Mathematics Contests the Australian Scene 2018 Combined MCYA and AMOC

Mathematics Contests The Australian Scene 2018 Combined MCYA and AMOC

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

AMT PUBLISHING

Australian Maths Trust 170 Haydon Drive Bruce ACT 2617 AUSTRALIA Telephone: +61 2 6201 5136 www.amt.edu.au

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 Ford April 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

Year City Gold Silver Bronze HM Rank 1981 Washington 1 23 out of 27 teams 1982 Budapest 1 21 out of 30 teams 1983 Paris 1 2 19 out of 32 teams 1984 Prague 1 2 15 out of 34 teams 1985 Helsinki 1 1 2 11 out of 38 teams 1986 Warsaw 5 15 out of 37 teams 1987 Havana 3 15 out of 42 teams 1988 Canberra 1 1 1 17 out of 49 teams 1989 Braunschweig 2 2 22 out of 50 teams 1990 Beijing 2 4 15 out of 54 teams 1991 Sigtuna 3 2 20 out of 56 teams 1992 Moscow 1 1 2 1 19 out of 56 teams 1993 Istanbul 1 2 3 13 out of 73 teams 1994 Hong Kong 2 3 3 12 out of 69 teams 1995 Toronto 1 4 1 21 out of 73 teams 1996 Mumbai 2 3 23 out of 75 teams 1997 Mar del Plata 2 3 1 9 out of 82 teams 1998 Taipei 4 2 13 out of 76 teams 1999 Bucharest 1 1 3 1 15 out of 81 teams 2000 Taejon 1 3 1 16 out of 82 teams 2001 Washington D.C. 1 4 25 out of 83 teams 2002 Glasgow 1 2 1 1 26 out of 84 teams

Mathematics Contests The Australian Scene 2018 | Background Notes on the EGMO, IMO and AMOC | 2 Year City Gold Silver Bronze HM Rank 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 Contests The Australian Scene 2018 | Background Notes on the EGMO, IMO and AMOC | 3 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 Committee Mr M Clapper, Chief Mathematician, Australian Maths Trust, Australian Capital Territory Mrs B Denney, New South Wales Mr A Edwards, Queensland Studies Authority Mr B Henry, Victoria Ms J McIntosh, AMSI, Victoria Mrs L Mottershead, New South Wales Ms A Nakos, Temple Christian College, South Australia Prof M Newman, Australian National University, Australian Capital Territory Dr I Roberts, Northern Territory Miss T Shaw, SCEGGS, New South Wales Ms K Sims, New South Wales Dr S Thornton, Australian Capital Territory Ms G Vardaro, Wesley College, Victoria Dr C Wetherell, Australian Capital Territory

Moderators Mr W Akhurst, New South Wales Mr L Bao, Victoria Mr R Blackman, Victoria Mr A Canning, Queensland Dr E Casling, Australian Capital Territory Mr B Darcy, Rose Park Primary School, South Australia Mr J Dowsey, University of Melbourne, Victoria Mr S Ewington, Sydney Grammar School Mr S Gardiner, University of Sydney Ms J Hartnett, Queensland Dr N Hoffman, Edith Cowan University, Western Australia Dr T Kalinowski, University of Newcastle Mr J Lawson, St Pius X School, New South Wales Ms K McAsey, Penleigh and Essendon Grammar School, Victoria Ms T McNamara, Victoria Mr G Meiklejohn, Department of Education, Queensland Mr M O’Connor, AMSI, Victoria Mr J Oliver, Northern Territory Mr A Peck, Tasmania Mr G Pointer, Marryatville High School, South Australia Dr H Sims, Victoria Ms C Smith, Queensland Ms D Smith, New South Wales Ms C Stanley, Queensland Studies Authority Ms R Stone, South Australia Dr M Sun, New South Wales Mr P Swain, Victoria

Mathematics Contests The Australian Scene 2018 | Membership of MCYA Committees | 6 Dr P Swedosh, The King David School, Victoria Ms K Trudgian, Queensland Dr D Wells, USA

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

Enrichment Editors Mr G R Ball, University of Sydney, New South Wales Mr M Clapper, Chief Mathematician, Australian Maths Trust, Australian Capital Territory Dr M Evans, International Centre of Excellence for Education in Mathematics, Victoria Mr K Hamann, South Australia Mr B Henry, Victoria Dr K McAvaney, Victoria Dr A M Storozhev, Attorney General’s Department, Australian Capital Territory Prof P Taylor, Australian Capital Territory Dr 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 Chair Prof C Praeger, University of Western Australia

Deputy Chair Prof A Hassall, Australian National University, ACT

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

Chief Mathematician Mr M Clapper, Australian Maths Trust, ACT

Treasurer Dr P Swedosh, The King David School, VIC

Chair, Senior Problems Committee Dr N Do, Monash University, VIC

Chair, Challenge Dr K McAvaney, VIC

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

IMO Deputy Team Leader Mr A Elvey Price, University of Melbourne, VIC

EGMO Team Leader Ms T Shaw, SCEGGS Darlinghurst, NSW

State Directors Dr K Dharmadasa, University of Tasmania Dr G Gamble, University of Western Australia Mr Nhat Anh Hoang, WA Deputy Dr I Roberts, Northern Territory Dr D Badziahin, University of Sydney, NSW Mr D Martin, South Australia Dr A Offer, Queensland Dr P Swedosh, The King David School, VIC Dr C Wetherell, St Francis Xavier College, ACT

Editorial Consultant Dr O Yevdokimov, Queensland

Representatives Ms J McIntosh, Challenge Committee Ms A Nakos, Challenge Committee Prof 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 • AMOC Selection School March • 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 2018 Mr M Clapper, Chief Mathematician, Australian Maths Trust, ACT Dr A Devillers, University of Western Australia, WA Dr A Di Pasquale, University of Melbourne, VIC Dr N Do, Monash University, VIC (Chair) Dr I Guo, University of Sydney, NSW Dr J Kupka, Monash University, VIC Dr K McAvaney, Deakin University, VIC Dr D Mathews, Monash University, VIC Dr A Offer, Queensland Dr C Rao, NEC Australia, VIC Dr B Saad, Monash University, VIC Assoc Prof J Simpson, Curtin University of Technology, WA Dr 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 CHALLENGE2018 Challenge PROBLEMS Problems – - MIDDLE Middle Primary 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 any cone, runs to another cone, touches it, reverses direction, runs to another cone, touches it, reverses direction, and so on until each cone is touched exactly once including the starting and ﬁnishing cones.

10 m 10 m 10 m 10 m •1 •2 •3 •4 •5 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 grid squares 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 ﬂips the tiles. For example, here is one way Tia could cover a 6 3 rectangle using two A tiles, three B tiles, and one C tile. ×

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 at least 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 show how 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 two diagrams to show how she might have done this.

Mathematics Contests The Australian Scene 2018 | Challenge Problems1 – Middle Primary | 11 MP3 Coloured Buckets

There are a number of buckets of diﬀerent colours. There are also counters with diﬀerent whole numbers written on them. 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 bucket obeys the rules but the red bucket doesn’t (2 1 = 2, 1 + 2 = 3, 2 + 3 = 5). ×

2 7 1 5 3 4 6

Red 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 in these 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 has moved a bit past the 9. The sum of the numbers closest to which the hands are pointing gives what is called the hand-sum. At 10 past 9, the hand-sum is 2+9 = 11.

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

Mathematics Contests The Australian Scene 2018 | Challenge Problems2 – Middle Primary | 12 When either hand is exactly halfway between two numbers the later number is used. For example, at 12:30 the minute hand is at 6 and the hour hand is exactly halfway between 12 and 1, so the hand-sum is 6 + 1 = 7.

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

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

11 12 1 10 2 9 3 8 4 7 6 5 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 are both 2 + 9 = 11. But a minute later at 1:48, the hand-sum becomes 12 because the minute hand has moved past the halfway 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?

Mathematics Contests The Australian Scene 2018 | Challenge Problems3 – Middle Primary | 13 CHALLENGE2018 Challenge PROBLEMS Problems – - UPPER Upper Primary PRIMARY

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

UP1 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 ﬂips the tiles. For example, here is one way Tia could cover a 5 4 rectangle using two A tiles, two B tiles, and two C tiles. ×

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 diagram to 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 each of the A, B, C tiles for each rectangle. Draw a diagram to show how she could do this. Explain why no other rectangle 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 ﬁve square sites:

Mathematics Contests The Australian Scene 2018 | Challenge Problems4 – Upper Primary | 14 Electronic transmitters provide protection from ﬁre ants by sending signals along the edges of caravan sites. An edge is 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.

• • • 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 transmitters. 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

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. •

Mathematics Contests The Australian Scene 2018 | Challenge Problems5 – 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 bucket obeys the rules but the red bucket doesn’t (2 1 = 2, 1 + 2 = 3, 2 + 3 = 5). ×

2 7 1 5 3 4 6

Red 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 in only one way. d You now have three buckets available: red, blue, green. Find an arrangement for placing the numbers 1 to 13 in these 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 grid point that is distance 1 cm from the vertex. We assume that all isopentagons are so drawn. Here is an example.

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

Mathematics Contests The Australian Scene 2018 | Challenge Problems6 – Upper Primary | 16 CHALLENGE2018 Challenge PROBLEMS Problems -– Junior 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 ﬁve square sites:

Electronic transmitters provide protection from ﬁre ants by sending signals along the edges of caravan sites. An edge is 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.

• • • a The caravan park owner thinks that if she relocates some transmitters, she can protect all ﬁve caravan sites with just 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 enough to do this. c Find the smallest number of transmitters needed to protect all edges of the park in Part b and explain your answer.

Mathematics Contests The Australian Scene 2018 | Challenge Problems7 – 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 the maximum 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 are not. Liam labels the vertices of a cube with eight diﬀerent positive integers so that the labels on each pair of adjacent vertices are coprime. This is called a Liam labelling. To make the edges of the cube easier to see, Liam uses this two-dimensional projection of the cube.

• •

• • 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

An isopentagon is a pentagon with the following properties:

Mathematics Contests The Australian Scene 2018 | Challenge Problems8 – Junior | 18 An isopentagon can be drawn on 1 cm isometric grid paper so that each side at the 60◦ vertex passes through a grid point that is distance 1 cm from the vertex. We assume that all isopentagons are so drawn. Here is an example.

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 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. For example, 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 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 a row, 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 , and if we apply the operation R to 3 we obtain 2 . Note that R can never be applied to the number 0. − 2 − 2 3 2 The operations T and R can be combined. For example, we can turn 0 into 5 by successively applying the operations T, T, T, R, T, T, R, T: 1 2 5 3 2 0 T 1 T 2 T 3 R T T R T . −→ −→ −→ −→ − 3 −→ 3 −→ 3 −→ − 5 −→ 5 a Starting with 2, list the numbers produced by successively applying the operations T, R, R, T, R, T, R, T, R. 3 2 b Find a sequence of operations which turns 4 into 3 . 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 sides that meet at the centre is shared with another rhombus. They all have the same size angle at the centre. Figure 1 shows 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 ring when the first ring contains 7 rhombuses. In this example, since it is not possible to draw any new rhombuses that share edges with two rhombuses in the third ring, there are only three rings in this rhombus ring pattern.

Mathematics Contests The Australian Scene 2018 | Challenge Problems9 – Junior | 19 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.

b b In a rhombus ring pattern there are 8 rhombuses in the ﬁrst ring. What are the angles in a rhombus in the last ring? c In a rhombus ring pattern, each rhombus in the last ring has an angle of 20◦ and each rhombus in the second last ring 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?

Mathematics Contests The Australian Scene 2018 | Challenge Problems10 – Junior | 20 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 12 scorecards numbered 1 to 12 is distributed amongst the judges so that each judge has 4 cards. Some judges are more important than others so some get some higher scorecards than others. After a routine, each judge holds up one scorecard. The player’s score is the sum of numbers on these three cards. a In one competition: Judge A had scorecards 1, 2, 3, 4 Judge B had scorecards 5, 6, 7, 8 Judge 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 6 distributed 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, were possible. Explain why there is no other way of distributing the scorecards except for swapping sets of 4 scorecards between judges.

Mathematics Contests The Australian Scene 2018 | Challenge Problems11 – Junior | 21 CHALLENGE2018 Challenge PROBLEMS Problems – - INTERMEDIATE Intermediate

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

I1 Isopentagons

An isopentagon is a pentagon with the following properties:

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

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. For example, the t-number of the (3, 2, 1, 2, 3) isopentagon is 17. 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 a row, 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 , and if we apply the operation R to 3 we obtain 2 . Note that R can never be applied to the number 0. − 2 − 2 3 2 The operations T and R can be combined. For example, we can turn 0 into 5 by successively applying the operations T, T, T, R, T, T, R, T: 1 2 5 3 2 0 T 1 T 2 T 3 R T T R T . −→ −→ −→ −→ − 3 −→ 3 −→ 3 −→ − 5 −→ 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 positive integer n can be turned back into 0 by applying 3n 1 operations. − d Explain how 0 can be turned into any negative integer.

Mathematics Contests The Australian Scene 2018 | Challenge Problems12 – Intermediate | 22 I3 Rail Cams

Two horizontal rails have been fixed near the ceiling of a rectangular stadium so that cameras running along them can film the action below. The stadium floor is 100 m by 70 m. One rail runs the length of the stadium. It is parallel to, and halfway between, the east and west walls and 20 m above the floor. The other rail runs the width of the stadium. It is parallel to, and 15 m from, the south wall and 25 m above the floor. The long rail carries camera A and the short rail carries camera B. Both cameras start at one end of their rails, as shown.

20m

25m 5m

A B 20m 85m 35m 25m W N 35m 15m S E

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 and at 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 is a 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 2 square. However, Tamara notices that some of the 2 2 patterns are just rotations of each other. She regards these× as the same and ﬁnds that there are only six diﬀerent× 2 2 patterns. ×

Mathematics Contests The Australian Scene 2018 | Challenge Problems13 – 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 sides that meet at the centre is shared with another rhombus. They all have the same size angle at the centre. Figure 1 shows 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 ring when the first ring contains 7 rhombuses. In this example, since it is not possible to draw any new rhombuses that share 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 second last 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 angles in 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?

Mathematics Contests The Australian Scene 2018 | Challenge Problems14 – Intermediate | 24 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 a row. 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 that player wins. If a guess is incorrect, the host, who knows where the prize is hidden, then says ‘higher’ or ‘lower’ to faithfully indicate in which direction the prize box is located. Both players always follow the host’s directions. For example, when there are 10 boxes numbered 1 to 10, with the prize in box 7, the game could progress like this:

Player 1: 3 Host: Higher Player 2: 5 Host: Higher Player 1: 8 Host: Lower Player 2: 7 Host: 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 of 1 2 winning is 3 , whereas if she chooses box 1 her chance of winning is 3 . b A game has boxes numbered 1 to 4. Show that no matter which box Player 1 chooses on the first turn, Player 2 1 has a strategy for ensuring his chance of winning is 2 . c A game has boxes numbered 1 to 9. Show that no matter which boxes Player 2 chooses, Player 1 has a strategy 1 which involves choosing box 5 at the first turn and ensures her chance of winning is more than 2 .

Mathematics Contests The Australian Scene 2018 | Challenge Problems15 – Intermediate | 25 CHALLENGE2018 Challenge SOLUTIONS Solutions – - MiddleMIDDLE Primary PRIMARY

MP1 Training a The following diagram shows the student’s path:

10 m 10 m 10 m 10 m •1 •2 •3 •4 •5

The total distance is 30 + 10 + 20 + 30 = 90 m. 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: 70 m, 100 m, 90 m, 100 m, 90 m. 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 their running 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: 70 m, 80 m, 80 m, 110 m, 90 m, 110 m, 90 m. Similarly, the only runs starting at cone 3 are: 31425, 31524, 32415, 32514, 34152, 34251, 35142, 35241. Their total distances are respectively: 100 m, 110 m, 100 m, 110 m, 110 m, 100 m, 110 m, 100 m. The runs starting at cone 4 are the reflections about cone 3 of the runs that start at cone 2. For example, 45231 is the 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 is the reflection of 13254. Hence the shortest run is 70 m and the longest run is 110 m. Alternative ii Label the line segments between the cones A, B, C, D.

A B C D •1 •2 •3 •4 •5 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 also has 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 the run immediately before each cone except for the starting cone. So the total tally of segment runs is at least 4 for each run. The student turns at three of the cones. At each such cone, the segment immediately before the cone in the run is repeated immediately after the turn. So, for each run, the total tally of segment runs is at least 4 + 3 = 7. This means the total distance for each exercise drill is at least 70 m. Since, for example, the run 21435 has total distance 70 m, the shortest running distance is 70 m.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions16 – Middle Primary | 26 We now ﬁnd the longest running distance. Consider any one of the cones 2, 3, 4. In any run, the student must either start, stop, or turn at the cone. In addition the student may pass the cone as well. In the next diagram, the cone is drawn a number of times. Each time, one or both of its attached line segments are also drawn to indicate each time the student runs on them. . . • • • • start• or stop 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 or 2. 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 most 4. 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 110 m. Since, for example, the run 25143 has total distance 110 m, the longest running distance is 110 m.

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 tile is 2 + 3 + 5 = 10. So the remaining 15 grid squares must be covered by seven tiles. This could be done with six A tiles 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× with three A tiles and ﬁve B tiles, as shown. There are other ways.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions17 – 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. A rectangle consisting of 30 grid squares must be 1 30, 2 15, 3 10, or 5 6. Since 1 30× and 2 15 rectangles cannot accommodate a C tile, the two rectangles× we require× are 3× 10 and× 5 6. Here× is one way to× cover each of these rectangles. There are other ways. × ×

MP3 Coloured Buckets a Red bucket: 1, 4 Blue 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 4 must 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, 9 Blue bucket: 4, 5, 6, 7 (This is the only arrangement.) d Here is one arrangement. There are many others. Red bucket: 1, 5, 8, 12 Blue bucket: 2, 6, 7, 11 Green 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. So the 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 3 but 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 3 but 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.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions18 – Middle Primary | 28 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 between them, the number allocated to the hour hand is either 3 or 4. If the number allocated to the hour hand is 4, then the minute hand is anywhere from 6 to 12 and the hand-sum is therefore greater than 7. If the number allocated to the hour hand is 3, then the number allocated to the minute hand is 4. 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.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions19 – Middle Primary | 29 CHALLENGE2018 Challenge SOLUTIONS Solutions – - UpperUPPER Primary 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 square covered 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, the required 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, the rectangles that we require could only be 3 5 or 4 4. × × Here are two such rectangles with a required covering.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions20 – Upper Primary | 30 UP2 Insect Barriers a Four extra transmitters are required. Here is one way to place them.

• • • •

b • • • • • •

c • • • • • • • •

d Alternative i • • One vertex has 4 edges attached to it. All other vertices have fewer edges attached. So the number of edges that could be protected with 3 transmitters is at most 4 + 3 + 3 = 10. But there are 12 edges to be protected. Hence 3 transmitters 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 one transmitter, then the transmitter must be at the middle vertex. Since there are three rows of horizontal edges, if there 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 least 4 transmitters are needed to protect all 8 edges. Hence 3 transmitters are not enough.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions21 – Upper Primary | 31 UP3 Coloured Buckets a Red bucket: 1, 4 Blue 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 4 must 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 in the red bucket, and 6 must be in the blue bucket (8 = 2 +× 6). × Red bucket: 2, 8 Blue 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, 9 Blue 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, 4 Blue bucket: 2, 3 Green 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 in red, and 12 in blue. Red bucket: 1, 4, 10, 13 Blue bucket: 2, 3, 11, 12 Green bucket: 5, 6, 7, 8, 9 There are only two other arrangements: Red bucket: 1, 4, 7, 10, 13 1, 4, 10, 13 Blue bucket: 2, 3, 11, 12 2, 3, 7, 11, 12 Green bucket: 5, 6, 8, 9 5, 6, 8, 9 Any one of the three arrangements is acceptable.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions22 – Upper Primary | 32 UP4 Isopentagons a The longest sides of an isopentagon are those meeting at the 60◦ vertex. There are 4 isopentagons with longest side 5 cm.

