002
or rather
166d + 10c = 65b + 667.
We conclude that b must be odd. If d = 2, then the LHS is at most 332 + 90,
which is impossible since the RHS is at least 667. It follows that d = 7.
Thus 1162 + 10c = 65b + 667, or rather 495 + 10c = 65b, or equivalently
99 + 2c = 13b. Now 0 < 2c < 18 so 99 < 99 + 2c < 117. The digit b is
odd, and the only odd multiple of 13 in the specified range is 117 = 9 x 13.
Thus c = 9 and b = 9.
Therefore, if there is any solution at all, it must be N = 1997. As we
calculated in the discussion, R(1997) = 4(1997) + 3 so there is a unique
solution to this problem.
Problem 2
For positive integers n, the sequence a\, a^, a^,.. . is defined by
/n + 1 \
an —
(^1 + ^2 + * ’ ’ +an_i)
\ n — 1)
for n > 1. Determine the value of #1997.
— 1 and
Chapter 6: BMO1 1996-97 Solutions
81
Discussion
Experiments will show that the sequence grows fairly rapidly. The first
few terms are 1,3,8, 20,48,112,.... In order to proceed, you really need to
guess what to prove. Once you have the correct conjecture for a formula
for an, then it is will be a technical matter to show that it is correct by
induction. It is suggestive that the later terms of the sequence are divisible
by large powers of 2. You may find it helpful to write the terms of the
sequence in factored form.
Solution
We will give two solutions.
(a) It turns out that an = (n + 1)2” . In order to make the induction
work nicely, it would be easier if the formula defining an were less messy.
Define bn = an/(n + 1), so then we simply have to show that bn = 2n~2.
(Equally well, we could define cn = an/2n~2, and then try to show that
cn = n + 1.)
We want to show that bn — 2”~2 for n > 1. We will do this by
the method of mathematical induction (or induction for short). Notice that
b^ = a\/2 = 1/2 so the base case is done. Now suppose that the result is
established for all values of n which are less than r, and consider br.
2br_y = 22br_2 = • • • = 2r-1&i = 2r 2.
The inductive step is established, and we conclude that bn = 2”
positive integers n. Therefore
an = (n + 1)2”-2
for all
A Mathematical Olympiad Primer
82
for all positive integers n and in particular
a1997 = 1998 ■ 21995.
Fortunately this number is so large that the question as to whether the ex­
aminers want you to write this out in base 10 notation has a ready answer.
(b)
You can manipulate the definition of an to discover that
2(n + l)aM-i
fln =
n
for n >■ ■ 2. This is a much easier recurrence with which tof work, and
you can prove that an = 2n~2(n + 1) by induction, without the need to
introduce an auxilary sequence such as (h„).
For any k > 1 we have
ak — I
) W + «2 + • • • + ak-l
Then we can proceed by means of a telescoping argument.
for every positive integer k (given that we check it also works when k = 1.)
Chapter 6: BMO1 1996-97 Solutions
83
Afterword
Note the appearance of the slightly disturbing
" in solution (b). Such
dots usually indicate, and in this case definitely do indicate, that an induc­
tion argument has been suppressed. It is not the function of mathematics
competitions to force people to write out their induction arguments in full
formal detail. Solution (b) provides all the ingredients to enable you to
write out the solution as an induction argument if you had to, but in a
mathematics competition, an argument such as (b) is good enough. It is
even possible that some students sitting the examination may not have
been trained in how to write out an induction argument such as the one
used in Solution (a).
Problem 3
The Dwarves in the Land-under-the-Mountain have just adopted a com­
pletely decimal currency system based on the Pippin, with gold coins to
the value of 1 Pippin, 10 Pippins, 100 Pippins and 1000 Pippins.
In how many ways is it possible for a Dwarf to pay, in exact coinage, a
bill of 1997 Pippins.
Discussion
This is a combinatorial problem. There are several ways to do it. Any
organized way of counting possibilities will work. Be diligent, and do not
make arithmetical errors.
Solution
The 7 Pippins indicated by the last digit can only be paid in 1 Pippin coins.
We consider the remaining 1990 Pippins. Using only 1 P and 10 P coins,
that bill can be paid in 200 different ways (since we must use between 0
and 199 10 P coins).
If one 100 P coin is used, the remaining 1890 Pippins can be paid in 190
ways. If two 100 P coins are used, the remaining 1790 Pippins can be paid
in 190 ways, and so on.
Thus the total number of ways of paying by using coins to the value of
A Mathematical Olympiad Primer
84
at most 100 P is
200 + 190 + 180 + ... + 10
2100.
If a 1000 P coin is used in the payment, similar arguments show that the
remaining bill of 990 P can be paid in
100 + 90 + • • • + 10 = 550 ways.
Thus the total number of different ways of paying the whole bill in
exact coinage is 2100 + 550 = 2650 ways.
Afterword
Another way to write this down is to let cn(x) be the number of ways of
paying a sum of money of x Pippins using denominations which are at
most 10” P coins. The proof give above can be neatly expressed using this
notation.
c3(1997) = c3(1990)
= c2 (1990) +c2 (990)
= ci (1990) + ci (1890) + • • • + ci (90)
+ ci (990) + ci (890) + • • • + ci (90)
= (200 + 190 + • • • + 10) + (100 + 90 + • • • + 10)
= 20(105) + 10(55) = 2100 + 550 = 2650.
Notice how handy it is to be able to sum a finite arithmetic progression in
your head. This is the number of terms times the average term, and the
average term is also the average of the first and last terms.
Problem 4
Let ABCD be a convex quadrilateral. The midpoints of AB, BC, CD and
DA are P, Q, R and S, respectively. Given that the quadrilateral PQRS has
area 1, prove that the area of ABCD is 2.
Discussion
There are lots of ways to do this, and we give one particularly neat solution
below. Notice that it is a general property of quadrilaterals (even nonconvex ones) that PQRS is a parallelogram. One easy way to prove this is
Chapter 6: BMO1 1996-97 Solutions
85
to use vectors (if you understand them). Choose an origin and let a, b, c
and d be the position vectors of A,B,C and D. Calculate the position
vectors of P, Q, R and S and then compare the vectors PQ and SR. A
proof without vectors is given in the solution below.
The solution we give uses nothing more than "half base times height"
for the area of a triangle, though some other solutions need the formula
^absinC.
Solution
Triangle ADB is an enlargement of triangle ASP from A (with scale factor
2). Therefore ST || DB. Similarly RQ || DB so SP || RQ. Similarly PQ || SR
A
Figure 6.1: Problem 4
so SPQR is a parallelogram. Alternatively one can avoid looking at PQ
and SR by observing that SP = ^DB = RQ.
Let the diagonals DB and AC meet at X. Using square bracket nota­
tion to denote area, note that [PSA] = [SPX] because both triangles have
the same base (SP) and have the same height (because of the similarity
mentioned earlier). There are three other similar results. Now
[SPX] + [PQX] + [QRX] + [RSX] = [PSRQ]
A Mathematical Olympiad Primer
86
so
2[PSRQ] = [PSRQ] + [SPX] + [PQX] + [QRX] + [RSX]
= [PSRQ] + [APS] + [BQP] + [CRQ] + [DSR]
= [ABCD].
Afterword
There is a short formula for the area of a convex quadrilateral, but it seems
difficult to use it directly in this problem. The area is the product of the
diagonals multiplied by the sine of (either) angle between them.
Problem 5
Let x, y and z be positive real numbers.
(i) If x + y + z > 3, is it necessarily true that
(ii) If x + y + z < 3, is it necessarily true that
Discussion
One of the two parts is easy, and the other half is equivalent to a stan­
dard inequality, the so-called Arithmetic Mean-Harmonic Mean inequal­
ity. In 2002 the UK introduced a continuous training programme for the
best young mathematicians that the coaches can find, so such a question
could no longer sensibly be set, as it would simply be a present of free
marks to the trained students on that scheme.
If you want to show that an inequality is true, you must find a proof,
which might be hard. On the other hand, to show that an inequality is
false, you simply need to supply values for the variables where things go
wrong. This is likely to be an easy matter.
Chapter 6: BMO1 1996-97 Solutions
87
Solution
(i) It is easy enough to show that this inequality is false. All you need to
do is to ensure that one of the variables is very small. Say x = 1 / f for f at
least 3. Let y = z = (3 — x) /2 so x, y, z > 0 and x + y + z = 3. Now
(ii) Indeed it is true. There are several solutions which are worth looking
at. We begin with a low-technology proof based on nothing more that a
toy form of the AM-GM inequality: if oc > 0, then oc + 1/x > 2.
(a)
so multiply up
>3 + 2 + 2 + 2
using AM-GM
(b) Apart from AM-GM, and the fact that sums of squares can never
be negative, the most well known inequality is probably that of CauchySchwarz. This states that the dot (scalar) product of two vectors has mod­
ulus which is less than or equal to the product of their lengths.
We use the vectors
u = (x/x, \/y, \fz] and v =
The Cauchy-Schwarz inequality tells us that
i
and therefore
.
\2
„ 2 „ 2
z
A Mathematical Olympiad Primer
88
(c) There is a standard inequality which solves this problem immedi­
ately. The harmonic mean of a collection of positive quantities is the inverse
of the arithmetic mean of their inverses. For three variables this is
3
1 । 1 i 1 ’
x + y + z
The AM-HM inequality states that HM < AM (with equality if and only
if the variables are equal). Therefore
Chapter 7
BMO1 1997-98 Solutions
Problem 1
A 5 x 5 square is divided into 25 unit squares. One of the numbers 1, 2, 3,
4, 5 is inserted into each of the unit squares in such a way that each row,
each column and each of the two diagonals contains each of the numbers
once and only once. The sum of the four numbers immediately below the
diagonal from top left to bottom right is called the score.
Show that it is impossible for the score to be 20. What is the highest
possible score?
Discussion
We introduce some terminology. The diagonal from top left to bottom
right is called the major diagonal. The diagonal from top right to bottom
left is called the minor diagonal.
There are at least two ways to solve this problem. We use A, B, C, D
and E to denote the numbers from 1 to 5 in some order. You could analyze
the whole square to find all possible distributions of letters which are con­
sistent with the conditions of the problem, and find the distribution which
maximizes the score. Alternatively you could focus on the four squares
immediately below the major diagonal, and see what the distribution of
letters can be like. In either event, to show that candidate for the maximum
score is achieved, you will have to give a specific grid which exhibits the
maximum value of the score. We adopt the second approach.
A Mathematical Olympiad Primer
90
Solution
If the score were 20, the numbers under the leading diagonal would all be
5s. The 5 in the top row must be in the top right position, and then there
can be no more 5s since there is one in every row and column. However
there is no 5 in the other long diagonal, so this is not a legitimate configu­
ration.
5
5
5
5
5
Next we strengthen the argument we have already.
(i) It is not possible to have the same number in the two middle two
positions of the scoreline. Suppose that A occurs in these two positions.
The A of the leading diagonal must be top left (or bottom right, but the
argument is then the same)1. The A of the minor diagonal must then be
in the second row, fourth column. Now there is no legal position for the
fifth A.
(ii) The same number cannot appear twice on the scoreline separated by
only one square (say in the first and third positions). This is because when
As are so placed, the A of the minor diagonal must be top right, and then
there is no place for an A on the leading diagonal.
’The professional way to point out that to understand what happens when A is top left
or bottom right of the leading diagonal, we need only consider the case that A is top left, is
to say that without loss of generality A is top left.
Chapter 7: BMO1 1997-98 Solutions
91
Now (ii) shows that it is impossible to have the same number three times
in the scoreline. Hence a score of 19, which could arise only from 5, 5, 5
and 4, is impossible. Similarly a score of 18 arising from 5, 5, 5 and 3 is
impossible.
(iii) If the same number A appears in the first (or last) two entries of the
scoreline, then the A entry of the minor diagonal must be in the top right
hand corner.
We are now in a position to finish the analysis. The possibility that a
score of 18 arising from two 5s and two 4s is yet to be eliminated. How­
ever 4455 and 5544 cannot arise because of (iii) (there being different rival
top right entries). 5445 and 4554 are eliminated by (i). 5454 and 4545 are
eliminated by (ii). Thus a score of 18 is impossible.
A score of 17 is possible as shown below.
92
A Mathematical Olympiad Primer
Problem 2
98. For n > 1 define an+2 to be the remainder of
Let
= 19, «2
an + a„+i when divided by 100. What is the remainder when
is divided by 8?
Discussion
We are interested in the remainder when a sum of numbers is divided by
8. Here is an ideal place to use modular arithmetic. Notice that for any
integers c and d, we have
(c + 4d)2 = c2 + 8(cd + 2d2).
Therefore c mod 4 determines c2 mod 8. Also observe that 4 divides 100.
Put these observations together in an intelligent way and you have the
outline of a solution.
If you were not using remainders when defining the sequence (flj,
then the recipe would be very similar to the Fibonacci sequence, except that
the starting values a-[ = 19 and r?2 = 98 are different from the usual Fi = 1
and F2 = 1- It is tempting to observe that working modulo 100 is the same
as looking at the last two digits when you write integers in the usual base
10 notation. We leave it to you to decide whether this is a sensible thing to
do.
Solution
Working with the representation of integers in a number base is usually a
bad idea, except when divisibility tests are relevant. Since o„+2 = an +
«n+i mod 100, then a„+2 = an + an+i mod 4. Since x2 mod 8 is deter­
mined by the value of x mod 4, it suffices to understand the sequence (a,)
modulo 4. This is
which repeats every six steps. Once the adjacent pair (3,2) is reached for a
second time then the recurrence ensures that this sequence repeats forever.
The sequence of (a2) modulo 8 is
Chapter 7: BMO1 1997-98 Solutions
93
Thus the sum of the repeating run of six quantities is 0 modulo 8. Since
1998 is a multiple of 6, the sum
------ 1- a 1998 mod 8
is a sum of Os and so is 0. Thus 8 divides the integer
Problem 3
ABP is an isosceles triangle with AB = AP and XPAB acute. PC is the
line through P perpendicular to BP, and C is a point on this line on the
same side of BP as A. (You may assume that C is not on the line AB.) D
completes the parallelogram ABCD. PC meets DA at M. Prove that M is
the midpoint of DA.
Solution
Drop the perpendicular from A to BP, meeting BC at X and BP at Y. To
Figure 7.1: Problem 3
BC .
show M is the midpoint of AD, it suffices to show 2 AM = AD
Notice that AXCM is a parallelogram so we need to show that X is the
midpoint of BC. However, this is clear because triangles BYX and BPC are
similar by angle considerations (ACBP is common, and ZBYX = ABPC
A Mathematical Olympiad Primer
94
because they are right angles. Since Y is the midpoint of BP, the sides
of the second triangle are twice the size of the first. Therefore X is the
midpoint of BC.
Problem 4
Show that there is a unique sequence of positive integers (a„) satisfying
the following conditions:
d\ = 1,
^n+l^n—1
fl2 =
#4 — 12
an ± 1 for n = 2,3,4,...
Discussion
This is quite a hard problem. We must do two things. We must show that
such a sequence exists, and moreover we must show that it is unique (in
other words, no other sequence will do the job). Existence is not so bad.
You follow the tried and tested route of numerical investigation. Calculate
small terms of the sequence, guessing a simple way to define the sequence,
and then proving that that your definition works nicely by means of an
inductive argument. Uniqueness is going to be a little delicate because of
the 'plus or minus' condition in the question. We are going to have to use
the fact that the largest integer which divides both a„ + 1 and a„ — 1 is
either 1 or 2. This is because (a^ + I) —
— 1) = 2.
