Mathematics Contests. The Australian Scene 2018

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Mathematics Contests The Australian Scene 2018 Combined MCYA and AMOC

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

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

Copyright © 2019 Australian Mathematics Trust AMTT Limited ACN 083 950 341 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

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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

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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

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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 � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 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� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��������� 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

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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

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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.

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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

2018

Florence

Gold

Silver

Bronze

HM

1

2

1

Bronze

HM

Rank 20 out of 52 teams

Summary of Australia’s achievements at previous IMOs Year

City

Gold

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

2

11 out of 38 teams

1986

Warsaw

5

15 out of 37 teams

1987

Havana

1988

Canberra

1989

Braunschweig

2

2

22 out of 50 teams

1990

Beijing

2

4

15 out of 54 teams

1991

Sigtuna

1992

Moscow

1

1993

Istanbul

1

1994

1

Silver

3 1

Rank

15 out of 42 teams 1

1

17 out of 49 teams

3

2

20 out of 56 teams

1

2

1

19 out of 56 teams

2

3

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

3

1

9 out of 82 teams

1998

Taipei

4

2

13 out of 76 teams

1999

Bucharest

1

1

3

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

2

1

13 out of 73 teams

1

1

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

15 out of 81 teams

26 out of 84 teams

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Year

City

Gold

2003

Tokyo

2004

Athens

2005

Merida

2006

Ljubljana

3

2

1

26 out of 90 teams

2007

Hanoi

1

4

1

22 out of 93 teams

2008

Madrid

5

1

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

2016

Hong Kong

2

4

2017

Rio de Janeiro

3

2

2018

Cluj–Napoca

3

1

1

Silver

Bronze

HM

2

2

2

26 out of 82 teams

1

2

1

27 out of 85 teams

6

2

Rank

25 out of 91 teams

19 out of 97 teams

6 out of 104 teams 25 out of 109 teams 1

34 out of 111 teams 11 out of 109 teams

* Perfect Score by Alexander Gunning

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

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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 

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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 

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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 

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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 

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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

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AMOC TIMETABLE FOR SELECTION OF THE TEAMS TO THE 2018 EGMO AND 2018 IMO August 2017—July 2018 Hundreds of students are involved in the AMOC programs which begin on a state basis. The students are given problem-solving experience and notes on various IMO topics not normally taught in schools. The students proceed through various programs with the top 45 students, including potential team members and other identified students, participating in a 10-day residential school in November/ December. Students are selected for EGMO after the AMO in February, while the IMO team is chosen at the conclusion of the Selection School in March. Team members then receive individual coaching by mentors prior to the EGMO or the IMO.

Month

Activity

August

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

September

• Australian Intermediate Mathematics Olympiad

November/December

• AMOC School of Excellence

January

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

February

• Australian Mathematical Olympiad

March

• Asian Pacific Mathematics Olympiad

March

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

May–June

• Personal Tutor Scheme for IMO team members

July

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

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

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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

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CHALLENGE PROBLEMS PRIMARY 2018 Challenge Problems–- MIDDLE Middle Primary Students may work on each of these four problems in groups of up to three, but must write their solutions individually.

MP1 Training Ms Fitz sets up five cones 10 metres apart in a straight line for her class’s daily exercise drill. A student starts at 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.

•1

10 m

•2

•3

10 m

10 m

•4

•5

10 m

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 Problems 1 – Middle Primary

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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 1

7 3

5

4

Red

6 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 Problems 2 – Middle Primary

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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 Problems 3 – Middle Primary

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CHALLENGE PROBLEMS PRIMARY 2018 Challenge Problems –- UPPER Upper Primary Students may work on each of these four problems in groups of up to three, but must write their solutions individually.

UP1 Bitripents Tia has a square grid and several of each of the tiles A, B, and C as shown below. The tiles consist of 2, 3, and 5 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 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 Problems 4 – Upper Primary

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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 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.

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

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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 1

7 3

5

4

Red

6 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. 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 less than 20 and explain why there are no others.

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

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CHALLENGE PROBLEMS JUNIOR 2018 Challenge Problems -–Junior Students may work on each of these six problems with a partner but each must write their solutions individually.

J1 Insect Barriers A caravan park is made up of square sites joined along their edges. For example, here is a caravan park with 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 Problems 7 – Junior

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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: • 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.

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

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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 − 12 , and if we apply the operation R to − 32 we obtain 23 . Note that R can never be applied to the number 0. The operations T and R can be combined. For example, we can turn 0 into T, T, T, R, T, T, R, T: T

T

T

R

0 −→ 1 −→ 2 −→ 3 −→ −

2 5

by successively applying the operations

1 T 2 T 5 R 3 T 2 −→ −→ −→ − −→ . 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 4

into 23 .

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 Problems 9 – Junior

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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.

a 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 Problems 10 – Junior

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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 Problems 11 – Junior

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CHALLENGE PROBLEMS 2018 Challenge 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: • 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. • •

• •

• •

•

•

• •

•

• •

• •

•

• •

• •

•

• •

•

• •

• •

• • •

•

•

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 − 12 , and if we apply the operation R to − 32 we obtain 23 . Note that R can never be applied to the number 0. The operations T and R can be combined. For example, we can turn 0 into T, T, T, R, T, T, R, T: T

T

T

R

0 −→ 1 −→ 2 −→ 3 −→ −

2 5

by successively applying the operations

1 T 2 T 5 R 3 T 2 −→ −→ −→ − −→ . 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 Problems 12 – Intermediate

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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

25m

35m

35m

W

N

S

E

15m

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 Problems 13 – Intermediate

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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 Problems 14 – Intermediate

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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 Host: Player Host: Player Host: Player Host:

1: 2: 1: 2:

3 Higher 5 Higher 8 Lower 7 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 winning is 31 , whereas if she chooses box 1 her chance of winning is 23 . b A game has boxes numbered 1 to 4. Show that no matter which box Player 1 chooses on the first turn, Player 2 has a strategy for ensuring his chance of winning is 12 . c A game has boxes numbered 1 to 9. Show that no matter which boxes Player 2 chooses, Player 1 has a strategy which involves choosing box 5 at the first turn and ensures her chance of winning is more than 12 .

Mathematics Contests The Australian Scene 2018 | Challenge Problems 15 – Intermediate

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CHALLENGE SOLUTIONS MIDDLE PRIMARY 2018 Challenge Solutions –- Middle Primary MP1 Training a The following diagram shows the student’s path:

•1

10 m

•2

10 m

•3

10 m

•4

10 m

•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.