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • b The (5, 3, 2, 3, 5) isopentagon is the bottom-left pentagon in the solution of Part a. By carefully counting the internal 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 and 17. 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 7 grid 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.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions23 – Upper Primary | 33 CHALLENGE2018 Challenge SOLUTIONS Solutions - – Junior JUNIOR

J1 Insect Barriers a • • •

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

• • • •

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

• • • • • •

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 since no transmitter on an internal vertex can protect a boundary edge. So none of these can protect any of the edges on the central square. At least two more transmitters are needed to do that. So at least 8 transmitters are needed to 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 protect all edges. Alternative iii Each of the corner squares requires at least 2 transmitters to protect its 4 edges, and none of these transmitters protect 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.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions24 – Junior | 34 d If we had 10 transmitters, then all sites could be protected if we placed them this way.

• • • • • • • •

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

• • • • • •

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 protect any boundary edge. All boundary sites and therefore the 14 boundary edges must be protected. This requires at least 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 any boundary edge. Then there are at most 5 transmitters remaining to protect the boundary edges. Since 5 transmitters can protect at most 10 boundary edges, at least 4 boundary edges must be unprotected. Therefore at least two boundary 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 must belong to one site and are therefore on the boundary. Since there are at most two boundary edges in a site, the number 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 at least 5 transmitters each protecting 4 distinct edges. The only vertices at which a transmitter can protect 4 edges are the interior (non-boundary) vertices. Since no such transmitter can protect a boundary edge, we have at most 3 transmitters to protect at least 12 boundary edges. This is impossible, since a single transmitter can protect at most 2 edges on the boundary. So the maximum number of sites that can be protected by 8 transmitters is 10.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions25 – 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.

8 7 • •

•3 2•

4 5 • •

1• •6 b Here is one possible labelling. There are others.

10 7 • •

•3 2•

4 5 • •

9• •8 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 are 5 and 7. Since each vertex has three other vertices adjacent to it, there are not enough integers to label all the vertices 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 or 3 on the right face. Either way, this would mean a pair of adjacent vertices both have even vertices, breaking the coprime rule. So a Liam labelling with composites requires at least 4 composite odd integers. The 4 smallest odd composites 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.

9 22 • •

•14 25•

15 8 • •

4• •21

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.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions26 – Junior | 36 J3 Isopentagons a The longest sides of an isopentagon are those meeting at the 60◦ vertex. There are 4 isopentagons with longest side 5 cm.

= a b

Mathematics Contests The Australian Scene 2018 | Challenge Solutions27 – Junior | 37 By direct counting we get the following table for the number of grid triangles inside an equilateral triangle with side length s.

s 1 2 3 4 5 6 7 8 9 10 11 t 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.

J4 Ts and Rs a 1 1 3 4 1 2 T 3 R R 3 T 4 R T R T R 3 −→ −→ − 3 −→ −→ −→ − 4 −→ 4 −→ − 3 −→ − 3 −→ b 3 R 4 T 1 T 2 4 −→ − 3 −→ − 3 −→ 3 There are longer, less eﬃcient, sequences. c 1 2 3 1 1 3 R T R T T R 2 T 1 T 0 −→ − 3 −→ 3 −→ − 2 −→ − 2 −→ 2 −→ − −→ − −→ There are longer, less eﬃcient, sequences. d Firstly we operate on smaller positive integers to see if there is a pattern. We stop each sequence when we reach a number that has occurred in the previous sequence.

1 R 1 T 0 −→ − −→ 1 1 2 R T R 2 T 1 −→ − 2 −→ 2 −→ − −→ − 1 2 3 1 3 R T R T −→ − 3 −→ 3 −→ − 2 −→ − 2 1 3 4 1 4 R T R T −→ − 4 −→ 4 −→ − 3 −→ − 3 Note that, from the second sequence, the last number in each of these sequences is the second number in the previous sequence. This suggests the following sequence of operations for changing 7 into 0.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions28 – Junior | 38 1 6 7 1 7 R T R T −→ − 7 −→ 7 −→ − 6 −→ − 6 T 5 R 6 T 1 −→ 6 −→ − 5 −→ − 5 T 4 R 5 T 1 −→ 5 −→ − 4 −→ − 4 T 3 R 4 T 1 −→ 4 −→ − 3 −→ − 3 T 2 R 3 T 1 −→ 3 −→ − 2 −→ − 2 1 T R 2 T 1 −→ 2 −→ − −→ − 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 same angle at the centre, we have these angles.

c c c c c

a a b b

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 move outwards.

C 90 135 45 90 45 45 135 90 90 B 45 135 45 90 45 90 135 A

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

Mathematics Contests The Australian Scene 2018 | Challenge Solutions29 – Junior | 39 c We have these angles (all in degrees).

160 20 20 160 60 x 60

Thus x = 360 160 60 60 = 80◦. Hence the angles in each rhombus of the third last ring are 80◦ and − − − 180◦ 80◦ = 100◦. − Extending the diagram above and inserting a few more angles, we get:

160 20 20 160 60 60 80

120 100 100 y

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

160 20 20 160 60 60 80

120 100 100 40

80 140 140

Since 140 + 140 + 80 = 360, no more rings are possible. So there are 4 rings in total and the number of rhombuses in 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 also possible. For example:

1+5+9=15 1+6+12=19 2+8+12=22 1+5+10=16 1+7+12=20 3+8+12=23 1+5+11=17 1+8+12=21 4+8+12=24 1+5+12=18

Mathematics Contests The Australian Scene 2018 | Challenge Solutions30 – Junior | 40 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 by the other two judges. c Any one of the following six distributions.

A 19101118101118911 B 2781227912271012 C 34 5 6 34 5 6 34 5 6 A 19101218101218912 B 2781127911271011 C 34 5 6 34 5 6 34 5 6 d As in Part b, there is only one distribution of scorecards that gives scores 6 to 9:

A 1 B 2 C 3456

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 two distributions. A 1 11 1 12 B 2 7 12 2 7 11 C 345 6 345 6

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

A 19101118101118911 B 2781227912271012 C 34 5 6 34 5 6 34 5 6

A 19101218101218912 B 2781127911271011 C 34 5 6 34 5 6 34 5 6

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

Mathematics Contests The Australian Scene 2018 | Challenge Solutions31 – Junior | 41 CHALLENGE2018 Challenge SOLUTIONS Solutions – - IntermediateINTERMEDIATE

I1 Isopentagons a The longest sides of an isopentagon are those meeting at the 60◦ vertex. There are 4 isopentagons with longest side 5 cm.

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • t = 34 t = 41

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • t = 46 t = 49 b 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 and 17. 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 7 grid 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. c One way to look at an isopentagon is to see it as two identical large equilateral triangles with side lengths a minus a smaller equilateral triangle with side length b, as shown. The smallest value of a is 2, and the largest value of b is a 1. −

Mathematics Contests The Australian Scene 2018 | Challenge Solutions32 – Intermediate | 42 = a b

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

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

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 = 199 − − 8 3 64 + 64 9 = 119 10 3 100 + 100 9 = 191 − − 8 5 64 + 64 25 = 103 10 5 100 + 100 25 = 175 − − 8 7 64 + 64 49 = 79 10 7 100 + 100 49 = 151 − − 9 1 81 + 81 1 = 161 10 9 100 + 100 81 = 119 − − 9 3 81 + 81 9 = 153 − 9 5 81 + 81 25 = 137 − 9 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 the isopentagon 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 39 2 10 14 18 22 26 30 34 38 3 13 17 21 25 29 33 37 4 16 20 24 28 32 36 5 19 23 27 31 35 6 22 26 30 34 7 25 29 33 8 28 32 9 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.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions33 – Intermediate | 43 I2 Ts and Rs a 1 1 3 4 1 2 T 3 R R 3 T 4 R T R T R 3 −→ −→ − 3 −→ −→ −→ − 4 −→ 4 −→ − 3 −→ − 3 −→ b 1 2 3 1 1 3 R T R T T R 2 T 1 T 0 −→ − 3 −→ 3 −→ − 2 −→ − 2 −→ 2 −→ − −→ − −→ There are longer, less eﬃcient, sequences. c Firstly we operate on small positive integers to see if there is a pattern. We stop each sequence when we reach a number that has occurred in the previous sequence.

1 R 1 T 0 −→ − −→ 1 1 2 R T R 2 T 1 −→ − 2 −→ 2 −→ − −→ − 1 2 3 1 3 R T R T −→ − 3 −→ 3 −→ − 2 −→ − 2 1 3 4 1 4 R T R T −→ − 4 −→ 4 −→ − 3 −→ − 3 For any integer n>1 we have:

R 1 T n 1 R n T 1 n − −→ − n −→ n −→ − n 1 −→ − n 1 − − Note that, from the second sequence, the last number in each of these sequences is the second number in the previous sequence. This suggests the following sequence of operations for changing n into 0.

R 1 T n 1 R n T 1 n − −→ − n −→ n −→ − n 1 −→ − n 1 − − T n 2 R n 1 T 1 − − −→ n 1 −→ − n 2 −→ − n 2 − − − T n 3 R n 2 T 1 − − −→ n 2 −→ − n 3 −→ − n 3 − − − and so on down to T 2 R 3 T 1 −→ 3 −→ − 2 −→ − 2 1 T R 2 T 1 −→ 2 −→ − −→ − T 0 −→ The number of operations used is 1 + 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.

0 T 1 R 1 −→ −→ − 1 1 1 R 1 T 2 R T R 2 − −→ −→ −→ − 2 −→ 2 −→ − 1 3 2 1 2 R T R T R 3 − −→ 2 −→ 2 −→ − 3 −→ 3 −→ − 1 4 3 1 3 R T R T R 4 − −→ 3 −→ 3 −→ − 4 −→ 4 −→ − For any integer n>0 we have: 1 n +1 n 1 n R T R T R (n + 1) − −→ n −→ n −→ − n +1 −→ n +1 −→ − Thus the sequence continues until any desired negative integer is obtained.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions34 – Intermediate | 44 I3 Rail Cams a Camera A is 15 m south and 35 m west of camera B . Let the horizontal distance between the two cameras be x metres.

35 B • 15 x

A•

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 between the two cameras. Then we have this right-angled triangle.−

B y •5 A • x

By Pythagoras’ theorem,

y2 = x2 +52 = 152 + 352 +52 =5232 +5272 +52 = 25(9 + 49 + 1) = 25(59)

So y =5√59 m or 3841 cm to the nearest centimetre. b After two seconds, camera A has moved 10 m north and camera B has moved 10 m west. Let the horizontal distance between the two cameras then be x metres.

25 B 10 • 5 x A • 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 two cameras now be y. Then we have this right-angled triangle.

B y •5 A • x

By Pythagoras’ theorem, y2 = x2 +52 =52 + 252 +52 = 25(1 + 25 + 1) = 25(27)

So y = 15√3 m. Therefore, after two seconds, the cameras are 5√59 15√3 metres closer. This is 1242 cm to the nearest centimetre. −

Mathematics Contests The Australian Scene 2018 | Challenge Solutions35 – Intermediate | 45 c Alternative i Since the cameras remain 5 m apart vertically, their direct distance apart is minimised when their horizontal distance d apart is minimised. At time t seconds, both cameras are 5t metres from their start positions. So we have these right-angled triangles.

35 5t B 5t − 15 5t • − A d 0 t 3 • ≤ ≤ 5t

A d 5t 15• − 3 t 7 35 5t B• 5t ≤ ≤ − 15

A •

d 5t 15 −

7 t 14 ≤ ≤ B 5t• 35 35 − 15

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 5 m apart vertically, the minimum direct distance between× them is √200 + 52 = √225 = 15 m and this occurs at 5 seconds. Alternative ii Since the cameras remain 5 m apart vertically, their direct distance apart is minimised when their horizontal distance d apart is minimised. At time t seconds, both cameras are 5t metres from their start positions. The next diagram shows 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-second time interval, both cameras travel 5 m.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions36 – Intermediate | 46 9 • 8 • 7 • 6 • 5 • 4 • 3 2 1 0 3 B 9• •8 •7 •6 •5 •4 • • • • 2 • 1 • 0 A•

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. The bottom-left dot is the origin of a cartesian system with axes along the ‘B’ dots and ‘A’ dots respectively. The gap between successive dots on the axes is 5 m. The point C is at (10,10). Triangles CY A and CXB are congruent right-angled triangles and angle XCY is 90◦. Hence ACB is a right-angled isosceles triangle.

7 •

6 A•

Y 5 C • •

d 4 •

3 •7 •6 B •5 •4 •3 X

By Pythagoras’ theorem, d2 = CA2 + CB2 =2CA2. So minimum d occurs when A is at Y , that is, when t =5 seconds. Then d2 =2 102 = 200. Since the cameras remain 5 m apart vertically, the minimum direct distance between them is √200 +× 52 = √225 = 15 m and this occurs at 5 seconds.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions37 – Intermediate | 47 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 5 occurring 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 are required 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 the probability that a random 3 3 placement contains a peanut is 8/512 = 1/64. × × ×

Mathematics Contests The Australian Scene 2018 | Challenge Solutions38 – Intermediate | 48 I5 Rhombus Rings a Each central angle is 360/8 = 45◦. Starting with those, we calculate the remaining angles in degrees as we move outwards.

C 90 135 45 90 45 45 135 90 90 B 45 135 45 90 45 90 135 A

Since ABC is a straight line, the third ring is the last (outer) ring. Thus the angles in a rhombus in the outer ring are 45◦ and 135◦ b We have these angles (all in degrees).

160 20 20 160 60 x 60

160 20 20 160 60 60 80

120 100 100 y

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

Mathematics Contests The Australian Scene 2018 | Challenge Solutions39 – Intermediate | 49 160 20 20 160 60 60 80

120 100 100 40

80 140 140

Since 140 + 140 + 80 = 360, no more rings are possible. So there are 4 rings in total and the number of rhombuses in 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 we calculate 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. − − − − −

5a 5a 180 4a − 4a 180 3a 4a − 4a

3a 3a 180 2a −

2a 2a 2a

a a 180 a −

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 the first to the 7th ring are: 24◦, 48◦, 72◦, 96◦, 120◦, 144◦, 168◦. Since 8 360/15 = 192 > 180, there is indeed no 8th ring. 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 the ﬁrst 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 8th ring. This conﬁrms there are exactly 7 rings. ×

Mathematics Contests The Australian Scene 2018 | Challenge Solutions40 – Intermediate | 50 I6 Higher or Lower

1 a If Player 1 chooses box 2 on the first turn, her chance of choosing correctly is 3 . 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 she 1 chooses box 2 on the first turn is 3 .

P1 chooses box 2

1 1 1 3 3 3

‘Lower’ ‘Correct’ ‘Higher’ (P2 wins) (P1 wins) (P2 wins)

1 If Player 1 chooses box 1 on the ﬁrst turn, her chance of choosing correctly is 3 . The game will continue with 2 1 probability 3 . If it does continue, then the probability of Player 2 winning on the second turn is 2 . If he doesn’t win, then Player 1 will win on the third turn. So the chance of Player 1 winning if she chooses box 1 on the ﬁrst turn is 1 + 2 1 = 2 . 3 3 × 2 3 P1 chooses box 1

1 2 3 3

‘Correct’ ‘Higher’ (P1 wins) (continues)

1 1 2 2

‘Correct’ ‘Higher/Lower’ (P2 wins) (P1 wins)

1 b If Player 1 chooses box 1 on the ﬁrst turn, her chance of choosing correctly is 4 . The game will continue with 3 probability 4 . 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 2 . Hence Player 2’s overall chance of winning with this strategy is 3 2 = 1 . 3 4 × 3 2 P1 chooses box 1

1 3 4 4

‘Correct’ ‘Higher’ (P1 wins) (continues)

2 3

(P2 wins, using a)

Similarly, if Player 1 chooses box 4 on the first turn and the game continues, then Player 2 can choose box 1 or 3 1 to ensure his overall chance of winning the game is 2 . 1 If Player 1 chooses box 2 on the first turn, her chance of choosing correctly is 4 . If she does not choose correctly, 1 there are two possibilities: either the prize is in box 1, with probability 4 , and Player 2 immediately wins; or the 1 1 prize is in box 3 or 4, with probability 2 , and Player 2’s chance of choosing correctly at that stage is 2 . Hence, if Player 1 chooses box 2 on the first turn, then Player 2’s chance of winning is 1 + 1 1 = 1 . 4 2 × 2 2

Mathematics Contests The Australian Scene 2018 | Challenge Solutions41 – Intermediate | 51 P1 chooses box 2

1 1 1 4 4 2

‘Lower’ ‘Correct’ ‘Higher’ (P2 wins) (P1 wins) (continues)

1 1 2 2

‘Correct’ ‘Higher/Lower’ (P2 wins) (P1 wins)

1 Similarly, if Player 1 chooses box 3 on the first turn, then Player 2’s chance of winning is 2 . So, no matter which box Player 1 chooses on the first turn, Player 2 has a strategy for ensuring his chance of 1 winning is 2 . 1 c If Player 1 chooses box 5 on the first turn, her chance of choosing correctly is 9 . The game will continue with 8 probability 9 . If it does continue, then there are 4 boxes left to choose from and Player 2 has the next turn. From 1 Part b, Player 1 now has a strategy for ensuring her chance of winning the 4-box game is 2 . Hence, if Player 1 chooses box 5 on the first turn, she has a strategy to ensure her chance of winning the game is 1 + 8 1 = 5 > 1 . 9 9 × 2 9 2

P1 chooses box 5

1 8 9 9

‘Correct’ ‘Higher/Lower’ (P1 wins) (continues)

1 2

(P1 wins, using b)

Mathematics Contests The Australian Scene 2018 | Challenge Solutions42 – Intermediate | 52 CHALLENGE STATISTICS – MIDDLE PRIMARY

Mean Score/School Year/Problem

Challenge Problem Number of Overall School Year Students Mean 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

Challenge Problem Score 1 2 3 4 Training Bitripents Coloured Buckets Hand-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 0.5 0.6 0.6 0.7 Factor 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

Mean Score/School Year/Problem

Challenge Problem Number of Overall School Year Students Mean 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 Please note: * This total includes students who did not provide their school year.

Score Distribution %/Problem

Challenge Problem Score 1 2 3 4 Bitripents Insect Barriers Coloured Buckets Isopentagons

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 0.4 0.4 0.5 0.7 Factor 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 – Upper Primary | 54 CHALLENGE STATISTICS – JUNIOR

Mean Score/School Year/Problem

Challenge Problem Number of Overall School Year Students Mean 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 Please note: * This total includes students who did not provide their school year.

Score Distribution %/Problem

Challenge Problem Score 1 2 3 4 5 6 Insect Cubic Isopentagons Ts and Rs Rhombus Parsecking Barriers Coprimes Rings 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 0.4 0.5 0.7 0.8 0.8 0.7 Factor 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 – Junior | 55 CHALLENGE STATISTICS – INTERMEDIATE

Mean Score/School Year/Problem

Challenge Problem Number of Overall School Year Students Mean 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 Please note: * This total includes students who did not provide their school year.