Solution
If such a sequence exists, a^a2 —
±1
24 and so a$ = 5 is the only
possibility since 23 is not a perfect square. We can calculate further terms
and discover that any such sequence must begin
1,2,5,12,29,70,...
Thus is very suggestive; it seems possible that the sequence is governed
by the recurrence bn+2 = 2b„+1 + bn with bi = 1 and b2 = 2. Then b2 =
lb2 + b-[ = 5 and b$ = 2t>3 + b2 = 12. Therefore the sequence (&,) has the
correct first, second and fourth terms.
Of course we have arranged that b^ = b2b2 — 1. Now we prove by
induction that bn+1fe„_1 = b^ ± 1 for n > 2. The base case n = 2 is
Chapter 7: BMO1 1997-98 Solutions
95
immediate since 5 x 1 = 22 4- 1. The inductive step is to assume that the
result holds when n = k, and try to deduce that the result holds when n =
k + 1- Thus k > 2 and fyt+ifyt-i = b2 ± 1 is true. Now bk+2bk = (2bfc+i 4bk)bk = 2bk+1bk + bl = bk+1(bk+1 - bk^) +b% = b£+1 - bfc+1bfc_i + bl =
bk+i - (bk ± !) + bk = bk+i =F IIt follows by induction that there is at least one sequence (Fz) which
satisfies the conditions of the problem.
Finally we must establish uniqueness. We claim that any such se­
quence must be strictly increasing. Certainly
< a2 < «3- Suppose the
sequence strictly increases up to ak where k > 3. Now ak+-i ak_-y = zz2 ± 1.
If we arc dealing with a2 + 1 then ak+1 > ak and («x) is a strictly increas­
ing sequence. However, if we are dealing with zz2 — 1, then we must be
slightly more careful.
Now ak+iak_i = zz2 — 1 so ak+1ak > zz2 — 1. However, ak is a positive
integer so we may divide by it. Therefore ak+^ > ak — l/ak. Now ak > 1
so
> akak+i = ak, then akak-t = ak ~ 1 so ak divides 1 which is
absurd because ak > 1. We conclude that any (zzj satisfying the conditions
of the problem is a strictly increasing sequence.
Now suppose that (q) and (dz) are two integer sequences satisfying
the conditions of the problem (so they agree for at least the first 4 terms).
We may assume that k > 4 so ck_i,dk_-[ > 2. Suppose these sequences
coincide up to the subscript k. Both sequences are strictly increasing.
ck+-[Ck_-[ and dk+]dk_i either coincide (and ck+j = dk+1) or differ by 2.
In the latter case
Q+1Q-1 — ^k+lck-l
(since Q-i = dk_{), so ck_-[ is a divisor of 2, but Q-i > 2 so this is ab­
surd. Therefore (cz ) = (d,) and there is exactly one sequence of integers
satisfying the conditions of the problem.
Afterword
The final argument that there is a unique sequence is really an induction.
As we have suggested elsewhere, as long as the key steps of an induc­
tive argument are spelled out, it is not necessary to couch the argument
as a formal induction. The exception to this rule is when you are taking
a course in induction, in which case you may be required to express in­
duction arguments in strict verse form using a ritual incantation. This is a
typical deskilling process which has considerable value for weak students.
96
A Mathematical Olympiad Primer
Problem 5
In triangle ABC, D is the midpoint of AB and E is the point of trisection
of BC nearer to C. Given that ZADC = ABAE, find ABAC.
Solution
There are a very wide variety of ways to solve this problem. We suggest
three elementary solutions, and then a short sophisticated answer which
a student in IMO training might produce.
Figure 7.2: Problem 5
(a) Let P be the point on BA produced so that DA = AP. Now triangle
BPC is an enlargement of triangle BAE with centre B and scale factor 3/2.
Therefore PC//AE so ACPD = AEAD = ACDP so CP = CD. Now
A is the midpoint of the base of this isosceles triangle with apex C, so
CA ± DP. Therefore ABAC = 90°.
(b) [Outline] Instead of choosing P on BA, choose / on BC so that DJ
AE. Now worry a lot about similar triangles.
(c) Please turn this into a proof. Let X be the intersection of A E and
CD. Let ft be the area of triangle ADX, /3 be the area of triangle AXC, y
be the area of DBEX and 3 be the area of triangle XEC. Now use area
considerations to discover both that oc + y = 2(^ + 3) and y = oc + 23.
Trade off these equations to learn that a = /L Then deduce that XD = XC.
Now A is on the circle with diameter DC and so (by the theorem of Thales)
ABAC is a right angle.
Chapter 7: BMO1 1997-98 Solutions
97
Afterword
Finally we give a natural solution to this question which will be inacces­
sible to most BMO contestants because it requires an excessive dose of
education. This will be quick because we deploy results of advanced Eu­
clidean geometry which are ideally tailored to the problem. Students in
serious training to get into a national IMO team should certainly know
these results, and for such students this should be extremely easy.
Let F be on AC so that AE, CD and BF concur at X. By Ceva's theorem,
AF : FC = 2 : 1. Now by one of van Aubel's theorems
CX _ CF
CE
XD ~ FA + EB ~
so the circle on diameter DC passes through A and therefore, by the theo­
rem of Thales, ABAC is a right angle. You should find out about these theorems
if you harbour a serious interest in Euclidean geometry, or want to try for your
IMO team. This theorem is attributed to van Aubel on the internet but is rarely
mentioned in texts. Of course it is but a breath away from the proof we gave in
Solution (c). However, it is a very handy supplement to Ceva's theorem, and is
well worth knowing.
Chapter 8
BMO1 1998-99 Solutions
Problem 1
I have four children. The age of each child in years is a positive integer
between 2 and 16 inclusive and all four ages are distinct. A year ago the
square of the age of the oldest child was equal to the sum of the squares
of the ages of the other three. In one year's time, the sum of the squares of
the oldest and the youngest will be equal to the sum of the squares of the
other two children.
Decide whether this information is enough to determine their ages
completely, and find all possibilities for their ages.
Discussion
It seems inevitable that we must introduce unknowns for the ages of the
children, but when? Since we are given information both about their ages
a year ago, and in one year's time, the algebra should work out as neatly
as possible if we decide not to break the symmetry, so it is best to declare
the unknowns to be the ages of the children today.
Solution
Let their ages be a, b, c and d years now. Without loss of generality we may
assume that a > b > c > d.
A Mathematical Olympiad Primer
100
We are given that
(a - I)2 - (b - I)2 - (c - l)2 - (d - l)2 = 0
and
(fl + l)2 — (b + l)2 — (c + l)2 + (d + I)2 = 0
so by subtracting we learn that
4fl — 4b — 4c + 2d2 + 2 = 0.
Now d2 = 2(b + c — a) — 1 which is odd, and so therefore d is odd.
Now
d2 = 2(b + c - fl) - 1 < 2(c - 1) - 1 < 2(14 - 1) - 1
25
so d = 3 or 5.
If d = 5, then b + c — a = 13 so b — 2 > 13 so b > 15. Thus a =
16, b = 15, d = 5 and so c = 14. By checking, these values do not satisfy
the original pair of equations.
If d = 3, then b + c — fl = 5 so fl + 1 = b + c — 4 and fl — 1 = b + c — 6.
Therefore (b + c — 6)2 — (b — I)2 — (c — I)2 = 4 so 2bc — 12(b + c) +
36 + 2b + 2c — 2 = 4. This tidies to be — 6(b + c) + 18 + b + c = 3 or rather
be — 5b — 5c + 15 = 0. Factorizing we learn that (b — 5)(c — 5) = 10. One
possibility isb — 5 = 10, c — 5 = 1, fl = b + c — 5 = 15 + 6 — 5 = 16 giving
(a, b, c, d) = (16,15,6,3). Alternatively b — 5 = 5, c — 5 = 2, fl = b + c —
5 = 12, yielding (fl, b, c, d) = (12,10,7,3). One can verify that both of these
candidates for solutions actually work. (This observation must be made.)
Note that neither b — 5 = — 1, c — 5 = —10 nor b — 5 = —2, c — 5 = —5
yield solutions which are all positive integers, so these possibilities may
be discounted.
Afterword
A solution using a search strategy is also possible. The condition about
the ages one year in the future gives rise to a positive integer which can be
expressed as the sum of two squares in two different ways (a distant echo
of Ramanujan's observation about 1729). There is an elegant theoretical
way to find such numbers using algebraic number theory, but in this case a
brute force search works well. There are eight numbers in the appropriate
size range which are the sum of two squares in two different ways, of
which two (185 and 305) lead to solutions to the problem.
Chapter 8: BMO1 1998-99 Solutions
101
Problem 2
A circle has diameter AB and X is a fixed point on AB lying between A
and B. A point P, distinct from A and B, lies on the circumference of the
circle. Prove that, for all possible positions of P, the quantity
tan Z APX
tan Z PA X
remains constant.
Discussion
A construction is needed. Since we want to be able to understand both
Figure 8.1: Problem 2
ZAPX and ZPAX in terms of the diagram, what should one add to the
picture?
Solution
Let M be the foot of the perpendicular from X to AP. Now ZXMA is a
right angle so (by the converse of Thales's theorem1) M lies on the circle
on diameter AX. Therefore M is inside the original circle.
’in correct British English the possessive form of a singular name ending in 's' ends "s
apostrophe s" (with the standard Biblical exception). In some quarters this practice is viewed
with such horror that the cowardly "the theorem of Pythagoras" is sometimes replaced by
"the Pythagorean theorem". I have yet to see "the Thalean theorem" in print, but we await
developments.
A Mathematical Olympiad Primer
102
Now tan AAPX = XM/MP, and tan PAX = XM/AM so
tan AAPX _ AM
tan Z PA X “ ~MP
which we want to be fixed. This ratio is also AX/XB because triangles
AMX and APB are similar by angle considerations (ZA is common, and
ZX MA = AB PA is a right angle). Therefore
AM _ AP
AP - AM
MP
AX “ AB ~ AB - AX ~ XB'
so the ratio of tangents is independent of the location of P.
Problem 3
Determine a positive constant c such that the equation
xy2 - y1 - x + y =c
has precisely three solutions (x,y) in positive integers.
Discussion
It is possible to do this problem by algebraic cunning, but this time a brute
force search turns out to be faster and easier. After all, you are only re­
quired to find a single value of c which does the job, and any way that you
can lay your hands on such a c will be fine, as long as you can prove that
it has the required property.
Solution
Let /(x,y) = xy2 — y2 — x + y. Now
and this is 0 if y = 1, but is strictly positive if y > 1. Also
/(x,y + l)
f(x,y) = 2yx + x — 2y,
but
2yx + x - 2y = (x - l)(2y + 1) + 1.
Chapter 8: BMO1 1998-99 Solutions
103
Therefore
f(x,y + 1) - f(x,y) > 0 for x > 1.
Now, when y = 1, f(x,y) = 0. For x > 1 and y > 2, if you increase
either argument by 1, then the value of the function increases by at least 1.
Now make a table (see Table 8.1), with values of x labelling the columns
and values of y labelling the rows. We need not fill in the whole table of
course. Notice that 10 appears at least three times in the table. However,
2
3
4
0 0
5
1
0
0
0
2
1
4
7 10 13
3
2 10 18
18
11
10
12 11
Table 8.1
we can show that it never appears again. Certainly it does not appear in
the row corresponding to y = 1. In any other row or column it can appear
at most once. We need only search the rows corresponding to y < 11,
because after that 10 < /(l,y) < f(x,y).
Similarly we need to search only columns corresponding to x < 4,
because after that 10 < /(%, 2) < f(x,y).
It remain to show that 10 does not appear in the column corresponding
to x = 3. Now /(3,3) = 18 and if y > 3, then 10 < 18 < /(3,y). By
inspection there are no other occurrences of 10 in this column.
Afterword
There are lots of possible values of c which do the job. The one we have
given is the smallest and therefore the easiest to find by brute force search.
It is possible to show that any positive integer appears as a value of
f(x, y) (for positive integers x and y) only finitely many times. Every pos­
itive integer occurs at least once because of the column labelled x = 1.
A Mathematical Olympiad Primer
104
Many numbers (such as 3) appear only once. Are there any numbers
which occur more than 3 times?
Problem 4
Any positive integer m can be written uniquely in base 3 form as a string
of Os, Is and 2s (not beginning with a zero). For example
98 = (1 x 81) + (0 x 27) + (1 x 9) + (2 x 3) + (2 x 1)
= (10122)3.
Let c(m) denote the sum of the cubes of the digits of the base 3 represen­
tation of m; thus, for instance
c(98) = I3 + 03 + I3 + 23 + 23 = 18.
Let n be any fixed positive integer. Define the sequence (ur) by iq = n and
ur = c(ur_i) for r > 2. Show that there is a positive integer r for which
ur = 1,2 or 17.
Discussion
As almost always with sequence problems, calculate some terms and the
pattern should reveal itself. What happens here is that for large n, the
quantity c(n) is smaller than n. This means that when showing that suc­
cessive applications of c will eventually yield 1,2 or 17, you can ignore
large values of n. You must sort out what large means in this context.
Then simply calculate c(n) accurately for small values of n, and the proof
will appear in front of you.
Solution
Suppose that the positive integer n has f digits when written in base 3, so
n > 3f-1 and c(n) < 23f = 8f. Now if f > 5, then 8f < 3f-1. We prove this
by induction on t. It is true when f = 5 by inspection. Suppose the result
is true when t = m (at least 5). Now 8(m + 1) = 8m + 8 < 3"1-1 + 8 <
3'”-i 4- 2 • 3"'-1 = 3m. Therefore if f > 5, then c(n) < n.
If t = 4, then c(n) < 32, so c(h) < n for n > 33. Indeed, if 27 < n < 54,
then the first base 3 digit of n is 1 so c(n) < 1 + 24, so c(n) < n.
Chapter 8: BMO1 1998-99 Solutions
105
We conclude that c(h) < n for all n > 27. Successive application of
the c function will eventually take every positive integer below 27. The
problem is therefore reduced to the study of 26 cases, shown in Table 8.2.
(n)3
c(n)
n
(n)3
c(n)
n
(n)3
c(n)
111
2
2
8
3
10
1
4
11
2
5
12
9
6
20
8
7
21
9
8
22
16
9
100
1
10
11
12
13
14
15
16
17
18
101
102
110
111
112
120
121
122
200
2
9
2
3
10
9
10
17
8
19
20
21
22
23
24
25
26
201
202
210
211
212
220
221
222
9
16
9
10
17
16
17
24
n
Table 8. 2
By inspection, for every n except 17, 8, 7, 6,5, 2 and 1 we have c(n) < n.
Successive application of c must therefore eventually give rise to one of
these numbers. Since we aim to show that 1,2 or 17 must arise, it suffices
to look at the iterates of the others under application of c.
8 — 16 h-» 10
2,
7^ 9w 1,
6
8 and we have dealt with 8,
5 *—> 9 and we have dealt with 9.
Afterword
Successive applications of c in every case eventually get stuck in one of
the three periodic cycles 2 i—» 8 i—> 16 >—> 10 i—> 2,1 i—> 1 or 17 h-> 17.
There is an echo here of the statistical card trick attributed to Martin
J. Kruskal. If you are unfamiliar with this, please search the internet for
"Kruskal's card trick".
106
A Mathematical Olympiad Primer
Problem 5
Consider all functions f from the positive integers to the positive integers
such that
(i) for each positive integer m, there is a unique positive integer n such
that f(n) = m;
(ii) for each positive integer n, we have /(n + 1) is either 4/(n) — 1 or
Find the set of positive integers p such that /(1999) = p for some function
f with the properties (i) and (ii).
Solution
Suppose that we have such a function f. We consider the sequence /(l),
/(2), /(3), ... and so on. By condition (i), every integer occurs in this
sequence exactly once.