•1

A

•2

B

•3

C

•4

D

•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 Solutions 16 – Middle Primary

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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 Solutions 17 – Middle Primary

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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 Solutions 18 – Middle Primary

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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 Solutions 19 – Middle Primary

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CHALLENGE SOLUTIONS UPPER PRIMARY 2018 Challenge Solutions – - Upper Primary UP1 Bitripents a The smallest square that can be covered using only B tiles is 6 × 6. Here is one such square.

Comment No square smaller than 6 × 6 is possible. Each B tile has 3 grid squares. So the number of grid squares in a 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 Solutions 20 – Upper Primary

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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 Solutions 21 – Upper Primary

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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: Blue bucket: Green bucket:

1, 4 2, 3 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: Blue bucket: Green bucket:

1, 4, 10, 13 2, 3, 11, 12 5, 6, 7, 8, 9

There are only two other arrangements: Red bucket: Blue bucket: Green bucket:

1, 4, 7, 10, 13 2, 3, 11, 12 5, 6, 8, 9

1, 4, 10, 13 2, 3, 7, 11, 12 5, 6, 8, 9

Any one of the three arrangements is acceptable.

Mathematics Contests The Australian Scene 2018 | Challenge Solutions 22 – Upper Primary

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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. • • • • • • • • •

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b The (5, 3, 2, 3, 5) isopentagon is the bottom-left pentagon in the solution of Part a. By carefully counting 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 Solutions 23 – Upper Primary

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CHALLENGE SOLUTIONS JUNIOR 2018 Challenge Solutions - –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 Solutions 24 – Junior

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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 Solutions 25 – Junior

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J2 Cubic Coprimes a Notice in particular that no two even numbers can be adjacent. Here is one possible labelling. There are others. 8

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• 1

3

2

4

5

7

•

• •

•

6

b Here is one possible labelling. There are others. 10

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• •

• 9

3

2

4

5

7

•

• •

•

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

•

• •

• 4

14

25

15

8

22

•

•

•

•

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 Solutions 26 – Junior

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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.

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b The (5, 3, 2, 3, 5) isopentagon is the bottom-left pentagon in the solution of Part a. By carefully counting 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. d 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.

=

a b

Mathematics Contests The Australian Scene 2018 | Challenge Solutions 27 – Junior

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

1 1

2 4

3 9

4 16

5 25

6 36

7 49

8 64

9 81

10 100

11 121

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

The t-number of an isopentagon is more than a × a and less than 2 × a × a. If an isopentagon has t-number 119, then a is at most 10 and at least 8. We are looking for values of a and b so that 2 × a × a − b × b = 119. Since 2 × a × a is even and 119 is odd, b must be odd.

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

b 1 3 5 7 1 3 5 7

t 64 + 64 − 1 = 127 64 + 64 − 9 = 119 64 + 64 − 25 = 103 64 + 64 − 49 = 79 81 + 81 − 1 = 161 81 + 81 − 9 = 153 81 + 81 − 25 = 137 81 + 81 − 49 = 113

a 10 10 10 10 10

b 1 3 5 7 9

t 100 + 100 − 1 = 199 100 + 100 − 9 = 191 100 + 100 − 25 = 175 100 + 100 − 49 = 151 100 + 100 − 81 = 119

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

J4 Ts and Rs a T

R

2 −→ 3 −→ −

1 R 1 T 3 R 4 T 1 R T R −→ 3 −→ 4 −→ − −→ −→ − −→ − −→ 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 R

3 −→ −

1 T 2 R 3 T 1 T 1 R T T −→ −→ − −→ − −→ −→ −2 −→ −1 −→ 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 1 R 2 −→ − 2 1 R 3 −→ − 3 1 R 4 −→ − 4

R

T

−→ −1 −→ 0 1 R T T −→ −→ −2 −→ −1 2 2 R 3 T 1 T −→ −→ − −→ − 3 2 2 3 R 4 T 1 T −→ −→ − −→ − 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 Solutions 28 – Junior

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R

7 −→ −

1 T 6 −→ 7 7 5 T −→ 6 4 T −→ 5 3 T −→ 4 2 T −→ 3 1 T −→ 2

R

7 6 6 − 5 5 − 4 4 − 3 3 − 2

T

−→ − R

−→ R

−→ R

−→ R

−→

1 6 1 − 5 1 − 4 1 − 3 1 − 2

−→ − T

−→ T

−→ T

−→ T

−→

R

T

−→ −2 −→ −1

T

−→ 0 There are 20 operations in this sequence, as required.

J5 Rhombus Rings a Since opposite angles in a rhombus are equal and the rhombuses in the first ring are identical and have the 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 45

135

45

135

90

90 45

45 90

90

B

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 Solutions 29 – Junior

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c We have these angles (all in degrees). 20

160 160 60 x 60

20

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: 20

160 160 60 60 80

20

100 y

120 100

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: 20

160 160 60 60 80

20

120 100 40

100

80 140 140

40

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 + 5 + 10 = 16 1 + 5 + 11 = 17 1 + 5 + 12 = 18

1 + 6 + 12 = 19 1 + 7 + 12 = 20 1 + 8 + 12 = 21

Mathematics Contests The Australian Scene 2018 | Challenge Solutions 30 – Junior

2 + 8 + 12 = 22 3 + 8 + 12 = 23 4 + 8 + 12 = 24

| 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 B C

1 2 3

9 7 4

10 8 5

11 12 6

1 2 3

8 7 4

10 9 5

11 12 6

1 2 3

8 7 4

9 10 5

11 12 6

A B C

1 2 3

9 7 4

10 8 5

12 11 6

1 2 3

8 7 4

10 9 5

12 11 6

1 2 3

8 7 4

9 10 5

12 11 6

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

1 2 3

4

5

6

If Judge A has scorecard 7, then score 11 is impossible. So Judge B has scorecard 7. To achieve score 29, we must place scorecards 11 and 12 distributions. A 1 11 B 2 7 12 C 3 4 5 6

with Judges A and B separately. So we have two 1 2 3

7 4

5

12 11 6

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

1 2 3

9 7 4

10 8 5

11 12 6

1 2 3

8 7 4

10 9 5

11 12 6

1 2 3

8 7 4

9 10 5

11 12 6

A B C

1 2 3

9 7 4

10 8 5

12 11 6

1 2 3

8 7 4

10 9 5

12 11 6

1 2 3

8 7 4

9 10 5

12 11 6

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

Mathematics Contests The Australian Scene 2018 | Challenge Solutions 31 – Junior

| 41

CHALLENGE SOLUTIONS INTERMEDIATE 2018 Challenge Solutions –- Intermediate 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 = 46

• • • • •

• • • • •

• • • •

• • • • •

• • • •

• • • • •

• • • •

• • • • •

• • • •

• • • • •

• • • •

• • • • •

• • • •

• • • • •

• • • •

• • • • •

• • • •

• • • •

• • • •

• • • •

• • • •

t = 41 • • • • •

• • • • •

• • • • •

• • • • •

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 Solutions 32 – 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 t

1 1

2 4

3 9

4 16

5 25

6 36

7 49

8 64

9 81

10 100

11 121

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

The t-number of an isopentagon is more than a × a and less than 2 × a × a. If an isopentagon has t-number 119, then a is at most 10 and at least 8. We are looking for values of a and b so that 2 × a × a − b × b = 119. Since 2 × a × a is even and 119 is odd, b must be odd.