Score Distribution %/Problem

Challenge Problem Score 1 2 3 4 5 6 Isopentagons Ts and Rs Rail Cams Curvy Tiles Rhombus Higher or Rings 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 0.6 0.6 0.7 0.7 0.7 0.7 Factor 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 – Intermediate | 56 AUSTRALIAN2018 Australian INTERMEDIATE Intermediate Mathematics MATHEMATICS Olympiad 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 area of 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 passes through the three trisection points closest to v. The resulting polyhedron has 24 vertices. How many diagonals joining 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 BA produced such that CQ bisects AD at M. Lines DP and CQ meet at O. If the area of parallelogram ABCD is 192, 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 and A such that CP =3PA, R is a point on side AB such that the area of P QR is twice the area of RBQ. Find the area of P QR. [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 the other number equals the product of all digits in the pair. Find all pairs of positive compatible numbers less than 100. [5 marks]

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

Mathematics Contests The Australian Scene 2018 | Australian Intermediate1 Mathematics Olympiad | 57 AUSTRALIAN2018 Australian INTERMEDIATE Intermediate Mathematics MATHEMATICS Olympiad OLYMPIAD - Solutions SOLUTIONS

1. Method 1 The table shows the product 2x7 39 for all values of x. × x 2x7 39 x 2x7 39 0 8073× 5 10023× 1 8463 6 10413 2 8853 7 10803 3 9243 8 11193 4 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 234 = 234 =2 64 + 3 8+4=156. b 8 × × 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, 4 remains the same, 30 increases by 3, but 200 increases by 2(b + 1)2 2b2 =4b + 2. b b b − So the increase in 234b+1 is greater than the increase in 234b 1. Hence 234b+1 234b 1 increases with increasing − − − b. This means 234b+1 234b 1 > 7010 for b>8. − − So 234 = 234 =2 64 + 3 8+4=156. b 8 × ×

Mathematics Contests The Australian Scene 2018 | Australian Intermediate2 Mathematics Olympiad Solutions | 58 3. Let O be the centre of the circumcircle.

A B

D C

Since OA = OB = OC = OD and AB = BC = CD = DA, triangles AOB, BOC, COD, DOA are isosceles and 1 congruent. So AOB = 360/4 = 90◦. Hence the area of AOB is 2 10 10 = 50 and the area of the sector 1 × × AOB = 4 π100 = 25π. 2 2 2 1 2 2 By Pythagoras, AB = AO +OB = 200. Hence the area of the semicircle on AB is 2 π(AB/2) = AB π/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:

12345678910 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 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 product of these factors: 1134167191 111213143161718191 212223241262728293 313233347363738391 414243449464748491

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.

Mathematics Contests The Australian Scene 2018 | Australian Intermediate3 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-zero digit 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) when divided 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 for 50!/512. So we want the remainder of the product of the following factors.

1234167892 111213143161718194 212223241262728296 313233347363738398 414243449464748492

Before we multiply these factors we may subtract from each any multiple of 5. So we need only multiply these numbers. 1234112342 1234312344 1234112341 1234212343 1234412342

After multiplying any two of these numbers, we may subtract any multiple of 5. Since 2 3 and 4 4 have remainder 1 when divided by 5, we need only multiply 3, 2, 4, 2, 4, 3, 2. Hence the required× remainder× is 2. So the 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 other vertices from one octagonal face, and 6 other vertices from the other octagonal face (the two octagons share an edge 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 vertex joins 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 the number 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 10 internal diagonals joined to v. By rotating the polyhedron about one or more of its axes of symmetry, v represents any of its 24 vertices. Since each of these diagonals joins two vertices, multiplying 10 by 24 counts each diagonal exactly twice. So the number of diagonals inside the polyhedron is 10 24/2=120. ×

Mathematics Contests The Australian Scene 2018 | Australian Intermediate4 Mathematics Olympiad Solutions | 60 Method 3 The polyhedron has 8 triangular faces and 6 octagonal faces. Since each edge of the polyhedron is shared by two faces, 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. × 24 The number of pairs of vertices of the polyhedron is 2 = 276. So the number of internal diagonals of the polyhedron is 276 36 120 = 120. − −

6. Draw MN.

D C

O M N

Q A B P

Method 1 Since BP and CD are parallel and BN = NC, triangles BNP and CND are congruent (ASA). Similarly, triangles AMQ and DMC are congruent. Since AM and BN are parallel and equal, MN and AB are parallel. So ABNM and MNCD are congruent parallelograms 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, and 4 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, triangles AMQ 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 congruent parallelograms 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. ×

Mathematics Contests The Australian Scene 2018 | Australian Intermediate5 Mathematics Olympiad Solutions | 61 Method 3 Parallelogram ABCD is unspeciﬁed so we may take it to be a rectangle with AD =2AB.

D C

M N

Q A B P

Thus ABNM and MNCD are congruent squares, triangles NBP and NCD are congruent, and triangles MAQ and MDC are congruent. Let denote area. Then || 1 192 ABNM = MNCD = ABCD = = 96 | | | | 2| | 2 1 NBP = NDC = MNCD = 48 | | | | 2| | 1 MAQ = MDC = MNCD = 48 | | | | 2| | 1 MNO = MNCD = 24 | | 4| | 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 − − 2 4 + 4(b2 + 2018) a = − ± = 1 b2 + 2019 2 − ± 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.

Mathematics Contests The Australian Scene 2018 | Australian Intermediate6 Mathematics Olympiad Solutions | 62 8. Method 1 Draw AQ, BP, and CR. Let denote area. ||

B Q C

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

1 1 1 1 300 BQR = BCR = ABC = | | 2| | 2 × 1+p ×| | 2 × 1+p 1 1 p 1 300p AP R = ACR = ABC = | | 4| | 4 × 1+p ×| | 4 × 1+p 1 1 3 3 CPQ = BCP = ABC = 300 | | 2| | 2 × 4 ×| | 8 × 300 P QR =2BQR = | | | | 1+p

Adding these areas gives

4 300 2 300p 3 300(p + 1) 8 300 300 = + + + 8 1+p 8 1+p 8 p +1 8 1+p 1 300 = (4 + 2p + 3(p + 1) + 8) 8 p +1 8(p + 1) = 15 + 5p 3p =7

300 300 3 900 Finally P QR = = × = = 90. | | 1+p 3+3p 10

Mathematics Contests The Australian Scene 2018 | Australian Intermediate7 Mathematics Olympiad Solutions | 63 Method 2

B Q C

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

300 = AP R + P QR + BQR + CQP | | | | | | | | 1 1 3 = ACR + P QR + P QR + AQC 4| | | | 2| | 4| | 1200 = ACR +6P QR +3AQC | | | | | | 3 = (300 BCR )+6P QR + ABC −| | | | 2| | = 300 2 BQR +6P QR + 450 − | | | | = 750 P QR +6P QR −| | | | = 750 + 5 P QR | | 1 1 P QR = (1200 750) = (450) = 90. | | 5 − 5

Mathematics Contests The Australian Scene 2018 | Australian Intermediate8 Mathematics Olympiad Solutions | 64 Method 3 First consider the following diagram and let denote area. ||

V Z

X Y M N U

XY Z MZ XY XZ XY XY XZ | | = × = = × (1) XUV NV XU XV × XU XU XV | | × × Next consider the given diagram with distances as shown.

b P d

R 3b

B a Q a C

From (1) we have

300 = BQR + AP R + CPQ + P QR | | | | | | | | =3BQR + AP R + CPQ | | | | | | 3c d 3 1= + + 2(c + d) 4(c + d) 8 8(c + d) = 12c +2d + 3(c + d)=15c +5d d =7c/3

300c 300c 90 So BQR = = = = 45. Hence P QR =2BQR = 90. | | 2(c + d) 20c/3 2 | | | |

Mathematics Contests The Australian Scene 2018 | Australian Intermediate9 Mathematics Olympiad Solutions | 65 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 summand is 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 three even 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 at least 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 composite and 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 five even 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 at least 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. Note that 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 numbers in 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 =1or2. 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)and9=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. If c = 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 . { } { }

Mathematics Contests The Australian Scene 2018 | Australian Intermediate10 Mathematics Olympiad Solutions | 66 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

Investigation Let s and p be compatible numbers, where s is the sum of their digits and p is the product of their digits. Note that 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 product 2cde, the last 1, 2, or 3 digits of 2bcde for b =1, 2, 3, 4, 5, 6, 7, 8, 9and2bcde < 1000, then a check whether any 2bcde 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 none 2, 7, 9 252 52, 4, 6 none 2, 8, 8 256 6, 12, 68 none 3, 6, 9 324 4, 8, 2 none 3, 7, 8 336 6, 2 none 4, 5, 9 360 0, 0, 0, 0, 0, 0, 0, 0, 0 none 4, 6, 8 384 384, 768 none 4, 7, 7 392 2, 84 none 5, 5, 8 400 0, 0, 0, 0, 0, 0, 0, 0, 0 none 5, 6, 7 420 0, 0, 0, 0, 0, 0, 0, 0, 0 none 6, 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, the product 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 be p = 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 none 1, 2, 6 12 012, 4, 36, 8, 0, 72, 4, 96, 8 none 1, 3, 5 15 015, 0, 45, 0, 75, 0, 05, 0, 135 135 1, 4, 4 16 6, 2, 8, 64, 0, 6, 2, 8, 144 144 2, 2, 5 20 0, 0, 0, 0, 0, 0, 0, 0, 0 none 2, 3, 4 24 024, 8, 72, 6, 0, 44, 8, 92, 6 none 3, 3, 3 27 7, 4, 1, 8, 5, 2, 9, 6, 43 none

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

Mathematics Contests The Australian Scene 2018 | Australian Intermediate11 Mathematics Olympiad Solutions | 67 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 digits of 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 with 12 n 27. Again n = 13, 17, 19, 22, 23, 26. Of the remainder, only n = 15, and n = 16 give a digit value for b =≤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 . { } { }

Mathematics Contests The Australian Scene 2018 | Australian Intermediate12 Mathematics Olympiad Solutions | 68 AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD STATISTICS

Distribution of Awards/School Year

Number of Awards Number of Year Students High Prize Distinction Credit Participation Distinction

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

Number Correct/Question Year 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

Question Number of School Year Overall Mean Students 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 AMOC SENIOR CONTEST

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

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest | 70 2018 AMOC SENIOR CONTEST Solutions

c 2018 AustralianAMOC Mathematics SENIOR Trust CONTEST SOLUTIONS

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.

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.

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions | 72 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 01201201201201201201 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.

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions | 73 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 be WW, WLor 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.

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions | 74 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, W LL, LW L 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.

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions | 75 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, ∠P AO = ∠PBO, but ∠PBO = ∠CBO = ∠BCO, since triangle BOC is isosceles. It follows that AOP C is a cyclic quadrilateral. Moreover, ∠OAQ = ∠OCQ = 90◦, so OAQC is also a cyclic quadrilateral. Therefore, OAQCP is a cyclic pentagon and ∠QP O = 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 = ∠P AB = ∠P BA.

Let this angle be α. Then AP B = 180 2α = AQC, so quadrilateral AQCP is cyclic. ∠ ◦ − ∠ Hence, AP Q = α = CPQ and we have AP D = 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.

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions | 76 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.

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 c xf(x)=a (1 + xf(x)) f(x)= , ⇒ x a where c = 1 a is a positive real constant. (Note that the equation xf(x)=a (1 + xf(x)) − implies that a = 1.) c Substituting f(x)= x into the functional equation yields 2y = y(1 + c).

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

Solution 2 (Angelo Di Pasquale) y Replace y with 2 in the functional equation to obtain 1 xf(xf(y)) = y (1 + xf(x)) . ( ) 2 ∗ 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 ∗ 1 f(y)(1 + xf(x)) f(y) f(x)f(y) xf(xf(f(y))) = f(y)(1 + xf(x)) f(axy)= = + . 2 ⇒ 2x 2x 2 Interchanging x with y in this equation yields f(x) f(x)f(y) f(ayx)= + . 2y 2 Equating the previous two expressions for f(axy) implies that f(y) f(x) = . 2x 2y Now substitute y = 1 into this equation to deduce that c f(x)= , x where c = f(1) is a constant. We may now ﬁnish the proof as in the previous solution.

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions | 77 AMOC SENIOR CONTEST RESULTS

Name School Year Perfect Score and Gold James Bang Baulkham Hills High School NSW 11 Andres Buritica Scotch College VIC 9 Yasiru Jayasooriya James Ruse Agricultural High School NSW 10 Sharvil Kesarwani Merewether High School NSW 11 Preet Patel Vermont Secondary College VIC 11 William Steinberg Scotch College WA 10 Hadyn Tang Trinity Grammar School VIC 9 Fengshuo (Fredy) Ye (Yip) Knox Grammar School NSW 8 Ziqi Yuan Narrabundah ACT 11 Gold Haowen Gao Knox Grammar School NSW 11 Ken Gene Quah Melbourne High School VIC 10 Silver Zefeng (Jeff) Li Caulfield Grammar School, Caulfield Campus VIC 11 Junhua Chen Caulfield Grammar School, Wheelers Hill VIC 10 Grace He Methodist Ladies' College VIC 10 Mikhail Savkin Gosford High School NSW 10 Harry Zhang Christian Brothers College VIC 10 Frank Zhao Geelong Grammar School VIC 11 David Lee James Ruse Agricultural High School NSW 11 Zijin (Aaron) Xu Caulfield Grammar School, Wheelers Hill VIC 10 Liam Coy Sydney Grammar School NSW 10 Anthony Pisani St Paul's Anglican Grammar VIC 11 Jason Wang Queensland Academy for Science, Mathematics and Technology 10 QLD Marcus Rees Hobart College TAS 11 Daniel Wiese Scotch College WA 10 Bronze Christopher Leak Perth Modern School WA 10 Huxley Berry Perth Modern School WA 10 Vicky Feng MLC School NSW 11 Evgeniya Artemova Presbyterian Ladies' College VIC 11 Yang Zhang St Joseph’s College, Gregory Terrace QLD 10 Reef Kitaeff Perth Modern School WA 11 Oliver Papillo Camberwell Grammar School VIC 11 Patrick Gleeson St Joseph’s College, Gregory Terrace QLD 10 Samuel Lam James Ruse Agricultural High School NSW 10 Leosha Trushin Perth Modern School WA 10 Peter Vowles Wesley College WA 11 Kevin Wu Scotch College VIC 11 Christopher Do Penleigh and Essendon Grammar School VIC 11 Angus Ritossa St Peter's College SA 11

Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Results | 78 Name School Year Emilie Wu James Ruse Agricultural High School NSW 11 Eva Ge James Ruse Agricultural High School NSW 9 Wenguan Lu Barker College NSW 11 Claire Huang Radford College ACT 10 Steve Wu Prince Alfred College SA 11 Chenxiao Zhou Caulfield Grammar School, Wheelers Hill VIC 9 Lachlan Rowe Canberra College ACT 11 Micah Sinclair Perth Modern School WA 9 Le Yao Zha St Edmund's College ACT 10 Xiaoyu Chen All Saints' College WA 8 Honourable Mention Christina Lee James Ruse Agricultural High School NSW 10 Adrian Lo Newington College NSW 10 David Lumsden Scotch College VIC 8 Oliver Cheng Hale School WA 9 Shevanka Dias All Saints' College WA 11 Ethan Ryoo Knox Grammar School NSW 9 Zian Shang Scotch College VIC 7 Lucinda Xiao Methodist Ladies' College VIC 11 Elizabeth Yevdokimov St Ursula’s College QLD 9 Matthew Cho St Joseph’s College, Gregory Terrace QLD 10 Remi Hart All Saints' College WA 10 Linda Lu The Mac.Robertson Girls' High School VIC 11 Andrey Lugovsky Perth Modern School WA 11 Oliver New Scotch College VIC 8 Angela Wang Lauriston Girl's School VIC 9 Leo Xu All Saints Anglican School QLD 10 William Cheah Penleigh and Essendon Grammar School VIC 4 Anagha Kanive-Hariharan James Ruse Agricultural High School NSW 10 Dhruv Hariharan Knox Grammar School NSW 9 Jocelin Hon James Ruse Agricultural High School NSW 11 Yikai Wu Prince Alfred College SA 11 Ryan Gray Brisbane State High School QLD 11 Mikaela Gray Brisbane State High School QLD 9 Kento Seki All Saints Anglican School QLD 10 Vivian Wang James Ruse Agricultural High School NSW 10 Gen Conway Methodist Ladies' College VIC 8 Tom Hauck All Saints Anglican School QLD 10 Hanyuan Li North Sydney Boys High School NSW 10 Bertrand Nheu Perth Modern School WA 11 Benjamin Davison-Petch Christ Church Grammar School WA 11 Zhenghao Hua Penleigh and Essendon Grammar School VIC 8 Sam Meredith Brisbane State High School QLD 8 Tony Teng Concordia College SA 10 Yale Cheng Hale School WA 11

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

Score Distribution/Problem

Number of Students/Score Problem Mean Number 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 Pasquale Director 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 A USTR A LI A N M ATHE MATIC A L O LY M PI A D C OMM ITTEE

A DEP A RT M ENT O F THEAUSTRALIAN A USTR A LI A N MATHE MATICS MATHEMATICAL TRUST OLYMPIAD

A USTR A LI A N M ATHE MATIC A L O LY M PI A D C OMM ITTEE

A DEP A RT M ENT O F THE A USTR A LI A N MATHE MATICS TRUST AUSTRALIAN MATHEMATICAL OLYMPIAD 2018

DAY 1 Tuesday, 6 FebruaryAUSTRALIAN 2018 MATHEMATICAL OLYMPIAD Time allowed: 4 hours No calculators are to be used. 2018 Each question is worth seven points.

DAY 2 Wednesday, 7 February 2018 Time1. allowed:Find all pairs 4 hours of positive integers (n, k) such that No calculators are to be used. n!+8=2k. Each question is worth seven points.

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

1 100 5.2. TheConsider sequence a linea with,a ,a2 (3,... is+ deﬁned 1) equally by a spaced= 1 and, points for markedn 2, on it. 1 2 3 1 ≥ Prove that 2100 of these marked points can be coloured red so that no red point is at the an =(a1 + a2 + + an 1) n. same distance from two other red points. ··· − ×

2 3. ProveLet ABCDEF that a2018 GHIJKLMNis divisible bybe 2018 a regular. tetradecagon. Prove that the three lines AE, BG and CK intersect at a point. 6. Let P , Q and R be three points on a circle , such that PQ = PR and PQ>QR. C Let(A regularbe the tetradecagon circle with centreis a convexP that polygon passes with through 14 sides,Q and suchR. Suppose that all thatsides the have circle the D withsame centre lengthQ andand all passing angles through are equal.)R intersects again at X and again at Y . C D Prove that P , X and Y lie on a line. 4. Find all functions f deﬁned for real numbers and taking real numbers as values such that

7. Let b1,b2,b3,... be a sequence off positive(xy + f integers(y)) = yf such(x) that, for each positive integer n, bn+1 is the square of the number of positive factors of bn (including 1 and bn). For 2 example,for all real if numbersb1 = 27, thenx andb2y=4. = 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?

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

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad | 83 The Mathematics/Informatics Olympiads are supported by the Australian Official sponsor of the Olympiad program Government through the National Innovation and Science Agenda.