The key observation is that if n > 2 and f{n) occurs before f(n) — 1 in
this sequence, then/(n +1) = /(n) — 1. We will explain why this happens
in our analysis below.
Suppose that /(I) = 1, then /(2) / 0 since 0 is not a positive integer,
but 0 = /(I) — 1. Therefore /(2) = 4/(1) — 1=3. Now 2 must occur in
the sequence, and if f(n + 1) = 2, then f(n) = 3. Therefore n = 2 since
3 occurs only once in the sequence, and so /(3) = 2. The sequence must
begin 1,3,2. Each term of the sequence is obtained from its predecessor
either by decrementing (that is, subtracting 1) or by multiplying by 4 and
subtracting 1. Therefore /(4) is either 1 or 7. However, 1 has already
appeared in the sequence so /(4) = 7, and the sequence begins 1,3,2,7.
The next term must be either 6 or 27. We now address the remark in
the second paragraph of this solution. It asserts that if decrementing the
sequence does not cause repetition of an entry, then you must do it. The reason
is perfectly illustrated by the situation in which we now find ourselves. If
we do not decrement the sequence now, the next term will be larger than 7.
Since it is required that eventually 6 must appear in the sequence, then a
later decrementation from 7 to 6 must occur. However, this would involve
7 appearing more than once in the sequence which is absurd.
The sequence therefore begins
Chapter 8: BMO1 1998-99 Solutions
107
and so on. In general f(2n + k) = 2"+1 — (fc + 1) where n = 0,1,2,...
and 0 < k < 2" — 1. In this case, since 1999 = 1024 + 975 = 210 + 975 so
/(1999) = 2048 - 976 = 1072.
The only other possibility is that f(l)
1. The predecessor of 1 in
the sequence must be 2 and the successor of 1 must be 3. Therefore no
predecessor of 2 is possible so the sequence must begin
Using the key observation, the next term must be 11. Next there must be
a decrementing run until 4 is reached. The sequence must therefore begin
2,1,3,11,10,9,8,7,6,5,4,15,14,13,12,47,46,...
and then a long decrementing run until 16 is reached. Thus
f(4n+k) = 3 x 4n - (k + l)
when n = 0,1,2,... and 0 < k < 2 x 4n — 1 and
f(3x4n+k) = 4”+1 — (fc +1)
when n = 0,1,2,... and 0 < k < 4" — 1.
Now 1999 = 1024 + 975 = 45 + 975 = 3 x 1024 - 976 = 2096. There­
fore the possible positive integer values of p are 1072 and 2096.
Chapter 9
BMO11999-2000 Solutions
Problem 1
Two intersecting circles Cy and C2 have a common tangent which touches
Ci at P and C2 and Q. The two circles intersect at M and N, where N is
nearer to PQ than M is. The line PN meets the circle C2 again at R. Prove
that MQ bisects APMR.
Discussion
If you know the circle theorems, you should be able to deal with this.
Figure 9.1: Problem 1
A Mathematical Olympiad Primer
110
Draw a diagram using good geometrical instruments. If you cannot see
your way to a solution directly, try working backwards; begin by assum­
ing that MQ bisects Z.PMR and try to deduce something you know to be
true. Then see if this argument reverses. If that doesn't work, try staring at
your diagram to see if you can spot a pair of triangles which look similar.
Solution
(a) ZQMR = AQNR by angles in the same segment and ZQNK =
ANPQ + APQN since AQNR is an exterior angle to triangle NPQ.
Now ANPQ = ZNA4P and APQN = ZQMN, both by angles in the
alternate segment. Next observe that ZQMP = AQMN + ANMP since
the angles are juxtaposed. Therefore AQMP = ARMQ.
(b) We invite the you to find another solution based on showing that
triangles RMQ and QMP are similar.
Problem 2
Show that, for every positive integer n.
121" -25" + 1900" - (-4)"
is divisible by 2000.
Discussion
If this result is true, then it must be possible to demonstrate this fact by
working modulo 16 = 24 and modulo 125 = 53 separately, and combining
the results. However, there is a more attractive solution, based on the
fact that for all integers x and y, (x — y) is a factor of x" — y". There are
two ways to group the terms so that our expression is the sum of two
things, each of which is the difference of two n-th powers. Use these two
possibilities wisely.
Solution
Observe that if x and y are integers, then
x"-y" = (x —y)(x" 1 + x" 2y + x" 3y2 + • • •-y"
Chapter 9: BMO1 1999-2000Solutions
111
and so is divisible by x — y. We use this observation twice. Notice that
16 divides 96 which is 121 — 25, and also that 16 divides 1904 which is
1900 — (—4). Therefore our expression is divisible by 16. Reorganizing,
we have that 25 divides 125 which 121 — (—4), and 25 divides 1875 which
is 1900 — 25. Therefore our expression is divisible by 25. Now 16 and 25
are coprime so, independent of n, our expression is divisible by 16 x 25 =
2000.
Afterword
One can also do this problem by modular arithmetic, using 16 and 25 as
moduli, but this is not necessary. However, if that is the route you discover
first, then so be it.
Problem 3
Triangle ABC has a right-angle at A. Among all points P on the perimeter
of the triangle, find the position of P such that AP + BP + CP is mini­
mized.
Solution
If P is not on the hypotenuse BC, then without loss of generality it is on the
side CA. AP + CP is independent of the position of P, so AP + BP + CP
is minimized when BP is minimized, and this happens when P is at A.
A
Figure 9.2: Problem 3
If P is on the hypotenuse BC, then BP + CP is independent of the posi­
tion of P, so AP + BP + CP is minimized when AP is minimized, and this
happens when P is at the foot of the altitude from A.
A Mathematical Olympiad Primer
112
Let the altitude from A have length d. Using the standard triangle no­
tation, we must compare b + c and a + d. Given that Pythagoras's theorem
is lurking not far away, it seems sensible to compare these lengths by com­
paring their squares.
(a + d)2 — (b + c)2 = (a2 — b2 — c2) -I- (2ad — 2bc) + d2 = d2.
The two terms in brackets both vanish, the first because of Pythagoras's
theorem, and the second because the area of the triangle is both be/2 and
ad/2.
Hence a + d > b + c and the solution is when P is at A.
Problem 4
For each positive integer k > 1, define the sequence (aM) by
ao = 1,
an = kn + (— 1
for each n > 1. Determine all values of k for which 2000 is a term of the
sequence.
Discussion
As usual with sequence questions, it will prove helpful to gather some
data. The condition is a little cumbersome, so it is important to take great
care to get the algebra correct.
Solution
The sequence
• • • begins as follows (and we group the terms by
looking at successive runs of four terms):
1,
4k+ 1,
8k + 1,
12k+ 1,
k-1,3k-1,
k-1,7k-1,
k-1,llk-1,
k-1,15k-1,
1,
1,
1,
1,
Chapter 9: BMO1 1999-2000Solutions
113
Thus it appears that the sequence takes the form
a$m = 4mA: + 1;
^4m+l — A'
1/
«4m+2 = (4m + 3)fc-l;
^4m+3 = I'­
ll is then a simple matter of technique to verify this. It is certainly true
when m = 0. We begin the inductive step by assuming that the formula is
correct when the subscript is 4m so
a4m = 4mA: + 1;
fl4w+i = k(4m + 1) — (4mk + 1)
= k - 1;
fl4m+2 = k(4m + 2) + k — 1
= (4m + 3)k — 1;
fl4w+3 = k(4m + 3) — ((4m + 3)k — 1)
= 1;
fl4m+4 = A(4m + 4) + 1
= 4(m + l)k + 1.
Once this is done it is easy to finish the problem. We conclude that 2000
arises as a term of the sequence if any one of the following four conditions
holds:
(i) 4mA: +1
2000; this is never satisfied.
(ii) k - 1 = 2000, that is, k = 2001.
(iii) (4m + 3)k - 1 = 2000, that is, k = 2001/(4m + 3).
(iv) 1 = 2000 which is never satisfied.
Condition (iii) is interesting. The prime factorization of 2001 is 3 x 23 x 29.
The only factors congruent to 3 modulo 4 are therefore 3,23,87 and 667.
These give rise to the following values of k: 667, 87, 23 and 3 [in addition
to k = 2001 arising from (ii)].
114
A Mathematical Olympiad Primer
Problem 5
The seven dwarfs decide to form four team to compete in the Millennium
Quiz. Of course, the sizes of the teams will not all be equal. For instance,
one team might consist of Doc alone, one of Dopey alone, one of Sleepy,
Happy and Grumpy, and one made up of Bashful and Sneezy. In how
many ways can the four teams be made up? (The order of the teams or the
order of the dwarfs within the team does not matter, but each dwarf must
be in exactly one of the teams.)
Suppose that Snow White agreed to take part as well. In how many
ways could the teams then be formed?
Discussion
Perhaps the most controversial aspect of this problem is the spelling of
the plural of dwarf. As usual with these enumeration problems, any suffi­
ciently accurate and systematic attempt at a solution is bound to succeed.
However, the use of binomial coefficients (") leads to a succinct solution.
Solution
First the case without Snow White. The distribution of team sizes must be
4,1,1,1; 3,2,1,1 or 2,2,2,1. The case first case arise in (4) = 35 ways. The
second arises in (3) x (f) = 35 x 6 = 210 ways. The third case arises in
105
ways. We divide by 3! because the three teams of 2 may occur in any of 6
orders. The total number ways to make up four sides is 35 + 210 + 105 =
350.
Now with Snow White, the distribution of team sizes must be 5,1,1,1;
4,2,1,1; 3,3,1,1; 3,2,2,1 and 2,2,2,2. The first case arises in (|) = 56 ways.
The second case arises in (®) x
= 420 ways. The third case arises in
(3) x (|)/2! = 280 ways. The fourth case arises in (|) x (|) x (3)/2! = 840
ways. The fifth case arises in (8) x (^) x (^)/4! = 105 ways. This gives a
total of 56 + 420 + 280 + 840 + 105 = 1701 ways.
Chapter 10
BMO1 2000-01 Solutions
Problem 1
Find all two-digit integers N for which the sum of the digits of 10N — N is
divisible by 170.
Discussion
There is a neat trick which simplifies the problem of having to worry about
all the "carrying" involved in the subtraction. Find it.
Solution
To eliminate carrying notice that
10N - N = (10N - 1) - (N- 1)
and let N — 1 = 10c + d where c,d are integers and 0 < c,d < 9. The
digit sum is 9(N — 2) + (9 — c) + (9 — d) — 9(10c + d — 1) + 18 — c — d =
89c + 8d + 9. We want this quantity to be divisible by 170. For this to
happen it is necessary that c = 2e + 1 is odd for an integer e in the range
0 < c < 4. Thus we need 170 to divide 178c + 8d + 98 or equivalently to
divide 8e + 8d + 98 = 2(4c + 4d + 49). This is the same as asking for 85 to
divide 4c + 4d + 49. Since 0 < c < 4 and 0 < d < 9 this happens if and
only if 4c + 4d + 49 = 85, that is, e + d = 9. Since c = 2c + 1 and d = 9 - c
we see that the possibilities for N are 20,39,58,77 and 96.
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A Mathematical Olympiad Primer
Problem 2
Circle S lies inside circle T and touches it at A. From a point P (distinct
from A) on T, chords PQ and PR of T are drawn touching S at X and Y
respectively. Show that /.QAR = 2AXAY.
Discussion
This problem solves itself. Draw the diagram carefully and let ZXAY = a.
Figure 10.1: Problem 2
You want to show that AQAR = 2a. What would it be necessary for other
angles to be in order that this be the case?
Solution
Let AXAY = a, Now by angles in the alternate segment (twice) we have
ZPXY = ZXYP = a. In triangle XYP, the angles sum to 180° so ZRPQ =
180° — 2a. Now QARP is a cyclic quadrilateral, so opposite angles sum to
180°. Therefore XQAR = 2oc = 2ZXAY.
Chapter 10: BMO1 2000-01 Solutions
117
Problem 3
A tetromino is a figure made up of four unit squares connected by common
edges.
1. If we do not distinguish between the possible rotations of a tetro­
mino within its plane, prove that there are seven distinct tetrominos.
2. Prove or disprove the statement: it is possible to pack all seven tetrominoes is a 4 x 7 rectangle without overlapping.
Discussion
These shapes will be familiar to anyone who has played the computer
game tetris. Writing out a proof that there are exactly 7 different tetrominoes may not be easy, even though fiddling around with pictures it is easy
to convince yourself that it is true. Of course a variety of arguments are
possible, but whatever you decide to do, be clear and systematic. Ideally
we want an argument which counts the shapes irrespective of rotations.
Solution
Consider a tetromino where no three unit squares occur in a line. Consider
an adjacent pair of unit squares in the tetromino. Choose a unit square of
tetromino which shares an edge with one of this pair. This must form a
capital ell shape. Now the fourth unit square must share an edge with this
capital ell. We can have a 2 x 2 square, or two types of zig-zag.
If three unit squares occur in a line, we can form a long 4x1 shape, or
two types of long corner, or a tee.
There are seven shapes. Suppose (for contradiction) that they could fit
together to form a 7 x 4 rectangle. Colour the rectangle black and white
like a chessboard, so there are 14 white squares and 14 black ones. Six of
the tetromino shapes cover the same number of blacks as whites, but the
tee shape does not. It covers three of one colour and one of the other. This
is the required contradiction, so the tetrominos can not but put together to
form a 7 x 4 rectangle.
A Mathematical Olympiad Primer
118
Problem 4
Define the sequence (an) by
an = n + {a/h}
where n is a positive integer and {%} denotes the nearest integer to x,
where halves are rounded up if necessary. Determine the smallest integer
k for which the terms
ak'ak+V ■ ■ -'ak+2000
form a sequence of 2001 consecutive integers.
Discussion
This "nearest integer" function is unusual. Before charging ahead, sort
out for which positive integers n we have {y/n} = m.
Solution
We first examine this "nearest integer" function of positive integers. Now
Therefore we get a consecutive sequence of integers an = n + m while
2
,1^
.
2
,
.
1
m — m H— < n < m + m 4—
4 —
4
for some integer m. The sequence increases by 2 when n < (m + 1 )2 < n +
1 since then
= n + m + 2 but an — n + m. A consecutive run therefore
has 2m integers from m2 -- m + 1 to m2 + m. We require 2m > 2001, and
the smallest value of m which will do this is m = 1001 and the smallest
k so that ak>ak+i>- ■ ->ak+2000 are consecutive is k = 10012 — 1001 + 1 =
1001001.
Problem 5
A triangle has sides of length a, b and c and its circumcircle has radius R.
Prove that the triangle is right-angled if and only if a2 + b2 + c2 = 8R2.
Chapter 10: BMO1 2000-01 Solutions
119
Discussion
This is a particularly sweet condition because the given condition is sym­
metric in a, b and c. If you ask for a condition on the sides for there to be
Figure 10.2: Problem 5
a right-angle at C, then of course up pops c2 = a2 + b2 or equivalently
c2 — a2 — b2 = 0. Therefore one of the three angles of ABC is a right angle
if, and only if,
(c2 — a2 — b2)(b2 — c2 — a2) (a2 — b2 — c2) = 0
which is a bit of a mess. The first time that you see it, it must be a sur­
prise that by introducing the circumradius, one can get a natural algebraic
condition for there being a right-angle which has degree 2 rather than de­
gree 6.
Solution
We first give a "broken" solution that contains a subtle error of logic. You
have the opportunity to detect the flaw for yourself. Let this be a lesson to
those who assert that mathematical proof gives the only form of absolute
knowledge. Any proof that has ever been written might be wrong, and all
its readers may have overlooked the error. Of course, if there are a wide
variety of different 'proofs' of a result, then it is highly unlikely that they
are all wrong. However, you can never be absolutely sure.