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

b 1 3 5 7 1 3 5 7

t 64 + 64 − 1 = 127 64 + 64 − 9 = 119 64 + 64 − 25 = 103 64 + 64 − 49 = 79 81 + 81 − 1 = 161 81 + 81 − 9 = 153 81 + 81 − 25 = 137 81 + 81 − 49 = 113

a 10 10 10 10 10

b 1 3 5 7 9

t 100 + 100 − 1 = 199 100 + 100 − 9 = 191 100 + 100 − 25 = 175 100 + 100 − 49 = 151 100 + 100 − 81 = 119

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 1 2 3 4 5 6 7 8 9

2 7

3 11 10

4 15 14 13

5 19 18 17 16

6 23 22 21 20 19

7 27 26 25 24 23 22

8 31 30 29 28 27 26 25

9 35 34 33 32 31 30 29 28

10 39 38 37 36 35 34 33 32 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 Solutions 33 – Intermediate

| 43

I2 Ts and Rs a T

R

2 −→ 3 −→ − b R

3 −→ −

1 R 1 T 3 R 4 T 1 R T R −→ 3 −→ 4 −→ − −→ −→ − −→ − −→ 3 3 4 4 3 3

1 T 2 R 3 T 1 T 1 R T T −→ −→ − −→ − −→ −→ −2 −→ −1 −→ 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 1 R 2 −→ − 2 1 R 3 −→ − 3 1 R 4 −→ − 4

R

T

−→ −1 −→ 0 1 R T T −→ −→ −2 −→ −1 2 2 R 3 T 1 T −→ −→ − −→ − 3 2 2 3 R 4 T 1 T −→ −→ − −→ − 4 3 3

For any integer n > 1 we have: R

n −→ −

1 T n−1 R n 1 T −→ −→ − −→ − 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

n −→ −

and so on down to

1 T n−1 R n 1 T −→ −→ − −→ − n n n−1 n−1 n−2 R n−1 T 1 T −→ −→ − −→ − n−1 n−2 n−2 n−3 R n−2 T 1 T −→ −→ − −→ − n−2 n−3 n−3 2 R 3 T 1 −→ − −→ − 3 2 2 1 R T T −→ −→ −2 −→ −1 2 T

−→

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. T

R

−1 −→ 1 1 2 1 R −3 −→ 3 R

−2 −→

R

0 −→ 1 −→ −1 1 T T R −→ 2 −→ − −→ 2 3 R 2 T T −→ −→ − −→ 2 3 4 R 3 T T −→ −→ − −→ 3 4

1 R −→ −2 2 1 R −→ −3 3 1 R −→ −4 4

For any integer n > 0 we have: R

−n −→

1 T n+1 R n 1 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 Solutions 34 – 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 15 A

•B

x •

By Pythagoras’ theorem, x2 = 152 + 352 . The vertical distance between the two cameras is 25 − 20 = 5 metres. Let y be the direct distance in metres between the two cameras. Then we have this right-angled triangle. •B 5

y A•

x

By Pythagoras’ theorem, y 2 = x 2 + 52 = 152 + 352 + 52 = 52 32 + 52 72 + 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 5 A•

B •

x

10

10

By Pythagoras’ theorem, x2 = 52 + 252 . The vertical distance between the two cameras is still 5 metres. Let the direct distance in metres between the two cameras now be y. Then we have this right-angled triangle. y A•

•B 5

x

By Pythagoras’ theorem, y 2 = x 2 + 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 Solutions 35 – 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 0≤t≤3

15 − 5t A•

B •

5t

d

5t

A 5t − 15

•

d 35 − 5t

3≤t≤7

• B

5t

15

A•

d 7 ≤ t ≤ 14

5t − 15

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 Solutions 36 – Intermediate

| 46

•9 •8 •7 6• 5• 4• • 9

• 8

3• 7

• 6

• 5

• 4

3 •

2 •

1 •

0 •B

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 X

• 4

• 3

By Pythagoras’ theorem, d2 = CA2 + CB 2 = 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 Solutions 37 – 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 Solutions 38 – 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 45

135

45

135

90

90

45 90

90

45

B

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). 20

160 160 60 x 60

20

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: 20

160 160 60 60 80

20

100 y

120 100

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 Solutions 39 – Intermediate

| 49

20

160 160 60 60 80

20

120 100 40

100

80 140 140

40

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 3a

4a

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 Solutions 40 – Intermediate

| 50

I6 Higher or Lower a If Player 1 chooses box 2 on the first turn, her chance of choosing correctly is 13 . If she doesn’t win on the first turn, then Player 2 will win on the second turn because of the host’s directions. So the chance of Player 1 winning if she chooses box 2 on the first turn is 13 . P1 chooses box 2 1 3

1 3

‘Lower’ (P2 wins)

1 3

‘Correct’ (P1 wins)

‘Higher’ (P2 wins)

If Player 1 chooses box 1 on the first turn, her chance of choosing correctly is 13 . The game will continue with probability 23 . If it does continue, then the probability of Player 2 winning on the second turn is 12 . 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 13 + 23 × 12 = 23 . P1 chooses box 1 1 3

2 3

‘Correct’ (P1 wins)

‘Higher’ (continues) 1 2

1 2

‘Correct’ (P2 wins)

‘Higher/Lower’ (P1 wins)

b If Player 1 chooses box 1 on the first turn, her chance of choosing correctly is 14 . The game will continue with probability 34 . If it does continue then, from Part a, Player 2 should choose either box 2 or 4 to ensure his chance of winning from that stage is 23 . Hence Player 2’s overall chance of winning with this strategy is 34 × 23 = 12 . P1 chooses box 1 1 4

‘Correct’ (P1 wins)

3 4

‘Higher’ (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 to ensure his overall chance of winning the game is 12 . If Player 1 chooses box 2 on the first turn, her chance of choosing correctly is 14 . If she does not choose correctly, there are two possibilities: either the prize is in box 1, with probability 14 , and Player 2 immediately wins; or the prize is in box 3 or 4, with probability 12 , and Player 2’s chance of choosing correctly at that stage is 12 . Hence, if Player 1 chooses box 2 on the first turn, then Player 2’s chance of winning is 14 + 12 × 12 = 12 .

