mathematics contests learn from yesterday, live for today, hope for tomorrow. the important thing is...

1

Lear n from yesterday, l ive for today, hope for tomor row. The impor tant thing is not to stop questioning.

Alber t Einstein

A u s t r A l i A n M A t h e M A t i c A l O l y M p i A d c O M M i t t e e

A depArtMent Of the AustrAliAn MAtheMAtics trust

MATHEMATICS CONTESTS THE AUSTRALIAN SCENE 2015

PART 2: INVITATIONAL COMPETITIONSA Di Pasquale and N Do

2

Published by

AMT Publishing

Australian Mathematics Trust

University of Canberra Locked Bag 1

Canberra GPO ACT 2601

Australia

Tel: 61 2 6201 5137

www.amt.edu.au

AMTT Limited

ACN 083 950 341

Copyright © 2015 by the Australian Mathematics Trust

National Library of Australia Card Number

and ISSN 1323-6490

http://www.amt.edu.au

3

SUPPORT FOR THE AUSTRALIAN MATHEMATICAL OLYMPIAD COMMITTEE TRAINING PROGRAM

The Australian Mathematical Olympiad Committee Training Program is an activity of the Australian Mathematical Olympiad Committee, a department of the Australian Mathematics Trust.

Trustee

The University of Canberra

Sponsors

The Mathematics/Informatics Olympiads are supported by the Australian Government Department of Education and Training through the Mathematics and Science Participation Program.

The Australian Mathematical Olympiad Committee (AMOC) also acknowledges the significant financial support it has received from the Australian Government towards the training of our Olympiad candidates and the participation of our team at the International Mathematical Olympiad (IMO).

The views expressed here are those of the authors and do not necessarily represent the views of the government.

Special Thanks

With special thanks to the Australian Mathematical Society, the Australian Association of Mathematics Teachers and all those schools, societies, families and friends who have contributed to the expense of sending the 2015 IMO team to Chiang Mai, Thailand.

http://www.canberra.edu.au/

4

ACKNOWLEDGEMENTS

The Australian Mathematical Olympiad Committee thanks sincerely all sponsors, teachers, mathematicians and others who have contributed in one way or another to the continued success of its activities.

The editors thank sincerely those who have assisted in the compilation of this book, in particular the students who have provided solutions to the 2015 IMO. Thanks also to members of AMOC and Challenge Problems Committees, Adjunct Professor Mike Clapper, staff of the Australian Mathematics Trust and others who are acknowledged elsewhere in the book.

5

PREFACE

After last year, there seemed little room for improvement, but 2015 has been even better, marked particularly by our best ever result at an IMO, where we were placed 6th out of the 104 competing countries, finishing ahead of all European countries (including Russia) and many other traditional powerhouses such as Singapore, Japan and Canada. For the first time ever, all six team members obtained Silver or better, with two team members (Alex Gunning and Seyoon Ragavan) claiming Gold. Alex finished fourth in the world (after his equal first place last year, and becomes Australia’s first triple Gold medallist in any academic Olympiad. Seyoon, who finished 19th, now has three IMOs under his belt with a year still to go. Once again, we had three Year 12 students in the team, so there will certainly be opportunities for new team members next year. It was also pleasing to, once again, see an Australian-authored question on the paper, this year, the prestigious Question 6, which was devised by Ivan Guo and Ross Atkins, based on the mathematics of juggling.

In the Mathematics Ashes we tied with the British team; however, we finished comfortably ahead of them in the IMO competition proper. Director of Training and IMO Team Leader, Dr Angelo Di Pasquale, along with his Deputy Andrew Elvey Price and their team of former Olympians continue to innovate and keep the training alive, fresh and, above all, of high quality. The policy of tackling very hard questions in training was daunting for team members at times but seems to have paid off.

The Mathematics Challenge for Young Australians (MCYA) also continues to attract strong entries, with the Challenge continuing to grow, helped by the gathering momentum of the new Middle Primary Division, which began in 2014. The Enrichment stage, containing course work, allows students to broaden their knowledge base in the areas of mathematics associated with the Olympiad programs and more advanced problem-solving techniques. We have continued running workshops for teachers to develop confidence in managing these programs for their more able students—this seems to be paying off with strong numbers in both the Challenge and Enrichment stages. The final stage of the MCYA program is the Australian Intermediate Mathematics Olympiad (AIMO). It is a delight to record that over the last two years the number of entries to AIMO has doubled. This has been partly due to wider promotion of the competition, but more specifically a result of the policy of offering free entry to AMC prize winners. There were some concerns in 2014 that some of the ‘new’ contestants were under-prepared for AIMO and there were more zero scores than we would have liked. However, this year, the number of zero scores was less than 1% and the quality of papers was much higher, revealing some significant new talent, some of whom will be rewarded with an invitation to the December School of Excellence.

There are many people who contribute to the success of the AMOC program. These include the Director of Training and the ex-Olympians who train the students at camps; the AMOC State Directors; and the Challenge Director, Dr Kevin McAvaney, and the various members of his Problems Committee, who develop such original problems, solutions and discussions each year. The AMOC Senior Problems Committee is also a major contributor and Norman Do is continuing with his good work. The invitational program saw some outstanding results from Australian students, with a number of perfect scores. Details of these achievements are provided in the appropriate section of this book. As was the case last year, we are producing Mathematics Contests—The Australian Scene in electronic form only and making it freely available through the website. We hope this will provide greater access to the problems and section reports. This book is also available in two sections, one containing the MCYA reports and papers and the other containing the Olympiad training program reports and papers.

Mike Clapper

November 2015

6

CONTENTS

Support for the Australian Mathematical Olympiad committee Training Program 3

Acknowledgements 4

Preface 5

Contents 6

Background Notes on the IMO and AMOC 7

Membership of AMOC Committees 10

AMOC Timetable for Selection of the Team to the 2016 IMO 11

Activities of AMOC Senior Problems Committee 12

AMOC Senior Contest 13

AMOC Senior Contest Solutions 14

AMOC Senior Contest Results 22

AMOC Senior Contest Statistics 23

AMOC School of Excellence 24

Australian Mathematical Olympiad 26

Australian Mathematical Olympiad Solutions 28

Australian Mathematical Olympiad Statistics 53

Australian Mathematical Olympiad Results 54

27th Asian Pacific Mathematics Olympiad 56

27th Asian Pacific Mathematics Olympiad Solutions 57

27th Asian Pacific Mathematics Olympiad Results 71

AMOC Selection School 73

IMO Team Preparation School 76

The Mathematics Ashes 77

The Mathematics Ashes Results 78

IMO Team Leader’s Report 79

International Mathematical Olympiad 83

International Mathematical Olympiad Solutions 85

International Mathematical Olympiad Results 111

Origin of Some Questions 116

Honour Roll 117

7

BACKGROUND NOTES ON THE IMO AND AMOC

The Australian Mathematical Olympiad Committee

In 1980, a group of distinguished mathematicians formed the Australian Mathematical Olympiad Committee (AMOC) to coordinate an Australian entry in the International Mathematical Olympiad (IMO).

Since then, AMOC has developed a comprehensive program to enable all students (not only the few who aspire to national selection) to enrich and extend their knowledge of mathematics. The activities in this program are not designed to accelerate students. Rather, the aim is to enable students to broaden their mathematical experience and knowledge.

The largest of these activities is the MCYA Challenge, a problem-solving event held in second term, in which thousands of young Australians explore carefully developed mathematical problems. Students who wish to continue to extend their mathematical experience can then participate in the MCYA Enrichment Stage and pursue further activities leading to the Australian Mathematical Olympiad and international events.

Originally AMOC was a subcommittee of the Australian Academy of Science. In 1992 it collaborated with the Australian Mathematics Foundation (which organises the Australian Mathematics Competition) to form the Australian Mathematics Trust. The Trust, a not-for-profit organisation under the trusteeship of the University of Canberra, is governed by a Board which includes representatives from the Australian Academy of Science, Australian Association of Mathematics Teachers and the Australian Mathematical Society.

The aims of AMOC include:

(1) giving leadership in developing sound mathematics programs in Australian schools (2) identifying, challenging and motivating highly gifted young Australian school students in mathematics(3) training and sending Australian teams to future International Mathematical Olympiads.

AMOC schedule from August until July for potential IMO team members

Each year hundreds of gifted young Australian school students are identified using the results from the Australian Mathematics Competition sponsored by the Commonwealth Bank, the Mathematics Challenge for Young Australians program and other smaller mathematics competitions. A network of dedicated mathematicians and teachers has been organised to give these students support during the year either by correspondence sets of problems and their solutions or by special teaching sessions.

It is these students who sit the Australian Intermediate Mathematics Olympiad, or who are invited to sit the AMOC Senior Contest each August. Most states run extension or correspondence programs for talented students who are invited to participate in the relevant programs. The 25 outstanding students in recent AMOC programs and other mathematical competitions are identified and invited to attend the residential AMOC School of Excellence held in December.

In February approximately 100 students are invited to attempt the Australian Mathematical Olympiad. The best 20 or so of these students are then invited to represent Australia in the correspondence Asian Pacific Mathematics Olympiad in March. About 12 students are selected for the AMOC Selection School in April and about 13 younger students are also invited to this residential school. Here, the Australian team of six students plus one reserve for the International Mathematical Olympiad, held in July each year, is selected. A personalised support system for the Australian team operates during May and June.

It should be appreciated that the AMOC program is not meant to develop only future mathematicians. Experience has shown that many talented students of mathematics choose careers in engineering, computing, and the physical and life sciences, while others will study law or go into the business world. It is hoped that the AMOC Mathematics Problem-Solving Program will help the students to think logically, creatively, deeply and with dedication and perseverance; that it will prepare these talented students to be future leaders of Australia.

The International Mathematical Olympiad

The IMO is the pinnacle of excellence and achievement for school students of mathematics throughout the world. The concept of national mathematics competitions started with the Eötvos Competition in Hungary during 1894. This idea was later extended to an international mathematics competition in 1959 when the first IMO was

8

held in Romania. The aims of the IMO include: (1) discovering, encouraging and challenging mathematically gifted school students (2) fostering friendly international relations between students and their teachers (3) sharing information on educational syllabi and practice throughout the world.

It was not until the mid-sixties that countries from the western world competed at the IMO. The United States of America first entered in 1975. Australia has entered teams since 1981.

Students must be under 20 years of age at the time of the IMO and have not enrolled at a tertiary institution. The Olympiad contest consists of two four-and-a-half hour papers, each with three questions.

Australia has achieved varying successes as the following summary of results indicate. HM (Honorable Mention) is awarded for obtaining full marks in at least one question.

The IMO will be held in Hong Kong in 2016.

9

Summary of Australia’s achievements at previous IMOs

Year City Gold Silver Bronze HM Rank

1981 Washington 1 23 out of 27 teams

1982 Budapest 1 21 out of 30 teams

1983 Paris 1 2 19 out of 32 teams

1984 Prague 1 2 15 out of 34 teams

1985 Helsinki 1 1 2 11 out of 38 teams

1986 Warsaw 5 15 out of 37 teams

1987 Havana 3 15 out of 42 teams

1988 Canberra 1 1 1 17 out of 49 teams

1989 Braunschweig 2 2 22 out of 50 teams

1990 Beijing 2 4 15 out of 54 teams

1991 Sigtuna 3 2 20 out of 56 teams

1992 Moscow 1 1 2 1 19 out of 56 teams

1993 Istanbul 1 2 3 13 out of 73 teams

1994 Hong Kong 2 3 3 12 out of 69 teams

1995 Toronto 1 4 1 21 out of 73 teams

1996 Mumbai 2 3 23 out of 75 teams

1997 Mar del Plata 2 3 1 9 out of 82 teams

1998 Taipei 4 2 13 out of 76 teams

1999 Bucharest 1 1 3 1 15 out of 81 teams

2000 Taejon 1 3 1 16 out of 82 teams

2001 Washington D.C. 1 4 25 out of 83 teams

2002 Glasgow 1 2 1 1 26 out of 84 teams

2003 Tokyo 2 2 2 26 out of 82 teams

2004 Athens 1 1 2 1 27 out of 85 teams

2005 Merida 6 25 out of 91 teams

2006 Ljubljana 3 2 1 26 out of 90 teams

2007 Hanoi 1 4 1 22 out of 93 teams

2008 Madrid 5 1 19 out of 97 teams

2009 Bremen 2 1 2 1 23 out of 104 teams

2010 Astana 1 3 1 1 15 out of 96 teams

2011 Amsterdam 3 3 25 out of 101 teams

2012 Mar del Plata 2 4 27 out of 100 teams

2013 Santa Marta 1 2 3 15 out of 97 teams

2014 Cape Town1

Perfect Score by Alexander

Gunning3 2 11 out of 101 teams

2015 Chiang Mai 2 4 6 out of 104 teams

10

MEMBERSHIP OF AMOC COMMITTEES

Australian Mathematical Olympiad Committee 2015ChairProf C Praeger, University of Western AustraliaDeputy ChairAssoc Prof D Hunt, University of New South Wales

Executive Director Adj Prof Mike Clapper, Australian Mathematics Trust, ACT

TreasurerDr P Swedosh, The King David School, VIC

Chair, Senior Problems CommitteeDr N Do, Monash University, VIC

Chair, ChallengeDr K McAvaney, Deakin University, VIC

Director of Training and IMO Team LeaderDr A Di Pasquale, University of Melbourne, VIC

IMO Deputy Team LeaderMr A Elvey Price, University of Melbourne, VIC

State DirectorsDr K Dharmadasa, University of TasmaniaDr G Gamble, University of Western AustraliaDr Ian Roberts, Northern TerritoryDr W Palmer, University of Sydney, NSWMr D Martin, South AustraliaDr V Scharaschkin, University of QueenslandDr P Swedosh, The King David School, VICDr Chris Wetherell, Radford College, ACT

RepresentativesMs A Nakos, Challenge CommitteeProf M Newman, Challenge CommitteeMr H Reeves, Challenge Committee

11

AMOC TIMETABLE FOR SELECTION OF THE TEAM TO THE 2016 IMO

August 2015—July 2016

Hundreds of students are involved in the AMOC programs which begin on a state basis. The students are given problem-solving experience and notes on various IMO topics not normally taught in schools.

The students proceed through various programs with the top 25 students, including potential team members and other identified students, participating in a ten-day residential school in December.

The selection program culminates with the April Selection School during which the team is selected.

Team members then receive individual coaching by mentors prior to assembling for last minute training before the IMO.

Month Activity

August

Outstanding students are identified from AMC results, MCYA, other competitions and recommendations; and eligible students from previous training programsAMOC state organisers invite students to participate in AMOC programsVarious state-based programsAMOC Senior Contest

September Australian Intermediate Mathematics Olympiad

December AMOC School of Excellence

January Summer Correspondence Program for those who attended the School of Excellence

February Australian Mathematical Olympiad

March Asian Pacific Mathematics Olympiad

April AMOC Selection School

May–June Personal Tutor Scheme for IMO team members

JulyShort mathematics school for IMO team members2016 IMO in Hong Kong.

12

ACTIVITIES OF AMOC SENIOR PROBLEMS COMMITTEE

This committee has been in existence for many years and carries out a number of roles. A central role is the collection and moderation of problems for senior and exceptionally gifted intermediate and junior secondary school students. Each year the Problems Committee provides examination papers for the AMOC Senior Contest and the Australian Mathematical Olympiad. In addition, problems are submitted for consideration to the Problem Selection Committees of the annual Asian Pacific Mathematics Olympiad and the International Mathematical Olympiad.

AMOC Senior Problems Committee October 2014–September 2015Dr A Di Pasquale, University of Melbourne, VICDr N Do, Monash University, VIC, (Chair)Dr M Evans, Australian Mathematical Sciences Institute, VICDr I Guo, University of Sydney, NSWAssoc Prof D Hunt, University of NSWDr J Kupka, Monash University, VICAssoc Prof H Lausch, Monash University, VICDr K McAvaney, Deakin University, VICDr D Mathews, Monash University, VICDr A Offer, QueenslandDr C Rao, NEC Australia, VICDr B B Saad, Monash University, VICAssoc Prof J Simpson, Curtin University of Technology, WAEmeritus Professor P J Taylor, Australian Capital TerritoryDr I Wanless, Monash University, VIC

1. 2015 Australian Mathematical Olympiad

The Australian Mathematical Olympiad (AMO) consists of two papers of four questions each and was sat on 10 and 11 February. There were 106 participants including 11 from New Zealand, seven more participants than 2014. Two students, Alexander Gunning and Seyoon Ragavan, achieved perfect scores and nine other students were awarded Gold certificates,15 students were awarded Silver certificates and 26 students were awarded Bronze certificates.

2. 2015 Asian Pacific Mathematics Olympiad

On Tuesday 10 March students from nations around the Asia-Pacific region were invited to write the Asian Pacific Mathematics Olympiad (APMO). Of the top ten Australian students who participated, there were 1 Gold, 2 Silver, 4 Bronze and 3 HM certificates awarded. Australia finished in 9th place overall.

3. 2015 International Mathematical Olympiad, Chiang Mai, Thailand.

The IMO consists of two papers of three questions worth seven points each. They were attempted by teams of six students from 104 countries on 8 and 9 July in Cape Town, South Africa. Australia was placed 6th of 104 countries, its most successful result since it began participating. The medals for Australia were two Gold and four Silver.

4. 2015 AMOC Senior Contest

Held on Tuesday 11 August, the Senior Contest was sat by 84 students (compared to 81 in 2014). There were three students who obtained perfect scores and two other students who were also Prize winners. Three studentent obtained High Distinctions and 14 students obtained Distinctions.

13

AMOC SENIOR CONTESTTHE 2015 AMOC SENIOR CONTEST

Tuesday, 11 August 2015

Time allowed: 4 hours

No calculators are to be used.

Each question is worth seven points.

1. A number is called k-addy if it can be written as the sum of k consecutive positive

integers. For example, the number 9 is 2-addy because 9 = 4+5 and it is also 3-addy

because 9 = 2 + 3 + 4.

(a) How many numbers in the set 1, 2, 3, . . . , 2015 are simultaneously 3-addy,

4-addy and 5-addy?

(b) Are there any positive integers that are simultaneously 3-addy, 4-addy, 5-addy

and 6-addy?

2. Consider the sequence a1, a2, a3, . . . defined by a1 = 1 and

am+1 =1a1 + 2a2 + 3a3 + · · ·+mam

amfor m ≥ 1.

Determine the largest integer n such that an < 1 000 000.

3. A group of students entered a mathematics competition consisting of five problems.

Each student solved at least two problems and no student solved all five problems.

For each pair of problems, exactly two students solved them both.

Determine the minimum possible number of students in the group.

4. Let ABCD be a rectangle with AB > BC. Let E be the point on the diagonal AC

such that BE is perpendicular to AC. Let the circle through A and E whose centre

lies on the line AD meet the side CD at F .

Prove that BF bisects the angle AFC.

5. For a real number x, let x be the largest integer less than or equal to x.

Find all prime numbers p for which there exists an integer a such that

⌊a

p

⌋+

⌊2a

p

⌋+

⌊3a

p

⌋+ · · ·+

⌊pa

p

⌋= 100.

c© 2015 Australian Mathematics Trust

14

THE 2015 AMOC SENIOR CONTEST

Solutions and cumulative marking scheme


1. A number is called k-addy if it can be written as the sum of k consecutive positive integers.

For example, the number 9 is 2-addy because 9 = 4 + 5 and it is also 3-addy because

9 = 2 + 3 + 4.

(a) How many numbers in the set 1, 2, 3, . . . , 2015 are simultaneously 3-addy, 4-addy

and 5-addy?

(b) Are there any positive integers that are simultaneously 3-addy, 4-addy, 5-addy and

6-addy?

Solution (Angelo Di Pasquale)

Since (a− 1) + a+ (a+ 1) = 3a, the 3-addy numbers are precisely those that are divisible

by 3 and greater than 3. 1

Since (a− 2)+ (a− 1)+ a+(a+1)+ (a+2) = 5a, the 5-addy numbers are precisely those

that are divisible by 5 and greater than 10.

Since (a− 1)+ a+ (a+1)+ (a+2) = 4a+2, the 4-addy numbers are precisely those that

are congruent to 2 modulo 4 and greater than 6. 2

(a) From the observations above, a positive integer is simultaneously 3-addy, 4-addy and

5-addy if and only if it is divisible by 3, divisible by 5, divisible by 2, and not divisible

by 4. Such numbers are of the form 30m, where m is a positive odd integer. 3

Since 67×30 = 2010, the number of elements of the given set that are simultaneously

3-addy, 4-addy and 5-addy is 682 = 34. 4

(b) Since (a− 2) + (a− 1) + a+ (a+ 1) + (a+ 2) + (a+ 3) = 6a+ 3, all 6-addy numbers

are necessarily odd. 5

On the other hand, we have already deduced that all 4-addy numbers are even. 6

Therefore, there are no numbers that are simultaneously 4-addy and 6-addy. 7

1

AMOC SENIOR CONTEST SOLUTIONS

15

2. Consider the sequence a1, a2, a3, . . . defined by a1 = 1 and

am+1 =1a1 + 2a2 + 3a3 + · · ·+mam

amfor m ≥ 1.

Determine the largest integer n such that an < 1 000 000.

Solution 1 (Norman Do)

First, we note that all terms of the sequence are positive rational numbers. Below, we

rewrite the defining equation for the sequence in both its original form and with the value

of m shifted by 1.

amam+1 = 1a1 + 2a2 + 3a3 + · · ·+mam

am+1am+2 = 1a1 + 2a2 + 3a3 + · · ·+mam + (m+ 1)am+1 1

Subtracting the first equation from the second yields

am+1am+2 − amam+1 = (m+ 1)am+1 ⇒ am+2 − am = m+ 1,

for all m ≥ 1. 2

So for m = 2k + 1 an odd positive integer,

a2k+1 − a1 = (a2k+1 − a2k−1) + (a2k−1 − a2k−3) + · · ·+ (a3 − a1)

= 2k + (2k − 2) + · · ·+ 2

= 2 [k + (k − 1) + · · ·+ 1]

= k(k + 1). 3

Similarly, for m = 2k an even positive integer,

a2k − a2 = (a2k − a2k−2) + (a2k−2 − a2k−4) + · · ·+ (a4 − a2)

= (2k − 1) + (2k − 3) + · · ·+ 3

= k2 − 1. 4

Using a1 = 1 and a2 = 1, we obtain the formula

am =

m2

4 , if m is even,

m2+34 , if m is odd.

5

If m is odd, then

am+1 − am =(m+ 1)2

4− m2 + 3

4=

2m− 2

4,

and if m is even, then

am+1 − am =(m+ 1)2 + 3

4− m2

4=

2m+ 4

4. 6

In particular, it follows that a2 < a3 < a4 < · · · . Since a2000 = 1000 000, the largest

integer n such that an < 1 000 000 is 1999. 7

2

16

Solution 2 (Angelo Di Pasquale)

We will prove by induction that a2k−1 = k2 − k + 1 and a2k = k2 for each positive integer

k. The base case k = 1 is true since a1 = a2 = 1. Now assume that the two formulas hold

for k = 1, 2, . . . , n. We will show that they also hold for k = n+ 1. 2

Consider the following sequence of equalities.

