The Australian Scene 2018 · 2019-05-24 · Mathematics Contests The Australian Scene 2018 |...

Mathematics Contests The Australian Scene 2018

Combined MCYA and AMOC

A DI PASQUALE, N DO, KL MCAVANEY AND T SHAW

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AMT PUBLISHING

Australian Maths Trust170 Haydon DriveBruce ACT 2617

AUSTRALIATelephone: +61 2 6201 5136

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Copyright © 2019 Australian Mathematics Trust

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National Library of Australia Card Number and ISSN 1323-6490

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 Maths Trust.

Sponsors

The Olympiad programs are funded through the Australian Government’s National Innovation and Science Agenda.

The AMT’s EGMO initiative received grant funding from the Australian Government through the Department of Industry, Innovation and Science under the Inspiring Australia – Science Engagement Programme.

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 European Girls’ Mathematical Olympiad (EGMO), and the International Mathematical Olympiad (IMO).

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

The Olympiad programs are also supported by the Trust’s National Sponsor of the Australian Informatics and Mathematical Olympiads, Optiver.

Optiver is a market maker, offering trading opportunities on major global financial markets using their own capital at their own risk. They hire top talent at graduate and undergraduate levels with science, technology, engineering and mathematics (STEM) qualifications from the world’s leading universities. Optiver employs over 900 people across offices in the United States, Europe and Asia Pacific—including past Olympians and students that have been involved in AMT competitions and programs.

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 2018 EGMO team to Florence and the 2018 IMO team to Cluj–Napoca.

Mathematics Contests The Australian Scene 2018 | Support for the AMOC Training Program | i

ACKNOWLEDGEMENTS

The Australian Mathematical Olympiad Committee (AMOC) sincerely thanks all sponsors, teachers, mathematicians and others who have contributed in one way or another to the continued success of its activities. The editors sincerely thank those who have assisted in the compilation of this book, in particular the students who have provided solutions to the 2018 EGMO and the 2018 IMO. Thanks also to members of AMOC and Challenge Problems Committee, Chief Mathematician Mike Clapper, Chief Executive Officer Nathan Ford, staff of the Australian Maths Trust and others who are acknowledged elsewhere in the book.

Mathematics Contests The Australian Scene 2018 | Acknowledgements | ii

FROM THE AMT CHIEF EXECUTIVE OFFICER

In 2018, the Australian team continued our strong performance at the IMO, ranking 11th in the competition, with 2 Gold medals, 3 Silvers and a Bronze. Australia has placed 15th or higher at the IMO 4 times in the last six years.

Our inaugural European Girls’ Mathematical Olympiad team also performed well, placing 20th out of 52 countries with a Silver medal, two Bronze medals and an Honourable Mention. This was a wonderful result for our first attempt, and we are looking forward to continuing our participation in this great competition.

These results are a testament to the incredible work of our Australian Mathematical Olympiad Committee (AMOC), our Director of Training, our Team Leaders and Deputies, our AMOC State Directors, the Senior Problems Committee and the dozens of other volunteers and staff who work to support our training program.

On behalf of the Australian Maths Trust, I would like to thank in particular:

Emeritus Professor Cheryl Praeger AM, AMOC, Chair

Dr Angelo Di Pasquale, AMOC Director of Training and International Mathematical Olympiad (IMO), Team Leader

Mr Andrew Elvey Price, IMO Deputy Team Leader

Miss Thanom Shaw, EGMO Team Leader

Miss Michelle Chen, EGMO Deputy Team Leader

Mr Mike Clapper, AMT Chief Mathematician

Dr Norman Do, Chair, AMOC Senior Problems Committee

Dr Kevin McAvaney, Chair, MCYA Challenge Committee

Members of the AMT Board and AMOC Committee

AMOC tutors, mentors, volunteers and ex-Olympians

Nathan FordApril 2019

Mathematics Contests The Australian Scene 2018 | From the AMT Chief Executive Officer | iii

CONTENTS

Support for the Australian Mathematical Olympiad Committee Training Program i

Acknowledgements ii

From the AMT Chief Executive Officer iii

Background Notes on the EGMO, IMO and AMOC 1

Summary of Australia’s achievements at EGMO 2

Summary of Australia’s achievements at previous IMOs 2

Mathematics Challenge for Young Australians 4

Membership of MCYA Committees 6

Membership of AMOC 8

AMOC Timetable for Selection of the Teams to the 2018 EGMO and 2018 IMO 9

Activities of AMOC Senior Problems Committee 10

Challenge Problems – Middle Primary 11

Challenge Problems – Upper Primary 14

Challenge Problems – Junior 17

Challenge Problems – Intermediate 22

Challenge Solutions – Middle Primary 26

Challenge Solutions – Upper Primary 30

Challenge Solutions – Junior 34

Challenge Solutions – Intermediate 42

Challenge Statistics – Middle Primary 53

Challenge Statistics – Upper Primary 54

Challenge Statistics – Junior 55

Challenge Statistics – Intermediate 56

Australian Intermediate Mathematics Olympiad 57

Australian Intermediate Mathematics Olympiad Solutions 58

Australian Intermediate Mathematics Olympiad Statistics 69

AMOC Senior Contest 70

AMOC Senior Contest Solutions 71

AMOC Senior Contest Results 78

AMOC Senior Contest Statistics 80

AMOC School of Excellence 81

Participants at the 2017 AMOC School of Excellence 82

Australian Mathematical Olympiad 83

Australian Mathematical Olympiad Solutions 85

Australian Mathematical Olympiad Results 109

Australian Mathematical Olympiad Statistics 112

Mathematics Contests The Australian Scene 2018 | Contents | iv

30th Asian Pacific Mathematics Olympiad 113

30th Asian Pacific Mathematics Olympiad Solutions 115

30th Asian Pacific Mathematics Olympiad Results 124

AMOC Selection School 126

2018 Australian EGMO team 127

2018 Australian IMO team 127

Participants at the 2018 AMOC Selection School 128

EGMO Team Leader’s Report 129

European Girls’ Mathematical Olympiad 134

European Girls’ Mathematical Olympiad Solutions 136

IMO Team Preparation School 151

The Mathematics Ashes 152

The Mathematics Ashes Results 153

IMO Team Leader’s Report 154

International Mathematical Olympiad 161

International Mathematical Olympiad Solutions 163

Origin of Some Questions 181

AMOC Honour Roll 1979–2018 182

Interim Committee 1979–1980 182

Australian Mathematical Olympiad Committee 182

Mathematics Challenge for Young Australians 185

Mathematics Enrichment development 187

Australian Intermediate Mathematics Olympiad Committee 188

AMOC Senior Problems Committee 189

Mathematics School of Excellence 189

International Mathematical Olympiad Selection School 189

Mathematics Contests The Australian Scene 2018 | Contents | v

BACKGROUND NOTES ON THE EGMO, IMO AND AMOC

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 Mathematics Challenge for Young Australians (MCYA) program. The Maths Challenge is a problem-solving event held in the first or second term, in which approximately 15,000 young Australians explore carefully developed mathematical problems. Students who wish to continue to extend their mathematical experience can then participate in the Maths Enrichment and after that, they may attempt the Australian Intermediate Mathematics Olympiad (AIMO), which is the third stage of this program.

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 is based in Canberra with representatives on our board from a range of mathematical organisations. We are supported by a vast network of passionate volunteers from Australia and around the world, including academics, members of professional mathematics societies, educators and working mathematicians.

The AMOC has responsibility for the MCYA program, and many open and invitational mathematics competitions, culminating in the selection of a team of students who represent Australia internationally each year.

The AMOC schedule from August until July for potential IMO and EGMO team members

Each year many hundreds of gifted young Australian school students are identified using the results from the Australian Mathematics Competition, the Mathematics Challenge for Young Australians program and other smaller mathematics competitions. A network of thirty or so dedicated mathematicians and teachers has been organised to give these students support during the year, either by correspondence programs or by special teaching sessions run in each state. These programs are known to the respective AMOC State Directors.

These students are among others who sit the AIMO paper in September, or who are invited to sit the AMOC Senior Contest each August. The outstanding students in these contests, programs and other mathematical competitions are then identified. Forty-five of these are then invited to attend the residential AMOC School of Excellence, which is held each November.

In February, approximately 130 students are invited to attempt the Australian Mathematical Olympiad, after which the Australian team of four female students is selected for the European Girls’ Mathematical Olympiad (EGMO), held in April each year. The best 20 or so of these students are then invited to represent Australia in the correspondence Asian Pacific Mathematics Olympiad in March. About 45 students are chosen for the AMOC Selection School, including younger students who are also invited to this residential school. Six students plus one reserve are then selected for the International Mathematical Olympiad, held in July annually. A personalised support system for the Australian teams operates prior to EGMO and the IMO.

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

Mathematics Contests The Australian Scene 2018 | Background Notes on the EGMO, IMO and AMOC | 1

International Mathematical Olympiads (IMO and EGMO)

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 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-1960s 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, and has achieved varying successes including a spectacular perfect score in 2014 by Alexander Gunning.

Australia has participated in the European Girls’ Mathematical Olympiad since 2018.

In 2019 the IMO will be held in Bath, United Kingdom, while EGMO will be held in Kyiv, Ukraine.

Students must be enrolled full time in primary or secondary education and be under 20 years of age at the time of the EGMO or IMO. The Olympiad contest consists of two four-and-a-half hour papers, each with three questions.

Summary of Australia’s achievements at EGMO

Year City Gold Silver Bronze HM Rank

2018 Florence 1 2 1 20 out of 52 teams

Summary of Australia’s achievements at previous IMOs


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 Town 1* 3 2 11 out of 101 teams

2015 Chiang Mai 2 4 6 out of 104 teams

2016 Hong Kong 2 4 25 out of 109 teams

2017 Rio de Janeiro 3 2 1 34 out of 111 teams

2018 Cluj–Napoca 2 3 1 11 out of 109 teams

* Perfect Score by Alexander Gunning


MATHEMATICS CHALLENGE FOR YOUNG AUSTRALIANS

The Mathematics Challenge for Young Australians (MCYA) started on a national scale in 1992. It was set up to cater for the needs of the top 10 percent of secondary students in years 7–10, especially in country schools and schools where the number of students may be quite small. Teachers with a handful of talented students spread over a number of classes and working in isolation can find it very difficult to cater for the needs of these students. The MCYA provides materials and an organised structure designed to enable teachers to help talented students reach their potential. At the same time, teachers in larger schools, where there are more of these students, are able to use the materials to better assist the students in their care.

The aims of the Mathematics Challenge for Young Australians include:

• encouraging and fostering

- a greater interest in and awareness of the power of mathematics - a desire to succeed in solving interesting mathematical problems - the discovery of the joy of solving problems in mathematics

• identifying talented young Australians, recognising their achievements nationally and providing support that will enable them to reach their own levels of excellence

• providing teachers with

- interesting and accessible problems and solutions as well as detailed and motivating teaching discussion and extension materials

- comprehensive Australia-wide statistics of students’ achievements in the Maths Challenge.

There are three independent stages in the Mathematics Challenge for Young Australians:

• Challenge (three to four weeks during the period March–June)• Enrichment (12–16 weeks between April–September) • Australian Intermediate Mathematics Olympiad (September).

Challenge

The Challenge consists of four levels. Middle Primary (years 3–4) and Upper Primary (years 5–6) present students with four problems each to be attempted over three to four weeks, students are allowed to work on the problems in groups of up to three participants, but each must write their solutions individually. The Junior (years 7–8) and Intermediate (Years 9–10) levels present students with six problems to be attempted over three to four weeks, students are allowed to work on the problems with a partner but each must write their solutions individually.

There were 12,345 submissions (1538 Middle Primary, 3255 Upper Primary, 5288 Junior, 2264 Intermediate) for the Challenge in 2018. The 2018 problems and solutions, together with some statistics, appear later in this book.

Enrichment

This is a six-month program running from April to September, which consists of seven different parallel stages of comprehensive student and teacher support notes. Each student participates in only one of these stages.

The materials for all stages are designed to be a systematic structured course over a flexible 12–16 week period between April and September. This enables schools to timetable the program at convenient times during their school year.

Enrichment is completely independent of the earlier Challenge; however, they have the common feature of providing challenging mathematics problems for students, as well as accessible support materials for teachers.

Ramanujan (years 4–5) includes estimation, special numbers, counting techniques, fractions, clock arithmetic, ratio, colouring problems, and some problem-solving techniques. There were 279 entries in 2018.

Newton (years 5–6) includes polyominoes, fast arithmetic, polyhedra, pre-algebra concepts, patterns, divisibility and specific problem-solving techniques. There were 606 entries in 2018.

Mathematics Contests The Australian Scene 2018 | Mathematics Challenge for Young Australians | 4

Dirichlet (years 6–7) includes mathematics concerned with tessellations, arithmetic in other bases, time/distance/speed, patterns, recurring decimals and specific problem-solving techniques. There were 912 entries in 2018.

Euler (years 7–8) includes primes and composites, least common multiples, highest common factors, arithmetic sequences, figurate numbers, congruence, properties of angles and pigeonhole principle. There were 1338 entries in 2018.

Gauss (years 8–9) includes parallels, similarity, Pythagoras’ Theorem, using spreadsheets, Diophantine equations, counting techniques and congruence. Gauss builds on the Euler program. There were 753 entries in 2018.

Noether (top 10% years 9–10) includes expansion and factorisation, inequalities, sequences and series, number bases, methods of proof, congruence, circles and tangents. There were 434 entries in 2018.

Pólya (top 10% year 10) includes angle chasing, combinatorics, number theory, graph theory and symmetric polynomials. There were 210 entries in 2018.

Australian Intermediate Mathematics Olympiad

This four-hour competition for students up to year 10 offers a range of challenging and interesting questions. It is suitable for students who have performed well in the AMC (Distinction and above), and is designed as an endpoint for students who have completed the Gauss or Noether stage. There were 2077 entries for 2018 and 15 perfect scores.

Mathematics Contests The Australian Scene 2018 | Mathematics Challenge for Young Australians | 5

MEMBERSHIP OF MCYA COMMITTEES

2018 Mathematics Challenge for Young Australians Committee

Director Dr K McAvaney, Victoria

Challenge CommitteeMr M Clapper, Chief Mathematician, Australian Maths Trust, Australian Capital TerritoryMrs B Denney, New South WalesMr A Edwards, Queensland Studies AuthorityMr B Henry, Victoria Ms J McIntosh, AMSI, VictoriaMrs L Mottershead, New South WalesMs A Nakos, Temple Christian College, South AustraliaProf M Newman, Australian National University, Australian Capital TerritoryDr I Roberts, Northern TerritoryMiss T Shaw, SCEGGS, New South WalesMs K Sims, New South WalesDr S Thornton, Australian Capital TerritoryMs G Vardaro, Wesley College, VictoriaDr C Wetherell, Australian Capital Territory

Moderators Mr W Akhurst, New South WalesMr L Bao, VictoriaMr R Blackman, VictoriaMr A Canning, QueenslandDr E Casling, Australian Capital TerritoryMr B Darcy, Rose Park Primary School, South AustraliaMr J Dowsey, University of Melbourne, VictoriaMr S Ewington, Sydney Grammar SchoolMr S Gardiner, University of SydneyMs J Hartnett, QueenslandDr N Hoffman, Edith Cowan University, Western AustraliaDr T Kalinowski, University of NewcastleMr J Lawson, St Pius X School, New South WalesMs K McAsey, Penleigh and Essendon Grammar School, VictoriaMs T McNamara, VictoriaMr G Meiklejohn, Department of Education, QueenslandMr M O’Connor, AMSI, VictoriaMr J Oliver, Northern TerritoryMr A Peck, TasmaniaMr G Pointer, Marryatville High School, South AustraliaDr H Sims, VictoriaMs C Smith, QueenslandMs D Smith, New South WalesMs C Stanley, Queensland Studies AuthorityMs R Stone, South AustraliaDr M Sun, New South WalesMr P Swain, Victoria

Mathematics Contests The Australian Scene 2018 | Membership of MCYA Committees | 6

Dr P Swedosh, The King David School, VictoriaMs K Trudgian, QueenslandDr D Wells, USA

2018 Australian Intermediate Mathematics Olympiad Problems CommitteeDr K McAvaney, Victoria (Chair)Mr M Clapper, Chief Mathematician, Australian Maths Trust, Australian Capital TerritoryDr M Evans, International Centre of Excellence for Education in Mathematics, VictoriaMr B Henry, VictoriaDr D Mathews, Monash University, Victoria

Enrichment

EditorsMr G R Ball, University of Sydney, New South WalesMr M Clapper, Chief Mathematician, Australian Maths Trust, Australian Capital TerritoryDr M Evans, International Centre of Excellence for Education in Mathematics, VictoriaMr K Hamann, South AustraliaMr B Henry, VictoriaDr K McAvaney, VictoriaDr A M Storozhev, Attorney General’s Department, Australian Capital TerritoryProf P Taylor, Australian Capital TerritoryDr O Yevdokimov, University of Southern Queensland

Mathematics Contests The Australian Scene 2018 | Membership of MCYA Committees | 7

MEMBERSHIP OF AMOC

2018 Australian Mathematical Olympiad Committee

ChairProf C Praeger, University of Western Australia

Deputy ChairProf A Hassall, Australian National University, ACT

Chief Executive Officer Mr N Ford, Australian Maths Trust, ACT

Chief MathematicianMr M Clapper, Australian Maths Trust, ACT

TreasurerDr P Swedosh, The King David School, VIC

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

Chair, ChallengeDr K McAvaney, 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

EGMO Team LeaderMs T Shaw, SCEGGS Darlinghurst, NSW

State DirectorsDr K Dharmadasa, University of TasmaniaDr G Gamble, University of Western AustraliaMr Nhat Anh Hoang, WA DeputyDr I Roberts, Northern TerritoryDr D Badziahin, University of Sydney, NSWMr D Martin, South AustraliaDr A Offer, QueenslandDr P Swedosh, The King David School, VICDr C Wetherell, St Francis Xavier College, ACT

Editorial ConsultantDr O Yevdokimov, Queensland

RepresentativesMs J McIntosh, Challenge CommitteeMs A Nakos, Challenge CommitteeProf M Newman, Challenge Committee

Mathematics Contests The Australian Scene 2018 | Membership of AMOC | 8

AMOC TIMETABLE FOR SELECTION OF THE TEAMS TO THE 2018 EGMO AND 2018 IMO

August 2017—July 2018

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 45 students, including potential team members and other identified students, participating in a 10-day residential school in November/December.

Students are selected for EGMO after the AMO in February, while the IMO team is chosen at the conclusion of the Selection School in March.

Team members then receive individual coaching by mentors prior to the EGMO or the IMO.

Month Activity

August • Outstanding students are identified from AMC results, MCYA, other competitions and recommendations; and eligible students from previous training programs

• AMOC state organisers invite students to participate in AMOC programs• Various state-based programs• AMOC Senior Contest

September • Australian Intermediate Mathematics Olympiad

November/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

March • AMOC Selection School• Team preparation for 2018 EGMO in Florence, Italy

May–June • Personal Tutor Scheme for IMO team members

July • Short mathematics school for IMO team members 2018 IMO in Cluj–Napoca, Romania

Mathematics Contests The Australian Scene 2018 | AMOC Timetable for Selection of the Teams to the 2018 EGMO and IMO | 9

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 2017–September 2018Mr M Clapper, Chief Mathematician, Australian Maths Trust, ACTDr A Devillers, University of Western Australia, WADr A Di Pasquale, University of Melbourne, VICDr N Do, Monash University, VIC (Chair)Dr I Guo, University of Sydney, NSWDr J Kupka, Monash University, VICDr K McAvaney, Deakin University, VICDr D Mathews, Monash University, VICDr A Offer, QueenslandDr C Rao, NEC Australia, VICDr B Saad, Monash University, VICAssoc Prof J Simpson, Curtin University of Technology, WADr I Wanless, Monash University, VIC

2018 Australian Mathematical Olympiad

The Australian Mathematical Olympiad (AMO) consists of two papers of four questions each and was sat on 6 and 7 February. There were 121 participants including 13 from New Zealand. Eleven students, James Bang, Jack Gibney, William Han, Grace He, Charles Li, Haobin Liu, Jerry Mao, William Steinberg, Ethan Tan, Fengshuo Ye, and Guowen Zhang, achieved perfect scores and four other students were awarded Gold certificates, 16 students were awarded Silver certificates and 30 students were awarded Bronze certificates.

30th Asian Pacific Mathematics Olympiad

On 13–14 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 3 Silver, 4 Bronze and 3 Honourable Mention certificates awarded. Australia finished in 12th place overall.

2018 European Girls’ Mathematical Olympiad, Florence, Italy

The first Australian team attended the 7th EGMO in Florence, Italy. The EGMO consists of two papers of three questions worth seven points each. They were attempted by teams of four students from 52 countries on 11 and 12 April. Australia placed 20th in the final ranking, gaining one Silver and two Bronze medals and an Honourable Mention.

2018 International Mathematical Olympiad, Cluj–Napoca, Romania

The IMO consists of two papers of three questions worth seven points each. They were attempted by teams of six students from 108 countries on 9 and 10 July in Cluj–Napoca, Romania, with Australia placing 11th. The results for Australia were two Gold, three Silver and one Bronze medals.

2018 AMOC Senior Contest

Held on Tuesday 21 August, the Senior Contest was sat by 96 students (compared to 99 in 2017). There were nine students who obtained Gold certificates with Perfect Scores and two other students who also obtained Gold certificates. Thirteen students obtained Silver certificates and 24 students obtained Bronze certificates.

Mathematics Contests The Australian Scene 2018 | Activities of AMOC Senior Problems Committee | 10

2018 Challenge Problems - Middle Primary

Students may work on each of these four problems in groups of up to three, but must write their solutions individually.

MP1 Training

Ms Fitz sets up five cones 10 metres apart in a straight line for her class’s daily exercise drill. A student starts at anycone, runs to another cone, touches it, reverses direction, runs to another cone, touches it, reverses direction, and soon until each cone is touched exactly once including the starting and finishing cones.

• • • • •1 2 3 4 5

10m 10m 10m 10m

a If a student touches the cones in the order 1, 4, 3, 5, 2, how far does that student run?

b List all the exercise drills that start at cone 2 and finish at cone 4.

c List all the exercise drills that start at cone 1 and give their running distances.

d Find the shortest and longest running distances among all exercise drills.

MP2 Bitripents

Tia has a square grid and several of each of the tiles A, B, and C as shown below. The tiles consist of 2, 3, and 5 gridsquares as shown by the dotted lines. For this reason she refers to these tiles collectively as bitripents.

A B C

Tia places her tiles to cover different shapes on the grid with no gaps and no overlaps. She sometimes rotates or flipsthe tiles. For example, here is one way Tia could cover a 6× 3 rectangle using two A tiles, three B tiles, and one Ctile.

a Tia covered a 5 × 5 square on the grid using 10 tiles that included at least one A tile, at least one B tile, and atleast one C tile. Draw a diagram to show how she might have done this.

b Tia covered a 6 × 6 square on the grid using exactly three C tiles and eight other tiles. Draw a diagram to showhow she might have done this.

c Tia covered two rectangles of different shapes, each with three A tiles, three B tiles, and three C tiles. Draw twodiagrams to show how she might have done this.

1

CHALLENGE PROBLEMS – MIDDLE PRIMARY

Mathematics Contests The Australian Scene 2018 | Challenge Problems – Middle Primary | 11

MP3 Coloured Buckets

There are a number of buckets of different colours. There are also counters with different whole numbers written onthem. We want to put the numbered counters into the buckets according to the following rules:

• the lowest number must be in the red bucket

• no bucket has three numbers where one is the sum of the other two

• no bucket has two numbers where one is double the other.

For example, if counters numbered 1 to 7 were placed in a red and a blue bucket as shown below, then the blue bucketobeys the rules but the red bucket doesn’t (2× 1 = 2, 1 + 2 = 3, 2 + 3 = 5).

1 35

2

Red

4 6

7

Blue

We start with two empty buckets, one red and the other blue.

a Show how the numbers 1 to 4 can be placed in the two buckets following the rules.

b Explain why it is not possible to place the numbers 1 to 5 in the two buckets following the rules.

c Show that the numbers 2 to 9 can be placed in the two buckets following the rules.

d You now have three buckets available: red, blue, green. Find an arrangement for placing the numbers 1 to 12 inthese buckets with exactly four numbers in each bucket.

MP4 Hand-Sum

When the time is 10 past 9, the minute hand of an analog clock is pointing straight at the 2 and the hour hand hasmoved a bit past the 9. The sum of the numbers closest to which the hands are pointing gives what is called thehand-sum. At 10 past 9, the hand-sum is 2+9 = 11.

12

3

4567

8

9

1011 12

2Mathematics Contests The Australian Scene 2018 | Challenge Problems – Middle Primary | 12

When either hand is exactly halfway between two numbers the later number is used. For example, at 12:30 the minutehand is at 6 and the hour hand is exactly halfway between 12 and 1, so the hand-sum is 6 + 1 = 7.

12

3

4567

8

9

1011 12

At 12:22:30, the minute hand is exactly halfway between 4 and 5, and the hour hand is between 12 and 1 but closerto 12, so the hand-sum is 5 + 12 = 17.

12

3

4567

8

9

1011 12

a What is the hand-sum for a quarter to 10?

b What is the hand-sum at 29 minutes past 2? What is the hand-sum 4 minutes later?

c Give four different times when the hand-sum is 5. Each pair of these times must differ by more than 30 minutes.

Two times that are close together often have the same hand-sum. For example, the hand-sums at 1:46 and 1:47 areboth 2 + 9 = 11. But a minute later at 1:48, the hand-sum becomes 12 because the minute hand has moved past thehalfway mark between 9 and 10.

d Consider the hour starting at 3:00 and finishing at 4:00. At what times during this hour is the hand-sum 7?

3Mathematics Contests The Australian Scene 2018 | Challenge Problems – Middle Primary | 13

2018 Challenge Problems - Upper Primary

Students may work on each of these four problems in groups of up to three, but must write their solutions individually.

UP1 Bitripents

Tia has a square grid and several of each of the tiles A, B, and C as shown below. The tiles consist of 2, 3, and 5 gridsquares as shown by the dotted lines. For this reason she refers to these tiles collectively as bitripents.

A B C

Tia places her tiles to cover different shapes on the grid with no gaps and no overlaps. She sometimes rotates or flipsthe tiles. For example, here is one way Tia could cover a 5 × 4 rectangle using two A tiles, two B tiles, and two Ctiles.

a Draw a diagram to show how Tia could cover the smallest square possible using only B tiles.

b Tia uses at least one each of the A, B, C tiles to cover a rectangle whose area is 18 grid squares. Draw a diagramto show how she could do this.

c Tia covers two rectangles that both have perimeter 16 but have different dimensions. She uses at least one eachof the A, B, C tiles for each rectangle. Draw a diagram to show how she could do this. Explain why no otherrectangle of perimeter 16 is possible.

UP2 Insect Barriers

A caravan park is made up of square sites joined along their edges. For example, here is a caravan park with fivesquare sites:

4

CHALLENGE PROBLEMS – UPPER PRIMARY

Mathematics Contests The Australian Scene 2018 | Challenge Problems – Upper Primary | 14

Electronic transmitters provide protection from fire ants by sending signals along the edges of caravan sites. An edgeis protected by having a transmitter on at least one of its ends (called vertices), as shown below. Dots are transmitters,solid lines are protected edges, and dashed lines are unprotected edges.

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a Copy the diagram above and include enough extra transmitters to protect all edges.

b The caravan park owner thinks that if she relocates the transmitters, she can protect all edges with just 5 trans-mitters. Show how to do this.

c There are 8 caravan sites in an arrangement like this:

Show where to place just 8 transmitters so that all 24 edges are protected.

d There are 4 caravan sites in a 2× 2 arrangement like this:

Explain why 3 transmitters are not enough to protect all 12 edges in this arrangement.

UP3 Coloured Buckets

There are a number of buckets of different colours. There are also counters with different whole numbers written onthem. We want to put the numbered counters into the buckets according to the following rules:

• the lowest number must be in the red bucket

• no bucket has three numbers where one is the sum of the other two

• no bucket has two numbers where one is double the other.

5Mathematics Contests The Australian Scene 2018 | Challenge Problems – Upper Primary | 15

For example, if counters numbered 1 to 7 were placed in a red and a blue bucket as shown below, then the blue bucketobeys the rules but the red bucket doesn’t (2× 1 = 2, 1 + 2 = 3, 2 + 3 = 5).

1 35

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Red

4 6

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We start with two empty buckets, one red and the other blue.

a Show how the numbers 1 to 4 can be placed in the two buckets following the rules.

b Explain why it is not possible to place the numbers 1 to 5 in the two buckets following the rules.

c Show that the numbers 2 to 9 can be placed in the two buckets in only one way.

d You now have three buckets available: red, blue, green. Find an arrangement for placing the numbers 1 to 13 inthese buckets.

UP4 Isopentagons

An isopentagon is a pentagon with the following properties:

• the internal angle at one vertex is 60

• the internal angle at each of the other four vertices is 120

• the length of each side is a whole number of centimetres

• the two sides that meet at the 60 vertex have the same length.

An isopentagon can be drawn on 1 cm isometric grid paper so that each side at the 60 vertex passes through a gridpoint that is distance 1 cm from the vertex. We assume that all isopentagons are so drawn. Here is an example.

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a Draw all isopentagons for which the longest side length is 5 cm.

An isopentagon can be specified by listing its side lengths clockwise from the 60 vertex. For example, (3, 2, 1, 2, 3)specifies the isopentagon above.

The number of equilateral grid triangles of side length 1 cm that lie inside an isopentagon is called its t-number. Forexample, the t-number of the (3, 2, 1, 2, 3) isopentagon is 17.

b Find the t-number of the (5, 3, 2, 3, 5) isopentagon.

c Find all isopentagons that have a t-number less than 20 and explain why there are no others.

6Mathematics Contests The Australian Scene 2018 | Challenge Problems – Upper Primary | 16

2018 Challenge Problems - Junior

Students may work on each of these six problems with a partner but each must write their solutions individually.

J1 Insect Barriers

A caravan park is made up of square sites joined along their edges. For example, here is a caravan park with fivesquare sites:

Electronic transmitters provide protection from fire ants by sending signals along the edges of caravan sites. An edgeis protected by having a transmitter on at least one of its ends (called vertices), as shown below. Dots are transmitters,solid lines are protected edges, and dashed lines are unprotected edges.

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a The caravan park owner thinks that if she relocates some transmitters, she can protect all five caravan sites withjust 5 transmitters. Show how to do this.

b There are 9 caravan sites in a 3× 3 arrangement like this:

Show how to protect all 12 outer boundary edges with 6 transmitters and explain why 5 transmitters are not enoughto do this.

c Find the smallest number of transmitters needed to protect all edges of the park in Part b and explain your answer.

7

CHALLENGE PROBLEMS – JUNIOR

Mathematics Contests The Australian Scene 2018 | Challenge Problems – Junior | 17

d There are 12 caravan sites in a 3× 4 arrangement like this:

A caravan site is protected if all four of its edges are protected. The owner has only 8 transmitters. Find themaximum number of sites that can be protected and explain your answer.

J2 Cubic Coprimes

Two integers are coprimes if their only common divisor is 1. For example, 10 and 21 are coprime but 15 and 20 arenot. Liam labels the vertices of a cube with eight different positive integers so that the labels on each pair of adjacentvertices are coprime. This is called a Liam labelling. To make the edges of the cube easier to see, Liam uses thistwo-dimensional projection of the cube.

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a Find a Liam labelling using the integers 1 to 8.

b Find a Liam labelling using integers in the range 2 to 10.

c Explain why there is no Liam labelling using the integers 2 to 9.

d Find a Liam labelling in which only composite numbers are used, and the largest label is as small as possible.Justify your answer.

J3 Isopentagons






8Mathematics Contests The Australian Scene 2018 | Challenge Problems – Junior | 18


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a Draw all isopentagons for which the longest side length is 5 cm.



b Find the t-number of the (5, 3, 2, 3, 5) isopentagon.

c Find all isopentagons that have a t-number smaller than 20 and explain why there are no others.

d There are two different isopentagons with t-number 119. Describe these isopentagons and find their perimeters.

J4 Ts and Rs

We define two numerical operations labelled T and R.

The effect of T is to add 1 to a number. For example, if we apply the operation T to the number 2 three times in arow, we obtain 3, then 4, then 5.

The effect of R is to find the negative reciprocal of a number. For example, if we apply the operation R to 2 we obtain− 1

2 , and if we apply the operation R to − 32 we obtain 2

3 . Note that R can never be applied to the number 0.

The operations T and R can be combined. For example, we can turn 0 into 25 by successively applying the operations

T, T, T, R, T, T, R, T:

0T−→ 1

T−→ 2T−→ 3

R−→ −1

3

T−→ 2

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T−→ 5

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R−→ −3

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T−→ 2

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a Starting with 2, list the numbers produced by successively applying the operations T, R, R, T, R, T, R, T, R.

b Find a sequence of operations which turns 34 into 2

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c Find a sequence of operations which turns 3 into 0.

d Find a sequence of 20 operations that turns 7 into 0.

J5 Rhombus Rings

Bruce draws rings of rhombuses about a common centre point. All rhombuses have the same side length.

Rhombuses in the first, or inner, ring are all identical. Each rhombus has a vertex at the centre and each of its sidesthat meet at the centre is shared with another rhombus. They all have the same size angle at the centre. Figure 1shows a first ring with 7 rhombuses.

Each rhombus in the second ring has two adjacent sides each of which is shared with a rhombus in the first ring.Figure 2 shows the second ring when the first ring contains 7 rhombuses.

Bruce continues adding rings of rhombuses in the same way for as long as possible. Figure 3 shows the third ringwhen the first ring contains 7 rhombuses. In this example, since it is not possible to draw any new rhombuses thatshare edges with two rhombuses in the third ring, there are only three rings in this rhombus ring pattern.


Figure 1

Figure 2

Figure 3

a This diagram shows a rhombus ring pattern with 5 rhombuses in each ring. Find the size of the angles a and b.

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b In a rhombus ring pattern there are 8 rhombuses in the first ring. What are the angles in a rhombus in the lastring?

c In a rhombus ring pattern, each rhombus in the last ring has an angle of 20 and each rhombus in the second lastring has an angle of 60.

What are the angles in each rhombus in the third last ring?

How many rhombuses are there in the first ring, and how many rings are there in total?


J6 Parsecking

In the sport of parsecking, a player performs a routine and is judged by a scoring panel of 3 judges. A set of 12scorecards numbered 1 to 12 is distributed amongst the judges so that each judge has 4 cards. Some judges are moreimportant than others so some get some higher scorecards than others. After a routine, each judge holds up onescorecard. The player’s score is the sum of numbers on these three cards.

a In one competition:Judge A had scorecards 1, 2, 3, 4Judge B had scorecards 5, 6, 7, 8Judge C had scorecards 9, 10, 11, 12.

List all possible player scores.

b In another competition, amongst other scores, each score from 6 to 9 was possible. How were the scorecards 1 to 6distributed amongst the judges?

c Show one way the scorecards could be distributed among the judges if all player scores from 6 to 29, except 10,were possible.

d Show all ways the scorecards could be distributed among the judges if all player scores from 6 to 29, except 10, werepossible. Explain why there is no other way of distributing the scorecards except for swapping sets of 4 scorecardsbetween judges.


2018 Challenge Problems - Intermediate

Students may work on each of these six problems with a partner but each must write their solutions individually.

I1 Isopentagons







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a Draw all isopentagons for which the longest side length is 5 cm and find their t-numbers.

b Find all isopentagons that have a t-number smaller than 20 and explain why there are no others.

c There are two different isopentagons with t-number 119. Describe these isopentagons and find their perimeters.

d Find the smallest perimeter that is the same for three different isopentagons and the t-numbers for such isopentagons.

I2 Ts and Rs

We define two numerical operations labelled T and R.

The effect of T is to add 1 to a number. For example, if we apply the operation T to the number 2 three times in arow, we obtain 3, then 4, then 5.

The effect of R is to find the negative reciprocal of a number. For example, if we apply the operation R to 2 we obtain− 1

2 , and if we apply the operation R to − 32 we obtain 2

3 . Note that R can never be applied to the number 0.

The operations T and R can be combined. For example, we can turn 0 into 25 by successively applying the operations

T, T, T, R, T, T, R, T:

0T−→ 1

T−→ 2T−→ 3

R−→ −1

3

T−→ 2

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T−→ 5

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R−→ −3

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T−→ 2

5.

a Starting with 2, list the numbers produced by successively applying the operations T, R, R, T, R, T, R, T, R.

b Find a sequence of operations which turns 3 into 0.

c Notice that 0 can be turned into any positive integer n by applying n successive Ts. Explain how any positiveinteger n can be turned back into 0 by applying 3n− 1 operations.

d Explain how 0 can be turned into any negative integer.

12

CHALLENGE PROBLEMS – INTERMEDIATE

Mathematics Contests The Australian Scene 2018 | Challenge Problems – Intermediate | 22

I3 Rail Cams

Two horizontal rails have been fixed near the ceiling of a rectangular stadium so that cameras running along themcan film the action below. The stadium floor is 100m by 70m.

One rail runs the length of the stadium. It is parallel to, and halfway between, the east and west walls and 20m abovethe floor. The other rail runs the width of the stadium. It is parallel to, and 15m from, the south wall and 25m abovethe floor. The long rail carries camera A and the short rail carries camera B. Both cameras start at one end of theirrails, as shown.

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85m35m

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For the following questions, give your answer in exact surd form or to the nearest centimetre.

a Find the direct distance between cameras A and B at their starting positions.

The cameras start moving simultaneously along their rails at 5 metres per second.

b How much closer are the cameras after 2 seconds?

c Within the first 14 seconds, what is the minimum distance between the cameras as they move along their rails andat what time does this occur?

I4 Curvy Tiles

Tamara has several square tiles of the same size and all have the pattern below. Each of the two curves on the tile isa quadrant of a circle which has its centre at a tile corner and its radius equal to half a tile edge.

She creates various patterns by joining these tiles edge-to-edge, after rotating some of them if necessary.

Since each of the four tiles has two orientations, there are 24 = 16 ways of placing four of her tiles to form a 2 × 2square. However, Tamara notices that some of the 2 × 2 patterns are just rotations of each other. She regards theseas the same and finds that there are only six different 2× 2 patterns.

13Mathematics Contests The Australian Scene 2018 | Challenge Problems – Intermediate | 23

a Draw the six different 2× 2 patterns.

b For each 2× 2 pattern in Part a, calculate the probability it will appear if Tamara places the four tiles at random.

c This shape is called a peanut.