A DEP A RT M ENT O F THE A USTR A LI A N MATHE MATICS TRUST

A USTR A LI A N M ATHE MATIC A L O LY M PI A D C OMM ITTEE

A DEP A RT M ENT O F THE A USTR A LI A N MATHE MATICS TRUST

AUSTRALIAN MATHEMATICAL OLYMPIAD 2018

DAY 2 AUSTRALIAN MATHEMATICAL OLYMPIAD Wednesday, 7 February 2018 2018 Time allowed: 4 hours No calculators are to be used. Each question is worth seven points. DAY 2 Wednesday, 7 February 2018 Time allowed: 4 hours No calculators are to be used. 5. The sequence a1,a2,a3,... is deﬁned by a1 = 1 and, for n 2, Each question is worth seven points. ≥

an =(a1 + a2 + + an 1) n. ··· − ×

2 Prove that a2018 is divisible by 2018 . 5. The sequence a ,a ,a ,... is deﬁned by a = 1 and, for n 2, 1 2 3 1 ≥ 6. Let P , Q and R be three points on a circle , such that PQ = PR and PQ>QR. an =(a1 + a2 + C+ an 1) n. Let be the circle with centre P that passes··· through− Q×and R. Suppose that the circle D with centre Q and passing through R2 intersects again at X and again at Y . Prove that a2018 is divisible by 2018 . C D Prove that P , X and Y lie on a line. 6. Let P , Q and R be three points on a circle , such that PQ = PR and PQ>QR. C 7. Let b1,bbe2,b the3,... circlebe witha sequence centre ofP positivethat passes integers through suchQ that,and forR. Supposeeach positive thatinteger the circlen, D withbn+1 centreis theQ squareand passing of the number through ofR positiveintersects factorsagain of atbnX(includingand again 1 and at Ybn.). For C D example, if b = 27, then b =42 = 16, since 27 has four positive factors: 1, 3, 9 and 27. Prove that P1, X and Y lie2 on a line. Prove that if b1 > 1, then the sequence contains a term that is equal to 9.

7. Let b1,b2,b3,... be a sequence of positive integers such that, for each positive integer n, 8. bAmyn+1 is has the a number square of of therocks number such that of positive the mass factors of each of rock,bn (including in kilograms, 1 and is ab positiven). For 2 example,integer. The if b1 sum= 27, of then the massesb2 =4 of= the 16, rockssince 27 is 2018 has four kilograms. positive Amy factors: realises 1, 3, that9 and it 27. is impossible to divide the rocks into two piles of 1009 kilograms. Prove that if b1 > 1, then the sequence contains a term that is equal to 9. What is the maximum possible number of rocks that Amy could have? 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?

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

© 2018 Australian Mathematics Trust The Mathematics/Informatics Olympiads are supported by the Australian Official sponsor of the Olympiad program Government through the National Innovation and Science Agenda. Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad | 84

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 positive integer 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 odd integer that is greater than 1. Hence there are no solutions in this case. Case 2 n 5 ≤ We simply tabulate the values of n!+8and check which ones are powers of 2.

nn!+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.

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 this implies 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, which is impossible. So there| are no solutions in this case. Case 3 n =4or 5 If n =4then k =5, and if n =5then k =7.

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

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 86 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. 1 n Claim For each positive integer n, if a line has 2 (3 + 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. 1 1 1 The base case n =1holds as we simply colour each of the 2 (3 +1)=2 points red. For the inductive step, suppose that the claim is true for some positive integer n. 1 n 1 n+1 Let k = 2 (3 + 1). Note that 3k 1= 2 (3 + 1). To prove the claim for n +1we divide the 3k 1 points into three− groups from left to right as follows. − k points k 1 points k points − group A group B group C

Using the inductive assumption we colour 2n points in group A red so that no red point 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 in C. All points in B are left uncoloured. A total of 2n +2n =2n+1 points have been coloured red in the union of the three groups. Suppose, for the sake of contradiction, that this colouring is not good. Then there 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 the inductive construction.

n =1 n =2 n =3

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 87 Solution 2 (Sharvil Kesarwani, year 11, Merewether High School, NSW) 1 100 Without loss of generality, we identify the 2 (3 + 1) equally spaced points with the integers from 0 up to 1 (3100 1) on the real number line. 2 − Colour red each integer of the form

99 N = d 3i i · i=0 100 where di 0, 1 for 0 i 99. In this way exactly 2 integers are coloured red. ∈{ } ≤ ≤ 1 100 Note that these are precisely the integers in the range from 0 to 2 (3 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 from two other red integers A and C, where A

99 99 2b 3i = (a + c )3i. i · i i i=0 i=0

Since ai,bi,ci 0, 1 , we have 2bi 0, 2 and ai + ci 0, 1, 2 . Therefore both sides of the above∈{ equation} are the∈{ ternary} representation∈{ of the} same number. It follows 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 C have the same ternary digits, and so A = C. This contradicts A

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 88 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.

I H J G

K F

L E

M D

N C A B

Since the vertices of the tetradecagon are equally spaced around the circle, and equal length 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 such they are concurrent at the incentre of AGC.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 89 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 around a circle. Form triangle BEK by joining segments BE, EK, and KB.

I H J G

K F

L E

M D

N C A B

By symmetry we have parallel chords BG CF DE. Since DK is a diameter of the 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 are concurrent at the orthocentre of BEK.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 90 Solution 3 (Ethan Tan, year 12, Cranbrook School, NSW) As in solution 1, we consider A,B,...,N as being 14 equally spaced points around a circle. Let X be the intersection of lines AE and BG. It suffices to show that C, X, and K are collinear.

I H J G

K F

L E

M X D

N C A B

Since the vertices of the tetradecagon are equally spaced around the circle, and equal length 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 because DK is a diameter of the circle. Thus CK CX, from⊥ which it follows⊥ that C, X, and K are collinear, as desired.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 91 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 around a 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◦.

I H J G

K F

L E

M X D

N C A B

Each of the 14 equal sides of the tetradecagon subtends an angle of 360◦/14 with the centre of the circle. Hence each of these sides subtends an angle of α = 180◦/14 with 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, it ∠ follows that ∠BXC = ∠XCB =5α. Finally, since ∠XCB =5α = ∠KCB, it follows that C, X, and K are collinear, as desired.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 92 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 around a circle. Form the hexagon ABCEGK.

I H J G

K F

L E

M D

N C A B

We quote a theorem that will help us solve the problem. Theorem If UV WXY Z is a convex cyclic hexagon, then its main diagonals UX, VY, and WZ are concurrent if and only if

UV WX YZ = 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 tetradecagon around 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

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 93 4. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA) Answers f(x)=0and f(x)=1 x − For reference the given functional equation is

f(xy + f(y)) = yf(x) (1)

for all real numbers x and y. Put y =0in (1) to find f(f(0)) = 0. (2)

Next put x =0and 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 =0in (1) to find f(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), we see 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)=0for all real numbers◦ x. This function obviously satisfies (1). Case 2 f(0) = 1 Putting x =0into (1) yields f(f(y)) = y (5) for all real numbers y. It follows that f(1) = f(f(0)) = 0. Putting x =1in (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. −

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 94 Solution 2 (William Steinberg, year 10, Scotch College, WA) For reference the given functional equation is

f(xy + f(y)) = yf(x) (1)

for all real numbers x and y. Put x =0in (1) to find f(f(y)) = yf(0), (2) for all real numbers y. Putting y =1in (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 becomes f(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 and y = c/f(c) in (5), we deduce that f(c)=0. This is a contradiction. Thus no such c exists. Hence f(x)=0for all real numbers x. Case 2 f(0) =0 It follows from (3) that f(1) = 0. Setting x =1in (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)=0and f(x)=1 x each satisfy the given − functional equation.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 95 5. Solution 1 (Vicky Feng, year 11, Methodist Ladies’ College, NSW) We are given an = 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) − − ··· − for any integer n 3. ≥ Substituting (2) into (1) yields

an = n(a1 + a2 + + an 2 + an 1) ··· − − = n(a1 + a2 + + an 2 +(n 1)(a1 + a2 + + an 2)) − − 2 ··· − ··· = n (a1 + a2 + + an 2). ··· − Hence a is divisible by n2 for each integer n 3. In particular, a is divisible by n ≥ 2018 20182.

Comment We did not use the initial condition a1 =1. Hence the result is true for any integer a1.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 96 Solution 2 (Xinyue (Alice) Zhang, year 12, A. B. Paterson College, QLD) 3 We prove an even stronger result, namely a2018 is divisible by 2018 .

First we prove by induction that a1 + a2 + + an 1 = n!/2 for all integers n 2. ··· − ≥ 2! The base case is true because a1 =1= 2 .

For the inductive step, assume that a1 + a2 + + an 1 = n!/2 for some integer − n 2. Then ··· ≥

a1 + a2 + + an 1 + an = a1 + a2 + + an 1 + n(a1 + a2 + + an 1) ··· − ··· − ··· − n! n n! = + · 2 2 (n + 1) n! = · 2 (n + 1)! = 2 which completes the induction. Using this result, it follows that

a = 2018(a + + a ) 2018 1 ··· 2017 = 2018(2018 2017 3) × ×···× = 20182(2017 2016 1009 4 3) × ×···× ×···× ×

which is obviously a multiple of 20182. However 2017 2016 1009 4 3 × ×···× ×···× × is divisible by both 2 and 1009, and so it is also divisible by 2018. Therefore a2018 is divisible by 20183.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 97 Solution 3 (Andres Buritica, year 9, Scotch College, VIC) We are given an = a1 + a2 + + an 1 (1) n ··· − for any integer n 2. ≥ Replacing n with n +1in (1), we find

an+1 = a1 + a2 + + an 1 + an n +1 ··· − a = n + a n n a (n + 1)2 n+1 = (2) ⇒ an n for any integer n 2. ≥ Multiplying all instances of equality (2) together for n = m 1,m 2,...,2, where m 3, yields a product that telescopes as follows. − − ≥ 2 2 2 am am 1 a3 m (m 1) 3 − = − am 1 × am 2 ×···× a2 m 1 × m 2 ×···× 2 − − − − a m2(m 1)! m = − ⇒ a2 4 m2(m 1)! a = − ⇒ m 2 In particular a = (2018)2 2017! , which is a multiple of 20182. 2018 × 2

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 98 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 .

D R

P Q

In circle , chords QY and QR have equal length. Hence they subtend equal angles at the centreD P of this circle. Thus

∠YPQ = ∠QP R. (1)

In circle , chords QX and QR have equal length. Hence they subtend equal angles at the pointC P on the circumference of this circle. Thus

∠XPQ = ∠QP R. (2)

From (1) and (2) we have ∠XPQ = ∠YPQ. Thus P , X, and Y are collinear.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 99 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 second time at X . It suffices to prove that P RQX is cyclic as this implies X = X.

P Q

As P and Q are centres of circles, we have PR = PQ = PY and QR = QX = QY . Hence P QR P QY (SSS). Thus ≡ ∠PRQ = ∠QY P = ∠QY X = ∠YX Q

where the last equality is due to QX = QY .

Since ∠PRQ = ∠YX Q, it follows that P RQX is cyclic, as desired.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 100 Solution 3 (Sharvil Kesarwani, year 11, Merewether High School, NSW)

Let Y be the intersection of ray PX with circle . It suffices to prove that Y = Y . D Let ∠QY X = ∠QY P = α.

D R

P Q

X α

Observe that PY = PQ = PR because they are radii of circle . It follows that D P QY = QY P = α. Considering the angle sum in P QY yields ∠ ∠

Y PQ = 180◦ 2α. ∠ − Observe that QX = QR because X lies on the circle centred at Q and passing through R. Since QX and QR are equal length chords in circle , they subtend equal angles at the point P on the circumference of this circle. HenceC

QP R = XPQ = Y PQ = 180◦ 2α. ∠ ∠ ∠ − Since P QR is isosceles with apex P , the angle sum in this triangle yields ∠PRQ = ∠RQP = α.

Since P RQX 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 on the circle centred at Q and passing through R. Since Y and R are on opposite sides of the line PQ, it follows that Y = Y .

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 101 Solution 4 (Alan Offer, AMOC Senior Problems Committee) Comment Let be the circle in the problem statement centred at Q and passing through R. If we relaxE the condition that Q is the centre of by requiring only that the centre of lies on , then the conclusion of the problemE is still true. E C Here is a proof of this generalisation of the problem.

D R

P Q X O

We have only considered the configuration where X lies between P and Y . Other configurations may be dealt with similarly. Let O be the centre of . Since PR = PY (radii of ), and OR = OY (radii of ), it follows that OPR E OPY (SSS). Thus D E ≡ ∠YPO = ∠OPR. (1)

In circle , chords OX and OR have equal length. Hence they subtend equal angles at the pointC P on the circumference of this circle. Thus

∠XPO = ∠OPR. (2)

From (1) and (2) we have ∠XPO = ∠YPO. Thus P , X, and Y are collinear.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 102 Comments (Angelo Di Pasquale, Director of Training, AMOC) The astute reader may have noticed that the point Q did not get mentioned at all in the previous proof. The reason is because, in this generalisation, the point Q is irrelevant to the proof. Hence the diagram could have been drawn without the point Q and without the line segments PQ and QR. This minimal statement of the generalisation of the problem is as follows.

Let P , O, and R be three points on a circle . Let be a circle centred at P and passing through R, and let be theC circleD centred at O and passing through R. Suppose that intersectsE again at X and again at Y . Then P , X, and Y lie on a line.E C D

In fact we can say even more! There is a combinatorial symmetry in the problem statement between and with respect to . Specifically, exchanging points P and O also exchanges Dand E. Thus if intersectsC again at Q, then O, Q, and Y also lie on a line. D E D C

C R

P Q E

X O

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 103 7. Solution 1 (Tianyue (Ellen) Zheng, year 12, Smith’s Hill High School, NSW) Since b > 1, it follows inductively that b > 1 for all n 2. 1 n ≥ Observe that b is a perfect square for all n 2. n ≥ For any perfect square m2, all of the factors of m2, except for m, come in pairs (d, m2/d) where 1 d 1, it follows that b 9 for all n 3. ≥ n ≥ ≥ 2 Consider any integer n 3. Then bn = m for some odd integer m 3. As explained in the previous paragraph,≥ apart from m, the factors of m2 come≥ in pairs (d, m2/d) where 1 d

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 104 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 an odd perfect square that is greater than or equal to 9. Consider any integer n 3. As in solution 1 it suffices to prove that b b with ≥ n+1 ≤ n equality if and only if bn =9. r Let b = p2ai , where p

pa 2a +1 (2) ≥ with equality if and only if p =3and a =1. Since pa > 3a whenever p>3, it suffices to prove (2) just for the case p =3. We proceed 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 +1for 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

2a 2a p2a (1 + 2)2a 1+ 2+ 22 =8a2 +1 (2a + 1)2. ≥ ≥ 1 · 2 · ≥ Alternative 2 (Ivan Guo, AMOC Senior Problems Committee) Factoring the difference of perfect ath powers, the inequality follows from

a a a 1 a 2 p 1 3 1=(3 1)(3 − +3 − + + 1) 2(1 + 1 + +1)=2a. − ≥ − − ··· ≥ ···

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 105 8. For ease of exposition, in each of the solutions that follow, whenever we say something has mass n, then we use this as shorthand for saying that it has mass n kilograms. 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 i with 1 i r, let ≥ ≤ ≤ S = m + m + + m . i 1 2 ··· i 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. Suppose that (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 answer − cannot be greater than or equal to 1010. To complete the proof, here is a construction for exactly 1009 rocks. Suppose that Amy has 1 rock of mass 1010, and 1008 rocks of mass 1. Then the pile that contains the rock of mass 1010 obviously has mass exceeding 1009. Comment 1 Some students found a somewhat geometric version of the above argument which basically goes as follows. Suppose Amy has at least 1010 rocks. Consider a circle of circumference 2018 and partition it into arcs whose lengths correspond to the masses of the rocks. The endpoints of the arcs are 1010 (or more) of the vertices of a regular 2018-gon, and so by the pigeonhole principle there is a diametrically opposite pair of endpoints. Comment 2 Some students stated and proved the following fairly well-known lemma, and used this to complete the proof. Lemma Any sequence of n integers contains a nonempty subsequence whose sum is a multiple of n. Proof Let the integers be m ,m ,...,m . Let S = m +m + +m for 1 i n. 1 2 n i 1 2 ··· i ≤ ≤ If S 0 (mod n) for some i, then we are done. i ≡ Otherwise, by the pigeonhole principle we have Si Sj (mod n) for some i

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 106 Solution 2 (William Steinberg, year 10, Scotch College, WA) Suppose that Amy has at least 1010 rocks. Suppose that a of the rocks have unit mass and b of the rocks have mass greater than 1. If b =0, then a = 2018, and in this 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 following inequalities. ≥

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 found by 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, continually choose rocks of non-unit mass and add them to the pile one at a time until one of the following two things occurs.

(i) It is not possible to add any more rocks of non-unit mass without the mass of the 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 is less than M. Since a M we can top up the pile with rocks of unit mass until the pile has mass exactly≥ 1009. If (ii) occurs, then all remaining rocks have unit mass. Hence we can top up the pile with these until the pile has mass exactly 1009. Either way a pile of mass 1009 has been created. This shows that the answer cannot be greater than or equal to 1010. To complete the proof, here is a construction for exactly 1009 rocks. Suppose that Amy has 1009 rocks, each of mass 2. If some of these rocks are put into a pile, then the total mass of the pile will be even, and so cannot have mass 1009.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 107 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 it is impossible to divide the rocks into two piles of equal mass. Hence the answer is greater than or equal to 1009. The proof that this is the largest possible number of rocks that Amy could have follows from the following lemma. Lemma Let n be any positive integer. Suppose that Amy has at least n +1rocks where the mass of each rock is a positive integer, and the total mass of all the rocks is 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 of Amy’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 the case n = k +1. Thus Amy has at least k +2rocks whose total mass is 2k +2. If all of Amy’s rocks have mass greater than or equal to 2, then the total mass of all 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 mass then the result follows immediately. Hence we may let the masses of the rocks be

1,m1,m2,...,mr (*)

where r k +1and m > 1. ≥ r 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 the total mass of these rocks is equal to 2k. From the inductive assumption we can divide these into two piles, each of total mass k. Next change m 1 to m and add r − r a single rock of unit mass to the pile not containing the rock of mass mr 1. This gives a division of the rocks with masses given in (*) into two piles, each− having mass 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 +1rocks where the mass of each rock is a positive integer, and the total mass of all the rocks is 2n. Then for any integer N with 0 N 2n it is always possible to divide the rocks into two piles, one of which has≤ mass ≤N. The proof is very similar to the one given above.

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Solutions | 108 AUSTRALIAN MATHEMATICAL OLYMPIAD RESULTS

Name School Year Perfect 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 Gold William 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 Silver Andres 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* Bronze Grace 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 Year Alan 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 Mention Vincent 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

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Results | 110 Name School Year Reef 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 Queensland Academy for Science, Mathematics and Technology Jianyi (Jason) Wang 10 QLD 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

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad Results | 111 AUSTRALIAN MATHEMATICAL OLYMPIAD STATISTICS

Score Distribution/Problem

Problem Number Number of Students/Score 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 30th ASIAN PACIFIC MATHEMATICS OLYMPIAD

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 oﬃcial APMO website http://apmo.ommenlinea.org. Please do not disclose nor discuss the problems over online until that date. The use of calculators is not allowed.

Problem 1. Let H be the orthocenter of the triangle ABC. Let M and N be the midpoints of the sides AB and AC, respectively. Assume that H lies inside the quadrilateral BMNC and that the circumcircles of triangles BMH and CNH are tangent to each other. The line through H parallel to BC intersects the circumcircles of the triangles BMH and CNH in the points K and L, respectively. Let F be the intersection point of MK and NL and let J be the incenter of triangle MHN. Prove that FJ = FA.