2n∑i=1

iai =

n∑i=1

(2i− 1) a2i−1 +

n∑i=1

2i a2i

=

n∑i=1

(2i− 1) (i2 − i+ 1) +

n∑i=1

(2i) (i2) (by the inductive assumption)

=

n∑i=1

4i3 − 3i2 + 3i− 1

=

n∑i=1

3i3 + (i− 1)3 3

= 3

n∑i=1

i3 +

n−1∑i=1

i3

=3n2(n+ 1)2

4+

(n− 1)2n2

4

(since

n∑i=1

i3 =n2(n+ 1)2

4

)

= n2(n2 + n+ 1)

= (n2 + n+ 1)a2n (by the inductive assumption)

It follows that

a2n+1 =1a1 + 2a2 + 3a3 + · · ·+ 2na2n

a2n= n2 + n+ 1. 4

Now consider the following sequence of equalities, which uses the facts derived above that

state that∑2n

i=1 iai = n2(n2 + n+ 1) and a2n+1 = n2 + n+ 1.

2n+1∑i=1

iai = n2(n2 + n+ 1) + (2n+ 1)a2n+1

= n2a2n+1 + (2n+ 1)a2n+1

= (n+ 1)2a2n+1

It follows that

a2n+2 =1a1 + 2a2 + 3a3 + · · ·+ (2n+ 1)a2n+1

a2n+1= (n+ 1)2. 5

So we have shown that the two formulas a2k−1 = k2 − k + 1 and a2k = k2 hold for

k = 1, 2, . . . , n+1. This completes the induction and the rest of the proof follows Solution

1. 7

3

17

3. A group of students entered a mathematics competition consisting of five problems. Each

student solved at least two problems and no student solved all five problems. For each

pair of problems, exactly two students solved them both.

Determine the minimum possible number of students in the group.

Solution (Norman Do)

It is possible that the group comprised six students, as demonstrated by the following

example.

Student 1 solved problems 1, 2, 3, 4.

Student 2 solved problems 1, 2, 3, 5.

Student 3 solved problems 1, 4, 5.

Student 4 solved problems 2, 4, 5.

Student 5 solved problems 3 and 4.

Student 6 solved problems 3 and 5. 2

Suppose that a students solved 4 problems, b students solved 3 problems, and c students

solved 2 problems. Therefore, a students solved(42

)= 6 pairs of problems, b students

solved(32

)= 3 pairs of problems and c students solved

(22

)= 1 pair of problems. Since we

have shown an example in which the number of students in the group is 6, let us assume

that a+ b+ c ≤ 5.

There are(52

)= 10 pairs of problems altogether and, for each pair of problems, exactly

two students solved them both, so we must have

6a+ 3b+ c = 20. 4

Reading the above equation modulo 3 yields c ≡ 2 (mod 3). If c ≥ 5, then we have

a+b+c ≥ 6, contradicting our assumption. Therefore, we must have c = 2 and 2a+b = 6.

For a+ b+ c ≤ 5, the only solution is given by (a, b, c) = (3, 0, 2). 5

However, it is impossible for 3 students to have solved 4 problems each. That would mean

that each of the 3 students did not solve exactly 1 problem. So there would exist a pair of

problems for which 3 students solved them both, contradicting the required conditions.

In conclusion, the minimum possible number of students in the group is 6. 7

4

18

4. Let ABCD be a rectangle with AB > BC. Let E be the point on the diagonal AC such

that BE is perpendicular to AC. Let the circle through A and E whose centre lies on the

line AD meet the side CD at F .

Prove that BF bisects the angle AFC.

Solution 1 (Alan Offer)

Let the circle through A and E whose centre lies on AD meet the line AD again at H.

A B

CD

E

F

H

Since FD is the altitude of the right-angled triangle AFH, we have AFH ∼ ADF .

Since triangles AEH and ADC are right-angled with a common angle at A, we have

AEH ∼ ADC. Since BE is the altitude of the right-angled triangle ABC, we have

ABC ∼ AEB. These three pairs of similar triangles lead respectively to the three

pairs of equal ratios

AF

AD=

AH

AF

AE

AD=

AH

AC

AB

AE=

AC

AB. 3

Putting these together, we have

AF 2 = AD ·AH = AC ·AE = AB2. 6

So triangle BAF is isosceles and we have

∠AFB = ∠ABF = 90 − ∠CBF = ∠CFB,

where the last equality uses the angle sum in triangle BCF . Since ∠AFB = ∠CFB, we

have proven that BF bisects the angle AFC. 7

Solution 2 (Angelo Di Pasquale)

First, note that the circumcircles of triangle AEF and triangle BEC are tangent to the

line AB at A and B, respectively. Considering the power of the point A with respect to

the circumcircle of triangle BEC, we have

AB2 = AE ·AC. 1

5

19

Next, by the alternate segment theorem and the fact that AB ‖ CD, we have ∠EFA =

∠EAB = ∠ECF . Hence, by the alternate segment theorem again, the circumcircle of

triangle EFC is tangent to the line AF at F . Considering the power of the point A with

respect to the circumcircle of triangle EFC, we have

AF 2 = AE ·AC. 4

Comparing the two equations above, we deduce that AB = AF . 6

Hence, ∠AFB = ∠ABF = ∠CFB, as required. 7

Solution 3 (Ivan Guo)

Let the circle through A and E whose centre lies on AD meet the line AD again at H.

Then ∠AEB = ∠AEH = 90, so the points B, E and H are collinear. Consider the

inversion f with centre A and radius AF . 2

Note that the circumcircle of the cyclic quadrilateral AEFH must map to a line parallel

to AB through F . Thus,

f(circle AEFH) = line CD. 4

This immediately gives f(C) = E and f(H) = D. Now the circumcircle of ABCD must

map to a line passing through f(C) = E and f(D) = H. This implies that f(B) = B.

Hence, AB = AF . 6

Therefore, we have ∠BFC = ∠ABF = ∠AFB. 7

Solution 4 (Chaitanya Rao)

Consider the following chain of equalities.

DF 2 = AD ·DH (ADF ∼ FDH) 1

= AD · (AH −AD)

= AD ·(AB2

BC−AD

)(ABC ∼ HAB) 3

= AB2 −AD2 (AD = BC) 4

Hence, AB2 = DF 2 +AD2 = AF 2, which implies that AB = AF . 6

So triangle BAF is isosceles and we have

∠AFB = ∠ABF = 90 − ∠CBF = ∠CFB,

where the last equality uses the angle sum in triangle BCF . Since ∠AFB = ∠CFB, we

have proven that BF bisects the angle AFC. 7

6

20

5. For a real number x, let x be the largest integer less than or equal to x.

Find all prime numbers p for which there exists an integer a such that⌊a

p

⌋+

⌊2a

p

⌋+

⌊3a

p

⌋+ · · ·+

⌊pa

p

⌋= 100.

Solution 1 (Norman Do)

The possible values for p are 2, 5, 17 and 197.

We divide the problem into the following two cases.

The number a is divisible by p.

If we write a = kp, the equation becomes

a(p+ 1)

2= 100 ⇒ kp(p+ 1) = 200.

So both p and p + 1 are positive divisors of 200. However, one can easily see that

there are no such primes p. 1

The number a is not divisible by p.

For a real number x, let x = x− x. Then we may write the equation as(a

p+

2a

p+

3a

p+ · · ·+ pa

p

)−

(a

p

+

2a

p

+

3a

p

+ · · ·+

pa

p

)= 100.

Summing the terms of the arithmetic progression on the left-hand side yields

a(p+ 1)

2−

[a

p

+

2a

p

+

3a

p

+ · · ·+

pa

p

]= 100. 3

We will prove that the sequence of numbersap

,2ap

,3ap

, . . . ,

pap

is a rearrange-

ment of the sequence of numbers 0p ,

1p ,

2p , . . . ,

p−1p . 4

Observe that if k is a positive integer, thenkap

is one of the numbers 0

p ,1p ,

2p , . . . ,

p−1p .

So it suffices to show that no two of the numbersap

,2ap

,3ap

, . . . ,

pap

are equal.

Suppose for the sake of contradiction thatiap

=

jap

, where 1 ≤ i < j ≤ p. Then

jap − ia

p = a(j−i)p must be an integer. It follows that either a is divisible by p or j− i is

divisible by p. However, since we have assumed that a is not divisible by p and that

1 ≤ i < j ≤ p, we obtain the desired contradiction. Hence, we may conclude that the

sequence of numbersap

,2ap

,3ap

, . . . ,

pap

is a rearrangement of the sequence

of numbers 0p ,

1p ,

2p , . . . ,

p−1p . 5

We may now write the equation as

a(p+ 1)

2− p− 1

2= 100 ⇒ (a− 1)(p+ 1) = 198. 6

Therefore, p+ 1 is a positive divisor of 198 — in other words, one of the numbers

1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198.

Since p is a prime, it follows that p must be equal to 2, 5, 17 or 197. This leads to

the possible solutions (p, a) = (2, 67), (5, 34), (17, 12), (197, 2). All four of these pairs

satisfy the given equation with a not divisible by p, so we obtain p = 2, 5, 17, 197. 7

7

21

Solution 2 (Ivan Guo, Angelo Di Pasquale and Ian Wanless)

The possible values for p are 2, 5, 17 and 197.

The case where a is divisible by p is handled in the same way as Solution 1. 1

Furthermore, one can check that the pair (p, a) = (2, 67) satisfies the conditions of the

problem. So assume that p is an odd prime and that a is not divisible by p.

For any integers 1 ≤ r, s ≤ p− 1 with r + s = p, we have p ra and p sa. Therefore,

ar

p− 1 +

as

p− 1 <

⌊ar

p

⌋+

⌊as

p

⌋<

ar

p+

as

p⇒ a− 2 <

⌊ar

p

⌋+

⌊as

p

⌋< a.

But since⌊arp

⌋+

⌊asp

⌋is an integer, we conclude that

⌊ar

p

⌋+

⌊as

p

⌋= a− 1. 3

The numbers 1, 2, . . . , p− 1 can be partitioned into p−12 pairs whose sum is p. 4

Using the equation above for each such pair and substituting into the original equation,

we obtain (p− 1

2

)(a− 1) + a = 100 ⇒ (a− 1)(p+ 1) = 198. 6

Therefore, p+ 1 is a positive divisor of 198 — in other words, one of the numbers

1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198.

Since p is a prime, it follows that p must be equal to 2, 5, 17 or 197. This leads to the

possible solutions (p, a) = (2, 67), (5, 34), (17, 12), (197, 2). All four of these pairs satisfy

the given equation with a not divisible by p, so we obtain p = 2, 5, 17, 197. 7

8

22

Name School Year Score

Prize

Yong See Foo Nossal High School VIC 11 35

Kevin Xian James Ruse Agricultural High School NSW 11 35

Wilson Zhipu Zhao Killara High School NSW 11 35

Ilia Kucherov Westall Secondary College VIC 11 34

Seyoon Ragavan Knox Grammar School NSW 11 34

High Distinction

Jongmin Lim Killara High School NSW 11 32

Matthew Cheah Penleigh and Essendon Grammar School VIC 10 29

Jerry Mao Caulfield Grammar School (Wheelers Hill) VIC 9 29

Distinction

Alexander Barber Scotch College VIC 11 28

Michelle Chen Methodist Ladies’ College VIC 11 28

Thomas Baker Scotch College VIC 11 27

Linus Cooper James Ruse Agricultural High School NSW 9 27

Steven Lim Hurlstone Agricultural High School NSW 9 27

Eric Sheng Newington College NSW 11 27

William Song Scotch College VIC 11 27

Jack Liu Brighton Grammar VIC 9 26

Leo Li Christ Church Grammar School WA 11 25

Bobby Dey James Ruse Agricultural High School NSW 10 24

Michael Robertson Dickson College ACT 11 24

Charles Li Camberwell Grammar School VIC 9 22

Isabel Longbottom Rossmoyne Senior High School WA 10 22

Guowen Zhang St Joseph’s College, Gregory Terrace QLD 9 22

AMOC SENIOR CONTEST RESULTS

23

Score Distribution/Problem

Problem Number

Number of Students/ScoreMean

0 1 2 3 4 5 6 7

1 1 0 0 0 1 5 46 31 6.2

2 16 9 8 1 2 8 15 25 4.1

3 25 4 11 2 4 5 8 25 3.5

4 54 13 2 0 0 0 0 15 1.5

5 65 2 1 0 0 4 6 6 1.2

AMOC SENIOR CONTEST STATISTICS

24

AMOC SCHOOL OF EXCELLENCE2014 AMOC School of Excellence

The 2014 AMOC School of Excellence was held 1–10 December at Newman College,University of Melbourne. The main qualifying exams to be invited to this are the AIMOand the AMOC Senior Contest.

This year AMOC went ahead with a new initiative in an attempt to widen the net ofidentification. In particular, any prize winner in the Australian Mathematics Competition(AMC) would be given free entry into the AIMO. In this way we hoped to identify topstudents from among schools that may not normally enter students in the AIMO. Thisturned out to be quite successful and prompted us to increase the number of invitationswe normally make. Consequently, 28 students from around Australia attended the school.A further student from New Zealand also attended.

The students are divided into a senior group and a junior group. There were 17 juniorstudents, 16 of whom were attending for the first time. There were 12 students makingup the senior group.

The program covered the four major areas of number theory, geometry, combinatoricsand algebra. Each day would start at 8am with lectures or an exam and go until 12noon or 1pm. After a one-hour lunch break they would have a lecture at 2pm. At4pm participants would usually have free time, followed by dinner at 6pm. Finally, eachevening would round out with a problem session, topic review, or exam review from 7pmuntil 9pm.

Another new initiative we tried was to invite two of our more experienced senior studentsto give a lecture. The rationale behind this is that teaching a subject is highly bene-ficial to the teacher because it can really solidify the foundations of the teacher’s ownunderstanding. A further fringe benefit is that they also get to hone their LATEX skills.Alexander Gunning was assigned the senior inequalities lecture, and Seyoon Ragavan wasassigned the senior transformation geometry lecture. I did some trial runs with themprior to the lectures so as to trouble shoot any problems, as well as to give them somepractice for the real thing. Overall it was successful, and I will likely try it again in thefuture.

Many thanks to Andrew Elvey Price, Ivan Guo, Victor Khou, and Sampson Wong, whoserved as live-in staff. Also my thanks go to Adrian Agisilaou, Norman Do, Patrick He,Alfred Liang, Daniel Mathews, Konrad Pilch, Chaitanya Rao, and Mel Shu who assistedin lecturing and marking.

Angelo Di PasqualeDirector of Training, AMOC

1

25

Name School Year

Seniors

Thomas Baker Scotch College VIC 10

Matthew Cheah Penleigh and Essendon Grammar School VIC 9

Yong See Foo Nossal High School VIC 10

Alexander Gunning Glen Waverley Secondary College VIC 11

Leo Li Christ Church Grammar School WA 10

Allen Lu Sydney Grammar School NSW 11

Seyoon Ragavan Knox Grammar School NSW 10

Kevin Shen St Kentigern College NZ 11*

Yang Song James Ruse Agricultural High School NSW 11

Kevin Xian James Ruse Agricultural High School NSW 10

Jeremy Yip Trinity Grammar School VIC 11

Henry Yoo Perth Modern School WA 11

Juniors

Adam Bardrick Whitefriars College VIC 8

Rachel Hauenschild Kenmore State High School QLD 9

William Hu Christ Church Grammar School WA 8

Shivasankaran Jayabalan Rossmoyne Senior High School WA 8

Tony Jiang Scotch College VIC 9

Sharvil Kesarwani Merewether High School NSW 7

Ilia Kucherov Westall Secondary College VIC 10

Adrian Law James Ruse Agricultural High School NSW 9

Charles Li Camberwell Grammar School VIC 8

Jack Liu Brighton Grammar School VIC 8

Isabel Longbottom Rossmoyne Senior High School WA 9

Hilton Nguyen Sydney Technical High School NSW 8

Madeline Nurcombe Cannon Hill Anglican College QLD 10

Zoe Schwerkolt Fintona Girls' School VIC 10

Katrina Shen James Ruse Agricultural High School NSW 7

Eric Sheng Newington College NSW 10

Wen Zhang St Joseph's College QLD 8

Participants at the 2014 AMOC School of Excellence

* Equivalent to year 10 in Australia.

26

AUSTRALIAN MATHEMATICAL OLYMPIADTHE 2015 AUSTRALIAN MATHEMATICALOLYMPIAD

DAY 1

Tuesday, 10 February 2015




1. Define the sequence a1, a2, a3, . . . by a1 = 4, a2 = 7, and

an+1 = 2an − an−1 + 2, for n ≥ 2.

Prove that, for every positive integer m, the number amam+1 is a term of the sequence.

2. For each positive integer n, let s(n) be the sum of its digits. We call a number nifty if it

can be expressed as n− s(n) for some positive integer n.

How many positive integers less than 10,000 are nifty?

3. Let S be the set of all two-digit numbers that do not contain the digit 0. Two numbers

in S are called friends if their largest digits are equal and the difference between their

smallest digits is 1. For example, the numbers 68 and 85 are friends, the numbers 78

and 88 are friends, but the numbers 58 and 75 are not friends.

Determine the size of the largest possible subset of S that contains no two numbers that

are friends.

4. Let Γ be a fixed circle with centre O and radius r. Let B and C be distinct fixed points

on Γ. Let A be a variable point on Γ, distinct from B and C. Let P be the point such

that the midpoint of OP is A. The line through O parallel to AB intersects the line

through P parallel to AC at the point D.

(a) Prove that, as A varies over the points of the circle Γ (other than B or C), D lies

on a fixed circle whose radius is greater than or equal to r.

(b) Prove that equality occurs in part (a) if and only if BC is a diameter of Γ.


27

THE 2015 AUSTRALIAN MATHEMATICALOLYMPIAD

DAY 2

Wednesday, 11 February 2015




5. Let ABC be a triangle with ∠ACB = 90. The points D and Z lie on the side AB such

that CD is perpendicular to AB and AC = AZ. The line that bisects ∠BAC meets CB

and CZ at X and Y , respectively.

Prove that the quadrilateral BXYD is cyclic.

6. Determine the number of distinct real solutions of the equation

(x− 1) (x− 3) (x− 5) · · · (x− 2015) = (x− 2) (x− 4) (x− 6) · · · (x− 2014).

7. For each integer n ≥ 2, let p(n) be the largest prime divisor of n.

Prove that there exist infinitely many positive integers n such that

(p(n+ 1)− p(n)

) (p(n)− p(n− 1)

)> 0.

8. Let n be a given integer greater than or equal to 3. Maryam draws n lines in the plane

such that no two are parallel.

For each equilateral triangle formed by three of the lines, Maryam receives three apples.

For each non-equilateral isosceles triangle formed by three of the lines, she receives one

apple.

What is the maximum number of apples that Maryam can obtain?


28

AUSTRALIAN MATHEMATICAL OLYMPIAD SOLUTIONS

1. For reference, the sequence a1, a2, a3, . . . is given by a1 = 4, a2 = 7, and

an+1 = 2an − an−1 + 2, for n ≥ 2.

Solution 1 (Linus Cooper, year 9, James Ruse Agricultural High School, NSW)

First we prove the following formula by induction.

an = n2 + 3, for n ≥ 1.

We require two base cases to get started. The formula is true for n = 1 and n = 2because a1 = 4 = 12 + 3 and a2 = 7 = 22 + 3.

For the inductive step, assume that the formula is true for n = k − 1 and n = k.Then for n = k + 1, we have

ak+1 = 2ak − ak−1 + 2 (given)

= 2(k2 + 3)− ((k − 1)2 + 3) + 2 (inductive assumption)

= k2 + 2k + 4

= (k + 1)2 + 3.

Hence the formula is also true for n = k + 1. This completes the induction.

Using the formula, we calculate

amam+1 = (m2 + 3)((m+ 1)2 + 3)

= (m2 + 3)(m2 + 2m+ 4)

= m4 + 2m3 + 7m2 + 6m+ 12

= (m2 +m+ 3)2 + 3

= am2+m+3.

Hence amam+1 is a term of the sequence.

17

29

Solution 2 (Jack Liu, year 9, Brighton Grammar School, VIC)

The given rule for determining an+1 from an and an−1 can be rewritten as

an+1 − an = an − an−1 + 2, for n ≥ 2.

Thus the difference between consecutive terms of the sequence increases by 2 as wego up. Hence starting from a2 − a1 = 3, we may write down the following.

a2 − a1 = 3

a3 − a2 = 5

a4 − a3 = 7

...

am − am−1 = 2m− 1

If we add all these equations together, we find that almost everything cancels onthe LHS, and we have

am − a1 = 3 + 5 + · · ·+ 2m− 1.

Using the standard formula1 for summing an arithmetic progression, we find

am − a1 =(2m+ 2)(m− 1)

2= m2 − 1.

Since a1 = 4, it follows that am = m2 + 3 for each m ≥ 1.

It is now a simple matter to calculate

amam+1 = (m2 + 3)((m+ 1)2 + 3)

= m4 + 2m3 + 7m2 + 6m+ 12

= (m2 +m+ 3)2 + 3.

Hence amam+1 = an, where n = m2 + m + 3. Therefore, amam+1 is a term of thesequence.

1The standard formula for the sum a + (a + d) + (a + 2d) + · · · + (a + kd) is (2a+kd)(k+1)2 . In our case

a = 3, d = 2 and k = m− 2.

18

1

1

30

Solution 3 (Michael Robertson, year 11, Dickson College, ACT)

Starting from am and am+1, let us compute the next few terms of the sequence interms of am and am+1.

am+2 = 2am+1 − am + 2

= 2(am+1 + 1)− am

am+3 = 2am+2 − am+1 + 2

= 2((2am+1 + 1)− am)− am+1 + 2

= 3(am+1 + 2)− 2am

Similarly, am+4 = 4(am+1 + 3)− 3am.

We prove the following general formula by induction.

am+k = k(am+1 + k − 1)− (k − 1)am, for k = 0, 1, 2, . . .. (1)

By inspection, formula (1) is true for the base cases k = 0 and k = 1.

For the inductive step, assume that formula (1) is true for k = r − 1 and k = r.Then for k = r + 1, we have

am+r+1 = 2am+r − am+r−1 + 2

= 2(r(am+1 + r − 1)− (r − 1)am)

− ((r − 1)(am+1 + r − 2)− (r − 2)am) + 2

= (r + 1)(am+1 + r)− ram.

Hence the formula is also true for k = r + 1. This completes the induction.

Let us substitute k = am into formula (1). We have

am+am = am(am+1 + am − 1)− (am − 1)am

= amam+1.

Hence amam+1 = ak, where k = m + am. Therefore, amam+1 is a term of thesequence.

Comment This proof shows that the conclusion of the problem remains true forany starting values a1 and a2 of the sequence, provided that m+ am ≥ 0 for all m.This is certainly true whenever a2 ≥ a1 ≥ 0.

19

31

2. Solution 1 (Yang Song, year 12, James Ruse Agricultural High School, NSW)

Answer: 1000

For each positive integer n, let f(n) = n − s(n). We seek the number of differentvalues that f(n) takes in the range from 1 up to 9999.

Lemma The function f has the following two properties.