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A Mathematical Olympiad Primer
(a) (Incorrect solution) If a triangle a, b, c has a right angle at ZA, then
by Pythagoras's theorem a2 = b2 + c2. The circumradius will be half the
length of the hypotenuse so
The symmetry of the formula means that the right angle can be at any of
the vertices and the formula still holds.
Now for the reverse implication. We suppose that 8R2 = a2 + b2 + c2.
Choose X on the circumcircle antipodal to B. Therefore \BX\ = 2R and
both ABAX and ZBCX are right angles. Let |AX| = h and |CX| = k.
The theorem of Pythagoras tells us that 4R2 = h2 + c2 and 4R2 = k2 + a2.
Adding these equations we learn that
8R2 = a2 + (h2 + k2) + c2.
Now b2 = h2 + k2 if and only if AAXC is a right angle, that is, if and only
if AC is a diameter which happens if and only if AABC is a right angle.
So, can you spot the mistake? The problem lies in the reverse implica­
tion. It must be wrong because it claims to show that if 8R2 = a2 + b2 + c2,
then AABC must be a right angle. This is wrong, because sometimes
8R2 = a2 + b2 + c2 when ZABC is not a right angle. It certainly hap­
pens when ABAC is a right angle or when ABC A is a right angle. Now, if
you have not already spotted the flaw in (a), return to it again armed with
this new insight and find the mistake. We correct it in the next solution.
(b) Look at the reverse implication carefully. The problem arises when
it says "Now b2 = h2 + k2 if and only if AAXC is a right angle, that is,
if and only if AC is a diameter". The problem is that b2 = h2 + k2 can
arise in degenerate circumstances, when h = 0 or k = 0. In the first case
X = A and the second case X = C. In either case the "triangle" AXC is
degenerate, and while you may deem it true that AAXC is a right angle,
it does not follow that AC is a diameter. Indeed, if h = 0, then AB is a
diameter and if k = 0, then BC is a diameter. Thus we conclude that one
of the angles of triangle ABC must be a right angle.
(c) This solution tells us more than the question requires. We learn that
8R2 — (a2 + b2 + c2) is negative if ABC is acute, positive if the triangle is
obtuse, and vanishes if the triangle is right-angled (a sort of transitional
Chapter 10: BMO1 2000-01 Solutions
121
case). In the unbroken section of (a), we showed that if ABC is right an­
gled, then 8R2 — (fl2 + b2 + c2) = 0.
If triangle ABC is obtuse, then the circumcentre lies outside the trian­
gle. Label the obtuse angle C and let X be the antipodal point to B. We
borrow the notation from (a), so h = |AX| and k = |CX|. Now ACBA is
acute, so by angles in the same segment, ACXA is acute. By the cosine
rule, b2 < h2 + k2 so
Note that the application of the theorem of Pythagoras is allowed since
ZBCX and ZB AX are angles in a semicircle (the theorem of Thales).
On the other hand, if triangle ABC is acute, then ZCXA is opposite
ZABC in the cyclic quadrilateral ABCX, so ZCXA is obtuse. Hence b2 >
h2 + k2 and, arguing in much the same way as in the obtuse case, we learn
that
(d) Here is a proof by trigonometry. You will need to know some stan­
dard trigonometry formulas to follow this. Let O be the circumcircle of
ABC; then by considering the cosine rule in triangle BOC, we learn that
a2 = 2X2(1 — cos2A).
There are two similar formulas. If we add we learn that the condition
8R2 = a2 + b2 + c2 is equivalent to
cos 2A + cos 2B + cos 2C + 1 — 0.
Now A + B + C = 180° so
cos2C = cos(360° - 2(A + B)) = cos2(A + B).
Also for all angles ft and fl we have
COS
A Mathematical Olympiad Primer
122
Hence
cos 2A + cos 2B + cos 2C + 1 = 0
2cos(A + B) cos(A — B) + 2cos2(A + B) = 0
2cos(A + B)(cos(A — B) + cos(A + B)) =0
2cos(A + B)(cos(A — B) + cos(A + B)) =0
cos(A + B) cos A cos B — 0
A = 90°, B = 90° or C = 90°.
Afterword
We finally give a proof by gratuitous use of advanced theorems. This ar­
gument is in the spirit of Mathematics made difficult [13]. If you go to the
right social gatherings, you will meet many people who know that
OH2 = 9R2
(a2 + b2 + c2).
Now R2 — OH2 (which is a2 + b2 + c2 — 8R2) is 0 if and only if H is on the
circumcircle, Thus we are asked to show that the triangle is right-angled
if and only if H is on the circumcircle. If a triangle is right-angled, then
the right angle vertex is H and is on the circumcircle by the converse of
the theorem of Thales. Conversely, if H is on the circumcircle, then by
the Simson-Wallace theorem, the points on the side lines which are the
feet of the altitudes are collinear. However, these three points lie on the
nine-point circle, so two of the points coincide. Thus two altitudes meet at
a triangle vertex. Therefore two altitudes are triangles side lines and the
triangle contains a right angle.
Chapter 11
BMO1 2001-02 Solutions
Problem 1
Find all positive integers m, n, where n is odd, that satisfy
Discussion
This is one of those number theory questions which depend on under­
standing a factorization. Modular arithmetic will not necessarily be very
helpful.
Solution
There are several methods which work, and we give the best. Multiply­
ing by 12mn we obtain the equivalent condition that 12n + 48m = mn or
equally well
mn — 12n — 48m = 0.
Equivalently
mn — 12n — 48m + 576 = 576,
or in other words
(m — 12)(n — 48) = 576.
A Mathematical Olympiad Primer
124
The original statement of the problem forces m > 12 and n > 48 since
both \/m < 1/12 and 4/n < 1/12, so both m — 12 and n — 48 are positive
integer factors of 576 = 2632. There is no need to fret about the possibil­
ity of factorizing 576 into negative integers. Since n is odd, n — 48 must
be 1, 3 or 9; corresponding values of m — 12 being 576, 192 and 64 re­
spectively. Therefore the only candidates for values of the pair (m, n) are
(588,49), (204,51) and (76,57). Substituting into the original problem, we
discover that each of them is a genuine solution.
Problem 2
The quadrilateral ABCD is inscribed in a circle. The diagonals AC, BD
meet at Q. The sides DA, extended beyond A, and CB, extended beyond
B, meet at P.
Given that CD = CP = DQ, prove that ACAD = 60°.
Discussion
This is a BMO1 geometry question, so nothing serious is likely to be going
on. Chasing angles by brute force is usually enough at this level, and it
will work here. Focus on the angles of triangle BDC.
P
Figure 11.1: Problem 2
Chapter 11: BMO1 2001-02 Solutions
125
Solution
Let ACPD = 0. Since CP = CD it follows that APDC = 0 and Z.DCP =
180° - 20. Let ZCQD = </>. Since DC = DQ it follows that ZDCQ = <p
and AQDC = 180° — 2(p. Next ACBD = ZCAD by angles in the same
segment. However APAC = Z0 + Z</> since this is an exterior angle to
triangle ADC. Therefore ACBD = 180° — 0 — (p.
Now the angles of triangle BDC sum to 180° so
180° - 2(p + 180° -20 + 180° - 0 - <p = 180°
so
A0 + A(p = 120°
and so ACAD = 60°.
Problem 3
Find all positive real solutions to the equation
x
x
2x
where [tJ denotes the largest integer less than or equal to the real num­
ber t.
Discussion
There are ways to solve this problem correctly but expensively, using up
time and paper constructing large tables of information. Please try to
avoid this if you can.
If you stare at the equation, you should learn something to your ad­
vantage. In particular, you may be able to avoid beginning the solution by
writing “Suppose that x — n + r where n is an integer and 0 < r < 1." An­
other tip: look for an integer k such that if % is a solution to the equation,
then so is x ± k.
Solution
First note that if x is a solution, then
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A Mathematical Olympiad Primer
is an integer.
Also observe that if the integer x is a solution, then so are x + 6 and
x — 6. This is because
whereas
Thus it remains to test a set of six consecutive integers. By inspection
0,2,3,4 and 5 are solutions but 1 is not. Therefore all integers are solutions
except those that leave remainder 1 on division by 6, and they are not
solutions.
Problem 4
Twelve people are seated around a circular table. In how many ways can
six pairs of people engage in handshakes so that no arms cross? (Nobody
is allowed to shake hands with more than one person at once.)
Discussion
We shall assume no funny business with people trying to shake hands
behind one another's backs. We replace 12 by an arbitrary even number
2n. Let us call the number of ways of shaking hands Cn. Pick on a per­
son. He or she must shake hands with someone so that their joined arms
divide the other groups into two groups of even size, otherwise no legal
arrangement is possible. Moreover, if they do this, then by induction the
remaining two groups can each come to legal handshaking arrangements.
Note that it is possible for immediate neighbours to shake hands, because
then they divide the remaining population into one group of size n — 2
and one group of size 0.
It helps to keep the mathematics clean if we agree that a group of size
0 can shake hands legally in exactly one way!
Solution
Given this convention, we see that
+ Cn_ ] Co.
Chapter 11: BMO1 2001-02 Solutions
127
This is a recurrence, but a slightly complicated one.
Ci = 1 clearly, and this is consistent with our convention about Co,
because Q = C0C0 = 1. Then C2 = C0Ci + CiC0 = 2, C3 = 2 + 1 + 2 = 5,
C4 = 5 + 2 + 2 + 5 = 14, C5 = 14 + 5 + 4 + 5 + 14 = 42 and finally
C6 = 42 + 14 + 10 + 10 + 14 + 42 = 132. Thus the answer is 132.
Afterword
This sequence of numbers Co, Q, C2,... is well-known, and is called the
Catalan sequence. There is a closed form (that is, a formula) for Cn, but it is
a little hard to spot. It is
The best way to understand the Catalan sequence is by introducing a gen­
erating function C(x) (a formal power series) with coefficients which are
the Catalan numbers. Then
C(x) = Co + Cix + C2x2 + • • • .
Therefore
because of the recurrence, so
xC(x)2 — C(x) + 1 = 0.
Now, and this needs more justification that we can afford, we treat this as
a quadratic equation in C(x) and solve.
1 + \/l - 4x2
C(x) =
2x
Those of you who know the binomial theorem for a fractional exponent
should press on and get their true reward.
The Catalan sequence occurs in many guises. For instance, if you are
given n letters as a string, and you want to compose them using a binary
product, how many ways are there to do this? For example xyz can be
bracketed in 2 ways — (xy)z and x(yz) — and in no other way. However,
10 xyz can be bracketed in 5 ways: ((wx)(yz)), ((w(xy))z), (((wx)yjz),
(w(x(yz))) and (w((xy)z)). What is the connection with handshakes?
A Mathematical Olympiad Primer
128
Problem 5
f is a function from Z+ to Z+, where Z+ is the set of non-negative integers, which has the following properties:-
(a) f(n + 1) > /(«) for each n E Z+,
(b) f(n + f (nz)) = /(n) 4- tn + 1 for all m,n E Z+.
Find all possible values of /(2001).
Discussion
Condition (a) looks very useful. It is easy to work with the statement that
f is a strictly increasing function. On the other hand, condition (b) looks
worryingly complicated. However, since this equation is supposed to hold
for all possible values of m and n, it must hold for particular values of m
or n. Which values of tn or n lead to a simple condition?
Solution
Condition (a) tells us that the function f is strictly increasing. Let m = 0
in (b), so we learn that
/n+/(0))=/(n)+l
for every n E Z+. From this we can immediately conclude that /(0)
0.
Next we suppose (for contradiction) that /(0) > 2. Then n < n + 1 <
n + /(0) so f(n) < f(n + l) < f(n) +1 which impossible because/(n +1)
must be an integer between to consecutive integers.
The only remaining possibility is that /(0) = 1, in which case f(n +
1) = /(h) + 1 for every n E Z+. By induction /(%) = x + 1 for all x E Z+.
We must verify that this / satisfies both the original conditions (this is
crucial, and marks would be deducted if this step were omitted). Certainly
(a) holds. Condition (b) becomes n + tn + 2 = (n + 1) + m + 1 for every
tn, n E 2ZP which is correct.
Therfore there is a unique function / satisfying the given conditions,
and the only possible value of /(2001) is 2002.
Chapter 12
BMO1 2002-03 Solutions
Problem 1
Given that
34! = 295 232 799 cd9 604 140 847 618 609 643 5ab 000 000,
determine the digits a, b, c and d.
[Mr Adrian Sanders, Trinity College, Cambridge]
Discussion
34! is divisible by lots of positive integers, so it must pass lots of divisibility
tests.
Solution
The factorization of 34! is
K • ll3 • 74 • 57 • 315
where K is a product of primes which are all at least 13. Thus 107 divides
34! but 108 does not. Therefore b = 0. Now 34!/107 is divisible by 8,
which happens if, and only if, its last three digits, viewed as a number, is
divisible by 8. Now 352 is divisible by 8 so a = 2.
130
A Mathematical Olympiad Primer
Now 34! is divisible by 9 so its digit sum is divisible by 9. Thus c + d =
3 or 12. Also 34! is divisible by 11. The test for divisibility by 11 is that the
difference between the sum of the odd placed digits and the sum of the
even placed digits should be divisible by 11. Thus d — c — 3 is a multiple of
11 so d = c + 3orc = d + 8. Solving each of the two possibilities obtained
from considerations modulo 9 against the two possibilities obtained from
considerations modulo 11, given that c and d are digits, yields a unique
solution c = 0, d = 3. Therefore (a, b, c, d) = (2,0,0,3) and we have a
reference to the year in which the competition was sat.
Problem 2
The triangle ABC, where AB < AC, has circumcircle S. The perpendicular
from A to BC meets S again at P. The point X lies on the line segment AC,
and BX meets S again at Q. Show that BX = CX if, and only if, PQ is a
diameter of S.
[Dr Gerry Leversha, St Paul's School, London]
Discussion
The condition AB < AC is there for two reasons. It AB = AC, then Q = A.
This does not make the conclusion wrong, but it would lead to complica-
Figure 12.1: Problem 2
tions in the proof if it were allowed. It would require a separate diagram,
Chapter 12: BMO1 2002-03 Solutions
131
possibly a separate proof (any proof using an angle such as ZAQB would
be invalid). If AB > AC then the conclusion would hold, but you would
have to allow X to be on the line AC, and not just on the line segment AC,
and a third diagram would be necessary.
In the IMO, geometry problems requiring multiple diagrams are cer­
tainly in order. For BMO1, it is probably asking a bit much of candidates
that they be sufficiently professional as to draw diagrams for all cases with
suitably adjusted proofs. It would also make the marking much more
complicated. The setting committee clearly decided that they just wanted
to see if students could do the geometry, and did not want the problem
any more complicated than as stated.
In the solution, you may find the theorem of Thales or its converse
useful. The direct form says that the angle in a semicircle is a right angle.
The converse form is that if AU VW is a right angle, then V is on the circle
with diameter UW.
Solution
XB = XC <4> ZCBX = ZBCX since X is not on BC. (If Y is a point in
the interior of the interval BC then ZCBY = ZBCY = 0° but in general
YB
YC.) Now ZCBX = ZBCX if, and only if, ACAQ = ABQA by
angles in the same segment (twice). This condition forces BC || AQ by
alternate angles. Conversely if BC || AQ, then ZCBX = ABQA (alternate
angles) and ABQA = ZBCA by angles in the same segment. Thus our
original condition is equivalent to AQ || BC. However AP ± BC so our
condition is equivalent to APAQ being a right angle. By the Theorem of
Thales (on angles in a semicircle) and its converse, this is equivalent to PQ
being a diameter of circle S.