Mathematics Contests The Australian Scene 2018 | Challenge Solutions 41 – Intermediate

| 51

P1 chooses box 2 1 4

‘Lower’ (P2 wins)

1 4

1 2

‘Correct’ (P1 wins)

‘Higher’ (continues) 1 2

‘Correct’ (P2 wins)

1 2

‘Higher/Lower’ (P1 wins)

Similarly, if Player 1 chooses box 3 on the first turn, then Player 2’s chance of winning is 12 . So, no matter which box Player 1 chooses on the first turn, Player 2 has a strategy for ensuring his chance of winning is 12 . c If Player 1 chooses box 5 on the first turn, her chance of choosing correctly is 19 . The game will continue with probability 89 . If it does continue, then there are 4 boxes left to choose from and Player 2 has the next turn. From Part b, Player 1 now has a strategy for ensuring her chance of winning the 4-box game is 12 . Hence, if Player 1 chooses box 5 on the first turn, she has a strategy to ensure her chance of winning the game is 19 + 89 × 12 = 59 > 12 . P1 chooses box 5 1 9

‘Correct’ (P1 wins)

8 9

‘Higher/Lower’ (continues) 1 2

(P1 wins, using b)

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

| 52

CHALLENGE STATISTICS – MIDDLE PRIMARY Mean Score/School Year/Problem School Year

Number of Students

Overall Mean

3

599

4 *ALL YEARS

Challenge Problem 1

2

3

4

9.1

2.2

2.5

2.6

2.2

917

10.4

2.5

2.9

2.8

2.5

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 Training

2 Bitripents

3 Coloured Buckets

4 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 Factor

0.5

0.6

0.6

0.7

Please note: The discrimination factor for a particular problem is calculated as follows: 1. The students are ranked in regard to their overall scores. 2. The mean score for the top 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean top score’. 3. The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean bottom score’. 4. The discrimination factor = mean top score – mean bottom score 4 Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is considered to be a good discriminator.

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

| 53

CHALLENGE STATISTICS – UPPER PRIMARY Mean Score/School Year/Problem School Year

Number of Students

Overall Mean

5

1426

6

Challenge Problem 1

2

3

4

11.3

3.1

3.1

2.7

2.5

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 Bitripents

2 Insect Barriers

3 Coloured Buckets

4 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 Factor

0.4

0.4

0.5

0.7

Please note: The discrimination factor for a particular problem is calculated as follows: 1. The students are ranked in regard to their overall scores. 2. The mean score for the top 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean top score’. 3. The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean bottom score’. 4. The discrimination factor = mean top score – mean bottom score 4 Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is considered to be a good discriminator.

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

| 54

CHALLENGE STATISTICS – JUNIOR Mean Score/School Year/Problem School Year

Number of Students

Overall Mean

7

2737

8 *ALL YEARS

Challenge Problem 1

2

3

4

5

6

13.1

2.8

2.6

2.4

2.6

1.7

1.9

2508

15.3

3.1

2.9

2.7

3.1

2.2

2.2

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 Insect Barriers

2 Cubic Coprimes

Isopentagons

3

4 Ts and Rs

5 Rhombus Rings

6 Parsecking

Did not attempt

1%

4%

8%

8%

13%

12%

0

2%

6%

10%

12%

22%

14%

1

7%

8%

10%

7%

15%

17%

2

21%

21%

17%

10%

17%

19%

3

33%

33%

26%

18%

13%

25%

4

36%

28%

29%

45%

21%

12%

Mean

3.0

2.7

2.6

2.8

1.9

2.1

Discrimination Factor

0.4

0.5

0.7

0.8

0.8

0.7

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 School Year

Number of Students

Overall Mean

9

1490

10 *ALL YEARS

Challenge Problem 1

2

3

4

5

6

13.8

2.8

2.5

2.4

2.7

2.0

2.0

757

14.8

3.0

2.8

2.6

2.9

2.2

2.3

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 Isopentagons

1

2 Ts and Rs

3 Rail Cams

4 Curvy Tiles

5 Rhombus Rings

6 Higher or Lower

Did not attempt

5%

4%

7%

7%

15%

11%

0

6%

6%

12%

7%

14%

16%

1

9%

8%

13%

7%

17%

17%

2

14%

30%

18%

20%

17%

18%

3

27%

23%

23%

23%

21%

20%

4

39%

29%

28%

36%

16%

19%

Mean

2.9

2.6

2.5

2.8

2.1

2.1

Discrimination Factor

0.6

0.6

0.7

0.7

0.7

0.7

Score

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

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AUSTRALIAN MATHEMATICS 2018 AustralianINTERMEDIATE Intermediate 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 [4 marks] 192, find the area of triangle P OQ. 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 = 3P A, 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]

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AUSTRALIAN MATHEMATICS 2018 AustralianINTERMEDIATE Intermediate Mathematics OlympiadOLYMPIAD - Solutions SOLUTIONS 1. Method 1 The table shows the product 2x7 × 39 for all values of x. x 0 1 2 3 4

2x7 × 39 8073 8463 8853 9243 9633

x 5 6 7 8 9

2x7 × 39 10023 10413 10803 11193 11583

Thus x = 8. Method 2 We have 2x7 × 39 = 2x7 × 30 + 2x7 × 9. The units digit in 2x7 × 30 is 0, and its tens digit is 1. The tens digit of 2x7 × 9 is the units digit of 6 + 9 × x.

Hence 1 + 6 + 9 × x ≡ 9 (mod 10), 9 × x ≡ 2 (mod 10), x = 8. Method 3 We have 2x7 × 39 = 207 × 39 + 390 × x. The units digit in 207 × 39 is 3, and its tens digit is 7. The tens digit of 390 × x is the units digit of 9 × x.

Hence 7 + 9 × x ≡ 9 (mod 10), 9 × x ≡ 2 (mod 10), x = 8. Method 4 We have 2x7 × 39 = 2x7 × 40 − 2x7. The units digit in 2x7 × 40 is 0, and its tens digit is 8. So the tens digit of 2x7 × 39 is the units digit of 8 − x − 1 or 18 − x − 1. Since x is non-negative, 17 − x = 9 and x = 8.

2. Method 1 We have 7010 = 234b+1 − 234b−1

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

= 2(b2 + 2b + 1) + 3(b + 1) − 2(b2 − 2b + 1) − 3(b − 1) = 8b + 6 b=8 So 234b = 2348 = 2 × 64 + 3 × 8 + 4 = 156. Method 2 The largest digit on the left side of the given equation is 4. Hence b − 1 is at least 5. So b ≥ 6.