(i) f(n+ 1) = f(n) if the last digit of n is not a 9.

(ii) f(n+ 1) > f(n) if the last digit of n is a 9.

Proof If the last digit of n is not a 9, then s(n+1) = s(n)+ 1. From this it easilyfollows that f(n+ 1) = f(n).

If the last digit of n is a 9, then suppose that the first k (k ≥ 1) digits from theright-hand end of n are 9s, but the (k + 1)th digit from the right-hand end of nis not a 9. In going from n to n + 1, the rightmost k digits all change from 9to 0, and the (k + 1)th digit from the right-hand end of n increases by 1. Hences(n+ 1) = s(n)− 9k + 1, and so

f(n+ 1) = n+ 1− s(n+ 1)

= n+ 1− (s(n)− 9k + 1)

= n− s(n) + 9k

= f(n) + 9k

> f(n).

From the lemma it follows that

f(1) = f(2) = · · · = f(9) < f(10) = f(11) = · · · = f(19)

< f(20) = f(21) = · · · = f(29)

...

< f(10000) = f(10001) = · · · = f(10009)

< f(10010).

Since, f(9) = 0, f(10) = 9, f(10000) = 9999 and f(10010) = 10008, the requirednumber of different values of f in the range from 1 up to 9999 is simply equalto the number of multiples of 10 from 10 up to 10000. The number of these is10000÷ 10 = 1000.

20

32

Solution 2 (Seyoon Ragavan, year 11, Knox Grammar School, NSW)

Suppose that the decimal representation of n is akak−1 . . . a1a0 where ak = 0. Thenwe have the following.

n = 10kak + 10k−1ak−1 + · · ·+ 101a1 + a0

s(n) = ak + ak−1 + · · ·+ a1 + a0

⇒ n− s(n) = (10k − 1)ak + (10k−1 − 1)ak−1 + · · ·+ (101 − 1)a1

If k ≥ 4, then n − s(n) ≥ (10k − 1)ak ≥ 9999. Note that 9999 is nifty, because ifn = 10000 then n − s(n) = 9999. So it only remains to deal with positive integersthat are less than 9999.

If k ≤ 3, then n − s(n) = 999a3 + 99a2 + 9a1. It follows that the nifty numbersless than 999 are precisely those numbers of the form 999a3 + 99a2 + 9a1, wherea1, a2, a3 ∈ 0, 1, 2, . . . , 9.

Lemma If 999a+99b+9c = 999d+99e+9f where a, b, c, d, e, f ∈ 0, 1, 2, . . . , 9,then a = d, b = e and c = f .

Proof If 999a+ 99b+ 9c = 999d+ 99e+ 9f , then this can be rearranged as

111(a− d) + 11(b− e) + (c− f) = 0. (1)

If a > d, then since b− e ≥ −9 and c−f ≥ −9, we have LHS(1) ≥ 111−99−9 > 0.A similar contradiction is reached if a < d. Hence a = d and equation (1) becomes

11(b− e) + (c− f) = 0. (2)

If b > e, then since c−f ≥ −9, we have LHS(2) ≥ 11−9 > 0. A similar contradictionis reached if b < e. Hence b = e, from which c = f immediately follows.

Returning to the problem, the number of combinations of a1, a2, a3 is 103 = 1000.The lemma guarantees that each of these combinations leads to a different niftynumber. However, we exclude the combination a1 = a2 = a3 = 0 because although0 is nifty, it is not positive.

In summary, there are 999 nifty numbers from the case k ≤ 3, and one nifty numberfrom the case k ≥ 4. This yields a total of 1000 nifty numbers.

21

33

3. Solution 1 (Alexander Gunning, year 12, Glen Waverley Secondary College, VIC)

Answer: 45

Call a subset T of S friendless if no two numbers in T are friends. We visualise afriendless subset T as follows. Draw a 9× 9 square grid. If the number 10a+ b is inT we shade in the square that lies in the ath row and bth column.

First we exhibit a friendless subset T of size 45.

Let T consist of all two-digit numbers whose smaller digit is odd, as depicted in thefirst diagram below. No two numbers in T are friends because their smaller digitsare both odd and hence cannot differ by 1. Since there are 45 shaded squares, wehave shown that |T | = 45 is possible.

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

Finally, we show that all friendless subsets T of S have at most 45 numbers.

To see this we pair up some of the squares in the 9× 9 square grid as shown in thesecond diagram above. Observe that the numbers in each pair are friends. Hence Tcan have at most one number from each pair. Since there are 9 unpaired numbersin the top row and 36 pairs, this shows that T has at most 45 numbers.

22

34

Solution 2 (Anand Bharadwaj, year 9, Trinity Grammar School, VIC)

Consider the following nine lines of numbers.

11

21, 22, 12

31, 32, 33, 23, 13

41, 42, 43, 44, 34, 24, 14

51, 52, 53, 54, 55, 45, 35, 25, 15

61, 62, 63, 64, 65, 66, 56, 46, 36, 26, 16

71, 72, 73, 74, 75, 76, 77, 67, 57, 47, 37, 27, 17

81, 82, 83, 84, 85, 86, 87, 88, 78, 68, 58, 48, 38, 28, 28

91, 92, 93, 94, 95, 96, 97, 98, 99, 89, 79, 69, 59, 49, 39, 29, 19

Each pair of neighbouring numbers on any given line are friends. So a subset T of Sthat contains no friends cannot include consecutive numbers on any of these lines.

On the ith line there are exactly 2i− 1 integers. The maximum number of integerswe can choose from the ith line without choosing neighbours is i. This shows thatT contains at most 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 integers.

Furthermore, |T | = 45 only if we choose exactly i numbers from the ith line withoutchoosing neighbours. There is only one way to do this, namely take every secondnumber starting from the left of each line.

It is still necessary to verify that T contains no friends because even some non-adjacent numbers from the same line, such as 31 and 23, can be friends. To do this,note that the smaller digit of each number in T is odd. Thus for any pair of integersin T , the difference between their smaller digits is even, and thus cannot be equalto 1. Hence T contains no friends.

Comment This proof shows that there is exactly one subset of S of maximal sizethat contains no friends.

23

35

4. Comment All solutions that were dependent on how the diagram was drawn re-ceived a penalty deduction of 1 point. The easiest way to avoid diagram dependencewas to use directed angles as in the three solutions we present here.

For any two lines m and n, the directed angle between them is denoted by ∠(m,n).This is the angle by which one may rotate m anticlockwise to obtain a line parallelto n.2

Solution 1 (Yang Song, year 12, James Ruse Agricultural High School, NSW)

Let Q be the intersection of lines OC and PD. Since AC ‖ PQ and A is themidpoint of OP , it follows that C is the midpoint of OQ.

B C

A

O

P

D

Q

(a) We know OD ‖ BA and DQ ‖ AC. It follows that ∠(OD,DQ) = ∠(BA,AC),which is fixed because A lies on Γ. This implies D lies on a fixed circle, Ω say,through O and Q.

Let s be the radius of Ω. Since the diameter is the largest chord length in acircle, we have 2s ≥ OQ. Since OQ = 2OC = 2r we have s ≥ r, as desired.

(b) From the preceding analysis, we have s = r if and only if OQ is a diameter ofΩ. This is achieved if and only if OD ⊥ DQ, which is equivalent to BA ⊥ AC.But the chords BA and AC of Γ are perpendicular if and only if BC is adiameter of Γ.

2For more details on how to work with directed angles, see the section Directed angles in chapter 17 ofProblem Solving Tactics published by the AMT.

24

2

2

36

Solution 2 (Ilia Kucherov, year 11, Westall Secondary College, VIC)

Let E be the intersection of lines AC and OD. Since AC ‖ PD and A is themidpoint of OP , it follows that E is the midpoint of OD.

B C

A

O

P

D

Q

E

(a) As in solution 1, we use directed angles.

Since OE ‖ BA, we have ∠(OE,EC) = ∠(BA,AC), which is fixed becauseA lies on Γ. This implies that E lies on a fixed circle, ω say, through O andC. Since E is the midpoint of OD, the point D is the image of E under thedilation centred at O and of enlargement factor 2. The image of ω under thedilation is a circle Ω that is twice as large as ω. Thus D lies on Ω.

Let s be the radius of Ω. Then ω has radius s2. But OC is a chord of ω,

hence its length r is less than or equal to the diameter s of ω. Hence s ≥ r, asrequired.

(b) From the preceding analysis we have s = r if and only if OC is a diameter ofω. This is achieved if and only if OE ⊥ EC, which is equivalent to BA ⊥ AC.But the chords BA and AC of Γ are perpendicular if and only if BC is adiameter of Γ.

25

37

Solution 3 (Kevin Xian, year 11, James Ruse Agricultural High School, NSW)

The motivation for this solution comes from considering a couple of special positionsfor A.

If A is diametrically opposite C, then D = O. If A is diametrically opposite B, thenthis gives a second position for D. With this in mind, let X be the point on Γ thatis diametrically opposite B, and let Y be the point such that X is the midpoint ofOY .

B C

A

O

P

D

X

Y

(a) Using directed angles, we have

∠(OY,OD) = ∠(BO,BA) (OD ‖ BA)

= ∠(AB,AO) (OB = OA)

= ∠(OD,OP ). (BA ‖ OD)

We also have OY = 2OX = 2OA = OP . Therefore, Y and P are symmetricin the line OD. It follows that

∠(Y D,OD) = ∠(OD,PD)

= ∠(OD,OP ) + ∠(OP, PD)

= ∠(BA,OA) + ∠(OA,AC)

(OD ‖ BA and PD ‖ AC)

= ∠(BA,AC).

Hence ∠(Y D,OD) is fixed because B, C and Γ are fixed.

Since O and Y are also fixed points, it follows that D lies on a fixed circle, Ωsay, that passes through O and Y .

Let s be the radius of Ω. Then by the extended sine rule applied to ODY

26

38

we have

2s =OY

sin∠(Y D,OD)

=2r

sin∠(BA,AC)

≥ 2r.

Therefore s ≥ r, as required.

(b) From the preceding analysis, we have s = r if and only if

sin(BA,AC) = 1.

This is equivalent to BA ⊥ AC, which occurs if and only if BC is the diameterof Γ.

27

39

5. This was the easiest problem of the competition. Of the 106 students who sat theAMO, 77 found a complete solution. There were many different routes to a solution,and we present some of them here.

Solution 1 (Zoe Schwerkolt, year 11, Fintona Girls’ School, VIC)

We are given ACZ is isosceles with AC = AZ. By symmetry, the angle bisectorat A is also the altitude from A, and so AX ⊥ CZ. Thus ∠CY A = 90 = ∠CDA.It follows that ADY C is a cyclic quadrilateral.

BA

C

D

X

Z

Y

xx

Since ADY C is cyclic, we may set ∠Y DC = ∠Y AC = x, as in the diagram. Hence

∠BDY = 90 − x. (1)

But from the exterior angle sum in CAX, we have ∠AXB = 90 + x, and so

∠Y XB = 90 + x. (2)

Adding (1) and (2) yields

∠BDY + ∠Y XB = 180,

and so BXYD is a cyclic quadrilateral.

28

40

Solution 2 (Alan Guo, year 12, Penleigh and Essendon Grammar School, VIC)

We deduce that ADY C is cyclic as in solution 1.

It follows that

∠AYD = ∠ACD (ADY C cyclic)

= 90 − ∠DAC (angle sum CAD)

= 90 − ∠BAC

= ∠CBA. (angle sum ABC)

Hence ∠AYD = ∠XBD, and so BXYD is cyclic.

BA

C

D

X

Z

Y

29

41

Solution 3 (Wen Zhang, year 9, St Joseph’s College, QLD)

We are given ACZ is isosceles with AC = AZ. Hence by symmetry, the anglebisector at A is both the median and the altitude from A. Thus Y is the midpointof CZ.

SinceCDZ is right-angled atD, the point Y is the circumcentre ofCDZ. Hencewe have Y C = Y D = Y Z.

Let ∠CBA = ∠XBD = 2x. We now compute some other angles in the diagram.

∠BAC = 90 − 2x (angle sum ABC)

⇒ ∠ZAY = 45 − x (AY bisects ∠BAC)

⇒ ∠Y ZA = 45 + x (angle sum AZY )

⇒ ∠ZDY = 45 + x (Y D = Y Z)

⇒ ∠DY Z = 90 − 2x (angle sum Y DZ)

⇒ ∠DYX = 180 − 2x (AX ⊥ CZ)

Hence ∠XBD + ∠DYX = 180, and so BXYD is cyclic.

BA

C

D

X

Z

Y

2x

30

42

Solution 4 (Seyoon Ragavan, year 11, Knox Grammar School, NSW)

As in solution 1, we deduce that AY ⊥ CZ.

Let lines AY and CD intersect at H. Since AY ⊥ CZ and CD ⊥ AZ, it follows thatH is the orthocentre of CAZ. Hence ZH ⊥ AC. But since AC ⊥ BC, we haveZH ‖ BC. Observe also that HDZY is cyclic because ∠ZDH = 90 = ∠HY Z.

We now have

∠CXH = ∠ZHX (BC ‖ ZH)

= ∠ZHY

= ∠ZDY. (HDZY cyclic)

Hence ∠CXY = ∠BDY , and so BXYD is cyclic.

BA

C

D

X

Z

Y

H

31

43

Solution 5 (Kevin Xian, year 11, James Ruse Agricultural High School, NSW)

As in solution 1, we deduce that AY ⊥ CZ. Hence using the angle sums in CAYand CAX, we have

∠ACY = 90 − ∠XAC = ∠CXY.

Hence by the alternate segment theorem, line AC is tangent to circle CY X at C.Using the power of point A with respect to circle CY X, we have

AC2 = AY · AX. (1)

In a similar way we may use the angle sums in CAD and ABC to find

∠ACD = 90 − ∠BAC = ∠CBD.

Hence by the alternate segment theorem, line AC is tangent to circle CDB at C.Using the power of point A with respect to circle CDB, we have

AC2 = AD · AB. (2)

Combining (1) and (2), it follows that

AY · AX = AD · AB,

and so BXYD is cyclic.

BA

C

D

X

Z

Y

32

44

6. Solution 1 (Michelle Chen, year 11, Methodist Ladies’ College, VIC)

Answer: 1008

Let f(x) = p(x)− q(x), where

p(x) = (x− 1)(x− 3)(x− 5) · · · (x− 2015),

q(x) = (x− 2)(x− 4)(x− 6) · · · (x− 2014).

We seek the number of distinct real solutions to the equation f(x) = 0. Note thatf(x) is a polynomial of degree 1008 because p(x) has degree 1008 while q(x) hasdegree 1007.

Observe that p(x) is the product of an even number of brackets and q(x) is theproduct of an odd number of brackets. Therefore,

f(0) = (−1)× (−3)× · · · × (−2015)

− (−2)× (−4)× · · · × (−2014)

= 1× 3× · · · × 2015 + 2× 4× · · · × 2014

> 0.

We also have

f(2016) = 2015× 2013× · · · × 1− 2014× 2012× · · · × 2

> 0,

where the last line is true because 2015 > 2014, 2013 > 2012, and so on down to3 > 2.

Next, for x = 2, 4, 6, . . . , 2014, we have q(x) = 0, and so f(x) = p(x). Hence

f(2) = 1× (−1)× (−3)× · · · × (−2013) < 0.

Note that by increasing the value of x from x = 2 to x = 4, we reduce the numberof negative brackets in the product for q(x) by 1. This has the effect of changingthe sign of f(x). In general, each time we increase x by 2 up to a maximum ofx = 2014, we change the sign of f(x). Therefore, we have the following inequalities.

f(0) > 0

f(2) < 0

f(4) > 0

f(6) < 0

...

f(2014) < 0

f(2016) > 0

Since f(x) is a polynomial it is a continuous function. Hence, by the interme-diate value theorem, f(x) = 0 has at least one solution in each of the intervals(0, 2), (2, 4), (4, 6), . . . , (2014, 2016). Therefore, there are at least 1008 different so-lutions to f(x) = 0.

Finally, since f(x) is a polynomial of degree 1008, the equation f(x) = 0 has atmost 1008 roots. Hence we may conclude that f(x) = 0 has exactly 1008 distinctreal solutions.

33

45

Solution 2 (Yong See Foo, year 11, Nossal High School, VIC)

The polynomials p(x), q(x) and f(x) are defined as in solution 1. We also deducef(2016) > 0, as in solution 1.

For any x ∈ 1, 3, 5, . . . , 2015, we have p(x) = 0, and so f(x) = q(x). Therefore,for any x ∈ 1, 3, 5, . . . , 2015, we have

f(x) = −(x− 2)(x− 4) · · · (x− (x− 1)) ∗ (x− (x+ 1)) · · · (x− 2014).

(We use the ∗ symbol merely as a placeholder for future reference.)

All bracketed terms to the left of ∗ are positive while the remaining bracketed termsto the right of ∗ are negative. Hence the total number of bracketed terms that arenegative is equal to 2014−(x+1)

2+ 1 = 2015−x

2.3 Taking into account the minus sign at

the front, it follows that for each x ∈ 1, 3, 5, . . . , 2015, we have

f(x) > 0 for x ≡ 1 (mod 4) and f(x) < 0 for x ≡ 3 (mod 4).

Hence we have the following inequalities.

f(1) > 0

f(3) < 0

f(5) > 0

...

f(2013) > 0

f(2015) < 0

f(2016) > 0

Since f(x) is a polynomial, it is a continuous function. Hence by the interme-diate value theorem, f(x) = 0 has at least one solution in each of the intervals(1, 3), (3, 5), . . . , (2013, 2015), (2015, 2016). Therefore, there are at least 1008 differ-ent solutions to the equation f(x) = 0.

However, as in solution 1, f(x) is a polynomial of degree 1008, and so f(x) = 0 hasat most 1008 roots. Thus we may conclude that f(x) = 0 has exactly 1008 distinctreal solutions.

Comment Some contestants found it helpful to do a rough sketch of p(x) and q(x).This helps to determine the signs of p(x), q(x) and f(x) for x = 0, 1, 2, . . . , 2016.From this one can further narrow down the location of the solutions to f(x) = 0.They are found in the intervals (1, 2), (3, 4), (5, 6), . . . , (2013, 2014), (2015, 2016).

y = p(x)

y = q(x)

2013 2014 2015 2016

y

x. . .1 2 3 4 5 6 7

3This is all still true in the extreme cases x = 1 and x = 2015. The case x = 1 corresponds to putting the∗ to the left of the first term (x−2) but to the right of the negative sign. The case x = 2015 correspondsto putting the ∗ to the right of the last term (x− 2014).

34

3

3

46

7. There are many different ways to solve this problem. Over 10 different methods ofsolution were found among the 24 contestants who completely solved the problem.Some of these are presented here.

Solution 1 (Charles Li, year 9, Camberwell Grammar School, VIC)

We proceed by contradiction. Assume that only finitely many integers n satisfy

(p(n+ 1)− p(n)

)(p(n)− p(n− 1)

)> 0.

Observe that p(n) = p(n + 1) for all positive integers n ≥ 2 because adjacentnumbers cannot be divisible by the same prime. Hence there is a positive integer Nsuch that

(p(n+ 1)− p(n)

)(p(n)− p(n− 1)

)< 0 for all integers n > N . (*)

Thus p(n) > p(n − 1) if and only if p(n + 1) < p(n) for all n > N . It follows thatthe graph of p(n) versus n alternates between peaks and valleys once n > N . Thiscan be schematically visualised as follows.

n

p(n)

Consider the number 2m where m > 1 is large enough so that 2m > N . Note thatp(2m) = 2. Since 2m−1 and 2m+1 are odd we have p(2m+1) > 2 and p(2m−1) > 2.Hence the above graph has a valley at 2m. Since peaks and valleys alternate forevery n > N , it follows that there is a valley at every even number and a peak atevery odd number once we pass N .

Consider the number 3m. Note that p(3m) = 3. However, from the precedingparagraph, there is a peak at 3m because it is odd and greater than N . Hencep(3m − 1) = 2 and p(3m + 1) = 2. This is possible if and only if 3m − 1 and 3m + 1are powers of 2. But the only powers of 2 that differ by 2 are 21 and 22. This impliesm = 1, which is a contradiction.

Comment 1 Some contestants found a second way to deduce that the graph ofp(n) versus n (for n > N) has peaks at odd n and valleys at even n.

They observed that if q > 3 is a prime number greater than N , then there is a peakat q. This is because p(q) = q, while p(q + 1) ≤ q+1

2< q and p(q − 1) ≤ q−1

2< q.

Comment 2 All solutions used the method of indirect proof.4 They all establishedthat the graph of p(n) versus n (for n > N) has peaks at odd values of n and valleys

4Also known as ‘proof by contradiction’.

35

4

4

47

at even values of n. We present some of the variations that contestants used toderive a contradiction from this point on.

Variation 1 (Isabel Longbottom, year 10, Rossmoyne Senior High School, WA)

Consider the number 32m+1, where m ≥ 1 satisfies 32m+1 > N .

Since 32m+1 is odd and p(32m+1) = 3, it follows that 32m+1 − 1 is a power of 2.

Since m > 1, the number 32m+1 − 1 is a power of 2 which is greater than 2. Hence32m+1 ≡ 1 (mod 4).

But 32m+1 ≡ (−1)2m+1 ≡ −1 ≡ 1 (mod 4), which is a contradiction.

Variation 2 (Richard Gong, year 10, Sydney Grammar School, NSW)

Consider the number 34m, where m ≥ 1 is large enough so that 34m > N .

Since 34m is odd and p(34m) = 3, it follows that 34m − 1 is a power of 2.

But 34m = 81m ≡ 1m (mod 5).

So 5 | 34m − 1, which is a contradiction.

Variation 3 (Anthony Pisani, year 8, St Paul’s Anglican Grammar School, VIC)

By Dirichlet’s theorem there are infinitely many primes of the form q = 6k + 1.Choose any such prime q > N .

There is a valley at 2q because it is even. Hence p(2q + 1) > p(2q) = q.

But 2q + 1 = 2(6k + 1) + 1 = 3(4k + 1), and so 2q + 1 is a multiple of 3.

Hence p(2q + 1) ≤ 2q+13

< q, which is a contradiction.

Variation 4 (Jack Liu, year 9, Brighton Grammar School, VIC)

Consider the number 2q, where q > 3 is prime greater than N . Note that p(2q) = q.

There is a valley at 2q because it is even.

However, 2q − 1, 2q and 2q + 1 are three consecutive numbers, and so one of themis a multiple of 3. Since q > 3 we know 3 2q.If 3 | 2q− 1, then p(2q− 1) ≤ 2q−1

3< q, and if 3 | 2q+1, then p(2q+1) ≤ 2q+1

3< q.

Either way, this contradicts that there is a valley at 2q.

Variation 5 (Kevin Shen, year 125, Saint Kentigern College, NZ)

Let q > N be any prime. There is a valley at 2q because it is even.

So we have p(2q − 1) > p(2q) = q. This implies that p(2q − 1) = 2q − 1.

Thus 2q−1 is prime whenever q is prime. Therefore, we also have 2(2q−1)−1 = 4q−3is prime.

A simple induction shows that 2r(q − 1) + 1 is prime for r = 0, 1, 2, . . ..