Afterword
The intersecting chord theorem is also at work here. Consider the chords
AC and BQ which intersect at X. Now BX = CX O QX = AX, and
for the usual reasons triangle XQA is similar to triangle XBC. This gives
another way to launch the argument.
A Mathematical Olympiad Primer
132
Problem 3
Q
Q
Let x, y, z be positive real numbers such that x + y + z
,
2
2
।
x yz + xy z + zyz
Q
1. Prove that
2
[Dr Gerry Leversha, St Paul's School, London]
Discussion
The arithmetic-geometric mean (AM-GM) inequality is quite well-known.
The three variable version is
tyxyz <
x+y+z
whenever x, y and z are positive reals.
Another important way to average a collection of numbers has two
names. The quadratic mean or root mean square of the n real numbers %i,
•••z
is
The AM-QM or AM-RMS inequality asserts that AM < QM (or AM
RMS). In our three variable case, this asserts that
You can just quote AM-GM and AM-QM. For proofs of these standard
facts, see Christopher Bradley's double text [2], You can get AM — QM
either by cooking up an appropriate Cauchy-Schwarz inequality, or by a
suitably cunning application of the rearrangement inequality to (x + y + z)2.
You may need to look up the rearrangement inequality.
Solution
GM < AM < QM (note that x,y,z are positive so AM-GM applies).
Therefore
3
3
Chapter 12: BMO1 2002-03 Solutions
133
so both
and
x + y + z < v3
Multiplying these last two inequalities between positive quantities, we ob­
tain the required result.
Afterword
Both AM-GM and AM-QM are equalities if, and only if, the variables are
equal. Equality is achieved in the inequality of this problem by setting
x = y = z = 4=.
Problem 4
Let m and n be integers greater than 1. Consider an m x n rectangular grid
of points in the plane. Some k of these points are coloured red in such a
way that no three red points are vertices of a right-angled triangle two of
whose sides are parallel to the sides of the grid. Determine the greatest
possible value of k.
[Dr Matthew Payers, Magdalene College, Cambridge]
Discussion
This is a hard BMO1 problem. The Dirichlet principle (the pigeon-hole
principle) must play an important role.
Solution
We may suppose without loss of generality that there are m rows and n
columns. The greatest possible value of k is at least m + n — 2 because you
can fill a row and a column with red points, except for their intersection.
In fact you cannot do better, and this is the greatest possible value for k as
we now show.
Suppose for contradiction, that one can find m + n — 1 red points sat­
isfying the condition.
We begin by showing that there is a pair of red points which are neither
in the same row nor the same column. Distinguish a red point. If there is
a red point not in the same row or column as the special point, we are
134
A Mathematical Olympiad Primer
done, so we may assume that all the other red points are in the union of
the row and the column of the special point. This violates the given rightangled triangle condition unless they are all in the same row or the same
column as the special point. Now apply the Dirichlet principle to learn
that m + n — 1 < max{m,n} which is absurd since m,n > 2. Therefore
there is a pair of red points which are neither in the same row nor the
same column.
That was the hard part, the start of an induction. The subsequent in­
duction steps are relatively easy. We have m + n — 3 remaining red points.
We focus on them one at a time. Each successive red point must go on a
place which either has no other red in the row, or no other red in the col­
umn (else you make a forbidden red triangle). This each time we consider
a new red point, we use up either a fresh row (with no reds in) or a fresh
column. After the selection of the first pair of red points, there are m — 2
fresh rows and n — 2 fresh columns, so we can add at most m + n — 4 red
points and so can have at most m + n — 2 red points altogether.
Afterword
This is intellectually hard. The Dirichlet principle is clearly going to play
an important role in any solution. The particular solution we give is an
example of a recurring theme in mathematics competitions; an induction
is used where the inductive step is fairly straightforward but the base case
of the induction is where the trouble lives. Indeed, in the booklet of solu­
tions which has been on sale for years from UKMT, the problem of finding
two red points which are neither in the same row nor column is glossed
over a little too lightly.
This problem is a classic combinatorial question. Ingenuity is your
only real weapon, and the only way to become proficient at such problems
is to have done hundreds of them before.
The role of the phrase 'suppose, for contradiction' in the solution is
subtle. At first glance you might imagine that you could omit it, since
the inductive argument shows that you cannot do better than m + n — 2
red points. However, you have to get the induction started (the base case)
by finding two points which are neither in the same row now the same
column. That is where the supposition is used.
Chapter 12: BMO1 2002-03 Solutions
135
Problem 5
Find all solutions in positive integers a, b, c to the equation
fl!b! = «! + b! + cl.
[Mr Robin Bhattacharyya, Trinity College, Cambridge]
Discussion
This is a hard and fiddly problem. Do not be deceived by the relatively
short solution given in this book. That was the result of considerable
thought and much polishing. In a sense it is marvellous that one can an­
swer this question, but it is not really a natural problem. Still, in its favour
one can say that there is a low-technology solution, and a student with
sufficient ingenuity might find it. As a competition problem it has the ad­
vantage that several ideas must be joined together to get the solution, and
one could devise a marking scheme which rewarded significant partial
progress.
To provide a hint: you may as well assume that a < b. Can you con­
trol c? Once you can say something intelligent about the size of c, then
move on to address the questions as to whether a and b can be different,
or whether they can be equal.
Solution
Without loss of generality we may assume a < b. There is clearly no solu­
1. When a = 2 the equation bl = 2 + cl has no solutions.
tion when a
3. Also
Therefore a
bl
so fl < c. Since the terms on the right are positive integers, b! > 3 and so
b > 3 (or just as well b > a > 3). If c < b, then al + bl + cl < 3bl but a > 3.
Therefore 3 < a < b < c.
Suppose that a
b, then a < b < c. Therefore bl/al is odd so b = a + 1
is odd. Dividing our original condition by fl! we learn that
(fl + 1)! = 1 + (fl + 1) +cl/al
but every term except 1 is divisible by fl + 1 which is absurd. Therefore
3 < a = b < c.
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A Mathematical Olympiad Primer
Dividing by al we obtain al — 2 = cl/al. The left hand side is not
divisible by 3, so the right hand side is either a + 1 or (a + l)(a + 2). These
equations, modulo a, tell us that either 3 = 0 mod a or 4 = 0 mod a so a
must divide 3 or 4. However a > 3 so a = 3 or a = 4. The only candidate
solutions are therefore (a, &,c) = (3,3,4) and (4,4,6). Checking with the
original condition we find that 3!3! = 3! + 3! + 4! but 4141
4! -H 4! -H 6!
because 41
2 + 30. Therefore (a,b,c) = (3,3,4) is the unique solution.
BMO1 2003-04 Solutions
Problem 1
Solve the simultaneous equations
ab + c + d = 3,
be + d + a = 5,
cd + a + b — 2,
da + b + c = 6
where a, b, c, d are real numbers.
[Mr Sam Maltby, Avoca Systems Ltd, Chesterfield]
Discussion
Be careful here. For example, if you subtract the first pair of equations,
then you learn that (a — c)(b — 1) = —2. However, you do not know that
a — c and b — 1 are integers, so extracting further information is not so
simple.
Solution
Subtract the first pair of equations, and then subtract the second pair. We
obtain b(a - c) + (c - a) = -2 and d(c - a) + (a - c) = -4. Therefore
(a - c)(b - 1) = -2 and (fl - c)(l -d) = -4. Thus fl
cand so (1 - d) =
2(b — 1) or rather 2b + d = 3
A Mathematical Olympiad Primer
138
Start again, but this time subtract the third equation from the second,
and the first from the fourth. We obtain c(b — d) + d — b = 3 and a(d —
b) + (b — d) = 3 so (b — d)(c — 1) =3 and (b — d)(l — a) = 3. We learn
that b
d and c — 1 = 1 — a so a + c = 2.
Now eliminate c and d.
ab+ (2-a) + (3 — 2&) = 3
b(2 - a) + a + (3 - 2b) =5
Adding we obtain
2b+ 2 + 6-4b = 8
so b = 0. Therefore d = 3. Using b = 0 in ab + c + d = 3 we find that
c + d = 3soc = 0. Using c = 0 in be+ d + « = 5 we find that d + a = 5 so
a = 2. Thus the only candidate for a simultaneous solution to the system
of equations is [a, b, c,d) = (2,0,0,3). By substituting into the original
equations, we find that this is indeed a solution.
Afterword
The solution is a reference to the year in which the problem was set. Note
the need to verify that the (2,0,0,3) is a solution by substituting back into
the four original equations. This is because of the way the logic works. We
begin by (implicitly) assuming that a, b, c and d are a solution, and deduce
consequences. Eventually we learn that if (a, b, c, d) is a solution, then
(a, b, c, d) = (2,0,0,3). We can not just deduce that (2,0,0,3) is a solution,
because of the danger that there might be no solution at all.
Problem 2
ABCD is a rectangle, P is the midpoint of AB, and Q is the point on PD
such that CQ is perpendicular to PD. Prove that triangle BQC is isosceles.
[Mr Howard Groves, Royal Grammar School, Worcester]
Discussion
This is an attractive geometry question with a variety of solutions. It may
be helpful to make a construction, that is to say, draw some extra bits of
diagram. The problem is generous in that it can be solved with various
Chapter 13: BMO1 2003-04 Solutions
139
A
B
Figure 13.1: Problem 2
constructions; there is not one "hard to find" construction that you ab­
solutely must find. Note that it would be a routine matter to produce a
solution by Cartesian co-ordinates, using D as the origin and DA, DC as
axes. Choose co-ordinates for B which determine the co-ordinates of P
and C. The slope of DP is known so the slope of CQ can be calculated.
The equations of the lines DP and CQ can be written down and solved
against one another to find the co-ordinates of Q. Then one can calculate
the distances BQ and BC and verify that they are equal.
There is much unwarranted snobbery directed at the use of co-ordinate
methods. A mathematics competition is a game, and your job is to collect
as many marks as possible within the time limit. A Euclidean proof will
usually be faster if you can see it, but if you don't see the way through a
Euclidean argument, but you do see an algebraic proof with co-ordinates,
then roll up your sleeves and let the algebra flow. There is also the mat­
ter of alternative diagrams. A co-ordinate proof eliminates any ambiguity
which might arise. For example, in this Problem, diagrams are possible
where Q is inside the rectangle, and where Q is outside the rectangle. At
BMO1 level, the examiners do not worry too much about alternative di­
agrams, but in more advanced competitions, this becomes a very serious
issue.
Solution
(a) Let R be the midpoint of DC so there is a circle PBCR with diameter
BR (because ABPR and ARCB are right angles). Now APQC is also a right
angle so Q is on this circle. Thus QC is a chord of the circle perpendicular
to the diameter BR PD. Therefore Q is the reflection of C in the line BR
so QB = CB
A Mathematical Olympiad Primer
140
(b) Let R be the midpoint of DC. Now the line segments DR and PB are
parallel and have the same length, so PBRD is a parallelogram. Therefore
DP || RB. Let RB meet QC at S. The corresponding angles of triangles
CSR and CQD are the same, so the triangles are similar. Since |DC =
2|RC\, it follows that |QC| = 2|SC|. Therefore S is the midpoint of QC so
BR is the perpendicular bisector of QC. Thus |BQ| = BC|.
(c) Draw lines parallel to DP through both B and C. Then the new line
through B crosses DC at R, the midpoint of CD (because triangles PDA
and BRC are congruent using right-angle, hypotenuse, side; '.hen |BC| =
AP\ = | DR |). Now we have three regularly spaced parallel lines through
D, R and C. Let S be the intersection of QC and RB. Now finish as in (b).
(d) Let DP meet CB at T. The corresponding angles of triangles TBP
and TCD are the same (ZT is common, and APBT and ADCT are right
angles) so the triangles are similar, and since |DC| = 2| PB | the scale factor
is 2. Therefore |TB
BC . The circle on diameter TC with centre B
passes through Q because ATQC is a right-angle (this is the converse of
the theorem of Thales). Therefore BQ = BC .
Note that none of these proofs rely on assuming that Q lies between P
and D, as it happens to do in our diagram.
Problem 3
Alice and Barbara play a game with a pack of 2n cards, on each of which
is written a positive integer. The pack is shuffled and the cards laid out in
a row, with the numbers facing upwards. Alice starts, and the girls take
turns to remove one card from either end of the row, until Barbara picks
up the final card. Each girl's score is the sum of the numbers on the chosen
cards at the end of the game.
Prove that Alice can always obtain a score at least as great as Barbara's.
[Dr Tony Gardiner, University of Birmingham]
Chapter 13: BMO1 2003-04 Solutions
141
Discussion
Note that at each turn, each girl may select a card from either end. She
does not have to pick an end, and always select cards from that end.
We are given no information about the positive integers on the card. It
is possible that the number on one of the cards is bigger than the sum of the
numbers on the rest, in which case the winner will be the person who takes
that large card. If there is such a card, all that Alice has to do is to either
take it if she can, but otherwise prevent it from being exposed for Barbara
to take. One can imagine a strategy to do this easily enough. However,
it is also possible that the numbers on the cards are not dominated by a
single card with a vast number written on it, in which case we will have
to think again. A strategy is needed which will enable Alice to look at all
the cards, and decide which card to take next. Given that this is a BMO1
problem 3, and not an IMO problem 6, the strategy is likely to be quite
simple.
Solution
Imagine the cards coloured black and white alternately. Alice adds up
the numbers on the black cards and compares it with the numbers on the
white cards. She then decides which is the larger sum (say white), and
always takes cards of that colour.
When the game starts the end cards have opposite colours, Alice takes
the white card so Barbara takes a black card, and leaves a single white card
exposed. The game continues in the same way with Alice always taking
the white card, and Barbara forced to take a black card, until the cards run
out. Alice has all the white cards, so her total is at least as big as Barbara's.
Afterword
This is another colouring argument. Always be alert for colouring argu­
ments. Consider an m player version of the game, with mn cards, and
players Plz P2, . . ., Pn who takes turns to select and remove a card from the
end of the row. Is it possible for P\, P2, • • •, Pn-1 to gang up on Pn, and to
devise a strategy so that the sum of the numbers on the cards chosen by
Pn is not greater than the sum of the numbers on the cards chosen by any
other player?
A Mathematical Olympiad Primer
142
This problem is ill suited to an induction argument, and you will likely
need some other approach.
Problem 4
A set of positive integers is defined to be wicked if it contains no three
consecutive integers. We count the empty set, which contains no elements
at all, as a wicked set. Find the number of wicked subsets of the set
{1,2,3,4,.. .,10}.
[Dr Gerry Leversha, St Paul's School, London]
Discussion
It seems likely that if W„ is the number of wicked subsets of {1,2,... ,n},
then W„ will be expressible in terms of a recurrence, especially since an
initial fragment of a wicked string is wicked. If you knew what the re­
currence was, then perhaps that might help you to find the mathematical
explanation of why it has to be the recurrence. If that is the way your mind
works, then some experiments are in order. The first four terms of the se­
quence are 1,2,4,7. That is probably not enough to inspire anything but a
guess. If you work out one or two more terms, then nature may whisper
in your ear. The justification of a recurrence is almost certainly going to be
an induction argument, since a recurrence is induction made concrete.
Solution
Let W„ be the number of wicked subsets of {1,2,..., n}. By direct calcula­
tion, Wo = 1, W] =2 and W2 = 4. Assume that n > 3. We will develop
a recurrence to enable us to work out W„. Suppose that S is a subset of
{1,2,..., n} then exactly one of the following must occur.