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

If b = 7, then the left side in base 10 is 2348 −2346 = (2×64+3×8+4)−(2×36+3×6+4) = (128+24)−(72+18) = 152 − 90 = 62 = 70. If b = 8, then the left side in base 10 is 2349 −2347 = (2×81+3×9+4)−(2×49+3×7+4) = (162+27)−(98+21) = 189 − 119 = 70.

Each time b increases by 1, 4b remains the same, 30b increases by 3, but 200b increases by 2(b + 1)2 − 2b2 = 4b + 2. So the increase in 234b+1 is greater than the increase in 234b−1 . Hence 234b+1 − 234b−1 increases with increasing b. This means 234b+1 − 234b−1 > 7010 for b > 8. So 234b = 2348 = 2 × 64 + 3 × 8 + 4 = 156.

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3. Let O be the centre of the circumcircle.

A

B

O

D

C

Since OA = OB = OC = OD and AB = BC = CD = DA, triangles AOB, BOC, COD, DOA are isosceles and congruent. So  AOB = 360/4 = 90◦ . Hence the area of AOB is 12 × 10 × 10 = 50 and the area of the sector AOB = 14 π100 = 25π. By Pythagoras, AB 2 = AO2 +OB 2 = 200. Hence the area of the semicircle on AB is 12 π(AB/2)2 = AB 2 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: 1 11 21 31 41

2 12 22 32 42

3 13 23 33 43

4 14 24 34 44

5 15 25 35 45

6 16 26 36 46

7 17 27 37 47

8 18 28 38 48

9 19 29 39 49

10 20 30 40 50

From this we see that in the prime factorisation of 50!, 5 occurs exactly 12 times. Then 50!/(212 512 ) is the product of these factors: 1 1 3 4 1 6 7 1 9 1 11 12 13 14 3 16 17 18 19 1 21 22 23 24 1 26 27 28 29 3 31 32 33 34 7 36 37 38 39 1 41 42 43 44 9 46 47 48 49 1 So the last digit of 50!/(212 512 ) 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.

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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!/212 512 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!/(212 512 ) when divided by 5 will reveal its last digit. Since the remainder of 212 is 1 when divided by 5, the remainder for 50!/212 512 is the same as the remainder for 50!/512 . So we want the remainder of the product of the following factors. 1 11 21 31 41

2 12 22 32 42

Before we multiply these factors we may numbers. 1 1 1 1 1

3 13 23 33 43

4 14 24 34 44

1 3 1 7 9

6 16 26 36 46

7 17 27 37 47

8 18 28 38 48

9 19 29 39 49

2 4 6 8 2

subtract from each any multiple of 5. So we need only multiply these 2 2 2 2 2

3 3 3 3 3

4 4 4 4 4

1 3 1 2 4

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

4 4 4 4 4

2 4 1 3 2

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 Intermediate Mathematics Olympiad Solutions 4

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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.   The number of pairs of vertices of the polyhedron is 24 = 276. So the number of internal diagonals of the 2 polyhedron is 276 − 36 − 120 = 120. 6. Draw M N . D

M

Q

A

C

O

N

B

P

Method 1 Since BP and CD are parallel and BN = N C, triangles BN P and CN D are congruent (ASA). Similarly, triangles AM Q and DM C are congruent. Since AM and BN are parallel and equal, M N and AB are parallel. So ABN M and M N CD are congruent parallelograms and their areas are half the area of ABCD, that is, 192/2 = 96. Since M N CD is a parallelogram, its area is twice the area of triangle CN D, twice the area of triangle DM C, and 4 times the area of triangle M N O. So the area of triangle P OQ is 96 + 2(96/2) + (96/4) = 216. Method 2 Since BP and CD are parallel and BN = N C, triangles BN P and CN D are congruent (ASA). Similarly, triangles AM Q and DM C are congruent. So QP = 3 × DC.

Since AM and BN are parallel and equal, M N and AB are parallel. So ABN M and M N CD are congruent parallelograms and their areas are half the area of ABCD, that is, 192/2 = 96. Since M N CD is a parallelogram, the area of triangle COD is 96/4 = 24.

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

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Method 3 Parallelogram ABCD is unspecified so we may take it to be a rectangle with AD = 2AB. D

C

O

M

Q

N

A

B

P

Thus ABN M and M N CD are congruent squares, triangles N BP and N CD are congruent, and triangles M AQ and M DC are congruent. Let | | denote area. Then |ABN M | = |M N CD| =

1 192 |ABCD| = = 96 2 2

1 |M N CD| = 48 2 1 |M AQ| = |M DC| = |M N CD| = 48 2 1 |M N O| = |M N CD| = 24 4 |N BP | = |N DC| =

So |P OQ| = 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) = −1 ± b2 + 2019 a= 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.

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8. Method 1 Draw AQ, BP , and CR. Let | | denote area. A

P

R

B

Q

C

Let BR = 1 and RA = p. Noting that BQ = QC and CP = 3P A, we have: 1 1 1 300 × × |ABC| = × 2 1+p 2 1+p 1 p 1 300p × × |ABC| = × 4 1+p 4 1+p 1 3 3 × × |ABC| = × 300 2 4 8 300 |P QR| = 2|BQR| = 1+p

1 |BCR| = 2 1 |AP R| = |ACR| = 4 1 |CP Q| = |BCP | = 2

|BQR| =

Adding these areas gives         4 2 300p 3 300(p + 1) 8 300 300 + + + 8 1+p 8 1+p 8 p+1 8 1+p   300 1 (4 + 2p + 3(p + 1) + 8) = 8 p+1

300 =

8(p + 1) = 15 + 5p 3p = 7 Finally

|P QR| =

300 × 3 900 300 = = = 90. 1+p 3 + 3p 10

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Method 2 A

P

R

B

Q

C

Let | | denote area. Noting that BQ = QC and CP = 3P A, we have: 300 = |AP R| + |P QR| + |BQR| + |CQP | 1 3 1 = |ACR| + |P QR| + |P QR| + |AQC| 4 2 4 1200 = |ACR| + 6|P QR| + 3|AQC| 3 = (300 − |BCR|) + 6|P QR| + |ABC| 2 = 300 − 2|BQR| + 6|P QR| + 450 = 750 − |P QR| + 6|P QR| = 750 + 5|P QR| 1 1 |P QR| = (1200 − 750) = (450) = 90. 5 5

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Method 3 First consider the following diagram and let | | denote area. V Z

X

Y

M

N

U

M Z × XY XZ XY XY × XZ |XY Z| = = × = |XU V | N V × XU XV XU XU × XV

(1)

Next consider the given diagram with distances as shown. A b P d

3b

R c

B

a

Q

a

C

From (1) we have

Combining these gives

ac c |BQR| = = 300 2a(c + d) 2(c + d) bd d |AP R| = = 300 4b(c + d) 4(c + d) 3ab 3 |CP Q| = = 300 (4b)(2a) 8 300 = |BQR| + |AP R| + |CP Q| + |P QR| = 3|BQR| + |AP R| + |CP Q| 3c d 3 1= + + 2(c + d) 4(c + d) 8 8(c + d) = 12c + 2d + 3(c + d) = 15c + 5d d = 7c/3

So |BQR| =

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

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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 = 1 or 2. If a = 2, then c + d = 18, c = d = 9, and the second equation gives 99 = 162b, which is impossible. If a = 1, then c + d = 9 and the second equation gives 9c + 9 = bc(9 − c)

and

9 = c(b(9 − c) − 9).