In particular 2q−1(q − 1) + 1 is prime.

By Fermat’s little theorem, 2q−1(q − 1) + 1 ≡ 1(q − 1) + 1 ≡ 0 (mod q).

Hence q | 2q−1(q − 1) + 1. Since 2q−1(q − 1) + 1 > q, we have contradicted that2q−1(q − 1) + 1 is prime.

5Equivalent to year 11 in Australia.

36

5

5

48

Variation 6 (Thomas Baker, year 11, Scotch College, VIC)

Let q1 < q2 < · · · be the list of all odd primes in order starting from q1 = 3.

By Dirichlet’s theorem there are infinitely many primes of the form 4k + 3. So wemay choose j > 3 such that qj > N and such that the list q1, q2, . . . , qj contains anodd number of primes of the form of 4k + 3.

Consider the number m = q1q2 · · · qj. Note that p(m) = qj. Furthermore, there is apeak at m because it is odd.

Note m ≡ 3 (mod 4). Thus m− 1 = 2r for some odd number r.

Furthermore, r is not divisible by any qi (1 ≤ i ≤ j). Hence p(m − 1) is a primethat is greater than pj. Thus p(m − 1) > p(m), which contradicts that there is apeak at m.

37

49

Solution 2 (Jeremy Yip, year 12, Trinity Grammar School, VIC)

We shall prove that p(n− 1) < p(n) < p(n+1) for infinitely many n. In particular,for each odd prime q, we show there exists a positive integer k such that

p(q2k − 1) < p(q2

k

) < p(q2k

+ 1). (1)

Lemma 1 For any odd prime q, we have gcd(q2a+ 1, q2

b+ 1) = 2 for all integers

1 ≤ a < b.

Proof Let d = gcd(q2a+1, q2

b+1). Using the difference of perfect squares factori-

sation we know x− 1 | x2 − 1 and x+1 | x2 − 1. From this we deduce the followingchain of divisibility.

q2a

+ 1 | q2a+1 − 1

q2a+1 − 1 | q2a+2 − 1

q2a+2 − 1 | q2a+3 − 1

...

q2b−1 − 1 | q2b − 1

It follows that q2a+ 1 | q2b − 1, and so d | q2b − 1. Combining this with d | q2b + 1

yields d | 2. Since q2a+ 1 and q2

b+ 1 are even, we conclude that d = 2.

Lemma 2 If (1) is false for all positive integers k, then

p(q2k − 1) < q and p(q2

k

+ 1) < q for k = 1, 2, . . . .

Proof We proceed by induction under the assumption that (1) is false.

For the base case k = 1, since q is odd we have q2 − 1 = 2(q − 1)(q+12

). Hence

p(q2 − 1) < q. However, since (1) is false for k = 2, we also have p(q2 + 1) < q.

For the inductive step, suppose we know that

p(q2j − 1) < q and p(q2

j

+ 1) < q,

for some positive integer j. Then since q2j+1 − 1 = (q2

j − 1)(q2j+ 1), we have

p(q2j+1−1) < q. However, since (1) is false for k = j+1, we also have p(q2

j+1+1) < q.

This completes the induction.

To complete the proof of the problem, let us focus on the following list of numbers.

L = p(q2 + 1), p(q4 + 1), p(q8 + 1), . . .

From lemma 1, q2 + 1, q4 + 1, q8 + 1, . . . all have pairwise greatest common divisorequal to 2. Hence at most one of the numbers in L is equal to 2, while the othersare distinct odd primes. This means that L contains infinitely many different primenumbers. This is a contradiction because lemma 2 implies that L contains onlyfinitely many different numbers.

38

50

Comment Jeremy’s proof is rather impressive because it establishes a muchstronger result than what was required. The AMO problem only required a proofthat at least one of p(n − 1) < p(n) < p(n + 1) and p(n − 1) > p(n) > p(n + 1)occurs infinitely often, but without pinning down which of the two alternatives doesoccur infinitely often. Jeremy’s proof establishes that p(n − 1) < p(n) < p(n + 1)definitely occurs infinitely often.6

6Erdos and Pomerance proved this result in their 1978 research paper On the largest prime factors of nand n+ 1. The question of whether p(n− 1) > p(n) > p(n+ 1) could occur infinitely often was finallyresolved in the positive by Balog in his 2001 research paper On triplets with descending largest primefactors.

39

6

6

51

8. This was the most difficult problem of the 2015 AMO. Just four contestants managedto solve it completely.

Solution (Seyoon Ragavan, year 11, Knox Grammar School, NSW)

Answer: n⌊n−12

⌋

We shall refer to a triangle ABC as being isosceles with base BC if AB = AC.Note that if ABC is equilateral then it is isosceles with base AB, isosceles withbase BC, and isosceles with base AC. In this way an equilateral triangle counts asthree isosceles triangles depending on which side is considered the base. Thus thenumber of apples Maryam receives is equal to the number of base-specified isoscelestriangles.

Let S denote the set of lines that Maryam draws. Consider a line a ∈ S. Wedetermine how many ways a can be the base of an isosceles triangle. Observe thatif a, b, c and a, d, e both form isosceles triangles with base a, then since no two linesare parallel, the pairs b, c and d, e are either equal or disjoint. Thus there are atmost

⌊n−12

⌋pairs b, c from S such that a, b, c forms an isosceles triangle with base

a. Since there are n possibilities for a, we conclude that the number of base-specifiedisosceles triangle is at most n

⌊n−12

⌋.

It remains to show that this is attainable. Take n equally spaced lines passingthrough the origin so that the angle between consecutive lines is 180

n. Next translate

each of the lines so that no three of them are concurrent. We claim this configurationof lines results in n

⌊n−12

⌋base-specified isosceles triangles.

For any line x, let x′ denote the line after it has been translated. Consider anytriangle bounded by lines a′, b′, c′. It is isosceles with base a′ if and only if before thetranslations, the lines b and c were symmetric in a. There are n ways of choosinga. Once a is chosen, then by our construction there are

⌊n−12

⌋pairs b, c such that

they are symmetric in a. Indeed, if n is odd, all the other lines are involved ina symmetric pair about a. If n is even then exactly one line is not involved in asymmetric pair about a because it is perpendicular to a. Thus the total number ofbase specified isosceles triangles for this construction is n

⌊n−12

⌋, as desired.

The following diagram illustrates the construction for n = 6. Each line is a base fortwo isosceles triangles, one of which is equilateral. In total there are two equilateraltriangles and six isosceles triangles, making twelve base-specified isosceles trianglesin all.

−→

40

52

Comment Other constructions that achieve the bound n⌊n−12

⌋are possible. The

following one is by Andrew Elvey Price, Deputy Leader of the 2015 Australian IMOteam.

If n is odd, Maryam can achieve this by drawing the lines to form the sides of aregular polygon with n sides. If n is even, Maryam can achieve this by drawing thelines to form all but one of the sides of a regular polygon with n+ 1 sides.

41

53

AUSTRALIAN MATHEMATICAL OLYMPIAD STATISTICS

Score Distribution/Problem

Number of Students/Score

Problem Number

1 2 3 4 5 6 7 8

0 10 14 29 80 17 60 47 86

1 7 2 1 8 0 1 30 8

2 1 7 18 0 6 3 1 1

3 2 10 4 0 5 4 4 4

4 13 10 11 1 0 5 0 0

5 6 9 9 0 0 1 0 3

6 17 26 11 8 1 7 0 0

7 50 28 23 9 77 25 24 4

Average Mark 5.2 4.6 3.4 1.2 5.4 2.5 2.0 0.6

54

AUSTRALIAN MATHEMATICAL OLYMPIAD RESULTS

Name School Year

Perfect Score and Gold



Gold






Alan Guo Penleigh and Essendon Grammar School VIC 12


Richard Gong Sydney Grammar School NSW 10


Silver



Kevin Shen Saint Kentigern College NZ 12 NZ

George Han Westlake Boys High School NZ 13 NZ

Wilson Zhao Killara High School NSW 11

Michelle Chen Methodist Ladies' College VIC 11




Anthony Pisani St Paul's Anglican Grammar School VIC 8

Ivan Zelich Anglican Church Grammar School QLD 12

Michael Robertson Dickson College ACT 11

Edward Chen Palmerston North Boys High School NZ 11 NZ


Martin Luk King's College NZ 13 NZ

55

Name School Year

Bronze

Jerry Mao Caulfield Grammar School VIC 9

Zoe Schwerkolt Fintona Girls' School VIC 11

Harish Suresh James Ruse Agricultural High School NSW 11

Justin Wu James Ruse Agricultural High School NSW 11

Austin Zhang Sydney Grammar School NSW 10


Anand Bharadwaj Trinity Grammar School VIC 9

Devin He Christ Church Grammar School WA 11

Miles Yee-Cheng Lee Auckland International College NZ 12 NZ

Bobby Dey James Ruse Agricultural High School NSW 10


Nitin Niranjan All Saints Anglican School QLD 12

William Song Scotch College VIC 11

Wen Zhang St Joseph's College, Gregory Terrace QLD 9

Prince Michael Balanay Botany Downs Secondary College NZ 13 NZ

Michael Chen Scotch College VIC 12

William Wang King's College NZ 12 NZ

Xuzhi Zhang Auckland Grammar School NZ 13 NZ

Keiran Lewellen Home Schooled NZ 10 NZ

Eryuan Sheng Newington College NSW 11

Linus Cooper James Ruse Agricultural High School NSW 9

David Steketee Hale School WA 12

Alexander Barber Scotch College VIC 11

Matthew Jones All Saints Anglican School QLD 12

Madeline Nurcombe Cannon Hill Anglican College QLD 11

Steven Lim Hurlstone Agricultural High School NSW 9

56

27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD

XXVII Asian Pacific Mathematics Olympiad

March, 2015

Time allowed: 4 hoursNo calculators are to be usedEach problem is worth 7 points

Problem 1. Let ABC be a triangle, and let D be a point on side BC. A linethrough D intersects side AB at X and ray AC at Y . The circumcircle of triangleBXD intersects the circumcircle ω of triangle ABC again at point Z = B. The linesZD and ZY intersect ω again at V and W , respectively. Prove that AB = VW .

Problem 2. Let S = 2, 3, 4, . . . denote the set of integers that are greater than orequal to 2. Does there exist a function f : S → S such that

f(a)f(b) = f(a2b2) for all a, b ∈ S with a = b?

Problem 3. A sequence of real numbers a0, a1, . . . is said to be good if the followingthree conditions hold.

(i) The value of a0 is a positive integer.

(ii) For each non-negative integer i we have ai+1 = 2ai + 1 or ai+1 =ai

ai + 2.

(iii) There exists a positive integer k such that ak = 2014.

Find the smallest positive integer n such that there exists a good sequence a0, a1, . . .of real numbers with the property that an = 2014.

Problem 4. Let n be a positive integer. Consider 2n distinct lines on the plane, notwo of which are parallel. Of the 2n lines, n are colored blue, the other n are coloredred. Let B be the set of all points on the plane that lie on at least one blue line,and R the set of all points on the plane that lie on at least one red line. Prove thatthere exists a circle that intersects B in exactly 2n − 1 points, and also intersects Rin exactly 2n− 1 points.

Problem 5. Determine all sequences a0, a1, a2, . . . of positive integers with a0 ≥ 2015such that for all integers n ≥ 1:

(i) an+2 is divisible by an;

(ii) |sn+1 − (n+ 1)an| = 1, where sn+1 = an+1 − an + an−1 − · · ·+ (−1)n+1a0.

57

27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD SOLUTIONS

1. Solution (Alan Guo, year 12, Penleigh and Essendon Grammar School, VIC)

Applying the Pivot theorem to AXY we see that circles ABC, XBD and CDYare concurrent at point Z.1 Hence CDZY is cyclic.

A

B C

X

Y

D

ZW

V

Let ∠(m,n) denote the directed angle between any two lines m and n.2 We have

∠(WZ, V Z) = ∠(Y Z,DZ)

= ∠(Y C,DC) (CDZY cyclic)

= ∠(AC,BC).

Therefore VW and AB subtend equal or supplementary angles in ω. It follows thatVW = AB.

Comment All solutions that were dependent on how the diagram was drawn re-ceived a penalty deduction of 1 point. The easiest way to avoid diagram dependencewas to use directed angles as in the solution presented above.

1The point Z is also the Miquel point of the four lines AB, AY , BC and XY . It is the common point ofthe circumcircles of ABC, AXY , BDX and CDY . See the sections entitled Pivot theorem andFour lines and four circles found in chapter 5 of Problem Solving Tactics published by the AMT.

2This is the angle by which one may rotate m anticlockwise to obtain a line parallel to n. For moredetails, see the section Directed angles in chapter 17 of Problem Solving Tactics published by the AMT.

43

1

2

1

2

58

2. Solution 1 (Henry Yoo, year 12, Perth Modern School, WA)

Answer: No

Assume such a function exists. For any positive integer n we have

f(2n)f(2n+4) = f(22(2n+4)

)= f(2n+1)f(2n+3),

and sof(2n)

f(2n+1)=

f(2n+3)

f(2n+4). (1)

Similarly,f(2n+1)f(2n+4) = f

(22(2n+5)

)= f(2n+2)f(2n+3),

and sof(2n+1)

f(2n+2)=

f(2n+3)

f(2n+4). (2)

Combining (1) and (2), we find that

f(2n)

f(2n+1)=

f(2n+1)

f(2n+2). (3)

Since (3) is true for all positive integers n, it follows that

f(21), f(22), f(23), . . .

is a geometric sequence. Thus f(2n) = arn−1 for some positive rational numbers aand r.

Since f(2)f(22) = f(26) we have

a · ar = ar5

⇒ a = r4. (4)

And since f(2)f(23) = f(28) we have

a · ar2 = ar7

⇒ a = r5. (5)

Comparing (4) and (5) we have r4 = r5. Since r > 0, we have r = 1. But then from(4) we have a = 1. Thus f(2) = 1 ∈ S, a contradiction. Hence there are no suchfunctions.

44

59

Solution 2 (Linus Cooper, year 9, James Ruse Agricultural High School, NSW)

Suppose such a function exists. Then for any integer b > 2 we have

f(8b)f(b) = f(64b4) = f(2)f(4b2) = f(2)f(2)f(b),

and sof(8b) = f(2)2.

Thus f(c) = f(2)2 whenever c ≥ 24 and c is a multiple of 8.

Next we have

f(24)f(32) = f(242 · 322)⇒ f(2)4 = f(2)2.

Since f(2) > 0, we have f(2) = 1 ∈ S. This contradiction shows that no suchfunction exists.

45

60

Solution 3 (Jeremy Yip, year 12, Trinity Grammar School, VIC)

Suppose such a function exists. Let a < b < c be positive integers. We work withthe quantity f(2a)f(2b)f(2c) in two different ways.

(f(2a)f(2c)

)f(2b) = f(22a+2c)f(2b) (since c > a)

= f(24a+2b+4c) (since 2a+ 2c > b) (1)(f(2b)f(2c)

)f(2a) = f(22b+2c)f(2a) (since c > b)

= f(22a+4b+4c) (since 2b+ 2c > a) (2)(f(2a)f(2b)

)f(2c) = f(22a+2b)f(2c) (since b > a)

= f(24a+4b+2c) (if c = 2a+ 2b) (3)

From (1), (2) and (3) it follows that

f(42a+2b+c) = f(42a+b+2c) = f(4a+2b+2c), (4)

for any three positive integers a < b < c satisfying c = 2a+ 2b.

In what follows we use (4) repeatedly.

(a, b, c) = (1, 2, 3) ⇒ f(49) = f(410) = f(411)

(a, b, c) = (1, 2, 4) ⇒ f(410) = f(412) = f(413)

(a, b, c) = (1, 3, 5) ⇒ f(413) = f(415) = f(417)

(a, b, c) = (1, 4, 7) ⇒ f(417) = f(420) = f(423)

Thus f(49) = f(420).

However, f(49)f(4) = f(420), and so f(4) = 1 ∈ S. This contradiction shows thatno such function exists.

46

61

Solution 4 (Angelo Di Pasquale, Director of Training, AMOC)

For any a, b ∈ S, choose an integer c > a, b. Then since bc > a and c > b, we have

f(a4b4c4) = f(a2)f(b2c2) = f(a2)f(b)f(c).

Furthermore, since ac > b and c > a, we have

f(a4b4c4) = f(b2)f(a2c2) = f(b2)f(a)f(c).

Comparing these two equations, we find that for all a, b ∈ S,

f(a2)f(b) = f(b2)f(a) ⇒ f(a2)

f(a)=

f(b2)

f(b).

It follows that there exists a positive rational number k such that

f(a2) = kf(a), for all a ∈ S. (1)

Substituting this into the functional equation yields

f(ab) =f(a)f(b)

k, for all a, b ∈ S with a = b. (2)

Now combine the functional equation with (1) and (2) to obtain for all a ∈ S,

f(a)f(a2) = f(a6) =f(a)f(a5)

k=

f(a)f(a)f(a4)

k2=

f(a)f(a)f(a2)

k.

It follows that f(a) = k for all a ∈ S.

Putting this into the functional equation yields k2 = k. But 0, 1 ∈ S and hencethere is no solution.

47

62

3. Solution 1 (Alexander Gunning, year 12, Glen Waverley Secondary College, VIC)

Answer: n = 60

Clearly all members of the sequence are positive rational numbers. For each non-negative integer i let ai =

piqiwhere pi and qi are positive integers with gcd(pi, qi) = 1.

Therefore,pi+1

qi+1

=2pi + qi

qior

pi+1

qi+1

=pi

pi + 2qi. (1)

If each of pi and qi are odd, then so are 2pi + qi, qi, pi, and pi + 2qi. Thus when theRHSs of (1) are reduced to lowest terms, the numerators and denominators are stillodd. Hence pi+1 and qi+1 are odd. It follows inductively that if pi and qi are odd,then pk and qk are odd for all k ≥ i. Since pn

qn= 2014 we cannot have pi and qi both

being odd for any i ≤ n. Since gcd(pi, qi) = 1, it follows that

pi and qi are of opposite parity for i = 0, 1, . . . , n. (2)

Suppose pi is odd for some i < n. We cannot have the second option in (1) becausethat implies pi+1 and qi+1 are both odd, which contradicts (2). So we must havethe first option in (1), namely, pi+1

qi+1= 2pi+qi

qi. From (2), qi is even, and so we have

gcd(2pi + qi, qi) = gcd(2pi, qi) = 2. Hence

pi odd ⇒ pi+1 = pi +qi2

and qi+1 =qi2

for i < n. (3)

On the other hand, a similar argument shows that if pi is even for some i < n, thenwe must take the second option in (1), namely pi+1

qi+1= pi

pi+2qi, and gcd(pi, pi+2qi) = 1.

Hence

pi even ⇒ pi+1 =pi2

and qi+1 =pi2+ qi for i < n. (4)

In both (3) and (4), we have pi+1 + qi+1 = pi + qi. Since pn + qn = 2015, we have

pi + qi = 2015 for i ≤ 2015. (5)

If pi is odd, we may combine (3) and (5) to find

(pi+1, qi+1) =(pi +

qi2,qi2

)

≡(pi2,qi2

)(mod 2015). (6)

A very similar calculation using (4) and (5) shows that (6) is also true if pi is even.

A simple induction on (6) yields,

(pn, qn) ≡(p02n

,q02n

)(mod 2015). (7)

We require qn = 1. However, from (5), this is true if and only if

qn ≡ 1 (mod 2015).

48

63

Note that q0 = 1 because a0 is a positive integer. Thus from (7) we require

2n ≡ 1 (mod 2015).

Since 2015 = 3 · 13 · 31, we require 5, 13, 31 | 2n − 1. Computing powers of 2, we seethat

5 | 2n − 1 ⇔ 4 | n13 | 2n − 1 ⇔ 12 | n31 | 2n − 1 ⇔ 5 | n.

Since 60 = lcm(4, 12, 5) we conclude that 2015 | 2n − 1 if and only if 60 | n. Thesmallest such positive integer is n = 60.

49

64

Solution 2 (Henry Yoo, year 12, Perth Modern School, WA)

Clearly all members of the sequence are positive rational numbers. For each positive

integer i, we have ai =ai+1 − 1

2or ai =

2ai+1

1− ai+1

. Since ai > 0 we deduce that

ai =

ai+1 − 1

2if ai+1 > 1

2ai+1

1− ai+1

if ai+1 < 1.

(1)

Thus ai is uniquely determined from ai+1. Hence we may start from an = 2014 andsimply run the sequence backwards until we reach a positive integer.

From (1), if ai+1 =uv, then

ai =

u− v

2vif u > v

2u

v − uif u < v.

Consequently, if we define the sequence of pairs of integers (u0, v0), (u1, v1), . . . by(u0, v0) = (2014, 1) and

(ui+1, vi+1) =

(ui − vi, 2vi) if ui > vi

(2ui, vi − ui) if ui < vi,(2)

then an−i =ui

vifor i = 0, 1, 2, . . ..

Observe from (2) that ui+1+ vi+1 = ui+ vi. So since u0 = 2014 and v0 = 1, we have

ui + vi = 2015 for i = 0, 1, 2, . . .. (3)

Suppose that d is a common factor of ui+1 and vi+1. Then (3) implies d | 2015, andso d is odd. If ui > vi, then from (2) we have d | ui − vi and d | 2vi. This easilyimplies d | ui and d | vi. If ui < vi, we similarly conclude from (2) that d | ui andd | vi. This inductively cascades back to give d | u0 and d | v0. Since gcd(u0, v0) = 1,we deduce that d | 1. Therefore,

gcd(ui, vi) = 1 for i = 0, 1, 2, . . .. (4)

From (3) and (4), we have ui = vi. Hence (2) yields ui+1, vi+1 > 0, and so from (3)we have

ui, vi ∈ 1, 2, . . . , 2014 for i = 0, 1, 2, . . .. (5)

Next we prove by induction that

(ui, vi) ≡ (−2i, 2i) (mod 2015) for i = 0, 1, 2, . . .. (6)

The base case is immediate. Also, if (ui, vi) ≡ (−2i, 2i) (mod 2015), then using (2)we have

ui > vi ⇒ (ui+1, vi+1) ≡ (−2i − 2i, 2 · 2i) (mod 2015)

≡ (−2i+1, 2i+1) (mod 2015)

50

65

and

ui < vi ⇒ (ui+1, vi+1) ≡ (2(−2i), 2i − (−2i)) (mod 2015)

≡ (−2i+1, 2i+1) (mod 2015).

Either way, this completes the inductive step.

We are given that a0 =un

vnis a positive integer. We know that un and vn are positive

integers with gcd(un, vn) = 1. Thus we seek n such that vn = 1. From (5) and (6),this occurs if and only if

2n ≡ 1 (mod 2015).

As in solution (1), the smallest such positive integer is n = 60.