Problem 2. Let f(x) and g(x) be given by 1 1 1 1 f(x)= + + + + x x 2 x 4 ··· x 2018 − − − and 1 1 1 1 g(x)= + + + + . x 1 x 3 x 5 ··· x 2017 − − − − Prove that f(x) g(x) > 2 | − | for any non-integer real number x satisfying 0

Problem 3. A collection of n squares on the plane is called tri-connected if the following 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 the squares.

(iii) Each square touches exactly three other squares.

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 exists a collection of n squares that is tri-connected? ≤ ≤

Problem 4. Let ABC be an equilateral triangle. From the vertex A we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle. When the ray reaches a side, it then bounces oﬀ following the law of reﬂection, that is, if it arrives with a directed angle α, it leaves with a directed angle 180◦ α. After n bounces, the ray returns to A without ever landing on any of the other− two vertices. Find all possible values of n.

Problem 5. Find all polynomials P (x) with integer coeﬃcients such that for all real numbers 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 30th ASIAN PACIFIC MATHEMATICS OLYMPIAD SOLUTIONS 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.

M N

Z J Y

K L H

B C

Since M and N are the midpoints of AB and AC, respectively, it follows that MN BC KL. Using this and cyclic BKMH yields ∠NMF = ∠LKF = ∠HKM = ∠HBM = x. Similarly, ∠FNM = ∠F LK = x. The above angle equalities imply FK = FL and FM = FN. Hence

FM FK = FN F L. · · Thus F has equal power with respect to circles BKMH and CHNL. Hence F lies on the radical axis of these two circles. Since the two circles are tangent, it follows that F lies on the common tangent at H. Hence FH is tangent to the two circles at 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

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 intersection of the circumcircle of MNH with the line HJ.1 In this configuration it is known that 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 centre of circle MJN. We also calculate

NJM = 180◦ JMN MNJ − ∠ − ∠ 1 = 180◦ ( HMN + MNH) − 2 ∠ ∠ 1 = 180◦ (180◦ NHM) − 2 − ∠ 1 = 180◦ (180◦ 2x) − 2 − = 90◦ + x.

Also ∠MAN = ∠BAC = 90◦ x, and so ∠NJM+∠MAN = 180◦. Thus MJNA is cyclic. 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.

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 116 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

y z M x x N

J P

x x K L H

x x y z B C

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 the image 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 . Hence H and F are isogonal conjugates in AMN. But the orthocentre and circumcentre of 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 because then F would be the centre of circle MJNA. Let P be any point on the common tangent at H of the two circles such that P lies on the same side of the line KL as 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 due ∠ ∠ − to ∠NJM + ∠MAN = 180◦, as desired.

2See problem 1 and the corresponding footnote in section 6.0 of Problem Solving Tactics published by the AMT.

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 117 2. Solution (Based on the solution by James Bang, year 11, Baulkham Hills High School, NSW) There are two cases: 2n 1 0 for each i. ≥ − − − Case 1 0 + . − x (x 1)(x 2) − − Thus it suffices to show 1 1 + > 2 (3) x (x 1)(x 2) − − (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 (2n 1))(x 2n) (x (2n + 1))(x (2n + 2)) − − − − − Let us use the substitution y = x 2n, so that 0 2 (6) y(y + 1) (y 1)(y 2) − − (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

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 118 Comment 1 A shortcut can be made in the preceding solution as follows.

1 1009 1 1 f(x) g(x) = − − x +1 x 2i − x (2i 1) i=0 − − − 1009 1 f(x) g(x) > ⇒ − (x (2i 1))(x 2i) i=0 − − − Having the above inequality removes the need to treat 0 2 y(y + 1) (y 1)(y 2) − − 1 1 1 1 + > 2 ⇔ y − y +1 y 2 − y 1 1 1 −1 −1 + > 2 ⇔ y − y 1 y 2 − y +1 1− −1 1 + > 2. ⇔ y(1 y) y 2 − y +1 − −

However since y and 1 y are both positive, the GM–HM inequality yields − 1 1 2 2 =4. y · 1 y ≥ y + (1 y) − − Also for 0 1 and > 1. y 2 − − y +1 − − Putting it all together yields 1 1 1 + > 4 1 1=2, y(1 y) y 2 − y +1 − − − − as desired.

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 119 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 possible if and only≤ if ≤n is even. For any collection of n tri-connected squares, consider the graph G obtained as follows. Each vertex of G corresponds to a square in the collection, and two vertices of 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 the 3n total number of edges of G is equal to 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 configuration on the right a B-piece. Five A-pieces can be linked together to make a tri-connected collection 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+16b for some integers a, b 0. For example,≤ ≤ 30, 60, 90, 120, 150, 180, 210, 240 represents all 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-square configuration on the right, making sure that all a + b configurations are mutually disjoint. This yields a tri-connected collection of n squares. Indeed we have shown that 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 aA-pieces and bB-pieces into a single closed loop 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≥ a tri-connected collection of n squares exists for any≥ even n 16. ≥ 56

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 120 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 triangle ABC in one of its three sides.

P C

A B

For each point P that is a vertex of the triangular lattice, we write P =(r, s) where r 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 black according to r s 0, 1, or 2 (mod 3), respectively. It is easy to show by induction that each reflected− ≡ image of A is red, each reflected image of B is white, and each reflected image of C is black. Consider a path in triangle ABC. Each time the path reflects off a side of the triangle, we reflect the triangle along with the remaining part of the path in the said side. In this way, since the original path obeys the law of reflection, it unfolds into a straight line through the above triangular lattice. A path that starts and ends at A in the original triangle now corresponds to a straight line segment through the triangular lattice that starts at the red vertex at A and ends at another red vertex P say. One such original path in triangle ABC along with its unfolded path AP is illustrated 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 any of the other two vertices. Thus the segment AP contains no lattice points strictly between A and P . This is true if and only if gcd(r, s)=1. (2)

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 121 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 1 lines 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 only remains to discuss n 1, 5 (mod 6). ≡ Case 1 n =6k +1for some integer k 0 ≥ It is straightforward to verify that P = (1, 3k + 1) satisfies (1), (2), and (4). Case 2 n = 12k + 11 for some integer k 0 ≥ It is straightforward to verify that P = (2, 6k + 5) satisfies (1), (2), and (4). Case 3 n = 24k +5for some integer k 1 ≥ It is straightforward to verify that P = (6k 1, 6k + 5) satisfies (1), (2), and (4). − Case 4 n = 24k + 17 for some integer k 1 ≥ 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, and together with case 2 cover all n 5 (mod 6) except for n =5and 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.

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 122 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 the problem. 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) is an integer. Suppose that P is a solution to the problem. We may write

P (x)=a + a x + + a xn 0 1 ··· n for some integers a ,a ,a ,...,a where a =0. Let S be the set of P -good real 0 1 2 n n numbers. Clearly Z S. Also the problem statement informs us that if r, s S then rs S. ⊆ ∈ ∈ Consider the polynomial

Q(x)=2nP (x) P (2x) − n 1 − = (2n 2i)a xi. − i i=0 If r S, then since 2 S, we have 2r S. Thus Q(r)=2rP (r) P (2r) Z. So we have∈ shown the following.∈ ∈ − ∈

Each real number that is P -good is also Q-good. (1)

n i We claim that a1 = a2 = = an 1 =0. Since 2 2 =0for i

1+ Q(u + 1) Q(u) 1+ P (u + 1) P (u) , (2) | − |≥ | − | for all sufficiently large u. Let P (x)=P (x+1) P (x) and Q (x)=Q(x+1) Q(x). 1 − 1 − Note that deg(P1)=n 1 >m 1 = deg(Q1). Hence P1(x) > Q1(x) for all sufficiently large x, which− contradicts− (2). | | | | n So we have shown that a1 = a2 = = an 1 =0. Hence P (x)=anx + a0. 1 ··· − 2 Let r = 1 . Since P (r)= 1+a0 Z, we have r S. Thus also r S. So an n ± ∈ ∈ ∈ 2 |1 | P (r )= + a0 Z. Hence an = 1, which concludes the proof. an ∈ ±

Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions | 123 30th ASIAN PACIFIC MATHEMATICS OLYMPIAD RESULTS

Top 10 Australian scores

Name School Year Total Silver 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 Award Ethan 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 Mention Haowen 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

Number of Total Gold Silver Bronze Honourable Rank Country Contestants Score Awards Awards Awards 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 Islamic Republic of 8 10 186 1 2 4 3 Iran 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 3 13 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 Number of Total Gold Silver Bronze Honourable Rank Country Contestants Score Awards Awards Awards 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 Trinidad and 35 10 24 0 0 0 1 Tobago 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 Pasquale Director 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

Mathematics Contests The Australian Scene 2018 | AMOC Selection School | 127 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 Queensland Academy for Science, Mathematics & Jianyi (Jason) Wang m 10 QLD Technology 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

Mathematics Contests The Australian Scene 2018 | AMOC Selection School | 128 The 7th EGMO, Florence, Italy The 7th EGMO, Florence, Italy

The 7th European Girls’ Mathematical Olympiad (EGMO) was held 9–15 April 2018,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 Olympiadin andthe 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,it was the biggest EGMO to date with 8 new countries participating in 2018: Australia, Austria, Bolivia, Canada,EGMO Germany, TEAM Greece, Mongolia, LEADER’S and Peru. REPORT For this 7th EGMO,Austria, a 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 1 total of 195 contestants from 52 countries participated (137 contestants from 36 official EuropeanThe 7th European countries) Girls’. Peru Mathematical had the youngest Olympiad team (EGMO) with was an averageheld 9–15 age April of 2018, 14 years inEuropean the and city of countries) 1. Peru had the youngest team with an average age of 14 years and 10Florence, months, Italy. Latvia This had was the the oldestfirst time team Italy with has anhosted average an Olympiad age of 18 and years it was and the 7 months,biggest10 months, EGMO Latvia to had the oldest team with an average age of 18 years and 7 months, anddate Lichtenstein, with 8 new countries a country participating of just 40000, in had2018: the Australia, smallest Austria, team ofBolivia, one. Canada, Germany,and Lichtenstein, Greece, a country of just 40000, had the smallest team of one. Mongolia, and Peru. For this 7th EGMO, a total of 195 contestants from 52 countries participated Each participating country may send a team of up to four students, a Team LeaderEach participating country may send a team of up to four students, a Team Leader (137 contestants from 36 official European countries)1. and a Deputy Team Leader. At the EGMO, the Team Leaders form what is calledand the a Deputy Team Leader. At the EGMO, the Team Leaders form what is called the Jury,Peru which had the in youngest 2018 was team chaired with by an Roberto average Dvornicich. age of 14 years For and decisions 10 months, such Latvia as approving Jury,had the which oldest in 2018 was chaired by Roberto Dvornicich. For decisions such as approving eachteam paper with asan wellaverage as marking age of 18 schemes, years and all 7 Leaders months, may and vote,Lichtenstein, and motions a country are carriedof justeach 40000, by paper had as well the as marking schemes, all Leaders may vote, and motions are carried by smallest team of one. a simple majority of those voting. For decisions such as determining medal cut-offsa simple and majority of those voting. For decisions such as determining medal cut-offs and coordinationEach participating appeals, country however, may only send Leaders a team of officialup to four European students, teams a Team may Leader vote. andcoordination a Deputy appeals, however, only Leaders of official European teams may vote. TeamParticipating Leader. At countries the EGMO, are the invited Team toLeaders submit form up to what six proposed is called the problems Jury, which to be in received 2018Participating was chaired countries are invited to submit up to six proposed problems to be received byby the Roberto Problem Dvornicich. Selection For Committee, decisions such which as inapproving 2018 was each chaired paper by as Massimowell as marking Gobbino.by theschemes, Problem all Selection Committee, which in 2018 was chaired by Massimo Gobbino. Leaders may vote, and motions are carried by a simple majority of those voting. For decisions such as This committee selects the contest problems and has alternative problems preparedThis if acommittee selects the contest problems and has alternative problems prepared if a problemdetermining turns medal out to cut-offs be already and known. coordination The selected appeals, problems however, were only presentedLeaders of to official the Jury European teams may vote. 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, problemsin their for first meeting on Tuesday 10 April. With no objections from the Jury, problems for bothParticipating Day 1 and countries Day 2 papers are invited were to approved, submit up including to six proposed translations problems into the to be 33 received languagesboth by Day the 1 and Day 2 papers were approved, including translations into the 33 languages Problem Selection Committee, which in 2018 was chaired by Massimo Gobbino. This committee selects required by the contestants. The Jury, Observers, and any others who have knowledgerequired by the contestants. The Jury, Observers, and any others who have knowledge ofthe the contest problems problems and solutions and has alternative before the problems examinations, prepared are if trusted a problem to doturns their out utmost to be already known. The selected problems were presented to the Jury in their first meeting on Tuesdayof the problems 10 April. and solutions before the examinations, are trusted to do their utmost to ensure that no contestant has any information, direct or indirect, about any proposedto ensure that no contestant has any information, direct or indirect, about any proposed problem.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,problem. and any others The six problems on the EGMO contest papers may be described as follows. who have knowledge of the problems and solutions before the examinations, are trustedThe to do six their problems on the EGMO contest papers may be described as follows. utmost1. A lovelyto ensure geometry that no problem contestant about has any a point information, lying on direct a circle, or indirect, proprosed about by any Velina proposed problem. 1. A lovely geometry problem about a point lying on a circle, proprosed by Velina Ivanova, Bulgaria. Ivanova, Bulgaria. The six problems on the EGMO contest papers may be described as follows. 1.2. AA numberlovely geometry theory problem problem about about expressinga point lying integers on a circle, as a productproposed of by numbers Velina Ivanova, of the Bulgaria. 1 2. A number theory problem about expressing integers as a product of numbers of the form 1+ k , proposed by Mihail Baluna, Romania. 1 2. A number theory problem about expressing integers as a product of numbers of the form 1+ k , proposed by Mihail Baluna, Romania. proposed by Mihail Baluna, Romania. 3. A fun combinatorics problem involving a Jury choosing and then rearranging3. theA fun combinatorics problem involving a Jury choosing and then rearranging the 3. orderA fun ofcombinatorics contestants problem in a queue. involving It was a proposedJury choosing by Hungary. and then rearranging the order of contestants in a queue. It was proposed by Hungary. order of contestants in a queue. It was proposed by Hungary. 4. AA combinatorics problem involving involving balancedbalanced configurationsconfigurations of of dominoes dominoes on an an n 4. nA board. combinatorics It problem involving balanced configurations of dominoes on an n n was proposed by Merlijn Staps, the Netherlands. × board. It was proposed by Merlijn Staps, the Netherlands. board. It was proposed by Merlijn Staps, the Netherlands. × 5. A classical circle geometry problem, proposed by Dominika Regiec, Poland. 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 of positive5. A classical integers. circle geometry problem, proposed by Dominika Regiec, Poland. It was proposed by Merlijn Staps, the Netherlands. 6. A tricky algebra problem involving the closeness of pairs of elements in a set6. ofA tricky algebra problem involving the closeness of pairs of elements in a set of Like for the International Mathematical Olympiad, Problems 1–3 appeared on the Day 1 Paper and positive integers. It was proposed by Merlijn Staps, the Netherlands. positive integers. It was proposed by Merlijn Staps, the Netherlands. problems 4–6 appeared on the Day 2 Paper. These papers were held on Wednesday 11 April and ThursdayLike for 12 the April International respectively, Mathematicalwith the contestants Olympiad, allowed Problems four and 1–3 a half appeared hours per on paper theLike to for work the International Mathematical Olympiad, Problems 1–3 appeared on the Dayindividually 1 Paper on and the problems problems 4–6 and appeared write their on attempted the Day 2proofs. Paper. Each These problem papers was were scoredDay held1 out Paper of a and problems 4–6 appeared on the Day 2 Paper. These papers were held maximum of seven points. on Wednesday 11 April and Thursday 12 April respectively, with the contestants allowedon Wednesday 11 April and Thursday 12 April respectively, with the contestants allowed fourPrior and to the a half first hours day of per competition, paper to workthe opening individually ceremony on thewas problemsheld in the and beautiful writefour Teatro their and Verdi a half hours per paper to work individually on the problems and write their attemptedand hosted proofs. by Alessandra Each problem Caraceni was and scored Massimo out ofGobbino. a maximum The occasion of seven included points. speechesattempted from proofs. the Each problem was scored out of a maximum of seven points. Vice Mayor of Florence and various sponsors for the event as well as several jokes about a lack of a 1Thistimetable is a new for record the week. for the The EGMO, parade with of the teams number was of complete participating with countries beautifully-designed increasing every background year. 1This is a new record for the EGMO, with the number of participating countries increasing every year. Inslides 2012 for for theeach first country edition and there an were accompaniment only 19 countries, of 16the of unique them official instrumental European music ones. of Lame Da Barba. Finally, it was declared that the 2018 EGMO was officially open. After the opening ceremonyIn 2012 for there the first edition there were only 19 countries, 16 of them official European ones. was also an opportunity for teams and their deputies to mingle and participate in a Treasure Hunt in 1 Florence before the two contest days. 1

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

Mathematics Contests The Australian Scene 2018 | EGMO Team Leader’s Report | 130 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.

Mathematics Contests The Australian Scene 2018 | EGMO Team Leader’s Report | 131 The table below shows the distribution of awards for each country at the 2018 EGMO.

Country Size Total Gold Silver Bronze HM Rank Albania 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

Mathematics Contests The Australian Scene 2018 | EGMO Team Leader’s Report | 132 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 Shaw EGMO Team Leader, Australia

Mathematics Contests The Australian Scene 2018 | EGMO Team Leader’s Report | 133 EUROPEAN GIRLS’ MATHEMATICAL OLYMPIAD

Language: English Day: 1

Wednesday, April 11, 2018

Problem 1. Let ABC be a triangle with CA = CB and ∠ACB = 120◦, and let M be the midpoint of AB. Let P be a variable point on the circumcircle of ABC, and let Q be the point on the segment CP such that QP =2QC. It is given that the line through P and perpendicular to AB intersects the line 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 1 A = 1+ : k =1, 2, 3,... . k (a) Prove that every integer x 2 can be written as the product of one or more elements of A, which ≥ are not necessarily diﬀerent.

(b) For every integer x 2, let f(x) denote the minimum integer such that x can be written as the ≥ product 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)

(Pairs (x ,y ) and (x ,y ) are diﬀerent if x = x or y = y .) 1 1 2 2 1 2 1 2

Problem 3. The n contestants of an EGMO are named C1,...,Cn. After the competition they queue 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 the Jury 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 and the 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 cunningly choosing the initial order and the sequence of moves.

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

Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad | 134 Language: English Day: 2

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 each ≥ column has a value of k. Prove that a balanced configuration exists for every n 3, and find the minimum number of ≥ dominoes needed in such a configuration.

Problem 5. Let Γ be the circumcircle of triangle ABC. A circle Ω is tangent to the line segment AB and is tangent to Γ at a point lying on the same side of the line AB as C. The angle bisector of ∠BCA intersects Ω at two diﬀerent points P and Q. Prove that ∠ABP = ∠QBC.

Problem 6.

1 (a) Prove that for every real number t such that 0

1 (b) Determine whether for every real number t such that 0 ty | − | for every pair of diﬀerent elements x and y of S and every positive integer m (i.e. m>0).

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

Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad | 135 EUROPEAN GIRLS’ MATHEMATICAL OLYMPIAD Solutions to the 2018 EuropeanSOLUTIONS Girls’ Mathematical Olympiad

1. Solution 1 (Xinyue Alice Zhang, year 12, A.B. Paterson College, QLD. Alice was a Bronze medallist with the 2018 Australian EGMO team.)