So c = 1, 3, or 9. If c = 1, then 18 = 8b, which is impossible. If c = 9, then 1 = 0 − 9, which is a contradiction. 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}.

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Method 2 Let s = 10a + b and p = 10c + d, where a, b, c, and d are positive digits. Then 10a + b = a + b + c + d

and

10c + d = abcd.

The second equation gives d = c(abd − 10). Since d is a positive digit, 10 < abd < 20. Since a, b, d are digits and 11, 13, 17, 19 are primes, abd = 12, 14, 15, 16, or 18. If abd = 18, then 18c = 10c + d, d = 8c, and 4abc = 9, which is impossible. If abd = 16, then 16c = 10c + d, d = 6c, and 3abc = 8, which is impossible. If abd = 15, then 15c = 10c + d, d = 5c. So c = 1, d = 5, and 9a = 6, which is impossible. If abd = 14, then 14c = 10c + d, d = 4c, and 2abc = 7, which is impossible. If abd = 12, then 12c = 10c + d, and d = 2c. So d is even and divides 12. Hence d = 2, 4, or 6, and correspondingly c = 1, 2, or 3. Since 10a + b = a + b + c + d, we have 9a = c + d. So 9 divides c + d. Hence d = 6, c = 3, a = 1, b = 2, and we have the compatible numbers s = 12 and p = 36. Hence there are only 2 pairs of compatible numbers less than 100, namely {9, 11} and {12, 36}.

Investigation Let s and p be compatible numbers, where s is the sum of their digits and p is the product of their digits. 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, 9 and 2bcde < 1000, then a check whether any 2bcde could be p = 100c + 10d + e. {c, d, e} 1, 8, 9 2, 7, 9 2, 8, 8 3, 6, 9 3, 7, 8 4, 5, 9 4, 6, 8 4, 7, 7 5, 5, 8 5, 6, 7 6, 6, 6

2cde 144 252 256 324 336 360 384 392 400 420 432

last few digits of 2bcde 4, 88, 2, 6, 0, 4 52, 4, 6 6, 12, 68 4, 8, 2 6, 2 0, 0, 0, 0, 0, 0, 0, 0, 0 384, 768 2, 84 0, 0, 0, 0, 0, 0, 0, 0, 0 0, 0, 0, 0, 0, 0, 0, 0, 0 2, 4

any p? none none none none none none none none none none 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} 1, 1, 7 1, 2, 6 1, 3, 5 1, 4, 4 2, 2, 5 2, 3, 4 3, 3, 3

cde 7 12 15 16 20 24 27

last few digits of bcde 07, 4, 21, 8, 5, 2, 9, 6, 3 012, 4, 36, 8, 0, 72, 4, 96, 8 015, 0, 45, 0, 75, 0, 05, 0, 135 6, 2, 8, 64, 0, 6, 2, 8, 144 0, 0, 0, 0, 0, 0, 0, 0, 0 024, 8, 72, 6, 0, 44, 8, 92, 6 7, 4, 1, 8, 5, 2, 9, 6, 43

any p? none none 135 144 none none none

Hence the only compatible pairs are {135, 19} and {144, 19}. Mathematics Contests The Australian Scene 2018 | Australian Intermediate Mathematics Olympiad Solutions 11

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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}.

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AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD STATISTICS Distribution of Awards/School Year

Year

Number of Students

8

Number of Awards Prize

High Distinction

Distinction

Credit

Participation

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 School Year

Number of Students

8

Question

Overall Mean

1–8

9

10

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

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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 P D 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 

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2018 AMOC SENIOR CONTEST Solutions AMOC SENIOR c 2018 Australian  Mathematics Trust

CONTEST SOLUTIONS

1. Determine the maximum possible value of a + b, where a and b are two different nonnegative real numbers that satisfy a+

√ √ b = b + a.

Solution (Angelo Di Pasquale) The equation may be rewritten as √

a−

√

b=a−b

⇔

√ √ √ √ √ √ a − b = ( a − b)( a + b)

⇔

√ √ a + b = 1,

since a = b. Squaring the last equation and rearranging yields √ a + b = 1 − 2 ab

⇒

a + b ≤ 1.

Since a = 0 and b = 1 satisfy the original equation and satisfy a + b = 1, it follows that the maximum possible value of a + b is 1.

1

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2. Prove that, among any ten consecutive positive integers, there are five numbers such that no two of them have a common factor larger than 1.

Solution 1 (Norman Do) Among any ten consecutive positive integers, five of them are odd and the remaining five are even. The five odd integers contain either one or two multiples of 3. We remove one multiple of 3 and select the remaining four integers. The five even integers contain at most two multiples of 3, at most one multiple of 5, and at most one multiple of 7. Therefore, there is at least one of them that is not a multiple of 3, 5 or 7. We select this integer. We now prove that, among the five integers selected, no two of them have a common factor larger than 1. Since the largest possible difference between two of the numbers is 9, the largest possible common factor that two of the numbers can have is 9. So it suffices to show that we have not selected a pair of numbers with a common factor of 2, 3, 5 or 7. By construction, we have chosen exactly one number that is divisible by 2. By construction, we have chosen at most one number that is divisible by 3. At most one of the odd integers selected is divisible by 5 and, by construction, the even integer selected is not divisible by 5. At most one of the odd integers selected is divisible by 7 and, by construction, the even integer selected is not divisible by 7. Therefore, we have selected five numbers such that no two of them have a common factor larger than 1.