Comment 1 In this solution, if we were to run the sequence backwards froman = 2014 until a positive integer is reached, the result would be20141, 2013

2, 2011

4, 2007

8, 1999

16, 1983

32, 1951

64, 1887

128, 1759

256, 1503

512, 9911024

, 198233

, 194966

, 1883132

, 1751264

, 1487528

, 9591056

,191897

, 1821194

, 1627388

, 1239776

, 4631552

, 9261089

, 1852163

, 1689326

, 1363652

, 7111304

, 1422593

, 8291186

, 1658357

, 1301714

, 5871428

, 1174841

, 3331682

,6661349

, 1332683

, 6491366

, 1298717

, 5811434

, 1162853

, 3091706

, 6181397

, 1236779

, 4571558

, 9141101

, 1828187

, 1641374

, 1267748

, 5191496

, 1038977

, 611954

,1221893

, 2441771

, 4881527

, 9761039

, 195263

, 1889126

, 1763252

, 1511504

, 10071008

, 20141.

Since there are 61 terms in the above list, this also shows that n = 60.

Comment 2 A corollary of both solutions is that the only value of a0 which yieldsa good sequence is a0 = 2014.

51

66

4. Solution 1 (Jeremy Yip, year 12, Trinity Grammar School, VIC)

Let θ be the maximal angle that occurs between a red line and a blue line. Let rbe a red line and b be a blue line such that the non-acute angle between them is θ.Note that r and b divide the plane into four regions.

In the following diagrams, the two regions that lie within the angular areas definedby θ are shaded. Also we orient our configuration so that the x-axis is the anglebisector of r and b that passes through the two shaded regions. Let O be theintersection of lines r and b.

Let be another line in the configuration. Four options arise that need to beconsidered.

θ

θ1

θ2

α

β

θα

β

In the first two diagrams, does not pass through O. Observe that , r and b enclosea triangle which is either completely shaded or completely unshaded. If the triangleis completely shaded, as in the first diagram, then θ1 = θ+α > θ and θ2 = θ+β > θ.But this is impossible because it implies that cannot be blue or red. Hence thetriangle is completely unshaded, as in the second diagram.

In the last two diagrams, passes through O. Observe that apart from the point O,either all of lies in the unshaded regions, or all of lies in the shaded regions. If were to lie in the unshaded regions, as in the third diagram, then since θ + α > θand θ + β > θ, it would follow that could not be red or blue. Hence lies in theshaded regions, as in the fourth diagram.

Let S be the set of intersection points of lines in the configuration that lie on r or b.Consider any circle Γ that lies to the right of all points of S and is tangent to r andb in the right-hand shaded region. We claim that Γ has the required property. Itsuffices to show that every line in the configuration, apart from r and b, intersectsΓ in two distinct points.

O

B

R

Let R and B be the points of tangency of Γ with r and b, respectively, and let T bethe union of the segments OB and OR.

52

67

If ( = r, b) is any line of the configuration, then the part of lying in the rightshaded region is an infinite ray ′.

Let F be the figure enclosed by T and the minor arc RB of Γ. Then ′ passes intothe interior of F , and so intersects the boundary of F at least twice. Since ′ cannotintersect T twice, it must intersect the minor arc RB.

Finally, ′ cannot be tangent to Γ because that would imply that ′ intersects bothOB and OR (not at O). It follows that ′ intersects Γ at two distinct points, asdesired.

53

68

Solution 2 (Yang Song, year 12, James Ruse Agricultural High School, NSW)

We may rotate the plane so that no red line or blue line is vertical. Let 1, 2, . . . , 2nbe the lines listed in order of increasing gradient. Then there is a k such that linesk and k+1 are oppositely coloured. By rotating our coordinate system and cycliclyrelabelling our lines we can ensure that 1, 2, . . . , 2n are listed in order of increasinggradient, 1 and 2n are oppositely coloured, and no line is vertical. Without loss ofgenerality 1 is red and 2n is blue. Let O = 1 ∩ 2n.

Let (not one of the 2n lines) be a variable vertical line to the right of O. LetR = 1 ∩ and B = 2n ∩ . Since the lines 2, 3, . . . , 2n−1 have gradients lyingin between those of 1 and 2n, we can move far enough to the right so that allthe intersection points of 2, 3, . . . , 2n−1 with lie between R and B. Let Γ be theexcircle of ORB opposite A. We claim that Γ has the required properties. Toprove this, it suffices to show that each line i (2 ≤ i ≤ 2n− 1) intersects Γ twice.

O

B′

R′

R

B

Let ′ be the vertical line which is tangent to Γ and lying to the right of Γ. LetR′ = ′ ∩ 1 and B′ = ′ ∩ 2n. Then Γ is the incircle of BRR′B′. The result nowfollows because any line that intersects the opposite sides of a quadrilateral havingan incircle, must also intersect the incircle twice.

54

69

5. Solution (APMO Problem Selection Committee)

There are two families of answers.

• an = c(n+ 2)n! for all n ≥ 1 and a0 = c+ 1 for some integer c ≥ 2014.

• an = c(n+ 2)n! for all n ≥ 1 and a0 = c− 1 for some integer c ≥ 2016.

Let a0, a1, a2, . . . be a sequence of positive integers satisfying the given conditions.We can rewrite (ii) as

sn+1 = (n+ 1)an + hn,

where hn ∈ −1, 1 . Substituting n with n− 1 yields

sn = (n− 1)an + hn−1,

for n ≥ 2. Adding together the two equations we find

an+1 = (n+ 1)an + nan−1 + δn (1)

for n ≥ 2 and where δn ∈ −2, 0, 2.We also have |s2 − 2a1| = 1, which yields a0 = 3a1 − a2 ± 1 ≤ 3a1, and thereforea1 ≥ a0

3≥ 671. Substituting n = 2 in (1), we find that a3 = 3a2 + 2a1 + δ2. Since

a1 | a3, we have a1 | 3a2 + δ2, and therefore a2 ≥ 223. Using (1), we obtain thatan ≥ 223 for all n ≥ 0.

Lemma 1 For n ≥ 4, we have an+2 = (n+ 1)(n+ 4)an.

Proof For n ≥ 3 we have

an = nan−1 + (n− 1)an−2 + δn−1

> nan−1 + 3. (2)

By applying (1) and (2) with n substituted by n− 1, we have for n ≥ 4,

an = nan−1 + (n− 1)an−2 + δn−1

< nan−1 + (an−1 − 3) + δn−1

< (n+ 1)an−1. (3)

Using (1) to write an+2 in terms of an and an−1 along with (2), we have for n ≥ 3,

an+2 = (n+ 3)(n+ 1)an + (n+ 2)nan−1 + (n+ 2)δn + δn+1

< (n+ 3)(n+ 1)an + (n+ 2)nan−1 + 3(n+ 2)

< (n2 + 5n+ 5)an.

Also for n ≥ 4,

an+2 = (n+ 3)(n+ 1)an + (n+ 2)nan−1 + (n+ 2)δn + δn+1

> (n+ 3)(n+ 1)an + nan

= (n2 + 5n+ 3)an.

Since an | an+2, we have an+2 = (n2 + 5n+ 4)an = (n+ 1)(n+ 4)an, as desired.

Lemma 2 For n ≥ 4, we have an+1 =(n+1)(n+3)

n+2an.

55

70

Proof Using the recurrence an+3 = (n + 3)an+2 + (n + 2)an+1 + δn+2 and writingan+3, an+2 in terms of an+1, an according to lemma 1 we obtain

(n+ 2)(n+ 4)an+1 = (n+ 3)(n+ 1)(n+ 4)an + δn+2.

Hence n+ 4 | δn+2, which yields δn+2 = 0 and an+1 =(n+1)(n+3)

n+2an, as desired.

Lemma 3 For n ≥ 1, we have an+1 =(n+1)(n+3)

n+2an.

Proof Suppose there exists n ≥ 1 such that an+1 = (n+1)(n+3)n+2

an. By lemma 2, there

is a greatest integer 1 ≤ m ≤ 3 with this property. Then am+2 =(m+2)(m+4)

m+3am+1.

If δm+1 = 0, then am+1 = (m+1)(m+3)m+2

am, which contradicts our choice of m. Thusδm+1 = 0.

Clearly m + 3 | am+1. So we have am+1 = (m + 3)k and am+2 = (m + 2)(m + 4)k,for some positive integer k. Then

(m+ 1)am + δm+1 = am+2 − (m+ 2)am+1 = (m+ 2)k.

So, am | (m+2)k− δm+1. But am also divides am+2 = (m+2)(m+4)k. Combiningthe two divisibility conditions, we obtain am | (m+ 4)δm+1.

Since δm+1 = 0, we have am | 2m + 8 ≤ 14, which contradicts the previous resultthat an ≥ 223 for all non-negative integers n.

So, an+1 =(n+1)(n+3)

n+2an for n ≥ 1. Substituting n = 1 yields 3 | a1. Letting a1 = 3c,

we have by induction that an = n!(n+ 2)c for n ≥ 1. Since |s2 − 2a1| = 1, we thenget a0 = c± 1, yielding the two families of solutions.

Finally, since (n+2)n! = n!+(n+1)!, it follows that sn+1 = c(n+2)!+(−1)n(c−a0).Hence both families of solutions satisfy the given conditions.

56

71

Top 10 Australian scores

Name School Year Total Award

Jeremy Yip Trinity Grammar School VIC 12 28 Gold

Alexander Gunning Glen Waverley Secondary College VIC 12 28 Silver

Yang Song James Ruse Agricultural High School NSW 12 23 Silver

Henry Yoo Perth Modern School WA 12 20 Bronze

Thomas Baker Scotch College VIC 11 20 Bronze

Allen Lu Sydney Grammar School NSW 12 20 Bronze

Ilia Kucherov Westall Secondary College VIC 11 19 Bronze

Seyoon Ragavan Knox Grammar School NSW 11 17

Yong See Foo Nossal High School VIC 11 16

Alan Guo Penleigh and Essendon Grammar School VIC 12 14

Country scores

Rank Country Number of Contestants Score Gold Silver Bronze Hon.Men

1 USA 10 298 1 2 4 3

2 Korea 10 279 1 2 4 3

3 Russia 10 266 1 2 4 3

4 Singapore 10 259 1 2 4 3

5 Japan 10 256 1 2 4 3

6 Canada 10 237 1 2 4 3

7 Thailand 10 228 1 2 4 3

8 Taiwan 10 222 1 2 4 3

9 Australia 10 205 1 2 4 3

10 Brazil 10 202 1 2 4 3

11 Peru 10 185 0 3 4 3

12 Mexico 10 169 1 1 5 3

13 Hong Kong 10 167 0 3 4 3

14 Kazakhstan 10 163 0 2 5 3

15 Indonesia 10 161 0 3 4 3

16 Malaysia 10 134 0 3 3 3

17 India 10 127 0 1 5 3

27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD RESULTS

72

Rank Country Number of Contestants Score Gold Silver Bronze Hon.Men

17 Tajikistan 10 127 0 0 6 4

19 Bangladesh 10 122 0 0 7 1

20 Philippines 10 105 0 3 0 2

21 Turkmenistan 10 99 0 0 4 3

22 Saudi Arabia 10 94 0 0 4 1

23 New Zealand 10 86 0 2 1 1

24 Argentina 10 73 0 0 1 3

24 Colombia 10 73 0 1 3 1

26 Syria 6 52 0 0 1 5

27 Sri Lanka 7 48 0 0 1 4

28 El Salvador 7 47 0 0 2 2

29 Trinidad and Tobago 10 31 0 0 1 0

30 Ecuador 10 27 0 0 1 1

31 Costa Rica 5 20 0 0 0 1

32 Panama 2 12 0 0 0 0

33 Cambodia 2 9 0 0 0 1

Total 299 4583 11 42 102 81

73

AMOC SELECTION SCHOOL2015 IMO Team Selection School

The 2015 IMO Selection School was held 5–14 April at Robert Menzies College, MacquarieUniversity, Sydney. The main qualifying exams are the AMO and the APMO from which25 students are selected for the school.

The routine is similar to that for the December School of Excellence; however, there isthe added interest of the actual selection of the Australian IMO team. This year the IMOwould be held in Chiang Mai, Thailand.

The students are divided into a junior group and a senior group. This year there were 10juniors and 15 seniors. It is from the seniors that the team of six for the IMO plus onereserve team member is selected. The AMO, the APMO and the final three senior examsat the school are the official selection criteria.

My thanks go to Andrew Elvey Price, Ivan Guo, Victor Khou, and Konrad Pilch, whoassisted me as live-in staff members. Also to Peter Brown, Vaishnavi Calisa, Mike Clapper,Nancy Fu, Declan Gorey, David Hunt, Vickie Lee, Peter McNamara, John Papantoniou,Christopher Ryba, Andy Tran, Gareth White, Rachel Wong, Sampson Wong, JonathanZheng, and Damon Zhong, all of whom came in to give lectures or help with the markingof exams.

Angelo Di PasqualeDirector of Training, AMOC

2015 Australian IMO team

Name Year School State

Alexander Gunning 12 Glen Waverley Secondary College VIC

Ilia Kucherov 11 Westall Secondary College VIC

Seyoon Ragavan 11 Knox Grammar School NSW

Yang Song 12 James Ruse Agricultural High School NSW

Kevin Xian 11 James Ruse Agricultural High School NSW

Jeremy Yip 12 Trinity Grammar School VIC

Reserve

Yong See Foo 11 Nossal High School VIC

3

74

2015 Australian IMO Team

Name School Year







Reserve


2015 Australian IMO Team, from left, Jeremy Yip, Alexander Gunning, Yang Song, Kevin Xian, Ilia Kucherov and Seyoon Ragavan.

75

Participants at the 2015 IMO Selection School

Name School Year

Seniors



Michelle Chen Methodist Ladies' College VIC 11



Alan Guo Penleigh and Essendon Grammar School VIC 12








Wilson Zhao Killara High School NSW 11

Juniors

Bobby Dey James Ruse Agricultural High School NSW 10

Rachel Hauenschild Kenmore State High School QLD 10






Hilton Nguyen Sydney Technical High School NSW 9

Tommy Wei Scotch College VIC 9

Wen Zhang St Joseph's College, Gregory Terrace QLD 9

76

IMO TEAM PREPARATION SCHOOLIMO Team Preparation School

The pre-IMO July school is always a great reality check when it comes to our perceptionof the team’s ability. This is of course because we train with the UK team. Our jointtraining school was held 2–8 July at Nexus International School, Singapore.

The routine for the teams each day consisted of an IMO trial exam in the morning, freetime in the afternoon while their papers were being assessed, a short debrief of the examlate in the afternoon followed by going out to dinner each evening. The results of trainingshowed that both teams were quite strong, with the UK having the edge.

The final exam also doubles as the annual Mathematics Ashes contest. Australia wonthe Ashes in its inaugural year, lost them the next year, and have not been able to winthem back since. There have been a few close calls, and even a tie in 2011. In anotherheart-breaking nail biter, the UK again retained the Ashes after both teams tied on 84points apiece.

Angelo Di PasqualeIMO Team Leader

1

77

THE MATHEMATICS ASHESThe 2015 Mathematical Ashes: AUS v UK

Exam

1. Does there exist a 2015× 2015 array of distinct positive integers such that the sumsof the entries on each row and on each column yield 4030 distinct perfect squares?

2. Let Ω and O be the circumcircle and the circumcentre of an acute-angled triangleABC with AB > BC. The angle bisector of ∠ABC intersects Ω at M = B. Let Γbe the circle with diameter BM . The angle bisectors of ∠AOB and ∠BOC intersectΓ at points P and Q, respectively. The point R is chosen on the line PQ so thatBR = MR.

Prove that BR ‖ AC.

(In this problem, we assume that an angle bisector is a ray.)

3. We are given an infinite deck of cards, each with a real number on it. For everyreal number x, there is exactly one card in the deck that has x written on it. Nowtwo players draw disjoint sets A and B of 100 cards each from this deck. We wouldlike to define a rule that declares one of them a winner. This rule should satisfy thefollowing conditions:

1. The winner only depends on the relative order of the 200 cards: if the cards arelaid down in increasing order face down and we are told which card belongs towhich player, but not what numbers are written on them, we can still decidethe winner.

2. If we write the elements of both sets in increasing order as A = a1, a2, . . . , a100and B = b1, b2, . . . , b100, and ai > bi for all i, then A beats B.

3. If three players draw three disjoint sets A,B,C from the deck, A beats B andB beats C, then A also beats C.

How many ways are there to define such a rule?

(In this problem, we consider two rules as different if there exist two sets A and Bsuch that A beats B according to one rule, but B beats A according to the other.)

Results

Q1 Q2 Q3 Σ

AUS 1 7 7 7 21

AUS 2 7 0 7 14

AUS 3 7 7 0 14

AUS 4 7 5 0 12

AUS 5 3 6 1 10

AUS 6 6 7 0 13

Σ 37 32 15 84

Q1 Q2 Q3 Σ

UNK 1 7 7 5 19

UNK 2 7 2 0 9

UNK 3 5 0 0 5

UNK 4 3 7 7 17

UNK 5 7 6 7 20

UNK 6 7 7 0 14

Σ 36 29 19 84

2

78

THE MATHEMATICS ASHES RESULTS

The 8th Mathematics Ashes competition at the joint pre-IMO training camp in Putrajaya, was tied; the results for the two teams were as follows, with each team scoring a total of 84:

Code Name Q1 Q2 Q3 Total

AUS1 Alexander Gunning 7 7 7 21

AUS2 Ilia Kucherov 7 0 7 14

AUS3 Seyoon Ragavan 7 7 0 14

AUS4 Yang Song 7 5 0 12

AUS5 Kevin Xian 3 6 1 10

AUS6 Jeremy Yip 6 7 0 13

TOTAL 37 32 15 84

United Kingdom

Code Name Q1 Q2 Q3 Total

UNK1 Joe Benton 7 7 5 19

UNK2 Lawrence Hollom 7 2 0 9

UNK3 Samuel Kittle 5 0 0 5

UNK4 Warren Li 3 7 7 17

UNK5 Neel Nanda 7 6 7 20

UNK6 Harvey Yau 7 7 0 14

TOTAL 36 29 19 84

Australia

79

IMO TEAM LEADER’S REPORTThe 56th International Mathematical Olympiad, Chiang Mai,

Thailand

The 56th International Mathematical Olympiad (IMO) was held 4–16 July in Chiang Mai,Thailand.

This was the largest IMO in history with a record number of 577 high school studentsfrom 104 countries participating. Of these, 52 were girls.

Each participating country may send a team of up to six students, a Team Leader and aDeputy Team Leader. At the IMO the Team Leaders, as an international collective, formwhat is called the Jury. This Jury was chaired by Soontorn Oraintara.

The first major task facing the Jury is to set the two competition papers. During thisperiod the Leaders and their observers are trusted to keep all information about the contestproblems completely confidential. The local Problem Selection Committee had alreadyshortlisted 29 problems from 155 problem proposals submitted by 53 of the participatingcountries from around the world. During the Jury meetings one of the shortlisted problemshad to be discarded from consideration due to being too similar to material already inthe public domain. Eventually, the Jury finalised the exam questions and then madetranslations into the more than 50 languages required by the contestants. Unfortunately,due to an accidental security breach, the second day’s paper had to be changed on thenight before that exam was scheduled. This probably resulted in a harder than intendedsecond day.

The six questions that ultimately appeared on the IMO contest are described as follows.

1. A relatively easy two-part problem in combinatorial geometry proposed by theNetherlands. It concerns finite sets of points in the plane in which the perpen-dicular bisector of any pair of points in such a set also contains another point of theset.

2. A medium classical number theory problem proposed by Serbia.

3. A difficult classical geometry problem in which one is asked to prove that a certaintwo circles are mutually tangent. It was proposed by Ukraine.

4. A relatively easy classical geometry problem proposed by Greece.

5. A medium to difficult functional equation proposed by Albania.

6. A difficult problem in which one is asked to prove an inequality about a sequenceof integers. Although it does not seem so at first sight, the problem is much morecombinatorial than algebraic. It was inspired by a notation used to describe juggling.The problem was proposed by Australia.

These six questions were posed in two exam papers held on Friday 10 July and Saturday11 July. Each paper had three problems. The contestants worked individually. They wereallowed four and a half hours per paper to write their attempted proofs. Each problemwas scored out of a maximum of seven points.

For many years now there has been an opening ceremony prior to the first day of compe-tition. HRH Crown Princess Sirindhorn presided over the opening ceremony. Followingthe formal speeches there was the parade of the teams and the 2015 IMO was declaredopen.

5

80

After the exams the Leaders and their Deputies spent about two days assessing the workof the students from their own countries, guided by marking schemes, which had beendiscussed earlier. A local team of markers called Coordinators also assessed the papers.They too were guided by the marking schemes but are allowed some flexibility if, forexample, a Leader brought something to their attention in a contestant’s exam scriptthat is not covered by the marking scheme. The Team Leader and Coordinators have toagree on scores for each student of the Leader’s country in order to finalise scores. Anydisagreements that cannot be resolved in this way are ultimately referred to the Jury.

The IMO paper turned out to be quite difficult. While the easier problems 1 and 4were quite accessible, the other four problems 2, 3, 5 and 6 were found to be the mostdifficult combination of medium and difficult problems ever seen at the IMO. There wereonly around 30 complete solutions to each of problems 2, 3 and 5. Problem 6 was verydifficult, averaging just 0.4 points. Only 11 students scored full marks on it.

The medal cuts were set at 26 for gold, 19 for silver and 14 for bronze.1 Consequently,there were 282 (=48.9%) medals awarded. The medal distributions2 were 39 (=6.8%)gold, 100 (=17.3%) silver and 143 (=24.8%) bronze. These awards were presented at theclosing ceremony. Of those who did not get a medal, a further 126 contestants receivedan honourable mention for solving at least one question perfectly.

Alex Song of Canada was the sole contestant who achieved the most excellent feat ofa perfect score of 42. He now leads the IMO hall of fame, being the most decoratedcontestant in IMO history. He is the only person to have won five IMO gold medals.3 Hewas given a standing ovation during the presentation of medals at the closing ceremony.

Congratulations to the Australian IMO team on an absolutely spectacular performancethis year. They smashed our record rank4 to come 6th, and they also smashed our recordmedal haul, bringing home two Gold and four Silver medals.5 This is the first time thateach team member has achieved Silver or better. The team finished ahead of many ofthe traditionally stronger teams. In particular, they finished ahead of Russia, whom wewould have considered as untouchable.

Congratulations to Gold medallist Alexander Gunning, year 12, Glen Waverley SecondaryCollege. He is now the most decorated Australian at the IMO, being the only Australianto have won three Gold medals at the IMO. On each of these occasions he also finishedin the top 10 in individual rankings.6 He is now equal 17th on the IMOs all-time hall offame.

Congratulations also to Gold medallist Seyoon Ragavan, year 11, Knox Grammar School.Seyoon solved four problems perfectly and was comfortably above the Gold medal cut.He was individually ranked 19th.

1This was the lowest ever cut for gold, and the equal lowest ever cut for silver. (This was indicative ofthe difficulty of the exam, not the standard of the contestants.)

2The total number of medals must be approved by the Jury and should not normally exceed half thetotal number of contestants. The numbers of gold, silver and bronze medals must be approximately inthe ratio 1:2:3.

3In his six appearances at the IMO, Alex Song won a bronze medal in 2010, and followed up with goldmedals in 2011, 2012, 2013, 2014 and 2015.

4The ranking of countries is not officially part of the IMO general regulations. However, countries areranked each year on the IMO’s official website according to the sum of the individual student scoresfrom each country.