1. n 0 S;
2. n G S and n — 1
S;
3. n — l,n G S and n — 2
S.
The number of sets of each type are as follows.
Chapter 13: BMO1 2003-04 Solutions
143
1- w„_i;
2. W„_2;
3. W„_3.
Therefore
+ W„_2 + Wn-3- We can now write down an initial
fragment of the sequence (W();
1,2,4,7,13,24,44,81,149,274,504
so there are 504 wicked subsets of {1,2,3,..., 10}.
Problem 5
Let p, q and r be prime numbers. It is given that p divides qr — 1, q divides
rp - 1 and r divides pq — 1. Determine all possible values of pqr.
[Dr Geoff Smith, University of Bath]
Discussion
It is natural to begin with some experiments. The problem does not stip­
ulate that p, q and r have to be different. There is an example with three
different small primes which you should find by experiment. Of course
the problem is cyclically symmetric (if (p,q,r) is a solution, then so are
(t/, r,s) and (r,s,q), but since you are only asked to supply the possible
values of the product pqr this complication is not important). Perhaps this
small solution is the only one, or there may be all sorts of large solutions
which you will never find by experiment.
Notice that the divisibility conditions tell us that
are all integers. Try to find a single useful fact which is a consequence of
the three conditions, keeping in mind that the quantity pqr is of particular
interest.
A Mathematical Olympiad Primer
144
Solution
Taking the product of the three divisibility conditions we obtain that
pqr | (qr — 1) (rp — l)(pq — 1).
Multiply out the right hand side, discarding multiples of pqr, we obtain
that
p«?r pq + rq + rp — 1
pt?r
must be a positive integer k. Since p,q,r > 2, it follows that k = 1 and so
pt? qr + rp — 1 = pqr. In an unusual move, we now deduce that p, q, r
must be distinct (if, for example, p = q, then p would be a divisor of 1).
Without loss of generality, p < q < r. Unless p = 2 and q = 3 we get
p
pt/r
q
Therefore p = 2, q = 3
r so
6r
so r = 5. Thus if there is any triple of primes which does the job, it must
be 2, 3 and 5. A quick check reveals that these three prime numbers do
indeed satisfy the stipulated conditions, so pqr = 30.
Afterword
This problem is number theory, but with a strong algebraic flavour.
Chapter 14
BMO1 2004-05 Solutions
Problem 1
Each of Paul and Jenny has a whole number of pounds. He says to her: "If
you give me £ 3,1 will have n times as much as you". She says to him: "If
you give me £ n, I will have 3 times as much as you". Given that all these
statements are true and that n is a positive integer, what are the possible
values for n?
[Dr Andrew Jobbings, Arbelos, Shipley]
Discussion
There are three unknown quantities, but only two equations can be ex­
tracted from the problem. However, the additional information that the
unknowns are integers allows us to determine a finite collection of possi­
ble solutions to this problem. The strategy must be to use the two equa­
tions to eliminate one unknown. Then make one of the remaining un­
knowns the object of an equation (that is, put it on one side, with a mess
involving the last unknown on the other side). Then worry a lot about that
fact that the unknowns take integral values.
Solution
There are three unknown quantities. Suppose that Paul starts with £ p,
and Jenny with £j. These are non-negative integers and we are also given
A Mathematical Olympiad Primer
146
that n is a positive integer.
The statements by Paul and Jenny give us that
p + 3 = n(/ — 3) and 3(p — n) = j + n.
We have three equations but only two unknowns, so there is reason to
worry that we have not got enough information. However, p, j and n are
not arbitrary real numbers. They are whole numbers, and this will help.
We tidy up each equation. The first equation is equivalent to p = nj —
3n — 3 and the second to 3p = 4n + j. We have chosen to tidy up in such a
way that p is isolated on one side of the equation. The reason that we have
done this is that if you try doing the same for n or j, then the expression
on the other side is more messy. Dividing these equations (why is p
0?)
we obtain
<”+/
=3
nj — 3n — 3
We have got rid of one unknown. We rearrange this expression making
one of the unknowns the object. This time it doesn't matter at all which
unknown we select, and we choose j. We deduce that
4(3n - 1)
3n - 1
Now if (n + 13)/(3n — 1) is to be a positive integer, then 3n — 1 < n + 13
son <7 (note that 3n — 1 > 0).
Now n determines j and together n and j determine p. For 1 < n < 7
we find the following integral solutions:
(1,11,5), (2,7,5), (3,6,6) and (7,5,5).
One can verify that these are all genuine solutions to the original problem
by checking directly. This is an important final step, and avoids having
to verify that our reasoning is reversible by stepping through the entire
argument backwards.
Problem 2
Let ABC be an acute-angled triangle, and let D, E be the feet of the per­
pendiculars from A,B to BC,CA respectively. Let P be the point where
the line AD meets the semicircle constructed outwardly on BC, and Q be
Chapter 14: BMO1 2004-05 Solutions
147
the point where the line BE meets the semicircle constructed outwardly
on AC. Prove that CP = CQ.
[Dr Gerry Leversha, St Paul's School, London]
Solutions
There are many instructive an interesting ways to approach this problem.
This problem is the author's favourite BMO1 geometry problem of recent
Q
years. One solution is a celebration of the dot (scalar) product of vectors.
Alternatively the tangent-secant theorem (the limiting case of the inter­
secting chords theorem) can be deployed. A more unusual method is to
embed the diagram in 3-dimensional space, and fold it. This is demon­
strated in (c).
(a) Perhaps the fastest is to use a property of the scalar product of vec­
tors: suppose that XY and XZ are vectors, and that W is the foot of the
perpendicular dropped from Y to the line XZ, then XY.XZ = XVV.XZ.
Using this principle, we get
CP2 = CP.CP = CP.CB = CD.CB = CA.CB.
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A Mathematical Olympiad Primer
However an entirely similar argument shows that CQ2 = CB.CA. Since
CB.CA = CA.CB the argument is complete.
(b) A variation is to use the tangent-secant theorem. The line CP is
tangent to the circle PDB since CP is perpendicular to the diameter PB.
Therefore CP2 = (CD)(CB). Since ABEA and ABDA are right angles,
there is a circle S passing through B, D, E and A. Therefore (CD)(CB) =
(CE)(CA). The final step is like the first, so CQ2 = (CE)(CA). Thus
CP2 = CQ2.
(c) We can use 3-dimensional geometry. Note that the intersection of the
altitudes of a triangle is (almost) invariably denoted H (and is called the
orthocentre). Imagine that the diagram is drawn on paper, and cut along
the lines QC, CP, PB, BA and AQ. Now fold the flaps AQC, CPB up along
the lines AC and CB. It suffices to show that P and Q meet (above H), for
then CP = CQ. It suffices to show that EQ2 — HE2 = PD2 — HD2.
Now EQ2 = AE ■ EC since EQ is an altitude of the right angled tri­
angle AQC. This is a well known result, but if you feel the need to jus­
tify it, consider the similar triangles EAQ and EQC. We can get at the
lengths CE, EA and HE by trigonometry. Let RE be the circumradius of
ABC. By the sine rule EC = 2Rsin A Then CE = 2RsinAcosC. Simi­
larly EA = 2R sin Ceos A. By chasing angles AEHC = AA, and so using
trigonometry in triangle HCE we obtain HE = 2R cos A cos C.
Now
EQ2 — EH2 = 4R2 cos A cos C(sin A sin C — cos A cos C)
= —4R2 cos A cos C cos(A + C)
= 4R2 cos A cos B cos C
since cos B = — cos( A + C).
This final expression is symmetric in A, B and C. This shows that the
flaps do meet over H as suggested, and indeed the same is true from the
unmentioned flap BAR which we might have constructed on the side AB.
This finishes the problem, but it is interesting to note that deeming
the flap ABR to be in play, the angles AAQC, ACPB and ABRA are all
right angles. When the flaps are folded up and P, Q and R identified, the
vertex at P = Q = R is the corner of a rectangular parallelipiped (a brick),
and ABC is obtained by slicing through the brick. Thus any acute angled
triangle can be obtained in this way.
Chapter 14: BMO1 2004-05 Solutions
149
Problem 3
Determine the least natural number n for which the following result holds:
No matter how the elements of the set {1,2, ...,n} are coloured red or
blue, there are integers x, y, z, w in the set (not necessarily distinct) of the
same colour such that x + y + z = w.
[Mr Adrian Sanders, Trinity College, Cambridge]
Discussion
A few experiments should enable you to find a colouring of {1, 2, 3, ...,
10} so that no three numbers of the same colour sum to a number of that
same colour. The same colouring can be used on subsets of this set of the
form {1,2,3,... ,m} (where m < 10) to show that n > 11. In fact 11 is the
correct answer, and one must give a detailed analysis of that case to show
that it has the required property.
Solution
When we look at the first 10 positive integers we may colour 1,2,9 and 10
red and colour 3,4,5 and 6 blue. Then the sum of any three numbers of
one colour is not of the same colour by inspection. We can use the same
colouring for any shorter initial fragment of the positive integers with the
same result. Therefore n > 10.
We now show that n = 11. Any colouring must either have 1 and 2
of the same colour or of different colours. We address these cases in turn.
Without loss of generality assume that 1 and 2 are both red. In seeking
a colouring for which no x, y, z of the same colour have sum of the same
colour, we are forced to colour the numbers 3 ,4, 5 and 6 all blue because
of 1+1+1,1+1+2,1+2+2 and 2 +2+ 2. Now 9 = 3 + 3 + 3 must be red, and
so 1 + 1 + 9 = 11 must be blue. However, 3 + 3 + 4 = 11 so 11 must be
red.
Therefore any colouring of the type we seek must have 1 and 2 of dif­
ferent colours. Without loss of generality we may assume that 1 is red and
2 is blue. Now 1 + 1 + 1 =3 must be blue, 2 + 2 + 2 = 6 must be red
and 1 + 1 + 6 = 8 must be blue. However 2 + 3 + 3 = 8 must also be
red. Therefore no colouring exists of the type we seek, the solution to the
problem is n = 11. Any colouring of 1,2,3,.. .,11 as red and blue must
150
A Mathematical Olympiad Primer
have the property that there are three numbers of one colour which sum
to a number of the same colour.
Problem 4
Determine the least possible value of the largest term in an arithmetic pro­
gression of seven distinct primes.
[Mr Adrian Sanders, Trinity College, Cambridge]
Discussion
The next problem was inspired by the theorem of Ben Green and Terry Tao
that there are arbitrarily long finite sequences of prime numbers in arith­
metic progression. This was a conjecture of G. H. Hardy. Terry Tao and
Ben Green represented Australia and the UK respectively at the Interna­
tional Mathematical Olympiad. Terry Tao went on to win a Fields Medal.
This BMO1 problem was proposed in the wake of the publication of the
Green-Tao theorem. I remember very well when the news of Green-Tao
arrived in my electronic mailbox. New facts about prime numbers which
can be stated in elementary terms are almost as rare as hen's teeth. It was
a scalp-tingling moment. Suddenly we lived in a new universe.
Solution
Let the first (prime) term be p and the common difference be d. Thus our
seven prime numbers are
First note that p is odd (else p + 2d is an even prime bigger than 2). There­
fore d is even (else p + d is an even prime bigger than 2). Also d is a
multiple of 3 (else one of the primes p + d,p + 2d, p + 3d is greater than 3
and a multiple of 3). Now p > 3 (else p + 3d would be a multiple of 3 and
bigger than 3). Therefore p, d > 3 (and p + d > 6). Similarly d is a multiple
of 5 (else one of the primes p + d,p + 2d, p + 3d, p + 4d,p + 5d is divisible
by 5). Thus p > 5 (else p + 5d would be a multiple of 5).
Now we know that p is at least 7 and d is a multiple of 30. If p > 7, then
for the usual reasons, d is divisible by 7 and so d is a multiple of 210. There­
fore any sequence with p = 7 and common difference 30,60,90,120,150
Chapter 14: BMO1 2004-05 Solutions
151
or 180 must have a smaller last term than any sequence beginning with a
prime larger than 7.
We have to avoid 187 = 11 x 17 which is not prime. Using p = 7
we examine the given list of candidate common differences. The integers
d = 30,60,90 are not viable because of 187. If d = 120 then p + 2d = 247 =
13 x 19 is not prime. When d = 150 the numbers
7,157, 307, 457, 607, 757 and 907
are all prime. Therefore 907 is the smallest possible largest term of such a
sequence.
Afterword
In this particular problem, it would be helpful to know the prime numbers
less than 1000. There are worse things to commit to memory.
Problem 5
Let S be a set of rational numbers with the following properties:
(ii) If x 6 S, then both
G S and
G S.
Prove that S contains all rational numbers in the interval 0 < x < 1.
[Dr Gerry Leversha, St Paul's School, London]
Discussion
This is a beautiful problem. There is an elegant induction argument which
will do it, but it relies have the idea of treating numbers differently de­
pending on whether they are greater than or less than, one half.
There are some technical points here. It is tempting to write let q = a/b
be a rational number and stop there. Such a statement is fine, but may not
be what you mean. After all,
2 _
3 “ 3\/3
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A Mathematical Olympiad Primer
so if you want a and b to be integers, then you have to say so. If the rational
number q is positive then you may care to insist that the integers a and b
are positive, but again you have to say so.
Now consider the statement let q = a/b be a rational number where a
and b are integers. We have not stated that the fraction is written in lowest
terms (that is, a and b are coprime). After all
1 _ -3 _ 1729
2
—6
3458'
This can always be arranged by cancelling out the gcd of a and b, but
unless you state that this is what you have done, then it may be that a and
b have a common factor bigger than 1.
Solution
Suppose that q = t/u with f,u positive integers and 0 < q < 1. Given
the remarks in the discussion, we had best assume that f and u are coprime
(that is, all non-trivial common factors have been cancelled). We prove by
induction on n = t + u that q E S. The smallest possible value of n is 3,
and that only occurs when q = 1/2 which is given to be in S.
Now for the inductive step. Take q = t/u for positive integers t, u
where t + u > 3 and u > t. We may assume that q
1/2. Define positive
integers a, b as follows.
If q > 1/2, let b = t and a = u — t > 0 so q = b/(a + b). Now
a + b = u<u + t = n so r = a/b E S by inductive hypothesis. Therefore
On the other hand, if q < 1 /2, let a = t and b = u - t so once again a
and b are positive integers. Now a + b = u < u + t = n so s = a/b E S by
inductive hypothesis. Therefore
Chapter 15
BMO1 2005-06 Solutions
Problem 1
Let n be an integer greater than 6. Prove that if n — 1 and n + 1 are both
prime, then n2(n2 + 16) is divisible by 720. Is the converse true?
[The late J. Wolstenholme]
Discussion
In this year, an extra question was introduced to BMO1. This additional
first question was designed to provide a problem with which inexperi­
enced students could engage. This reform has had the desired effect of
dramatically reducing the number of very low scores on the exam. An
welcome bonus has been that introducing an extra question has put a lit­
tle more time pressure on the experienced students who are trying to get
full marks.
Solution
(a) Let N = n2(n2 + 16). Since n - 1 and n + 1 are both prime, n = 2m is
an even integer where m > 3. Therefore
N = 4m2(4m2 + 16) = 16m2(m2 + 4)
is divisible by 16.
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A Mathematical Olympiad Primer
Now, one of the consecutive integers n — 1, n, n + 1 must be divisible by
3. However n — 1 and n + 1 are prime and at least 6, so neither is divisible
by 3. Therefore 3 divides n, so 9 divides N.