A M B Q

Consider triangles QNP and QMC. Vertically opposite angles NQP and MQC are equal. We also have that CM is perpendicular to AB since ABC is isosceles and M is the midpoint of AB. Given that NP is also perpendicular to AB, CM and NP are parallel. Therefore ∠NPQ = ∠MCQ and ∠PNQ = ∠CMQ. Hence triangles 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 midpoint 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 through the 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, and therefore N also does not lie on the line CO. The possible positions of N are in fact all the points of ω with the exception of the two points lying on the line CO.

8 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’s Hill School, NSW. Ellen was a Bronze medallist with the 2018 Australian EGMO team.)

A M B Q

As in solution 1 (with comments), NP is parallel to CM and hence CO (C, M, O are 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 adjacent sides OC and OP are equal, OCNP is actually a rhombus. Hence CN = CO, the radius of the circumcircle of ABC. Therefore N lies on the circle with centre C and radius CO.

9 Mathematics 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 √3 1 √3 1 1 A = , ,B= , ,C= (0, 1) ,M= 0, . − 2 2 2 2 2

Let P =(a, b). Since Q lies on CP such that QP =2QC, we get that

a b +2 Q = , . 3 3 The equation of the line through P and perpendicular to AB is x = a and the equation of the line MQ (if a = 0), is 2b +1 1 y = x + . 2a 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 result is independent of the position→ of P (provided that a = 0, because otherwise N is not well-deﬁned). Hence, when P lies on the circumcircle of ABC, with the exception of the two points with a = 0, N lies on the translated circle which is the circle with centre C and radius 1.

10 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 138 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 1 k +1 A = 1+ : k =1, 2, 3,... = : k =1, 2, 3,... . k k For every integer x 2, ≥ x 1 2 3 x − k +1 x = = . 1 × 2 ×···× x 1 k − k=1 That is, every integer x 2, can be written as a product of one or more elements of 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 = 3 2n+1, the product of n + 2 elements of A, f(3 2n)=n + 2. × 2 × × Similarly, 33 2n > 2n+5 f(33 2n) n +6 × ⇒ × ≥ and since 33 2n = 33 2n+5, the product of n+6 elements of A, f(33 2n)=n+6. × 32 × × Lastly, f(11) 5 since 11 cannot be written as the product of four or fewer elements 4 ≥ 3 3 of A:2 > 11, 2 2 > 11, and any other product of at most four elements of A 3× 4 2 11 5 does not exceed 2 3 = 10 3 < 11. Since 11 = 10 4 2 2 2, we have that f(11) = 5. × × × × × Putting these results together,

f(33 2n)=n +6

11 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 139 Solution 2 (2018 EGMO Problem Selection Committee) First we show that f(xy) 7, and≥ 2 3 any other product of at most three elements of A does not exceed 2 2 =6< 7. Also f(49) 7 since 49 = 2 2 2 2 2 3 49 . Hence × ≤ × × × × × 2 × 48 f(49) 7 < 8 f(7) + f(7). ≤ ≤ Now suppose by contradiction that there exists only finitely many pairs (x, y) that satisfy f(xy) Mor b>Mholds we have f(ab)=f(a)+f(b). (Note that f(ab) f(a)+f(b) is always satisfied.) ≤ Now take any pair (x, y) that satisfies f(xy) M be any integer. Then

f(n)+f(xy)=f(nxy)=f(nx)+f(y)=f(n)+f(x)+f(y),

which contradicts f(xy)

12 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 140 3. Solution 1 (Cunning initial order and sequence of moves for maximum number of euros found by all medallists of the 2018 Australian EGMO Team)

n i (a) We prove by induction that each contestant Ci can move at most 2 − 1 times and note that this also shows that each contestant can move only a ﬁnite− number of times and therefore the process cannot continue indeﬁnitely. For this induction we work backwards and use the key observation that when a contestant Ci moves forward, past i other contestants, she must pass at least one Cj where j>i.

In the base case, Cn cannot move at all because there are not n other contestants in the queue. As for Cn 1, when she moves, she must past every other of − the n 1 contestants, including C . Once she has passed C there is no way − n n Cn can get back in front of Cn 1 (because Cn cannot move). Hence there is no − way Cn 1 can move again and so Cn 1 can move at most once. − − n j Assume that each contestant C , for j>i, can move at most 2 − 1 times j − and consider how many times contestant Ci can pass Cj where j>i. If Cj n j moves at most 2 − 1 times then, by passing Ci every move, she can pass Ci n j − at most 2 − 1 times. Since Cj may have started ahead of Ci, we have that − n j Ci can pass Cj at most 2 − times. 0 1 Therefore we have that Ci can pass Cn at most 2 times, Cn 1 at most 2 times, 2 n i 1 − Cn 2 at most 2 times, ..., Ci+1 at most 2 − − times. − Each time Ci moves, she must pass at least one Cj where j>i, and in the extreme case, she passes all of C1,C2,...,Ci 1 and one Cj. Therefore, the − maximum number of times Ci can move is

2 n i 1 n i 1+2+2 + +2 − − =2 − 1. ··· − (b) What remains is to show how the Jury can arrange the queue initially, and what moves they can make, to achieve this maximal number of moves for each 0 1 2 contestant. That is, 2 1 moves for Cn,2 1 moves for Cn 1,2 1 moves n 1 − − − − for Cn 2, ...,2 − 1 moves for C1 and so a total number of moves (i.e. euros) − of − 0 1 2 n 1 n (2 1) + (2 1) + (2 1) + + (2 − 1) = 2 1 n. − − − ··· − − − We prove by induction that by lining the contestants up in reverse order, n Cn,Cn 1,...,C1, the jury can collect 2 1 n euros before the contestants − − − are 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 the Jury makes 21 1 1 = 0 euros. Indeed, for 2 contestants lined up in reverse − − order C2,C1, only a single move can occur to give C1,C2 before the restaurant opens 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 can make 2n 1 n euros reversing the order of the remaining contestants, − −

Cn+1,C1,C2,...,Cn 1,Cn. −

13 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 141 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 euros reversing 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 the restaurant 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 collect a maximum number of 2n 1 n euros. − −

Solution 2 (2018 EGMO Problem Selection Committee) This solution is a diﬀerent argument for proving that the process cannot continue indeﬁnitely.

As in the previous solution, the key observation is that when a contestant Ci moves forward, 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 this is not the case. Then there must be at least one contestant who moves inﬁnitely

many 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 that

Ci0 passes some ﬁxed Cj0 inﬁnitely many times with j0 >i0. However, Cj0 moves a

ﬁnite number of times and so can only precede Ci0 in the queue a ﬁnite number of

times. Hence it is impossible for Ci0 to move inﬁnitely many times and the process must 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 continue indeﬁnitely. We prove by induction on the number of contestants that the process must eventually end.

For the base case n = 1, the only contestant is C1 and there are no contestants in front 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 contestant Ck+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 only move a ﬁnite number of times, and so there will be a point after which Ck+1 does not move anymore.

After this point, all contestants behind Ck+1 will remain behind Ck+1, and all contestants in front of Ck+1 will remain in front of Ck+1. If we now remove Ck+1, we

14 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 142 are left with a valid initial order for k contestants, and moreover, we may observe that any move which occurs after this point remains a valid move even with Ck+1 removed: any move by C will move them ahead of exactly i of C ,C ,...,C i { 1 2 k} since no contestant jumps over Ck+1. The problem has now been reduced to the case with k contestants, so by the inductive hypothesis the process will eventually end.

Solution 4 (2018 EGMO Problem Selection Committee) 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 an − − arrangement of the n contestants, for each pair in which Ci is behind Cj in the queue i 1 and i

Now consider the total weight of an arrangement before and after Ci moves forward in the queue. Since Ci moves forward past i other contestants, she must pass at least one contestant Cj where j>i, and at most i 1 contestants Cj where j

15 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 143 4. Solution 1 (Yifan Guo, year 12, Glen Waverley Secondary College, VIC. Yifan was awarded an Honourable Mention with the 2018 Australian EGMO team.) Consider a balanced conﬁguration with D dominoes on an n n board in which each row and each column has a value k. On one hand, the total× value of all rows and columns is 2nk. On the other hand, since each of D dominoes contributes 2 + 1 = 3 to the total value of all rows and columns, this total value is 3D. Hence we have the equality 2nk 2nk =3D D = . ⇒ 3 2n First consider the case where n is a multiple of 3. Since k 1, we have that D 3 . The following diagram shows a balanced conﬁguration for≥ n = 3 with k = 1≥ and 2 3 ×3 = 2 dominoes on the board.

n If n is a multiple of 3, we can obtain a balanced conﬁguration with k = 1 and 3 of 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 balanced 2n conﬁguration is 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 and 2 4 = 8 dominoes on the board. ×

16 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 144 Balanced conﬁgurations for n 1(mod 3) and n 2(mod 3) with 2n dominoes for n 5 can be constructed as described≡ below using≡ the following 3 3 blocks. ≥ ×

Block A Block B

n 1 For n 1(mod 3) with n 7, −3 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 dominoes contribute− × 2− to the value of each of the ﬁrst n 1 rows and n 1 columns. Then n 4 − − −3 Block A’s are placed along a main diagonal of the (n 7) (n 7) sub-board as well as in the bottom-right corner. These Block A’s with− a× further− 2 dominoes placed 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 to the value of the nth row and nth column. The− total number− of dominoes placed on n 1 n 4 the board in this construction is 4 − +2 − +4=2n. × 3 × 3

. . .

n 2 For n 2(mod 3) with n 5, −3 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 dominoes − × − n 2 contribute 2 to the value of each of the ﬁrst n 2 rows and n 2 columns. Then −3 Block A’s are placed along a main diagonal of− the (n 5) (n− 5) sub-board as well as in the bottom-right corner. These Block A’s with− a× further− 2 dominoes placed

17 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 145 in the bottom two rows and 2 dominoes placed in the rightmost two columns, as shown, each contribute 1 to the value of the ﬁrst n 2 rows and n 2 columns, and 3 to the value of the bottom two rows and rightmost− two columns.− The total number n 2 n 2 of dominoes placed on the board in this construction is 4 − +2 − +4=2n. × 3 3 Note that this construction does indeed work for n = 5. Despite Block× B and Block A overlapping in the middle row and column in this case, only Block B has a domino placed in this middle cell.

. . .

Therefore the minimum number of dominoes required in a balanced configuration 2n is therefore 3 if n is a multiple of 3, and 2n otherwise. Solution 2 (Xinyue Alice Zhang, year 12, A.B. Paterson College, QLD. Alice was a Bronze medallist with the 2018 Australian EGMO team.) This solution presents alternative balanced configurations for n 0 (mod 3) with k = 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 and r 4, 5≥, 6, 7 . Hence a balanced conﬁguration can be obtained for n 8 and k =3 by∈{ 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.

18 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 146 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, and let M be the midpoint of the arc AB that does not contain C. Note that the angle bisector of ∠BCA intersects Γ at M.

Ω Q

V O1 O2

P U A B

We ﬁrst show that V , U, and M are collinear. Let O1 and O2 be the centres of Ω and Γ respectively, and let VU intersect Γ at M . We have ∠O1UV = ∠O1VU = O VM = O M V and hence O U O M . Since O U AB (AB is tangent ∠ 2 ∠ 2 1 2 1 ⊥ to Ω at U), O M AB and hence M is the midpoint of the arc AB (that does 2 ⊥ not contain C) and M = M. (Indeed, the dilation with centre V that sends Ω to Γ sends U to the point of Γ where the tangent of Γ is parallel to AB and this point is 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 and MP MB MB = MQ. 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.

19 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 147 Solution 2 (2018 EGMO Problem Selection Committee) 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 , that the 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 through M and V . Also, since Ω is tangent to AB (at U), Ω is tangent to Γ at a point on the line through M and U. To summarise, Ω is a circle tangent to both AB and Γ that passes through fixed points W1 and W2. So Ω must be Ω itself. The inversion swaps points of tangency V and U, and these points and M, the centre of the inversion, are collinear. Now P and Q are intersections between the fixed line MC and Ω, and since the only 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. ×

20 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 148 6. Solution (part a) (2018 EGMO Problem Selection Committee) 1 Let t be a real number such that 0

~~tsi~~

~~0 si 2si 3si 4si sj~~

If si+1 si tsi, for some 1 i n 1, then we have that si+1 is within tsi of the ﬁrst multiple− ≤ of s , namely 1≤ s≤(m−= 1). Hence the inequality is satisﬁed. i × i Otherwise, s s > ts s > (1 + t)s for all 1 i n 1. That is, i+1 − i i ⇒ i+1 i ≤ ≤ − 2 3 n 1 sn > (1 + t)sn 1 > (1 + t) sn 2 > (1 + t) sn 3 > > (1 + t) − s1. − − − ··· This gives that n 1 1 − sn > (1 + t) s1 s1 < n 1 sn. ⇒ (1 + t) − Finally, due to Bernoulli’s inequality, we know that for n large enough,

1 n 1 1 − < (1 + t) n 1 t. t ⇒ (1 + t) − ≤ Hence we have that s 0s < ts , that is, s is within ts of the zero-th multiple 1 − n n 1 n of sn (m = 0). Hence the inequality is satisfied. Solution (part b) (Stephen Farrar, long-time friend of the Olympiad program in Australia. Stephen was also a Gold and Bronze medallist with the 1997 and 1998 Australian IMO team.) 1 Let t be a real number such that 0 ts for every pair of different elements s and s of S and for every | j − i| i i j positive integer m. In other words, no element of S (different from si) is within tsi of 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 such that ts < (s 1)/2. This is equivalent to s > 1/(1 2t). 1 1 − 1 − Having chosen s1, certain numbers are excluded from being in S. S cannot contain any multiple of s1, like Eratosthenes’ sieve for ﬁnding primes. The diﬀerence in this question is that the numbers close to a multiple of s1 are also excluded from S. The exclusion radius is ts1, and when t is close to 1/2 it could be that most integers are excluded from S. The highlighted regions on the numberline below show those numbers excluded by s1.

~~21 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 149 s1 1 ts1 < 2−~~

~~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 not integers. However, near (m +1/2)s , we have integers x =(m +1/2)s 1/2. Call 1 1 ± these 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 > ts1 and (m + 1)s ((m +1/2)s 1/2) = (s 1)/−2 > ts−and these half-multiples− 1 − 1 − 1 − 1 are even further from other multiples of s1.

~~1 s1 1 (m + 2 )s1 ts1 < 2−~~

~~ms1 (m + 1)s1 half-multiples of s1: (m + 1 )s 1 ,(m + 1 )s + 1 2 1 − 2 2 1 2~~

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 that s1 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 then this gives s (m ≥t) >s ms s > ts s −ms > ts . 2 − 1 ⇒ 2 − 1 2 ⇒| 1 − 2| 2 So in particular, we will choose s2 to be whichever of the half-multiples (5/2)s1 1/2 is odd and note that since s > 1, s = (5/2)s 1/2=2s +(s 1)/2 > 2s±. 1 2 1 ± 1 1 ± 1 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 s2 is 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 are also half-multiples of both a and b. Suppose h =(k +1/2)ab 1/2 is a half-multiple of ab. Then h =( +1/2)a 1/2 where = kb +(b 1)/2 is± an integer. Therefore h 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 an odd half-multiple of s s s with s > 2s for all 1 i k. In this way, s 1 2 ··· k k+1 i ≤ ≤ k+1 is an odd half-multiple of each of s1,s2,...,sk and so not excluded by any of these elements. Furthermore, sk+1 > 2si sk+1(m t) >si msk+1 si > tsk+1 s ms > ts and so for all 1⇒ i k, s−is not excluded⇒ by −s . ⇒ | i − k+1| k+1 ≤ ≤ i k+1 Specifically, we choose sk+1 to be an odd integer given by s =5s s s /2 1/2. k+1 1 2 ··· k ± Note that since each of s1,s2,...,sk is greater than 1, sk+1 > 5si/2 1/2=2si + (s 1)/2 > 2s for all 1 i k. Also note that the s sequence is increasing.± i ± i ≤ ≤ i Therefore, we have found such an infinite set S of integers s1 1/(1 2t) and s =5s s s /2 ···1/2. 1 − k+1 1 2 ··· k ±

~~22 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions | 150 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 Price IMO Deputy Leader~~

~~Mathematics Contests The Australian Scene 2018 | IMO Team Preparation School | 151 THE MATHEMATICS ASHES~~

~~F5 F5F5 2018 IMO20182018 Team IMO IMO Training Team Team Training Training Exam Exam Exam AUS andAUSAUS UNK and and UNK UNK 2018 Mathematical20182018 Mathematical Mathematical Ashes Ashes Ashes~~

~~Friday,Friday,Friday, July 6, July July 2018 6, 6, 2018 2018~~

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Language:Language:Language: English English English Time:Time: 4Time: hours 4 4 and hours hours 30 and andminutes 30 30 minutes minutes Each problemEachEach problem problem is worth is is 7worth worth points 7 7 points points

~~Mathematics Contests The Australian Scene 2018 | The Mathematics Ashes | 152 THE MATHEMATICS ASHES RESULTS~~

~~Australia P1 P2 P3 Total William 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 Total Agnijo 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 Total Number 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.

Mathematics Contests The Australian Scene 2018 | IMO Team Leader's Report | 155 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

~~Mathematics Contests The Australian Scene 2018 | IMO Team Leader's Report | 156 Some Country Totals~~

Rank Country Total 1 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

~~Mathematics Contests The Australian Scene 2018 | IMO Team Leader's Report | 157 Distribution of Awards at the 2018 IMO~~

Country Total Gold Silver Bronze HM Albania 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 0 Austria 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

Mathematics Contests The Australian Scene 2018 | IMO Team Leader's Report | 158 Country Total Gold Silver Bronze HM Ireland 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

Mathematics Contests The Australian Scene 2018 | IMO Team Leader's Report | 159 Country Total Gold Silver Bronze HM Switzerland 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 Pasquale IMO Team Leader, Australia~~

~~Mathematics Contests The Australian Scene 2018 | IMO Team Leader's Report | 160 INTERNATIONAL MATHEMATICAL OLYMPIAD~~

Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad | 161 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad | 162 INTERNATIONAL MATHEMATICAL OLYMPIAD SOLUTIONS Solutions to the 2018 International Mathematical Olympiad

1. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA. William was 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 line through G and E intersect Γ for a second time at Y .

~~Y E D G F~~

~~B C~~

Since F AXB is cyclic, we have ∠DXA = ∠FXA = ∠FBA = ∠FBD. Since also ∠FDB = ∠XDA, it follows that FBD AXD (AA). Since F lies on the perpendicular bisector of BD, we have FBD∼ is isosceles with FB = FD. Thus AXD is isosceles with AX = AD. A similar argument shows that AE = AY . But since AD = AE, it follows that AX = AD = AE = AY , and so XEDY is cyclic with centre A. Using this and the fact that F GXY 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 Grammar School, 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 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 163 Solution 2 (William Steinberg, year 10, Scotch College, WA. William was a Silver medallist 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 AB ⊥ and GG AC. ⊥ Let X and Y be the projections of O onto the lines FF and GG , respectively.

~~A A~~

~~D E D E M N G Q G P F F O G G F F X Y~~

~~B C B C~~

~~Note that MOXF is a rectangle due to all its right angles. Hence we have 1 1 1 OX = MF = MB F B = AB DB = AD. − 2 − 2 2~~

1 Similarly OY = 2 AE. Since AD = AE, we deduce that OX = OY . But OF = OG (radii of Γ), and OX XF and OY YG. Hence OXF OY G (RHS). Thus ⊥ ⊥ ≡ ∠XFO = ∠OGY . Since also ∠OFG = ∠F GO, we have ∠F FG = ∠F GG . Let FG intersect lines AB and AC at P and Q, respectively. From the angle sums in PFF and QGG , we deduce ∠FPF = ∠G QG, and so ∠QP A = ∠AQP . Hence AP Q is isosceles with apex A. Since ADE is also isosceles with apex A, it follows that DE and FG are parallel.