5Australia’s best performance prior to this was the dream team of 1997. They came 9th, with a medaltally of two Gold, three Silver and one Bronze.

6In his four appearances at the IMO, Alexander won a bronze medal in 2012, and followed up with goldmedals in 2013 (8th), 2014 (1st) and 2015 (4th).

6

12

3

45

6

1

2

3

4

5

6

81

And congratulations to our four Silver medallists: Ilia Kucherov, year 11, Westall Sec-ondary College; Yang Song, year 12, James Ruse Agricultural High School; Kevin Xian,year 11, James Ruse Agricultural High School; and Jeremy Yip, year 12, Trinity GrammarSchool.

Three members of this year’s team are eligible for selection to the 2016 IMO team. Sowhile it is unlikely we will be able to repeat this year’s stellar performance, the outlookseems promising.

Congratulations also to Ross Atkins and Ivan Guo, who were IMO medallists with theAustralian team when they were students.7 They were the authors of the juggling-inspiredIMO problem number six. In fact Ross is a proficient juggler.

The 2015 IMO was organised by: The Institute for the Promotion of Teaching Scienceand Technology, Chiang Mai University, The Mathematical Association of Thailand underthe Patronage of His Majesty the King, and The Promotion of Academic Olympiad andDevelopment of Science Education Foundation.

The 2016 IMO is scheduled to be held July 6-16 in Hong Kong. Venues for future IMOshave been secured up to 2019 as follows.

2017 Brazil

2018 Romania

2019 United Kingdom

Much of the statistical information found in this report can also be found at the officialwebsite of the IMO.

www.imo-official.org

Angelo Di PasqualeIMO Team Leader, Australia

7Ross and Ivan won Bronze at the 2003 IMO, and Ivan won Gold at the 2004 IMO.

7

7

7

82

Ross Atkins demonstrates his juggling skills. (Photo credit: Gillian Bolsover)

84

83

INTERNATIONAL MATHEMATICAL OLYMPIAD

Language: English

Day: 1

S

A B S C S AC = BC S

A B C S P S PA = PB = PC

n 3 n

n 3 n

(a, b, c)

ab− c, bc− a, ca− b

2

2 2n n

ABC AB > AC Γ H

F A M BC Q

Γ ∠HQA = 90 K Γ ∠HKQ = 90

A B C K Q Γ

KQH FKM

IMO Papers

Day: 1

Friday, July 10, 2015

Problem 1. We say that a finite set S of points in the plane is balanced if, for any twodifferent points A and B in S, there is a point C in S such that AC = BC. We say thatS is centre-free if for any three different points A, B and C in S, there is no point P inS such that PA = PB = PC.

(a) Show that for all integers n ≥ 3, there exists a balanced set consisting of n points.

(b) Determine all integers n ≥ 3 for which there exists a balanced centre-free set con-sisting of n points.

Problem 2. Determine all triples (a, b, c) of positive integers such that each of thenumbers


is a power of 2.

(A power of 2 is an integer of the form 2n, where n is a non-negative integer.)

Problem 3. Let ABC be an acute triangle with AB > AC. Let Γ be its circumcircle,H its orthocentre, and F the foot of the altitude from A. Let M be the midpoint of BC.Let Q be the point on Γ such that ∠HQA = 90, and let K be the point on Γ such that∠HKQ = 90. Assume that the points A, B, C, K and Q are all different, and lie on Γin this order.

Prove that the circumcircles of triangles KQH and FKM are tangent to each other.

Language: English Time: 4 hours and 30 minutesEach problem is worth 7 points

8

IMO Papers

Day: 1

Friday, July 10, 2015

Problem 1. We say that a finite set S of points in the plane is balanced if, for any twodifferent points A and B in S, there is a point C in S such that AC = BC. We say thatS is centre-free if for any three different points A, B and C in S, there is no point P inS such that PA = PB = PC.

(a) Show that for all integers n ≥ 3, there exists a balanced set consisting of n points.

(b) Determine all integers n ≥ 3 for which there exists a balanced centre-free set con-sisting of n points.

Problem 2. Determine all triples (a, b, c) of positive integers such that each of thenumbers


is a power of 2.

(A power of 2 is an integer of the form 2n, where n is a non-negative integer.)

Problem 3. Let ABC be an acute triangle with AB > AC. Let Γ be its circumcircle,H its orthocentre, and F the foot of the altitude from A. Let M be the midpoint of BC.Let Q be the point on Γ such that ∠HQA = 90, and let K be the point on Γ such that∠HKQ = 90. Assume that the points A, B, C, K and Q are all different, and lie on Γin this order.

Prove that the circumcircles of triangles KQH and FKM are tangent to each other.


8

84

Language: English

Day: 2

ABC Ω O Γ A

BC D E B D E C

BC F G Γ Ω A F

B C G Ω K

BDF AB L

CGE CA

FK GL X X

AO

R f : R → R

f(x+ f(x+ y)

)+ f(xy) = x+ f(x+ y) + yf(x)

x y

a1, a2, . . .

1 aj 2015 j 1

k + ak = ℓ+ aℓ 1 k < ℓ

b N

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ 10072

m n n > m N

Day: 2

Saturday, July 11, 2015

Problem 4. Triangle ABC has circumcircle Ω and circumcentre O. A circle Γ withcentre A intersects the segment BC at points D and E, such that B, D, E and C are alldifferent and lie on line BC in this order. Let F and G be the points of intersection ofΓ and Ω, such that A, F , B, C and G lie on Ω in this order. Let K be the second pointof intersection of the circumcircle of triangle BDF and the segment AB. Let L be thesecond point of intersection of the circumcircle of triangle CGE and the segment CA.

Suppose that the lines FK and GL are different and intersect at the point X. Prove thatX lies on the line AO.

Problem 5. Let R denote the set of real numbers. Determine all functions f : R → Rsatisfying the equation

f(x+ f(x+ y)) + f(xy) = x+ f(x+ y) + yf(x)

for all real numbers x and y.

Problem 6. The sequence a1, a2, . . . of integers satisfies the following conditions:

(i) 1 ≤ aj ≤ 2015 for all j ≥ 1;

(ii) k + ak = + a for all 1 ≤ k < .

Prove that there exist two positive integers b and N such that

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ ≤ 10072

for all integers m and n satisfying n > m ≥ N .


9

Day: 2

Saturday, July 11, 2015

Problem 4. Triangle ABC has circumcircle Ω and circumcentre O. A circle Γ withcentre A intersects the segment BC at points D and E, such that B, D, E and C are alldifferent and lie on line BC in this order. Let F and G be the points of intersection ofΓ and Ω, such that A, F , B, C and G lie on Ω in this order. Let K be the second pointof intersection of the circumcircle of triangle BDF and the segment AB. Let L be thesecond point of intersection of the circumcircle of triangle CGE and the segment CA.

Suppose that the lines FK and GL are different and intersect at the point X. Prove thatX lies on the line AO.

Problem 5. Let R denote the set of real numbers. Determine all functions f : R → Rsatisfying the equation

f(x+ f(x+ y)) + f(xy) = x+ f(x+ y) + yf(x)

for all real numbers x and y.

Problem 6. The sequence a1, a2, . . . of integers satisfies the following conditions:

(i) 1 ≤ aj ≤ 2015 for all j ≥ 1;

(ii) k + ak = + a for all 1 ≤ k < .

Prove that there exist two positive integers b and N such that

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ ≤ 10072

for all integers m and n satisfying n > m ≥ N .


9

85

INTERNATIONAL MATHEMATICAL OLYMPIAD SOLUTIONS

1. Solution (All members of the 2015 Australian IMO team solved this problem usingsimilar methods. Here we present the solution by Yang Song, year 12, James RuseAgricultural High School, NSW. Yang was Silver medallist with the 2015 AustralianIMO team.)

(a) For n odd, we may take S to be the set of vertices of a regular n-gon P .

It seems obvious that S is balanced but we shall prove it anyway. Let A andB be any two vertices of P . Since n is odd, one side of the line AB contains anodd number of vertices of P . Thus if we enumerate the vertices of P in orderfrom A around to B on that side of AB, one of them will be the middle one,and hence be equidistant from A and B.

For n even, say n = 2k we may take S to be the set of vertices of a collectionof k unit equilateral triangles, all of which have a common vertex, O say, andexactly one pair of them has a second common vertex. Note that apart fromO, all of the vertices lie on the unit circle centred at O.

The reason why this works is as follows. Let A and B be any two points in S.If they are both on the circumference of the circle, then OA = OB and O ∈ S.If one of them is not on the circumference, say B = O, then by constructionthere is a third point C ∈ S such that ABC is equilateral, and so AC = BC.

The cases for n = 11 and n = 12 are illustrated below.

O

(b) We claim that a balanced centre-free set of n points exists if and only if n isodd.

Note that the construction used in the solution to part (a) is centre-free. Weshall show that there is no balanced centre-free set of n points if n is even.

For any three points A,X, Y ∈ S, let us write A → X, Y to mean AX = AY .We shall estimate the number of instances of A → X, Y in two different ways.

First, note that if A → X, Y and A → X,Z where Y = Z, then S cannotbe centre-free because AX = AY = AZ. Hence for a given point A, there areat most

⌊n−12

⌋pairs X, Y such that A → X, Y . Since there are n choices

for A, the total number of instances of A → X, Y is at most n⌊n−12

⌋.

On the other hand, since S is balanced, for each pair of points X, Y ∈ S, thereis at least one point A such that A → X, Y . Since the number of pairsX, Y is

(n2

), the total number of instances of A → X, Y is at least

(n2

).

If we combine our estimates, we obtain n⌊n−12

⌋≥

(n2

), which simplifies to⌊

n−12

⌋≥ n−1

2. This final inequality is impossible if n is even.

58

86

Comment 1 A nice graph theoretical interpretation of the solution to part (b) wasgiven by Kevin Xian, year 11, James Ruse Agricultural High School, NSW. Kevinwas a Silver medallist with the 2015 Australian IMO team.

Let S ′ denote the set of unordered pairs of elements of S. Form a directed bipartitegraph G as follows. The vertex set of G is S ∪ S ′. The directed edges of G aresimply all the instances of A → X, Y as per the solution to part (b) above. Thenthe total outdegree of G is at most n

⌊n−12

⌋, while the total indegree of G is at least(

n2

). The inequality found in the solution to part (b) is simply a consequence of the

total indegree being equal to the total outdegree.

Comment 2 Alternative constructions for n odd in part (a) were found by YangSong, year 12, James Ruse Agricultural High School, NSW, and Jeremy Yip, year12, Trinity Grammar School, VIC. Yang and Jeremy were Silver medallists with the2015 Australian IMO team.

For n = 2k + 1, let S be the set of vertices of a collection of k unit equilateraltriangles, all of which have exactly a common vertex, O say, and no two of whichhave any other common vertices besides O. Note that apart from O, all of thevertices lie on the unit circle centred at O. The case for n = 13 is illustrated below.

O

Comment 3 At the time of writing, the classification of balanced sets is an openproblem. From the foregoing, we see that for n odd there are two infinite families ofbalanced sets, and for n even there is one infinite family. It is unknown if there areany other infinite families which do not fit into this scheme. Note however, thereare sporadic constructions which do not fit into either of these families. Some ofthese are shown below. (All drawn segments are of unit length.)

59

87

2. Solution 1 (Jeremy Yip, year 12, Trinity Grammar School, VIC. Jeremy was aSilver medallist with the 2015 Australian IMO team.)

The answers are (a, b, c) = (2, 2, 2), (2, 2, 3), (2, 6, 11), (3, 5, 7) and their permuta-tions. It is straightforward to verify that they all work.

Note that ab− c = bc− a ⇔ (b+ 1)(a− c) = 0 ⇔ a = c.

We divide our solution into five cases as follows.

Case 1 At least two of a, b, c are equal.

Case 2 All of a, b, c are different and none of them are even.

Case 3 All of a, b, c are different and one of them is even.

Case 4 All of a, b, c are different and two of them are even.

Case 5 All of a, b, c are different and all three of them are even.

Case 1 Without loss of generality a = b. Then b2 − c and bc− b are powers of 2.That is,

b2 − c = 2x (1)

b(c− 1) = 2y (2)

for some non-negative integers x and y. Equation (2) implies that b = 2p andc− 1 = 2q for some non-negative integers p and q. Putting these into (1) we find

22p = 2x + 2q + 1. (3)

Since RHS(3) ≥ 3, we have p ≥ 1, and so 4 | 2x + 2q + 1.

If q = 0, then (3) becomes 22p − 2x = 2. The only powers of two that differ by 2 are2 and 4. Hence x = p = 1, which quickly leads to (a, b, c) = (2, 2, 2).

Similarly, if x = 0, then (3) becomes 22p − 2q = 2. Hence q = p = 1, and so(a, b, c) = (2, 2, 3).

Finally, if q, x ≥ 1, then (3) cannot hold because the LHS is even while the RHS isodd.

Case 2 Without loss of generality we may suppose that a > b > c, where a, b, care all odd.

It follows that ab− c > ac− b and a ≥ 3. Since ab− c and ac− b are powers of two,the smaller divides the larger. Therefore,

ab− c ≡ 0 (mod ac− b)

⇔ a(ac)− c ≡ 0 (mod ac− b)

⇔ (a− 1)(a+ 1) ≡ 0 (mod ac− b). (c is odd)

Since a− 1 and a+ 1 are consecutive even positive integers, exactly one of them isdivisible by 4, while the other is even but not divisible by 4. Since ac− b is a powerof two, it follows that either ac − b | 2(a − 1) or ac − b | 2(a + 1). Either way, wehave

ac− b ≤ 2(a+ 1).

Since a > b, we have ac− a < 2(a+ 1), which can be rearranged as

(c− 3)a < 2. (4)

60

88

Observe that a ≥ 5, b ≥ 3 and c ≥ 1. This is because a, b, c are all odd anda > b > c ≥ 1.

If c > 3 then (4) implies a < 2. This contradicts a ≥ 5.

If c = 1, then a − b and b − a are powers of two, which is impossible because theysum to zero.

If c = 3, then 3b− a and 3a− b are powers of two. Since 3b− a < 3a− b, we have

3a− b ≡ 0 (mod 3b− a)

⇔ 3(3b)− b ≡ 0 (mod 3b− a)

⇔ 8 ≡ 0 (mod 3b− a). (b is odd)

Hence 3b− a ∈ 1, 2, 4, 8.Certainly 3b− a = 1, because a and b are odd.

If 3b−a = 2, then a = 3b−2, and so 3a− b = 8b−6 ≡ 2 (mod 4). Hence 8b−6 = 2,and b = 1. But this impossible because b > c ≥ 1.

If 3b − a = 4, then a = 3b − 4, and so 3a − b = 8b − 12 ≡ 4 (mod 8). Hence8b− 12 = 4, and b = 2, which contradicts that b is odd.

If 3b− a = 8, then a = 3b− 8, and so 3a− b = 8(b− 3). Hence b− 3 is a power oftwo. Thus b = 3 + 2k for some positive integer k. Also

ab− c = b(3b− 8)− 3

= (3b+ 1)(b− 3)

is a power of two. Hence 3b+1 = 3 ·2k+10 is also a power of two. This is impossiblefor k ≥ 2 because 3 · 2k + 10 ≡ 2 (mod 4) and 3 · 2k + 10 = 2. However, for k = 1,we find 3b+1 = 16. Thus b = 5 and a = 3b− 8 = 7, and we have found the solution(a, b, c) = (3, 5, 7).

Case 3 Without loss of generality a is even, while b and c are odd.

Then ab− c and ac− b are both odd. Hence ab− c = ac− b = 1. This implies b = c,which is a contradiction. So this case does not occur.

Case 4 Without loss of generality c is odd, while a and b are even with a > b.

This immediately implies that a ≥ 4. Also, since ab− c is odd, we have ab− c = 1.Thus c = ab− 1 ≡ 3 (mod 4). Hence c ≥ 3.

Let 2k (note k ≥ 1) be the greatest power of two dividing both a and b. Thena = 2km and b = 2kn for some integers m > n ≥ 1.

We have ac− b = 2k(mc−n). Thus mc−n is a power of two. Thus m and n are ofthe same parity because otherwise mc− n would be an odd number that is greaterthan 1. From the choice of k, this implies that m and n are both odd.

Since a > b we have bc − a < ac − b. Because they are both powers of two, thesmaller divides the larger. Therefore,

ac− b ≡ 0 (mod bc− a) (5)

⇔ (bc)c− b ≡ 0 (mod bc− a)

⇔ 2kn(c2 − 1) ≡ 0 (mod 2k(nc−m))

⇔ (c− 1)(c+ 1) ≡ 0 (mod nc−m). (n is odd) (6)

61

89

Using similar reasoning as in case 2, since nc−m is a power of two, and c− 1 andc+ 1 are consecutive even positive integers, we deduce that

nc−m ≤ 2(c+ 1)

⇔ n(4kmn− 1)−m ≤ 2 · 4kmn (c = ab− 1 = 4kmn− 1)

⇔ 4k(n2 − 2n) ≤ n

m< 1. (m > n)

Since n is odd, this implies n = 1. Putting this into (6), we find

(c− 1)(c+ 1) ≡ 0 (mod c−m)

⇔ (m− 1)(m+ 1) ≡ 0 (mod c−m).

Using similar reasoning as in case 2, since c−m is a power of two, and m− 1 andm+ 1 are consecutive even positive integers, we deduce that

c−m ≤ 2(m+ 1)

⇔ 4km−m ≤ 2m+ 2 (c=4km− 1)

⇔ 4km < 3m+ 2

⇒ k = 1.

To summarise, we have found (a, b, c) = (2m, 2, 4m− 1). In particular, c = 2a− 1.Putting this into (5) we find,

a(2a− 1)− 2 ≡ 0 (mod 3a− 2)

⇔ 3a(6a− 3) ≡ 18 (mod 3a− 2)

⇔ 2 ≡ 18 (mod 3a− 2).

Hence 3a − 2 | 16. Since a ≥ 4, we have 3a − 2 = 16. Thus a = 6, and so(a, b, c) = (6, 2, 11).

Case 5 Without loss of generality we may suppose that a = 2αA, b = 2βB andc = 2γC, where α ≥ β ≥ γ ≥ 1, and A, B and C are odd positive integers. Itfollows that

ab− c = 2γ(2α+β−γAB − C) and ac− b = 2β(2α+γ−βAC − B).

Since α ≥ β ≥ γ ≥ 1, we have that 2α+β−γAB − C and 2α+γ−βAC − B are odd.But since ab− c and ac− b are powers of two, it follows that

ab− c = 2γ and ac− b = 2β.

Since 2γ | c, we have 2γ ≤ c, and similarly 2β ≤ b. Hence

ab ≤ 2c and ac ≤ 2b. (7)

Multiplying these inequalities together yields

a2bc ≤ 4bc.

Hence a ≤ 2, and thus a = 2 because a is even. But now the two inequalities in(7) become 2b ≤ 2c and 2c ≤ 2b. Hence b = c, which contradicts that a, b, c are alldifferent. So this case does not occur, and the proof is complete.

62

90

Solution 2 (Alex Gunning, year 12, Glen Waverley Secondary College, VIC. Alexwas a Gold medallist with the 2015 Australian IMO team.)

The case where two of the variables are equal is handled in the same way as insolution 1.

Without loss of generality we may assume that a > b > c. If c = 1, then we wouldrequire both a − b and b − a to be powers of two. But this is impossible becausetheir sum is zero.

Hence a > b > c ≥ 2.

It easily follows that ab− c ≥ ca− b ≥ bc− a. Since a smaller power of two alwaysdivides a larger power of two, we have

bc− a | ca− b | ab− c.

Thus ca− b | (a(ab− c)− (ca− b)), that is, ca− b | (a+ 1)(a− 1)b.

If a is even, then since ca− b is a power of two and (a− 1)(a+ 1) is odd, we musthave ca−b | b. Thus ca ≤ 2b < 2a, which yields the contradiction c < 2. Henceforthwe may assume that a is odd. Also a > b > c ≥ 2 implies a ≥ 4, and hence a ≥ 5.

If 4 | a − 1, then since ca − b is a power of two and a + 1 is divisible by 2 but notby 4, we have

ca− b | 2(a− 1)b

⇒ ca− b | 2(a− 1)b− 2(ab− c) (ca− b | ab− c)

⇔ ca− b | 2(b− c)

⇒ ca− b ≤ 2(b− c). (b = c)

If 4 | a+ 1, then since a− 1 is divisible by 2 but not by 4, we have

ca− b | 2(a+ 1)b

⇒ ca− b | 2(a+ 1)b− 2(ab− c) (ca− b | ab− c)

⇔ ca− b | 2(b+ c)

⇒ ca− b ≤ 2(b+ c).

Thus whatever odd number a is, we can be sure that ca− b ≤ 2(b+ c). That is,

ca ≤ 3b+ 2c.

Since b, c < a, we have ca < 3a+ 2a = 5a, and so c < 5.

If c = 4, then 4a < 3a + 8. Thus a < 8. But a > b > c = 4 implies a ≥ 6. Thusa = 7 because a is odd. But since cb − a = 4b − 7 is a power of two and it is odd,we must have 4b− 7 = 1. Thus b = 2, which contradicts b > c.

If c = 3, then 3a ≤ 3b+ 6 < 3a+ 6. Thus a ≤ b+ 2 < a+ 2.

If b = a−1, then ca− b = 3a− (a−1) = 2a+1 cannot be a power of two becauseit is odd and greater than 1.

If b = a−2, then bc−a = 3(a−2)−a = 2a−6 and ca− b = 3a− (a−2) = 2a+2are two powers of two that differ by 8. The only such powers of two are 8 and16. This quickly yields a = 7 and implies (a, b, c) = (7, 5, 3).

63

91

If c = 2, then bc− a = 2b− a = 1 because it is odd (a is odd) and is a power of two.But

ca− b | ab− c

⇒ 3b− 2 | (2b− 1)b− 2 (a = 2b− 1, c = 2)

⇒ 3b− 2 | 3((2b− 1)b− 2)− 2b(3b− 2)

⇒ 3b− 2 | b− 6.

Since 3b− 2 > b− 6 > 0 for b ≥ 7, we must have b ≤ 6.

Checking 3b− 2 | b− 6 directly for b = 3, 4, 5, 6, we find that only b = 6 works. Thisimplies (a, b, c) = (11, 6, 2).

64

92

3. Solution 1 (Alex Gunning, year 12, Glen Waverley Secondary College, VIC. Alexwas a Gold medallist with the 2015 Australian IMO team.)

Let A′ be the point diametrically opposite A on Γ and let E be the second point ofintersection of the line AHF with Γ.

Lemma The points A′, M , H and Q are collinear.

Proof First note that QA ⊥ QA′ because AA′ is a diameter of Γ. Also sinceQA ⊥ QH, it follows that QHA′ is a straight line. Thus Q lies on the line A′H.

A

B C

H

F

Q

EA′

Next we show that CHBA′ is a parallelogram.