What is the probability that a random 3× 3 placement of tiles will contain a peanut?

I5 Rhombus Rings

Bruce draws rings of rhombuses about a common centre point. All rhombuses have the same side length.

Rhombuses in the first, or inner, ring are all identical. Each rhombus has a vertex at the centre and each of its sidesthat meet at the centre is shared with another rhombus. They all have the same size angle at the centre. Figure 1shows a first ring with 7 rhombuses.

Each rhombus in the second ring has two adjacent sides each of which is shared with a rhombus in the first ring.Figure 2 shows the second ring when the first ring contains 7 rhombuses.

Bruce continues adding rings of rhombuses in the same way for as long as possible. Figure 3 shows the third ringwhen the first ring contains 7 rhombuses. In this example, since it is not possible to draw any new rhombuses thatshare an edge with two rhombuses in the third ring, there are only three rings in this rhombus ring pattern.

Figure 1

Figure 2

Figure 3

a If a rhombus ring pattern has 8 rhombuses in the first ring, what are the angles in a rhombus in the last ring?

b In another rhombus ring pattern, each rhombus in the last ring has an angle of 20 and each rhombus in the secondlast ring has an angle of 60.

How many rhombuses are in each ring, and how many rings are there in this rhombus ring pattern?

c A rhombus ring pattern has r rings and the rhombuses in its first ring have central angle a. What are the anglesin a rhombus in the rth ring?

d A rhombus ring pattern has 7 rings in total. How many rhombuses could there be in each ring?


I6 Higher or Lower

Two contestants are playing a game called Higher or Lower. A prize is hidden inside one of several boxes placed in arow. The boxes are numbered from left to right 1, 2, 3, . . . . All boxes are equally likely to contain the prize.

The players take turns to guess where the prize is hidden. If a guess is correct, then the host says ‘correct’ and thatplayer wins. If a guess is incorrect, the host, who knows where the prize is hidden, then says ‘higher’ or ‘lower’ tofaithfully indicate in which direction the prize box is located. Both players always follow the host’s directions. Forexample, when there are 10 boxes numbered 1 to 10, with the prize in box 7, the game could progress like this:

Player 1: 3Host: HigherPlayer 2: 5Host: HigherPlayer 1: 8Host: LowerPlayer 2: 7Host: Correct. Player 2 wins.

a A game has three boxes numbered 1, 2, 3. Show that if Player 1 chooses box 2 on the first turn, her chance ofwinning is 1

3 , whereas if she chooses box 1 her chance of winning is 23 .

b A game has boxes numbered 1 to 4. Show that no matter which box Player 1 chooses on the first turn, Player 2has a strategy for ensuring his chance of winning is 1

2 .

c A game has boxes numbered 1 to 9. Show that no matter which boxes Player 2 chooses, Player 1 has a strategywhich involves choosing box 5 at the first turn and ensures her chance of winning is more than 1

2 .


2018 Challenge Solutions - Middle Primary

MP1 Training

a The following diagram shows the student’s path:

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10m 10m 10m 10m

The total distance is 30 + 10 + 20 + 30 = 90m.

b A run that starts at cone 2 and finishes at cone 4 must be one of the following: 21354, 21534, 23154, 23514, 25134,25314. Of these, the only legitimate runs are: 21534, 23154. In all other cases, the student does not turn at cone 3.

c A run that starts at cone 1 has its first turn at cone 3, 4, or 5.

If the first turn is at cone 3, then the second turn must be at cone 2 and the third turn at cone 5.

If the first turn is at cone 4, then the second turn must be at cone 2 or 3 and the third turn at cone 5.

If the first turn is at cone 5, then the second turn must be at cone 2 or 3 and the third turn at cone 4.

So the only runs that start at cone 1 are:13254, 14253, 14352, 15243, 15342.

Their total distances are respectively:70m, 100m, 90m, 100m, 90m.

d Alternative i

An exercise drill must start at one of the five cones. Part c gives all drills that start at cone 1 together with theirrunning distances.

Using the method in Part c, the only runs starting at cone 2 are:21435, 21534, 23154, 24153, 24351, 25143, 25341.

Their total distances are respectively:70m, 80m, 80m, 110m, 90m, 110m, 90m.

Similarly, the only runs starting at cone 3 are:31425, 31524, 32415, 32514, 34152, 34251, 35142, 35241.

Their total distances are respectively:100m, 110m, 100m, 110m, 110m, 100m, 110m, 100m.

The runs starting at cone 4 are the reflections about cone 3 of the runs that start at cone 2. For example, 45231 isthe reflection of 21435.

The runs starting at cone 5 are the reflections about cone 3 of the runs that start at cone 1. For example, 53412 isthe reflection of 13254.

Hence the shortest run is 70m and the longest run is 110m.

Alternative ii

Label the line segments between the cones A, B, C, D.

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A B C D

Each exercise drill can be represented not only by a sequence of cones but also a sequence of these line segments.We can combine these sequences to make this clearer. For example, the run which has the cone sequence 23154 alsohas the segment sequence BBAABCDD, and the combined sequence is 2B3BA1ABCD5D4.

For each segment we keep a tally of the number of times the student runs along it. There is one line segment in therun immediately before each cone except for the starting cone. So the total tally of segment runs is at least 4 foreach run.

The student turns at three of the cones. At each such cone, the segment immediately before the cone in the run isrepeated immediately after the turn. So, for each run, the total tally of segment runs is at least 4 + 3 = 7. Thismeans the total distance for each exercise drill is at least 70m. Since, for example, the run 21435 has total distance70m, the shortest running distance is 70m.

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CHALLENGE SOLUTIONS – MIDDLE PRIMARY

Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Middle Primary | 26

We now find the longest running distance. Consider any one of the cones 2, 3, 4. In any run, the student musteither start, stop, or turn at the cone. In addition the student may pass the cone as well. In the next diagram, thecone is drawn a number of times. Each time, one or both of its attached line segments are also drawn to indicateeach time the student runs on them.

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start or stop

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turn

We can see from the diagram that the tally of runs on the segments either side of the cone will differ by either 1 or2. The tally for each of segments A and D is at most 2. So the tally for each of the segments B and C is at most4. The tallies for B and C must differ by 1 or 2. So the total tally of segment runs is at most 2 + 2 + 4 + 3 = 11.This means the total distance for each exercise drill is at most 110m. Since, for example, the run 25143 has totaldistance 110m, the longest running distance is 110m.

MP2 Bitripents

a A total of 25 grid squares need to be covered. The number of grid squares covered by one A, one B, and one C tileis 2 + 3 + 5 = 10. So the remaining 15 grid squares must be covered by seven tiles. This could be done with six Atiles and one B tile, as shown. There are other ways.

b A total of 36 grid squares need to be covered. The number of grid squares covered by three C tiles is 3 × 5 = 15.So the remaining 21 grid squares must be covered by eight tiles, none of which is a C tile. This could be done withthree A tiles and five B tiles, as shown. There are other ways.

17Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Middle Primary | 27

c The number of grid squares covered by three A tiles, three B tiles, and three C tiles is 3 × (2 + 3 + 5) = 30. Arectangle consisting of 30 grid squares must be 1× 30, 2× 15, 3× 10, or 5× 6. Since 1× 30 and 2× 15 rectanglescannot accommodate a C tile, the two rectangles we require are 3× 10 and 5× 6. Here is one way to cover each ofthese rectangles. There are other ways.

MP3 Coloured Buckets

a Red bucket: 1, 4Blue bucket: 2, 3

(This is the only arrangement.)

b As stipulated, 1 must go in the red bucket. Then 2 must go in the blue bucket since 2 is the double of 1. Then 4must go in the red bucket since 4 is the double of 2. Then 3 must go in the blue bucket since 1 + 3 = 4. However,5 = 1 + 4 and 5 = 2 + 3, so 5 cannot be placed in either bucket.

c Red bucket: 2, 3, 8, 9Blue bucket: 4, 5, 6, 7


d Here is one arrangement. There are many others.

Red bucket: 1, 5, 8, 12Blue bucket: 2, 6, 7, 11Green bucket: 3, 4, 9, 10

MP4 Hand-Sum

a At quarter to 10, the minute hand points directly at 9 and the hour hand is between 9 and 10 but closer to 10. Sothe hand-sum at quarter to 10 is 9 + 10 = 19.

b At 29 minutes past 2, the minute hand is between 5 and 6 but closer to 6, while the hour hand is between 2 and 3but closer to 2. So the hand-sum at 29 minutes past 2 is 6 + 2 = 8.

At 33 minutes past 2, the minute hand is between 6 and 7 but closer to 7, while the hour hand is between 2 and 3but closer to 3. So the hand-sum at 33 minutes past 2 is 7 + 3 = 10.

c We first note that 5 = 1 + 4 = 2 + 3 = 3 + 2 = 4 + 1. This gives us a clue as to which times to look for.The sum 1 + 4 suggests 4:05 (or any time from 4:02:30 and before 4:07:30).The sum 2 + 3 suggests 3:10 (or any time from 3:07:30 and before 3:12:30).The sum 3 + 2 suggests 2:15 (or any time from 2:12:30 and before 2:17:30).The sum 4 + 1 suggests 1:20 (or any time from 1:17:30 and before 1:22:30).Each pair of these times differ by more than 30 minutes.


d Alternative i

From 3:00 and before 3:02:30 the hand-sum is 12 + 3 = 15.From 3:02:30 and before 3:07:30 the hand-sum is 1 + 3 = 4.From 3:07:30 and before 3:12:30 the hand-sum is 2 + 3 = 5.From 3:12:30 and before 3:17:30 the hand-sum is 3 + 3 = 6.From 3:17:30 and before 3:22:30 the hand-sum is 4 + 3 = 7.From 3:22:30 up to 4:00 the hand-sum is greater than 7.So between 3:00 and 4:00 the hand-sum is 7 only at the times from 3:17:30 and before 3:22:30.

Alternative ii

We first note that 7 = 1 + 6 = 2 + 5 = 3 + 4 = 4 + 3 = 5 + 2 = 6 + 1. Since the hour hand is at 3 or 4 or betweenthem, the number allocated to the hour hand is either 3 or 4. If the number allocated to the hour hand is 4, thenthe minute hand is anywhere from 6 to 12 and the hand-sum is therefore greater than 7. If the number allocatedto the hour hand is 3, then the number allocated to the minute hand is 4. So between 3:00 and 4:00 the hand-sumis 7 only at the times from 3:17:30 and before 3:22:30.


2018 Challenge Solutions - Upper Primary

UP1 Bitripents

a The smallest square that can be covered using only B tiles is 6× 6. Here is one such square.

Comment

No square smaller than 6× 6 is possible. Each B tile has 3 grid squares. So the number of grid squares in a squarecovered by only B tiles is a multiple of 3. Hence the covered square must be one of 3× 3, 6× 6, 9× 9 etc.

If the covered square is 3× 3, then there are essentially only two ways to cover its centre grid square.

In both cases, the top-left grid square cannot be covered by a B tile. So the covered square must be at least 6× 6.

b Since the rectangle covering includes a C tile, it must have all side lengths at least 3. So, to have area 18, therequired rectangle must be 3× 6. Here is such a rectangle with a required covering.

c If a rectangle covering includes a C tile, then it must have all side lengths at least 3. So, to have perimeter 16, therectangles that we require could only be 3× 5 or 4× 4.

Here are two such rectangles with a required covering.

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CHALLENGE SOLUTIONS – UPPER PRIMARY

Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Upper Primary | 30

UP2 Insect Barriers

a Four extra transmitters are required. Here is one way to place them.

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One vertex has 4 edges attached to it. All other vertices have fewer edges attached. So the number of edges thatcould be protected with 3 transmitters is at most 4 + 3 + 3 = 10. But there are 12 edges to be protected. Hence 3transmitters are not enough.

Alternative ii

Each row of horizontal edges requires at least one transmitter to protect its two edges. If a row has only onetransmitter, then the transmitter must be at the middle vertex. Since there are three rows of horizontal edges, ifthere are only three transmitters, then they are all at the row centres. Then none of the side edges are protected.Hence 3 transmitters are not enough.

Alternative iii

Each boundary edge must have a transmitter on at least one of its vertices. There are 8 boundary edges, so at least4 transmitters are needed to protect all 8 edges. Hence 3 transmitters are not enough.

21Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Upper Primary | 31

UP3 Coloured Buckets

a Red bucket: 1, 4Blue bucket: 2, 3


b As stipulated, 1 must go in the red bucket. Then 2 must go in the blue bucket since 2 is the double of 1. Then 4must go in the red bucket since 4 is the double of 2. Then 3 must go in the blue bucket since 1 + 3 = 4. However,5 = 1 + 4 and 5 = 2 + 3, so 5 cannot be placed in either bucket.

c As stipulated, 2 must go in the red bucket. Then 4 (= 2 × 2) must be in the blue bucket, 8 (= 2 × 4) must be inthe red bucket, and 6 must be in the blue bucket (8 = 2 + 6).Red bucket: 2, 8Blue bucket: 4, 6

Then 3 must be in the red bucket (6 = 2× 3), 5 must be in the blue bucket (5 = 2+3), 9 must be in the red bucket(9 = 4 + 5), and 7 must be in the blue bucket (9 = 2 + 7).Red bucket: 2, 3, 8, 9Blue bucket: 4, 5, 6, 7

d A useful strategy is to start with a run of consecutive integers as long as possible in one bucket. For example:

Red bucket:Blue bucket:Green bucket: 5, 6, 7, 8, 9

Then 1 is in red, 2 in blue, 4 in red, 3 in blue:

Red bucket: 1, 4Blue bucket: 2, 3Green bucket: 5, 6, 7, 8, 9

If 10 is in blue, then 12 and 13 must both be in red, which is not allowed. So 10 must be in red, 11 in blue, 13 inred, and 12 in blue.

Red bucket: 1, 4, 10, 13Blue bucket: 2, 3, 11, 12Green bucket: 5, 6, 7, 8, 9

There are only two other arrangements:

Red bucket: 1, 4, 7, 10, 13 1, 4, 10, 13Blue bucket: 2, 3, 11, 12 2, 3, 7, 11, 12Green bucket: 5, 6, 8, 9 5, 6, 8, 9

Any one of the three arrangements is acceptable.


UP4 Isopentagons

a The longest sides of an isopentagon are those meeting at the 60 vertex. There are 4 isopentagons with longest side5 cm.

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b The (5, 3, 2, 3, 5) isopentagon is the bottom-left pentagon in the solution of Part a. By carefully counting theinternal triangles, we see that its t-number is 46.

c There are no isopentagons with longest side 1 cm.

There is only one isopentagon with longest side 2 cm, (2, 1, 1, 1, 2), which has t-number 7.

There are two isopentagons with longest side 3 cm, (3, 1, 2, 1, 3) and (3, 2, 1, 2, 3), which have t-numbers 14 and17.

If the longest side of an isopentagon is 4 cm or more, then it contains an equilateral triangle of side length 4 cm,which in turn contains 16 grid triangles. The isopentagon also contains a trapezium which has at least another 7grid triangles, a total of at least 23.

So the only isopentagons with t-number less than 20 are:(2, 1, 1, 1, 2) with t-number 7,(3, 1, 2, 1, 3) with t-number 14,(3, 2, 1, 2, 3) with t-number 17.


2018 Challenge Solutions - Junior

J1 Insect Barriers

a

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b A single transmitter can protect at most two edges on the boundary. So at least 6 transmitters are needed toprotect all 12 edges on the boundary. Here is one way to do that.

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c Here is one way to protect all edges with 8 transmitters.

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We now show that at least 8 transmitters are always needed.

Alternative i

From Part b, six transmitters are needed to protect all boundary edges. They must all be on the boundary sinceno transmitter on an internal vertex can protect a boundary edge. So none of these can protect any of the edgeson the central square. At least two more transmitters are needed to do that. So at least 8 transmitters are neededto protect all edges.

Alternative ii

There are 4 rows of horizontal edges. Each row requires at least two transmitters to protect all three of its edges,and none of these transmitters protect an edge in any other row. So at least 8 transmitters are needed to protectall edges.

Alternative iii

Each of the corner squares requires at least 2 transmitters to protect its 4 edges, and none of these transmittersprotect an edge in any other corner square. So at least 8 transmitters are needed to protect all edges.

So the smallest number of transmitters needed to protect all edges is 8.

24

CHALLENGE SOLUTIONS – JUNIOR

Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Junior | 34

d If we had 10 transmitters, then all sites could be protected if we placed them this way.

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If we removed the two transmitters in the top corners, 10 sites would still be protected.

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Could we relocate the eight transmitters so that more than 10 sites are protected? We show that this is impossible.

Alternative i

Suppose 11 or more sites are protected. Then at least one of the central sites is protected.

Suppose just one of the central sites is protected. This requires at least 2 transmitters, none of which can protectany boundary edge. All boundary sites and therefore the 14 boundary edges must be protected. This requires atleast 7 transmitters, a total of at least 9 transmitters.

So both central sites must be protected. This requires at least 3 transmitters, none of which can protect anyboundary edge. Then there are at most 5 transmitters remaining to protect the boundary edges. Since 5 transmitterscan protect at most 10 boundary edges, at least 4 boundary edges must be unprotected. Therefore at least twoboundary sites are unprotected and we are left with at most 10 protected sites.

So the maximum number of sites that can be protected by 8 transmitters is 10.

Alternative ii

Suppose 11 or more sites are protected, that is, at most one site is unprotected. Then all unprotected edges mustbelong to one site and are therefore on the boundary. Since there are at most two boundary edges in a site, thenumber of protected edges is at least 31− 2 = 29.

A single transmitter can protect at most 4 edges. So to protect 29 or more edges with 8 transmitters, we need atleast 5 transmitters each protecting 4 distinct edges. The only vertices at which a transmitter can protect 4 edgesare the interior (non-boundary) vertices. Since no such transmitter can protect a boundary edge, we have at most3 transmitters to protect at least 12 boundary edges. This is impossible, since a single transmitter can protect atmost 2 edges on the boundary.

So the maximum number of sites that can be protected by 8 transmitters is 10.

25Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Junior | 35

J2 Cubic Coprimes

a Notice in particular that no two even numbers can be adjacent. Here is one possible labelling. There are others.

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1 6

78

4 5

23

b Here is one possible labelling. There are others.

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9 8

710

4 5

23

c There are only 8 integers from 2 to 9, so all of them must be used. The only labels that can be adjacent to 6 are5 and 7. Since each vertex has three other vertices adjacent to it, there are not enough integers to label all thevertices adjacent to 6. So there is no Liam labelling with labels 2 to 9.

d At most 4 vertices have an even label otherwise there would be at least 3 even labels on the left face of the cube or3 on the right face. Either way, this would mean a pair of adjacent vertices both have even vertices, breaking thecoprime rule. So a Liam labelling with composites requires at least 4 composite odd integers. The 4 smallest oddcomposites are 9, 15, 21, 25. Hence the largest label is at least 25.

Here is a Liam labelling with all labels composite and the largest label 25.

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4 21

229

15 8

2514

So 25 is indeed the least largest label for a Liam labelling with all labels composite.

There are other possible labellings. For example, replacing any even number with 16.


J3 Isopentagons


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b The (5, 3, 2, 3, 5) isopentagon is the bottom-left pentagon in the solution of Part a. By carefully counting theinternal triangles, we see that its t-number is 46.

c There are no isopentagons with longest side 1 cm.





d One way to look at an isopentagon is to see it as two identical large equilateral triangles with side lengths a minusa smaller equilateral triangle with side length b, as shown. The smallest value of a is 2, and the largest value of bis a− 1.

=b

a


By direct counting we get the following table for the number of grid triangles inside an equilateral triangle withside length s.

s 1 2 3 4 5 6 7 8 9 10 11t 1 4 9 16 25 36 49 64 81 100 121

Notice that t = s× s for each value of s.

The t-number of an isopentagon is more than a × a and less than 2 × a × a. If an isopentagon has t-number 119,then a is at most 10 and at least 8.

We are looking for values of a and b so that

2× a× a− b× b = 119.

Since 2× a× a is even and 119 is odd, b must be odd.

The next table shows t-numbers of all isopentagons with odd b for a = 8, 9, 10.

a b t a b t

8 1 64 + 64− 1 = 127 10 1 100 + 100− 1 = 1998 3 64 + 64− 9 = 119 10 3 100 + 100− 9 = 1918 5 64 + 64− 25 = 103 10 5 100 + 100− 25 = 1758 7 64 + 64− 49 = 79 10 7 100 + 100− 49 = 1519 1 81 + 81− 1 = 161 10 9 100 + 100− 81 = 1199 3 81 + 81− 9 = 1539 5 81 + 81− 25 = 1379 7 81 + 81− 49 = 113

So the only isopentagons with t-number 119 are:(8, 5, 3, 5, 8) which has perimeter 29,(10, 1, 9, 1, 10) which has perimeter 31.

J4 Ts and Rs

a

2T−→ 3

R−→ −1

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R−→ 3T−→ 4

R−→ −1

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T−→ 3

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R−→ −4

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T−→ −1

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R−→ 3

b3

4

R−→ −4

3

T−→ −1

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T−→ 2

3

There are longer, less efficient, sequences.

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T−→ 2

3

R−→ −3

2

T−→ −1

2

T−→ 1

2

R−→ −2T−→ −1

T−→ 0


d Firstly we operate on smaller positive integers to see if there is a pattern. We stop each sequence when we reach anumber that has occurred in the previous sequence.

1R−→ −1

T−→ 0

2R−→ −1

2

T−→ 1

2

R−→ −2T−→ −1

3R−→ −1

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T−→ 2

3

R−→ −3

2

T−→ −1

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4R−→ −1

4

T−→ 3

4

R−→ −4

3

T−→ −1

3

Note that, from the second sequence, the last number in each of these sequences is the second number in the previoussequence. This suggests the following sequence of operations for changing 7 into 0.


7R−→ −1

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T−→ 6

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R−→ −7

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T−→ −1

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T−→ 5

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R−→ −6

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T−→ −1

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T−→ 4

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R−→ −5

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T−→ −1

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T−→ 3

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R−→ −4

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T−→ −1

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T−→ 2

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R−→ −3

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T−→ −1

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T−→ 1

2

R−→ −2T−→ −1

T−→ 0

There are 20 operations in this sequence, as required.

J5 Rhombus Rings

a Since opposite angles in a rhombus are equal and the rhombuses in the first ring are identical and have the sameangle at the centre, we have these angles.

cc

ccc

aa

bb

Since the sum of angles at a point is 360, c = 360/5 = 72.

Since the sum of adjacent angles in a rhombus is 180, a = 180− c = 108.

Since the sum of angles at a point is 360, b = 360− 2× a = 360− 216 = 144.

b Each central angle is 360/8 = 45. Starting with those, we calculate the remaining angles in degrees as we moveoutwards.

45

45

45

45

45

45

135

135

135

135

90 90

90

90

90

90

A

B

C

Since ABC is a straight line, the third ring is the last (outer) ring. Thus the angles in a rhombus in the outer ringare 45 and 135.


c We have these angles (all in degrees).

160

1602020

6060 x

Thus x = 360 − 160 − 60 − 60 = 80. Hence the angles in each rhombus of the third last ring are 80 and180 − 80 = 100.

Extending the diagram above and inserting a few more angles, we get:

160

1602020

606080

100100

120

y

Thus y = 360− 120− 100− 100 = 40. Hence there is a fourth last ring of rhombuses. Extending the diagram andadding a few more angles gives:

160

1602020

606080

100100

120

40

40

140140

80

Since 140 + 140 + 80 = 360, no more rings are possible. So there are 4 rings in total and the number of rhombusesin the inner ring is 360/40 = 9.

J6 Parsecking

a The minimum score is 1 + 5 + 9 = 15 and the maximum score is 4 + 8 + 12 = 24. All scores between are alsopossible. For example:

1 + 5 + 9 = 15 1 + 6 + 12 = 19 2 + 8 + 12 = 221 + 5 + 10 = 16 1 + 7 + 12 = 20 3 + 8 + 12 = 231 + 5 + 11 = 17 1 + 8 + 12 = 21 4 + 8 + 12 = 241 + 5 + 12 = 18


b The only way to express 6 as the sum of three different numbers is 6 = 3 + 2 + 1.So scorecards 1, 2, 3 are with separate judges. (1)

The only way to express 7 as the sum of three different numbers is 7 = 4 + 2 + 1.So scorecards 1, 2, 4 are with separate judges. (2)

The only ways to express 8 as the sum of three different numbers are 8 = 5 + 2 + 1 = 4 + 3 + 1.From (1) and (2), 4 + 3 + 1 is impossible.So scorecards 1, 2, 5 are with separate judges. (3)

The only ways to express 9 as the sum of three different numbers are 9 = 6 + 2 + 1 = 5 + 3 + 1 = 4 + 3 + 2.From (1) and (3), 5 + 3 + 1 is impossible.From (1) and (2), 4 + 3 + 2 is impossible.So scorecards 1, 2, 6 are with separate judges. (4)

From (1), (2), (3), (4), scorecards 3, 4, 5, 6 are with the same judge and scorecards 1 and 2 are held separately bythe other two judges.

c Any one of the following six distributions.

A 1 9 10 11 1 8 10 11 1 8 9 11B 2 7 8 12 2 7 9 12 2 7 10 12C 3 4 5 6 3 4 5 6 3 4 5 6

A 1 9 10 12 1 8 10 12 1 8 9 12B 2 7 8 11 2 7 9 11 2 7 10 11C 3 4 5 6 3 4 5 6 3 4 5 6

d As in Part b, there is only one distribution of scorecards that gives scores 6 to 9:

A 1B 2C 3 4 5 6

If Judge A has scorecard 7, then score 11 is impossible. So Judge B has scorecard 7.

To achieve score 29, we must place scorecards 11 and 12 with Judges A and B separately. So we have twodistributions.

A 1 11 1 12B 2 7 12 2 7 11C 3 4 5 6 3 4 5 6

In each case, one of the scorecards 8, 9, 10 must be placed with Judge B. So we have six distributions.

A 1 9 10 11 1 8 10 11 1 8 9 11B 2 7 8 12 2 7 9 12 2 7 10 12C 3 4 5 6 3 4 5 6 3 4 5 6

A 1 9 10 12 1 8 10 12 1 8 9 12B 2 7 8 11 2 7 9 11 2 7 10 11C 3 4 5 6 3 4 5 6 3 4 5 6

In each case every score from 6 to 29, except 10, is possible.


2018 Challenge Solutions - Intermediate

I1 Isopentagons


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•t = 34 t = 41

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•t = 46 t = 49

b There are no isopentagons with longest side 1 cm.





c One way to look at an isopentagon is to see it as two identical large equilateral triangles with side lengths a minusa smaller equilateral triangle with side length b, as shown. The smallest value of a is 2, and the largest value of bis a− 1.

32

CHALLENGE SOLUTIONS – INTERMEDIATE

Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Intermediate | 42

=b

a

By direct counting we get the following table for the number of grid triangles inside an equilateral triangle withside length s.

s 1 2 3 4 5 6 7 8 9 10 11t 1 4 9 16 25 36 49 64 81 100 121

Notice that t = s× s for each value of s.

The t-number of an isopentagon is more than a × a and less than 2 × a × a. If an isopentagon has t-number 119,then a is at most 10 and at least 8.

We are looking for values of a and b so that

2× a× a− b× b = 119.

Since 2× a× a is even and 119 is odd, b must be odd.

The next table shows t-numbers of all isopentagons with odd b for a = 8, 9, 10.

a b t a b t

8 1 64 + 64− 1 = 127 10 1 100 + 100− 1 = 1998 3 64 + 64− 9 = 119 10 3 100 + 100− 9 = 1918 5 64 + 64− 25 = 103 10 5 100 + 100− 25 = 1758 7 64 + 64− 49 = 79 10 7 100 + 100− 49 = 1519 1 81 + 81− 1 = 161 10 9 100 + 100− 81 = 1199 3 81 + 81− 9 = 1539 5 81 + 81− 25 = 1379 7 81 + 81− 49 = 113

So the only isopentagons with t-number 119 are:(8, 5, 3, 5, 8) which has perimeter 29,(10, 1, 9, 1, 10) which has perimeter 31.

d Let a and b be the side lengths of an isopentagon as shown in the solution to Part c. Then the perimeter of theisopentagon is 4× a− b.

The following table gives values of 4× a− b for small values of a and b.

b\a 2 3 4 5 6 7 8 9 10

1 7 11 15 19 23 27 31 35 392 10 14 18 22 26 30 34 383 13 17 21 25 29 33 374 16 20 24 28 32 365 19 23 27 31 356 22 26 30 347 25 29 338 28 329 31

From the table, the smallest common perimeter for three isopentagons is 31.The isopentagons are (8, 7, 1, 7, 8), (9, 4, 5, 4, 9), (10, 1, 9, 1, 10).Their t-numbers are 127, 137, 119.

33Mathematics Contests The Australian Scene 2018 | Challenge Solutions – Intermediate | 43

I2 Ts and Rs

a

2T−→ 3

R−→ −1

3

R−→ 3T−→ 4

R−→ −1

4

T−→ 3

4

R−→ −4

3

T−→ −1

3

R−→ 3

b

3R−→ −1

3

T−→ 2

3

R−→ −3

2

T−→ −1

2

T−→ 1

2

R−→ −2T−→ −1

T−→ 0


c Firstly we operate on small positive integers to see if there is a pattern. We stop each sequence when we reach anumber that has occurred in the previous sequence.

1R−→ −1

T−→ 0

2R−→ −1

2

T−→ 1

2

R−→ −2T−→ −1

3R−→ −1

3

T−→ 2

3

R−→ −3

2

T−→ −1

2

4R−→ −1

4

T−→ 3

4

R−→ −4

3

T−→ −1

3

For any integer n > 1 we have:

nR−→ − 1

n

T−→ n− 1

n

R−→ − n

n− 1

T−→ − 1

n− 1

Note that, from the second sequence, the last number in each of these sequences is the second number in the previoussequence. This suggests the following sequence of operations for changing n into 0.

nR−→ − 1

n

T−→ n− 1

n

R−→ − n

n− 1

T−→ − 1

n− 1

T−→ n− 2

n− 1

R−→ −n− 1

n− 2

T−→ − 1

n− 2

T−→ n− 3

n− 2

R−→ −n− 2

n− 3

T−→ − 1

n− 3

and so on down toT−→ 2

3

R−→ −3

2

T−→ −1

2

T−→ 1

2

R−→ −2T−→ −1

T−→ 0

The number of operations used is1 + 3(n− 1) + 1 = 3n− 1.

d We use a similar approach to that in Part c, this time using the last number in a sequence to start the next sequence.

0T−→ 1

R−→ −1

−1R−→ 1

T−→ 2R−→ −1

2

T−→ 1

2

R−→ −2

−2R−→ 1

2

T−→ 3

2

R−→ −2

3

T−→ 1

3

R−→ −3

−3R−→ 1

3

T−→ 4

3

R−→ −3

4

T−→ 1

4

R−→ −4

For any integer n > 0 we have:

−nR−→ 1

n

T−→ n+ 1

n

R−→ − n

n+ 1

T−→ 1

n+ 1

R−→ −(n+ 1)

Thus the sequence continues until any desired negative integer is obtained.


I3 Rail Cams

a Camera A is 15m south and 35m west of camera B . Let the horizontal distance between the two cameras be xmetres.

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B

15

35

x

By Pythagoras’ theorem, x2 = 152 + 352.

The vertical distance between the two cameras is 25−20 = 5 metres. Let y be the direct distance in metres betweenthe two cameras. Then we have this right-angled triangle.

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x

y

A

B

By Pythagoras’ theorem,

y2 = x2 + 52

= 152 + 352 + 52

= 5232 + 5272 + 52

= 25(9 + 49 + 1)

= 25(59)

So y = 5√59m or 3841 cm to the nearest centimetre.

b After two seconds, camera A has moved 10m north and camera B has moved 10m west. Let the horizontal distancebetween the two cameras then be x metres.

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B

5

25

x

10

10

By Pythagoras’ theorem, x2 = 52 + 252.

The vertical distance between the two cameras is still 5 metres. Let the direct distance in metres between the twocameras now be y. Then we have this right-angled triangle.

••5

x

y

A

B

By Pythagoras’ theorem,

y2 = x2 + 52

= 52 + 252 + 52

= 25(1 + 25 + 1)

= 25(27)

So y = 15√3m. Therefore, after two seconds, the cameras are 5

√59− 15

√3 metres closer. This is 1242 cm to the

nearest centimetre.


c Alternative i

Since the cameras remain 5m apart vertically, their direct distance apart is minimised when their horizontal distanced apart is minimised. At time t seconds, both cameras are 5t metres from their start positions. So we have theseright-angled triangles.

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B

15− 5t

35− 5t

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5t

5t

0 ≤ t ≤ 3

•

•

A

B

5t− 15

35− 5t

d

15

5t3 ≤ t ≤ 7

•

•

A

B

5t− 35

5t− 15d

15

35

7 ≤ t ≤ 14

In each case, by Pythagoras’ theorem,

d2 = (35− 5t)2 + (15− 5t)2

= 25(7− t)2 + 25(3− t)2

= 25(2t2 − 20t+ 58)

= 50(t2 − 10t+ 29)

= 50((t− 5)2 + 4)

So minimum d2 = 50 × 4 = 200 when t = 5 seconds. Since the cameras remain 5m apart vertically, the minimumdirect distance between them is

√200 + 52 =

√225 = 15m and this occurs at 5 seconds.

Alternative ii

Since the cameras remain 5m apart vertically, their direct distance apart is minimised when their horizontal distanced apart is minimised. At time t seconds, both cameras are 5t metres from their start positions. The next diagramshows the horizontal positions of cameras A and B at times t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 seconds. In each one-secondtime interval, both cameras travel 5m.


• • • • • • • • • •

•

•

•

•

•

•

•

•

•

•

A

B0

0

1

1

2

2

33

4

4

5

5

6

6

7

7

8

8

9

9

So, if 0 ≤ t ≤ 3 or 7 ≤ t ≤ 14, then d ≥ 5(7 − 3) = 20. If t = 5, then, by Pythagoras’ theorem, d2 = 102 + 102 =200 < 400 = 202. So minimum d occurs in the time interval 3 ≤ t ≤ 7.

The next diagram shows the horizontal positions of cameras A and B for any value of t between 3 and 7. Thebottom-left dot is the origin of a cartesian system with axes along the ‘B’ dots and ‘A’ dots respectively. The gapbetween successive dots on the axes is 5m. The point C is at (10,10). Triangles CY A and CXB are congruentright-angled triangles and angle XCY is 90. Hence ACB is a right-angled isosceles triangle.

• • • • ••

•

•

•

•

A

B

CY

X

d

33

4

4

5

5

6

6

7

7

•

By Pythagoras’ theorem, d2 = CA2 + CB2 = 2CA2. So minimum d occurs when A is at Y , that is, when t = 5seconds. Then d2 = 2 × 102 = 200. Since the cameras remain 5m apart vertically, the minimum direct distancebetween them is

√200 + 52 =

√225 = 15m and this occurs at 5 seconds.


I4 Curvy Tiles

a The six different 2× 2 tiling patterns are:

1 2 3

4 5 6

Rotations of these patterns are acceptable.

b Each random placement of four tiles has probability 1/16 of occurring.

There is only one random placement that gives pattern 1, so the probability of pattern 1 occurring is 1/16. Similarly,the probability of pattern 6 occurring is 1/16.

There are exactly 4 random placements that give pattern 2: the one shown and its rotations through 90, 180,and 270.

So the probability of pattern 2 occurring is 4/16 = 1/4. Similarly, the probability of each of patterns 4 and 5occurring is 1/4.

There are exactly two random placements that give pattern 3: the one shown and its 90 rotation.

So the probability of pattern 3 occurring is 2/16 = 1/8.

c Since each tile has two orientations, the number of ways of placing nine tiles to form a 3× 3 square is 29 = 512.

Each end of a peanut is an arc of a circle that is greater than a semicircle. So three tiles that share a vertex arerequired to form each end. Hence there are only two ways in which a peanut can occur in a 3× 3 placement of tiles.

In each case, the orientations of the 7 tiles that form the peanut are fixed but the 2 remaining tiles (shown blank)have 2 possible orientations. Hence the number of placements that contain a peanut is 2 × (2 × 2) = 8. So theprobability that a random 3× 3 placement contains a peanut is 8/512 = 1/64.


I5 Rhombus Rings

a Each central angle is 360/8 = 45. Starting with those, we calculate the remaining angles in degrees as we moveoutwards.

45

45

45

45

45

45

135

135

135

135

90 90

90

90

90

90

A

B

C

Since ABC is a straight line, the third ring is the last (outer) ring. Thus the angles in a rhombus in the outer ringare 45 and 135

b We have these angles (all in degrees).

160

1602020

6060 x

Thus x = 360 − 160 − 60 − 60 = 80. Hence the angles in each rhombus of the third last ring are 80 and180 − 80 = 100.

Extending the diagram above and inserting a few more angles, we get:

160

1602020

606080

100100

120

y

Thus y = 360− 120− 100− 100 = 40. Hence there is a fourth last ring of rhombuses. Extending the diagram andadding a few more angles gives:


160

1602020

606080

100100

120

40

40

140140

80

Since 140 + 140 + 80 = 360, no more rings are possible. So there are 4 rings in total and the number of rhombusesin the inner ring is 360/40 = 9.

c We start with the first ring. Each of its rhombuses has central angle a. For a rhombus in each subsequent ring wecalculate as follows the angle that is closest to the pattern centre.

Second ring: angle = 360− 2(180− a) = 2a.Third ring: angle = 360− 2(180− 2a)− a = 3a.Fourth ring: angle = 360− 2(180− 3a)− 2a = 4a.rth ring: angle = 360− 2(180− (r − 1)a)− (r − 2)a = ra.

a a

2a2a2a

3a 3a

4a

4a4a

5a 5a

180− a

180− 3a

180− 4a

180− 2a

Hence the angles in a rhombus in the rth ring are ra and 180− ra.

d Let n be the number of rhombuses in each ring. Then the central angles for the first ring are a = 360/n.

From Part c, the angle in each rhombus in the 7th ring that is closest to the pattern centre is 7a = 7 × 360/n.Similarly, if there was an 8th ring, the angle in each rhombus closest to the pattern centre would be 8× 360/n.

Since there are 7 rings, 7× 360/n < 180. Hence n > 14. Since there is no 8th ring, 8× 360/n ≥ 180. Hence n ≤ 16.

If n = 15, the central angles are 360/15 = 24. So the angles closest to the pattern centre for each ring, from thefirst to the 7th ring are: 24, 48, 72, 96, 120, 144, 168. Since 8× 360/15 = 192 > 180, there is indeed no 8thring. This confirms there are exactly 7 rings.