62 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 164 2. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA. William was a Bronze medallist with the 2018 Australian IMO team.) Answer All positive integers n that are multiples of 3

For each integer i>nlet ai+2 = aiai+1 +1. In this way we have an infinite sequence a1,a2,... with the property that ai+2 = aiai+1 +1for each positive integer i and a = a whenever i j (mod n). Thus n is a period of the sequence. i j ≡ 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 positive integer i. It follows that ai+2 = aiai+1 +1> 1. Similarly ai+3 > 1. It follows that ai+4 = ai+2ai+3 +1>ai+3 +1>ai+3. Similarly ai+5 >ai+4, and inductively we see that the sequence is strictly increasing from ai+3 onward. But this is impossible for a periodic sequence.

~~Lemma 2 The sequence does not contain a number ai such that ai =0.~~

~~Proof If ai =0for some positive integer i, then ai+1 = ai+2 =1> 0. But this contradicts 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 denote positive and negative terms by + and , respectively. We have shown that neither +, + nor , , occur in the sequence.− Thus any two consecutive + terms are separated− 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− − 0. Observe that aa= 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 implies that n is a multiple of 3. − − − − − −

~~It only remains to observe that the sequence that starts with a1 = a2 = 1 follows the pattern 1, 1, 2, 1, 1, 2,... which yields a valid sequence for any−n that is − − − − a multiple of 3.~~

63 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 165 Solution 2 (Ross Atkins, Leader of the 2018 New Zealand IMO team. Ross was also 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~~

b =2a a a (a 1)2 (a 1)2 (a 1)2. i i i+1 i+2 − i − − i+1 − − i+2 − We show that this sequence is increasing by considering the term b b . i+1 − i b b =2a a a 2a a a (a 1)2 +(a 1)2 i+1 − i i+1 i+2 i+3 − i i+1 i+2 − i+3 − i − =2ai+1ai+2(ai+3 ai) (ai 3 ai)(ai+3 + ai 2) − − − − − =(a a )(2a a (a + a 2)) i+3 − i i+1 i+2 − i+3 i − =(a a )(2(a 1) (a + a 2)) i+3 − i i+3 − − i+3 i − =(a a )2 i+3 − i 0. ≥

Since the sequence is increasing and b1 = bn+1, the sequence must be constant. However, if b = b , then we have the equality (a a )2 =0and therefore i i+1 i+3 − i ai = 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. 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. • − − − −

64 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 166 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~~

~~2 ai+2 = aiai+1ai+2 + ai+2 = a (a 1) + a i i+3 − i+2 a2 a a = a a . (1) ⇔ i+2 − i i+3 i+2 − i Summing the equalities at (1) for i =1, 2,...,n yields~~

n n (a2 a a )= (a a ) i+2 − i i+3 i+2 − i i=1 i=1 n (a a )2 =0. (2) ⇔ i − i+3 i=1 2 2 2 Here we have used ai+2 = ai and ai+2 = ai = ai+3. It follows from (2) that ai = ai+3 for alli. Hence the sequence has period 3. Since the 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. • − − − −

65 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 167 3. This was the hardest problem of the 2018 IMO. Only 11 of the 594 contestants were able to solve this problem completely. Solution (Ethan Tan, year 12, Cranbrook School, NSW. Ethan was a Gold medallist 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-Pascal triangle T that is not in the bottom row, the two numbers below it are y and x + y for some positive integer y. Thus starting from any integer x1 in T , we may generate the 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 the larger 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~~

~~x = x + y + y + + y n 1 2 3 ··· n for each positive integer n.~~

~~For example, in the following four-row anti-Pascal triangle, starting with x1 =4we generate (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 existence of an anti-Pascal triangle T that has 2018 rows and which contains the numbers from 1 to 1+2+ + 2018. ··· Let x1 be the top number in T , and consider the sequences x1,x2,x3,...,x2018 and y2,y3,...,y2018 generated from x1 as described at (*). We have

~~x = x + y + y + + y . 2018 1 2 3 ··· 2018~~

~~Since x1,y2,y3,...,y2018 are distinct positive integers, it follows that~~

~~x 1+2+ + 2018. 2018 ≥ ··· But the largest number in T is 1+2+ + 2018. Thus x =1+2+ + 2018 ··· 2018 ··· and (x1,y2,...,y2018) is a permutation of (1, 2,...,2018).~~

Next, let T1 and T2 be the sub-equilateral triangular arrays whose bases consist of the 1009 leftmost numbers and 1009 rightmost numbers, respectively, in the bottom row of T . Note that T1 and T2 are disjoint and that their union covers the entire bottom 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 T1 be the sub-equilateral triangular array whose base consists of the 1008 leftmost numbers in the bottom row of T . Since yn is adjacent to xn in each row of the triangle, and each xn lies outside of T1, it follows that T1 contains no xn and no yn. In particular none of the numbers 1, 2,...,2018 lies inside T1 .

~~66 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 168 T~~

~~T1 T2~~

~~x2018~~

Let x1 be the top number in T1 . Generate the sequences x1 ,x2 ,x3 ,...,x1008 and y2 ,y3 ,...,y1008 as described at (*). These numbers all lie in T1 . Since T1 does not contain any of 1, 2,...,2018, it follows that

x = x + y + y + + y 1008 1 2 3 ··· 1008 2019 + 2020 + + 3026 ≥ ··· 1 = 1008 (2019 + 3026) × 2 1009 2019 × =1+2+ + 2018. ··· This is an obvious contradiction because the numbers in T are less than or equal to 1+2+ + 2018. ···

67 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 169 Discussion (Angelo Di Pasquale, Leader of the 2018 Australian IMO team) n(n+1) Suppose that n is any positive integer, N =1+2+ + n = 2 , and T is an anti-Pascal triangle with n rows that contains the numbers··· from 1 to N. In what follows, we present a refinement of the argument given in the previous solution to prove that n 8. ≤ Following the previous solution, we generate the sequences x1,x2,x3,...,xn and y2,y3,...,yn of numbers in T , and deduce that xn = N and (x1,y2,y3,...,yn) is a permutation 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 on the bottom row of T . And let T be the sub-equilateral triangle whose base consists of all numbers to the right of xn and yn on the bottom row of T .

~~For any i~~

Suppose that the base of T contains a numbers and the base of T contains b numbers. Note that a + b = n 2. − As described at (*) in the previous solution, generate the sequences x1 ,x2 ,x3 ,...,xa and y2 ,y3 ,...,ya for T and the sequences x1 ,x2 ,x3 ,...,xb and y2 ,y3 ,...,yb for T . As in the previous solution, we have

x + y + y + + y = x and x + y + y + + y = x . 1 2 3 ··· a a 1 2 3 ··· b b Adding these two equalities together and using the fact that all of the above terms are different numbers that are greater than n but less than N yields

(n +1)+(n +2)+ +(n +(n 2)) N 1+N 2 ··· − ≤ − − 1 (n 2)(3n 1) n(n + 1) 3 ⇒ 2 − − ≤ − n2 9n 8. ⇒ ≤ − Hence n 8. ≤ Remark The above argument can be further refined to show that no anti-Pascal triangles exist for n =8and n =7, although the arguments, especially for n =7, bifurcate into a number of cases. The case for n =6can be ruled out by a simple parity argument as follows. One may 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 such a triangle cannot contain the numbers from 1 to 21 because their sum is odd.

68 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 170 Comment It can be shown that anti-Pascal triangles with n rows which use each of the positive integers from 1 up to 1+2+ + n exactly once exist only for the 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 at http://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 5 rows are shown below.

~~n =2 • 1 2 3 2 3 1~~

~~n =3 • 1 2 3 3 4 3 5 3 4 1 5 2 6 2 5 6 1 4 2 6 5 1 6 4~~

~~n =4 • 3 3 4 4 4 7 5 2 2 6 5 1 5 9 2 4 9 7 5 7 1 2 7 6 6 1 10 8 6 10 1 8 8 3 10 9 8 10 3 9~~

~~n =5 • 5 9 4 2 11 7 10 12 1 8 13 3 15 14 6~~

69 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 171 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 viewed as the 400 unit squares of a 20 20 chessboard. Amy and Ben take turns placing their stones in unoccupied unit× squares such that no two red stones are a knight’s move apart, that is, no two red stones are at diagonally opposite corners of a 2 3 or 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 chessboard colouring. Note that if two squares are separated by a knight’s move, then one is white 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 red stones, 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 each such small chessboard into four groups of× four squares as follows.

~~A1 B1 C1 D2 A2 B2 C2 D1 A2 B2 C2 D1 A1 B1 C1 D2 Group A Group B Group C Group D~~

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 in each group. Then Amy can play at most once in each group, resulting in at most four red stones in each small chessboard, which is at most 100 stones in total.

70 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 172 5. Solution (Hadyn Tang, year 9, Trinity Grammar School, VIC. Hadyn was a Silver medallist 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. ≥ a a a n + n+1 − n an+1 a1

~~Proof The result follows by calculating the difference of the two integers~~

~~a1 a2 an 1 an a1 a2 an 1 an an+1 + + + − + and + + + − + + . a2 a3 ··· an a1 a2 a3 ··· an an+1 a1~~

Lemma 2 Suppose that a, b, c, d are integers with b, d > 0. If each of the fractions a c a c b and d is in lowest terms and b + d is an integer, then b = d. a c ad+bc Proof We have b + d = 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(a ,a ) gcd(a ,a ). 1 n ≤ 1 n+1

Proof Suppose, for the sake of contradiction, that gcd(a1,an) > gcd(a1,an+1) for k k some integer n N. Then there exists a prime power p such that p a1,an but k ≥ k | p an+1. It follows that p an+1 an. Hence in lowest terms, the denominator a − an+1 an of n is not divisible by p, but the denominator of − is divisible by p. This an+1 a1 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 because gcd(a ,a ) can increase only≤ finitely many times for n N. 1 n ≥ Lemma 5 For all integers n G we have a a . ≥ n+1 ≤ n Proof From lemma 4, there is a positive integer g such that gcd(a1,an)=g for all n G. Hence the numbers b = an are integers for n =1and for all integers ≥ n g n G. Moreover gcd(b ,b )=1for all n G. ≥ 1 n ≥ From lemma 1 it follows that

~~2 b b b b1bn + b bnbn+1 n + n+1 − n = n+1 − bn+1 b1 b1bn+1~~

is an integer for all n G. Thus b b b + b2 b b , and so b b b . ≥ n+1 | 1 n n+1 − n n+1 n+1 | 1 n 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 a a for all n |G. ≤ ≥ n+1 ≤ n ≥ Lemma 6 There is a positive integer M such that a = a for all n M. n n+1 ≥ Proof Clearly an 1 for all n. So the result follows from lemma 5 because an can decrease only finitely≥ many times for n G. ≥

~~71 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 173 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 any positive integer, then νp(n) denotes the exponent of p in the prime factorisation of n. 3 1 For example, if n =24=2 3 , then ν2(n)=3, ν3(n)=1, and νp(n)=0for any prime p>3. We also have by· convention ν (0) = + . p ∞ It is straightforward to prove the following standard properties of νp for any positive integers m and n.

~~ν (mn)=ν (m)+ν (n) • p p p ν (m n) min ν (m),ν (n) • p ± ≥ { p p } ν (m n) = min ν (m),ν (n) whenever ν (m) =ν (n) • p ± { p p } p p~~

72 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 174 Solution 2 (Guowen Zhang, year 12, St Joseph’s College, QLD. Guowen was a Gold medallist with the 2018 Australian IMO team.)

~~a an+1 an As in solution 1, n + − is an integer for each n N. Hence, so also is an+1 a1 ≥ a a a (a a )(a a ) n + n+1 − n 1= 1 − n+1 n − n+1 . (1) an+1 a1 − a1an+1~~

~~Thus an+1 (a1 an+1)(an an+1), and so an+1 a1an. It follows inductively that for any integer| n− N, the set− of prime divisors of| a is a subset of the set of prime ≥ n divisors 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 fixed prime 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(a )+ν(a ) ν(a a )+ν(a a ). (2) 1 n+1 ≤ 1 − n+1 n − n+1 Case 1 ν(a )=ν(a ) for some n N n 1 ≥ Let ν(a1)=ν(an)=a and ν(an+1)=c. If c = a, then (2) yields the following contradiction. a + c min a, c + min a, c

Thus c = a. So if ν(an)=ν(a1) for some n N, then ν(an+1)=ν(a1). This inductively implies that ν(a ) is constant for all≥m n, as desired. m ≥ Case 2 ν(a ) = ν(a ) for all n N n 1 ≥ Let ν(a )=a, ν(a )=b, and ν(a )=c. Note that a = b and a = c. 1 n n+1 If b

~~a + c min a, c + min b, c a + b~~

73 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 175 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), and diagonals e, f. Then e2 + f 2 = b2 + d2 +2ac. (1)

Proof This is straightforward using vectors. Let the trapezium be ABCD where a = 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 both equal 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 trapezium (a parallel to c), and e, f be its diagonals, with f>e. Then

~~f 2 > ac + bd > e2.~~

~~Proof Using lemma 1, we have~~

~~e2 + f 2 b2 + d2 +2ac 2bd +2ac f 2 = = ac + bd. ≥ 2 2 ≥ 2 Also from Ptolemy’s inequality we have~~

~~2 ac + bd>ef>f .~~

~~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 the symmetry 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 the longer diagonal of trapezium AY BX. From the trapezium inequality we have

~~AB2 > XA BY + XB AY. (4) · · From ABY DCX, we have ∼ BY AB XC AB = BY = · , CX DC ⇒ CD~~

74 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 176 and AY AB XD AB = AY = · . DX DC ⇒ CD Substituting these values into (4) and rearranging yields

~~AB CD > XA XC + XB XD. (5) · · ·~~

~~B C Y~~

~~A D~~

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 the shorter diagonal of trapezium BZCX. From the trapezium inequality we have

BC2 < XB CZ + XC BZ. (6) · · From BCZ ADX, we have ∼ CZ BC XD BC = CZ = · , DX AD ⇒ AD and BZ BC XA BC = BZ = · . AX AD ⇒ 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. · · ~~

75 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 177 Solution 2 (Andrew Elvey Price, Deputy Leader of the 2018 Australian IMO team) 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 may assume without loss of generality that ray AB intersects ray DC at some point P and ray AD intersects ray BC at some point Q. It follows from the given angle conditions that quadrilaterals DXBQ and PCXA are cyclic.

~~B C~~

~~A D Q~~

~~Now we invert the diagram around the point A, sending each point Z =A to the point 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 and • vice 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,Qlie on a line in 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 intersection of lines BQ and DP, its image C is the intersection of circles AB Q and AD P . The given length condition can be rewritten as BC CD = . AB AD

~~Converting this to the inverted diagram using the similarities ABC AC B ∼ and ADC AC D yields ∼ B C C D = AC AC~~

~~and so B C = C D .~~

76 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 178 Finally, X is an intersection of circles DBQ and PCA, so X is an intersection of line P C with circle D B Q . In order to determine which of these two intersection points is X , we will need to use the condition that X lies inside quadrilateral ABCD. In the inverted diagram, this means that X lies within angle B AD , but outside circles AB C and AC D .

~~Y B~~

~~X1 = X C~~

~~P X2~~

~~A Q D~~

~~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 the ∼ \ \ reflection 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 and X2 be the reflections of Y and Q in the perpendicular bisector of B D . Since the perpendicular bisector of B D passes through the centre of circle B D Q , the points X1 and X2 lie on this circle, so either X = X1 or X = X2. We will show that the latter is impossible.

Suppose, for the sake of contradiction, that X2 = X . Then X2 lies within angle B AD . Since X2 is the reflection of Q in the perpendicular bisector of B D , we have X Q B D , and so X lies within angle AQ B . Together these imply that 2 2 X = X2 lies inside triangle AB Q , which is inside circle AB C Q . This contradicts the 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)~~

~~77 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 179 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 YX + YX C (exterior angle C X Y ) ∠ ∠ ∠ = ∠AB Q + ∠Q B X + ∠YX C (circle B Q D X ) = ∠AB X + ∠YX C . Thus (3) is equivalent to

~~∠YX C + ∠X D Q = 180◦. (4)~~

~~Recall that lines P C X and Q C Y are symmetric with respect to the perpendicular bisector of B D , which passes through the centre of circle B Q D X Y . Hence~~

~~YX C = C YX = Q YX = 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 are parallel. 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, and the 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.