Since AA′ is a diameter of Γ we have AE ⊥ A′E. However AE ⊥ BC. ThusBC ‖ A′E. Since A′EBC is cyclic, it follows that A′ECB is an isosceles trapeziumwith CE = A′B. Thus

∠BCA′ = ∠EBC (A′B = CE)

= ∠EAC (ECAB cyclic)

= 90 − ∠ACB (AE ⊥ BC)

= ∠CBH. (BH ⊥ AC)

Thus BH ‖ A′C. A similar argument shows that CH ‖ A′B, and so CHBA′ is aparallelogram.

Finally, since the diagonals of any parallelogram bisect each other, and M is themidpoint of BC, it follows that M is also the midpoint of A′H. Thus M lies on theline A′H. Since we already know that Q lies on the line A′H, it follows that A′, M ,H and Q are collinear.

Let R be the intersection of the line A′E with the line through H that is perpendic-ular to A′Q. Observe that RH is tangent to circles KQH, FHM and EHA′. Thisis because each of these circles has a diameter which lies on the line A′Q. Applyingthe radical axis theorem to the three circles Γ, KQH and EHA′, we deduce thatlines RH, A′E and QK are concurrent. Hence R lies on QK.

65

93

A

B C

H

F

Q

EA′

K

M

R

S

Recall that M is the midpoint of A′H and that BC ‖ A′E. Thus the midpoint, Ssay, of RH lies on the line BC. Since HK ⊥ KR we have that S is the centre ofcircle RKH, and so SH = SK. But since SH is tangent to circle KQH at H andSH = SK, it follows that SK is tangent to circle KQH at K. Considering thepower of point S with respect to circle FHM , we have

SF · SM = SH2 = SK2.

It follows that SK is tangent to circle FKM at K. Since circle KQH is also tangentto SK at K, we conclude that circles KQH and FKM are tangent to each otherat K.

66

94

Solution 2 (Andrew Elvey Price, Deputy Leader of the 2015 Australian IMOteam)

Point A′ is defined as in solution 1. Furthermore, as in solution 1, we establish thatM is the midpoint of A′H, and A′, M , H and Q are collinear.

Since ∠A′AK = ∠A′QK = ∠HQK and ∠AKA′ = 90 = ∠QKH, we haveKAA′ ∼ KQH. Hence there is a spiral symmetry, f say, centred at K, suchthat f(A′) = A and f(H) = Q. Note that f is the composition of a 90 rotationabout K with a dilation of factor KA

KA′ about K.

Let M ′ and F ′ be the respective images of M and F under f . Thus M ′F ′ ⊥ MF .But since AH ⊥ MF , it follows that M ′F ′ ‖ AH.

Since M is the midpoint of A′H, it follows that M ′ is the midpoint of AQ. SinceM ′F ′ ‖ AH, it follows that M ′F ′ passes through the midpoint, S ′ say, of HQ. Notethat S ′ is the centre of circle KQH.

A

B C

H

F

M

Q

K

A′

S′

M ′

F ′

S

Let S be the preimage of S ′ under f . (Note that S happens to lie on the line MF ,as shown in the diagram, because S ′ lies on line M ′F ′. But we will not need thisfact.)

Since S ′F ′Q ∼ S ′QM ′ (AA), we have

S ′F ′

S ′Q=

S ′Q

S ′M ′ .

It follows thatS ′F ′ · S ′M ′ = S ′Q2 = S ′K2.

Therefore, S ′K is tangent to circle F ′KM ′ at K.

Applying the inverse of f , we have SK is tangent to circle FKM at K. But sinceS is the preimage of S ′ under f , we also have ∠S ′KS = 90, and so SK is tangentto circle KQH. Hence circles FKM and KQH are tangent at K.

67

95

Solution 3 (IMO Problem Selection Committee)

Points A′ and E are defined as in solution 1. Moreover, as in solution 1, we establishthat M is the midpoint of A′H, and A′, M , H and Q are collinear. Note that sinceMF ‖ A′E, it follows that F is the midpoint of EH.

Let Q′ be the point diametrically opposite Q on Γ. Then KQ ⊥ KQ′ because QQ′

is a diameter of Γ. Also since KQ ⊥ KH, it follows that KHQ′ is a straight line.

Let T be a point on the tangent to circle KQH at K, such that T and Q lieon the same side of the line KH. By the alternate segment theorem we have∠TKQ = ∠KHQ.

By the alternate segment theorem, it is sufficient to prove that

∠KFM = ∠TKM.

We have

∠KFM = ∠TKM

⇔ 90 + ∠KFA = ∠TKQ+ 90 + ∠HKM

⇔ ∠KFA = ∠Q′HA′ + ∠HKM. (1)

Thus it suffices to establish (1).

Let J be the midpoint of HQ′. Observe that triangles KHE and AHQ′ are similarwith F and J being the midpoints of corresponding sides. Hence ∠KFA = ∠HJA.

A

B C

H

FM

Q

K

EA′

Q′

J

T

Observe also that triangles KHA′ and QHQ′ are similar with M and J being themidpoints of corresponding sides. Hence ∠HKM = ∠JQH.

Thus our task is reduced to proving

∠HJA = ∠Q′HA′ + ∠JQH.

Let us draw a new diagram that will help us focus on the task at hand.

68

96

A

H

Q

A′

Q′

J

O

Note that QAQ′A′ is a rectangle. Let O be its centre. We also know that H lieson side A′Q and that J is the midpoint of Q′H. Thus J and O both lie on themid-parallel of QA′ and Q′A. Hence

∠HJA = ∠HJO + ∠OJA

= ∠Q′HA+ ∠Q′AJ. (A′Q ‖ JO ‖ Q′A)

Thus it suffices to prove that ∠JQH = ∠Q′AJ . However, this is an immediateconsequence of the fact that JO a line of reflective symmetry of the rectangle.

69

97

Solution 4 (Jacob Tsimerman, Leader of the 2015 Canadian IMO team)

Points A′, E and Q′ are defined as in solution 3. Furthermore, as in solution 3, wehave M is the midpoint of A′H; points A′, M , H and Q are collinear; points Q′, Hand K are collinear; and F is the midpoint of HE.

Since the three chords AE, QA′ and KQ′ are concurrent we have

HA ·HE = HQ ·HA′ = HK ·HQ′ = r2,

for some positive real number r. Let f be the inversion with centre H and radiusr, followed by the reflection1 in H. Let A′′ and Q′′ be the reflections of H in A andQ, respectively. Then f has the effect of exchanging the following pairs of points.

A ↔ E A′′ ↔ F Q ↔ A′ Q′′ ↔ M K ↔ Q′

A

B C

H

FM

Q

K

EA′

Q′

A′′

Q′′

Hence circle KQH is transformed into line Q′A′, and circle FKM is transformedinto circleA′′Q′Q′′. So it suffices to show thatQ′A′ is tangent to circleA′′Q′Q′′. SinceQ′A′ ‖ AQ ‖ A′′Q′′, it suffices to show that A′′Q′Q′′ is isosceles with Q′A′′ = Q′Q′′.

Observe Q′A ‖ A′Q because AQA′Q′ is a rectangle. So Q′A ⊥ A′′Q′′. But A is thecentre of circle A′′HQ′′ because it is the midpoint of A′′H and ∠A′′Q′′H = 90. ThusQ′A is the perpendicular bisector of A′′Q′′. From this it follows that Q′A′′ = A′Q′′,as desired.

1Reflection in a point is the same as a 180 rotation about that point.

70

1

1

98

Comment Solutions 1 and 3 first establish that Q, H, M and A′ are collinear, andthat M and F are the respective midpoints of A′H and EH. After this the pointsB and C are no longer relevant to the solution. The crux of matter boils down tothe following.

Let AQA′Q′ be a rectangle inscribed in a circle Γ. Let H be any point onthe line A′Q. Let E and K be the respective second points of intersectionof the lines AH and Q′H with Γ. Let M and F be the midpoints of A′Hand EH, respectively. Then circles KQH and KMF are tangent to eachother.

A

H

FM

Q

K

EA′

Q′

71

99

Here is an even more minimal way of looking at the same thing.

Let A′EKQ be a cyclic quadrilateral. Let H be a point on the line A′Qsuch that ∠HEA′ = ∠QKH = 90. Let M and F be the midpoints ofA′H and EH, respectively. Then circles KQH and KMF are tangent toeach other. (*)

H

FM

Q

K

EA′

72

100

Solution 5 (Panupong Pasupat, one of the Coordinators at the 2015 IMO)

It is sufficient to prove statement (*) found in the comments on the previous page.

Let T be a point on the tangent to circle KQH at K, such that T and Q lie on thesame side of the line KH. Let W be the midpoint of QH. Since ∠QKH = 90, itfollows that W is the centre of circle KQH. Hence WH = WK, and we may let

∠WKH = ∠KHW = α.

From the angle sum in KQH, we have ∠HKQ = 90−α. Since A′EKQ is cyclic,we have ∠AEK = 90 + α, and so ∠KEH = α. Thus by the alternate segmenttheorem, circle HEK is tangent to line A′Q at H. Then since MH = ME, itfollows by symmetry that MH and ME are the common tangents from M to circleHEK.

H

FM

Q

K

EA′

T

W

The configuration where MH and ME are common tangents to the circumcircle ofHEK, is a standard one. In particular it yields an alternative characterisation ofthe symmedian.2 Thus we may let

∠HKM = ∠FKE = β.

From the exterior angle sum in EKF , we have

∠KFH = α + β = ∠WKM. (1)

Since WK is the radius of circle KQH, we have TK ⊥ KW . Thus adding 90 toboth sides of (1) yields

∠KFM = ∠TKM.

Hence from the alternate segment theorem, TK is tangent to circle FKM at K.Thus circles FKM and KQH are tangent at K.

2For more details see the section Alternative characterisations of symmedian found in chapter 5 of ProblemSolving Tactics published by the AMT.

73

2

2

101

4. Solution 1 (Kevin Xian, year 11, James Ruse Agricultural High School, NSW.Kevin was a Silver medallist with the 2015 Australian IMO team.)

A

B CD E

F

G

K

L

X

O

Observe that AF = AG (radius of Γ), and OF = OG (radius of Ω). So F andG are symmetric in AO. Thus lines FK and GL intersect on AO if and only if∠KFA = ∠AGL. We have,

∠KFA = ∠FKB − ∠FAB (exterior angle AFK)

= ∠FDB − ∠FGB (BDKF and AFBG cyclic)

= ∠FGE − ∠FGB (FDEG cyclic)

= ∠BGE

= ∠CEG− ∠CBG (exterior angle GEB)

= ∠CLG− ∠CAG (ECGL and ABCG cyclic)

= ∠AGL. (exterior angle GLA)

Comment A careful analysis of this solution shows that the result is still true ifwe only assume that the centre of Γ lies on the line AO.

74

102

Solution 2 (Found independently by Ilia Kucherov, year 11, Westall SecondaryCollege, VIC, and Jeremy Yip, year 12, Trinity Grammar School, VIC. Ilia andJeremy were Silver medallists with the 2015 Australian IMO team.)

As in solution 1 it suffices to show that ∠KFA = ∠AGL.

Let the lines DF and EG intersect Ω for a second time at points P and Q, respec-tively. Then

∠PDE = ∠FGQ (FDEG cyclic)

= ∠FPQ. (FQPG cyclic)

Thus BC ‖ QP . Since BQPC is cyclic, it follows that BQPC is an isoscelestrapezium with BQ = PC. Hence ∠BGQ = ∠PFC, that is,

∠BGE = ∠DFC. (1)

A

B CD E

F

G

PQ

75

103

A

B CD E

F

G

K

L

X

O

Since ABCG is cyclic, we have ∠LAG = ∠CAG = ∠CBG = ∠EBG. Since ECGLis cyclic, ∠CLG = ∠CEG, and so ∠GLA = ∠GEB. Thus GAL ∼ GBE (AA).Hence

∠AGL = ∠BGE. (2)

Similarly∠KFA = ∠DFC. (3)

Combining (1), (2) and (3), it follows that

∠KFA = ∠AGL.

Comment There are two pairs of similar triangles associated with circles ECGLand Ω. They are GAL ∼ GBE and GLE ∼ GAB. This is a standardconfiguration which can help fast track the route to a solution.3

3For more details see the section Similar Switch found in chapter 5 of Problem Solving Tactics publishedby the AMT.

76

3

3

104

5. Solution 1 (Seyoon Ragavan, year 11, Knox Grammar School, NSW. Seyoon wasa Gold medallist with the 2015 Australian IMO team.)

We show that the only answers are: f(x) = x for all x ∈ R and f(x) = 2− x for allx ∈ R.

We are asked to find all functions f : R → R such that for all x, y ∈ R,

f(x+ f(x+ y)) + f(xy) = x+ f(x+ y) + yf(x). (1)

Set y = 1 in (1) to find that for all x ∈ R,

x+ f(x+ 1) is a fixed point of f . (2)

With (2) in mind, set x = 0 and y = z + f(z + 1) in (1) to find that for all z ∈ R,

f(0) = f(0)(z + f(z + 1)). (3)

Case 1 f(0) = 0

Equation (3) implies z + f(z + 1) = 1 for all z ∈ R. Putting z = x− 1, this may berearranged to f(x) = 2− x for all x ∈ R. We verify this is a solution.

LHS = 2− (x+ 2− (x+ y)) + 2− xy = 2 + y − xy

RHS = x+ 2− (x+ y) + y(2− x) = 2 + y − xy = LHS

Case 2 f(0) = 0

Set x = 0 in (1) to find that for all y ∈ R,

f(f(y)) = f(y). (4)

Set y = 0 in (1) to find that for all x ∈ R,

x+ f(x) is a fixed point of f . (5)

Let S denote the set of fixed points of f . Suppose that u ∈ S. Then (5) with x = utells us that 2u ∈ S. And (2) with x = u− 1 tells us that 2u− 1 ∈ S. Hence

u ∈ S ⇒ 2u, 2u− 1 ∈ S. (6)

Since 0 ∈ S, applying (6) tells us that −1 ∈ S. Applying (6) again tells us that−2,−3 ∈ S. Continuing inductively, we find that all negative integers are in S.

On the other hand, if x is any positive integer, choose an integer y such that y < 0,x+ y < 0, 2x+ y < 0 and xy < 0 (any y < −2x will do). Using these values in (1)yields f(x) = x. Hence

Z ⊆ S. (7)

Since f(1) = 1, we may set x = 1 in (1) to find that for all y ∈ R,

f(1 + f(y + 1)) + f(y) = y + 1 + f(y + 1). (8)

Suppose that u, u + 1 ∈ S. Then setting y = u in (8) tells us that u + 2 ∈ S. Alsosetting y = u− 1 in (8) tells us that u− 1 ∈ S. An immediate corollary of this is

u, u+ 1 ∈ S ⇒ u+ n ∈ S for any n ∈ Z. (9)

77

105

Let y ∈ R. Then from (2) with x = y−1 we have y+ f(y)−1 ∈ S, and from (5) wehave y+ f(y) ∈ S. Hence with u = y+ f(y)− 1 in (7), we find that for each n ∈ Z,

y + f(y) + n ∈ S. (10)

Using the change of variables y = x+m and n = −m, we have for each m ∈ Z andx ∈ R,

x+ f(x+m) ∈ S. (11)

If we set y = m in (1) and use (11), then for each m ∈ Z and x ∈ R we have,

f(mx) = mf(x). (12)

If we replace y with f(y) in (10) and remember that f(f(y)) = f(y) from (4), wefind

2f(y) + n ∈ S. (13)

Let y ∈ R and let y = 2x. Then applying (12) and (13), we are able to deduce thatf(y) + 1 = f(2x) + 1 = 2f(x) + 1 ∈ S. Hence for all y ∈ R we have

f(y) + 1 ∈ S. (14)

Finally, put x = 1 in (1). Then using f(1 + f(1 + y)) = 1+ f(1 + y) from (14), andf(1) = 1 from (7), we deduce that f(y) = y for all y ∈ R. This is easily seen to bea solution.

78

106

Solution 2 (Based on the presentation by Alex Gunning, year 12, Glen WaverleySecondary College, VIC. Alex was a Gold medallist with the 2015 Australian IMOteam.)

We are asked to find all functions f : R → R such that for all x, y ∈ R,

f(x+ f(x+ y)) + f(xy) = x+ f(x+ y) + yf(x). (1)

The case f(0) = 0 is handled as in solution 1.

It remains to deal with the case f(0) = 0. All equations that follow will hold truefor all z ∈ R.Replacing x with y and y with x in (1) yields

f(y + f(x+ y)) + f(xy) = y + f(x+ y) + xf(y). (2)

Computing the difference (1)− (2), we have for all x, y ∈ R,

f(x+ f(x+ y))− f(y + f(x+ y)) = x− y + yf(x)− xf(y). (3)

Putting y = −x in (3) yields

f(x)− f(−x) = 2x− xf(x)− xf(−x). (4)

Suppose that f(x) = x. Then (4) can be rearranged to yield

(x− 1)(f(−x) + x)) = 0.

If x = 1, then f(−x) = −x. If x = 1, then putting (x, y) = (−1, 1) in (1), yieldsf(−1) = −1. Hence if S is the set of fixed points of f , we have shown that

x ∈ S ⇒ −x ∈ S. (5)

Putting (x, y) = (z, 0) in (1), we find,

z + f(z) ∈ S. (6)

Putting (x, y) = (0, z) in (1), we find,

f(z) ∈ S. (7)

Putting (x, y) = (f(z), 0) in (1) and using (7), we find,

2f(z) ∈ S. (8)

Put (x, y) = (z + f(z),−f(z)) in (3). The LHS of (3) is f(z + 2f(z)). We alsohave x = z + f(z) ∈ S from (6). And y = −f(z) ∈ S from (7) and (5). Thusyf(x) − xf(y) = yx − xy = 0, and so the RHS of (3) is just x − y = z + 2f(z).Hence

z + 2f(z) ∈ S. (9)

79

107

Put (x, y) = (z,−z − f(z)) in (3). Note that y = −z − f(z) ∈ S from (6) and (5).And x+ y = −f(z) ∈ S from (7) and (5). Thus (3) simplifies to

f(2x+ y)− f(x+ 2y) = x− y + yf(x)− xy.

However, x+ 2y = −z − 2f(z) ∈ S from (9) and (5). So (3) simplifies further to

f(2x+ y) = 2x+ y + yf(x)− xy.

Writing x and y in term of z, and tidying up yields

f(z − f(z)) = z − f(z) + z2 − f(z)2. (10)

Put (x, y) = (z−f(z), f(z)) in (3). The LHS of (3) equals f(z)−f(2f(z)) = −f(z)from (8). Hence

−f(z) = x− y + yf(x)− xf(y)

= z − 2f(z) + yf(x)− xy (since y = f(z) ∈ S)

⇒ f(z)− z = f(z)(f(x)− x) (y = f(z))

= f(z)(z2 − f(z)2). (by (10) as x = z − f(z))

This can be written in the form

(f(z)− z)(f(z)2 + zf(z) + 1) = 0. (11)

Solving for f(z) in (11), we find f(z) ∈z, −z+

√z2−4

2, −z−

√z2−4

2

. Since f(z) ∈ R,

we have f(z) = z for all |z| < 2. Hence (−2, 2) ⊆ S. Finally, from (8) we see thatz ∈ S implies 2z ∈ S. This allows us to deduce that (−2k, 2k) ⊆ S for all positiveintegers k. Hence S = R, and f(x) = x for all x ∈ R.

80

108

6. This was the hardest problem of the 2015 IMO. Only 11 of the 577 contestants wereable to solve this problem completely.

The authors of this problem were Ross Atkins and Ivan Guo of Australia. Ross andIvan were Bronze medallists with the 2003 Australian IMO Team and Ivan was aGold medallist with the 2004 Australian IMO Team. The problem was inspired by anotation for juggling (Ross is also a juggler) in which each ai represents the airtimeof a ball thrown at time i, and b is the total number of balls.

Solution 1 (Alex Gunning, year 12, Glen Waverley Secondary College, VIC. Alexwas a Gold medallist with the 2015 Australian IMO team.)

Let S be the set of positive integers which are not of the form n + an for somepositive integer n. Note that S is nonempty because 1 ∈ S. Let s1 < s2 < · · · bethe elements of S listed in increasing order.

Lemma |S| ≤ 2015.

Proof Assume that |S| ≥ 2016. Choose n so that an+n ≥ s2016. Since an ≤ 2015,this implies that s1, s2, . . . , s2016 ∈ 1, 2, . . . , n + 2015. However, the n numbers1 + a1, 2 + a2, . . . , n + an are not equal to any si and are also members of the set1, 2, . . . , n+ 2015. Hence 1, 2, . . . , n+ 2015 contains at least n+ 2016 differentnumbers, contradiction.

We claim that if b = |S| and if N is larger than all members of S, then the inequalityposed in the problem statement is true.

Let n be any integer satisfying n ≥ N . We shall find bounds for∑n

j=1(j + aj) andhence also for

∑nj=1(aj − b). In what follows, let L be the following list of n + b

distinct positive integers.

1 + a1, 2 + a2, . . . , n+ an, s1, s2, . . . , sb

For the lower bound, since the n+ b numbers in L are distinct, their sum is greaterthan or equal to

∑n+bj=1 j. Hence we have

n∑j=1

(j + aj) +b∑

j=1

sj ≥n+b∑j=1

j

⇒n∑

j=1

aj ≥n+b∑

j=n+1

j −b∑

j=1

sj

=b(2n+ b+ 1)

2− s

⇒n∑

j=1

(aj − b) ≥ b2 + b

2− s, (1)

where s =∑b

j=1 sj.

For the upper bound, observe that s1, s2, . . . , sb are b members belonging to the setT = 1, 2, . . . , n + 1. The remaining n + 1− b members of T must be of the formj + aj where j ≤ n, and so are in L. The sum of these n+ 1− b numbers is exactly

n+1∑j=1

j −b∑

j=1

sb.

81

109

All together there are exactly n numbers of the form j + aj in L and so far we haveaccounted for n+ 1− b of them.

Consider the remaining b−1 numbers of the form j+aj which are in L. When listedin decreasing order, they can be no larger than n+2015, n+2014, . . . , n+2015−b+2,respectively. Hence their sum is at most

∑b−1j=1(n+ 2016− j). Thus

n∑j=1

(j + aj) ≤n+1∑j=1

j −b∑

j=1

sb +b−1∑j=1

(n+ 2016− j)

⇒n∑

j=1

aj ≤ n+ 1 +(b− 1)(2n+ 4032− b)

2− s

⇒n∑

j=1

(aj − b) ≤ 4033b− b2 − 4030

2− s. (2)

Summarising (1) and (2), we have established the following bounds for any n ≥ N .

4033b− b2 − 4030

2− s ≤

n∑j=1

(aj − b) ≤ b2 + b

2− s. (3)

Now let m,n be any two integers satisfying n > m ≥ N . Since also m ≥ N , (3) isalso satisfied if n is replaced by m. Thus

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ =∣∣∣∣∣

n∑j=1

(aj − b)−m∑j=1

(aj − b)

∣∣∣∣∣

≤∣∣∣∣(b2 + b

2− s

)−

(4033b− b2 − 4030

2− s

)∣∣∣∣

= (b− 1)(2015− b)

≤((b− 1) + (2015− b)

2

)2

(AM–GM)

= 10072.