Solution 2 (Alice Devillers, Angelo Di Pasquale and Daniel Mathews) Among the ten consecutive positive integers, let n be the smallest number that is not divisible by 2 or 3. Either the smallest or the second smallest odd number among the ten numbers satisfies this condition. So n is one of the four smallest integers and the five numbers n, n + 2, n + 3, n + 4, n + 6 are all among the ten consecutive integers. We proceed to calculate all possible greatest common divisors between two of these five numbers. We begin with the observation that for any integer n, we have gcd(n + 2, n + 3) = gcd(n + 3, n + 4) = 1. Since n is not divisible by 2, we have gcd(n, n + 2) = gcd(n, n + 4) = gcd(n + 2, n + 4) = gcd(n + 2, n + 6) = gcd(n + 4, n + 6) = 1. Since n is not divisible by 3, we have gcd(n, n + 3) = gcd(n + 3, n + 6) = 1. 2 Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions

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Since n is not divisible by 2 or 3, we have gcd(n, n + 6) = 1. This exhausts all possibilities, so no two of the five numbers n, n + 2, n + 3, n + 4, n + 6 have a common factor larger than 1.

Solution 3 (Kevin McAvaney) Any common factor of two integers must divide their difference. So any common factor of two of the ten integers is at most 9. Therefore, for such a common factor to be greater than 1, it must have 2, 3, 5 or 7 as a prime factor. The sequence of remainders of the ten consecutive integers after division by 2 must be one of 0101010101 or 1010101010. The sequence of remainders of the ten consecutive integers after division by 3 must be one of 0120120120 or 1201201201 or 2012012012. For the resulting six cases, we choose the five numbers from among the ten original integers corresponding to the underlined remainders shown below. 0101010101

0101010101

0101010101

0120120120

1201201201

2012012012

1010101010

1010101010

1010101010

0120120120

1201201201

2012012012

One can check that in each of the six cases, none of the primes 2, 3, 5 or 7 is a factor of more than one of the chosen integers. So in each case, no two of the five chosen numbers have a common factor lager than 1.

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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

W LW

W LL

LW W

LW L

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 W W , W L or LW . By the pigeonhole principle, at least three of these seven players had the same record for matches 2 and 3. Therefore, these three players were on the same team as each other for all three matches. 4 Mathematics Contests The Australian Scene 2018 | AMOC Senior Contest Solutions

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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.

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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 P D 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 P D. 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 = ∠P BO, but ∠P BO = ∠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 P D, 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 = ∠CP Q and we have ∠AP D = ∠BP D = 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.

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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 , 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.) ⇒

xf (x) = a (1 + xf (x))

Substituting f (x) =

c x

f (x) =

into the functional equation yields 2y = y(1 + c).

Therefore, c = 1 and we deduce that f (x) = solution to the functional equation.

1 x.

It is easily verified that this is indeed a

Solution 2 (Angelo Di Pasquale) Replace y with

y 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 xf (xf (f (y))) = f (y)(1 + xf (x)) 2

⇒

f (axy) =

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

Interchanging x with y in this equation yields f (ayx) =

f (x) f (x)f (y) + . 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.

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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 QLD

10

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

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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

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AMOC SENIOR CONTEST STATISTICS Score Distribution/Problem Problem Number

Number of Students/Score

Mean

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

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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

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Participants at the 2017 AMOC School of Excellence Name

m/f

Year

School

State

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

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

Hadyn Tang

m

8

Guowen Zhang

m

Stanley Zhu

Senior

NSW

NSW

Trinity Grammar School

VIC

11

St Joseph’s College, Gregory Terrace

QLD

m

11

Melbourne Grammar School

VIC

Andres Buritica

m

8

Scotch College

Liam Coy

m

9

Sydney Grammar School

Yifan Guo

f

11

Grace He

f

9

Jeffrey Li

m

10

Adrian Lo

m

9

Preet Patel

m

10

Vermont Secondary College

VIC

Ken Gene Quah

m

9

Melbourne High School

VIC

Mikhail Savkin

m

9

Gosford High School

Ruiqian Tong

f

11

Daniel Wiese

m

Fengshuo (Fredy) Ye (Yip)

Intermediate VIC NSW

Glen Waverley Secondary College

VIC

Methodist Ladies' College

VIC

North Sydney Boys High School

NSW

Newington College

NSW

NSW

Presbyterian Ladies' College

VIC

9

Scotch College

WA

m

7

Chatswood High School

Ziqi Yuan

m

10

Lyneham High School

ACT

Xinyue (Alice) Zhang

f

11

A. B. Paterson College

QLD

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

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

NSW

Junior

Presbyterian Ladies’ College

VIC

*Equivalent to year 12 in Australia.

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A U S T R A L I A N M AT H E M AT I C A L O LY M P I A D C O M M I T T E E

AUSTRALIAN MATHEMATICAL OLYMPIAD

A DEPARTMENT OF THE AUSTRALIAN MATHEMATICS TRUST

A U S T R A L I A N M AT H E M AT I C A L O LY M P I A D C O M M I T T E E A DEPARTMENT OF THE AUSTRALIAN MATHEMATICS TRUST

AUSTRALIAN MATHEMATICAL OLYMPIAD 2018

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

OLYMPIAD

DAY 2 Wednesday, 7 February 2018 1. allowed: Find all pairs of positive integers (n, k) such that Time 4 hours 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 2. The Consider a line 1) equally points marked on it. 5. sequence a1with , a2 , a23(3 , . . . is+defined by aspaced 1 = 1 and, for n ≥ 2, Prove that 2100 of these marked points can be coloured red so that no red point is at the an = 1 + a2 + · · · + an−1 ) × n. same distance from two other red(apoints. 2. is divisible N by be 2018 that a2018GHIJKLM 3. Prove Let ABCDEF 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 C, such that P Q = P R and P Q > QR. (A regular tetradecagon is a convex polygon 14 sides, that all that sidesthe have the Let D be the circle with centre P that passes with through Q andsuch R. Suppose circle samecentre lengthQand angles through are equal.) with andallpassing R intersects C again at X and D again at Y . 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 of fpositive (xy + fintegers (y)) = yfsuch (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 example, if numbers b1 = 27, then for all real x andb2y.= 42 = 16, since 27 has four positive factors: 1, 3, 9 and 27. Prove that if b1 > 1, then the sequence contains a term that is equal to 9. 8. Amy has a number of rocks such that the mass of each rock, in kilograms, is a positive integer. The sum of the masses of the rocks is 2018 kilograms. Amy realises that it is impossible to divide the rocks into two piles of 1009 kilograms. What is the maximum possible number of rocks that Amy could have? The Mathematics/Informatics Olympiads are supported by the Australian Government through the National Innovation and Science Agenda.