~~78 Mathematics Contests The Australian Scene 2018 | International Mathematical Olympiad Solutions | 180 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 1979–2018~~

~~Because of changing titles and affiliations, the most senior title achieved and later affiliations are generally used, except for the Interim committee, where they are listed as they were at the time.~~

Interim Committee 1979–1980 Mr P J O’Halloran Canberra College of Advanced Education, Chair Prof A L Blakers University of Western Australia Dr J M Gani Australian Mathematical Society Prof B H Neumann Australian National University Prof G E Wall University of Sydney Mr J L Williams University of Sydney

Australian Mathematical Olympiad Committee The 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; 1980–1986 Prof G B Preston Monash University 10 years; 1986–1995 Prof A P Street University of Queensland 6 years; 1996–2001 Prof C Praeger University of Western Australia 17 years; 2002–2018 Deputy Chair* Prof P J O’Halloran University of Canberra 15 years; 1980–1994 Prof A P Street University of Queensland 1 year; 1995 Prof C Praeger University of Western Australia 6 years; 1996–2001 Assoc Prof D Hunt University of New South Wales 14 years; 2002–2015 Prof Andrew Hassell Australian National University 3 years; 2016–2018 Executive Director* Prof P J O’Halloran University of Canberra 15 years; 1980–1994 Prof P J Taylor University of Canberra 18 years; 1994–2012 Adj Prof M Clapper University of Canberra 4 years; 2013–2016 Mr Nathan Ford University of Canberra 2 years; 2017–2018 Chief Mathematician Mr Mike Clapper Melbourne 2 years; 2017–2018 Secretary Prof J C Burns Australian Defence Force Academy 9 years; 1980–1988 Vacant 4 years; 1989–1992 Mrs K Doolan Victorian Chamber of Mines 6 years; 1993–1998 Treasurer* Prof J C Burns Australian Defence Force Academy 8 years; 1981–1988 Prof P J O’Halloran University of Canberra 2 years; 1989–1990 Ms J Downes CPA 5 years; 1991–1995 Dr P Edwards Monash University 8 years; 1995–2002 Prof M Newman Australian National University 6 years; 2003–2008 Dr 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; 1990–2006 Dr K McAvaney Deakin University 13 years; 2006–2018 Chair, Senior Problems Committee Prof B C Rennie James Cook University 1 year; 1980 Mr J L Williams University of Sydney 6 years; 1981–1986 Assoc Prof H Lausch Monash University 27 years; 1987–2013 Dr 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; 1980–1986 Mr G Ball University of Sydney 3 years; 1987–1989 Dr D Paget University of Tasmania 6 years; 1990–1995 Dr M Evans Scotch College, Victoria 3 months; 1995 Assoc Prof D Hunt University of New South Wales 5 years; 1996–2000 Dr A Di Pasquale University of Melbourne 18 years; 2001–2018 IMO Team Leader Mr J L Williams University of Sydney 5 years; 1981–1985 Assoc Prof D Hunt University of New South Wales 9 years; 1986, 1989, 1990, 1996–2001 Dr E Strzelecki Monash University 2 years; 1987, 1988 Dr D Paget University of Tasmania 5 years; 1991–1995 Dr A Di Pasquale University of Melbourne 16 years; 2002–2010, 2012–2018 Mr I Guo University of New South Wales 1 year; 2011 IMO Deputy Team Leader Prof G Szekeres University of New South Wales 2 years; 1981–1982 Mr G Ball University of Sydney 7 years; 1983–1989 Dr D Paget University of Tasmania 1 year; 1990 Mr J Graham University of Sydney 3 years; 1991–1993 Dr M Evans Scotch College, Melbourne 3 years; 1994–1996 Dr A Di Pasquale University of Melbourne 5 years; 1997–2001 Mr D Mathews University of Melbourne 3 years; 2002–2004 Mr N Do University of Melbourne 4 years; 2005–2008 Mr I Guo University of New South Wales 4 years; 2009–2010, 2012–2013 Mr G White University of Sydney 1 year; 2011 Mr A Elvey Price Melbourne University 5 years; 2014–2018 EGMO Team Leader Miss T Shaw SCEGGS Darlinghurst 1 year, 2018 EGMO Deputy Team Leader Miss M Chen Melbourne University 1 year, 2018 State Directors Australian Capital Territory Prof M Newman Australian National University 1 year; 1980 Mr D Thorpe ACT Department of Education 2 years; 1981–1982 Dr R A Bryce Australian National University 7 years; 1983–1989 Mr R Welsh Canberra Grammar School 1 year; 1990 Mrs J Kain Canberra Grammar School 5 years; 1991–1995 Mr J Carty ACT Department of Education 17 years; 1995–2011 Mr J Hassall Burgmann Anglican School 2 years; 2012–2013 Dr C Wetherell St. Francis Xavier College 5 years; 2014–2018 New South Wales Dr M Hirschhorn University of New South Wales 1 year; 1980 Mr G Ball University of Sydney 16 years; 1981–1996 Dr W Palmer University of Sydney 19 years; 1997–2015 A/Prof Daniel Daners University of Sydney 1 year; 2016 Dr Dzmitry Badziahin University of Sydney 2 years; 2017–2018 Northern Territory Dr I Roberts Charles Darwin University 6 years; 2013–2018 Queensland Dr N H Williams University of Queensland 21 years; 1980–2000 Dr G Carter Queensland University of Technology 10 years; 2001–2010 Dr V Scharaschkin University of Queensland 4 years; 2011–2014 Dr Alan Offer Queensland 4 years; 2015–2018 South Australia Mr K Hamann Rostrevor 19 years; 1980–1982, 1991–2005; 2013 Mr V Treilibs SA Department of Education 8 years; 1983–1990 Dr M Peake Adelaide 7 years; 2006–2012 Mr D Martin Adelaide 5 years; 2014–2018

Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 183 Tasmania Mr J Kelly Tasmanian Department of Education 8 years; 1980–1987 Dr D Paget University of Tasmania 8 years; 1988–1995 Mr W Evers St Michael’s Collegiate School 9 years; 1995–2003 Dr K Dharmadasa University of Tasmania 15 years; 2004–2018 Victoria Dr D Holton University of Melbourne 3 years; 1980–1982 Mr B Harridge Melbourne High School 1 year; 1982 Ms J Downes CPA 6 years; 1983–1988 Mr L Doolan Melbourne Grammar School 9 years; 1989–1998 Dr P Swedosh The King David School 21 years; 1998–2018 Western Australia Dr N Hoffman WA Department of Education 3 years; 1980–1982 Assoc Prof P Schultz University of Western Australia 14 years; 1983–1988, 1991–1994, 1996–1999 Assoc Prof W Bloom Murdoch University 2 years; 1989–1990 Dr E Stoyanova WA Department of Education 7 years; 1995, 2000–2005 Dr G Gamble University of Western Australia 13 years; 2006–2018 Editor Prof P J O’Halloran University of Canberra 1 year; 1983 Dr A W Plank University of Southern Queensland 11 years; 1984–1994 Dr A Storozhev Australian Maths Trust 15 years; 1994–2008 Editorial Consultant Dr 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; 1995–2012 Mr M Bammann South Australia 1 year; 2018 Dr S Britton University of Sydney 8 years; 1990–1998 Prof G Brown Australian Academy of Science 10 years; 1980, 1986–1994 Dr R A Bryce Australian Mathematical Society 8 years; 1991–1998 Mathematics Challenge for Young Australians 14 years; 1999–2012 Mr G Cristofani Department of Education and Training 2 years; 1993–1994 Mr Josh Dean Tasmania 2 years; 2017–2018 Ms L Davis IBM Australia 4 years; 1991–1994 Dr W Franzsen Australian Catholic University 9 years; 1990–1998 Dr J Gani Australian Mathematical Society 1980 Assoc Prof T Gagen ANU AAMT Summer School 6 years; 1993–1998 Ms P Gould Department of Education and Training 2 years; 1995–1996 Mr Keith Hamann South Australia 4 years; 2015–2018 Mr Nhat Hoang Western Australia 1 year; 2018 Prof G M Kelly University of Sydney 6 years; 1982–1987 Ms J McIntosh Mathematics Challenge for Young Australians 7 years; 2012–2018 Prof R B Mitchell University of Canberra 5 years; 1991–1995 Ms Anna Nakos Mathematics Challenge for Young Australians 16 years; 2003–2018 Mr S Neal Department of Education and Training 4 years; 1990–1993 Prof M Newman Australian National University 13 years; 1986–1998 Mathematics Challenge for Young Australians 20 years; 1999–2018, (Treasurer during the interim) 2003–2008 Prof R B Potts University of Adelaide 1980 Mr H Reeves Australian Association of Maths Teachers/ 16 years; 1988–1998, Australian Mathematics Foundation 2014–2018 Mr N Reid IBM Australia 3 years; 1988–1990 Mr Martin Roberts Tasmania 4 years; 2014–2017 Mr R Smith Telecom Australia 5 years; 1990–1994 Prof P J Taylor Australian Mathematics Foundation 6 years; 1990–1994, 2013 Prof N S Trudinger Australian Mathematical Society 3 years; 1986–1988 Assoc Prof I F Vivian University of Canberra 1 year; 1990 Dr M W White IBM Australia 9 years; 1980–1988

Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 184 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 Stage Dr K McAvaney Deakin University (Director) 13 years; 2006–2018; Member 2 years; 2005–2006 Mr B Henry Victoria (Director) 17 years; 1990–2006 Member 12 years; 2007–2018 Prof P J O’Halloran University of Canberra 5 years; 1990–1994 Dr R A Bryce Australian National University 23 years; 1990–2012 Mr M Clapper Australian Maths Trust 6 years; 2013–2018 Ms L Corcoran Australian Capital Territory 3 years; 1990–1992 Ms B Denney New South Wales 9 years; 2010–2018 Mr J Dowsey University of Melbourne 8 years; 1995–2002 Mr A R Edwards Department of Education, Queensland 29 years; 1990–2018 Dr M Evans Scotch College, Victoria 6 years; 1990–1995 Assoc Prof H Lausch Monash University 24 years; 1990–2013 Ms J McIntosh Australian Mathematical Sciences Institute 17 years; 2002–2018 Mrs L Mottershead New South Wales 27 years; 1992–2018 Miss A Nakos Temple Christian College, South Australia 26 years; 1993–2018 Dr M Newman Australian National University 29 years; 1990–2018 Ms F Peel St Peter’s College, South Australia 2 years; 1999–2000 Dr I Roberts Charles Darwin University, NT 6 years; 2013–2018 Miss T Shaw SCEGGS, NSW 6 years; 2013–2018 Ms K Sims Blue Mountains Grammar School 20 years; 1999–2018 Dr A Storozhev Attorney General’s Department 23 years; 1994–2016 Prof P Taylor Australian Maths Trust 20 years; 1995–2014 Mrs A Thomas New South Wales 18 years; 1990–2007 Dr S Thornton Exec. Dir. reSolve 21 years; 1998–2018 Miss G Vardaro Wesley College, Victoria 25 years; 1993–2006, 2008–2018 Visiting members Prof E Barbeau University of Toronto, Canada 1991, 2004, 2008 Prof G Berzsenyi Rose Hulman Institute of Technology, USA 1993, 2002 Dr L Burjan Department of Education, Slovakia 1993 Dr V Burjan Institute for Educational Research, Slovakia 1993 Mrs A Ferguson Canada 1992 Prof B Ferguson University of Waterloo, Canada 1992, 2005 Dr D Fomin St Petersburg State University, Russia 1994 Prof F Holland University College, Cork, Ireland 1994 Dr A Liu University of Alberta, Canada 1995, 2006, 2009 Prof Q Zhonghu Academy of Science, China 1995 Dr A Gardiner University of Birmingham, United Kingdom 1996 Prof P H Cheung Hong Kong 1997 Prof R Dunkley University of Waterloo, Canada 1997 Dr S Shirali India 1998 Mr M Starck New Caledonia 1999 Dr R Geretschlager Austria 1999, 2013 Dr A Soifer United States of America 2000 Prof M Falk de Losada Colombia 2000 Mr H Groves United Kingdom 2001 Prof J Tabov Bulgaria 2001, 2010 Prof A Andzans Latvia 2002 Prof Dr H-D Gronau University of Rostock, Germany 2003 Prof J Webb University of Cape Town, South Africa 2003, 2011

Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 185 Mr A Parris Lynwood High School, Christchurch, NZ 2004 Dr A McBride University of Strathclyde, United Kingdom 2007 Prof P Vaderlind Stockholm University, Sweden 2009, 2012 Prof A Jobbings United Kingdom 2014 Assoc Prof D Wells United States of America 2015 Emer. Fellow P Neumann Queen’s College Oxford UK 2016 Dr C Jones Alabama State University, USA 2018 Dr P Couch Lamar University, USA 2018 Moderators for Mathematics Challenge Stage Mr W Akhurst New South Wales Ms N Andrews ACER, Camberwell, Victoria Mr L Bao Leopold Primary School, Victoria Prof E Barbeau University of Toronto, Canada Mr R Blackman Victoria Ms J Breidahl St Paul’s Woodleigh, Victoria Ms S Brink Glen Iris, Victoria Prof J C Burns Australian Defence Force Academy Mr J Carty ACT Department of Education, Australian Capital Territory Mr A Canning Queensland Mrs F Cannon New South Wales Dr E Casling Australian Capital Territory Mr B Darcy South Australia Ms B Denney New South Wales Mr J Dowsey Victoria Mr S Ewington Sydney Grammar School, New South Wales Br K Friel Trinity Catholic College, New South Wales Dr D Fomin St Petersburg University, Russia Mrs P Forster Penrhos College, Western Australia Mr T Freiberg Queensland Mr W Galvin University of Newcastle, New South Wales Mr S Gardiner University of Sydney, New South Wales Mr M Gardner North Virginia, USA Ms P Graham Tasmania Mr B Harridge University of Melbourne Ms J Hartnett Queensland Mr G Harvey Australian Capital Territory Ms I Hill South Australia Ms N Hill Victoria Dr N Hoffman Edith Cowan University, Western Australia Prof F Holland University College, Cork, Ireland Mr D Jones Coff’s Harbour High School, New South Wales Ms R Jorgensen Australian Capital Territory Dr T Kalinowski University of Newcastle, New South Wales Assoc Prof H Lausch Victoria Mr J Lawson St Pius X School, Chatswood, New South Wales Mr R Longmuir China Dr K McAvaney Deakin University Ms K McAsey Penleigh and Essendon Grammar School, Victoria Ms J McIntosh University of Melbourne Ms N McKinnon Victoria Ms T McNamara Victoria Mr G Meiklejohn Queensland School Curriculum Council, Queensland Mr M O’Connor AMSI, Victoria Mr J Oliver Northern Territory Mr S Palmer New South Wales Dr W Palmer University of Sydney Mr A Peck Mt Carmel College, Tasmania Mr G Pointer Marryatville High School, South Australia Prof H Reiter University of North Carolina, USA Mr M Richardson Yarraville Primary School, Victoria Mr G Samson Nedlands Primary School, Western Australia Mr J Sattler Parramatta High School, New South Wales

Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 186 Mr A Saunder Victoria Mr W Scott Seven Hills West Public School, New South Wales Mr R Shaw Hale School, Western Australia Miss T Shaw New South Wales Dr B Sims University of Newcastle Dr H Sims Victoria Ms K Sims Blue Mountains Grammar School, New South Wales Prof J Smit The Netherlands Ms C Smith St Paul’s School, Queensland Ms D Smith Inquisitive Minds, New South Wales Mrs M Spandler New South Wales Mr G Spyker Curtin University Ms C Stanley Queensland Study Authority (Primary), Queensland Ms R Stone Plympton Primary School, South Australia Dr E Strzelecki Monash University, Victoria Dr M Sun New South Wales Mr P Swain Ivanhoe Girls Grammar School, Victoria Dr P Swedosh The King David School, Victoria Prof J Tabov Academy of Sciences, Bulgaria Mrs A Thomas New South Wales Ms K Trudgian Queensland Prof J Webb University of Capetown, South Africa Assoc Prof D Wells North Carolina, USA Ms 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 Sydney Dr M Evans Scotch College, Victoria Mr K Hamann South Australia Assoc Prof H Lausch Monash University Dr 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 Academy Mr J Carty Merici College, Australian Capital Territory Dr H Gastineau-Hill University of Sydney Mr B Henry Victoria Assoc Prof H Lausch Monash University Prof P O’Halloran University of Canberra Dr 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 Zealand Dr A Di Pasquale University of Melbourne Dr R Geretschlager Austria Dr D Mathews Monash University Dr 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, Victoria Mr K Hamann South Australia Assoc Prof H Lausch Monash University Prof P O’Halloran University of Canberra Mrs A Thomas Meriden School, New South Wales

Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 187 Gauss Enrichment Series — Development Team (1993–1995) Dr M Evans Scotch College, Victoria (Editor) Mr B Henry Victoria (Editor) Mr W Atkins University of Canberra Mr G Ball University of Sydney Prof J Burns Australian Defence Force Academy Mr L Doolan Melbourne Grammar School, Victoria Mr A Edwards Mildura High School, Victoria Mr N Gale Hornby High School, New Zealand Dr N Hoffman Edith Cowan University Prof P O’Halloran University of Canberra Dr W Pender Sydney Grammar School, New South Wales Mr 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 Victoria Dr D Fomin St Petersburg University, Russia Mr G Harvey New South Wales Newton Series — Development Team (2001–2002) Mr B Henry Victoria (Editor) Mr J Dowsey University of Melbourne Mrs L Mottershead New South Wales Ms G Vardaro Annesley College, South Australia Ms A Nakos Temple Christian College, South Australia Mrs A Thomas New South Wales Dirichlet Series — Development Team (2001–2003) Mr B Henry Victoria (Editor) Mr A Edwards Ormiston College, Queensland Ms A Nakos Temple Christian College, South Australia Mrs L Mottershead New South Wales Mrs K Sims Chapman Primary School, Australian Capital Territory Mrs A Thomas New South Wales Ramanujan Enrichment Series — Development Team (2014–2016) Mr B Henry Victoria (Editor) Adj Prof M Clapper Australian Maths Trust Ms A Nakos Temple Christian College, South Australia Mr A Edwards Department of Education, Queensland Dr K McAvaney Australian Maths Trust Dr I Roberts Charles Darwin University, NT

Australian Intermediate Mathematics Olympiad Committee Dr K McAvaney Deakin University (Chair) 12 years; 2007–2018 Mr J Dowsey University of Melbourne 18 years; 1999–2016 Dr M Evans Australian Mathematical Sciences Institute 20 years; 1999–2018 Mr B Henry Victoria (Chair) 8 years; 1999–2006 Member 12 years; 2007–2018 Assoc Prof H Lausch Monash University 17 years; 1999–2015 Mr R Longmuir China 2 years; 1999–2000 Adj Prof M Clapper Australian Maths Trust 5 years; 2014–2018 Dr D Mathews Monash University 2 years; 2017–2018

Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 188 AMOC Senior Problems Committee Current members Mr M Clapper Australian Maths Trust 6 years; 2013–2018 Dr Alice Devillers University of Western Australia 3 years; 2016–2018 Dr A Di Pasquale University of Melbourne 18 years; 2001–2018 Dr N Do Monash University (Chair) 5 years; 2014–2018 (Member) 11 years; 2003–2013 Mr I Guo University of Sydney 11 years; 2008–2018 Dr J Kupka Monash University 16 years; 2003–2018 Dr K McAvaney Deakin University 23 years; 1996–2018 Dr D Mathews Monash University 18 years; 2001–2018 Dr A Offer Queensland 7 years; 2012–2018 Dr C Rao Victoria 19 years; 1999–2018 Dr B Saad Monash University 25 years; 1994–2018 Dr J Simpson Curtin University 20 years; 1999–2018 Dr I Wanless Monash University 19 years; 2000–2018 Previous members Mr G Ball University of Sydney 16 years; 1982–1997 Mr M Brazil LaTrobe University 5 years; 1990–1994 Dr M S Brooks University of Canberra 8 years; 1983–1990 Dr G Carter Queensland University of Technology 10 years; 2001–2010 Dr M Evans Australian Mathematical Sciences Institute 28 years; 1990–2016 Mr J Graham University of Sydney 1 year; 1992 Dr M Herzberg Telecom Australia 1 year; 1990 Assoc Prof D Hunt University of New South Wales 29 years; 1986–2014 Assoc Prof H Lausch Monash University (Chair) 28 years; 1987–2013 (Member) 2 years; 2014–2015 Dr L Kovacs Australian National University 5 years; 1981–1985 Dr D Paget University of Tasmania 7 years; 1989–1995 Prof P Schultz University of Western Australia 8 years; 1993–2000 Dr L Stoyanov University of Western Australia 5 years; 2001–2005 Dr E Strzelecki Monash University 5 years; 1986–1990 Dr E Szekeres University of New South Wales 7 years; 1981–1987 Prof G Szekeres University of New South Wales 7 years; 1981–1987 Emerit Prof P Taylor Australian Maths Trust 1 year; 2013 Dr N H Williams University of Queensland 20 years; 1981–2000

Mathematics School of Excellence Dr S Britton University of Sydney (Coordinator) 2 years; 1990–1991 Mr L Doolan Melbourne Grammar (Coordinator) 6 years; 1992, 1993–1997 Mr W Franzsen Australian Catholic University (Coordinator) 2 years; 1990–1991 Dr D Paget University of Tasmania (Director) 6 years; 1990–1995 Assoc Prof D Hunt University of New South Wales (Director) 5 years; 1996–2000 Dr A Di Pasquale University of Melbourne (Director) 17 years; 2001–2018

International Mathematical Olympiad Selection School Mr G Ball University of Sydney (Director) 6 years; 1984–1989 Mr L Doolan Melbourne Grammar (Coordinator) 3 years; 1989–1991 Dr S Britton University of Sydney (Coordinator) 7 years; 1992–1998 Mr W Franzsen Australian Catholic University (Coordinator) 8 years; 1992–1996; 1999–2001 Dr D Paget University of Tasmania (Director) 6 years; 1990–1995 Mr J L Williams University of Sydney (Director) 2 years; 1982–1983 Assoc Prof D Hunt University of New South Wales (Director) 5 years; 1996–2000 Dr A Di Pasquale University of Melbourne (Director) 18 years; 2001–2018

~~Mathematics Contests The Australian Scene 2018 | AMOC Honour Roll 1979–2018 | 189~~