If n = 16, the central angles are 360/16 = 22.5. So the angles closest to the pattern centre for each ring, from thefirst to the 7th ring are: 22.5, 45, 67.5, 90, 112.5, 135, 157.5. Since 8× 360/16 = 180, there is indeed no 8thring. This confirms there are exactly 7 rings.


I6 Higher or Lower

a If Player 1 chooses box 2 on the first turn, her chance of choosing correctly is 13 . If she doesn’t win on the first turn,

then Player 2 will win on the second turn because of the host’s directions. So the chance of Player 1 winning if shechooses box 2 on the first turn is 1

3 .

P1 chooses box 2

‘Lower’(P2 wins)

13

‘Higher’(P2 wins)

13

‘Correct’(P1 wins)

13

If Player 1 chooses box 1 on the first turn, her chance of choosing correctly is 13 . The game will continue with

probability 23 . If it does continue, then the probability of Player 2 winning on the second turn is 1

2 . If he doesn’twin, then Player 1 will win on the third turn. So the chance of Player 1 winning if she chooses box 1 on the firstturn is 1

3 + 23 × 1

2 = 23 .

P1 chooses box 1


13

‘Higher’(continues)

23


12

‘Higher/Lower’(P1 wins)

12

b If Player 1 chooses box 1 on the first turn, her chance of choosing correctly is 14 . The game will continue with

probability 34 . If it does continue then, from Part a, Player 2 should choose either box 2 or 4 to ensure his chance

of winning from that stage is 23 . Hence Player 2’s overall chance of winning with this strategy is 3

4 × 23 = 1

2 .

P1 chooses box 1


14


34

(P2 wins, using a)

23

Similarly, if Player 1 chooses box 4 on the first turn and the game continues, then Player 2 can choose box 1 or 3to ensure his overall chance of winning the game is 1

2 .

If Player 1 chooses box 2 on the first turn, her chance of choosing correctly is 14 . If she does not choose correctly,

there are two possibilities: either the prize is in box 1, with probability 14 , and Player 2 immediately wins; or the

prize is in box 3 or 4, with probability 12 , and Player 2’s chance of choosing correctly at that stage is 1

2 . Hence, ifPlayer 1 chooses box 2 on the first turn, then Player 2’s chance of winning is 1

4 + 12 × 1

2 = 12 .


P1 chooses box 2

‘Lower’(P2 wins)

14


12


14


12

‘Higher/Lower’(P1 wins)

12

Similarly, if Player 1 chooses box 3 on the first turn, then Player 2’s chance of winning is 12 .

So, no matter which box Player 1 chooses on the first turn, Player 2 has a strategy for ensuring his chance ofwinning is 1

2 .

c If Player 1 chooses box 5 on the first turn, her chance of choosing correctly is 19 . The game will continue with

probability 89 . If it does continue, then there are 4 boxes left to choose from and Player 2 has the next turn. From

Part b, Player 1 now has a strategy for ensuring her chance of winning the 4-box game is 12 . Hence, if Player 1

chooses box 5 on the first turn, she has a strategy to ensure her chance of winning the game is 19 + 8

9 × 12 = 5

9 > 12 .

P1 chooses box 5


19

‘Higher/Lower’(continues)

89

(P1 wins, using b)

12


CHALLENGE STATISTICS – MIDDLE PRIMARY

Mean Score/School Year/Problem

School Year Number of Students

Overall Mean

Challenge Problem

1 2 3 4

3 599 9.1 2.2 2.5 2.6 2.2

4 917 10.4 2.5 2.9 2.8 2.5

*ALL YEARS 1538 9.9 2.4 2.8 2.7 2.4

Please note:* This total includes students who did not provide their school year.

Score Distribution %/Problem

Score

Challenge Problem

1Training

2Bitripents

3Coloured Buckets

4Hand-Sum

Did not attempt 1% 1% 3% 5%

0 5% 6% 6% 10%

1 22% 12% 10% 15%

2 27% 20% 21% 24%

3 24% 20% 31% 23%

4 21% 40% 30% 23%

Mean 2.4 2.8 2.7 2.4

Discrimination Factor 0.5 0.6 0.6 0.7

Please note:The discrimination factor for a particular problem is calculated as follows:

1. The students are ranked in regard to their overall scores.2. The mean score for the top 25% of these overall ranked students is calculated for that particular

problem including no attempts. Call this mean score the ‘mean top score’.3. The mean score for the bottom 25% of these overall ranked students is calculated for that

particular problem including no attempts. Call this mean score the ‘mean bottom score’.4. The discrimination factor = mean top score – mean bottom score 4

Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is considered to be a good discriminator.

Mathematics Contests The Australian Scene 2018 | Challenge Statistics – Middle Primary | 53

CHALLENGE STATISTICS – UPPER PRIMARY



Overall Mean

Challenge Problem

1 2 3 4

5 1426 11.3 3.1 3.1 2.7 2.5

6 1727 12.4 3.3 3.4 2.9 2.9

7 77 12.7 3.5 3.4 3.0 3.1

*ALL YEARS 3255 11.9 3.2 3.3 2.8 2.7



Score

Challenge Problem

1Bitripents

2Insect Barriers

3Coloured Buckets

4Isopentagons

Did not attempt 0% 1% 1% 4%

0 1% 3% 3% 10%

1 4% 4% 8% 11%

2 18% 9% 23% 14%

3 24% 29% 38% 23%

4 53% 54% 27% 39%

Mean 3.2 3.3 2.8 2.7

Discrimination Factor 0.4 0.4 0.5 0.7






Mathematics Contests The Australian Scene 2018 | Challenge Statistics – Upper Primary | 54

CHALLENGE STATISTICS – JUNIOR



Overall Mean

Challenge Problem

1 2 3 4 5 6

7 2737 13.1 2.8 2.6 2.4 2.6 1.7 1.9

8 2508 15.3 3.1 2.9 2.7 3.1 2.2 2.2

*ALL YEARS 5288 14.2 3.0 2.7 2.6 2.8 1.9 2.1



Score

Challenge Problem

1Insect

Barriers

2Cubic

Coprimes

3Isopentagons

4Ts and Rs

5Rhombus

Rings

6Parsecking

Did not attempt 1% 4% 8% 8% 13% 12%

0 2% 6% 10% 12% 22% 14%

1 7% 8% 10% 7% 15% 17%

2 21% 21% 17% 10% 17% 19%

3 33% 33% 26% 18% 13% 25%

4 36% 28% 29% 45% 21% 12%

Mean 3.0 2.7 2.6 2.8 1.9 2.1

Discrimination Factor 0.4 0.5 0.7 0.8 0.8 0.7






Mathematics Contests The Australian Scene 2018 | Challenge Statistics – Junior | 55

CHALLENGE STATISTICS – INTERMEDIATE



Overall Mean

Challenge Problem

1 2 3 4 5 6

9 1490 13.8 2.8 2.5 2.4 2.7 2.0 2.0

10 757 14.8 3.0 2.8 2.6 2.9 2.2 2.3

*ALL YEARS 2264 14.2 2.9 2.6 2.5 2.8 2.1 2.1



Score

Challenge Problem

1Isopentagons

2Ts and Rs

3Rail Cams

4Curvy Tiles

5Rhombus

Rings

6Higher or

Lower

Did not attempt 5% 4% 7% 7% 15% 11%

0 6% 6% 12% 7% 14% 16%

1 9% 8% 13% 7% 17% 17%

2 14% 30% 18% 20% 17% 18%

3 27% 23% 23% 23% 21% 20%

4 39% 29% 28% 36% 16% 19%

Mean 2.9 2.6 2.5 2.8 2.1 2.1

Discrimination Factor 0.6 0.6 0.7 0.7 0.7 0.7

Please note:

The discrimination factor for a particular problem is calculated as follows:





Mathematics Contests The Australian Scene 2018 | Challenge Statistics – Intermediate | 56

2018 Australian Intermediate Mathematics Olympiad - Questions

Time allowed: 4 hours. NO calculators are to be used.

Questions 1 to 8 only require their numerical answers all of which are non-negative integers less than 1000.Questions 9 and 10 require written solutions which may include proofs.The bonus marks for the Investigation in Question 10 may be used to determine prize winners.

1. Let x denote a single digit. The tens digit in the product of 2x7 and 39 is 9. Find x. [2 marks]

2. If 234b+1 − 234b−1 = 7010, what is 234b in base 10? [3 marks]

3. The circumcircle of a square ABCD has radius 10. A semicircle is drawn on AB outside the square. Find the areaof the region inside the semicircle but outside the circumcircle. [3 marks]

4. Find the last non-zero digit of 50! = 1× 2× 3× · · · × 50. [3 marks]

5. Each edge of a cube is marked with its trisection points. Each vertex v of the cube is cut off by a plane that passesthrough the three trisection points closest to v. The resulting polyhedron has 24 vertices. How many diagonalsjoining pairs of these vertices lie entirely inside the polyhedron? [3 marks]

6. Let ABCD be a parallelogram. Point P is on AB produced such that DP bisects BC at N . Point Q is on BAproduced such that CQ bisects AD at M . Lines DP and CQ meet at O. If the area of parallelogram ABCD is192, find the area of triangle POQ. [4 marks]

7. Two different positive integers a and b satisfy the equation a2 − b2 = 2018− 2a.What is the value of a+ b? [4 marks]

8. The area of triangle ABC is 300. In triangle ABC, Q is the midpoint of BC, P is a point on AC between C andA such that CP = 3PA, R is a point on side AB such that the area of PQR is twice the area of RBQ. Findthe area of PQR. [4 marks]

9. Prove that 38 is the largest even integer that is not the sum of two positive odd composite numbers. [4 marks]

10. A pair of positive integers is called compatible if one of the numbers equals the sum of all digits in the pair and theother number equals the product of all digits in the pair. Find all pairs of positive compatible numbers less than100. [5 marks]

Investigation

Find all pairs of positive compatible numbers less than 1000 with at least one number greater than 99.[3 bonus marks]

1

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD

Mathematics Contests The Australian Scene 2018 | Australian Intermediate Mathematics Olympiad | 57

2018 Australian Intermediate Mathematics Olympiad - Solutions

1. Method 1

The table shows the product 2x7× 39 for all values of x.

x 2x7× 39 x 2x7× 390 8073 5 100231 8463 6 104132 8853 7 108033 9243 8 111934 9633 9 11583

Thus x = 8.

Method 2

We have 2x7× 39 = 2x7× 30 + 2x7× 9.The units digit in 2x7× 30 is 0, and its tens digit is 1.The tens digit of 2x7× 9 is the units digit of 6 + 9× x.

Hence 1 + 6 + 9× x ≡ 9 (mod 10), 9× x ≡ 2 (mod 10), x = 8.

Method 3

We have 2x7× 39 = 207× 39 + 390× x.The units digit in 207× 39 is 3, and its tens digit is 7.The tens digit of 390× x is the units digit of 9× x.

Hence 7 + 9× x ≡ 9 (mod 10), 9× x ≡ 2 (mod 10), x = 8.

Method 4

We have 2x7× 39 = 2x7× 40− 2x7.The units digit in 2x7× 40 is 0, and its tens digit is 8.So the tens digit of 2x7× 39 is the units digit of 8− x− 1 or 18− x− 1.

Since x is non-negative, 17− x = 9 and x = 8.

2. Method 1

We have

7010 = 234b+1 − 234b−1

= 2(b+ 1)2 + 3(b+ 1) + 4− 2(b− 1)2 − 3(b− 1)− 4

= 2(b2 + 2b+ 1) + 3(b+ 1)− 2(b2 − 2b+ 1)− 3(b− 1)

= 8b+ 6

b = 8

So 234b = 2348 = 2× 64 + 3× 8 + 4 = 156.

Method 2

The largest digit on the left side of the given equation is 4. Hence b− 1 is at least 5. So b ≥ 6.

If b = 6, then the left side in base 10 is 2347−2345 = (2×49+3×7+4)−(2×25+3×5+4) = (98+21)−(50+15) =119− 65 = 54 = 70.

If b = 7, then the left side in base 10 is 2348−2346 = (2×64+3×8+4)−(2×36+3×6+4) = (128+24)−(72+18) =152− 90 = 62 = 70.

If b = 8, then the left side in base 10 is 2349−2347 = (2×81+3×9+4)−(2×49+3×7+4) = (162+27)−(98+21) =189− 119 = 70.

Each time b increases by 1, 4b remains the same, 30b increases by 3, but 200b increases by 2(b+1)2 − 2b2 = 4b+2.So the increase in 234b+1 is greater than the increase in 234b−1. Hence 234b+1 − 234b−1 increases with increasingb. This means 234b+1 − 234b−1 > 7010 for b > 8.

So 234b = 2348 = 2× 64 + 3× 8 + 4 = 156.

2

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD SOLUTIONS

Mathematics Contests The Australian Scene 2018 | Australian Intermediate Mathematics Olympiad Solutions | 58

3. Let O be the centre of the circumcircle.

A B

D C

O

Since OA = OB = OC = OD and AB = BC = CD = DA, triangles AOB, BOC, COD, DOA are isosceles andcongruent. So AOB = 360/4 = 90. Hence the area of AOB is 1

2 × 10 × 10 = 50 and the area of the sectorAOB = 1

4π100 = 25π.

By Pythagoras, AB2 = AO2+OB2 = 200. Hence the area of the semicircle on AB is 12π(AB/2)2 = AB2π/8 = 25π.

So the required area is 25π − (25π − 50) = 50.

4. Method 1

We first arrange the factors 1, 2, 3, . . . , 50 in a table:

1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 2021 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 4041 42 43 44 45 46 47 48 49 50

From this we see that in the prime factorisation of 50!, 5 occurs exactly 12 times. Then 50!/(212512) is the productof these factors:

1 1 3 4 1 6 7 1 9 111 12 13 14 3 16 17 18 19 121 22 23 24 1 26 27 28 29 331 32 33 34 7 36 37 38 39 141 42 43 44 9 46 47 48 49 1

So the last digit of 50!/(212512) is the last digit in the product

113.24.37.45.65.76.84.96 = (2.3.4.6.7.8.9)4 × (3.7.9)2 × (3.4.6)

The last digit of 2.3.4.6.7.8.9 is 6. So the last digit of (2.3.4.6.7.8.9)4 is 6.The last digit of 3.7.9 is 9. So the last digit of (3.7.9)2 is 1.The last digit of 3.4.6 is 2.So the last non-zero digit of 50! is the last digit of 6× 1× 2, which is 2.

Comment

There are many other workable groupings of factors.

3Mathematics Contests The Australian Scene 2018 | Australian Intermediate Mathematics Olympiad Solutions | 59

Method 2

In the prime factorisation of 50!, 5 occurs exactly 12 times and 2 occurs more than 12 times. So the last non-zerodigit of 50! is the last digit of 50!/212512 and it is even.

The remainders of 2, 4, 6, 8 when divided by 5 are respectively 2, 4, 1, 3. So the remainder of 50!/(212512) whendivided by 5 will reveal its last digit.

Since the remainder of 212 is 1 when divided by 5, the remainder for 50!/212512 is the same as the remainder for50!/512. So we want the remainder of the product of the following factors.

1 2 3 4 1 6 7 8 9 211 12 13 14 3 16 17 18 19 421 22 23 24 1 26 27 28 29 631 32 33 34 7 36 37 38 39 841 42 43 44 9 46 47 48 49 2

Before we multiply these factors we may subtract from each any multiple of 5. So we need only multiply thesenumbers.

1 2 3 4 1 1 2 3 4 21 2 3 4 3 1 2 3 4 41 2 3 4 1 1 2 3 4 11 2 3 4 2 1 2 3 4 31 2 3 4 4 1 2 3 4 2

After multiplying any two of these numbers, we may subtract any multiple of 5. Since 2 × 3 and 4 × 4 haveremainder 1 when divided by 5, we need only multiply 3, 2, 4, 2, 4, 3, 2. Hence the required remainder is 2. Sothe last non-zero digit of 50! is 2.

5. Method 1

Each vertex is on three faces: a triangular face and two octagonal faces. So each vertex shares a face with 7 othervertices from one octagonal face, and 6 other vertices from the other octagonal face (the two octagons share anedge and 2 vertices). The vertices from the triangular face have already been counted.

A vertex can be joined to 23 other vertices. Of these, 7 + 6 = 13 lie on the faces of the polyhedron. So each vertexjoins to 23− 13 = 10 vertices by diagonals that are internal to the polyhedron.

Since each of these diagonals joins two vertices, multiplying 10 by 24 counts each diagonal exactly twice. So thenumber of diagonals inside the polyhedron is 10× 24/2 = 120.

Method 2

The diagram is a projection of the polyhedron.

•

•

•

•• ••

•

•

••

v

As indicated by dots, there are exactly 10 vertices that are not on a face containing v. So there are exactly 10internal diagonals joined to v.

By rotating the polyhedron about one or more of its axes of symmetry, v represents any of its 24 vertices. Sinceeach of these diagonals joins two vertices, multiplying 10 by 24 counts each diagonal exactly twice. So the numberof diagonals inside the polyhedron is 10× 24/2 = 120.


Method 3

The polyhedron has 8 triangular faces and 6 octagonal faces. Since each edge of the polyhedron is shared by twofaces, its total number of edges is (8× 3 + 6× 8)/2 = 36.

Each octagonal face has 20 diagonals. So the number of diagonals of the polyhedron on its faces is 6× 20 = 120.

The number of pairs of vertices of the polyhedron is(242

)= 276. So the number of internal diagonals of the

polyhedron is 276− 36− 120 = 120.

6. Draw MN .

Q A B P

CD

M NO

Method 1

Since BP and CD are parallel and BN = NC, triangles BNP and CND are congruent (ASA). Similarly, trianglesAMQ and DMC are congruent.

Since AM and BN are parallel and equal, MN and AB are parallel. So ABNM and MNCD are congruentparallelograms and their areas are half the area of ABCD, that is, 192/2 = 96.

Since MNCD is a parallelogram, its area is twice the area of triangle CND, twice the area of triangle DMC, and4 times the area of triangle MNO.

So the area of triangle POQ is 96 + 2(96/2) + (96/4) = 216.

Method 2

Since BP and CD are parallel and BN = NC, triangles BNP and CND are congruent (ASA). Similarly, trianglesAMQ and DMC are congruent. So QP = 3×DC.

Since AM and BN are parallel and equal, MN and AB are parallel. So ABNM and MNCD are congruentparallelograms and their areas are half the area of ABCD, that is, 192/2 = 96.

Since MNCD is a parallelogram, the area of triangle COD is 96/4 = 24.

Since PQ and CD are parallel, triangles POQ and DOC are similar. So the area of triangle POQ is 9×24 = 216.


Method 3

Parallelogram ABCD is unspecified so we may take it to be a rectangle with AD = 2AB.

Q A B P

CD

M N

O

Thus ABNM and MNCD are congruent squares, triangles NBP and NCD are congruent, and triangles MAQand MDC are congruent.

Let | | denote area. Then

|ABNM | = |MNCD| = 1

2|ABCD| = 192

2= 96

|NBP | = |NDC| = 1

2|MNCD| = 48

|MAQ| = |MDC| = 1

2|MNCD| = 48

|MNO| = 1

4|MNCD| = 24

So |POQ| = 96 + 48 + 48 + 24 = 216.

7. Method 1

We have

2018 = a2 + 2a− b2

2019 = a2 + 2a+ 1− b2

= (a+ 1)2 − b2

= (a+ 1− b)(a+ 1 + b)

We know a and b are positive, so a+ 1 + b is positive, hence a+ 1− b is positive.Since a and b are integers, both factors are integers.Since 2019 = 3× 673 and 673 is prime, the only positive integer factor pairs are (1, 2019) and (3, 673).Since b is positive, a+ 1 + b > a+ 1− b. So a+ 1− b equals 1 or 3.

If a+ 1− b = 1, then a = b, which contradicts the requirement that a and b are different.

So a+ 1− b = 3 and a+ 1 + b = 673. Hence a+ b = 672.

Method 2

Solving a2 + 2a− b2 − 2018 = 0 using the quadratic formula yields

a =−2±

√4 + 4(b2 + 2018)

2= −1±

√b2 + 2019

Since a is positive, a =√b2 + 2019− 1.

Since a+ 1 is an integer, we have b2 + 2019 = c2 for some positive integer c.So a = c− 1 and (c+ b)(c− b) = 2019.

The only positive factorisations of 2019 are 1× 2019 and 3× 673.Since b is positive, we have c+ b = 2019 and c− b = 1, or c+ b = 673 and c− b = 3.

So 2b = 2018 or 670.Hence, respectively, b = 1009 or 335, c = 1010 or 338, and a = 1009 or 337.Since a and b are different, a+ b = 337 + 335 = 672.


8. Method 1

Draw AQ, BP , and CR. Let | | denote area.

B C

A

Q

P

R

Let BR = 1 and RA = p. Noting that BQ = QC and CP = 3PA, we have:

|BQR| = 1

2|BCR| = 1

2× 1

1 + p× |ABC| = 1

2× 300

1 + p

|APR| = 1

4|ACR| = 1

4× p

1 + p× |ABC| = 1

4× 300p

1 + p

|CPQ| = 1

2|BCP | = 1

2× 3

4× |ABC| = 3

8× 300

|PQR| = 2|BQR| = 300

1 + p

Adding these areas gives

300 =4

8

(300

1 + p

)+

2

8

(300p

1 + p

)+

3

8

(300(p+ 1)

p+ 1

)+

8

8

(300

1 + p

)

=1

8

(300

p+ 1

)(4 + 2p+ 3(p+ 1) + 8)

8(p+ 1) = 15 + 5p

3p = 7

Finally |PQR| = 300

1 + p=

300× 3

3 + 3p=

900

10= 90.


Method 2

B C

A

Q

P

R

Let | | denote area. Noting that BQ = QC and CP = 3PA, we have:

300 = |APR|+ |PQR|+ |BQR|+ |CQP |

=1

4|ACR|+ |PQR|+ 1

2|PQR|+ 3

4|AQC|

1200 = |ACR|+ 6|PQR|+ 3|AQC|

= (300− |BCR|) + 6|PQR|+ 3

2|ABC|

= 300− 2|BQR|+ 6|PQR|+ 450

= 750− |PQR|+ 6|PQR|= 750 + 5|PQR|

|PQR| = 1

5(1200− 750) =

1

5(450) = 90.


Method 3

First consider the following diagram and let | | denote area.

X U

V

Y

Z

M N

|XY Z||XUV |

=MZ ×XY

NV ×XU=

XZ

XV× XY

XU=

XY ×XZ

XU ×XV(1)

Next consider the given diagram with distances as shown.

B C

A

Q

P

R

a a

b

3b

c

d

From (1) we have

|BQR|300

=ac

2a(c+ d)=

c

2(c+ d)

|APR|300

=bd

4b(c+ d)=

d

4(c+ d)

|CPQ|300

=3ab

(4b)(2a)=

3

8Combining these gives

300 = |BQR|+ |APR|+ |CPQ|+ |PQR|= 3|BQR|+ |APR|+ |CPQ|

1 =3c

2(c+ d)+

d

4(c+ d)+

3

8

8(c+ d) = 12c+ 2d+ 3(c+ d) = 15c+ 5d

d = 7c/3

So |BQR| = 300c

2(c+ d)=

300c

20c/3=

90

2= 45. Hence |PQR| = 2|BQR| = 90.


9. The odd composites less than 38 are 9, 15, 21, 25, 27, 33, 35. No two of these have their sum equal to 38.

Method 1

Now 40 = 15 + 25, 42 = 9+ 33, 44 = 9+ 35, where each summand is an odd composite and at least one summandis a multiple of 3.

Adding 6 to an odd multiple of 3 gives an odd number that is also a multiple of 3. So we can express the next threeeven integers as the sum of two odd composites at least one of which is a multiple of 3: 46 = 21+25, 48 = 15+33,50 = 15 + 35.

We can therefore add 6 to express the next three even integers after these as the sum of two odd composites atleast one of which is a multiple of 3, and repeat indefinitely.

So every even integer greater than 38 is the sum of two odd composites.

Method 2

Now 40 = 15+25, 42 = 15+27, 44 = 9+35, 46 = 21+25, 48 = 15+33, where each summand is an odd compositeand at least one summand is a multiple of 5.

Adding 10 to an odd multiple of 5 gives an odd number that is also a multiple of 5. So we can express the next fiveeven integers as the sum of two odd composites at least one of which is a multiple of 5: 50 = 25+25, 52 = 25+33,54 = 9 + 45, 56 = 21 + 35, 58 = 25 + 33.

We can therefore add 10 to express the next five even integers after these as the sum of two odd composites atleast one of which is a multiple of 5, and repeat indefinitely.

So every even integer greater than 38 is the sum of two odd composites.

Comment

There are similar arguments adding 12, 18, 20, 24, 30, etc. at a time.

10. Let s and p be compatible numbers, where s is the sum of their digits and p is the product of their digits. Notethat all of the digits are non-zero since p = 0. We consider three cases.

Case 1. s has a single digit.

Since p has a positive digit, s ≥ s+ 1, a contradiction. So s has two digits.

Case 2. s has two digits and p has a single digit.

We have s = 10a+ b, where a and b are positive digits. Then

10a+ b = a+ b+ p and p = abp.

The second equation gives a = b = 1. Hence p = 9 from the first equation. So the only pair of compatible numbersin this case are s = 11 and p = 9.

Case 3. s and p both have two digits.

Method 1

Let s = 10a+ b and p = 10c+ d, where a, b, c, and d are positive digits. Then

10a+ b = a+ b+ c+ d and 10c+ d = abcd.

The first equation gives 9a = c+ d. So a = 1 or 2.

If a = 2, then c+ d = 18, c = d = 9, and the second equation gives 99 = 162b, which is impossible.

If a = 1, then c+ d = 9 and the second equation gives

9c+ 9 = bc(9− c) and 9 = c(b(9− c)− 9).

So c = 1, 3, or 9. If c = 1, then 18 = 8b, which is impossible. If c = 9, then 1 = 0− 9, which is a contradiction. Ifc = 3, then 3 = 6b− 9, b = 2, d = 6 and we have the compatible numbers s = 12 and p = 36.

Hence there are only 2 pairs of compatible numbers less than 100, namely 9, 11 and 12, 36.


Method 2

Let s = 10a+ b and p = 10c+ d, where a, b, c, and d are positive digits. Then

10a+ b = a+ b+ c+ d and 10c+ d = abcd.

The second equation gives d = c(abd− 10). Since d is a positive digit, 10 < abd < 20. Since a, b, d are digits and11, 13, 17, 19 are primes, abd = 12, 14, 15, 16, or 18.

If abd = 18, then 18c = 10c+ d, d = 8c, and 4abc = 9, which is impossible.


If abd = 15, then 15c = 10c+ d, d = 5c. So c = 1, d = 5, and 9a = 6, which is impossible.


If abd = 12, then 12c = 10c+d, and d = 2c. So d is even and divides 12. Hence d = 2, 4, or 6, and correspondinglyc = 1, 2, or 3. Since 10a + b = a + b + c + d, we have 9a = c + d. So 9 divides c + d. Hence d = 6, c = 3, a = 1,b = 2, and we have the compatible numbers s = 12 and p = 36.

Hence there are only 2 pairs of compatible numbers less than 100, namely 9, 11 and 12, 36.

Investigation

Let s and p be compatible numbers, where s is the sum of their digits and p is the product of their digits. Notethat all of the digits are non-zero since p = 0.

As in the solution above, s has at least 2 digits. Since the sum of six digits is at most 54, s has exactly 2 digits.So p has exactly 3 digits. Let s = 10a+ b and p = 100c+ 10d+ e, where a, b, c, d, e are positive digits. Then

10a+ b = a+ b+ c+ d+ e and p = abcde.

The first of these equations gives 9a = c+ d+ e, so 9 divides c+ d+ e. Hence c+ d+ e = 9, 18, or 27.

If c+ d+ e = 27, then a = 3, c = d = e = 9, p = 999 = 3b× 9× 9× 9 = 2187b, which is impossible.

Method 1

If c+ d+ e = 18, then a = 2 and p = 2bcde. The table shows for each combination of digits for c, d, e, the product2cde, the last 1, 2, or 3 digits of 2bcde for b = 1, 2, 3, 4, 5, 6, 7, 8, 9 and 2bcde < 1000, then a check whether any2bcde could be p = 100c+ 10d+ e.

c, d, e 2cde last few digits of 2bcde any p?

1, 8, 9 144 4, 88, 2, 6, 0, 4 none2, 7, 9 252 52, 4, 6 none2, 8, 8 256 6, 12, 68 none3, 6, 9 324 4, 8, 2 none3, 7, 8 336 6, 2 none4, 5, 9 360 0, 0, 0, 0, 0, 0, 0, 0, 0 none4, 6, 8 384 384, 768 none4, 7, 7 392 2, 84 none5, 5, 8 400 0, 0, 0, 0, 0, 0, 0, 0, 0 none5, 6, 7 420 0, 0, 0, 0, 0, 0, 0, 0, 0 none6, 6, 6 432 2, 4 none

If c + d + e = 9, then a = 1 and p = bcde. The table shows for each combination of digits for c, d, e, theproduct cde, the last 1, 2, or 3 digits of bcde for b = 1, 2, 3, 4, 5, 6, 7, 8, 9, then a check whether any bcde could bep = 100c+ 10d+ e.

c, d, e cde last few digits of bcde any p?

1, 1, 7 7 07, 4, 21, 8, 5, 2, 9, 6, 3 none1, 2, 6 12 012, 4, 36, 8, 0, 72, 4, 96, 8 none1, 3, 5 15 015, 0, 45, 0, 75, 0, 05, 0, 135 1351, 4, 4 16 6, 2, 8, 64, 0, 6, 2, 8, 144 1442, 2, 5 20 0, 0, 0, 0, 0, 0, 0, 0, 0 none2, 3, 4 24 024, 8, 72, 6, 0, 44, 8, 92, 6 none3, 3, 3 27 7, 4, 1, 8, 5, 2, 9, 6, 43 none

Hence the only compatible pairs are 135, 19 and 144, 19.


Method 2

Since 9 divides c+ d+ e and c, d, e are the digits of p, 9 divides p.

If c + d + e = 18, then a = 2 and p = 2bcde. So p is even, hence its last digit e is even. Therefore 4 divides p,hence p = 36n. Since 100 ≤ p ≤ 999, 3 ≤ n ≤ 27. Since p is a product of digits, all prime factors of n are digits.So n = 11, 13, 17, 19, 22, 23, 26. Of the remainder, only n = 8, 16, 18, 21, 24, 27 give 18 for the sum of the digitsof 36n. None of these give an integer value for b = p/2cde.

If c + d + e = 9, then a = 1 and p = bcde. By inspection, the maximum value for cde is 27. So p = 9n with12 ≤ n ≤ 27. Again n = 13, 17, 19, 22, 23, 26. Of the remainder, only n = 15, and n = 16 give a digit value forb = p/cde. If n = 15, then p = 135 and b = 9. If n = 16, then p = 144 and b = 9.

Hence the only compatible pairs are 135, 19 and 144, 19.


AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD STATISTICS

Distribution of Awards/School Year

Year Number of Students

Number of Awards

Prize High Distinction Distinction Credit Participation

8 532 4 35 49 163 281

9 618 20 70 92 211 225

10 584 29 79 93 198 185

Other 343 4 19 16 56 248

All Years 2077 57 203 250 628 939

The award distribution is based on approximately the top 10% for High Distinction, next 15% for Distinction and the following 25% for Credit.

Number of Correct Answers Questions 1–8

YearNumber Correct/Question

1 2 3 4 5 6 7 8

8 399 257 202 105 50 84 85 29

9 519 366 357 122 94 183 148 78

10 515 355 382 138 101 195 159 85

Other 231 121 67 49 20 31 36 22

All Years 1664 1099 1008 414 265 493 428 214

Mean Score/Question/School Year


QuestionOverall Mean

1–8 9 10

8 532 7.9 0.6 0.6 8.4

9 618 10.5 0.8 1.1 11.5

10 584 11.7 0.8 1.1 12.7

Other 343 5.5 0.4 0.6 6.3

All Years 2077 9.4 0.7 0.9 10.2

Mathematics Contests The Australian Scene 2018 | Australian Intermediate Mathematics Olympiad Statistics | 69

2018 AMOC SENIOR CONTEST

Tuesday, 21 August 2018

Time allowed: 4 hours

No calculators are to be used.

Each question is worth seven points.

1. Determine the maximum possible value of a + b, where a and b are two different

non-negative real numbers that satisfy

a+√b = b+

√a.

2. Prove that, among any ten consecutive positive integers, there are five numbers such

that no two of them have a common factor larger than 1.

3. Fourteen people meet one day to play three matches of netball. For each match, they

divide themselves into two teams of seven players. In each match, one team wins

while the other team loses. After all three matches, no person has been on a losing

team three times.

Prove that there are at least three players who were on the same team as each other

for all three matches.

4. Let K1 and K2 be circles that intersect at two points A and B. The tangents to

K1 at A and B intersect at a point P inside K2, and the line BP intersects K2

again at C. The tangents to K2 at A and C intersect at a point Q, and the line QA

intersects K1 again at D.

Prove that QP is perpendicular to PD if and only if the centre of K2 lies on K1.

5. Determine all functions f defined for positive real numbers and taking positive real

numbers as values such that

xf(xf(2y)) = y + xyf(x)

for all positive real numbers x and y.

c© 2018 Australian Mathematics Trust

AMOC SENIOR CONTEST

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest | 70

2018 AMOC SENIOR CONTEST

Solutions

c© 2018 Australian Mathematics Trust

1. Determine the maximum possible value of a + b, where a and b are two different non-

negative real numbers that satisfy

a+√b = b+

√a.

Solution (Angelo Di Pasquale)

The equation may be rewritten as

√a−

√b = a− b ⇔

√a−

√b = (

√a−

√b)(

√a+

√b) ⇔

√a+

√b = 1,

since a = b.

Squaring the last equation and rearranging yields

a+ b = 1− 2√ab ⇒ a+ b ≤ 1.

Since a = 0 and b = 1 satisfy the original equation and satisfy a + b = 1, it follows that

the maximum possible value of a+ b is 1.

1

AMOC SENIOR CONTEST SOLUTIONS

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions | 71

2. Prove that, among any ten consecutive positive integers, there are five numbers such that

no two of them have a common factor larger than 1.

Solution 1 (Norman Do)

Among any ten consecutive positive integers, five of them are odd and the remaining five

are even.

The five odd integers contain either one or two multiples of 3. We remove one multiple

of 3 and select the remaining four integers.

The five even integers contain at most two multiples of 3, at most one multiple of 5,

and at most one multiple of 7. Therefore, there is at least one of them that is not a

multiple of 3, 5 or 7. We select this integer.

We now prove that, among the five integers selected, no two of them have a common factor

larger than 1. Since the largest possible difference between two of the numbers is 9, the

largest possible common factor that two of the numbers can have is 9. So it suffices to

show that we have not selected a pair of numbers with a common factor of 2, 3, 5 or 7.

By construction, we have chosen exactly one number that is divisible by 2.

By construction, we have chosen at most one number that is divisible by 3.

At most one of the odd integers selected is divisible by 5 and, by construction, the

even integer selected is not divisible by 5.

At most one of the odd integers selected is divisible by 7 and, by construction, the

even integer selected is not divisible by 7.

Therefore, we have selected five numbers such that no two of them have a common factor

larger than 1.

Solution 2 (Alice Devillers, Angelo Di Pasquale and Daniel Mathews)

Among the ten consecutive positive integers, let n be the smallest number that is not

divisible by 2 or 3. Either the smallest or the second smallest odd number among the

ten numbers satisfies this condition. So n is one of the four smallest integers and the five

numbers n, n+ 2, n+ 3, n+ 4, n+ 6 are all among the ten consecutive integers.

We proceed to calculate all possible greatest common divisors between two of these five

numbers. We begin with the observation that for any integer n, we have

gcd(n+ 2, n+ 3) = gcd(n+ 3, n+ 4) = 1.

Since n is not divisible by 2, we have

gcd(n, n+2) = gcd(n, n+4) = gcd(n+2, n+4) = gcd(n+2, n+6) = gcd(n+4, n+6) = 1.

Since n is not divisible by 3, we have

gcd(n, n+ 3) = gcd(n+ 3, n+ 6) = 1.

2


Since n is not divisible by 2 or 3, we have

gcd(n, n+ 6) = 1.

This exhausts all possibilities, so no two of the five numbers n, n + 2, n + 3, n + 4, n + 6

have a common factor larger than 1.

Solution 3 (Kevin McAvaney)

Any common factor of two integers must divide their difference. So any common factor

of two of the ten integers is at most 9. Therefore, for such a common factor to be greater

than 1, it must have 2, 3, 5 or 7 as a prime factor.

The sequence of remainders of the ten consecutive integers after division by 2 must be one

of

0101010101 or 1010101010.

The sequence of remainders of the ten consecutive integers after division by 3 must be one

of

0120120120 or 1201201201 or 2012012012.

For the resulting six cases, we choose the five numbers from among the ten original integers

corresponding to the underlined remainders shown below.

0101010101 0101010101 0101010101

0120120120 1201201201 2012012012

1010101010 1010101010 1010101010

0120120120 1201201201 2012012012

One can check that in each of the six cases, none of the primes 2, 3, 5 or 7 is a factor of

more than one of the chosen integers. So in each case, no two of the five chosen numbers

have a common factor lager than 1.

3


3. Fourteen people meet one day to play three matches of netball. For each match, they

divide themselves into two teams of seven players. In each match, one team wins while

the other team loses. After all three matches, no person has been on a losing team three

times.

Prove that there are at least three players who were on the same team as each other for

all three matches.

Solution 1 (Norman Do)

For each person, we record the results of their matches with a string of W s and Ls, where

W denotes a win and L denotes a loss. The record of each player’s results is one of the

following seven possibilities.

WWW WWL WLW WLL LWW LWL LLW

It is impossible for each of these records to be obtained by exactly two players. That

would imply that the sum of the number of wins for each player is 24, while the sum of

the number of losses for each player is 18. However, there should be an equal number of

wins and losses overall.

Hence, there must exist three players who have the same record. It follows that these three

players were on the same team as each other for all three matches.

Solution 2 (Alice Devillers and Ivan Guo)

By renaming players, we can assume that the losing players in match 1 are 1, 2, 3, 4, 5, 6, 7.

The conditions of the problem imply that players 1, 2, 3, 4, 5, 6, 7 must be on the winning

team in at least one of match 2 or match 3.

By renaming players 1, 2, 3, 4, 5, 6, 7 again and swapping the names of matches 2 and 3, we

can assume by the pigeonhole principle that players 1, 2, 3, 4 won match 2.