A similar argument works with 5. One of the consecutive integers n —
2, n — 1, n, n + 1 and n + 2 is divisible by 5. As before, n — 1 and n + 1 are
prime and at least 6, so they are not divisible by 5. If 5 divides n, then 5
divides N. It remains to consider the possibility that 5 divides n ± 2. In
this case 5 divides (n — 2)(n + 2) = n2 — 4. Now
N = n2([n2 — 4] + 20)
is divisible by 5. The numbers 16,9 and 5 are pairwise coprime, so their
product 720 divides N.
The converse is false. If n is a multiple of 60 = 22 x 3 x 5, then rr
(and therefore N) will be divisible by 720 = 24 x 32 x 5. Choosing n = 60
does not produce a counter-example, because 59 and 61 are both prime.
However, when n = 120, at least one of n — 1 and n + 1 is not prime (in
fact, neither is prime: 119 = 7 x 19 and 121 = ll2).
Afterword
It would be natural to write up a solution to this problem using modular
arithmetic notation. There is a place where this would be slight advantage.
There is a point in the argument where we deduced that 5 divided either
n — 2 or n + 2, and we concluded that 5 divided their product. A student
using modular arithmetic could avoid having to spot this trick by writing
n = ±2 mod 5, so n2 = 4 mod 5, and therefore n2 + 16 = 20 = 0 mod 5.
Note the significance of the condition that n > 6. When n = 6 then 5 is
not a factor of n2(n2 + 16) so 720 is certainly not a factor either.
Problem 2
Adrian teaches a class of six pairs of twins. He wishes to set up teams for
a quiz, but wants to avoid putting any pair of twins into the same team.
Subject to this condition:
(i) In how many ways can he split them into two teams of six?
(ii) In how many ways can he split them into three teams of four?
[Mr Paul Jefferys, Trinity College, Cambridge]
Chapter 15: BMO1 2005-06 Solutions
155
Discussion
As usual with enumeration problems, be organized. If you are happy and
confident with the use of binomial coefficients ("), then you may find that
the solution will fall out more easily.
Solution
(a) (i) Call the teams A and B. For each pair of twins, make a choice as
to which one goes into team A. The make-up of team B is then forced.
The possible memberships of team A and team B correspond to 6 choices
giving a total of 26 outcomes. However, we do not care which team is
A and which is B, so each possible division of the players into two sides
of size 6 has happened twice (with the roles of A and B swapped). The
correct answer is therefore 64/2 = 32.
(ii) Call the teams A, B and C. Suppose that Adrian picks team A first. He
has 12 choices at first, then 10, then 8 and finally 6. However, it does not
matter in which order the people were picked. Therefore the number of
ways of forming team A is
12 x 10 x 8 x 6
4 x3 x 2 x 1
-4! = 240.
Now Adrian addresses team B. He must choose one each of the remaining
two pairs of twins, and 2 of the remaining people. Team C will then be
determined. The can do this in
22 x ($) = 22 x 6 = 24
ways.
It is tempting to think that the answer is 240 x 24, but this would not be
correct. The question does not specify the team names, and our method
counts each division into three teams 6(= 3!) times, because that is the
number of ways of arranging the letters A, B and C in a row. The correct
answer is therefore
240 x 24
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A Mathematical Olympiad Primer
156
(b) We present an alternative solution to the second part of the problem.
We suppose that that the twins are actually all members of triplets, and
that in every case the unannounced third sibling has an unusual character­
istic not shared by the original twins; they all have green noses. We solve a
new problem. Given six pairs of triplets (one of each triplet having a green
nose), how many ways can you form three teams of six, with one of each
triplet in each team, so that exactly two people in each team have green
noses? Once again we suppose that the teams are labelled A, B and C. We
chose the green nosed people for Team A in (6) = (6x5)/2 ways. Then
we chose the green noses for Team B in (^) = (4 x 3)/2 = 6 ways. The
green noses for Team C are then determined. For each remaining pair of
normal nosed twins, the teams they are in are determined, but not which
twin is which. Therefore we must multiply by 26. Finally the names of
the teams was not specified, so as before we must divide by 3!(= 6). The
answer to the new problem is therefore
15 X 6 X 64 = 960.
6
Finally we confess that the green nosed twins were imaginary, and we
erase them from the teams. The new problem is the same as the old prob­
lem, because the teams of four in the original problem can be padded to
teams of six using green nosed siblings in exactly one way, provided that
no siblings are allowed in the same team.
Problem 3
In the cyclic quadrilateral ABCD, the diagonal AC bisects the angle DAB.
The side AD is extended beyond D to a point E. Show that CE = CA if,
and only if, DE = AB.
[Mr Howard Groves, Royal Grammar School, Worcester]
Discussion
This is an 'if and only if' problem, so you must establish the implication
both ways round. No cunning construction is needed.
Chapter 15: BMO1 2005—06 Solutions
157
E
Figure 15.1: Problem 3
Solution
First suppose that CE = CA. We will show that triangles ABC and EDC
are congruent.
The triangle ACE is isosceles with apex C. Therefore ADAC = ACED.
However ACAB = ADAC by the angle bisecting property of AC. There­
fore ACAB = ACED. Also AABC = AEDC since the latter is an ex­
terior angle of the cyclic quadrilateral ABCD. Finally we are given that
CA = CE. The triangle congruence is established (AAS corresponding).
Therefore AB = ED.
We begin again, but this time we suppose that AB = ED. Again
we will show that triangles ABC and EDC are congruent. Once again
AABC — AEDC since the latter is an exterior angle of the cyclic quadri­
lateral ABCD. Finally BC = CD since these line segments subtend equal
angles (or by the sine rule). The congruence is established by SAS. There­
fore AB — ED.
Afterword
There are attractive ways to get this argument wrong. In the first half,
where we assume that CE = CA, we have not used the fact that BC = DC
(which can be deduced from the fact that ACAB = ADAC). If you do
use it, beware of quoting ASS as a reason for congruence, because it is not
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A Mathematical Olympiad Primer
one. You do have enough angle information to show that ZBCA = ZDCA,
and if you observe that, then SAS becomes a legitimate justification for the
congruence of ABC and EDC.
Problem 4
The equilateral triangle ABC has sides of integer length N. The triangle is
completely divided (by drawing lines parallel to the sides of the triangle)
into equilateral triangular cells of side length 1.
A continuous route is chosen, starting inside the cell with vertex A and
always crossing from one cell to another through an edge shared by the
two cells. No cell is visited more than once. Find, with proof, the greatest
number of cells which can be visited.
[Dr Andrew Jobbings, Arbelos, Shipley]
Discussion
It is easy enough to find a path which visits n2 — n + 1 cells. However,
it may not be so easy to prove that this is the maximum possible number
of cells which may be visited, unless you hit upon the idea of using a
colouring.
Solution
Colour the small triangular cell with vertex A black. Now colour all the
Figure 15.2: Problem 4, a colouring
cells in chessboard style, so that any two cells sharing a common edge are
of different colour. There is a unique way to do this is shown in the dia­
gram. The number of black cells is 1 + 2 + • • • + n = n(n +l)/2 whereas
Chapter 15: BMO1 2005-06 Solutions
159
the number of white cells is 1 + 2 -I------- F n - 1 = n(n -1)/2. The white
cells are therefore slightly more scarce. The length of the legal path as
specified in the problem can not exceed n(n - 1) + 1 = n2-n + l, the
length of a path which starts on a black cell, visits every white cell once on
alternate moves, and finishes in a black cell.
Figure 15.3: Problem 4, a route
Finally we show that this bound can be achieved by exhibiting a path
of the required type. Start in the triangular cell with vertex A. Move down,
then right, then down, then across left until one can get no further, then
down, then across right until one can get no further, then down and so on.
This zig-zag path does indeed have length n2 — n + 1.
Problem 5
Let G be a convex quadrilateral. Show that there is a point X in the plane
of G with the property that every straight line through X divides G into
two regions of equal area if, and only if, G is a parallelogram.
[Dr Geoff Smith, University of Bath]
Discussion
If G is a parallelogram, it has an uncontroversial centre and it is not hard
to show that this point has the appropriate area bisection property. The re­
verse argument is more subtle. We will give a proof which involves draw­
ing a line through X which does not pass through a vertex of G, and then
rotating the line in both directions through an angle which is sufficiently
small that the new lines still cut G on the same sides.
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A Mathematical Olympiad Primer
Solution
If G is a parallelogram, then let X be the point where the diagonals meet.
Any line through X will then divide the parallelogram into two congruent
pieces (using rotation through 180° about X), which must therefore have
equal area.
Now for the other implication. We suppose that there is an X with this
area bisection property, and we will show that G is a parallelogram.
First we observe that X must be inside G, for if X were outside the
convex G, then there would be a straight line through X which did not
meet G at all. Look at straight line segments BE through X joining two
points B and E on the boundary of G. By rotating the line segment about
X, we can ensure that neither B nor E is a vertex of the quadrilateral. Now
we can choose points A and C on the same side as B, and points D and F
on the same side as E, so that X is on both AD and CF. This is possible by
choosing A and C sufficiently close to B. We may as well assume that A
and C are on different sides of B.
Using a square bracket notation to indicate area, we have [AEX] =
[DEX], [BCX] = [EFX] and [ACX] = [DFX]. Now using the formula
Chapter 15: BMO1 2005-06 Solutions
161
-1
2ab sin C" for the area of a triangle, we conclude that
(AX)(CX) = (DX)(FX)
(AX)(BX) = (DX)(EX)
(BX)(CX) = (EX)(FX)
Multiply the first and third equations and divide by the second to learn
that CX2 = FX2 so CX = FX. Similarly AX = DX and BX = XE. Now
triangles ABX and DEX are congruent SAS. Therefore KXAB = AXDE
and by alternate angles AB || ED. Thus the two sides being intercepted by
these three lines are opposite and parallel. Therefore both pairs of opposite
sides of G are parallel and G is a parallelogram.
Problem 6
Let T be a set of 2005 coplanar points with no three collinear. Show that,
for any of the 2005 points, the number of triangles it lies strictly within,
whose vertices are points in T, is even.
[Mr Nathan Kettle, Trinity College, Cambridge]
Solution
Suppose that t E T, and put S = T \ {t} so |S|= 2004. Move t to a distant
part of the plane so that it is contained in no triangle with vertices in S.
Now slide t into position, avoiding the points of S. Keep track of how the
number of triangles which surround t changes. This number alters only
when t crosses a line segment joining two points of S. Suppose that t enters
x triangles when crossing such a segment, then it must leave 2002 — x
triangles. The difference of x and 2002 — x is 2% — 2002 which is even.
Therefore, when t is eventually in the correct position, it will be contained
in an even number of triangles.
Afterword
Many other proofs are possible, and Dr Joseph Myers of the BMO marking
team assembled a collection of 26 other proofs.
Chapter 16
BMO1 2006-07 Solutions
Problem 1
Find four prime numbers less than 100 which are factors of 332 — 232.
[Dr Geoff Smith, University of Bath]
Discussion
The last thing you should do here is to work out 332 and 232 and subtract
them. Both numbers are enormous, and even if you accurately calculated
their difference, there would be no sensible way to go looking for prime
factors, except perhaps for 2, 3 and 5 which would be easy to detect. One
possible route is to use congruences in conjunction with Fermat's little
theorem. However, there is a better way. The shape of the expression in
the problem actually tells how to factorize it.
Solution
Recall the factorization x2 — y2
edly.
o32 _ O32
(x + y) (x — y). We can use this repeat-
)(316
We have found two factors of 332 - 232 but they are both certainly much
greater than 100. However the shape of the second factor is just like the
A Mathematical Olympiad Primer
164
shape of the original difference, with 16 replacing 32. We can apply the
difference of two squares trick again and again, so 332 — 232
= (316 + 216)(38 + 28)(34 + 24)(32 + 22)(31 +21)(31 -21).
Therefore
16 , 2I6 )(38 + 28) • 97 • 13 ■ 5.
We have obtained three small prime factors 5, 13 and 97. The smaller
remaining factor is 6817, but it is easy to spot that 6817 = 17 • 401 so we
have found a fourth prime factor 17.
This method is far superior to any other route, but even so we explore
different possibilities. Clearly 2 and 3 are not prime factors of 332 — 232 so
the smallest possible prime factor is 5. Now working modulo 5 and using
Fermat's Little Theorem we have
332 _ 232 = (34)8 _ (24)8 = 18 _
= 0 mod 5
so 5 is one of the required factors. Following this promising start, things
go downhill rather quickly. The problem is that guessing a prime factor,
and then testing its correctness by modular arithmetic, is a random and
expensive process in terms of time frittered away on pointless calculation.
Problem 2
In the convex quadrilateral ABCD, points M, N lie on the side AB such
that AM = MN = NB, and points P, Q lie on the side CD such that
CP = PQ = QD. Prove that
Area of AMCP = Area of MNPQ
= i Area of ABCD.
[Dr Andrew Jobbings, Arbelos, Shipley]
Discussion
There is actually a nice formula for the area of a convex quadrilateral.
It is half the product of the lengths of the diagonals multiplied by the
sine of the angle between them. Well, there are two possible angles be­
tween them, but those angles add to 180° so their sines are the same. This
Chapter 16: BMO1 2006-07 Solutions
165
area formula is clearly connected in some way with the similar formula
^ab sin C for the area of a triangle. Please work out the connection.
Sadly this area formula does not seem to provide a fast way forward
in this problem.
Figure 16.1: Problem 2
We will give two solutions. The first consists of cutting the quadrilat­
eral into convenient pieces. The second approach is to produce the rays
BA and DC until they meet at E (if they do). In this case you can express
the area of a quadrilateral as the difference of the areas of two triangles.
You have two bites at that particular cherry of course, and the analysis
works almost the same if it the rays AB and CD which meet. This tech­
nique fails if AB and DC are parallel, but in that case the problem is easy.
Solution
(a)
[AMCP] = [ACP] + [AMC]
= 1[ACD] + 1[ABC]
= i[ABCD],
Also
[MNPQ] = [MNP] + [PQM]
= [AMP] + [CPM]
= [AMCP].
A Mathematical Olympiad Primer
166
All equalities are justified by partitioning a region into non-overlapping
parts, or using area = | base x height.
(b) We note that the results are trivially true if A B and DC are parallel but in a mathematics competition, you should supply the details -a mark
may depend on it. We will assume that the rays CD and BA meet at E
(there is a similar argument if the rays DC and AB meet). Suppose that
ACEB = oc. Then
[AMCP] = 1 sina(EC)(EM) - 1 sina(EP)(EA)
and
[MNPQ] = i sina(EP)(EN) - i sin«(£Q)(£M).
To establish our first equation, it suffices to show that
(£C)(EM) - (£P)(£A) = (EP)(EN) - (EQ)(£M)
or rather that
(EC + EQ)(EM) - (EA + EN)(EP)
but both sides are 2(EM) (EP) so this is correct.
For the second part of the question, let X be the midpoint of CD and
Y the midpoint of AB. Let s and t be positive quantities. Consider the
expression
((EX) + t)((£Y) + s) - ((EX) - f)((£Y) - s) = 2s(£X) + 2f(£Y).
This expression is trebled if you treble both s and t. The key observation
is that 3(XP) = (XC) and 3(YN) = (YB). Given that the areas are related
to our expressions via the same constant of proportionality, it follows that
[MNPQ] = 1[ABCD].
Afterword
These proofs are convincing enough, but perhaps we are missing some­
thing. This author has the feeling that an underlying simplification is be­
ing overlooked.
Chapter 16: BMO1 2006-07 Solutions
167
Problem 3
The number 916238457 is an example of a nine-digit number which con­
tains each of the digits 1 to 9 exactly once. It also has the property that
the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not.
How many such numbers are there?