82

110

Solution 2 (A juggling interpretation solution by the authors Ross Atkins andIvan Guo)

Suppose you are juggling several balls using only one hand. At the ith second, if aball lands in your hand, it is thrown up immediately. If no ball lands, you insteadreach for an unused ball (that is, a ball that has not been thrown yet) and throw itup. In both cases, a ball is thrown so that it will stay in the air for ai seconds. Thecondition ai + i = aj + j ensures that no two balls land at the same time.

Let b be the total number of balls used. If b > 2015, then eventually a ball muststay in the air for more than 2015 seconds, contradicting ai ≤ 2015. So b is finiteand bounded by 2015.

Select N so that all b balls have been introduced by the Nth second. For all i ≥ N ,denote by Ti the total remaining airtime of the current balls, immediately after theith throw is made. (That is, we calculate the remaining airtime for each currentball, and add these values together.) Consider what happens during the next second.The airtime of each of the b balls is reduced by 1. At the same time a ball is thrown,increasing its airtime by ai+1. Thus we have the equality Ti+1 − Ti = ai+1 − b. Thisgives a nice representation of the required sum,

n∑i=m+1

(ai − b) = Tn − Tm.

To complete the problem, it suffices to identify the maximal and minimal possiblevalues of the total remaining airtime Ti. Since no two balls can land at the sametime, the minimal value is 1 + 2 + · · · + b. On the other hand, the maximal valueis 1 + 2015 + 2014 + · · · + (2015 − b + 2). (Note that there must be a ball with aremaining airtime of 1 since something must be caught and thrown every second.)Taking the difference between these two sums, we find that

|Tn − Tm| ≤(4032− b)(b− 1)

2− (b+ 2)(b− 1)

2

= (2015− b)(b− 1)

≤ 10072 (GM ≤ AM)

as required.

83

111

INTERNATIONAL MATHEMATICAL OLYMPIAD RESULTS

Mark Q1 Q2 Q3 Q4 Q5 Q6

0 93 256 408 91 153 521

1 89 151 122 36 255 11

2 5 77 12 61 34 15

3 21 27 1 18 90 6

4 72 8 3 11 8 3

5 12 13 0 1 4 3

6 20 14 1 8 3 7

7 265 31 30 351 30 11

Total 577 577 577 577 577 577

Mean 4.3 1.4 0.7 4.8 1.5 0.4

Name Q1 Q2 Q3 Q4 Q5 Q6 Score Award

Alex Gunning 7 6 7 7 2 7 36 Gold

Ilia Kucherov 7 2 0 7 3 0 19 Silver

Seyoon Ragavan 7 7 1 7 7 0 29 Gold

Yang Song 7 2 1 7 3 0 20 Silver

Kevin Xian 7 3 1 7 3 0 21 Silver

Jeremy Yip 7 6 1 7 2 0 23 Silver

Totals 42 26 11 42 20 7 148

Australian average 7.0 4.3 1.8 7.0 3.3 1.2 24.7

IMO average 4.3 1.4 0.7 4.8 1.5 0.4 13.0

The medal cuts were set at 26 for Gold, 19 for Silver and 14 for Bronze.

Australian scores at the IMO

Mark distribution by question

112

Rank Country Total

1 United States of America 185

2 China 181

3 South Korea 161

4 North Korea 156

5 Vietnam 151

6 Australia 148

7 Iran 145

8 Russia 141

9 Canada 140

10 Singapore 139

11 Ukraine 135

12 Thailand 134

13 Romania 132

14 France 120

15 Croatia 119

16 Peru 118

17 Poland 117

18 Taiwan 115

19 Mexico 114

20 Hungary 113

20 Turkey 113

22 Brazil 109

22 Japan 109

22 United Kingdom 109

25 Kazakhstan 105

26 Armenia 104

27 Germany 102

28 Hong Kong 101

29 Bulgaria 100

29 Indonesia 100

29 Italy 100

29 Serbia 100

Some country totals

113

Distribution of awards at the 2015 IMO

Country Total Gold Silver Bronze HM

Albania 37 0 0 0 3

Algeria 60 0 1 1 2

Argentina 70 0 0 1 4

Armenia 104 0 1 5 0

Australia 148 2 4 0 0

Austria 63 0 0 3 1

Azerbaijan 73 0 0 2 4

Bangladesh 97 0 1 4 1

Belarus 84 0 0 3 3

Belgium 67 0 1 0 3

Bolivia 5 0 0 0 0

Bosnia and Herzegovina 76 0 0 2 4

Botswana 1 0 0 0 0

Brazil 109 0 3 3 0

Bulgaria 100 0 2 1 2

Cambodia 24 0 0 0 2

Canada 140 2 0 4 0

Chile 12 0 0 0 1

China 181 4 2 0 0

Colombia 72 0 0 4 0

Costa Rica 53 0 0 2 2

Croatia 119 1 3 1 0

Cuba 15 0 0 1 0

Cyprus 58 0 1 0 2

Czech Republic 74 0 0 3 3

Denmark 52 0 0 2 1

Ecuador 27 0 0 0 2

El Salvador 14 0 0 0 0

Estonia 51 0 0 1 3

Finland 26 0 0 0 1

France 120 0 3 3 0

Georgia 80 0 1 3 1Germany 102 0 2 3 0

Ghana 5 0 0 0 0

Greece 71 0 1 2 2

114

Country Total Gold Silver Bronze HMHong Kong 101 0 2 3 1

Hungary 113 0 3 3 0

Iceland 41 0 0 0 3

India 86 0 1 2 3

Indonesia 100 0 2 4 0

Iran 145 3 2 1 0

Ireland 37 0 0 0 3

Israel 83 1 0 2 2

Italy 100 1 2 0 0

Japan 109 0 3 3 0

Kazakhstan 105 1 1 2 2

Kosovo 24 0 0 0 1

Kyrgyzstan 17 0 0 0 0

Latvia 36 0 0 0 3

Liechtenstein 18 0 0 1 0

Lithuania 54 0 0 1 1

Luxembourg 12 0 0 0 1

Macau 88 0 1 2 3

Macedonia (FYR) 45 0 0 1 1

Malaysia 66 0 0 3 1

Mexico 114 1 2 3 0

Moldova 85 0 1 2 3

Mongolia 74 0 0 2 4

Montenegro 19 0 0 1 0

Morocco 27 0 0 0 1

Netherlands 76 0 0 3 1

New Zealand 72 0 0 2 4

Nicaragua 26 0 0 0 3

Nigeria 22 0 0 0 2

North Korea 156 3 3 0 0

Norway 54 0 1 0 2

Pakistan 25 0 0 1 0

Panama 9 0 0 0 0

Paraguay 53 0 0 3 0

Peru 118 2 2 1 0

Philippines 87 0 2 2 1

Poland 117 1 1 4 0

115

Country Total Gold Silver Bronze HM

Portugal 70 0 0 3 1

Puerto Rico 18 0 0 1 0

Romania 132 1 4 1 0

Russia 141 0 6 0 0

Saudi Arabia 81 0 1 3 2

Serbia 100 1 1 2 2

Singapore 139 1 4 1 0

Slovakia 97 0 2 3 0

Slovenia 46 0 0 1 1

South Africa 68 0 0 1 2

South Korea 161 3 1 2 0

Spain 47 0 0 1 2

Sri Lanka 51 0 0 0 4

Sweden 63 0 0 2 2

Switzerland 74 0 0 3 2

Syria 69 0 1 1 3

Taiwan 115 0 4 1 1

Tajikistan 57 0 1 1 2

Tanzania 0 0 0 0 0

Thailand 134 2 3 1 0

Trinidad and Tobago 26 0 1 0 0

Tunisia 41 0 0 1 2

Turkey 113 0 5 0 0

Turkmenistan 64 0 0 2 2

Uganda 6 0 0 0 0

Ukraine 135 2 3 1 0

United Kingdom 109 0 4 1 1

United States of America 185 5 1 0 0

Uruguay 16 0 0 0 1

Uzbekistan 64 0 0 3 2

Venezuela 13 0 0 0 1

Vietnam 151 2 3 1 0

Total (104 teams, 577 contestants) 39 100 143 126

116

ORIGIN OF SOME QUESTIONS

Senior ContestQuestion 1 was submitted by Angelo Di Pasquale.Questions 2, 3 and 5 were submitted by Norman Do.Question 4 was submitted by Alan Offer.

Australian Mathematical OlympiadQuestions 1, 2, 5 and 6 were submitted by Norman Do.Question 3 was submitted by Andrei Storozhev.Questions 4 and 7 were submitted by Angelo Di Pasquale.Question 8 was submitted by Andrew Elvey Price.

Asian Pacific Mathematical Olympiad 2015Question 2 was composed by Angelo Di Pasquale and submitted by the AMOC Senior Problems Com-mittee.

International Mathematical Olympiad 2015Question 6 was composed by Ross Atkins and Ivan Guo, and submitted by the AMOC Senior ProblemsCommittee. Ivan provided the following background information on the problem.

The original idea for this problem came about while Ross was reading the paper PositroidVarieties: Juggling and Geometry by Knutson, Lam and Speyer, in which the “excitationnumber” of a periodic juggling sequence was discovered. It seemed obvious that this wassimilar to some specific elementary result that could be proven using elementary methods.We had some difficulties in phrasing the problem in a concise self-contained way. Intuitively,each term ai in the sequence corresponds to throwing a ball at the ith second with an air timeof ai. The inequality condition ensures that no two balls land simultaneously.

The first formulation of the problem was to show that the long-term average of the sequenceconverges to an integer b, which is the total number of balls. However, the usage of limits wasinappropriate for an olympiad problem. We then came up with three more versions whichinvolved bounding the partial sums. Eventually we settled on the most difficult version, withthe explicit bound of |

∑(ai − b)| ≤ 10072. Interestingly, the term ai − b can be interpreted

as the change in the “total air time” on the ith second, while 10072 is the difference betweenmaximal and minimal possible “total air times”, after the introduction of all b balls. The finalwording may be a little difficult for students who are unfamiliar with the construct: “thereexists an N such that for all m > n > N”.

It is possible to solve the problem combinatorially without invoking any physical interpreta-tions, juggling or otherwise. Furthermore, as demonstrated by some at the IMO, the problemcan also be tackled using purely algebraic approaches. Overall, we are very happy with theproblem and we hope everyone enjoyed it.

It is worth noting that one of the authors of the paper that inspired this problem was Thomas Lam, amember of the 1997 Australian IMO team and recipient of an IMO gold medal.

1

ORIGIN OF SOME QUESTIONS

117

HONOUR ROLL

Because of changing titles and affiliations, the most senior title achieved and later affiliations are generally used, except for the Interim committee, where they are listed as they were at the time.

Interim Committee 1979–1980Mr P J O’Halloran Canberra College of Advanced Education, ACT, ChairProf A L Blakers University of Western AustraliaDr J M Gani Australian Mathematical Society, ACT,Prof B H Neumann Australian National University, ACT,Prof G E Wall University of Sydney, NSWMr J L Williams University of Sydney, NSW

Australian Mathematical Olympiad CommitteeThe Australian Mathematical Olympiad Committee was founded at a meeting of the Australian Academy of Science at its meeting of 2–3 April 1980.

* denotes Executive Position

Chair*Prof B H Neumann Australian National University, ACT 7 years; 1980–1986Prof G B Preston Monash University, VIC 10 years; 1986–1995Prof A P Street University of Queensland 6 years; 1996–2001Prof C Praeger University of Western Australia 14 years; 2002–2015Deputy Chair*Prof P J O’Halloran University of Canberra, ACT 15 years; 1980–1994Prof A P Street University of Queensland 1 year; 1995Prof C Praeger, University of Western Australia 6 years; 1996–2001Assoc Prof D Hunt University of New South Wales 14 years; 2002–2015Executive Director*Prof P J O’Halloran University of Canberra, ACT 15 years; 1980–1994Prof P J Taylor University of Canberra, ACT 18 years; 1994–2012Adj Prof M G Clapper University of Canberra, ACT 3 years; 2013–2015

SecretaryProf J C Burns Australian Defence Force Academy, ACT 9 years; 1980–1988Vacant 4 years; 1989–1992Mrs K Doolan Victorian Chamber of Mines, VIC 6 years; 1993–1998

Treasurer*Prof J C Burns Australian Defence Force Academy, ACT 8 years; 1981–1988Prof P J O’Halloran University of Canberra, ACT 2 years; 1989–1990Ms J Downes CPA 5 years; 1991–1995Dr P Edwards Monash University, VIC 8 years; 1995–2002Prof M Newman Australian National University, ACT 6 years; 2003–2008Dr P Swedosh The King David School, VIC 7 years; 2009–2015

Director of Mathematics Challenge for Young Australians*Mr J B Henry Deakin University, VIC 17 years; 1990–2006Dr K McAvaney Deakin University, VIC 10 years; 2006–2015

Chair, Senior Problems CommitteeProf B C Rennie James Cook University, QLD 1 year; 1980Mr J L Williams University of Sydney, NSW 6 years; 1981–1986Assoc Prof H Lausch Monash University, VIC 27 years; 1987–2013Dr N Do Monash University, VIC 2 years; 2014–2015

118

Director of Training*Mr J L Williams University of Sydney, NSW 7 years; 1980–1986Mr G Ball University of Sydney, NSW 3 years; 1987–1989Dr D Paget University of Tasmania 6 years; 1990–1995Dr M Evans Scotch College, VIC 3 months; 1995Assoc Prof D Hunt University of New South Wales 5 years; 1996–2000Dr A Di Pasquale University of Melbourne, VIC 15 years; 2001–2015

Team LeaderMr J L Williams University of Sydney, NSW 5 years; 1981–1985Assoc Prof D Hunt University of New South Wales 9 years; 1986, 1989, 1990, 1996–2001Dr E Strzelecki Monash University, VIC 2 years; 1987, 1988Dr D Paget University of Tasmania 5 years; 1991–1995Dr A Di Pasquale University of Melbourne, VIC 13 years; 2002–2010, 2012–2015Dr I Guo University of New South Wales 1 year; 2011

Deputy Team LeaderProf G Szekeres University of New South Wales 2 years; 1981–1982Mr G Ball University of Sydney, NSW 7 years; 1983–1989Dr D Paget University of Tasmania 1 year; 1990Dr J Graham University of Sydney, NSW 3 years; 1991–1993Dr M Evans Scotch College, VIC 3 years; 1994–1996Dr A Di Pasquale University of Melbourne, VIC 5 years; 1997–2001Dr D Mathews University of Melbourne, VIC 3 years; 2002–2004Dr N Do University of Melbourne, VIC 4 years; 2005–2008Dr I Guo University of New South Wales 4 years; 2009–10, 2012–2013Mr G White University of Sydney, NSW 1 year; 2011Mr A Elvey Price Melbourne University, VIC 2 years; 2014–2015

State DirectorsAustralian Capital TerritoryProf M Newman Australian National University 1 year; 1980Mr D Thorpe ACT Department of Education 2 years; 1981–1982Dr R A Bryce Australian National University 7 years; 1983–1989Mr R Welsh Canberra Grammar School 1 year; 1990Mrs J Kain Canberra Grammar School 5 years; 1991–1995Mr J Carty ACT Department of Education 17 years; 1995–2011Mr J Hassall Burgmann Anglican School 2 years; 2012–2013Dr C Wetherell Radford College 2 years; 2014–2015New South WalesDr M Hirschhorn University of New South Wales 1 year; 1980Mr G Ball University of Sydney, NSW 16 years; 1981–1996Dr W Palmer University of Sydney, NSW 19 years; 1997–2015Northern TerritoryDr I Roberts Charles Darwin University 2 years; 2014–2015QueenslandDr N H Williams University of Queensland 21 years; 1980–2000Dr G Carter Queensland University of Technology 10 years; 2001–2010Dr V Scharaschkin University of Queensland 4 years; 2011–2014Dr A Offer Queensland 1 year; 2015South Australia/Northern TerritoryMr K Hamann SA Department of Education 19 years; 1980–1982, 1991–2005, 2013Mr V Treilibs SA Department of Education 8 years; 1983–1990Dr M Peake Adelaide 8 years; 2006–2013

119

Dr D Martin Adelaide 2 years; 2014–2015TasmaniaMr J Kelly Tasmanian Department of Education 8 years; 1980–1987Dr D Paget University of Tasmania 8 years; 1988–1995Mr W Evers St Michael’s Collegiate School 9 years; 1995–2003Dr K Dharmadasa University of Tasmania 12 years; 2004–2015VictoriaDr D Holton University of Melbourne 3 years; 1980–1982Mr B Harridge Melbourne High School 1 year; 1982Ms J Downes CPA 6 years; 1983–1988Mr L Doolan Melbourne Grammar School 9 years; 1989–1998Dr P Swedosh The King David School 18 years; 1998–2015Western AustraliaDr N Hoffman WA Department of Education 3 years; 1980–1982Assoc Prof P Schultz University of Western Australia 14 years; 1983–1988, 1991–1994, 1996–1999Assoc Prof W Bloom Murdoch University 2 years; 1989–1990Dr E Stoyanova WA Department of Education 7 years; 1995, 2000–2005Dr G Gamble University of Western Australia 10 years; 2006–2015

EditorProf P J O’Halloran University of Canberra, ACT 1 year; 1983Dr A W Plank University of Southern Queensland 11 years; 1984–1994Dr A Storozhev Australian Mathematics Trust, ACT 15 years; 1994–2008Editorial ConsultantDr O Yevdokimov University of Southern Queensland 7 years; 2009–2015

Other Members of AMOC (showing organisations represented where applicable)Mr W J Atkins Australian Mathematics Foundation 18 years; 1995–2012Dr S Britton University of Sydney, NSW 8 years; 1990–1998Prof G Brown Australian Academy of Science, ACT 10 years; 1980, 1986–1994Dr R A Bryce Australian Mathematical Society, ACT 10 years; 1991–1998 Mathematics Challenge for Young Australians 13 years; 1999–2012Mr G Cristofani Department of Education and Training 2 years; 1993–1994Ms L Davis IBM Australia 4 years; 1991–1994Dr W Franzsen Australian Catholic University, ACT 9 years; 1990–1998Dr J Gani Australian Mathematical Society, ACT 1980Assoc Prof T Gagen ANU AAMT Summer School 6 years; 1993–1998Ms P Gould Department of Education and Training 2 years; 1995–1996Prof G M Kelly University of Sydney, NSW 6 years; 1982–1987Prof R B Mitchell University of Canberra, ACT 5 years; 1991–1995Ms Anna Nakos Mathematics Challenge for Young Australians 13 years; 2003–2015Mr S Neal Department of Education and Training 4 years; 1990–1993Prof M Newman Australian National University, ACT 15 years; 1986–1998 Mathematics Challenge for Young Australians 10 years; 1999–2002, (Treasurer during the interim) 2009–2014Prof R B Potts University of Adelaide, SA 1 year; 1980Mr H Reeves Australian Association of Maths Teachers 11 years; 1988–1998 Australian Mathematics Foundation 2014–2015Mr N Reid IBM Australia 3 years; 1988–1990Mr R Smith Telecom Australia 5 years; 1990–1994Prof P J Taylor Australian Mathematics Foundation 6 years; 1990–1994, 2013Prof N S Trudinger Australian Mathematical Society, ACT 3 years; 1986–1988Assoc Prof I F Vivian University of Canberra, ACT 1 year; 1990Dr M W White IBM Australia 9 years; 1980–1988

120

Associate Membership (inaugurated in 2000)Ms S Britton 16 years; 2000–2015Dr M Evans 16 years; 2000–2015Dr W Franzsen 16 years; 2000–2015Prof T Gagen 16 years; 2000–2015Mr H Reeves 16 years; 2000–2015Mr G Ball 16 years; 2000–2015

AMOC Senior Problems CommitteeCurrent membersDr N Do Monash University, VIC (Chair) 2 years; 2014–2015 (member) 11 years; 2003–2013M Clapper Australian Mathematics Trust 3 years; 2013-2015Dr A Di Pasquale University of Melbourne, VIC 15 years; 2001–2015Dr M Evans Australian Mathematical Sciences Institute, VIC 26 years; 1990–2015Dr I Guo University of Sydney, NSW 8 years; 2008–2015Dr J Kupka Monash University, VIC 13 years; 2003–2015Dr K McAvaney Deakin University, VIC 20 years; 1996–2015Dr D Mathews Monash University, VIC 15 years; 2001–2015Dr A Offer Queensland 4 years; 2012–2015Dr C Rao IBM Australia 16 years; 2000–2015Dr B B Saad Monash University, VIC 22 years; 1994–2015Dr J Simpson Curtin University, WA 17 years; 1999–2015Dr I Wanless Monash University, VIC 16 years; 2000–2015Previous membersMr G Ball University of Sydney, NSW 16 years; 1982–1997Mr M Brazil LaTrobe University, VIC 5 years; 1990–1994Dr M S Brooks University of Canberra, ACT 8 years; 1983–1990Dr G Carter Queensland University of Technology 10 years; 2001–2010Dr J Graham University of Sydney, NSW 1 year; 1992Dr M Herzberg Telecom Australia 1 year; 1990Assoc Prof D Hunt University of New South Wales 29 years; 1986–2014Dr L Kovacs Australian National University, ACT 5 years; 1981–1985Assoc Prof H Lausch Monash University, VIC (Chair) 27 years; 1987–2013 (member) 2 years; 2014-2015Dr D Paget University of Tasmania 7 years; 1989–1995Prof P Schultz University of Western Australia 8 years; 1993–2000 Dr L Stoyanov University of Western Australia 5 years; 2001–2005Dr E Strzelecki Monash University, VIC 5 years; 1986–1990Dr E Szekeres University of New South Wales 7 years; 1981–1987Prof G Szekeres University of New South Wales 7 years; 1981–1987Em Prof P J Taylor Australian Capital Territory 1 year; 2013Dr N H Williams University of Queensland 20 years; 1981–2000

Mathematics School of ExcellenceDr S Britton University of Sydney, NSW (Coordinator) 2 years; 1990–1991Mr L Doolan Melbourne Grammar, VIC (Coordinator) 6 years; 1992, 1993–1997Mr W Franzsen Australian Catholic University, ACT (Coordinator) 2 years; 1990–1991Dr D Paget University of Tasmania (Director) 5 years; 1990–1994Dr M Evans Scotch College, VIC 1 year; 1995Assoc Prof D Hunt University of New South Wales (Director) 4 years; 1996–1999Dr A Di Pasquale University of Melbourne, VIC (Director) 16 years; 2000–2015

121

International Mathematical Olympiad Selection SchoolMr J L Williams University of Sydney, NSW (Director) 2 years; 1982–1983Mr G Ball University of Sydney, NSW (Director) 6 years; 1984–1989Mr L Doolan Melbourne Grammar, VIC (Coordinator) 3 years; 1989–1991Dr S Britton University of Sydney, NSW (Coordinator) 7 years; 1992–1998Mr W Franzsen Australian Catholic University, ACT (Coordinator) 8 years; 1992–1996, 1999–2001Dr D Paget University of Tasmania (Director) 6 years; 1990–1995Assoc Prof D Hunt University of New South Wales (Director) 5 years; 1996–2000Dr A Di Pasquale University of Melbourne, VIC (Director) 15 years; 2001–2015

mathematics contests learn from yesterday, live for today, hope for tomorrow. the important thing is...

Documents