If player 5 also won match 2, then the pigeonhole principle guarantees that at least three of

the players 1, 2, 3, 4, 5 were on the same team for match 3 as well. Then these three players

were on the same team as each other for all three matches, which is what we wanted to

prove. The same argument may also be applied if player 6 or player 7 won match 2.

The only remaining case to analyse is if players 5, 6, 7 all lost match 2. But in that case,

they must all have won match 3 and were on the same team as each other for all three

matches.

Solution 3 (Chaitanya Rao and Ian Wanless)

Consider the seven players who lost the first match. Then the record of each of these

player’s results for matches 2 and 3 must beWW , WL or LW . By the pigeonhole principle,

at least three of these seven players had the same record for matches 2 and 3. Therefore,

these three players were on the same team as each other for all three matches.

4


Solution 4 (Thanom Shaw)

The 14 people play 3 matches each, which implies that there is a total of 21 wins and 21

losses between them. Suppose that the 21 losses are obtained by x people recording 1 loss

overall, and y people recording 2 losses overall. Since no one loses all 3 matches, this yields

x+ 2y = 21, where x+ y ≤ 14.

The only integer solutions for (x, y) are (1, 10), (3, 9), (5, 8) and (7, 7). Note that in each

case, we have y ≥ 7. That is, 7 or more people recorded 2 losses overall. There are only

three ways to record 2 losses overall — namely, WLL, LWL and LLW . Since at least

7 people had one of these three win–loss records, the pigeonhole principle asserts that at

least 3 of them had the same win–loss record. It follows that these three players were on

the same team for all three matches.

5


4. Let K1 and K2 be circles that intersect at two points A and B. The tangents to K1 at A

and B intersect at a point P inside K2, and the line BP intersects K2 again at C. The

tangents to K2 at A and C intersect at a point Q, and the line QA intersects K1 again

at D.

Prove that QP is perpendicular to PD if and only if the centre of K2 lies on K1.

Solution (Andrew Elvey Price)

We first prove that if the centre O of K2 lies on K1, then QP is perpendicular to PD. Since

DA is tangent to K2, we have ∠DAO = 90, so OD is a diameter of K1. So by symmetry,

D, O and P all lie on the line joining the centres of K1 and K2. Observe that A is the

reflection of B in this line. Hence, ∠PAO = ∠PBO, but ∠PBO = ∠CBO = ∠BCO,

since triangle BOC is isosceles. It follows that AOPC is a cyclic quadrilateral. Moreover,

∠OAQ = ∠OCQ = 90, so OAQC is also a cyclic quadrilateral. Therefore, OAQCP is a

cyclic pentagon and ∠QPO = 90, as required.

We now prove that if QP is perpendicular to PD, then the centre O of K2 lies on K1. By

the alternate segment theorem, we have ∠QCA = ∠CBA, and it follows that

∠QAC = ∠QCA = ∠PAB = ∠PBA.

Let this angle be α. Then ∠APB = 180−2α = ∠AQC, so quadrilateral AQCP is cyclic.

Hence, ∠APQ = α = ∠CPQ and we have ∠APD = ∠BPD = 90 − α. Therefore, D,

O, P and the centre X of K1 are collinear. Since ∠DAO = 90, we can deduce that

∠XOA = 90 − ∠XDA = 90 − ∠XAD = ∠XAO, so XO = XA. It follows that O lies

on K1, as required.

6


5. Determine all functions f defined for positive real numbers and taking positive real num-

bers as values such that

xf(xf(2y)) = y + xyf(x)

for all positive real numbers x and y.

Solution 1 (Angelo Di Pasquale)

Substitute x = 1 into the functional equation to obtain

f(f(2y)) = y(1 + f(1)).

Since 1 + f(1) > 0, it follows that the right side of this equation covers all positive real

numbers as y varies over the positive real numbers. Therefore, f is surjective and there

exists a positive real number a such that f(2a) = 1.

Now substitute y = a into the functional equation to obtain

xf(x) = a (1 + xf(x)) ⇒ f(x) =c

x,

where c = a1−a is a positive real constant. (Note that the equation xf(x) = a (1 + xf(x))

implies that a = 1.)

Substituting f(x) = cx into the functional equation yields

2y = y(1 + c).

Therefore, c = 1 and we deduce that f(x) = 1x . It is easily verified that this is indeed a

solution to the functional equation.

Solution 2 (Angelo Di Pasquale)

Replace y with y2 in the functional equation to obtain

xf(xf(y)) =1

2y (1 + xf(x)) . (∗)

Substitute x = 1 into this equation to deduce that f(f(y)) = ay, where a is a constant.

Now replace y with f(y) in equation (∗) to obtain

xf(xf(f(y))) =1

2f(y)(1 + xf(x)) ⇒ f(axy) =

f(y)(1 + xf(x))

2x=

f(y)

2x+

f(x)f(y)

2.

Interchanging x with y in this equation yields

f(ayx) =f(x)

2y+

f(x)f(y)

2.

Equating the previous two expressions for f(axy) implies that

f(y)

2x=

f(x)

2y.

Now substitute y = 1 into this equation to deduce that

f(x) =c

x,

where c = f(1) is a constant. We may now finish the proof as in the previous solution.

7


AMOC SENIOR CONTEST RESULTS

Name School YearPerfect Score and Gold

James Bang Baulkham Hills High School NSW 11Andres Buritica Scotch College VIC 9Yasiru Jayasooriya James Ruse Agricultural High School NSW 10Sharvil Kesarwani Merewether High School NSW 11Preet Patel Vermont Secondary College VIC 11William Steinberg Scotch College WA 10Hadyn Tang Trinity Grammar School VIC 9Fengshuo (Fredy) Ye (Yip) Knox Grammar School NSW 8Ziqi Yuan Narrabundah ACT 11

GoldHaowen Gao Knox Grammar School NSW 11Ken Gene Quah Melbourne High School VIC 10

SilverZefeng (Jeff) Li Caulfield Grammar School, Caulfield Campus VIC 11Junhua Chen Caulfield Grammar School, Wheelers Hill VIC 10Grace He Methodist Ladies' College VIC 10Mikhail Savkin Gosford High School NSW 10Harry Zhang Christian Brothers College VIC 10Frank Zhao Geelong Grammar School VIC 11David Lee James Ruse Agricultural High School NSW 11Zijin (Aaron) Xu Caulfield Grammar School, Wheelers Hill VIC 10Liam Coy Sydney Grammar School NSW 10Anthony Pisani St Paul's Anglican Grammar VIC 11Jason Wang Queensland Academy for Science, Mathematics and Technology

QLD10

Marcus Rees Hobart College TAS 11Daniel Wiese Scotch College WA 10

BronzeChristopher Leak Perth Modern School WA 10Huxley Berry Perth Modern School WA 10Vicky Feng MLC School NSW 11Evgeniya Artemova Presbyterian Ladies' College VIC 11Yang Zhang St Joseph’s College, Gregory Terrace QLD 10Reef Kitaeff Perth Modern School WA 11Oliver Papillo Camberwell Grammar School VIC 11Patrick Gleeson St Joseph’s College, Gregory Terrace QLD 10Samuel Lam James Ruse Agricultural High School NSW 10Leosha Trushin Perth Modern School WA 10Peter Vowles Wesley College WA 11Kevin Wu Scotch College VIC 11Christopher Do Penleigh and Essendon Grammar School VIC 11Angus Ritossa St Peter's College SA 11

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Results | 78

Name School YearEmilie Wu James Ruse Agricultural High School NSW 11Eva Ge James Ruse Agricultural High School NSW 9Wenguan Lu Barker College NSW 11Claire Huang Radford College ACT 10Steve Wu Prince Alfred College SA 11Chenxiao Zhou Caulfield Grammar School, Wheelers Hill VIC 9Lachlan Rowe Canberra College ACT 11Micah Sinclair Perth Modern School WA 9Le Yao Zha St Edmund's College ACT 10Xiaoyu Chen All Saints' College WA 8

Honourable MentionChristina Lee James Ruse Agricultural High School NSW 10Adrian Lo Newington College NSW 10David Lumsden Scotch College VIC 8Oliver Cheng Hale School WA 9Shevanka Dias All Saints' College WA 11Ethan Ryoo Knox Grammar School NSW 9Zian Shang Scotch College VIC 7Lucinda Xiao Methodist Ladies' College VIC 11Elizabeth Yevdokimov St Ursula’s College QLD 9Matthew Cho St Joseph’s College, Gregory Terrace QLD 10Remi Hart All Saints' College WA 10Linda Lu The Mac.Robertson Girls' High School VIC 11Andrey Lugovsky Perth Modern School WA 11Oliver New Scotch College VIC 8Angela Wang Lauriston Girl's School VIC 9Leo Xu All Saints Anglican School QLD 10William Cheah Penleigh and Essendon Grammar School VIC 4Anagha Kanive-Hariharan James Ruse Agricultural High School NSW 10Dhruv Hariharan Knox Grammar School NSW 9Jocelin Hon James Ruse Agricultural High School NSW 11Yikai Wu Prince Alfred College SA 11Ryan Gray Brisbane State High School QLD 11Mikaela Gray Brisbane State High School QLD 9Kento Seki All Saints Anglican School QLD 10Vivian Wang James Ruse Agricultural High School NSW 10Gen Conway Methodist Ladies' College VIC 8Tom Hauck All Saints Anglican School QLD 10Hanyuan Li North Sydney Boys High School NSW 10Bertrand Nheu Perth Modern School WA 11Benjamin Davison-Petch Christ Church Grammar School WA 11Zhenghao Hua Penleigh and Essendon Grammar School VIC 8Sam Meredith Brisbane State High School QLD 8Tony Teng Concordia College SA 10Yale Cheng Hale School WA 11

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Results | 79

AMOC SENIOR CONTEST STATISTICS

Score Distribution/Problem

Problem Number

Number of Students/ScoreMean

0 1 2 3 4 5 6 7

1 12 9 6 2 4 4 8 51 4.9

2 24 9 0 2 8 1 5 47 4.3

3 24 2 0 0 0 1 1 68 5.1

4 62 8 4 1 1 2 1 17 1.6

5 32 29 8 4 0 2 2 19 2.2

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Statistics | 80

AMOC SCHOOL OF EXCELLENCE

The 2017 AMOC School of Excellence was held 23 November – 2 December at Newman College, University of Melbourne. The main qualifying exams for this are the AIMO and the AMOC Senior Contest.

AMOC had made the decision to start participating in the European Girls’ Mathematical Olympiad (EGMO), starting with the 2018 contest. With this in mind, starting with this year’s School of Excellence, each of the two annual AMOC training schools has permanently increased its intake. This is in order to accommodate the need to have a sufficient pool of girls from which to select an EGMO team each year in addition to an IMO team. The School also replaced the old Junior-Senior system of two streams with a three streams model of Junior, Intermediate, and Senior. The new Junior would be easier than the old Junior, the new Intermediate would be in between the old Junior and the old Senior, and the new Senior would be about the same as the old Senior.

A total of 44 students from around Australia plus one student from New Zealand attended the school. The breakdowns of the Australian students into the three streams, were as follows.

Stream Female Male Total

Senior 0 14 14

Intermediate 4 10 14

Junior 8 8 16

Total 12 32 44

The program covered the four major areas of number theory, geometry, combinatorics and algebra. Each day would start at 8:30am with lectures or an exam and go until 12:30pm. After a one-hour lunch break they would resume the lecture program at 1:30pm. By 4pm participants would usually have free time, followed by dinner at 5:30pm. Finally, each evening would round out with a problem session, topic review, or exam review from 6:30pm until 8:30pm.

Two highly experienced senior students were assigned to give a lecture each. William Hu was assigned the senior Constructions—Reverse and Inspired geometry lecture, and Guowen Zhang was assigned the senior Functional Equations lecture. They both did an excellent job!

Many thanks to Adrian Agisilaou, Ross Atkins, Michelle Chen, Alexander Chua, Andrew Elvey Price, Jongmin Lim, and Thanom Shaw who served as live-in staff.

My thanks also go to Natalie Aisbett, Matthew Cheah, Aaron Chong, Norman Do, Yong See Foo, Ivan Guo, Ilia Kucherov, Alfred Liang, Daniel Mathews, Chaitanya Rao, Sally Tsang, and Jeremy Yip, who assisted in lecturing and marking.

Angelo Di PasqualeDirector of Training, AMOC

Mathematics Contests The Australian Scene 2018 | AMOC School of Excellence | 81

Participants at the 2017 AMOC School of Excellence

Name m/f Year School State

Senior

James Bang m 10 Baulkham Hills High School NSW

Linus Cooper m 11 James Ruse Agricultural High School NSW

Haowen Gao m 10 Knox Grammar School NSW

William Han m 11* Macleans College NZ

William Hu m 11 Christ Church Grammar School WA

Sharvil Kesarwani m 10 Merewether High School NSW

Charles Li m 11 Camberwell Grammar School VIC

Jeff (Zefeng) Li m 10 Caulfield Grammar School VIC

Jack Liu m 11 Brighton Grammar School VIC

William Steinberg m 9 Scotch College WA

Michael Sui m 11 Caulfield Grammar School, Wheelers Hill VIC

Ethan Tan m 10 Cranbrook School NSW

Hadyn Tang m 8 Trinity Grammar School VIC

Guowen Zhang m 11 St Joseph’s College, Gregory Terrace QLD

Stanley Zhu m 11 Melbourne Grammar School VIC

Intermediate

Andres Buritica m 8 Scotch College VIC

Liam Coy m 9 Sydney Grammar School NSW

Yifan Guo f 11 Glen Waverley Secondary College VIC

Grace He f 9 Methodist Ladies' College VIC

Jeffrey Li m 10 North Sydney Boys High School NSW

Adrian Lo m 9 Newington College NSW

Preet Patel m 10 Vermont Secondary College VIC

Ken Gene Quah m 9 Melbourne High School VIC

Mikhail Savkin m 9 Gosford High School NSW

Ruiqian Tong f 11 Presbyterian Ladies' College VIC

Daniel Wiese m 9 Scotch College WA

Fengshuo (Fredy) Ye (Yip) m 7 Chatswood High School NSW

Ziqi Yuan m 10 Lyneham High School ACT

Xinyue (Alice) Zhang f 11 A. B. Paterson College QLD

Junior

Evgeniya Artemova f 10 Presbyterian Ladies’ College VIC

Huxley Berry m 9 Perth Modern School WA

Junhua Chen m 9 Caulfield Grammar School, Wheelers Hill VIC

Matthew Cho m 9 St Joseph’s College, Gregory Terrace QLD

Vicky Feng f 10 Methodist Ladies’ College NSW

Eva Ge f 8 James Ruse Agricultural High School NSW

Dhruv Hariharan m 8 Knox Grammar School NSW

Jocelin Hon f 10 James Ruse Agricultural High School NSW

Samuel Lam m 9 James Ruse Agricultural High School NSW

Christina Lee f 9 James Ruse Agricultural High School NSW

Andy Li f 10 Presbyterian Ladies’ College VIC

Ethan Ryoo m 8 Knox Grammar School NSW

Marc Silins m 7 Australian Christian College Marsden Park NSW

Emilie Wu f 10 James Ruse Agricultural High School NSW

Yang Zhang m 9 St Joseph’s College, Gregory Terrace QLD*Equivalent to year 12 in Australia.

Mathematics Contests The Australian Scene 2018 | AMOC School of Excellence | 82

AUSTRAL IAN MATHEMAT ICAL OLYMP IAD COMMITTEE

A DEPARTMENT OF THE AUSTRAL IAN MATHEMAT ICS TRUST

The Mathematics/Informatics Olympiads are supported by the Australian Government through the National Innovation and Science Agenda.Official sponsor of the Olympiad program

© 2018 Australian Mathematics Trust

AUSTRALIAN MATHEMATICAL OLYMPIAD

2018

DAY 1

Tuesday, 6 February 2018




1. Find all pairs of positive integers (n, k) such that

n! + 8 = 2k.

(If n is a positive integer, then n! = 1× 2× 3× · · ·× (n− 1)× n.)

2. Consider a line with 12(3

100 + 1) equally spaced points marked on it.

Prove that 2100 of these marked points can be coloured red so that no red point is at the

same distance from two other red points.

3. Let ABCDEFGHIJKLMN be a regular tetradecagon.

Prove that the three lines AE, BG and CK intersect at a point.

(A regular tetradecagon is a convex polygon with 14 sides, such that all sides have the

same length and all angles are equal.)

4. Find all functions f defined for real numbers and taking real numbers as values such that

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

for all real numbers x and y.






2018

DAY 2

Wednesday, 7 February 2018




5. The sequence a1, a2, a3, . . . is defined by a1 = 1 and, for n ≥ 2,

an = (a1 + a2 + · · ·+ an−1)× n.

Prove that a2018 is divisible by 20182.

6. Let P , Q and R be three points on a circle C, such that PQ = PR and PQ > QR.

Let D be the circle with centre P that passes through Q and R. Suppose that the circle

with centre Q and passing through R intersects C again at X and D again at Y .

Prove that P , X and Y lie on a line.

7. Let b1, b2, b3, . . . be a sequence of positive integers such that, for each positive integer n,

bn+1 is the square of the number of positive factors of bn (including 1 and bn). For

example, if b1 = 27, then b2 = 42 = 16, since 27 has four positive factors: 1, 3, 9 and 27.

Prove that if b1 > 1, then the sequence contains a term that is equal to 9.

8. Amy has a number of rocks such that the mass of each rock, in kilograms, is a positive

integer. The sum of the masses of the rocks is 2018 kilograms. Amy realises that it is

impossible to divide the rocks into two piles of 1009 kilograms.

What is the maximum possible number of rocks that Amy could have?


Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad | 83






2018

DAY 2






an = (a1 + a2 + · · ·+ an−1)× n.



















2018

DAY 2






an = (a1 + a2 + · · ·+ an−1)× n.














Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad | 84

Solutions to the 2018 Australian Mathematical Olympiad

1. Solution 1 (Evgeniya Artemova, year 11, Presbyterian Ladies’ College, VIC)

Answers (n, k) = (4, 5) and (5, 7).

For reference the given equation is

n! + 8 = 2k.

Case 1 n ≥ 6

Observe that n! is a multiple of 6! = 24 × 32 × 5. Hence n! = 16x for some positiveinteger x. Therefore

n! + 8 = 8(2x+ 1).

But the RHS of the above equation cannot be a power of 2 because 2x+1 is an oddinteger that is greater than 1. Hence there are no solutions in this case.

Case 2 n ≤ 5

We simply tabulate the values of n! + 8 and check which ones are powers of 2.

n n! + 8 power of 2?

1 9 no

2 10 no

3 14 no

4 32 25

5 128 27

This yields the solutions given at the outset.

27

AUSTRALIAN MATHEMATICAL OLYMPIAD SOLUTIONS

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 85

Solution 2 (Andres Buritica, year 9, Scotch College, VIC)

For reference the given equation is

n! + 8 = 2k.

Case 1 n ≥ 6

We have 24 | n! and 23 ‖ 8. Thus 23 ‖ n! + 8.1 Hence 23 ‖ 2k, and so k = 3. But thisimplies n! = 0, which is impossible. So there are no solutions in this case.

Case 2 n ≤ 3

We have 23 n! and 23 | 8. Hence 23 2k, and so k < 3. It follows that n! < 0, whichis impossible. So there are no solutions in this case.

Case 3 n = 4 or 5

If n = 4 then k = 5, and if n = 5 then k = 7.

1For a prime number p and integers k ≥ 0 and N ≥ 1, the notation pk ‖ N means that pk | N butpk+1 N . Put another way, it means that the exponent of p in the prime factorisation of N is k.

28


2. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA)

Given some equally spaced points marked on a line, if we colour some of them red,we say the colouring is good if no red point is equidistant from two other red points.The result is the special case n = 100 of the following more general claim.

Claim For each positive integer n, if a line has 12(3n + 1) equally spaced points

marked on it, then there is a good colouring of 2n of those points.

The proof is by induction on n.

The base case n = 1 holds as we simply colour each of the 12(31+1) = 21 points red.

For the inductive step, suppose that the claim is true for some positive integer n.Let k = 1

2(3n + 1). Note that 3k− 1 = 1

2(3n+1 + 1). To prove the claim for n+ 1 we

divide the 3k − 1 points into three groups from left to right as follows.

k points

group A

k − 1 points

group B

k points

group C

Using the inductive assumption we colour 2n points in group A red so that no redpoint in A is equidistant from two other red points in A. Similarly we colour 2n

points in C red so that no red point in C is equidistant from two other red points inC. All points in B are left uncoloured.

A total of 2n + 2n = 2n+1 points have been coloured red in the union of the threegroups. Suppose, for the sake of contradiction, that this colouring is not good. Thenthere are three red points W,X, Y in that order from left to right, where WX = XY .Without loss of generality X is in A. Thus W is also in A. However,

• Y is not in A, from the inductive assumption,

• Y is not in B, because no point in Y is coloured red, and

• Y is not in C, because WX ≤ k − 1 and XY ≥ k.

This contradiction completes the induction and the proof.

Comment In order to illustrate the above proof, here are some iterations of theinductive construction.

n = 1

n = 2

n = 3

29


Solution 2 (Sharvil Kesarwani, year 11, Merewether High School, NSW)

Without loss of generality, we identify the 12(3100 + 1) equally spaced points with

the integers from 0 up to 12(3100 − 1) on the real number line.

Colour red each integer of the form

N =99∑i=0

di · 3i

where di ∈ 0, 1 for 0 ≤ i ≤ 99. In this way exactly 2100 integers are coloured red.Note that these are precisely the integers in the range from 0 to 1

2(3100 − 1) which

do not contain the digit 2 in their ternary (base-3) representations.

Suppose, for the sake of contradiction, that one red integer B is equidistant fromtwo other red integers A and C, where A < B < C. Thus C −B = B −A, which isthe same as 2B = A+ C.

Let

A =99∑i=0

ai · 3i, B =99∑i=0

bi · 3i, and C =99∑i=0

ci · 3i

be the ternary representations of A, B, and C, where ai, bi, ci ∈ 0, 1 for 0 ≤ i ≤ 99.From 2B = A+ C we have

99∑i=0

2bi · 3i =99∑i=0

(ai + ci)3i.

Since ai, bi, ci ∈ 0, 1, we have 2bi ∈ 0, 2 and ai + ci ∈ 0, 1, 2. Therefore bothsides of the above equation are the ternary representation of the same number. Itfollows that 2bi = ai + ci for each i.

If bi = 0, then ai = ci = 0. And if bi = 1, then ai = ci = 1. Either way, A and Chave the same ternary digits, and so A = C. This contradicts A < C, and completesthe proof.

30


3. Solution 1 (Elizabeth Yevdokimov, year 9, St Ursula’s College, QLD)

Since the tetradecagon is regular, it has a circumcircle.

Form triangle AGC by joining the segments AC, CG, and GA.

A B

C

D

E

F

G

HI

J

K

L

M

N

Since the vertices of the tetradecagon are equally spaced around the circle, and equallength arcs subtend equal angles, we have

∠CAE = ∠EAG, ∠AGB = ∠BGC, and ∠GCK = ∠KCA.

Hence AE, BG, and CK are the internal bisectors of the angles of AGC. As suchthey are concurrent at the incentre of AGC.

31


Solution 2 (Based on the solution by Ethan Ryoo, year 9, Knox Grammar School,NSW)

As in solution 1, we consider A,B, . . . , N as being 14 equally spaced points arounda circle. Form triangle BEK by joining segments BE, EK, and KB.

A B

C

D

E

F

G

HI

J

K

L

M

N

By symmetry we have parallel chords BG ‖ CF ‖ DE. Since DK is a diameter ofthe circle, we have DE ⊥ EK. Hence BG ⊥ EK.

Similarly we have EA ‖ DB and DB ⊥ BK. Hence EA ⊥ BK.

Finally we have KC ‖ LB and LB ⊥ BE (LE is a diameter). Hence KC ⊥ BE.

We have shown that BG, EA, and KC are altitudes of BEK. As such they areconcurrent at the orthocentre of BEK.

32


Solution 3 (Ethan Tan, year 12, Cranbrook School, NSW)

As in solution 1, we consider A,B, . . . , N as being 14 equally spaced points arounda circle. Let X be the intersection of lines AE and BG. It suffices to show that C,X, and K are collinear.

A B

C

D

E

F

G

HI

J

K

L

M

N

X

Since the vertices of the tetradecagon are equally spaced around the circle, and equallength arcs subtend equal angles, we have

∠XEB = ∠AEB = ∠BEC and ∠CBE = ∠EBG = ∠EBX.

It follows that BEX ≡ BEC (ASA). Hence BCEX is a kite with BE ⊥ CX.

We also have parallel chords BE ‖ CD. Hence CD ⊥ CX. But CD ⊥ CK becauseDK is a diameter of the circle. Thus CK ‖ CX, from which it follows that C, X,and K are collinear, as desired.

33


Solution 4 (Yifan Guo, year 12, Glen Waverley Secondary College, VIC)

As in solution 1, we consider A,B, . . . , N as being 14 equally spaced points arounda circle. Let X be the intersection of lines AE and BG. It suffices to show that C,X, and K are collinear. Let α satisfy 14α = 180.

A B

C

D

E

F

G

HI

J

K

L

M

N

X

Each of the 14 equal sides of the tetradecagon subtends an angle of 360/14 withthe centre of the circle. Hence each of these sides subtends an angle of α = 180/14with a point on the major arc opposite the side. Using this we have

∠BAE = ∠BAC + ∠CAD + ∠DAE = 3α.

Similar calculations yield

∠GBA = 8α and ∠KCB = 5α.

A consideration of the angle sum in ABX yields ∠BXA = 3α. Hence BX = BA.However BA = BC. Hence BXC is isosceles with apex B. Since ∠CBX = 4α, itfollows that ∠BXC = ∠XCB = 5α.

Finally, since ∠XCB = 5α = ∠KCB, it follows that C, X, and K are collinear, asdesired.

34


Solution 5 (Mikhail Savkin, year 10, Gosford High School, NSW)

As in solution 1, we consider A,B, . . . , N as being 14 equally spaced points arounda circle. Form the hexagon ABCEGK.

A B

C

D

E

F

G

HI

J

K

L

M

N

We quote a theorem that will help us solve the problem.

Theorem If UVWXY Z is a convex cyclic hexagon, then its main diagonals UX,V Y , and WZ are concurrent if and only if

UV ·WX · Y Z = VW ·XY · ZA.

This result is found in configuration C6 in chapter 5 of Problem Solving Tactics.2

We apply the theorem as follows. From the equal spacing of points of the tetradecagonaround the circle, we deduce that

AB = BC, CE = EG, and GK = KA.

Since ABCEGK is a convex cyclic hexagon with the property that

AB · CE ·GK = BC · EG ·KA,

the result follows.

2Published by the AMT

35


4. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA)

Answers f(x) = 0 and f(x) = 1− x

For reference the given functional equation is

f(xy + f(y)) = yf(x) (1)


Put y = 0 in (1) to findf(f(0)) = 0. (2)

Next put x = 0 and y = f(0) in (1), and use (2) to deduce

f(0) = f(0)2.

Thus f(0) = 0 or 1.

Case 1 f(0) = 0

Put x = 0 in (1) to findf(f(y)) = 0 (3)

for all y ∈ R.

Suppose there exists a real number c with f(c) = 0. Then if we put x = c in (1), wesee that RHS(1) covers all real numbers, and so f is surjective. But if f is surjective,then so is f f , which contradicts (3). Hence no such c exists. Thus f(x) = 0 for allreal numbers x. This function obviously satisfies (1).

Case 2 f(0) = 1

Putting x = 0 into (1) yieldsf(f(y)) = y (5)

for all real numbers y. It follows that f(1) = f(f(0)) = 0.

Putting x = 1 in (1) yields

f(y + f(y)) = 0 = f(1) ⇒ f(f(y + f(y))) = f(f(1)).

With the help of (5) this implies y + f(y) = 1, and so f(y) = 1− y.

It only remains to verify that f(x) = 1− x satisfies (1). Indeed we have

LHS(1) = 1− (xy + (1− y)) = y(1− x) = RHS(1).

Hence f(x) = 1− x is also a solution to the problem.

36


Solution 2 (William Steinberg, year 10, Scotch College, WA)

For reference the given functional equation is

f(xy + f(y)) = yf(x) (1)


Put x = 0 in (1) to findf(f(y)) = yf(0), (2)

for all real numbers y.

Putting y = 1 in (1) yields

f(x+ f(1)) = f(x)

⇒ f(f(x+ f(1))) = f(f(x))

⇒ (x+ f(1))f(0) = xf(0) (from (2))

⇒ f(0)f(1) = 0. (3)

Two cases follow from (3).

Case 1 f(0) = 0

Equation (2) now becomesf(f(y)) = 0 (4)

for all real numbers y.

Applying f to both sides of (1) and then using (4) yields

f(yf(x)) = 0. (5)

Suppose that there exists a real number c with f(c) = 0. Then putting x = c andy = c/f(c) in (5), we deduce that f(c) = 0. This is a contradiction. Thus no such cexists. Hence f(x) = 0 for all real numbers x.

Case 2 f(0) = 0

It follows from (3) that f(1) = 0.

Setting x = 1 in (1), and using f(1) = 0 yields

f(y + f(y)) = 0

⇒ f(f(y + f(y))) = f(0)

⇒ (y + f(y))f(0) = f(0) (from (2))

⇒ y + f(y) = 1. (since f(0) = 0)

Thus f(y) = 1− y.

As in solution 1, we check that f(x) = 0 and f(x) = 1 − x each satisfy the givenfunctional equation.

37


5. Solution 1 (Vicky Feng, year 11, Methodist Ladies’ College, NSW)

We are givenan = n(a1 + a2 + · · ·+ an−1) (1)

for any integer n ≥ 2.

Replacing n with n− 1 in (1), we find

an−1 = (n− 1)(a1 + a2 + · · ·+ an−2) (2)


Substituting (2) into (1) yields

an = n(a1 + a2 + · · ·+ an−2 + an−1)

= n(a1 + a2 + · · ·+ an−2 + (n− 1)(a1 + a2 + · · ·+ an−2))

= n2(a1 + a2 + · · ·+ an−2).

Hence an is divisible by n2 for each integer n ≥ 3. In particular, a2018 is divisible by20182.

Comment We did not use the initial condition a1 = 1. Hence the result is truefor any integer a1.

38


Solution 2 (Xinyue (Alice) Zhang, year 12, A. B. Paterson College, QLD)

We prove an even stronger result, namely a2018 is divisible by 20183.

First we prove by induction that a1 + a2 + · · ·+ an−1 = n!/2 for all integers n ≥ 2.

The base case is true because a1 = 1 = 2!2.

For the inductive step, assume that a1 + a2 + · · · + an−1 = n!/2 for some integern ≥ 2. Then

a1 + a2 + · · ·+ an−1 + an = a1 + a2 + · · ·+ an−1 + n(a1 + a2 + · · ·+ an−1)

=n!

2+

n · n!2

=(n+ 1) · n!

2

=(n+ 1)!

2

which completes the induction.

Using this result, it follows that

a2018 = 2018(a1 + · · ·+ a2017)

= 2018(2018× 2017× · · · × 3)

= 20182(2017× 2016× · · · × 1009× · · · × 4× 3)

which is obviously a multiple of 20182. However 2017×2016×· · ·×1009×· · ·×4×3is divisible by both 2 and 1009, and so it is also divisible by 2018. Therefore a2018 isdivisible by 20183.

39


Solution 3 (Andres Buritica, year 9, Scotch College, VIC)

We are givenann

= a1 + a2 + · · ·+ an−1 (1)


Replacing n with n+ 1 in (1), we find

an+1

n+ 1= a1 + a2 + · · ·+ an−1 + an

=ann

+ an

⇒ an+1

an=

(n+ 1)2

n(2)


Multiplying all instances of equality (2) together for n = m− 1,m− 2, . . . , 2, wherem ≥ 3, yields a product that telescopes as follows.

amam−1

× am−1

am−2

× · · · × a3a2

=m2

m− 1× (m− 1)2

m− 2× · · · × 32

2

⇒ ama2

=m2(m− 1)!

4

⇒ am =m2(m− 1)!

2

In particular a2018 = (2018)2 × 2017!2

, which is a multiple of 20182.

40


6. Solution 1 (William Steinberg, year 10, Scotch College, WA)

Since R, X, and Y all lie on a circle centred at Q, we have QR = QX = QY .

P Q

R

Y

X

D

C

In circle D, chords QY and QR have equal length. Hence they subtend equal anglesat the centre P of this circle. Thus

∠Y PQ = ∠QPR. (1)

In circle C, chords QX and QR have equal length. Hence they subtend equal anglesat the point P on the circumference of this circle. Thus

∠XPQ = ∠QPR. (2)

From (1) and (2) we have ∠XPQ = ∠Y PQ. Thus P , X, and Y are collinear.

41


Solution 2 (Matthew Kerr, year 12, St Anthony’s Catholic College, QLD)

Let the circle centred at Q and passing through R intersect the line PY for a secondtime at X ′. It suffices to prove that PRQX ′ is cyclic as this implies X ′ = X.

P Q

R

Y

X ′

As P and Q are centres of circles, we have PR = PQ = PY and QR = QX ′ = QY .Hence PQR ≡ PQY (SSS). Thus

∠PRQ = ∠QY P = ∠QYX ′ = ∠Y X ′Q

where the last equality is due to QX ′ = QY .

Since ∠PRQ = ∠Y X ′Q, it follows that PRQX ′ is cyclic, as desired.

42


Solution 3 (Sharvil Kesarwani, year 11, Merewether High School, NSW)

Let Y ′ be the intersection of ray PX with circle D. It suffices to prove that Y = Y ′.

Let ∠QY ′X = ∠QY ′P = α.

P Q

R

X

Y ′

D

C

α

Observe that PY ′ = PQ = PR because they are radii of circle D. It follows that∠PQY ′ = ∠QY ′P = α. Considering the angle sum in PQY yields

∠Y ′PQ = 180 − 2α.

Observe that QX = QR because X lies on the circle centred at Q and passingthrough R. Since QX and QR are equal length chords in circle C, they subtendequal angles at the point P on the circumference of this circle. Hence

∠QPR = ∠XPQ = ∠Y ′PQ = 180 − 2α.

Since PQR is isosceles with apex P , the angle sum in this triangle yields

∠PRQ = ∠RQP = α.

Since PRQX is a cyclic quadrilateral, we have

∠Y ′XQ = ∠PRQ = α = ∠QY ′X.

Hence QXY ′ is isosceles with apex Q. Thus QY ′ = QX = QR, and so Y ′ lies onthe circle centred at Q and passing through R. Since Y ′ and R are on opposite sidesof the line PQ, it follows that Y = Y ′.

43


Solution 4 (Alan Offer, AMOC Senior Problems Committee)

Comment Let E be the circle in the problem statement centred at Q and passingthrough R. If we relax the condition that Q is the centre of E by requiring only thatthe centre of E lies on C, then the conclusion of the problem is still true.

Here is a proof of this generalisation of the problem.

P Q

R

O

Y

X

D

C

We have only considered the configuration where X lies between P and Y . Otherconfigurations may be dealt with similarly.

Let O be the centre of E . Since PR = PY (radii of D), and OR = OY (radii of E),it follows that OPR ≡ OPY (SSS). Thus

∠Y PO = ∠OPR. (1)

In circle C, chords OX and OR have equal length. Hence they subtend equal anglesat the point P on the circumference of this circle. Thus

∠XPO = ∠OPR. (2)

From (1) and (2) we have ∠XPO = ∠Y PO. Thus P , X, and Y are collinear.

44


Comments (Angelo Di Pasquale, Director of Training, AMOC)

The astute reader may have noticed that the point Q did not get mentioned at allin the previous proof. The reason is because, in this generalisation, the point Qis irrelevant to the proof. Hence the diagram could have been drawn without thepoint Q and without the line segments PQ and QR. This minimal statement of thegeneralisation of the problem is as follows.

Let P , O, and R be three points on a circle C. Let D be a circle centredat P and passing through R, and let E be the circle centred at O andpassing through R. Suppose that E intersects C again at X and D againat Y . Then P , X, and Y lie on a line.

In fact we can say even more! There is a combinatorial symmetry in the problemstatement between D and E with respect to C. Specifically, exchanging points P andO also exchanges D and E . Thus if D intersects C again at Q, then O, Q, and Yalso lie on a line.

P Q

R

O

Y

X

D

C

E

45


7. Solution 1 (Tianyue (Ellen) Zheng, year 12, Smith’s Hill High School, NSW)

Since b1 > 1, it follows inductively that bn > 1 for all n ≥ 2.

Observe that bn is a perfect square for all n ≥ 2.

For any perfect square m2, all of the factors of m2, except for m, come in pairs(d,m2/d) where 1 ≤ d < m. Hence square numbers have an odd number of factors.Thus bn is an odd perfect square for all n ≥ 3. Moreover since bn > 1, it follows thatbn ≥ 9 for all n ≥ 3.

Consider any integer n ≥ 3. Then bn = m2 for some odd integer m ≥ 3. As explainedin the previous paragraph, apart from m, the factors of m2 come in pairs (d,m2/d)where 1 ≤ d < m. Since m2 is odd, each such d is odd. The number of odd positiveintegers less than m is equal to m−1

2. It follows that the number of factors of m2 is

at most 2× m−12

+ 1 = m. Therefore bn+1 ≤ bn.

However, for bn+1 = bn to occur, all of the odd positive integers less than m mustbe factors of m2. In particular m− 2 | m2. But

gcd(m,m− 2) = gcd(m, 2) = 1 ⇒ gcd(m2,m− 2) = 1.

Since m− 2 | m2, it follows that m − 2 = 1. Thus m = 3, and bn = 9. So from b3onward, the sequence is strictly decreasing until it reaches the fixed point bn = 9.

46


Solution 2 (James Bang, year 11, Baulkham Hills High School, NSW)

As in solution 1 we deduce that each member of the sequence from b3 onward is anodd perfect square that is greater than or equal to 9.

Consider any integer n ≥ 3. As in solution 1 it suffices to prove that bn+1 ≤ bn withequality if and only if bn = 9.

Let bn =r∏

i=1

p2aii , where p1 < p2 < · · · < pr are the prime factors of bn.

The number of factors of bn is given byr∏

i=1

(2ai + 1). Hence bn+1 =r∏

i=1

(2ai + 1)2.

It follows that bn+1 ≤ bn if and only if

r∏i=1

paii ≥r∏

i=1

(2ai + 1). (1)

To finish the proof it suffices to show that

pa ≥ 2a+ 1 (2)

with equality if and only if p = 3 and a = 1.

Since pa > 3a whenever p > 3, it suffices to prove (2) just for the case p = 3. Weproceed by induction on a.