,
[Dr Andrew Jobbings, Arbelos, Shipley]
Discussion
These counting arguments can be tricky. There are lots of ways to get the
correct answer (and also many ways to get the wrong answer). A useful
check on yourself is to find two different methods which give the same
answer. If you are sufficiently good at these arguments, then this would
be a waste of time. However, if you are nervous, it is a sensible strategy.
Solution
(a) The easy way to address this question is this. Imagine the digits 1
to 5 in the right order, but with flexible spaces between, before and after
the numbers. There are six spaces. Now place the remaining digits in the
gaps, one at a time, creating new gaps as we go along. Let us put in the
number 6 first because it is special. It cannot go in the gap behind the 5, so
it can go in any one of 5 spaces. Now there are 7 spaces. Place the number
7 in any one of them. Now there are 8 spaces. Place the number 8 in any
one of them. Now there are 9 spaces. Place the number 9 in any one of
them.
In each case there was no restriction on our choices, and if two se­
quence of choices differ in any respect, then the final sequences must be
different. Finally, every possible legal sequence can be built using our
procedure. Therefore the number of legal configurations of numbers is
the same as the number of different sequences of choices that we might
make. This number is5x7x8x9 = 2520.
(b) Any other procedure which satisfies the conditions of the previous
paragraph will also work. For example, imagine two queues. The first
queue consists of the numbers 1 to 6, with 1 to 5 in order, but 6 out of
order. There are 5 such queues. The second queue consists of the numbers
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A Mathematical Olympiad Primer
7,8,9 in any order. There are 6 such queues. Now we wish to merge the
queues into a queue of 9 numbers, so that if members of either original
queue were to evaporate, then the merged queue would have its surviving
members in the same order as the other original queue.
We can choose the 3 positions among the 9 which will be occupied
by the second queue members in (j) ways. This time the calculation is:
choose which first queue, choose which second queue, and then choose
which three places among the 9 will be occupied by members of the sec­
ond original queue. One can perform a similar check to ensure that the
possible choices correspond exactly to the possible final arrangements.
Therefore the number of legal arrangements is
9!
5 x 6 x----- = 5x9x8x7 = 2520.
6!3!
(c) Here is another way. We count the number of strings where 1,2,3,4
and 5 are in the correct order, and subtract the number of strings where
1, 2, 3, 4, 5 and 6 are in the correct order. Using our method of flexible
spaces, we see that this is
6x7x8x9 — 7x8x9 = 5x7x8x9
= 2520.
Problem 4
Two touching circles S and T share a common tangent which meets S at
A and T at B. Let AP be a diameter of S and let the tangent from P to T
touch it at Q. Show that AP = PQ.
[Dr Gerry Leversha, St Paul's School, London]
Discussion
There is a routine method of solution which requires nothing but the the­
orem of Pythagoras and a clear head. The more educated student may
like to look for a beautiful solution by inversion. Inversion is a mathemat­
ics competition dark art. It is definitely outside the scope of the informal
IMO syllabus, but even so it is a powerful method which you can some­
time deploy with devastating effectiveness in mathematics competitions.
Chapter 16: BMO1 2006-07 Solutions
169
P
Figure 16.2: Problem 4
There are other methods in this category of course; the use of areal (or
barycentric) co-ordinates and the use of theorems of projective geometry.
However, the important thing is to become very effective at Euclidean ge­
ometry, spiced from time to time by vectors or co-ordinate methods.
Solution
(a) Let circle S have centre C*i and radius a, and let circle T have radius
b and centre O2. Drop a perpendicular from Di to O2B to make a rightangled triangle. By Pythagoras we have
(a + b)2
so AB2 = 4ab. Dropping a perpendicular instead from O2 to AP we have
a second right-angled triangle and Pythagoras yields that
pol = 2a -b2 + AB2 = 4o2 + b2.
Now consider the right-angled triangle PO2Q. The theorem of Pythagoras
tells us that PO2 = PQ2 + QO2 so
4a2 + b2 = PQ2 + b2
and PQ = 2a = AP.
(b) Next we give a proof by inversion. We will not develop the theory
here, so until you acquire the background knowledge, ignore this solution.
Consider the circle U with centre P which passes through A. Invert
with respect to U. The line AB exchanges with circle S. The line PQ inverts
A Mathematical Olympiad Primer
170
to itself. The circle T is tangent to the circle S, the line AB and the line PQ,
so T inverts to itself. The point Q inverts to a point on the line PQ and on
the circle T, so Q inverts to Q. Therefore PQ = AP.
Afterword
You might wonder what kind of mind would come up with (b), when the
perfectly sensible method (a) is available.
Problem 5
For positive real numbers a, b, c, prove that
(a2 + b2)2 > (a + b + c)(a + b — c)(b + c — a)(c + a — b).
[Dr Gerry Leversha, St Paul's School, London]
Discussion
If you are familiar with Heron's formula for the area of the triangle, then
the right-hand side is very suggestive. This formula asserts that if ABC is
a triangle, then using standard notation, its area [ABC] is given by
16[ABC]2 = (a + b + c)(a -I- b — c)(b + c — a)(c + a — b).
However, there is a trap here. We are not given the information that a, b
and c form the sides of a triangle, so this formula is of restricted applica­
bility. If, for example, a > b + c then there is certainly no triangle with
sides a, b and c.
A naive method (though one which works well) is to ignore Heron's
formula, throw everything onto one side, and try to show that the result­
ing polynomial is positive by whatever methods come to hand.
Solution
(a)
It suffices to show that LHS - RHS is positive. The RHS is
((a + b)2 — c2)(c2 - (a — b)2) = (a2 + b2 — c2 + 2ab)(c2 - a2 — b2 + 2ob)
Chapter 16: BMO1 2006-07 Solutions
171
N.B. the symmetry in a, b and c is reassuring, it means that it is unlikely we have
made an algebraic error. Therefore LHS — RHS is
= (a2 — b2)2 + (a2 + b2 — c2)2
> 0.
The final inequality is because the expression is the sum of squares.
We were not asked to analyze the case when equality occurs, but this
is always worth looking at. It happens when a = b and a2 + b2 = c2, so
c = aC^.
(b) We give a solution based on Heron's formula. If a, b and c are the
sides of triangle ABC, then
[ABC] = ^flbsinC < \ab
The last inequality is by AM-GM. Equality occurs when sinC = 1, so
C = 90°, and a = b. This gives rise to an isosceles right-angled triangle
with c = a\/2 as in (a).
It remains to discuss what happens when a, b and c are not the sides of
a triangle. Consider the three factors (a + b — c), (b + c — (?) and (c + a — b).
If exactly one factor is negative, then
(a + b + c)(a + b
c)(b + c — (?) (c + a — b)
is not positive and the required inequality is clear. If at least two of them
are negative, then by permuting a, b, c if necessary we may assume that
a + b - c < 0 and b + c - a < 0. Adding we deduce that b < 0 which is
absurd.
Problem 6
Let n be an integer. Show that, if 2 + 2 ^1 + 12n2 is an integer, then it is a
perfect square.
[Dr Gerry Leversha, St Paul's School, London]
A Mathematical Olympiad Primer
172
Discussion
Sadly, this kind of question has become a little familiar to people who take
an interest in mathematics competitions. However, to a student seeing
such a problem for the first time, it must seem magical. Why on earth
must this particular expression have this peculiar property, when lots of
similar ones certainly do not? There is some Number Theoretic machin­
ery associated with expressions of the form a2 + kb2 called the theory of
Pell's Equation. However, this question was posed at BMO1 level, so it is
inconceivable that you need to understand the theory of Pell's equation in
order to solve the problem. Rather there must be an entirely naive argu­
ment which will take you home. We give a solution below which is short
and natural. However, it may not be easy to find, and there are many
opportunities to step away from a path to the answer.
Solution
If 2 + 2\/l + 12n2 is an integer, then 2\/l + 12n2 = a, say, is an integer.
Squaring, we see that a2 is an even integer so a = 2b is even. Therefore
\/l + 12n2 = b is an integer and b2 — 1 = 12n2. This forces b = 2c + 1
to be odd, so 4c2 + 4c = 12n2 and c2 + c = 3n2 for an integer c. Now
c + c = c(c + 1) and c, c + 1 are coprime (they have no common prime
factor). Therefore either c is a perfect square and c + 1 is 3 times a perfect
square, or c + 1 is a perfect square and c is 3 times a perfect square. Thus
there are integers x and y such that either 3x2 — y2 — 1 or x2 — 3y2 = 1. The
first case cannot occur, because — 1 is not a square modulo 3, so c 4-1 = x2
and c = 3y2. Now
O
/
\
2 + 2v/l + 12n2 = 2 + 2b
= 2 + 2(2c + l)
= 4c+ 4
= 4(c + 1)
= 4x
(2x)2.
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algebra, 13
alternate segment, angle in the
39
alternating sequence, 22
altitudes of a triangle, 30
AM-GM inequality, 14
AM-HM inequality, 88
angle
~ at the centre, 39
~ bisectors of a triangle, 30
exterior of a triangle, 31
~ in a semi-circle, 40
~ in the alternate segment, 39
reflex
39
angles
base
32
~ in opposite segments, 38
~ in the same segment, 37
~ subtended by arcs, 38
supplementary
38
apex, 32
arcs, angles subtended by
38
area
~ of a convex quadrilateral, 86
triangle ~ formula, 36
arithmetic
fundamental theorem of
46
~ mean, 14
modular
47
base angles, 32
binomial
~ coefficient, 26
~ expansion, 26
bisectors
angle ~ of a triangle, 30
perpendicular — of a triangle, 30
BMO, 3
bounded sequence, 22
British Mathematical Olympiad, 3
card trick, Kruskal's
105
casting out nines, 50
Catalan
~ numbers, 127
~ sequence, 127
Cauchy-Schwarz inequality, 15, 87
centre, angle at the
39
centres of a triangle, 30
centroid, 30
Ceva's theorem, 97
choose, 25
choosing, 23
chords, intersecting ~ theorem, 41
circle theorem, 37
circumcentre, 13, 30
circumcircle, 30
circumradius, 13, 31
coefficient, binomial
26
colouring arguments, 27
combinatorics, 21
congruent
~ numbers, 47
~ triangles, 32
constant sequence, 22
A Mathematical Olympiad Primer
176
contradiction, proof by
converse, 4
coprime, 47
cosine
~ function, 34
~ rule, 36
counting, double
27
cyclic quadrilateral, 38
8
decreasing sequence, 22
degree, 17
descent, proof by
8
difference of two squares, 16
direct similarity, 33
Dirichlet principle, 21
divisibility tests, 49
divisor
common
46
greatest common
46
dot product, 15
double counting, 27
Elements, Euclid's
11
equal tangents, 41
equation, Pell's
172
equilateral, 32
Erdos, Paul
6
Euclid, 11
Euclid's Elements, 11
Euler's inequality, 13
expansion, binomial
26
exterior angle of a triangle, 31
eyeball theorem, 42
factor
highest common
47
~ theorem, 19
factorial, 25
Fermat's little theorem, 49
Fibonacci sequence, 22
formula
Heron's
36,170
triangle area
36
trigonometric
35
French heresy, 11
function
cosine
34
sine
34
tangent
34
fundamental theorem of arithmetic, 46
gcd, 46
generality, without loss of
geometric mean, 14
geometry, 29
Green, Ben, 150
Green-Tao theorem, 150
90
handshaking lemma, 27
Hardy, G. H., 150
harmonic mean, 88
hcf, 47
heresy, French
11
Heron's formula, 36,170
highest common factor, 47
identity, 16
incentre, 13, 30
incircle, 30
increasing sequence, 22
indirect similarity, 33
induction
principle of mathematical
proof by
4
inequality, 13
AM-GM
14
AM-HM
88
Cauchy-Schwarz
15, 87
Euler's
13
inradius, 13, 31
integers, 10
intersecting chords theorem, 41
isosceles, 32
Kruskal's card trick, 105
Kruskal, Martin J
105
4
Index
lemma, handshaking
177
27
proof, 3, 6,11,12
~ by contradiction, 8
~ by descent, 8
~ by induction, 4
~ by parity, 10
proportion, theory of - , 11
Ptolemy's theorem, 42
Pythagoras, 7
theorem of
7
mean
arithmetic
14
geometric
14
harmonic
88
quadratic
132
root ~ square, 132
medians of a triangle, 30
modular arithmetic, 47
modulus, 15, 48
quadratic mean, 132
quadrilateral
area of a convex
cyclic
38
quotient, 18
natural numbers, 10
nines, casting out
50
non-negative, 13
notation, triangle
30
number theory, 43
numbers, 10
Catalan
127
congruent
47
natural
10
prime
44
rational
11
real
10
opposite segments, angles in
orthocentre, 31
38
parity, proof by
10
Pascal's triangle, 25
Pell's equation, 172
periodic sequence, 22
perpendicular bisectors of a triangle, 30
pigeon-hole principle, 21
polynomials, 16
prime numbers, 44
principle
Dirichlet
21
~ of mathematical induction, 4
pigeon-hole
21
product
dot
15
scalar
15
86
rational numbers, 11
real numbers, 10
reflex angle, 39
remainder, 18
~ theorem, 19
root
~ mean square, 132
square ~ of 2, 7
rule
cosine
36
sine
36
same segment, angles in the
scalar product, 15
semi-circle, angle in a
40
semiperimeter, 36
sequence, 21
alternating
22
bounded
22
Catalan
127
constant
22
decreasing
22
Fibonacci
22
increasing
22
periodic
22
weakly decreasing 22
weakly increasing 22
37
A Mathematical Olympiad Primer
178
similar triangles, 32
similarity, 32
direct ~, 33
indirect
33
sine
~ function, 34
~ rule, 36
square root of 2, 7
squares, difference of two
supplementary angles, 38
16
tangent function, 34
tangent-secant theorem, 42
tangents, equal
41
Tao, Terry, 150
tests, divisibility
49
tetris, 117
Thales, theorem of
40
theorem
Ceva's
97
circle
37
eyeball
42
factor
19
Fermat's little
49
fundamental ~ of arithmetic, 46
Green-Tao
150
intersecting chords ~, 41
Ptolemy's
42
~ of Pythagoras, 7
remainder
19
tangent-secant
42
~ of Thales, 40
van Aubel's
97
theory
number
43
~ of proportion, 11
triangle
altitudes of a
30
angle bisectors of a
30
~ area formula, 36
centres of a
30
exterior angle of a
31
medians of a
30
~ notation, 30
Pascal's
25
perpendicular bisectors of a
special
32
triangles
congruent
32
similar
32
trigonometric formula, 35
trigonometry, 34
30
UKMT, 3
United Kingdom Mathematics Trust, 3
van Aubel's theorem, 97
weakly
~ decreasing sequence, 22
~ increasing sequence, 22
without loss of generality, 90
A Mathematical Olympiad Primer
This accessible text will enable enthusiastic
students to enter the world of secondary school
mathematics competitions with confidence. In a
chronological sense, the book is a sequel to
Gardiner’s The Mathematical Olympiad Handbook.
Books in the Handbooks series are aimed
particularly at students in secondary school who
are interested in acquiring knowledge and skills
which are useful for tackling problems, such as
those posed in the competitions administered by
the UKMT and similar organisations.
Like the previous volumes in the Handbooks series
this book provides ready access to directly relevant
material. All these books are characterized by the
large number of carefully constructed exercises for
the reader to attempt.
From The Mathematical Gazette review of Plane Euclidean
Geometry, the first book in the Handbooks series:
“'This is essentiaC reading for anybody -who
wants to huiCdup the confidence to tachde
the sort of questions set at the 'British
Mathematical Olympiad.”
The United Kingdom
Mathematics Trust
UKMT
www.ukmt.org.uk
ISBN 978-1-906001-05-