For a = 1, it is readily seen that inequality (2) is in fact an equality.

For the inductive step, suppose that 3a ≥ 2a+ 1 for some a ≥ 1. It follows that

3a+1 ≥ 3(2a+ 1) > 2(a+ 1) + 1.

This concludes the induction and the proof.

Comment Here are two more alternative ways of proving inequality (2).

Alternative 1 (Angelo Di Pasquale, Director of Training AMOC)

Using the binomial theorem, the inequality follows from

p2a ≥ (1 + 2)2a ≥ 1 +

(2a

1

)· 2 +

(2a

2

)· 22 = 8a2 + 1 ≥ (2a+ 1)2.

Alternative 2 (Ivan Guo, AMOC Senior Problems Committee)

Factoring the difference of perfect ath powers, the inequality follows from

pa − 1 ≥ 3a − 1 = (3− 1)(3a−1 + 3a−2 + · · ·+ 1) ≥ 2(1 + 1 + · · ·+ 1) = 2a.

47


8. For ease of exposition, in each of the solutions that follow, whenever we say some-thing has mass n, then we use this as shorthand for saying that it has mass nkilograms.

Solution 1 (James Bang, year 11, Baulkham Hills High School, NSW)

Answer 1009

Suppose that Amy has r ≥ 1010 rocks of masses m1,m2, . . . ,mr. For each integer iwith 1 ≤ i ≤ r, let

Si = m1 +m2 + · · ·+mi.

Note that S1, S2, . . . , Sr is a strictly increasing sequence of r ≥ 1010 positive integers,each of which is in the range 1, 2, . . . , 2018.

Consider the 1009 pairs

(1, 1010), (2, 1011), (3, 1012), . . . , (1009, 2018).

By the pigeonhole principle one of the above pairs contains two of the Si. Supposethat (t, t+ 1009) = (Si, Sj). Then j > i, and

1009 = Sj − Si = mj +mj−1 + · · ·+mi+1.

Thus mj,mj−1, . . . ,mi+1 have total mass equal to 1009. This shows that the answercannot be greater than or equal to 1010.

To complete the proof, here is a construction for exactly 1009 rocks. Suppose thatAmy has 1 rock of mass 1010, and 1008 rocks of mass 1. Then the pile that containsthe rock of mass 1010 obviously has mass exceeding 1009.

Comment 1 Some students found a somewhat geometric version of the aboveargument which basically goes as follows.

Suppose Amy has at least 1010 rocks. Consider a circle of circumference 2018 andpartition it into arcs whose lengths correspond to the masses of the rocks. Theendpoints of the arcs are 1010 (or more) of the vertices of a regular 2018-gon, andso by the pigeonhole principle there is a diametrically opposite pair of endpoints.

Comment 2 Some students stated and proved the following fairly well-knownlemma, and used this to complete the proof.

Lemma Any sequence of n integers contains a nonempty subsequence whose sumis a multiple of n.

Proof Let the integers be m1,m2, . . . ,mn. Let Si = m1+m2+· · ·+mi for 1 ≤ i ≤ n.

If Si ≡ 0 (mod n) for some i, then we are done.

Otherwise, by the pigeonhole principle we have Si ≡ Sj (mod n) for some i < j. Itfollows that Sj − Si = mi+1 +mi+2 + · · ·+mj ≡ 0 (mod n).

To complete the proof of the given problem, if we have at least 1010 rocks, considerany 1009 of them. By the lemma, a nonempty sub-collection of those rocks has massequal to a multiple of 1009. Since the total mass of the sub-collection is strictly lessthan 2018, it is equal to 1009.

48


Solution 2 (William Steinberg, year 10, Scotch College, WA)

Suppose that Amy has at least 1010 rocks. Suppose that a of the rocks have unitmass and b of the rocks have mass greater than 1. If b = 0, then a = 2018, and inthis case it is obvious that the rocks can be divided into two piles of equal mass.Hence b > 0. Let M ≥ 2 be the mass of the heaviest rock. We have the followinginequalities.

a+ b ≥ 1010 (1)

a+ 2(b− 1) +M ≤ 2018 (2)

Inequality (1) simply states that there are at least 1010 rocks. Inequality (2) is foundby estimating the total mass. It is true because among the b rocks of non-unit mass,one of them has mass M , and the rest each have mass at least 2.

Inequality (1) implies b ≥ 1010− a. Substituting this into inequality (2) yields

2018 ≥ M + a+ 2(b− 1) ≥ M + a+ 2(1010− a− 1) = M + 2018− a.

It follows that a ≥ M.

Next we create a pile of rocks with total mass 1009 as follows. First, continuallychoose rocks of non-unit mass and add them to the pile one at a time until one ofthe following two things occurs.

(i) It is not possible to add any more rocks of non-unit mass without the mass ofthe pile under construction exceeding 1009.

(ii) We run out of rocks of non-unit mass and the mass of the pile is at most 1009.

If (i) occurs, then the difference between the current mass of the pile and 1009 isless than M . Since a ≥ M we can top up the pile with rocks of unit mass until thepile has mass exactly 1009.

If (ii) occurs, then all remaining rocks have unit mass. Hence we can top up the pilewith these until the pile has mass exactly 1009.

Either way a pile of mass 1009 has been created. This shows that the answer cannotbe greater than or equal to 1010.

To complete the proof, here is a construction for exactly 1009 rocks. Suppose thatAmy has 1009 rocks, each of mass 2. If some of these rocks are put into a pile, thenthe total mass of the pile will be even, and so cannot have mass 1009.

49


Solution 3 (Guowen Zhang, year 12, St Joseph’s College, QLD)

We claim the answer is 1009.

As in solution 1, if Amy has 1 rock of mass 1010 and 1008 rocks of mass 1, then itis impossible to divide the rocks into two piles of equal mass. Hence the answer isgreater than or equal to 1009. The proof that this is the largest possible number ofrocks that Amy could have follows from the following lemma.

Lemma Let n be any positive integer. Suppose that Amy has at least n+ 1 rockswhere the mass of each rock is a positive integer, and the total mass of all the rocksis 2n. Then it is always possible to divide the rocks into two piles, each of mass n.

Proof We proceed by induction on n.

For the case n = 1, Amy has two rocks with total mass 2. This implies that each ofAmy’s rocks has unit mass, from which the result immediately follows.

For the inductive step, let us assume that the lemma is true for n = k. Consider thecase n = k + 1. Thus Amy has at least k + 2 rocks whose total mass is 2k + 2.

If all of Amy’s rocks have mass greater than or equal to 2, then the total mass ofall the rocks is greater than or equal to 2(k+ 2) > 2k+ 2, which is a contradiction.Hence at least one of Amy’s rocks has unit mass. If all of the rocks have unit massthen the result follows immediately. Hence we may let the masses of the rocks be

1,m1,m2, . . . ,mr (*)

where r ≥ k + 1 and mr > 1.

Consider the situation where we have r rocks of masses

m1,m2, . . . ,mr−1,mr − 1.

Observe that the number of rocks above is greater than or equal to k + 1, and thetotal mass of these rocks is equal to 2k. From the inductive assumption we candivide these into two piles, each of total mass k. Next change mr − 1 to mr and adda single rock of unit mass to the pile not containing the rock of mass mr − 1. Thisgives a division of the rocks with masses given in (*) into two piles, each havingmass k + 1. This concludes the induction, and the proof.

Comment It is possible to strengthen the lemma as follows.

Lemma Let n be any positive integer. Suppose that Amy has at least n+ 1 rockswhere the mass of each rock is a positive integer, and the total mass of all the rocksis 2n. Then for any integer N with 0 ≤ N ≤ 2n it is always possible to divide therocks into two piles, one of which has mass N .

The proof is very similar to the one given above.

50


AUSTRALIAN MATHEMATICAL OLYMPIAD RESULTS

Name School YearPerfect Score and Gold

James Bang Baulkham Hills High School NSW 11

Jack Gibney Penleigh and Essendon Grammar School VIC 12

William Han Macleans College NZ 12*

Grace He Methodist Ladies' College VIC 10

Charles Li Camberwell Grammar School VIC 12

Haobin (Jack) Liu Brighton Grammar School VIC 12

Jerry Mao Caulfield Grammar School, Wheelers Hill VIC 12

William Steinberg Scotch College WA 10

Ethan Tan Cranbrook School NSW 11

Fengshuo (Fredy) Ye (Yip) Chatswood High School NSW 8

Guowen Zhang St Joseph's College QLD 12

GoldWilliam Hu Christ Church Grammar School WA 12

Yasiru Jayasooriya James Ruse Agricultural High School NSW 10

Hadyn Tang Trinity Grammar School VIC 9

Ziqi Yuan Narrabundah College ACT 11

SilverAndres Buritica Scotch College VIC 9

Andrew Chen St Kentigern College NZ 13*

Linus Cooper James Ruse Agricultural High School NSW 12

Liam Coy Sydney Grammar School NSW 10

Haowen Gao Knox Grammar NSW 11

Sharvil Kesarwani Merewether High School NSW 11

Johnathan Leung King’s College NZ 10*

Keiran James Lewellen Te Aho o Te Kura Pounamu NZ 13*

Steven Lim Hurlstone Agricultural High School NSW 12

Adrian Lo Newington College NSW 10

Forbes Mailler Canberra Grammar School ACT 12

Oliver Papillo Camberwell Grammar School VIC 11

Preet Patel Vermont Secondary College VIC 11

Tang (Michael) Sui Caulfield Grammar School, Wheelers Hill VIC 12

Anthony Tew Pembroke School SA 12

Xutong Wang Auckland International College NZ 13*

BronzeGrace Chang St Kentigern College NZ 11*

Kieren Connor Sydney Grammar School NSW 12

Carl Gu Melbourne High School VIC 12

Yifan Guo Glen Waverley Secondary College VIC 12

Rick Han Macleans College NZ 10*

Matthew Kerr St Anthony’s Catholic College QLD 12

David Lee James Ruse Agricultural High School NSW 11

*NZ school year

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Results | 109

Name School YearAlan Li Lincoln High School NZ 12*

William Li Barker College NSW 12

Zefeng (Jeff) Li Caulfield Grammar School, Caulfield Campus VIC 11

Ishan Nath John Paul College NZ 11*

Anthony Pisani St Paul’s Anglican Grammar VIC 11

Ken Gene Quah Melbourne High School VIC 10

Marcus Rees Hobart College TAS 11

Lachlan Rowe Canberra College ACT 11

Mikhail Savkin Gosford High School NSW 10

Albert Smith Christ Church Grammar School WA 12

Ryan Stocks Radford College ACT 12

Brian Su James Ruse Agricultural High School NSW 12

Ruiqian Tong Presbyterian Ladies’ College VIC 12

Andrew Virgona Smith’s Hill High School NSW 12

Daniel Wiese Scotch College WA 10

Kevin Wu Scotch College VIC 11

Shine Wu Newlands College NZ 13*

Zijin (Aaron) Xu Caulfield Grammar School, Wheelers Hill VIC 10

Xinyue Alice Zhang A. B. Paterson College QLD 12

Yang Zhang St Joseph’s College QLD 10

Linan (Frank) Zhao Geelong Grammar School VIC 11

Tianyue (Ellen) Zheng Smith’s Hill High School NSW 12

Stanley Zhu Melbourne Grammar School VIC 12

Honourable MentionVincent Abbott Hale School WA 12

Evgeniya Artemova Presbyterian Ladies’ College VIC 11

Huxley Berry Perth Modern School WA 10

Junhua Chen Caulfield Grammar School, Wheelers Hill VIC 10

Jonathan Chew Christ Church Grammar School WA 11

Matthew Cho St Joseph’s College QLD 10

Shevanka Dias All Saints’ College WA 11

Christopher Do Penleigh and Essendon Grammar School VIC 11

Vicky Feng Methodist Ladies’ College NSW 11

Eva Ge James Ruse Agricultural High School NSW 9

Mikaela Gray Brisbane State High School QLD 9

Ryan Gray Brisbane State High School QLD 11

Dhruv Hariharan Knox Grammar NSW 9

Remi Hart All Saints’ College WA 10

Tom Hauck All Saints Anglican School QLD 10

Jocelin Hon James Ruse Agricultural High School NSW 11

Claire Huang Radford College ACT 10

Hollis Huang Tintern Grammar School VIC 12

Shivasankaran Jayabalan Rossmoyne Senior High School WA 12

Anagha Kanive-Hariharan James Ruse Agricultural High School NSW 10

*NZ school year


Name School YearReef Kitaeff Perth Modern School WA 11

Samuel Lam James Ruse Agricultural High School NSW 10

Andy Li Presbyterian Ladies’ College VIC 11

Jeffrey Li North Sydney Boys High School NSW 11

Yueqi (Rose) Lin Shenton College WA 12

Linda Lu The Mac.Robertson Girls’ High School VIC 11

Wenquan Lu Barker College NSW 11

Nathaniel Masfen-Yan King’s College NZ 11*

Hilton Nguyen Sydney Technical High School NSW 12

Angus Ritossa St Peter’s College SA 11

Ethan Ryoo Knox Grammar NSW 9

Jianyi (Jason) Wang Queensland Academy for Science, Mathematics and Technology QLD 10

Ziang (Tommy) Wei Scotch College VIC 12

Emilie Wu James Ruse Agricultural High School NSW 11

Lucinda Xiao Methodist Ladies’ College VIC 11

Jason Yang James Ruse Agricultural High School NSW 11

Xinrong Yao Auckland International College NZ 12*

Christine Ye Qs School VIC 8

Elizabeth Yevdokimov St Ursula’s College QLD 9

Jasmine Zhang Macleans College NZ 11*

Zirui (Harry) Zhang Christian Brothers College VIC 9

Yufei (Phoebe) Zuo Tara Anglican School for Girls NSW 12

*NZ school year


AUSTRALIAN MATHEMATICAL OLYMPIAD STATISTICS

Score Distribution/Problem

Number of Students/Score

Problem Number

1 2 3 4 5 6 7 8

0 2 49 59 58 14 25 44 18

1 2 4 10 15 0 4 11 65

2 14 3 1 6 5 23 13 7

3 2 2 0 3 0 0 6 1

4 0 0 0 2 0 1 1 1

5 0 2 0 3 0 1 6 1

6 20 12 1 4 4 2 4 3

7 81 49 50 30 98 65 36 25

Average 6 3.6 3 2.4 6 4.3 3 2.3

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Statistics | 112

XXX Asian Pacific Mathematics Olympiad

March, 2018

Time allowed: 4 hours Each problem is worth 7 points

The contest problems are to be kept confidential until they are posted on the officialAPMO website http://apmo.ommenlinea.org.Please do not disclose nor discuss the problems over online until that date. The useof calculators is not allowed.

Problem 1. Let H be the orthocenter of the triangle ABC. Let M and N bethe midpoints of the sides AB and AC, respectively. Assume that H lies inside thequadrilateral BMNC and that the circumcircles of triangles BMH and CNH aretangent to each other. The line through H parallel to BC intersects the circumcirclesof the triangles BMH and CNH in the points K and L, respectively. Let F be theintersection point of MK and NL and let J be the incenter of triangle MHN . Provethat FJ = FA.

Problem 2. Let f(x) and g(x) be given by

f(x) =1

x+

1

x− 2+

1

x− 4+ · · ·+ 1

x− 2018

and

g(x) =1

x− 1+

1

x− 3+

1

x− 5+ · · ·+ 1

x− 2017.

Prove that|f(x)− g(x)| > 2

for any non-integer real number x satisfying 0 < x < 2018.

Problem 3. A collection of n squares on the plane is called tri-connected if thefollowing criteria are satisfied:

(i) All the squares are congruent.

(ii) If two squares have a point P in common, then P is a vertex of each of thesquares.

(iii) Each square touches exactly three other squares.

30th ASIAN PACIFIC MATHEMATICS OLYMPIAD

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad | 113

How many positive integers n are there with 2018 ≤ n ≤ 3018, such that there existsa collection of n squares that is tri-connected?

Problem 4. Let ABC be an equilateral triangle. From the vertex A we draw aray towards the interior of the triangle such that the ray reaches one of the sides ofthe triangle. When the ray reaches a side, it then bounces off following the law ofreflection, that is, if it arrives with a directed angle α, it leaves with a directed angle180 − α. After n bounces, the ray returns to A without ever landing on any of theother two vertices. Find all possible values of n.

Problem 5. Find all polynomials P (x) with integer coefficients such that for all realnumbers s and t, if P (s) and P (t) are both integers, then P (st) is also an integer.

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad | 114

Solutions to the 2018 Asian Pacific Mathematical Olympiad

1. Solution 1 (Tianyue (Ellen) Zheng, year 12, Smith’s Hill High School, NSW)

Let Y and Z be the feet of the perpendiculars from B and C to AC and AB,respectively. Hence BZY C is cyclic due to ∠BZC = 90 = ∠BY C. So we may let

∠HBZ = ∠Y CH = x.

Our plan is to calculate many angles in terms of x.

A

B C

H

M N

Z

Y

LK

F

J

Since M and N are the midpoints of AB and AC, respectively, it follows thatMN ‖ BC ‖ KL. Using this and cyclic BKMH yields

∠NMF = ∠LKF = ∠HKM = ∠HBM = x.

Similarly,∠FNM = ∠FLK = x.

The above angle equalities imply FK = FL and FM = FN . Hence

FM · FK = FN · FL.

Thus F has equal power with respect to circles BKMH and CHNL. Hence F lieson the radical axis of these two circles. Since the two circles are tangent, it followsthat F lies on the common tangent at H. Hence FH is tangent to the two circlesat H. Using the alternate segment theorem we deduce

∠FHM = ∠HBM = x and ∠NHF = ∠NCH = x.

In particular FH bisects ∠NHM , and so F , J , and H are collinear. Another conse-quence is that FMHN is cyclic due to ∠NMF = x = ∠NHF. But now we have a

51

30th ASIAN PACIFIC MATHEMATICS OLYMPIAD SOLUTIONS

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 115

standard configuration where J is the incentre of MNH and F is the intersectionof the circumcircle of MNH with the line HJ .1 In this configuration it is knownthat FM = FJ = FN , but here is a proof anyway. We calculate

∠FJM = ∠JHM + ∠HMJ = x+ ∠JMN = ∠JMF.

Thus FM = FJ . A similar argument shows that FN = FJ . Thus F is the centreof circle MJN .

We also calculate

NJM = 180 − ∠JMN − ∠MNJ

= 180 − 1

2(∠HMN + ∠MNH)

= 180 − 1

2(180 − ∠NHM)

= 180 − 1

2(180 − 2x)

= 90 + x.

Also ∠MAN = ∠BAC = 90−x, and so ∠NJM+∠MAN = 180. Thus MJNA iscyclic. Recall F is the centre of circle MJN . Thus F is the centre of circle MJNA,and so FA = FJ , as desired.

1See configuration B6 in chapter 5 of Problem Solving Tactics published by the AMT.

52


Solution 2 (William Hu, year 12, Christ Church Grammar School, WA)

As in solution 1, we have KL ‖ BC ‖ MN , and chase out the following equal angles.

∠NMF = ∠HKM = ∠HBM = ∠NCH = ∠NLH = ∠FNM = x

x x

x x

x x

y

y

z

z

A

B C

H

M N

LK

F

J

P

Let ∠CBH = y and ∠HCB = z. From MN ‖ BC we have ∠NMA = x + y and∠ANM = x+ z. Since ∠NMF = ∠FNM = x, it follows that

∠FMA = y and ∠ANF = z.

Observe that AMN is related to ABC by a dilation centred at A. Let H ′ be theimage of H under this dilation. Then H ′ is the orthocentre of AMN . Also we have∠NMH ′ = ∠CBH = y = ∠FMA, and ∠H ′NM = ∠HCB = z = ∠ANF . HenceH ′ and F are isogonal conjugates in AMN . But the orthocentre and circumcentreof any triangle are isogonal conjugates, and so F is the circumcentre of AMN . 2

In order to complete the proof it only remains to show that MJNA is cyclic becausethen F would be the centre of circle MJNA. Let P be any point on the commontangent at H of the two circles such that P lies on the same side of the line KLas A. Using the alternate segment theorem we find

∠NHM = ∠NHP + ∠PHM = ∠NLH + ∠HKM = 2x.

Using the same angle calculations near the end of solution 1, we deduce that∠NJM = 90 + x, and ∠MAN = 90 − x. Consequently AMJN is cyclic dueto ∠NJM + ∠MAN = 180, as desired.

2See problem 1 and the corresponding footnote in section 6.0 of Problem Solving Tactics published bythe AMT.

53


2. Solution (Based on the solution by James Bang, year 11, Baulkham Hills HighSchool, NSW)

There are two cases: 2n− 1 < x < 2n and 2n < x < 2n+ 1. Note that

f(2018− x) = −f(x) and g(2018− x) = −g(x),

that is, a half turn about the point (1009, 0) preserves the graphs of f and g. Soit suffices to consider only the case 2n < x < 2n + 1 where n is an integer with0 ≤ n ≤ 1008. We have the following.

f(x)− g(x) =1

x+

1009∑i=1

(1

x− 2i− 1

x− (2i− 1)

)(1)

=1

x+

1009∑i=1

1

(x− (2i− 1))(x− 2i)(2)

Observe that x − (2i − 1) and x − 2i are both positive if i ≤ n, and they are bothnegative if i ≥ n+ 1. Thus 1

(x−(2i−1))(x−2i)> 0 for each i.

Case 1 0 < x < 1

By dropping all terms in (2) with i ≥ 2, we see that

f(x)− g(x) >1

x+

1

(x− 1)(x− 2).

Thus it suffices to show

1

x+

1

(x− 1)(x− 2)> 2 (3)

⇔ (x− 1)(x− 2) + x > 2x(x− 1)(x− 2) (4)

⇔ −2x3 + 7x2 − 6x+ 2 > 0

⇔ x2 + 2(1− x)3 > 0. (5)

Note that (4) follows from (3) because x(x− 1)(x− 2) > 0 for 0 < x < 1.

It remains to observe that (5) is obviously true for 0 < x < 1.

Case 2 2n < x < 2n+ 1 where n is an integer with 1 ≤ n ≤ 1008

By dropping all terms in (2) except for those with i = n and i = n+ 1, we see that

f(x)− g(x) >1

(x− (2n− 1))(x− 2n)+

1

(x− (2n+ 1))(x− (2n+ 2)).

Let us use the substitution y = x− 2n, so that 0 < y < 1. It suffices to prove that

1

y(y + 1)+

1

(y − 1)(y − 2)> 2 (6)

⇔ (y − 1)(y − 2) + y(y + 1) > 2y(y + 1)(y − 1)(y − 2) (7)

⇔ −y4 + 2y3 + 2y2 − 3y + 1 > 0

⇔ y3(1− y) + y(y − 1)2 + (2y − 1)2 > 0. (8)

Note that (7) follows from (6) because y(y + 1)(y − 1)(y − 2) > 0 for 0 < y < 1.

It remains to observe that (8) is obviously true for 0 < y < 1.

54


Comment 1 A shortcut can be made in the preceding solution as follows.

f(x)− g(x)− 1

x+ 1=

1009∑i=0

(1

x− 2i− 1

x− (2i− 1)

)

⇒ f(x)− g(x) >1009∑i=0

1

(x− (2i− 1))(x− 2i)

Having the above inequality removes the need to treat 0 < x < 1 as a separate case.

Comment 2 Here is an alternative proof of inequality (6) from the precedingsolution.

1

y(y + 1)+

1

(y − 1)(y − 2)> 2

⇔ 1

y− 1

y + 1+

1

y − 2− 1

y − 1> 2

⇔ 1

y− 1

y − 1+

1

y − 2− 1

y + 1> 2

⇔ 1

y(1− y)+

1

y − 2− 1

y + 1> 2.

However since y and 1− y are both positive, the GM–HM inequality yields

1

y· 1

1− y≥

(2

y + (1− y)

)2

= 4.

Also for 0 < y < 1 we have

1

y − 2> −1 and − 1

y + 1> −1.

Putting it all together yields

1

y(1− y)+

1

y − 2− 1

y + 1> 4− 1− 1 = 2,

as desired.

55


3. Solution (Guowen Zhang, year 12, St Joseph’s College, QLD)

Answer 501

For 2018 ≤ n ≤ 3018 we claim that a collection of n tri-connected squares is possibleif and only if n is even.

For any collection of n tri-connected squares, consider the graph G obtained asfollows. Each vertex of G corresponds to a square in the collection, and two verticesof G are joined by an edge if and only if the two corresponding squares touch.Observe that the sum of the degrees of the vertices of G is equal to 3n. Thus thetotal number of edges of G is equal to 3n

2. This implies that n is even.

Consider the following two configurations of squares.

Call the 6-square configuration on the left an A-piece, and the 4-square configurationon the right a B-piece. Five A-pieces can be linked together to make a tri-connectedcollection of 30 squares. And four B-pieces can be linked together to make a tri-connected collection of 16 squares. These are shown below.

Each even integer n with 2018 ≤ n ≤ 3018 can be written in the form n = 30a+16bfor some integers a, b ≥ 0. For example, 30, 60, 90, 120, 150, 180, 210, 240 representsall even congruence classes modulo 16, and then we can top up with multiples of 16.

Take a copies of the 30-square configuration on the left and b copies of the 16-squareconfiguration on the right, making sure that all a + b configurations are mutuallydisjoint. This yields a tri-connected collection of n squares. Indeed we have shownthat a tri-connected collection of n squares exists for all even n ≥ 240.

Comment (Guowen Zhang, year 12, St Joseph’s College, QLD)

Alternatively, we could link together a A-pieces and b B-pieces into a single closedloop of a + b pieces whenever a and b are non-negative integers with a + b ≥ 4.Thus any n = 6a+ 4b = 2a+ 4(a+ b) with a+ b ≥ 4 is possible. This shows that atri-connected collection of n squares exists for any even n ≥ 16.

56


4. Solution (Hadyn Tang, year 9, Trinity Grammar School, VIC)

Answer All n ≡ 1, 5 (mod 6) with the exceptions of 5 and 17.

Consider the equilateral triangular lattice obtained by repeatedly reflecting triangleABC in one of its three sides.

A B

CP

For each point P that is a vertex of the triangular lattice, we write P = (r, s) wherer and s are the unique integers r and s such that

−→AP= r

−→AB +s

−→AC .

Note that A = (0, 0), B = (1, 0) and C = (0, 1) under this system.

Next we colour all points as follows. The point (r, s) is coloured red, white, or blackaccording to r− s ≡ 0, 1, or 2 (mod 3), respectively. It is easy to show by inductionthat each reflected image of A is red, each reflected image of B is white, and eachreflected image of C is black.

Consider a path in triangle ABC. Each time the path reflects off a side of thetriangle, we reflect the triangle along with the remaining part of the path in thesaid side. In this way, since the original path obeys the law of reflection, it unfoldsinto a straight line through the above triangular lattice. A path that starts and endsat A in the original triangle now corresponds to a straight line segment through thetriangular lattice that starts at the red vertex at A and ends at another red vertexP say. One such original path in triangle ABC along with its unfolded path AP isillustrated in the diagram. In the example P = (4, 1).

In general, if P has coordinates (r, s), then P is red if and only if

r ≡ s (mod 3). (1)

We are told that the ray returns to A after n bounces without ever landing on anyof the other two vertices. Thus the segment AP contains no lattice points strictlybetween A and P . This is true if and only if

gcd(r, s) = 1. (2)

57


With (1) in view, this implies that

r ≡ s ≡ 1 (mod 3) or r ≡ s ≡ 2 (mod 3). (3)

Let us calculate the number of bounces n in terms of r and s. Since AP crosses r−1lines parallel to AC, s− 1 lines parallel to AB, and r + s− 1 lines parallel to BC,we have

n = 2(r + s)− 3. (4)

Equation (4) forces n to be odd. Also from (3) we have 3 r+ s and so 3 n. It onlyremains to discuss n ≡ 1, 5 (mod 6).

Case 1 n = 6k + 1 for some integer k ≥ 0

It is straightforward to verify that P = (1, 3k + 1) satisfies (1), (2), and (4).


It is straightforward to verify that P = (2, 6k + 5) satisfies (1), (2), and (4).


It is straightforward to verify that P = (6k − 1, 6k + 5) satisfies (1), (2), and (4).


It is straightforward to verify that P = (6k − 1, 6k + 11) satisfies (1), (2), and (4).

Observe that cases 3 and 4 together cover all n = 12k+5 for all integers k ≥ 2, andtogether with case 2 cover all n ≡ 5 (mod 6) except for n = 5 and n = 17.

Case 5 n = 5

From (4), we require r + s = 4. And r ≡ s (mod 3) from (1). Thus (r, s) = (2, 2).But this violates (2). So there are no solutions in this case.

Case 6 n = 17

From (4), we require r + s = 10. And r ≡ s (mod 3) from (1). Thus (r, s) = (5, 5),(2, 8), or (8, 2). But these violate (2). So there are no solutions in this case either.

Having covered all cases, the proof is complete.

58


5. Solution (Based on the presentation by William Hu, year 12, Christ Church Gram-mar School, WA)

Answer P (x) = xn + c and P (x) = −xn + c for any integers c and n with n ≥ 0.

It is straightforward to verify that the above stated answers are solutions to theproblem. We claim that there are no further solutions.

For any integer polynomial P we shall say that a real number r is P -good if P (r) isan integer.

Suppose that P is a solution to the problem. We may write

P (x) = a0 + a1x+ · · ·+ anxn

for some integers a0, a1, a2, . . . , an where an = 0. Let S be the set of P -good realnumbers. Clearly Z ⊆ S. Also the problem statement informs us that if r, s ∈ Sthen rs ∈ S.

Consider the polynomial

Q(x) = 2nP (x)− P (2x)

=n−1∑i=0

(2n − 2i)aixi.

If r ∈ S, then since 2 ∈ S, we have 2r ∈ S. Thus Q(r) = 2rP (r) − P (2r) ∈ Z. Sowe have shown the following.

Each real number that is P -good is also Q-good. (1)

We claim that a1 = a2 = · · · = an−1 = 0. Since 2n − 2i = 0 for i < n, this is thesame as saying that Q is a constant polynomial.

Suppose, for the sake of contradiction, that deg(Q) = m where 1 ≤ m ≤ n−1. SinceP and Q are polynomials of positive degree, both are strictly monotonic continuousfunctions for all sufficiently large real numbers u. But any strictly monotonic contin-uous real function is also a bijection. So P attains each integer value between P (u)and P (u+ 1) inclusive exactly once. Hence there are exactly 1 + |P (u+ 1)− P (u)|P -good real numbers in the interval [u, u + 1]. We similarly reason that there areexactly 1 + |Q(u + 1) − Q(u)| Q-good real numbers in the interval [u, u + 1]. Butfrom (1), each P -good number in [u, u+ 1] is also Q-good. Hence

1 + |Q(u+ 1)−Q(u)| ≥ 1 + |P (u+ 1)− P (u)|, (2)

for all sufficiently large u. Let P1(x) = P (x+1)−P (x) and Q1(x) = Q(x+1)−Q(x).Note that deg(P1) = n − 1 > m − 1 = deg(Q1). Hence |P1(x)| > |Q1(x)| for allsufficiently large x, which contradicts (2).

So we have shown that a1 = a2 = · · · = an−1 = 0. Hence P (x) = anxn + a0.

Let r = 1

|an|1n. Since P (r) = ±1 + a0 ∈ Z, we have r ∈ S. Thus also r2 ∈ S. So

P (r2) = 1an

+ a0 ∈ Z. Hence an = ±1, which concludes the proof.

59


30th ASIAN PACIFIC MATHEMATICS OLYMPIAD RESULTS

Top 10 Australian scores

Name School Year TotalSilver Award

Guowen Zhang St Joseph's College QLD 12 21

William Hu Christ Church Grammar School WA 12 20

Charles Li Camberwell Grammar School VIC 12 20

Bronze AwardEthan Tan Cranbrook School NSW 11 20

Hadyn Tang Trinity Grammar School VIC 9 17

James Bang Baulkham Hills High School NSW 11 17

William Steinberg Scotch College WA 10 14

Honourable MentionHaowen Gao Knox Grammar NSW 11 12

Jerry Mao Caulfield Grammar School, Wheelers Hill VIC 12 12

Fengshuo (Fredy) Ye (Yip) Chatswood High School NSW 8 12

Country Scores

Rank Country Number of Contestants

Total Score

Gold Awards

Silver Awards

Bronze Awards

Honourable Mentions

1 Republic of Korea 10 320 1 2 4 3

2 USA 10 306 1 2 4 3

3 Japan 10 254 1 2 4 3

4 Singapore 10 231 1 2 4 3

5 Canada 10 220 1 2 4 3

6 Russia 10 210 1 2 4 3

7 Taiwan 10 208 1 2 4 3

8 Islamic Republic of Iran 10 186 1 2 4 3

9 Thailand 10 185 1 2 4 3

10 Indonesia 10 173 1 2 4 3

11 Peru 10 166 1 2 4 3

12 Australia 10 165 0 3 4 313 Philippines 10 153 1 2 4 3

14 Brazil 10 148 0 2 5 3

15 Hong Kong 10 145 0 3 4 2

16 Kazakhstan 10 141 0 1 6 3

17 Bangladesh 10 135 0 2 5 3

18 Mexico 10 132 0 1 6 3

19 India 10 123 0 0 7 3

20 Argentina 10 122 0 2 4 3

21 Malaysia 10 100 0 2 2 3

22 Saudi Arabia 10 97 0 0 3 7

23 New Zealand 10 88 0 1 3 1

24 Macedonia 10 85 0 0 2 7

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Results | 124

Rank Country Number of Contestants

Total Score

Gold Awards

Silver Awards

Bronze Awards

Honourable Mentions

25 Turkmenistan 10 85 0 0 2 8

26 Tajikistan 10 75 0 0 0 10

27 Bolivia 5 66 0 2 1 1

28 Colombia 9 60 0 0 2 3

29 Syria 10 60 0 0 2 4

30 Kyrgyzstan 8 54 0 0 1 5

31 Pakistan 10 47 0 0 1 1

32 Sri Lanka 6 40 0 0 0 5

33 El Salvador 4 28 0 0 1 1

34 Nicaragua 10 24 0 0 0 3

35 Trinidad and Tobago 10 24 0 0 0 1

36 Panama 3 21 0 0 0 3

37 Cambodia 5 20 0 0 0 1

38 Costa Rica 10 12 0 0 0 1

39 Guatemala 2 7 0 0 0 0

Total 352 4716 12 43 109 124

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Results | 125

AMOC SELECTION SCHOOL

The 2018 AMOC Selection School was held 21–30 March at Robert Menzies College, Macquarie University, Sydney. The qualifying exam was the 2018 AMO.

A total of 43 students from around Australia attended the school. The breakdowns of the students into the three streams, were as follows.

Stream Female Male Total

Senior 1 15 16

Intermediate 6 10 16

Junior 9 2 11

Total 16 27 43

The routine is similar to that for the AMOC School of Excellence; however, there is the added interest of the actual selection of the Australian IMO team. This year the IMO would be held in Cluj-Napoca, Romania in the month of July. It is from the seniors that the team of six for the 2018 IMO plus one reserve team member would be selected.

The team of four girls for the 2018 EGMO had already been selected on the basis of the AMO. This year’s EGMO would be held in Florence Italy in the month of April. So their presence at the AMOC Selection School would constitute the main part of their final preparations for EGMO.

Unfortunately I was unable to attend the School in person this year. I am grateful to Ivan Guo who ably took the lead in my absence.

Many thanks to Adrian Agisilaou, Ross Atkins, Michelle Chen, Jongmin Lim, Thanom Shaw, and Andy Tran who assisted Ivan as live-in staff members.

My thanks also go to Alexander Babidge, Dzmitry Badziahin, Stephen Farrar, Sean Gardiner, Victor Khou, Vickie Lee, Vinoth Nandakumar, Gareth White, Rachel Wong, Sampson Wong, and Jonathan Zheng, all of whom came in to give lectures or help with the marking of exams.

Angelo Di PasqualeDirector of Training, AMOC

Mathematics Contests The Australian Scene 2018 | AMOC Selection School | 126

2018 Australian EGMO team

Name Year School State

Yifan Guo 12 Glen Waverley Secondary College VIC

Grace He 10 Methodist Ladies’ College VIC

Xinyue (Alice) Zhang 12 A. B. Paterson College QLD

Tianyue (Ellen) Zheng 12 Smith’s Hill High School NSW

Reserve

Vacant in 2018

2018 Australian IMO team

Name Year School State

William Hu 12 Christ Church Grammar School WA

Charles Li 12 Camberwell Grammar School VIC

William Steinberg 10 Scotch College WA

Ethan Tan 12 Cranbrook School NSW

Hadyn Tang 9 Trinity Grammar School VIC

Guowen Zhang 12 St Joseph’s College, Gregory Terrace QLD

Reserve

Yasiru Jayasooriya 10 James Ruse Agricultural High School NSW


Participants at the 2018 AMOC Selection School

Name m/f Year School State

Senior

James Bang m 11 Baulkham Hills High School NSW

Linus Cooper m 12 James Ruse Agricultural High School NSW

Haowen Gao m 11 Knox Grammar School NSW

Jack Gibney m 12 Penleigh and Essendon Grammar School VIC

Grace He f 10 Methodist Ladies’ College VIC

William Hu m 12 Christ Church Grammar School WA

Yasiru Jayasooriya m 10 James Ruse Agricultural High School NSW

Sharvil Kesarwani m 11 Merewether High School NSW

Charles Li m 12 Camberwell Grammar School VIC

Haobin (Jack) Liu m 12 Brighton Grammar School VIC

William Steinberg m 10 Scotch College WA

Ethan Tan m 12 Cranbrook School NSW

Hadyn Tang m 9 Trinity Grammar School VIC

Fengshuo (Fredy) Ye (Yip) m 8 Chatswood High School NSW

Ziqi Yuan m 11 Narrabundah College ACT

Guowen Zhang m 12 St Joseph’s College, Gregory Terrace QLD

Intermediate

Evgeniya Artemova f 11 Presbyterian Ladies’ College VIC

Andres Buritica m 9 Scotch College VIC

Liam Coy m 10 Sydney Grammar School NSW

Vicky Feng f 11 Methodist Ladies’ College NSW

Yifan Guo f 12 Glen Waverley Secondary College VIC

Jocelin Hon f 11 James Ruse Agricultural High School NSW

David Lee m 11 James Ruse Agricultural High School NSW

Adrian Lo m 10 Newington College NSW

Preet Patel m 11 Vermont Secondary College VIC

Ken Gene Quah m 10 Melbourne High School VIC

Mikhail Savkin m 10 Gosford High School NSW

Daniel Wiese m 10 Scotch College WA

Zijin (Aaron) Xu m 10 Caulfield Grammar School, Wheelers Hill VIC

Xinyue (Alice) Zhang f 12 A. B. Paterson College QLD

Yang Zhang m 10 St Joseph’s College, Gregory Terrace QLD

Tianyue (Ellen) Zheng f 12 Smith’s Hill High School NSW

Junior

Genevieve Conway f 8 Methodist Ladies’ College VIC

Eva Ge f 9 James Ruse Agricultural High School NSW

Mikaela Gray f 9 Brisbane State High School QLD

Claire Huang f 10 Radford College ACT

Anagha Kanive-Hariharan f 10 James Ruse Agricultural High School NSW

Linda Lu f 11 The Mac.Robertson Girls’ High School VIC

Ying Tan f 10 Presbyterian Ladies’ College VIC

Jianyi (Jason) Wang m 10 Queensland Academy for Science, Mathematics & Technology QLD

Christine Ye f 8 Lauriston Girls' School VIC

Elizabeth Yevdokimov f 9 St Ursula’s College QLD

Zirui (Harry) Zhang m 9 Christian Brothers College VIC


EGMO TEAM LEADER’S REPORT

The 7th European Girls’ Mathematical Olympiad (EGMO) was held 9–15 April 2018, in the city of Florence, Italy. This was the first time Italy has hosted an Olympiad and it was the biggest EGMO to date with 8 new countries participating in 2018: Australia, Austria, Bolivia, Canada, Germany, Greece, Mongolia, and Peru. For this 7th EGMO, a total of 195 contestants from 52 countries participated (137 contestants from 36 official European countries)1.

Peru had the youngest team with an average age of 14 years and 10 months, Latvia had the oldest team with an average age of 18 years and 7 months, and Lichtenstein, a country of just 40000, had the smallest team of one.

Each participating country may send a team of up to four students, a Team Leader and a Deputy Team Leader. At the EGMO, the Team Leaders form what is called the Jury, which in 2018 was chaired by Roberto Dvornicich. For decisions such as approving each paper as well as marking schemes, all Leaders may vote, and motions are carried by a simple majority of those voting. For decisions such as determining medal cut-offs and coordination appeals, however, only Leaders of official European teams may vote.

Participating countries are invited to submit up to six proposed problems to be received by the Problem Selection Committee, which in 2018 was chaired by Massimo Gobbino. This committee selects the contest problems and has alternative problems prepared if a problem turns out to be already known. The selected problems were presented to the Jury in their first meeting on Tuesday 10 April. With no objections from the Jury, problems for both Day 1 and Day 2 papers were approved, including translations into the 33 languages required by the contestants. The Jury, Observers, and any others who have knowledge of the problems and solutions before the examinations, are trusted to do their utmost to ensure that no contestant has any information, direct or indirect, about any proposed problem.

The six problems on the EGMO contest papers may be described as follows.

1. A lovely geometry problem about a point lying on a circle, proposed by Velina Ivanova, Bulgaria.

2. A number theory problem about expressing integers as a product of numbers of the form

The 7th EGMO, Florence, Italy

The 7th European Girls’ Mathematical Olympiad (EGMO) was held 9–15 April 2018,in the city of Florence, Italy. This was the first time Italy has hosted an Olympiad andit was the biggest EGMO to date with 8 new countries participating in 2018: Australia,Austria, Bolivia, Canada, Germany, Greece, Mongolia, and Peru. For this 7th EGMO, atotal of 195 contestants from 52 countries participated (137 contestants from 36 officialEuropean countries)1. Peru had the youngest team with an average age of 14 years and10 months, Latvia had the oldest team with an average age of 18 years and 7 months,and Lichtenstein, a country of just 40000, had the smallest team of one.

Each participating country may send a team of up to four students, a Team Leaderand a Deputy Team Leader. At the EGMO, the Team Leaders form what is called theJury, which in 2018 was chaired by Roberto Dvornicich. For decisions such as approvingeach paper as well as marking schemes, all Leaders may vote, and motions are carried bya simple majority of those voting. For decisions such as determining medal cut-offs andcoordination appeals, however, only Leaders of official European teams may vote.

Participating countries are invited to submit up to six proposed problems to be receivedby the Problem Selection Committee, which in 2018 was chaired by Massimo Gobbino.This committee selects the contest problems and has alternative problems prepared if aproblem turns out to be already known. The selected problems were presented to the Juryin their first meeting on Tuesday 10 April. With no objections from the Jury, problems forboth Day 1 and Day 2 papers were approved, including translations into the 33 languagesrequired by the contestants. The Jury, Observers, and any others who have knowledgeof the problems and solutions before the examinations, are trusted to do their utmostto ensure that no contestant has any information, direct or indirect, about any proposedproblem.


1. A lovely geometry problem about a point lying on a circle, proprosed by VelinaIvanova, Bulgaria.

2. A number theory problem about expressing integers as a product of numbers of theform

(1 + 1

k

), proposed by Mihail Baluna, Romania.

3. A fun combinatorics problem involving a Jury choosing and then rearranging theorder of contestants in a queue. It was proposed by Hungary.

4. A combinatorics problem involving balanced configurations of dominoes on an n×nboard. It was proposed by Merlijn Staps, the Netherlands.

5. A classical circle geometry problem, proposed by Dominika Regiec, Poland.

6. A tricky algebra problem involving the closeness of pairs of elements in a set ofpositive integers. It was proposed by Merlijn Staps, the Netherlands.

Like for the International Mathematical Olympiad, Problems 1–3 appeared on theDay 1 Paper and problems 4–6 appeared on the Day 2 Paper. These papers were heldon Wednesday 11 April and Thursday 12 April respectively, with the contestants allowedfour and a half hours per paper to work individually on the problems and write theirattempted proofs. Each problem was scored out of a maximum of seven points.

1This is a new record for the EGMO, with the number of participating countries increasing every year.In 2012 for the first edition there were only 19 countries, 16 of them official European ones.

1

, proposed by Mihail Baluna, Romania.

3. A fun combinatorics problem involving a Jury choosing and then rearranging the order of contestants in a queue. It was proposed by Hungary.

4. A combinatorics problem involving balanced configurations of dominoes on an

The 7th EGMO, Florence, Italy

The 7th European Girls’ Mathematical Olympiad (EGMO) was held 9–15 April 2018,in the city of Florence, Italy. This was the first time Italy has hosted an Olympiad andit was the biggest EGMO to date with 8 new countries participating in 2018: Australia,Austria, Bolivia, Canada, Germany, Greece, Mongolia, and Peru. For this 7th EGMO, atotal of 195 contestants from 52 countries participated (137 contestants from 36 officialEuropean countries)1. Peru had the youngest team with an average age of 14 years and10 months, Latvia had the oldest team with an average age of 18 years and 7 months,and Lichtenstein, a country of just 40000, had the smallest team of one.

Each participating country may send a team of up to four students, a Team Leaderand a Deputy Team Leader. At the EGMO, the Team Leaders form what is called theJury, which in 2018 was chaired by Roberto Dvornicich. For decisions such as approvingeach paper as well as marking schemes, all Leaders may vote, and motions are carried bya simple majority of those voting. For decisions such as determining medal cut-offs andcoordination appeals, however, only Leaders of official European teams may vote.

Participating countries are invited to submit up to six proposed problems to be receivedby the Problem Selection Committee, which in 2018 was chaired by Massimo Gobbino.This committee selects the contest problems and has alternative problems prepared if aproblem turns out to be already known. The selected problems were presented to the Juryin their first meeting on Tuesday 10 April. With no objections from the Jury, problems forboth Day 1 and Day 2 papers were approved, including translations into the 33 languagesrequired by the contestants. The Jury, Observers, and any others who have knowledgeof the problems and solutions before the examinations, are trusted to do their utmostto ensure that no contestant has any information, direct or indirect, about any proposedproblem.


1. A lovely geometry problem about a point lying on a circle, proprosed by VelinaIvanova, Bulgaria.

2. A number theory problem about expressing integers as a product of numbers of theform

(1 + 1

k

), proposed by Mihail Baluna, Romania.

3. A fun combinatorics problem involving a Jury choosing and then rearranging theorder of contestants in a queue. It was proposed by Hungary.

4. A combinatorics problem involving balanced configurations of dominoes on an n×nboard. It was proposed by Merlijn Staps, the Netherlands.


6. A tricky algebra problem involving the closeness of pairs of elements in a set ofpositive integers. It was proposed by Merlijn Staps, the Netherlands.

Like for the International Mathematical Olympiad, Problems 1–3 appeared on theDay 1 Paper and problems 4–6 appeared on the Day 2 Paper. These papers were heldon Wednesday 11 April and Thursday 12 April respectively, with the contestants allowedfour and a half hours per paper to work individually on the problems and write theirattempted proofs. Each problem was scored out of a maximum of seven points.

1This is a new record for the EGMO, with the number of participating countries increasing every year.In 2012 for the first edition there were only 19 countries, 16 of them official European ones.

1

board. It was proposed by Merlijn Staps, the Netherlands.


6. A tricky algebra problem involving the closeness of pairs of elements in a set of positive integers. It was proposed by Merlijn Staps, the Netherlands.

Like for the International Mathematical Olympiad, Problems 1–3 appeared on the Day 1 Paper and problems 4–6 appeared on the Day 2 Paper. These papers were held on Wednesday 11 April and Thursday 12 April respectively, with the contestants allowed four and a half hours per paper to work individually on the problems and write their attempted proofs. Each problem was scored out of a maximum of seven points.

Prior to the first day of competition, the opening ceremony was held in the beautiful Teatro Verdi and hosted by Alessandra Caraceni and Massimo Gobbino. The occasion included speeches from the Vice Mayor of Florence and various sponsors for the event as well as several jokes about a lack of a timetable for the week. The parade of teams was complete with beautifully-designed background slides for each country and an accompaniment of the unique instrumental music of Lame Da Barba. Finally, it was declared that the 2018 EGMO was officially open. After the opening ceremony there was also an opportunity for teams and their deputies to mingle and participate in a Treasure Hunt in Florence before the two contest days.

1 This is a new record for EGMO, with the number of participating countries increasing every year. In 2012 for the first edition there were only 19 countries, 16 of them official European ones.

Mathematics Contests The Australian Scene 2018 | EGMO Team Leader’s Report | 129

On each day of the contest, the Jury considered written questions raised by contestants during the first half-hour and decided on replies. After this, proposals for marking schemes were presented to the Jury and approved. By late afternoon on each contest day, Leaders and their Deputies received the scripts of their students and were able to spend the evening assessing their work in accordance with the approved marking schemes. A local team of markers (Coordinators) also assessed the scripts independently. Coordination, the process of determining official scores for each contestant, took place the following day, Friday 13 April (while the students were out enjoying a trip to Pisa and Lucca). The live, online coordination schedule, which adjusted meeting times according to the average time taken by each team, was impressive and extremely helpful! Coordinators along with the Leader and Deputy of a team have to agree on scores for each student of that team. Disagreements that cannot not be resolved in this way are first referred to the Problem Captain, then the Chief Coordinator, and ultimately to the Jury if there is still no agreement. This year there was an unusually high number of six disagreements that were referred to the Jury. For the two disagreements in Problem 1 and one disagreement in Problem 3, the Jury voted in favour of the Coordinators, and for the three disagreements in Problem 4, including Australia’s, the Jury voted against the Coordinators. After a long day of coordination, it was a very long night resolving matters referred to the Jury, and an even longer night for the Coordinators of Problem 4 who reviewed all 194 scripts in light of the three disputes that did not go in their favour. No one got much sleep before an early Jury meeting on Saturday 14 April to finalise Problem 4 scores and to vote for medal cut-offs, which were necessary for organising the Closing Ceremony later that afternoon. The EGMO Advisory Board has now introduced a new way of handling coordination issues at the EGMO which involves an Appeal Committee consisting of five members of the Jury, who will make decisions to resolve any coordination disputes.

The contestants found Problem 1 (geometry) to be the easiest with an average score of 5.754, a similar average to Problem 1 in 2017 and the second-highest average for an EGMO Problem 1. Problem 4 (combinatorics) had the next-highest average score of 4.313 and this was the highest average for an EGMO Problem 4. Problem 6 was the most difficult with an average score of just 0.58, the lowest average for an EGMO Problem 6 (though two problem 3s have seen lower averages at the EGMO). The score distributions by problem number were as follows.

Mark P1 P2 P3 P4 P5 P6

0 11 20 102 30 86 163

1 7 70 28 23 46 8

2 8 37 26 16 5 12

3 12 14 16 8 6 0

4 3 8 8 1 2 0

5 7 11 1 1 1 1

6 13 7 1 13 1 1

7 134 28 13 87 48 10

Mean 5.754 2.621 1.344 4.313 2.200 0.579


The medal cut-offs were set at 32 points for Gold, 22 for Silver and 15 for Bronze. The medal distributions 2 were as follows.

Gold Silver Bronze Total

All teams (195 contestants) 17 39 52 108

Official European teams (137 contestants) 12 24 35 71

Proportion from official European teams 8.8% 17.5% 25.5% 51.8%

These awards were presented at the closing ceremony. Of those who did not get a medal, a further 45 contestants (37 from official European teams) received an Honourable Mention for scoring full marks on at least one problem. Five contestants achieved perfect scores of 42: Jelena Ivancic (Serbia), Alina Harbuzova (Ukraine), Emily Beatty (UK), Catherine Wu (USA), and Wanlin Li (USA).

A big congratulations goes to the Australian team on such a fantastic performance in Australia’s first ever appearance at an EGMO. The team finished 20th in the rankings,3 bringing home one Silver medal, two Bronze medals, and an Honourable Mention.

The Silver medallist was

• Grace He, year 10, Methodist Ladies’ College, VIC

The Bronze medallists were

• Xinyue Alice Zhang, year 12, A.B. Paterson College, QLD• Tianyue (Ellen) Zheng, year 12, Smith’s Hill High School, NSW

The Honourable Mention recipient was

• Yifan Guo, year 12, Glen Waverley Secondary College, VIC

The following table shows their scores in detail.

Name P1 P2 P3 P4 P5 P6 Score Award

Grace He 7 5 3 7 1 0 23 Silver Medal

Xinyue Alice Zhang 7 3 2 7 0 0 19 Bronze Medal

Tianyue (Ellen) Zheng 7 1 2 6 3 0 19 Bronze Medal

Yifan Guo 2 2 1 7 0 0 12 Honourable Mention

Australian Average 5.75 2.75 2.00 6.75 1.00 0.00 18.25

EGMO average 5.75 2.62 1.34 4.31 2.20 0.58 16.81

2 The total number of medals is approved by the Jury and is approximately half the number of contestants. The numbers of Gold, Silver, and Bronze medals are in the approximate ratio 1:2:3 and are chosen on the basis of the performances of members of official European teams. Medals are awarded to participants from guest teams and any additional teams on the basis of the boundaries set by the Jury.

3 Countries are ranked each year on the EGMO’s official website according to the sum of the individual student scores from each country.


The table below shows the distribution of awards for each country at the 2018 EGMO.

Country Size Total Gold Silver Bronze HM RankAlbania 4 43 0 0 0 4 35

Australia 4 73 0 1 2 1 20

Austria 3 34 0 0 1 0 42

Azerbaijan 4 18 0 0 0 0 48

Belarus 4 99 0 3 0 0 10

Belgium 4 46 0 0 1 1 34

Bolivia 3 4 0 0 0 0 52

Bosnia and Herzegovina 4 76 0 1 2 1 18

Brazil 4 84 0 2 2 0 13

Bulgaria 4 83 0 2 1 1 14

Canada 4 72 0 1 2 1 21

Costa Rica 4 19 0 0 0 0 46

Cyprus 4 30 0 0 0 2 44

Czech Republic 4 63 0 1 1 2 28

Ecuador 4 52 0 0 2 1 30

Finland 4 19 0 0 0 2 46

France 4 76 0 2 2 0 18

Georgia 4 66 0 2 0 1 25

Germany 4 50 0 0 1 1 33

Greece 4 36 0 0 1 1 41

Hungary 4 102 0 2 2 0 7

India 2 33 0 0 2 0 43

Ireland 4 37 0 0 0 3 40

Israel 4 80 1 1 0 1 15

Italy 4 64 0 0 4 0 27

Italy B 1 12 0 0 0 1 50

Japan 4 94 1 1 1 0 12

Kazakhstan 4 97 1 1 1 1 11

Latvia 4 52 0 0 1 3 30

Liechtenstein 1 14 0 0 0 1 49

Lithuania 4 78 1 0 2 1 17

Luxembourg 2 8 0 0 0 0 51

Macedonia (FYR) 3 43 0 0 1 2 35

Mexico 4 102 0 4 0 0 7

Moldova 4 56 0 1 1 1 29

Mongolia 4 80 0 1 3 0 15

Netherlands 4 68 0 1 2 1 24

Norway 4 30 0 0 1 1 44

Peru 4 71 1 1 0 1 22

Poland 4 108 1 2 1 0 4

Romania 4 101 2 1 0 0 9


Country Size Total Gold Silver Bronze HM Rank

Russian Federation 4 145 4 0 0 0 1

Saudi Arabia 4 71 0 2 1 0 22

Serbia 4 103 1 1 1 1 6

Slovenia 4 43 0 0 1 2 35

Spain 4 41 0 0 1 1 38

Switzerland 4 52 0 0 2 2 30

Tunisia 4 40 0 0 1 2 39

Turkey 4 66 0 0 3 1 25

Ukraine 4 104 1 2 0 0 5

United Kingdom 4 111 1 2 1 0 3

United States of America 4 129 2 1 1 0 2

Total (52 teams, 195 contestants) 17 39 52 45

The 2018 EGMO was organised by the Unione Matematica Italiana with the patronage and backing of the Italian Ministry of Education.

The 2019 EGMO is scheduled to be held 7–13 April in Kyiv, Ukraine, and the 2020 EGMO will be hosted by the Netherlands.

Much of the statistical information found in this report can also be found on the official website of the EGMO, https://www.egmo.org. More details can also be found on the EGMO 2018-specic site, https://www.egmo2018.org, and official photos found on Flickr, https://www.flickr.com/photos/egmo2018.

Many stories and photos of Australia’s first ever adventure to the European Girls’ Mathematical Olympiad can be found on the Australian EGMO team blog, https://ausegmo.wordpress.com. May these stories and photos that capture the incredibly wonderful and invaluable experience of girls competing in mathematics on an international stage be just the first of many more for years to come.

Thanom ShawEGMO Team Leader, Australia


https://www.egmo.org

https://www.egmo2018.org

https://www.egmo2018.org

https://www.flickr.com/photos/egmo2018

https://ausegmo.wordpress.com

Wednesday, April 11, 2018

Problem 1. Let ABC be a triangle with CA = CB and ∠ACB = 120, and let M be the midpointof AB. Let P be a variable point on the circumcircle of ABC, and let Q be the point on the segment CPsuch that QP = 2QC. It is given that the line through P and perpendicular to AB intersects theline MQ at a unique point N .

Prove that there exists a fixed circle such that N lies on this circle for all possible positions of P .

Problem 2. Consider the set

A =

1 + 1k

: k = 1, 2, 3, . . .

.

(a) Prove that every integer x ≥ 2 can be written as the product of one or more elements of A, whichare not necessarily different.

(b) For every integer x ≥ 2, let f(x) denote the minimum integer such that x can be written as theproduct of f(x) elements of A, which are not necessarily different.Prove that there exist infinitely many pairs (x, y) of integers with x ≥ 2, y ≥ 2, and

f(xy) < f(x) + f(y).

(Pairs (x1, y1) and (x2, y2) are different if x1 = x2 or y1 = y2.)

Problem 3. The n contestants of an EGMO are named C1, . . . , Cn. After the competition theyqueue in front of the restaurant according to the following rules.

• The Jury chooses the initial order of the contestants in the queue.

• Every minute, the Jury chooses an integer i with 1 ≤ i ≤ n.

– If contestant Ci has at least i other contestants in front of her, she pays one euro to theJury and moves forward in the queue by exactly i positions.

– If contestant Ci has fewer than i other contestants in front of her, the restaurant opens andthe process ends.

(a) Prove that the process cannot continue indefinitely, regardless of the Jury’s choices.

(b) Determine for every n the maximum number of euros that the Jury can collect by cunninglychoosing the initial order and the sequence of moves.

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

Language: English

Day: 1

EUROPEAN GIRLS’ MATHEMATICAL OLYMPIAD

Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad | 134

Thursday, April 12, 2018

Problem 4. A domino is a 1 × 2 or 2 × 1 tile.Let n ≥ 3 be an integer. Dominoes are placed on an n × n board in such a way that each domino

covers exactly two cells of the board, and dominoes do not overlap.The value of a row or column is the number of dominoes that cover at least one cell of this row or

column. The configuration is called balanced if there exists some k ≥ 1 such that each row and eachcolumn has a value of k.

Prove that a balanced configuration exists for every n ≥ 3, and find the minimum number ofdominoes needed in such a configuration.

Problem 5. Let Γ be the circumcircle of triangle ABC. A circle Ω is tangent to the line segment ABand is tangent to Γ at a point lying on the same side of the line AB as C. The angle bisector of ∠BCAintersects Ω at two different points P and Q.

Prove that ∠ABP = ∠QBC.

Problem 6.

(a) Prove that for every real number t such that 0 < t < 12 there exists a positive integer n with the

following property: for every set S of n positive integers there exist two different elements x andy of S, and a non-negative integer m (i.e. m ≥ 0), such that

|x − my| ≤ ty.

(b) Determine whether for every real number t such that 0 < t < 12 there exists an infinite set S of

positive integers such that|x − my| > ty

for every pair of different elements x and y of S and every positive integer m (i.e. m > 0).


Language: English

Day: 2

Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad | 135

Solutions to the 2018 European Girls’ Mathematical Olympiad

1. Solution 1 (Xinyue Alice Zhang, year 12, A.B. Paterson College, QLD. Alice wasa Bronze medallist with the 2018 Australian EGMO team.)

O

C

BA M

P

N

Q

Consider triangles QNP and QMC. Vertically opposite angles NQP and MQCare equal. We also have that CM is perpendicular to AB since ABC is isoscelesand M is the midpoint of AB. Given that NP is also perpendicular to AB, CMand NP are parallel. Therefore ∠NPQ = ∠MCQ and ∠PNQ = ∠CMQ. Hencetriangles QNP and QMC are similar.

Since QP = 2QC, it follows that NP = 2MC. Furthermore, NP is parallel to MC.

The point N is therefore a translation of P by the fixed vector 2−−→MC. Therefore, as

P varies on the circumcircle of ABC, there is a fixed circle on which N lies.

Comment From the condition that ∠ACB = 120, it follows that M is the mid-point of CO, where O is the circumcentre of ABC. That is, 2MC = OC. Therefore,N lies on the circle (ω, say) that is the image of the circumcircle of ABC throughthe translation that sends O to C; that is, the circle with centre C and radius CO.

Also note that since N is defined uniquely, P does not lie on the line CO, andtherefore N also does not lie on the line CO. The possible positions of N are in factall the points of ω with the exception of the two points lying on the line CO.

8

EUROPEAN GIRLS’ MATHEMATICAL OLYMPIAD SOLUTIONS

Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 136

Solution 2 (Based on the solution by Tianyue (Ellen) Zheng, year 12, Smith’sHill School, NSW. Ellen was a Bronze medallist with the 2018 Australian EGMOteam.)

O

C

BA M

P

N

Q

As in solution 1 (with comments), NP is parallel to CM and hence CO (C, M , Oare collinear with M the midpoint of CO), triangles QNP and QMC are similar,and NP = 2MC = OC.

Since NP and CO are equal and parallel, OCNP is a parallelogram. Since adjacentsides OC and OP are equal, OCNP is actually a rhombus. Hence CN = CO, theradius of the circumcircle of ABC. Therefore N lies on the circle with centre C andradius CO.

9Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 137

Solution 3 (2018 EGMO Problem Selection Committee)

A computational approach could be set up as follows.

Set the centre of the circumcircle of ABC at (0, 0) and set

A =

(−√3

2,1

2

), B =

(√3

2,1

2

), C = (0, 1) , M =

(0,

1

2

).

Let P = (a, b). Since Q lies on CP such that QP = 2QC, we get that

Q =

(a

3,b+ 2

3

).

The equation of the line through P and perpendicular to AB is x = a and theequation of the line MQ (if a = 0), is

y =2b+ 1

2ax+

1

2.

The point of intersection of the two lines is therefore

N = (a, b+ 1) = P + (0, 1).

This shows that the map P → N is a translation by the vector (0, 1) and this resultis independent of the position of P (provided that a = 0, because otherwise N isnot well-defined).

Hence, when P lies on the circumcircle of ABC, with the exception of the twopoints with a = 0, N lies on the translated circle which is the circle with centre Cand radius 1.


2. Solution 1 (Based on the partial solution by Grace He, year 10, Methodist Ladies’College, VIC. Grace was a Silver medallist with the 2018 Australian EGMO team.)

We have the set

A =

1 +

1

k: k = 1, 2, 3, . . .

=

k + 1

k: k = 1, 2, 3, . . .

.

For every integer x ≥ 2,

x =2

1× 3

2× · · · × x

x− 1=

x−1∏k=1

k + 1

k.

That is, every integer x ≥ 2, can be written as a product of one or more elementsof A which establishes the result for part (a).

The largest element of A is 2, hence for integers n ≥ 0,

3× 2n > 2n+1 ⇒ f(3× 2n) ≥ n+ 2.

Since 3× 2n = 32× 2n+1, the product of n+ 2 elements of A, f(3× 2n) = n+ 2.

Similarly,33× 2n > 2n+5 ⇒ f(33× 2n) ≥ n+ 6

and since 33×2n = 3332×2n+5, the product of n+6 elements of A, f(33×2n) = n+6.

Lastly, f(11) ≥ 5 since 11 cannot be written as the product of four or fewer elementsof A: 24 > 11, 23 × 3

2> 11, and any other product of at most four elements of A

does not exceed 23 × 43= 102

3< 11. Since 11 = 11

10× 5

4× 2 × 2 × 2, we have that

f(11) = 5.

Putting these results together,

f(33× 2n) = n+ 6 < n+ 2 + 5 = f(3× 2n) + f(11)

for all integers n ≥ 0. Hence, there are infinitely many pairs, x = 3 × 2n, y = 11,with x ≥ 2, y ≥ 2 and f(xy) < f(x) + f(y).



First we show that f(xy) < f(x) + f(y) for (x, y) = (7, 7). We have that f(7) ≥ 4since 7 cannot be written as the product of three or fewer elements of A: 23 > 7, andany other product of at most three elements of A does not exceed 22 × 3

2= 6 < 7.

Also f(49) ≤ 7 since 49 = 2× 2× 2× 2× 2× 32× 49

48. Hence

f(49) ≤ 7 < 8 ≤ f(7) + f(7).

Now suppose by contradiction that there exists only finitely many pairs (x, y) thatsatisfy f(xy) < f(x) + f(y). This implies that there exists an M large enough suchthat whenever a > M or b > M holds we have f(ab) = f(a) + f(b). (Note thatf(ab) ≤ f(a) + f(b) is always satisfied.)

Now take any pair (x, y) that satisfies f(xy) < f(x) + f(y) and let n > M be anyinteger. Then

f(n) + f(xy) = f(nxy) = f(nx) + f(y) = f(n) + f(x) + f(y),

which contradicts f(xy) < f(x) + f(y).

Therefore there exist infinitely many pairs (x, y) with x ≥ 2, y ≥ 2, and f(xy) <f(x) + f(y).


3. Solution 1 (Cunning initial order and sequence of moves for maximum number ofeuros found by all medallists of the 2018 Australian EGMO Team)

(a) We prove by induction that each contestant Ci can move at most 2n−i − 1times and note that this also shows that each contestant can move only a finitenumber of times and therefore the process cannot continue indefinitely.

For this induction we work backwards and use the key observation that whena contestant Ci moves forward, past i other contestants, she must pass at leastone Cj where j > i.

In the base case, Cn cannot move at all because there are not n other contes-tants in the queue. As for Cn−1, when she moves, she must past every other ofthe n − 1 contestants, including Cn. Once she has passed Cn there is no wayCn can get back in front of Cn−1 (because Cn cannot move). Hence there is noway Cn−1 can move again and so Cn−1 can move at most once.

Assume that each contestant Cj, for j > i, can move at most 2n−j − 1 timesand consider how many times contestant Ci can pass Cj where j > i. If Cj

moves at most 2n−j − 1 times then, by passing Ci every move, she can pass Ci

at most 2n−j − 1 times. Since Cj may have started ahead of Ci, we have thatCi can pass Cj at most 2n−j times.

Therefore we have that Ci can pass Cn at most 20 times, Cn−1 at most 21 times,Cn−2 at most 22 times, . . ., Ci+1 at most 2n−i−1 times.

Each time Ci moves, she must pass at least one Cj where j > i, and in theextreme case, she passes all of C1, C2, . . . , Ci−1 and one Cj. Therefore, themaximum number of times Ci can move is

1 + 2 + 22 + · · ·+ 2n−i−1 = 2n−i − 1.

(b) What remains is to show how the Jury can arrange the queue initially, andwhat moves they can make, to achieve this maximal number of moves for eachcontestant. That is, 20 − 1 moves for Cn, 2

1 − 1 moves for Cn−1, 22 − 1 moves

for Cn−2, . . ., 2n−1−1 moves for C1 and so a total number of moves (i.e. euros)

of(20 − 1) + (21 − 1) + (22 − 1) + · · ·+ (2n−1 − 1) = 2n − 1− n.

We prove by induction that by lining the contestants up in reverse order,Cn, Cn−1, . . . , C1, the jury can collect 2n − 1 − n euros before the contestantsare lined up in the order C1, C2, . . . , Cn and the restaurant must open.

For the base case, with just C1 in the queue, no moves can be made and theJury makes 21 − 1− 1 = 0 euros. Indeed, for 2 contestants lined up in reverseorder C2, C1, only a single move can occur to give C1, C2 before the restaurantopens and the Jury makes a total of 22 − 1− 2 = 1 euro.

Starting with n+ 1 contestants in reverse order

Cn+1, Cn, Cn−1, . . . , C2, C1,

keeping Cn+1 in its first position, by the inductive hypothesis, the Jury canmake 2n − 1− n euros reversing the order of the remaining contestants,

Cn+1, C1, C2, . . . , Cn−1, Cn.


Then in the next n moves, all contestants C1, C2, . . . , Cn, in this order, move(past Cn+1) and end up in the first n positions of the line in reverse order

Cn, Cn−1, . . . , C2, C1, Cn+1.

Finally, again by the inductive hypothesis, the Jury can make 2n− 1−n eurosreversing the order of the first n contestants

C1, C2, . . . , Cn, Cn+1.

This series of moves results in all n + 1 contestants queued in order and therestaurant opening after a total of

(2n − 1− n) + n+ (2n − 1− n) = 2n+1 − 1− (n+ 1)

euros are made by the Jury.

In this way, given n contestants queuing for the restaurant, the Jury can collecta maximum number of 2n − 1− n euros.


This solution is a different argument for proving that the process cannot continueindefinitely.

As in the previous solution, the key observation is that when a contestant Ci movesforward, past i other contestants, she must pass at least one Cj where j > i.

To show that the process ends after a finite number of moves, we assume that thisis not the case. Then there must be at least one contestant who moves infinitelymany times. Let i0 be the largest index such that Ci0 moves infinitely many times(that is, for all i > i0, Ci moves a finite number of times). It must be the case thatCi0 passes some fixed Cj0 infinitely many times with j0 > i0. However, Cj0 moves afinite number of times and so can only precede Ci0 in the queue a finite number oftimes. Hence it is impossible for Ci0 to move infinitely many times and the processmust necessarily end after a finite number of moves.

Solution 3 (Based on the partial solution by Grace He, year 10, Methodist Ladies’College, VIC. Grace was a Silver medallist with the 2018 Australian EGMO team.)

This solution is a different argument for proving that the process cannot continueindefinitely.

We prove by induction on the number of contestants that the process must eventuallyend.

For the base case n = 1, the only contestant is C1 and there are no contestants infront of her so the process immediately ends.

Now suppose that we know the process will end if there are k contestants in the line.Consider the situation where there are k + 1 contestants. Notice that contestantCk+1 can never move forward in the line since there are only k other contestants,hence she can only move backwards, and at most k times. Therefore Ck+1 can onlymove a finite number of times, and so there will be a point after which Ck+1 doesnot move anymore.

After this point, all contestants behind Ck+1 will remain behind Ck+1, and all con-testants in front of Ck+1 will remain in front of Ck+1. If we now remove Ck+1, we


are left with a valid initial order for k contestants, and moreover, we may observethat any move which occurs after this point remains a valid move even with Ck+1

removed: any move by Ci will move them ahead of exactly i of C1, C2, . . . , Cksince no contestant jumps over Ck+1. The problem has now been reduced to thecase with k contestants, so by the inductive hypothesis the process will eventuallyend.


This solution is a different proof of the upper bound for the number of euros col-lected.

We show that at most 2n − n− 1 moves are possible in the following way. Given anarrangement of the n contestants, for each pair in which Ci is behind Cj in the queueand i < j (we will call this a reverse pair), we assign a weight of 2i−1. We definethe total weight of an arrangement as the sum of the weights for each reverse pair.In this way, the maximum possible total weight of an arrangement occurs when Ci

is behind exactly all contestants with a larger number than herself, and so in frontof all contestants with a smaller number than herself. That is, for the arrangementCn, Cn−1, Cn−2, . . . , C2, C1. This maximum total weight of an arrangement is

(n− 1)20 + (n− 2)21 + (n− 3)22 + · · ·+ (2)2n−3 + (1)2n−2 = 2n − n− 1.

Now consider the total weight of an arrangement before and after Ci moves forwardin the queue. Since Ci moves forward past i other contestants, she must pass atleast one contestant Cj where j > i, and at most i− 1 contestants Cj where j < i.This gives a reduction of at least 2i−1 in the total weight, and an addition of at most20 +21 +22 + · · ·+2i−2 = 2i−1 − 1. Hence the total weight of an arrangement mustdecrease when any contestant Ci moves forward in the queue.

Since the maximum total weight of an arrangement is 2n − n − 1, and this totalweight decreases every move, at most 2n − n− 1 are possible.


4. Solution 1 (Yifan Guo, year 12, Glen Waverley Secondary College, VIC. Yifanwas awarded an Honourable Mention with the 2018 Australian EGMO team.)

Consider a balanced configuration with D dominoes on an n×n board in which eachrow and each column has a value k. On one hand, the total value of all rows andcolumns is 2nk. On the other hand, since each of D dominoes contributes 2+1 = 3to the total value of all rows and columns, this total value is 3D. Hence we havethe equality

2nk = 3D ⇒ D =2nk

3.

First consider the case where n is a multiple of 3. Since k ≥ 1, we have that D ≥ 2n3.

The following diagram shows a balanced configuration for n = 3 with k = 1 and2×33

= 2 dominoes on the board.

If n is a multiple of 3, we can obtain a balanced configuration with k = 1 and n3of

these 3×3 blocks (each containing 2 dominoes) along a main diagonal of the board.

. . .

Hence, for n a multiple of 3, the minimum number of dominoes needed in a balancedconfiguration is 2n

3.

If n is not a multiple of 3, then k is a multiple of 3. Hence k ≥ 3 and D ≥ 2n.

The following diagram shows a balanced configuration for n = 4 with k = 3 and2× 4 = 8 dominoes on the board.


Balanced configurations for n ≡ 1(mod 3) and n ≡ 2(mod 3) with 2n dominoes forn ≥ 5 can be constructed as described below using the following 3× 3 blocks.

Block A Block B

For n ≡ 1(mod 3) with n ≥ 7, n−13

Block B’s are placed along a main diagonal of the(n−1)× (n−1) sub-board, as shown in the diagram below. Each of these dominoescontribute 2 to the value of each of the first n − 1 rows and n − 1 columns. Thenn−43

Block A’s are placed along a main diagonal of the (n− 7)× (n− 7) sub-boardas well as in the bottom-right corner. These Block A’s with a further 2 dominoesplaced in the bottom row and 2 dominoes placed in the rightmost column, as shown,each contribute 1 to the value of the first n − 1 rows and n − 1 columns, and 3 tothe value of the nth row and nth column. The total number of dominoes placed onthe board in this construction is 4× n−1

3+ 2× n−4

3+ 4 = 2n.

...

...

For n ≡ 2(mod 3) with n ≥ 5, n−23

Block B’s are placed along a main diagonal of the(n−2)× (n−2) sub-board, as shown in the diagram below. Each of these dominoescontribute 2 to the value of each of the first n−2 rows and n−2 columns. Then n−2

3

Block A’s are placed along a main diagonal of the (n−5)×(n−5) sub-board as wellas in the bottom-right corner. These Block A’s with a further 2 dominoes placed


in the bottom two rows and 2 dominoes placed in the rightmost two columns, asshown, each contribute 1 to the value of the first n−2 rows and n−2 columns, and 3to the value of the bottom two rows and rightmost two columns. The total numberof dominoes placed on the board in this construction is 4× n−2

3+2× n−2

3+4 = 2n.

Note that this construction does indeed work for n = 5. Despite Block B and BlockA overlapping in the middle row and column in this case, only Block B has a dominoplaced in this middle cell.

...

...

Therefore the minimum number of dominoes required in a balanced configurationis therefore 2n

3if n is a multiple of 3, and 2n otherwise.

Solution 2 (Xinyue Alice Zhang, year 12, A.B. Paterson College, QLD. Alice wasa Bronze medallist with the 2018 Australian EGMO team.)

This solution presents alternative balanced configurations for n ≡ 0 (mod 3) withk = 3 and 2n dominoes.

The following diagrams show balanced configurations with k = 3 and n ∈ 4, 5, 6, 7.Note that 2n dominoes are used in each case.

Any n ≥ 8 can be written in the form 4x + r where x is a positive integer andr ∈ 4, 5, 6, 7. Hence a balanced configuration can be obtained for n ≥ 8 and k = 3by using x 4 × 4 blocks and one r × r block along a main diagonal of the board.Such a configuration uses 8x+ 2r = 2(4x+ r) = 2n dominoes.


5. Solution 1 (Based on the partial solutions by Tianyue (Ellen) Zheng, year 12,Smith’s Hill School, NSW and Grace He, year 10, Methodist Ladies’ College, VIC.Ellen was a Bronze medallist and Grace was a Silver medallist with the 2018 Aus-tralian EGMO team.)

Let V be the intersection of Ω and Γ, let U be the intersection of Ω and AB, andlet M be the midpoint of the arc AB that does not contain C. Note that the anglebisector of ∠BCA intersects Γ at M .

O2

A B

V

M

U

O1

C

P

QΩ

Γ

We first show that V , U , and M are collinear. Let O1 and O2 be the centres of Ωand Γ respectively, and let V U intersect Γ at M ′. We have ∠O1UV = ∠O1V U =∠O2VM ′ = ∠O2M

′V and hence O1U ‖ O2M′. Since O1U ⊥ AB (AB is tangent

to Ω at U), O2M′ ⊥ AB and hence M ′ is the midpoint of the arc AB (that does

not contain C) and M ′ = M . (Indeed, the dilation with centre V that sends Ω toΓ sends U to the point of Γ where the tangent of Γ is parallel to AB and this pointis M).

Triangles MAV and MUA are similar since ∠AMV = ∠UMA and ∠MVA =∠MCA = ∠MCB = ∠MAB = ∠MAU . It follows that MU × MV = MA2 =MB2.

Computing the power of M with respect to Ω we have that

MP ×MQ = MU ×MV = MB2.

Now we have that triangles MBP and MQB are similar as ∠BMP = ∠QMB andMPMB

= MBMQ

. In particular, we then have ∠MBP = ∠MQB. We already have that∠MCB = ∠MCA = ∠MBA. Putting these together, we can conclude

∠ABP = ∠MBP − ∠MBA = ∠MQB − ∠MCB = ∠QBC,

as required.



This solution is a different proof to the first part of Solution 1.

Consider the inversion with respect to the circle with centre M and radius MA =MB.

The inversion swaps AB and Γ. Let Ω swap to Ω′. We will prove that Ω = Ω′, thatthe inversion swaps V and U , and that V , U , and M are collinear.

First consider the points where Ω intersects the inversion circle, W1 and W2, say.We know that the inversion fixes W1 and W2 and so these must lie on Ω′. Now,since Ω is tangent to Γ (at V ), Ω′ is tangent to AB at a point on the line throughM and V . Also, since Ω is tangent to AB (at U), Ω′ is tangent to Γ at a pointon the line through M and U . To summarise, Ω′ is a circle tangent to both ABand Γ that passes through fixed points W1 and W2. So Ω′ must be Ω itself. Theinversion swaps points of tangency V and U , and these points and M , the centre ofthe inversion, are collinear.

Now P and Q are intersections between the fixed line MC and Ω, and since theonly fixed point on the line segment MC is its intersection with the inversion circle,P and Q are swapped. This implies that MP ×MQ = MB2.


6. Solution (part a) (2018 EGMO Problem Selection Committee)

Let t be a real number such that 0 < t < 12.

We wish to find a positive integer n, such that for every set S of n integers, denoted

s1 < s2 < s3 < · · · < sn,

there exist two elements si and sj such that |sj − msi| ≤ tsi for a non-negativeinteger m. In other words sj is within tsi of a multiple of si.

tsi

sj0 si 2si 3si 4si

If si+1 − si ≤ tsi, for some 1 ≤ i ≤ n− 1, then we have that si+1 is within tsi of thefirst multiple of si, namely 1× si (m = 1). Hence the inequality is satisfied.

Otherwise, si+1 − si > tsi ⇒ si+1 > (1 + t)si for all 1 ≤ i ≤ n− 1. That is,

sn > (1 + t)sn−1 > (1 + t)2sn−2 > (1 + t)3sn−3 > · · · > (1 + t)n−1s1.

This gives that

sn > (1 + t)n−1s1 ⇒ s1 <1

(1 + t)n−1sn.

Finally, due to Bernoulli’s inequality, we know that for n large enough,

1

t< (1 + t)n−1 ⇒ 1

(1 + t)n−1≤ t.

Hence we have that s1 − 0sn < tsn, that is, s1 is within tsn of the zero-th multipleof sn (m = 0). Hence the inequality is satisfied.

Solution (part b) (Stephen Farrar, long-time friend of the Olympiad program inAustralia. Stephen was also a Gold and Bronze medallist with the 1997 and 1998Australian IMO team.)

Let t be a real number such that 0 < t < 12.

This time, we wish to construct an infinite set S of positive integers such that|sj − msi| > tsi for every pair of different elements si and sj of S and for everypositive integer m. In other words, no element of S (different from si) is within tsiof every positive multiple of si. And this holds for all infinitely many si.

We can construct such an infinite set S using a sieving process.

To begin, we choose the smallest element s1 of S to be the smallest odd integer suchthat ts1 < (s1 − 1)/2. This is equivalent to s1 > 1/(1− 2t).

Having chosen s1, certain numbers are excluded from being in S. S cannot containany multiple of s1, like Eratosthenes’ sieve for finding primes. The difference in thisquestion is that the numbers close to a multiple of s1 are also excluded from S. Theexclusion radius is ts1, and when t is close to 1/2 it could be that most integersare excluded from S. The highlighted regions on the numberline below show thosenumbers excluded by s1.


ts1 <s1−12

0 s1 2s1 3s1 4s1

It’s easier to name a few numbers which are not excluded by s1. The numbers(m + 1/2)s1, where m is a positive integer, are not excluded, though these are notintegers. However, near (m+ 1/2)s1, we have integers x = (m+ 1/2)s1 ± 1/2. Callthese half-multiples of s1 and note that one of each pair must be odd. These half-multiples are not excluded because ((m+ 1/2)s1 − 1/2) −ms1 = (s1 − 1)/2 > ts1and (m + 1)s1 − ((m+ 1/2)s1 − 1/2) = (s1 − 1)/2 > ts1 and these half-multiplesare even further from other multiples of s1.

ms1 (m+ 1)s1

ts1 <s1−12

(m+ 12)s1

half-multiples of s1:(m+ 1

2)s1 − 1

2, (m+ 1

2)s1 +

12

We now choose the next smallest element of S, s2, to be an odd half-multiple of s1.We know that s2 is not excluded by s1 and claim that if s2 > 2s1, we also have thats1 is not excluded by s2 (that is, not within ts2 of every positive multiple of s2).Indeed, since m ≥ 1 and t < 1/2, we have that m− t > 1/2. And if s2 > 2s1 thenthis gives s2(m− t) > s1 ⇒ ms2 − s1 > ts2 ⇒ |s1 −ms2| > ts2.

So in particular, we will choose s2 to be whichever of the half-multiples (5/2)s1±1/2is odd and note that since s1 > 1, s2 = (5/2)s1 ± 1/2 = 2s1 + (s1 ± 1)/2 > 2s1.

Now this excludes even more numbers from S. Again we can be sure that the half-multiples of s2 are not excluded on account of being too close to a multiple of s2.And so we can be sure that any number which is a half-multiple of both s1 and s2is still a possibility for inclusion in S.

Now let’s prove that if a and b are odd integers, then the half-multiples of ab arealso half-multiples of both a and b. Suppose h = (k+1/2)ab±1/2 is a half-multipleof ab. Then h = (+ 1/2)a± 1/2 where = kb+ (b− 1)/2 is an integer. Thereforeh is a half-multiple of a. Similarly it is a half-multiple of b.

We now choose infinitely many more elements in S according to this rule: sk+1 is anodd half-multiple of s1s2 · · · sk with sk+1 > 2si for all 1 ≤ i ≤ k. In this way, sk+1

is an odd half-multiple of each of s1, s2, . . . , sk and so not excluded by any of theseelements. Furthermore, sk+1 > 2si ⇒ sk+1(m − t) > si ⇒ msk+1 − si > tsk+1 ⇒|si −msk+1| > tsk+1 and so for all 1 ≤ i ≤ k, si is not excluded by sk+1.

Specifically, we choose sk+1 to be an odd integer given by

sk+1 = 5s1s2 · · · sk/2± 1/2.

Note that since each of s1, s2, . . . , sk is greater than 1, sk+1 > 5si/2 ± 1/2 = 2si +(si ± 1)/2 > 2si for all 1 ≤ i ≤ k. Also note that the si sequence is increasing.

Therefore, we have found such an infinite set S of integers s1 < s2 < s3 < · · · , eachchosen to be odd and given by s1 > 1/(1− 2t) and sk+1 = 5s1s2 · · · sk/2± 1/2.


IMO TEAM PREPARATION SCHOOL

For the week preceding the IMO, our team met with our British counterparts in Budapest for a final dose of training. We were also joined by an Estonian student, Richard, for the duration of the camp. I was accompanied by Mike Clapper and Jo Cockwill for the entire camp, and Angelo Di Pasquale for the first few days, after which he left to do his leader duties at the IMO.

This year the plan for the Australians was to meet in Perth then fly together from there to Budapest. Upon landing in Perth we learnt that Wen would arrive a day late due to a delayed flight. I got the job of staying for an extra day in Perth while everyone else flew to Budapest. After enjoying a surprisingly warm winter’s morning in Perth, I met Wen at the airport and then accompanied him to Budapest.

At the team preparation school, each of the five days starts with a 4.5 hour IMO style exam for all 13 students. For the first exam only 12 students were present as Wen (and I) were still in transit. Rather than miss the exam entirely, Wen spent 3 hours on it while we waited at Perth airport, after which he boasted the highest point to time ratio of all 13 students. After the exam each morning, the teams were given free time for the rest of the day while the adults marked their exams. During the free time they mostly played games or visited some local attractions.

For the third exam each team set the problems for the other team. (For this purpose, Richard was designated an honorary Brit.) In past years the Australians have used problems from their training and sometimes constructed one or two themselves. This year they were committed to composing all three problems themselves, and even came prepared with a shortlist of about 30 problems which they had compiled over the preceding months. The three problems they finally chose were particularly appealing. The British team didn’t disappoint — they set two lovely original problems for the Australians as well as an interesting problem which they came across in training. In the afternoon after this exam, the students did the marking, then coordinated with us. This has two upshots: it teaches the students about what can make a mathematical argument better or worse, and it gives the adults an afternoon free of marking.

As tradition dictates, the fifth and final exam was designated the Mathematics Ashes, in which our teams compete for glory and an urn containing burnt remains of the British scripts from 2008. Before this year our only victory had been in the first Ashes, while in each of the 9 years since we either lost or tied. This year looked like it would be tight. Australia cleaned up on problem 1 putting us ahead 42-40. The United Kingdom came back by beating us 37-28 on problem 2, leaving us 7 points behind with only the most difficult problem 3 to mark. The UK scored a solid 21 on this problem, but Australia claimed the victory with 29 making the final score 99-98 in our favour.

Once again, this camp was a great way to wind up both teams’ training before the IMO. The co-training experience was hugely beneficial for all involved, as the students learnt a lot from each other, and as trainers we learnt something too. Hopefully this tradition will continue well into the future. Many thanks to everyone who made this a success, in particular, the UKMT and the entire UK delegation.

Andrew Elvey PriceIMO Deputy Leader

Mathematics Contests The Australian Scene 2018 | IMO Team Preparation School | 151

2018 IMO Team Training ExamAUS and UNK

F5

2018 Mathematical Ashes

Friday, July 6, 2018

Problem 1. Let ABCD be a cyclic quadrilateral. Rays AD and BC meet at P . In the interior ofthe triangle DCP a point M is given, such that the line PM bisects \CMD. Line CM meets thecircumcircle of triangle DMP again at Q. Line DM meets the circumcircle of triangle CMP againat R. The circumcircles of triangles APR and BPQ meet for a second time at S.

Prove that PS bisects \BSA.

Problem 2. Let N0 denote the set of non-negative integers. Find all functions f : N0 ! N0 suchthat

f f(m)(n) = n+ 2f(m)

for all m,n 2 N0 with m n.

(Here fk denotes the kth iterate of f .)

Problem 3. Let n ≥ 3 be a given integer. For each convex n-gon A1A2 . . . An with no two sidesparallel to each other, we construct a labelled graph G as follows.

• Its vertices V1, V2, . . . , Vn correspond to A1, A2, . . . , An, respectively.

• An edge connects Vi to Vj whenever it is possible to draw two parallel lines, one passing throughAi and the other passing through Aj, such that apart from Ai and Aj, the whole n-gon liesstrictly between these two lines.

Two such obtained graphs G = V1, V2, . . . , Vn and H = W1,W2, . . . ,Wn are said to be the same if

ViVj is an edge of G if and onlyWiWj is an edge of H for all integers i, j with 1 i < j n.

Find the number of di↵erent labelled graphs that can be obtained as described above, expressingyour answer as a simple formula in terms of n.


THE MATHEMATICS ASHES


F5






f f(m)(n) = n+ 2f(m)











F5






f f(m)(n) = n+ 2f(m)










Mathematics Contests The Australian Scene 2018 | The Mathematics Ashes | 152

THE MATHEMATICS ASHES RESULTS

Australia P1 P2 P3 TotalWilliam Hu 7 7 7 21

Charles Li 7 1 7 15

William Steinberg 7 5 1 13

Ethan Tan 7 7 6 20

Hadyn Tang 7 1 2 10

Guowen Zhang 7 7 6 20

Total 42 28 29 99

United Kingdom P1 P2 P3 TotalAgnijo Banerjee 7 7 7 21

Sam Bealing 5 7 0 12

Tom Hillman 7 2 0 9

Benedict Randall Shaw 7 7 6 20

Aron Thomas 7 7 1 15

Harvey Yau 7 7 7 21

Total 40 37 21 98

Mathematics Contests The Australian Scene 2018 | The Mathematics Ashes Results | 153

IMO TEAM LEADER’S REPORT

The 59th International Mathematical Olympiad (IMO) was held 3–14 July 2018 in the city of Cluj-Napoca, Romania. This year was the sixth time that Romania1 has hosted the IMO. A total of 594 high school students from 107 countries participated. Of these, 60 were girls.

Each participating country may send a team of up to six students, a Team Leader and a Deputy Team Leader. At the IMO the Team Leaders, as an international collective, form what is called the Jury. This Jury was ably chaired by Mihai Bălună.

The first major task facing the Jury is to set the two competition papers. During this period the Leaders and their observers are trusted to keep all information about the contest problems completely confidential. The local Problem Selection Committee had already shortlisted 28 problems from the 168 problem proposals submitted by 49 of the participating countries from around the world. During the Jury meetings four of the shortlisted problems had to be discarded from consideration due to being too similar to material already in the public domain. Eventually, the Jury finalised the exam problems and then made translations into the 57 languages required by the contestants.

The six problems that ultimately appeared on the IMO contest papers may be described as follows.

1. A relatively easy classical geometry problem proposed by Greece. It is remarkable that new easy problems in geometry are still being composed that have a relatively uncluttered diagram. This is a fine example of such a problem!

2. A medium sequence problem proposed by Slovakia.

3. A deceptively difficult combinatorics problem concerning the existence of a triangular array of positive integers. It was proposed by Iran.

4. An easy combinatorial problem in the form of a two-player game that may be played on a 20 × 20 chessboard. It was proposed by Armenia.

5. A medium number theory problem proposed by Mongolia.

6. A difficult geometry problem, whose diagram consists merely of five points and eight line segments. It was proposed by Poland.

These six problems were posed in two exam papers held on Monday 9 July and Tuesday 10 July. Each paper had three problems. The contestants worked individually. They were allowed four and a half hours per paper to write their attempted proofs. Each problem was scored out of a maximum of seven points.

For many years now there has been an opening ceremony prior to the first day of competition. The President of Romania attended this in person and addressed the audience. Following this and other formal speeches there was the parade of the teams, a dance show, and the 2018 IMO was declared open.

After the exams the Leaders and their Deputies spent about two days assessing the work of the students from their own countries, guided by marking schemes, which had been agreed to 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, for example, a Leader brought something to their attention in a contestant’s exam script that is not covered by the marking scheme. The Team Leader and Coordinators have to agree on scores for each student of the Leader’s country in order to finalise scores. Any disagreements that cannot be resolved in this way are ultimately referred to the Jury. No such referrals occurred this year.

1 Romania hosted the very first IMO in 1959, where just seven countries participated.

Mathematics Contests The Australian Scene 2018 | IMO Team Leader's Report | 154

The contestants found Problem 1 to be the easiest with an average score of 4.93. Problem 3 was the hardest, averaging just 0.28. Only 11 contestants scored full marks on it. The score distributions by problem number were as follows.

Mark P1 P2 P3 P4 P5 P6

0 96 158 548 148 175 419

1 54 85 7 13 184 108

2 24 87 9 106 31 26

3 15 66 14 18 7 11

4 10 18 4 18 6 5

5 7 16 1 15 8 2

6 7 7 0 5 11 5

7 381 157 11 271 172 18

Mean 4.93 2.95 0.28 3.96 2.70 0.64

The medal cuts were set at 31 points for Gold, 25 for Silver and 16 for Bronze. The medal distributions2 were as follows.

Gold Silver Bronze TotalNumber 48 98 143 289

Proportion 8.1% 16.5% 24.1% 48.7%

These awards were presented at the closing ceremony. Of those who did not get a medal, a further 138 contestants received an Honourable Mention for scoring full marks on at least one problem.

The following two contestants achieved the most excellent feat of a perfect score of 42.

• Agnijo Banerjee, United Kingdom• James Lin, United States

They were given a standing ovation during the presentation of medals at the closing ceremony. Fittingly their Gold medals were awarded by Ciprian Manolescu.3

A big congratulations to the Australian IMO team on their exceptional performance this year! They finished 11th in the rankings,4 bringing home two Gold5 medals, three Silver medals, and one Bronze medal.

The Gold medallists were

• Ethan Tan6, year 12, Cranbrook School, NSW• Guowen Zhang , year 12, St Joseph’s College, Gregory Terrace, QLD

The Silver medallists were

• Charles Li , year 12, Camberwell Grammar School, VIC• William Steinberg, year 10, Scotch College, WA• Hadyn Tang, year 9, Trinity Grammar School, VIC

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

3 Ciprian Manolescu, of Romania, has the singular distinction of achieving three perfect scores at the IMO. He competed at the IMO in 1995, 1996, and 1997.

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

5 Australia has now reached the milestone of 20 Gold medals at the IMO since first participating in 1981.6 This was Ethan’s first and only appearance at the IMO. Only three other members of Australian IMO teams have had the

distinction of winning only gold at IMOs. They were Andrew Hassell in 1985, Benjamin Burton in 1991, and Thomas Lam in 1997.


The Bronze medallist was

• William Hu, year 12, Christ Church Grammar School , WA

Their scores in detail are shown below.

Name P1 P2 P3 P4 P5 P6 Score Award

William Hu 7 7 0 7 1 1 23 Bronze

Charles Li 7 7 2 7 2 1 26 Silver

William Steinberg 7 7 0 7 7 1 29 Silver

Ethan Tan 7 3 7 7 7 0 31 Gold

Hadyn Tang 7 3 0 7 7 1 25 Silver

Guowen Zhang 7 7 4 7 7 3 35 Gold

Australian Average 7.00 5.67 2.17 7.00 5.17 1.17 28.17

IMO average 4.93 2.95 0.28 3.96 2.70 0.64 15.45

Two members of the 2018 Australian IMO are eligible for the 2019 IMO team.

The 2018 IMO was organised by the Romanian Mathematical Society with support from the Romanian government.

The 2019 IMO is scheduled to be held 10–22 July in Bath, United Kingdom. Hosts for future IMOs have been secured up to 2023 as follows.

• 2020 Russia• 2021 United States• 2022 Norway• 2023 Japan

Much of the statistical information found in this report can also be found on the official website of the IMO, www.imo-official.org


http://www.imo-official.org

Some Country Totals

Rank Country Total1 United States of America 212

2 Russia 201

3 China 199

4 Ukraine 186

5 Thailand 183

6 Taiwan 179

7 South Korea 177

8 Singapore 175

9 Poland 174

10 Indonesia 171

11 Australia 169

12 United Kingdom 161

13 Japan 158

13 Serbia 158

15 Hungary 157

16 Canada 156

17 Italy 154

18 Kazakhstan 151

19 Iran 150

20 Vietnam 148

21 Bulgaria 146

22 Croatia 145

23 Slovakia 140

24 Sweden 138

24 Turkey 138

26 Israel 136

27 Georgia 133

28 Brazil 132

28 India 132

28 Mongolia 132

31 Germany 131

32 Armenia 130

33 France 129

33 Romania 129

35 Peru 125

36 Mexico 123

36 Netherlands 123

38 Philippines 121

39 Argentina 115

39 Czech Republic 115


Distribution of Awards at the 2018 IMO

Country Total Gold Silver Bronze HMAlbania 37 0 0 0 1

Algeria 18 0 0 0 0

Argentina 115 0 1 4 0

Armenia 130 0 2 4 0

Australia 169 2 3 1 0Austria 72 0 0 3 1

Azerbaijan 50 0 0 0 5

Bangladesh 114 1 0 3 2

Belarus 102 0 0 4 1

Belgium 92 0 0 4 1

Bolivia 33 0 0 0 3

Bosnia and Herzegovina 103 0 0 4 2

Botswana 12 0 0 0 1

Brazil 132 1 0 4 1

Bulgaria 146 1 3 1 1

Cambodia 11 0 0 0 0

Canada 156 0 5 1 0

Chile 19 0 0 0 2

China 199 4 2 0 0

Colombia 59 0 0 1 2

Costa Rica 65 0 0 2 2

Croatia 145 0 4 1 0

Cyprus 45 0 0 1 1

Czech Republic 115 0 2 2 2

Denmark 71 0 0 3 0

Ecuador 48 0 0 0 3

Egypt 10 0 0 0 1

El Salvador 20 0 0 0 2

Estonia 80 0 1 0 3

Finland 70 0 0 2 2

France 129 1 1 4 0

Georgia 133 0 1 5 0

Germany 131 1 2 1 2

Ghana 13 0 0 0 0

Greece 74 0 0 2 2

Guatemala 11 0 0 0 1

Honduras 6 0 0 0 0

Hong Kong 89 0 0 2 4

Hungary 157 0 4 2 0

Iceland 56 0 0 1 3

India 132 0 3 2 1

Indonesia 171 1 5 0 0

Iran 150 1 3 1 1

Iraq 9 0 0 0 0


Country Total Gold Silver Bronze HMIreland 43 0 0 1 1

Israel 136 0 2 4 0

Italy 154 0 4 2 0

Ivory Coast 8 0 0 0 1

Japan 158 1 3 2 0

Kazakhstan 151 0 4 2 0

Kosovo 21 0 0 0 2

Kyrgyzstan 41 0 0 0 4

Latvia 40 0 0 0 2

Lithuania 77 0 0 2 3

Luxembourg 14 0 0 0 1

Macau 61 0 0 1 3

Macedonia (FYR) 27 0 0 0 2

Malaysia 90 0 0 2 3

Mexico 123 0 1 4 1

Moldova 86 0 0 3 3

Mongolia 132 0 1 5 0

Montenegro 20 0 0 0 1

Morocco 46 0 0 0 3

Myanmar 23 0 0 0 2

Nepal 5 0 0 0 0

Netherlands 123 0 1 4 1

New Zealand 102 0 1 2 3

Nigeria 26 0 0 0 2

Norway 73 0 0 2 1

Pakistan 35 0 0 0 3

Panama 21 0 0 0 2

Paraguay 12 0 0 0 0

Peru 125 0 2 3 1

Philippines 121 1 1 2 2

Poland 174 1 5 0 0

Portugal 77 0 0 2 3

Puerto Rico 46 0 0 1 1

Romania 129 1 1 2 2

Russia 201 5 1 0 0

Saudi Arabia 69 0 1 1 1

Serbia 158 2 2 2 0

Singapore 175 2 3 1 0

Slovakia 140 0 3 3 0

Slovenia 104 0 1 1 4

South Africa 66 0 0 1 4

South Korea 177 3 3 0 0

Spain 74 0 0 2 4

Sri Lanka 47 0 0 1 3

Sweden 138 1 2 2 1


Country Total Gold Silver Bronze HMSwitzerland 52 0 0 1 1

Syria 69 0 0 2 2

Taiwan 179 3 1 2 0

Tajikistan 103 0 0 5 1

Tanzania 1 0 0 0 0

Thailand 183 3 3 0 0

Trinidad and Tobago 26 0 0 0 1

Tunisia 49 0 0 0 3

Turkey 138 1 1 4 0

Turkmenistan 65 0 0 1 4

Uganda 9 0 0 0 0

Ukraine 186 4 2 0 0

United Kingdom 161 1 4 0 1

United States of America 212 5 1 0 0

Uruguay 7 0 0 0 0

Uzbekistan 21 0 0 0 2

Venezuela 2 0 0 0 0

Vietnam 148 1 2 3 0

Total (107 teams, 594 contestants) 48 98 143 138

N.B. Not all countries sent a full team of six students.

Angelo Di PasqualeIMO Team Leader, Australia


INTERNATIONAL MATHEMATICAL OLYMPIAD

Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad | 161

Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad | 162

Solutions to the 2018 International Mathematical Olympiad

1. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA. Williamwas a Bronze medallist with the 2018 Australian IMO team.)

Let the line through F and D intersect Γ for a second time at X, and let the linethrough G and E intersect Γ for a second time at Y .

A

B C

DE

F

G

X

Y

Since FAXB is cyclic, we have ∠DXA = ∠FXA = ∠FBA = ∠FBD. Since also∠FDB = ∠XDA, it follows that FBD ∼ AXD (AA). Since F lies on theperpendicular bisector of BD, we have FBD is isosceles with FB = FD. ThusAXD is isosceles with AX = AD.

A similar argument shows that AE = AY . But since AD = AE, it follows thatAX = AD = AE = AY , and so XEDY is cyclic with centre A. Using this and thefact that FGXY is also cyclic, we deduce

∠Y ED = ∠Y XD = ∠Y XF = ∠Y GF,

and so DE is parallel to FG.

Comment (Based on the solution by Charles Li, year 12, Camberwell GrammarSchool, VIC. Charles was a Silver medallist with the 2018 Australian IMO team.)

A further interesting property of the above diagram is

FB = FD = FY and GC = GE = GX.

We leave this as a challenge for the reader to demonstrate.

61

INTERNATIONAL MATHEMATICAL OLYMPIAD SOLUTIONS

Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 163

Solution 2 (William Steinberg, year 10, Scotch College, WA. William was a Silvermedallist with the 2018 Australian IMO team.)

Let O be the centre of Γ, and let M and N be the midpoints of AB and AC,respectively. Note that OM ⊥ AB and ON ⊥ AC.

Let F ′ and G′ be the midpoints of BD and CE, respectively. Note that FF ′ ⊥ ABand GG′ ⊥ AC.

Let X and Y be the projections of O onto the lines FF ′ and GG′, respectively.

A

B C

D E

F

G

O

M N

F ′ G′

X Y

A

B C

D E

F

G

F ′ G′

PQ

Note that MOXF ′ is a rectangle due to all its right angles. Hence we have

OX = MF ′ = MB − F ′B =1

2AB − 1

2DB =

1

2AD.

Similarly OY = 12AE. Since AD = AE, we deduce that OX = OY . But OF = OG

(radii of Γ), and OX ⊥ XF and OY ⊥ Y G. Hence OXF ≡ OY G (RHS). Thus∠XFO = ∠OGY . Since also ∠OFG = ∠FGO, we have ∠F ′FG = ∠FGG′.

Let FG intersect lines AB and AC at P and Q, respectively. From the angle sumsin PFF ′ and QGG′, we deduce ∠FPF ′ = ∠G′QG, and so ∠QPA = ∠AQP .Hence APQ is isosceles with apex A. Since ADE is also isosceles with apex A,it follows that DE and FG are parallel.

62Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 164

2. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA. Williamwas a Bronze medallist with the 2018 Australian IMO team.)

Answer All positive integers n that are multiples of 3

For each integer i > n let ai+2 = aiai+1+1. In this way we have an infinite sequencea1, a2, . . . with the property that ai+2 = aiai+1 + 1 for each positive integer i andai = aj whenever i ≡ j (mod n). Thus n is a period of the sequence.

Lemma 1 The sequence does not contain two consecutive positive numbers.

Proof Suppose, for the sake of contradiction, that ai, ai+1 > 0 for some positiveinteger i. It follows that ai+2 = aiai+1 + 1 > 1. Similarly ai+3 > 1. It follows thatai+4 = ai+2ai+3 + 1 > ai+3 + 1 > ai+3. Similarly ai+5 > ai+4, and inductively we seethat the sequence is strictly increasing from ai+3 onward. But this is impossible fora periodic sequence.

Lemma 2 The sequence does not contain a number ai such that ai = 0.

Proof If ai = 0 for some positive integer i, then ai+1 = ai+2 = 1 > 0. But thiscontradicts lemma 1.

Lemma 3 The sequence does not contain three consecutive negative numbers.

Proof If ai, ai+1 < 0, then ai+2 = aiai+1 + 1 > 0.

Consider the sequence where we only remember the signs of the terms. We denotepositive and negative terms by + and −, respectively. We have shown that neither+,+ nor −,−,− occur in the sequence. Thus any two consecutive + terms areseparated by either one or two − terms.

Case 1 The sequence does not contain +,−,−,+.

It follows that the sequence is a pure alternating sequence . . . ,+,−,+,−,+,−, . . ..Consider any fragment . . . , a,−b, c,−d, . . . of the sequence such that a, b, c, d > 0.We have the following two deductions.

1− ab = c > 0 ⇒ ab < 1

1− bc = −d < 0 ⇒ bc > 1

It follows that ab < bc, and so a < c. Since this is true of any two consecutive positiveterms, it follows that the positive terms form a strictly increasing subsequence. Butthis is impossible for a periodic sequence. So this case does not occur.

Case 2 The sequence contains both +,−,+ and +,−,−,+.

Choose a place where the sequence is +,−,−,+ and follow it forward until it reaches+,−,+ for the first time. In this way we find a part of the sequence that looks like+,−,−,+,−,+,−. Thus we may consider a fragment . . . ,−y,−z, a,−b, c,−d, . . .where y, z, a, b, c, d > 0. Observe that a < c may be deduced in exactly the sameway as in case 1. Thus

c > a = yz + 1 > 1.

But c = 1− ab < 1 which contradicts the above. So this case does not occur either.

Case 3 The sequence does not contain +,−,+.

It follows that the sequence proceeds as . . . ,+,−,−,+,−,−,+,−,−, . . .. This im-plies that n is a multiple of 3.

It only remains to observe that the sequence that starts with a1 = a2 = −1 followsthe pattern −1,−1, 2,−1,−1, 2, . . . which yields a valid sequence for any n that isa multiple of 3.


Solution 2 (Ross Atkins, Leader of the 2018 New Zealand IMO team. Ross wasalso a Bronze medallist with the 2003 Australian IMO team.)

As in solution 1, we extend to the infinite periodic sequence a1, a2, . . ..

Consider the sequence b1, b2, b3, . . . defined by

bi = 2aiai+1ai+2 − (ai − 1)2 − (ai+1 − 1)2 − (ai+2 − 1)2.

We show that this sequence is increasing by considering the term bi+1 − bi.

bi+1 − bi = 2ai+1ai+2ai+3 − 2aiai+1ai+2 − (ai+3 − 1)2 + (ai − 1)2

= 2ai+1ai+2(ai+3 − ai)− (ai−3 − ai)(ai+3 + ai − 2)

= (ai+3 − ai)(2ai+1ai+2 − (ai+3 + ai − 2))

= (ai+3 − ai)(2(ai+3 − 1)− (ai+3 + ai − 2))

= (ai+3 − ai)2

≥ 0.

Since the sequence is increasing and b1 = bn+1, the sequence must be constant.However, if bi = bi+1, then we have the equality (ai+3 − ai)

2 = 0 and thereforeai = ai+3 for all i. Therefore the sequence a1, a2, a3, . . . is periodic with period 3 and(since an+i = ai) periodic with period n.

• If n is not a multiple of 3, then this would have to be a constant sequence. Butthis is not possible because the equation x2 + 1 = x has no real solutions.

• If n is a multiple of 3, then −1,−1, 2,−1,−1, 2, . . . satisfies the problem.


Solution 3 (2018 IMO Problem Selection Committee)

The problem statement simply states that we have n simultaneous equations

ai+2 = aiai+1 + 1 for i = 1, 2, . . . , n

where the indices are considered modulo n. It follows that

a2i+2 = aiai+1ai+2 + ai+2

= ai(ai+3 − 1) + ai+2

⇔ a2i+2 − aiai+3 = ai+2 − ai. (1)

Summing the equalities at (1) for i = 1, 2, . . . , n yields

n∑i=1

(a2i+2 − aiai+3) =n∑

i=1

(ai+2 − ai)

⇔n∑

i=1

(ai − ai+3)2 = 0. (2)

Here we have used∑

ai+2 =∑

ai and∑

a2i+2 =∑

a2i =∑

a2i+3.

It follows from (2) that ai = ai+3 for all i. Hence the sequence has period 3. Sincethe sequence also has period n, it follows that it has period gcd(3, n).

• If n is not a multiple of 3, then the sequence has period 1, and so is constant.But this is not possible because the equation x2 + 1 = x has no real solutions.

• If n is a multiple of 3, then −1,−1, 2,−1,−1, 2, . . . satisfies the problem.


3. This was the hardest problem of the 2018 IMO. Only 11 of the 594 contestants wereable to solve this problem completely.

Solution (Ethan Tan, year 12, Cranbrook School, NSW. Ethan was a Gold medal-list with the 2018 Australian IMO team.)

Answer Such a proposed anti-Pascal triangle with 2018 rows does not exist!

We start with the following observations. For any positive integer x in an anti-Pascaltriangle T that is not in the bottom row, the two numbers below it are y and x+ yfor some positive integer y. Thus starting from any integer x1 in T , we may generatethe sequences x1, x2, x3, . . . and y2, y3, . . . of positive integers inductively as follows.

Whenever xn is not in the bottom row of T , let xn+1 and yn+1 be thelarger and smaller of the two numbers, respectively, below xn in T . (*)

Thus xn+1 = xn + yn+1 for each positive integer n. From this it easily follows that

xn = x1 + y2 + y3 + · · ·+ yn

for each positive integer n.

For example, in the following four-row anti-Pascal triangle, starting with x1 = 4 wegenerate (x1, x2, x3, x4) = (4, 6, 7, 10) and (y2, y3, y4) = (2, 1, 3).

4

2 6

5 7 1

8 3 10 9

For the problem at hand let us assume, for the sake of contradiction, the existenceof an anti-Pascal triangle T that has 2018 rows and which contains the numbersfrom 1 to 1 + 2 + · · ·+ 2018.

Let x1 be the top number in T , and consider the sequences x1, x2, x3, . . . , x2018 andy2, y3, . . . , y2018 generated from x1 as described at (*). We have

x2018 = x1 + y2 + y3 + · · ·+ y2018.

Since x1, y2, y3, . . . , y2018 are distinct positive integers, it follows that

x2018 ≥ 1 + 2 + · · ·+ 2018.

But the largest number in T is 1 + 2 + · · ·+ 2018. Thus x2018 = 1 + 2 + · · ·+ 2018and (x1, y2, . . . , y2018) is a permutation of (1, 2, . . . , 2018).

Next, let T1 and T2 be the sub-equilateral triangular arrays whose bases consist ofthe 1009 leftmost numbers and 1009 rightmost numbers, respectively, in the bottomrow of T . Note that T1 and T2 are disjoint and that their union covers the entirebottom row. Without loss of generality we may suppose that x2018 lies outside T1.It follows that all of x2017, x2016, . . . , x1 lie outside T1.

Let T ′1 be the sub-equilateral triangular array whose base consists of the 1008 left-

most numbers in the bottom row of T . Since yn is adjacent to xn in each row of thetriangle, and each xn lies outside of T1, it follows that T ′

1 contains no xn and no yn.In particular none of the numbers 1, 2, . . . , 2018 lies inside T ′

1.


T

T1 T2

T ′1

x2018

x′1

x1

Let x′1 be the top number in T ′

1. Generate the sequences x′1, x

′2, x

′3, . . . , x

′1008 and

y′2, y′3, . . . , y

′1008 as described at (*). These numbers all lie in T ′

1. Since T ′1 does not

contain any of 1, 2, . . . , 2018, it follows that

x′1008 = x′

1 + y′2 + y′3 + · · ·+ y′1008

≥ 2019 + 2020 + · · ·+ 3026

= 1008× 1

2(2019 + 3026)

1009× 2019

= 1 + 2 + · · ·+ 2018.

This is an obvious contradiction because the numbers in T are less than or equal to1 + 2 + · · ·+ 2018.


Discussion (Angelo Di Pasquale, Leader of the 2018 Australian IMO team)

Suppose that n is any positive integer, N = 1 + 2 + · · · + n = n(n+1)2

, and T is ananti-Pascal triangle with n rows that contains the numbers from 1 to N . In whatfollows, we present a refinement of the argument given in the previous solution toprove that n ≤ 8.

Following the previous solution, we generate the sequences x1, x2, x3, . . . , xn andy2, y3, . . . , yn of numbers in T , and deduce that xn = N and (x1, y2, y3, . . . , yn) is apermutation of (1, 2, . . . , n).

Observe that xn and yn are adjacent on the bottom row of T . Let T ′ be the sub-equilateral triangle whose base consists of all numbers to the left of xn and yn onthe bottom row of T . And let T ′′ be the sub-equilateral triangle whose base consistsof all numbers to the right of xn and yn on the bottom row of T .

For any i < n, the number xi is directly above xi+1 and yi+1, and yi is next toxi. It follows that T ′ and T ′′ are disjoint and neither contains any of the numbersx1, x2, x3, . . . , xn or y2, y3, . . . , yn. Hence neither contains any of 1, 2, . . . , n, or N .

Suppose that the base of T ′ contains a numbers and the base of T ′′ contains bnumbers. Note that a+ b = n− 2.

As described at (*) in the previous solution, generate the sequences x′1, x

′2, x

′3, . . . , x

′a

and y′2, y′3, . . . , y

′a for T ′ and the sequences x′′

1, x′′2, x

′′3, . . . , x

′′b and y′′2 , y

′′3 , . . . , y

′′b for T ′′.

As in the previous solution, we have

x′1 + y′2 + y′3 + · · ·+ y′a = x′

a and x′′1 + y′′2 + y′′3 + · · ·+ y′′b = x′′

b .

Adding these two equalities together and using the fact that all of the above termsare different numbers that are greater than n but less than N yields

(n+ 1) + (n+ 2) + · · ·+ (n+ (n− 2)) ≤ N − 1 +N − 2

⇒ 1

2(n− 2)(3n− 1) ≤ n(n+ 1)− 3

⇒ n2 ≤ 9n− 8.

Hence n ≤ 8.

Remark The above argument can be further refined to show that no anti-Pascaltriangles exist for n = 8 and n = 7, although the arguments, especially for n = 7,bifurcate into a number of cases.

The case for n = 6 can be ruled out by a simple parity argument as follows. Onemay show that the sum of the numbers in each indicated region is even.

∗ ∗∗

∗∗

∗

∗∗

∗∗

∗∗

∗∗

∗

∗∗

∗∗

∗∗

Thus the sum of all the numbers in a six-row anti-Pascal triangle is even. But sucha triangle cannot contain the numbers from 1 to 21 because their sum is odd.


Comment It can be shown that anti-Pascal triangles with n rows which use eachof the positive integers from 1 up to 1 + 2 + · · · + n exactly once exist only forthe cases n ≤ 5. See the research paper Exact Difference Triangles by Chang, Hu,Lih, and Shieh in the Bulletin of the Institute of Mathematics, Academia Sinica,Volume 5, Number 1, June 1977, pages 191–197. This paper can be found online athttp://w3.math.sinica.edu.tw/bulletin/bulletin_old/d51/5120.pdf.

Up to reflective symmetry, the complete list of anti-Pascal triangles with at most 5rows are shown below.

• n = 2

1

3 2

2

3 1

• n = 3

1

4 3

6 2 5

2

5 3

6 1 4

3

4 1

2 6 5

3

5 2

1 6 4

• n = 4

3

4 7

5 9 2

6 1 10 8

3

5 2

4 9 7

6 10 1 8

4

2 6

5 7 1

8 3 10 9

4

5 1

2 7 6

8 10 3 9

• n = 5

5

9 4

2 11 7

10 12 1 8

13 3 15 14 6


4. Solution (Found by all members of the 2018 Australian IMO team)

Answer K = 100

The problem has a natural reinterpretation as follows. The 400 sites can be viewedas the 400 unit squares of a 20 × 20 chessboard. Amy and Ben take turns placingtheir stones in unoccupied unit squares such that no two red stones are a knight’smove apart, that is, no two red stones are at diagonally opposite corners of a 2× 3or 3× 2 sub-board.

The proof falls naturally into the two parts that follow.

Part 1 Show that K ≥ 100

Colour each of the 400 squares either black or white using the usual chessboardcolouring. Note that if two squares are separated by a knight’s move, then one iswhite and one is black.

Suppose that Amy adopts the strategy of placing her stones only on white squares.Since there are 200 white squares, this ensures that she will place at least 100 redstones, and no two will be a knight’s move apart.

Part 2 Show that K ≤ 100

Partition the chessboard into 25 small 4× 4 chessboards. We further partition eachsuch small chessboard into four groups of four squares as follows.

Group A

A1

A1

A2

A2

Group B

B1

B1

B2

B2

Group C

C1

C1

C2

C2

Group D

D1

D1

D2

D2

Note that each group of four squares forms a four-cycle of knight’s moves.

Suppose that Ben adopts the strategy of playing symmetrically opposite Amy ineach group. Then Amy can play at most once in each group, resulting in at mostfour red stones in each small chessboard, which is at most 100 stones in total.


5. Solution (Hadyn Tang, year 9, Trinity Grammar School, VIC. Hadyn was a Silvermedallist with the 2018 Australian IMO team.)

We present a sequence of lemmas, the last of which completes the proof.

Lemma 1 For any integer n ≥ N the following number is an integer.

anan+1

+an+1 − an

a1

Proof The result follows by calculating the difference of the two integers

a1a2

+a2a3

+ · · ·+ an−1

an+

ana1

anda1a2

+a2a3

+ · · ·+ an−1

an+

anan+1

+an+1

a1.

Lemma 2 Suppose that a, b, c, d are integers with b, d > 0. If each of the fractionsab

and cd

is in lowest terms and ab+ c

dis an integer, then b = d.

Proof We have ab+ c

d= ad+bc

bd. Thus b | ad+ bc, and so b | ad. Since gcd(a, b) = 1,

it follows that b | d. A similar argument shows that d | b. Hence b = d.

Lemma 3 For all integers n ≥ N we have

gcd(a1, an) ≤ gcd(a1, an+1).

Proof Suppose, for the sake of contradiction, that gcd(a1, an) > gcd(a1, an+1) forsome integer n ≥ N . Then there exists a prime power pk such that pk | a1, an butpk an+1. It follows that pk an+1 − an. Hence in lowest terms, the denominatorof an

an+1is not divisible by p, but the denominator of an+1−an

a1is divisible by p. This

contradicts lemmas 1 and 2.

Lemma 4 There is a positive integer G such that for all integers n ≥ G we have

gcd(a1, an) = gcd(a1, an+1).

Proof Clearly gcd(a1, an) ≤ a1 for all n. So the result follows from lemma 3 becausegcd(a1, an) can increase only finitely many times for n ≥ N .

Lemma 5 For all integers n ≥ G we have an+1 ≤ an.

Proof From lemma 4, there is a positive integer g such that gcd(a1, an) = g forall n ≥ G. Hence the numbers bn = an

gare integers for n = 1 and for all integers

n ≥ G. Moreover gcd(b1, bn) = 1 for all n ≥ G.

From lemma 1 it follows that

bnbn+1

+bn+1 − bn

b1=

b1bn + b2n+1 − bnbn+1

b1bn+1

is an integer for all n ≥ G. Thus bn+1 | b1bn + b2n+1 − bnbn+1, and so bn+1 | b1bn.Since gcd(b1, bn+1) = 1, it follows that bn+1 | bn, and so bn+1 ≤ bn for all n ≥ G.Multiplying by g yields an+1 ≤ an for all n ≥ G.

Lemma 6 There is a positive integer M such that an = an+1 for all n ≥ M .

Proof Clearly an ≥ 1 for all n. So the result follows from lemma 5 because an candecrease only finitely many times for n ≥ G.


Comment The next solution uses the following machinery of p-adic valuations.

For each prime number p, the p-adic valuation νp is defined as follows. If n is anypositive integer, then νp(n) denotes the exponent of p in the prime factorisation of n.

For example, if n = 24 = 23 · 31, then ν2(n) = 3, ν3(n) = 1, and νp(n) = 0 for anyprime p > 3. We also have by convention νp(0) = +∞.

It is straightforward to prove the following standard properties of νp for any positiveintegers m and n.

• νp(mn) = νp(m) + νp(n)

• νp(m± n) ≥ minνp(m), νp(n)• νp(m± n) = minνp(m), νp(n) whenever νp(m) = νp(n)


Solution 2 (Guowen Zhang, year 12, St Joseph’s College, QLD. Guowen was aGold medallist with the 2018 Australian IMO team.)

As in solution 1, anan+1

+ an+1−ana1

is an integer for each n ≥ N . Hence, so also is

anan+1

+an+1 − an

a1− 1 =

(a1 − an+1)(an − an+1)

a1an+1

. (1)

Thus an+1 | (a1 − an+1)(an − an+1), and so an+1 | a1an. It follows inductively thatfor any integer n ≥ N , the set of prime divisors of an is a subset of the set of primedivisors of a1aN .

To solve the given problem it suffices to show that for each prime divisor p of a1aN ,the sequence νp(a1), νp(a2), . . . is eventually constant. In what follows p is a fixedprime divisor of a1aN , and ν means νp.

Since RHS(1) is an integer, we require ν(a1an+1) ≤ ν((a1 − an+1)(an − an+1)).Therefore

ν(a1) + ν(an+1) ≤ ν(a1 − an+1) + ν(an − an+1). (2)

Case 1 ν(an) = ν(a1) for some n ≥ N

Let ν(a1) = ν(an) = a and ν(an+1) = c.

If c = a, then (2) yields the following contradiction.

a+ c ≤ mina, c+mina, c < a+ c

Thus c = a. So if ν(an) = ν(a1) for some n ≥ N , then ν(an+1) = ν(a1). Thisinductively implies that ν(am) is constant for all m ≥ n, as desired.

Case 2 ν(an) = ν(a1) for all n ≥ N

Let ν(a1) = a, ν(an) = b, and ν(an+1) = c. Note that a = b and a = c.

If b < c, then (2) yields the following contradiction.

a+ c ≤ mina, c+minb, c ≤ a+ b < a+ c

Thus b ≥ c. So if ν(an) = ν(a1) for all n ≥ N , then ν(an) ≥ ν(an+1) for all n ≥ N .

But the sequence ν(an) ≥ ν(an+1) ≥ ν(an+2) ≥ · · · strictly decreases only finitelymany times. Hence it is eventually constant, as desired.


6. Solution 1 (Dan Carmon, Leader of the 2018 Israeli IMO team)

We start with some results about trapeziums that we will use later.

Lemma 1 Let a, b, c, d be the sides of any trapezium (with a parallel to c), anddiagonals e, f . Then

e2 + f 2 = b2 + d2 + 2ac. (1)

Proof This is straightforward using vectors. Let the trapezium be ABCD wherea = AB, b = BC, c = CD, and d = AD. For an arbitrary origin O, let A,B,C,D

also represent the vectors−→OA,

−→OB,

−→OC,

−→OD. We have

LHS(1) = (A− C) · (A− C) + (B −D) · (B −D). (2)

Also since a and c are parallel, we have

RHS(1) = (B − C) · (B − C) + (A−D) · (A−D) + 2(A− B) · (D − C). (3)

It is straightforward to expand RHS(2) and RHS(3) and verify that they are bothequal to

|A|2 + |B|2 + |C|2 + |D|2 − 2A · C − 2B ·D.

Lemma 2 (Trapezium inequality) Let a, b, c, d be the sides of a non-cyclic trapez-ium (a parallel to c), and e, f be its diagonals, with f > e. Then

f2 > ac+ bd > e2.

Proof Using lemma 1, we have

f 2 ≥ e2 + f 2

2=

b2 + d2 + 2ac

2≥ 2bd+ 2ac

2= ac+ bd.

Also from Ptolemy’s inequality we have

ac+ bd > ef > f2.

With lemma 2 in hand, we proceed to the proof of the given problem.

Assume, for the sake of contradiction, that ∠BXA + ∠DXC = 180. Due to thesymmetry in the problem statement, we may suppose without loss of generality that

∠BXA+ ∠DXC > 180 > ∠CXB + ∠AXD.

Construct ABY externally on the side AB of ABCD so that ABY ∼ DCX.Since ∠Y BA = ∠XCD = ∠XAB, we have AY BX is a trapezium with AY ‖ XB.

From ∠BXA+∠DXC > 180, we have ∠BXA+∠AY B > 180. Hence AB is thelonger diagonal of trapezium AY BX. From the trapezium inequality we have

AB2 > XA · BY +XB · AY. (4)

From ABY ∼ DCX, we have

BY

CX=

AB

DC⇒ BY =

XC · ABCD

,


andAY

DX=

AB

DC⇒ AY =

XD · ABCD

.

Substituting these values into (4) and rearranging yields

AB · CD > XA ·XC +XB ·XD. (5)

A

B

C

D

X

Y

Z

Next construct BCZ externally on ABCD so that BCZ ∼ ADX. Since∠ZCB = ∠XDA = ∠XBC, we have BZCX is a trapezium with CZ ‖ XB.

From ∠CXB+∠AXD < 180, we have ∠CXB+∠BZC > 180 and so BC is theshorter diagonal of trapezium BZCX. From the trapezium inequality we have

BC2 < XB · CZ +XC · BZ. (6)

From BCZ ∼ ADX, we have

CZ

DX=

BC

AD⇒ CZ =

XD · BC

AD,

andBZ

AX=

BC

AD⇒ BZ =

XA · BC

AD.

Substituting these values into (6) and rearranging a little yields

BC · AD > XB ·XD +XA ·XC. (7)

Finally, comparing (5) and (7) yields

AB · CD > BC ·DA,

which contradicts AB · CD = BC ·DA.


Solution 2 (Andrew Elvey Price, Deputy Leader of the 2018 Australian IMOteam)

This solution is via an inversion about the point A.

First we deal with the case where no two sides of ABCD are parallel. We mayassume without loss of generality that ray AB intersects ray DC at some point Pand ray AD intersects ray BC at some point Q. It follows from the given angleconditions that quadrilaterals DXBQ and PCXA are cyclic.

A

BC

D

X

P

Q

Now we invert the diagram around the point A, sending each point Z = A to thepoint Z ′ on the ray AZ which satisfies AZ · AZ ′ = 1.

We will utilise the following standard properties of inversion.

• Any line passing through A is sent to itself.

• Any circle passing through A is sent to a line not passing through A andvice versa.

• Any circle not passing through A is sent to a circle not passing through A.

• If Y and Z are sent to Y ′ and Z ′, respectively, then AY Z ∼ AZ ′Y ′.

We now analyse the diagram resulting from the inversion. Since A,D,Q lie on a linein that order, the point Q′ lies on segment AD′. Similarly, P ′ lies on segment AB′.

Since C lies within angle BAD, this is also true of C ′. Moreover, as C is the inter-section of lines BQ and DP , its image C ′ is the intersection of circles AB′Q′ andAD′P ′.

The given length condition can be rewritten as

BC

AB=

CD

AD.

Converting this to the inverted diagram using the similarities ABC ∼ AC ′B′

and ADC ∼ AC ′D′ yields

B′C ′

AC ′ =C ′D′

AC ′

and so B′C ′ = C ′D′.


Finally, X is an intersection of circles DBQ and PCA, so X ′ is an intersection of lineP ′C ′ with circle D′B′Q′. In order to determine which of these two intersection pointsis X ′, we will need to use the condition that X lies inside quadrilateral ABCD. Inthe inverted diagram, this means that X ′ lies within angle B′AD′, but outside circlesAB′C ′ and AC ′D′.

A

B′

D′

C ′

Q′

P ′

X1 = X ′

X2

Y

Using circles AP ′C ′D′ and AQ′C ′B′, we deduce

\CP ′B′ = \CD′Q′ and \CB′P ′ = \CQ′D′.

Hence C ′B′P ′ ∼ C ′Q′D′, and so \B′C ′P ′ = \Q′C ′D′. Thus line C ′P ′ is thereflection of line C ′Q′ in the angle bisector of B′C ′D′. However, since B′C ′ = C ′D′,this angle bisector is also the perpendicular bisector of B′D′.

Let Y be the second intersection of line C ′Q′ with circle B′D′Q′, and let X1 andX2 be the reflections of Y and Q′ in the perpendicular bisector of B′D′. Since theperpendicular bisector of B′D′ passes through the centre of circle B′D′Q′, the pointsX1 and X2 lie on this circle, so either X ′ = X1 or X ′ = X2. We will show that thelatter is impossible.

Suppose, for the sake of contradiction, that X2 = X ′. Then X2 lies within angleB′AD′. Since X2 is the reflection of Q′ in the perpendicular bisector of B′D′, wehave X2Q

′ ‖ B′D′, and so X2 lies within angle AQ′B′. Together these imply thatX ′ = X2 lies inside triangle AB′Q′, which is inside circle AB′C ′Q′. This contradictsthe fact that X lies inside the quadrilateral ABCD. Hence X ′ = X1.

Now we prove the statement in question, that is,

\BXA+ \DXC = 180. (1)

But \DXC = 360 − \CXA− \AXD, and so (1) is equivalent to

\BXA+ 180 = \CXA+ \AXD. (2)

Converting this to the inverted diagram using the similarities AXB ∼ AB′X ′,AXC ∼ AC ′X ′, and AXD ∼ AD′X ′, this is the same as

\AB′X ′ + 180 = \AC ′X ′ + \X ′D′A. (3)


The proof of (3) is an angle chase as follows.

We have

∠AC ′X ′ = ∠AC ′Q′ + ∠Q′C ′X ′

= ∠AB′Q′ + ∠Q′C ′X ′ (circle AB′C ′Q′)

= ∠AB′Q′ + ∠Q′Y X ′ + ∠Y X ′C ′ (exterior angle C ′X ′Y )

= ∠AB′Q′ + ∠Q′B′X ′ + ∠Y X ′C ′ (circle B′Q′D′X ′)

= ∠AB′X ′ + ∠Y X ′C ′.

Thus (3) is equivalent to

∠Y X ′C ′ + ∠X ′D′Q = 180. (4)

Recall that lines P ′C ′X ′ and Q′C ′Y are symmetric with respect to the perpendicularbisector of B′D′, which passes through the centre of circle B′Q′D′X ′Y . Hence

∠Y X ′C ′ = ∠C ′Y X ′ = ∠Q′Y X ′ = 180 − ∠X ′D′Q′,

where the last equality follows from circle Q′D′X ′Y . Hence we have established (4),as desired. This concludes the proof in the case that no two sides of ABCD areparallel.

For the case where either AB ‖ DC or AD ‖ BC, essentially the same proof applies.The only difference is that if AB ‖ DC, we set P ′ = A in the inverted diagram, andthe circle AC ′D′ is tangent to AB′ at this point, while if AD ‖ BC, we set Q′ = A,and the circle AB′C ′ is tangent to AD′ at this point.


ORIGIN OF SOME QUESTIONS

2018 AMOC Senior Contest

Questions 1 and 5 were submitted by Angelo Di Pasquale.

Questions 2 and 3 were submitted by Norman Do.

Question 4 was submitted by Andrew Elvey Price.

2018 Australian Mathematical Olympiad

Questions 1, 3, 4, 7 and 8 were submitted by Angelo Di Pasquale.

Question 2 was submitted by Mike Clapper.

Questions 5 and 6 were submitted by Norman Do.

2018 Asian Pacific Mathematics Olympiad

Questions 2 and 3 were composed by Angelo Di Pasquale and submitted by the AMOC Senior Problems Committee.

2018 International Mathematical Olympiad

Although no problem submitted by Australia appeared on the final IMO papers, one appeared on the IMO 2018 shortlist, which will be available to the public after the conclusion of IMO 2019.

Problem G2 on the 2018 IMO shortlist was composed by Andrew Elvey Price. Andrew is a member of the AMOC Senior Problems Committee. He represented Australia at the 2008 IMO and the 2009 IMO, where he was awarded a Silver medal and a Gold medal, respectively.

Mathematics Contests The Australian Scene 2018 | Origin of Some Questions | 181

AMOC HONOUR ROLL 1979–2018Because 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, ChairProf A L Blakers University of Western AustraliaDr J M Gani Australian Mathematical SocietyProf B H Neumann Australian National UniversityProf G E Wall University of SydneyMr J L Williams University of Sydney

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 7 years; 1980–1986Prof G B Preston Monash University 10 years; 1986–1995Prof A P Street University of Queensland 6 years; 1996–2001Prof C Praeger University of Western Australia 17 years; 2002–2018

Deputy Chair*Prof P J O’Halloran University of Canberra 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; 2002–2015Prof Andrew Hassell Australian National University 3 years; 2016–2018

Executive Director*Prof P J O’Halloran University of Canberra 15 years; 1980–1994Prof P J Taylor University of Canberra 18 years; 1994–2012Adj Prof M Clapper University of Canberra 4 years; 2013–2016Mr Nathan Ford University of Canberra 2 years; 2017–2018

Chief MathematicianMr Mike Clapper Melbourne 2 years; 2017–2018

SecretaryProf J C Burns Australian Defence Force Academy 9 years; 1980–1988Vacant 4 years; 1989–1992Mrs K Doolan Victorian Chamber of Mines 6 years; 1993–1998

Treasurer*Prof J C Burns Australian Defence Force Academy 8 years; 1981–1988Prof P J O’Halloran University of Canberra 2 years; 1989–1990Ms J Downes CPA 5 years; 1991–1995Dr P Edwards Monash University 8 years; 1995–2002Prof M Newman Australian National University 6 years; 2003–2008Dr P Swedosh The King David School, Victoria 10 years; 2009–2018

Director of Mathematics Challenge for Young Australians*Mr J B Henry Deakin University 17 years; 1990–2006Dr K McAvaney Deakin University 13 years; 2006–2018

Chair, Senior Problems CommitteeProf B C Rennie James Cook University 1 year; 1980Mr J L Williams University of Sydney 6 years; 1981–1986Assoc Prof H Lausch Monash University 27 years; 1987–2013Dr N Do Monash University 5 years; 2014–2018

Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 182

Director of Training*Mr J L Williams University of Sydney 7 years; 1980–1986Mr G Ball University of Sydney 3 years; 1987–1989Dr D Paget University of Tasmania 6 years; 1990–1995Dr M Evans Scotch College, Victoria 3 months; 1995Assoc Prof D Hunt University of New South Wales 5 years; 1996–2000Dr A Di Pasquale University of Melbourne 18 years; 2001–2018

IMO Team LeaderMr J L Williams University of Sydney 5 years; 1981–1985Assoc Prof D Hunt University of New South Wales 9 years; 1986, 1989, 1990, 1996–2001Dr E Strzelecki Monash University 2 years; 1987, 1988Dr D Paget University of Tasmania 5 years; 1991–1995Dr A Di Pasquale University of Melbourne 16 years; 2002–2010, 2012–2018Mr I Guo University of New South Wales 1 year; 2011

IMO Deputy Team LeaderProf G Szekeres University of New South Wales 2 years; 1981–1982Mr G Ball University of Sydney 7 years; 1983–1989Dr D Paget University of Tasmania 1 year; 1990Mr J Graham University of Sydney 3 years; 1991–1993Dr M Evans Scotch College, Melbourne 3 years; 1994–1996Dr A Di Pasquale University of Melbourne 5 years; 1997–2001Mr D Mathews University of Melbourne 3 years; 2002–2004Mr N Do University of Melbourne 4 years; 2005–2008Mr I Guo University of New South Wales 4 years; 2009–2010, 2012–2013Mr G White University of Sydney 1 year; 2011Mr A Elvey Price Melbourne University 5 years; 2014–2018

EGMO Team LeaderMiss T Shaw SCEGGS Darlinghurst 1 year, 2018

EGMO Deputy Team LeaderMiss M Chen Melbourne University 1 year, 2018

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 St. Francis Xavier College 5 years; 2014–2018

New South WalesDr M Hirschhorn University of New South Wales 1 year; 1980Mr G Ball University of Sydney 16 years; 1981–1996Dr W Palmer University of Sydney 19 years; 1997–2015A/Prof Daniel Daners University of Sydney 1 year; 2016Dr Dzmitry Badziahin University of Sydney 2 years; 2017–2018

Northern TerritoryDr I Roberts Charles Darwin University 6 years; 2013–2018

QueenslandDr 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 Alan Offer Queensland 4 years; 2015–2018

South AustraliaMr K Hamann Rostrevor 19 years; 1980–1982, 1991–2005; 2013Mr V Treilibs SA Department of Education 8 years; 1983–1990Dr M Peake Adelaide 7 years; 2006–2012Mr D Martin Adelaide 5 years; 2014–2018


TasmaniaMr 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 15 years; 2004–2018

VictoriaDr 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 21 years; 1998–2018

Western 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 13 years; 2006–2018

EditorProf P J O’Halloran University of Canberra 1 year; 1983Dr A W Plank University of Southern Queensland 11 years; 1984–1994Dr A Storozhev Australian Maths Trust 15 years; 1994–2008

Editorial ConsultantDr O Yevdokimov University of Southern Queensland 10 years; 2009–2018

Other Members of AMOC (showing organisations represented where applicable)Mr W J Atkins Australian Mathematics Foundation 18 years; 1995–2012Mr M Bammann South Australia 1 year; 2018Dr S Britton University of Sydney 8 years; 1990–1998Prof G Brown Australian Academy of Science 10 years; 1980, 1986–1994Dr R A Bryce Australian Mathematical Society 8 years; 1991–1998 Mathematics Challenge for Young Australians 14 years; 1999–2012Mr G Cristofani Department of Education and Training 2 years; 1993–1994Mr Josh Dean Tasmania 2 years; 2017–2018Ms L Davis IBM Australia 4 years; 1991–1994Dr W Franzsen Australian Catholic University 9 years; 1990–1998Dr J Gani Australian Mathematical Society 1980Assoc Prof T Gagen ANU AAMT Summer School 6 years; 1993–1998Ms P Gould Department of Education and Training 2 years; 1995–1996Mr Keith Hamann South Australia 4 years; 2015–2018Mr Nhat Hoang Western Australia 1 year; 2018Prof G M Kelly University of Sydney 6 years; 1982–1987Ms J McIntosh Mathematics Challenge for Young Australians 7 years; 2012–2018Prof R B Mitchell University of Canberra 5 years; 1991–1995Ms Anna Nakos Mathematics Challenge for Young Australians 16 years; 2003–2018Mr S Neal Department of Education and Training 4 years; 1990–1993Prof M Newman Australian National University 13 years; 1986–1998 Mathematics Challenge for Young Australians 20 years; 1999–2018, (Treasurer during the interim) 2003–2008Prof R B Potts University of Adelaide 1980Mr H Reeves Australian Association of Maths Teachers/ 16 years; 1988–1998, Australian Mathematics Foundation 2014–2018Mr N Reid IBM Australia 3 years; 1988–1990Mr Martin Roberts Tasmania 4 years; 2014–2017Mr 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 3 years; 1986–1988Assoc Prof I F Vivian University of Canberra 1 year; 1990Dr M W White IBM Australia 9 years; 1980–1988


Associate Membership (inaugurated in 2000)Ms S Britton 17 years; 2000–2016 Dr M Evans 19 years; 2000–2018 Dr W Franzsen 17 years; 2000–2016 Prof T Gagen 17 years; 2000–2016 Mr H Reeves 15 years; 2000–2014 Mr G Ball 13 years; 2004–2016 Mr I Guo 1 year; 2018

Mathematics Challenge for Young Australians

Problems Committee for Mathematics Challenge StageDr K McAvaney Deakin University (Director) 13 years; 2006–2018; Member 2 years; 2005–2006Mr B Henry Victoria (Director) 17 years; 1990–2006 Member 12 years; 2007–2018Prof P J O’Halloran University of Canberra 5 years; 1990–1994Dr R A Bryce Australian National University 23 years; 1990–2012Mr M Clapper Australian Maths Trust 6 years; 2013–2018Ms L Corcoran Australian Capital Territory 3 years; 1990–1992Ms B Denney New South Wales 9 years; 2010–2018Mr J Dowsey University of Melbourne 8 years; 1995–2002Mr A R Edwards Department of Education, Queensland 29 years; 1990–2018Dr M Evans Scotch College, Victoria 6 years; 1990–1995Assoc Prof H Lausch Monash University 24 years; 1990–2013Ms J McIntosh Australian Mathematical Sciences Institute 17 years; 2002–2018Mrs L Mottershead New South Wales 27 years; 1992–2018Miss A Nakos Temple Christian College, South Australia 26 years; 1993–2018Dr M Newman Australian National University 29 years; 1990–2018Ms F Peel St Peter’s College, South Australia 2 years; 1999–2000Dr I Roberts Charles Darwin University, NT 6 years; 2013–2018Miss T Shaw SCEGGS, NSW 6 years; 2013–2018Ms K Sims Blue Mountains Grammar School 20 years; 1999–2018Dr A Storozhev Attorney General’s Department 23 years; 1994–2016Prof P Taylor Australian Maths Trust 20 years; 1995–2014Mrs A Thomas New South Wales 18 years; 1990–2007Dr S Thornton Exec. Dir. reSolve 21 years; 1998–2018Miss G Vardaro Wesley College, Victoria 25 years; 1993–2006, 2008–2018

Visiting membersProf E Barbeau University of Toronto, Canada 1991, 2004, 2008Prof G Berzsenyi Rose Hulman Institute of Technology, USA 1993, 2002Dr L Burjan Department of Education, Slovakia 1993Dr V Burjan Institute for Educational Research, Slovakia 1993Mrs A Ferguson Canada 1992Prof B Ferguson University of Waterloo, Canada 1992, 2005Dr D Fomin St Petersburg State University, Russia 1994Prof F Holland University College, Cork, Ireland 1994Dr A Liu University of Alberta, Canada 1995, 2006, 2009Prof Q Zhonghu Academy of Science, China 1995Dr A Gardiner University of Birmingham, United Kingdom 1996Prof P H Cheung Hong Kong 1997Prof R Dunkley University of Waterloo, Canada 1997Dr S Shirali India 1998Mr M Starck New Caledonia 1999Dr R Geretschlager Austria 1999, 2013Dr A Soifer United States of America 2000Prof M Falk de Losada Colombia 2000Mr H Groves United Kingdom 2001 Prof J Tabov Bulgaria 2001, 2010Prof A Andzans Latvia 2002Prof Dr H-D Gronau University of Rostock, Germany 2003Prof J Webb University of Cape Town, South Africa 2003, 2011


Mr A Parris Lynwood High School, Christchurch, NZ 2004Dr A McBride University of Strathclyde, United Kingdom 2007Prof P Vaderlind Stockholm University, Sweden 2009, 2012Prof A Jobbings United Kingdom 2014Assoc Prof D Wells United States of America 2015Emer. Fellow P Neumann Queen’s College Oxford UK 2016Dr C Jones Alabama State University, USA 2018Dr P Couch Lamar University, USA 2018

Moderators for Mathematics Challenge StageMr W Akhurst New South WalesMs N Andrews ACER, Camberwell, VictoriaMr L Bao Leopold Primary School, VictoriaProf E Barbeau University of Toronto, CanadaMr R Blackman VictoriaMs J Breidahl St Paul’s Woodleigh, VictoriaMs S Brink Glen Iris, VictoriaProf J C Burns Australian Defence Force AcademyMr J Carty ACT Department of Education, Australian Capital TerritoryMr A Canning QueenslandMrs F Cannon New South WalesDr E Casling Australian Capital TerritoryMr B Darcy South AustraliaMs B Denney New South WalesMr J Dowsey VictoriaMr S Ewington Sydney Grammar School, New South WalesBr K Friel Trinity Catholic College, New South WalesDr D Fomin St Petersburg University, RussiaMrs P Forster Penrhos College, Western AustraliaMr T Freiberg QueenslandMr W Galvin University of Newcastle, New South WalesMr S Gardiner University of Sydney, New South WalesMr M Gardner North Virginia, USAMs P Graham TasmaniaMr B Harridge University of MelbourneMs J Hartnett QueenslandMr G Harvey Australian Capital TerritoryMs I Hill South AustraliaMs N Hill VictoriaDr N Hoffman Edith Cowan University, Western AustraliaProf F Holland University College, Cork, IrelandMr D Jones Coff’s Harbour High School, New South WalesMs R Jorgensen Australian Capital TerritoryDr T Kalinowski University of Newcastle, New South WalesAssoc Prof H Lausch VictoriaMr J Lawson St Pius X School, Chatswood, New South WalesMr R Longmuir ChinaDr K McAvaney Deakin UniversityMs K McAsey Penleigh and Essendon Grammar School, VictoriaMs J McIntosh University of MelbourneMs N McKinnon VictoriaMs T McNamara VictoriaMr G Meiklejohn Queensland School Curriculum Council, QueenslandMr M O’Connor AMSI, VictoriaMr J Oliver Northern TerritoryMr S Palmer New South WalesDr W Palmer University of SydneyMr A Peck Mt Carmel College, TasmaniaMr G Pointer Marryatville High School, South AustraliaProf H Reiter University of North Carolina, USAMr M Richardson Yarraville Primary School, VictoriaMr G Samson Nedlands Primary School, Western AustraliaMr J Sattler Parramatta High School, New South Wales


Mr A Saunder VictoriaMr W Scott Seven Hills West Public School, New South WalesMr R Shaw Hale School, Western AustraliaMiss T Shaw New South WalesDr B Sims University of NewcastleDr H Sims VictoriaMs K Sims Blue Mountains Grammar School, New South WalesProf J Smit The NetherlandsMs C Smith St Paul’s School, QueenslandMs D Smith Inquisitive Minds, New South WalesMrs M Spandler New South WalesMr G Spyker Curtin UniversityMs C Stanley Queensland Study Authority (Primary), QueenslandMs R Stone Plympton Primary School, South AustraliaDr E Strzelecki Monash University, VictoriaDr M Sun New South WalesMr P Swain Ivanhoe Girls Grammar School, VictoriaDr P Swedosh The King David School, VictoriaProf J Tabov Academy of Sciences, BulgariaMrs A Thomas New South WalesMs K Trudgian QueenslandProf J Webb University of Capetown, South AfricaAssoc Prof D Wells North Carolina, USAMs J Vincent Melbourne Girls Grammar School, Victoria

Mathematics Enrichment development

Mathematics Enrichment Series Committee — Development Team (1992–1995)Mr B Henry Victoria (Chairman)Prof P O’Halloran University of Canberra (Director)Mr G Ball University of SydneyDr M Evans Scotch College, VictoriaMr K Hamann South AustraliaAssoc Prof H Lausch Monash UniversityDr A Storozhev Australian Maths Trust, Australian Capital Territory

Pólya Enrichment Series — Development Team (1992–1995)Mr G Ball University of Sydney (Editor)Mr K Hamann South Australia (Editor)Prof J Burns Australian Defence Force AcademyMr J Carty Merici College, Australian Capital TerritoryDr H Gastineau-Hill University of SydneyMr B Henry VictoriaAssoc Prof H Lausch Monash UniversityProf P O’Halloran University of CanberraDr A Storozhev Australian Maths Trust, Australian Capital Territory

Pólya Enrichment Series — Development Team (2013–2015)Adj Prof M Clapper Australian Maths Trust (Editor)Dr R Atkins New ZealandDr A Di Pasquale University of MelbourneDr R Geretschlager AustriaDr D Mathews Monash UniversityDr K McAvaney Australian Maths Trust

Euler Enrichment Series — Development Team (1992–1995)Dr M Evans Scotch College, Victoria (Editor)Mr B Henry Victoria (Editor)Mr L Doolan Melbourne Grammar School, VictoriaMr K Hamann South AustraliaAssoc Prof H Lausch Monash UniversityProf P O’Halloran University of CanberraMrs A Thomas Meriden School, New South Wales


Gauss Enrichment Series — Development Team (1993–1995)Dr M Evans Scotch College, Victoria (Editor)Mr B Henry Victoria (Editor)Mr W Atkins University of CanberraMr G Ball University of SydneyProf J Burns Australian Defence Force AcademyMr L Doolan Melbourne Grammar School, VictoriaMr A Edwards Mildura High School, VictoriaMr N Gale Hornby High School, New ZealandDr N Hoffman Edith Cowan UniversityProf P O’Halloran University of CanberraDr W Pender Sydney Grammar School, New South WalesMr R Vardas Dulwich Hill High School, New South Wales

Noether Enrichment Series — Development Team (1994–1995)Dr M Evans Scotch College, Victoria (Editor)Dr A Storozhev Australian Maths Trust, Australian Capital Territory (Editor)Mr B Henry VictoriaDr D Fomin St Petersburg University, RussiaMr G Harvey New South Wales

Newton Series — Development Team (2001–2002)Mr B Henry Victoria (Editor)Mr J Dowsey University of MelbourneMrs L Mottershead New South WalesMs G Vardaro Annesley College, South AustraliaMs A Nakos Temple Christian College, South AustraliaMrs A Thomas New South Wales

Dirichlet Series — Development Team (2001–2003)Mr B Henry Victoria (Editor)Mr A Edwards Ormiston College, QueenslandMs A Nakos Temple Christian College, South AustraliaMrs L Mottershead New South WalesMrs K Sims Chapman Primary School, Australian Capital TerritoryMrs A Thomas New South Wales

Ramanujan Enrichment Series — Development Team (2014–2016)Mr B Henry Victoria (Editor)Adj Prof M Clapper Australian Maths TrustMs A Nakos Temple Christian College, South AustraliaMr A Edwards Department of Education, Queensland Dr K McAvaney Australian Maths TrustDr I Roberts Charles Darwin University, NT

Australian Intermediate Mathematics Olympiad CommitteeDr K McAvaney Deakin University (Chair) 12 years; 2007–2018Mr J Dowsey University of Melbourne 18 years; 1999–2016Dr M Evans Australian Mathematical Sciences Institute 20 years; 1999–2018Mr B Henry Victoria (Chair) 8 years; 1999–2006 Member 12 years; 2007–2018Assoc Prof H Lausch Monash University 17 years; 1999–2015Mr R Longmuir China 2 years; 1999–2000Adj Prof M Clapper Australian Maths Trust 5 years; 2014–2018Dr D Mathews Monash University 2 years; 2017–2018


AMOC Senior Problems Committee

Current membersMr M Clapper Australian Maths Trust 6 years; 2013–2018Dr Alice Devillers University of Western Australia 3 years; 2016–2018Dr A Di Pasquale University of Melbourne 18 years; 2001–2018Dr N Do Monash University (Chair) 5 years; 2014–2018 (Member) 11 years; 2003–2013Mr I Guo University of Sydney 11 years; 2008–2018Dr J Kupka Monash University 16 years; 2003–2018Dr K McAvaney Deakin University 23 years; 1996–2018Dr D Mathews Monash University 18 years; 2001–2018Dr A Offer Queensland 7 years; 2012–2018Dr C Rao Victoria 19 years; 1999–2018Dr B Saad Monash University 25 years; 1994–2018Dr J Simpson Curtin University 20 years; 1999–2018Dr I Wanless Monash University 19 years; 2000–2018

Previous membersMr G Ball University of Sydney 16 years; 1982–1997Mr M Brazil LaTrobe University 5 years; 1990–1994Dr M S Brooks University of Canberra 8 years; 1983–1990Dr G Carter Queensland University of Technology 10 years; 2001–2010Dr M Evans Australian Mathematical Sciences Institute 28 years; 1990–2016Mr J Graham University of Sydney 1 year; 1992Dr M Herzberg Telecom Australia 1 year; 1990Assoc Prof D Hunt University of New South Wales 29 years; 1986–2014Assoc Prof H Lausch Monash University (Chair) 28 years; 1987–2013 (Member) 2 years; 2014–2015Dr L Kovacs Australian National University 5 years; 1981–1985Dr 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 5 years; 1986–1990Dr E Szekeres University of New South Wales 7 years; 1981–1987Prof G Szekeres University of New South Wales 7 years; 1981–1987Emerit Prof P Taylor Australian Maths Trust 1 year; 2013Dr N H Williams University of Queensland 20 years; 1981–2000

Mathematics School of ExcellenceDr S Britton University of Sydney (Coordinator) 2 years; 1990–1991Mr L Doolan Melbourne Grammar (Coordinator) 6 years; 1992, 1993–1997Mr W Franzsen Australian Catholic University (Coordinator) 2 years; 1990–1991Dr 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 (Director) 17 years; 2001–2018

International Mathematical Olympiad Selection SchoolMr G Ball University of Sydney (Director) 6 years; 1984–1989Mr L Doolan Melbourne Grammar (Coordinator) 3 years; 1989–1991Dr S Britton University of Sydney (Coordinator) 7 years; 1992–1998Mr W Franzsen Australian Catholic University (Coordinator) 8 years; 1992–1996; 1999–2001Dr D Paget University of Tasmania (Director) 6 years; 1990–1995Mr J L Williams University of Sydney (Director) 2 years; 1982–1983Assoc Prof D Hunt University of New South Wales (Director) 5 years; 1996–2000Dr A Di Pasquale University of Melbourne (Director) 18 years; 2001–2018


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