Official sponsor of the Olympiad program

©

2018 Australian Mathematics Trust

Mathematics Contests The Australian Scene 2018 | Australian Mathematical Olympiad

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A U S T R A L I A N M AT H E M AT I C A L O LY M P I A D C O M M I T T E E A DEPARTMENT OF THE AUSTRALIAN MATHEMATICS TRUST

A U S T R A L I A N M AT H E M AT I C A L O LY M P I A D C O M M I T T E E A DEPARTMENT OF THE AUSTRALIAN MATHEMATICS TRUST

AUSTRALIAN MATHEMATICAL OLYMPIAD 2018 AUSTRALIAN MATHEMATICAL OLYMPIAD DAY 2 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. Prove that a2018 is divisible by 20182 . 5. The sequence a1 , a2 , a3 , . . . is defined by a1 = 1 and, for n ≥ 2, 6. Let P , Q and R be three points on a circle C, such that P Q = P R and P Q > QR. an = (a1 + a2 + · · · + an−1 ) × n. Let D be the circle with centre P that passes through Q and R. Suppose that the circle with centre Q and passing through R intersects C again at X and D again at Y . Prove that a2018 is divisible by 20182 . Prove that P , X and Y lie on a line. 6. Let P , Q and R be three points on a circle C, such that P Q = P R and P Q > QR. b1 ,be b2 ,the b3 , .circle . . be with a sequence integers suchQthat, each positive n, 7. Let D centre of P positive that passes through and for R. Suppose that integer the circle bn+1 centre is the Q square of the number of at bnX(including 1 and with and passing through of R positive intersectsfactors C again and D again at Ybn.). For 2 example, if b = 27, then b = 4 = 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, hasthe a number of the rocksnumber such that the massfactors of eachofrock, in kilograms, is abpositive 8. Amy bn+1 is square of of positive bn (including 1 and n ). For 2 integer. The sum of the masses of the rocks is 2018 kilograms. Amy realises that is example, if b1 = 27, then b2 = 4 = 16, since 27 has four positive factors: 1, 3, 9 andit27. 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?

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2018 Australian Mathematics Trust The Mathematics/Informatics Olympiads are supported by the Australian Innovation and Science Agenda.

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AUSTRALIAN MATHEMATICAL OLYMPIAD SOLUTIONS Solutions to the 2018 Australian Mathematical Olympiad 1. Solution 1 (Evgeniya Artemova, year 11, Presbyterian Ladies’ College, VIC) Answers (n, k) = (4, 5) and (5, 7). For reference the given equation is n! + 8 = 2k . Case 1 n ≥ 6

Observe that n! is a multiple of 6! = 24 × 32 × 5. Hence n! = 16x for some 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! + 8 and check which ones are powers of 2. n 1 2 3 4 5

n! + 8 9 10 14 32 128

power of 2? no no no 25 27

This yields the solutions given at the outset.



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

1



For 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.

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2. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA) Given some equally spaced points marked on a line, if we colour some of them red, we say the colouring is good if no red point is equidistant from two other red points. The result is the special case n = 100 of the following more general claim. Claim For each positive integer n, if a line has 12 (3n + 1) equally spaced points marked on it, then there is a good colouring of 2n of those points. The proof is by induction on n. The base case n = 1 holds as we simply colour each of the 12 (31 + 1) = 21 points red. For the inductive step, suppose that the claim is true for some positive integer n. Let k = 12 (3n + 1). Note that 3k − 1 = 12 (3n+1 + 1). To prove the claim for n + 1 we divide the 3k − 1 points into three groups from left to right as follows. k points group A

k − 1 points group B

k points group C

Using the inductive assumption we colour 2n points in group A red so that no 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 W X = 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 W X ≤ 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

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Solution 2 (Sharvil Kesarwani, year 11, Merewether High School, NSW) Without loss of generality, we identify the 12 (3100 + 1) equally spaced points with the integers from 0 up to 12 (3100 − 1) on the real number line. Colour red each integer of the form

N=

99  i=0

d i · 3i

where di ∈ {0, 1} for 0 ≤ i ≤ 99. In this way exactly 2100 integers are coloured red. Note that these are precisely the integers in the range from 0 to 12 (3100 − 1) which do not contain the digit 2 in their ternary (base-3) representations. Suppose, for the sake of contradiction, that one red integer B is equidistant from two other red integers A and C, where A < B < C. Thus C − B = B − A, which is the same as 2B = A + C. Let A=

99  i=0

i

ai · 3 ,

B=

99  i=0

i

bi · 3 ,

and C =

99  i=0

ci · 3i

be the ternary representations of A, B, and C, where ai , bi , ci ∈ {0, 1} for 0 ≤ i ≤ 99. From 2B = A + C we have 99  i=0

i

2bi · 3 =

99 

(ai + ci )3i .

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 < C, and completes the proof. 

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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 F

K

E

L

D

M 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. 

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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 F

K

E

L

D

M 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. 

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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. H

I J

G F

K

E

L

D

X

M 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. 

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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 F

K

E

L

D

X

M 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. 

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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 F

K

E

L

D

M N

C A

B

We quote a theorem that will help us solve the problem. Theorem If U V W XY Z is a convex cyclic hexagon, then its main diagonals U X, V Y , and W Z are concurrent if and only if U V · W X · Y Z = V W · 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.

2

Published by the AMT

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4. Solution 1 (William Hu, year 12, Christ Church Grammar School, WA) Answers f (x) = 0 and f (x) = 1 − x

For reference the given functional equation is f (xy + f (y)) = yf (x)

(1)

f (f (0)) = 0.

(2)

for all real numbers x and y. Put y = 0 in (1) to find Next put x = 0 and y = f (0) in (1), and use (2) to deduce f (0) = f (0)2 . Thus f (0) = 0 or 1. Case 1 f (0) = 0 Put x = 0 in (1) to 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) = 0 for all real numbers x. This function obviously satisfies (1). Case 2 f (0) = 1 Putting x = 0 into (1) yields f (f (y)) = y

(5)

for all real numbers y. It follows that f (1) = f (f (0)) = 0. Putting x = 1 in (1) yields f (y + f (y)) = 0 = f (1)

⇒

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

With the help of (5) this implies y + f (y) = 1, and so f (y) = 1 − y.

It only remains to verify that f (x) = 1 − x satisfies (1). Indeed we have LHS(1) = 1 − (xy + (1 − y)) = y(1 − x) = RHS(1). Hence f (x) = 1 − x is also a solution to the problem.



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Solution 2 (William Steinberg, year 10, Scotch College, WA) For reference the given functional equation is f (xy + f (y)) = yf (x)

(1)

f (f (y)) = yf (0),

(2)

for all real numbers x and y. Put x = 0 in (1) to find for all real numbers y. Putting y = 1 in (1) yields f (x + f (1)) = f (x) ⇒ f (f (x + f (1))) = f (f (x)) ⇒ (x + f (1))f (0) = xf (0) (from (2)) ⇒ f (0)f (1) = 0.

(3)

Two cases follow from (3). Case 1 f (0) = 0 Equation (2) now 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) = 0 for all real numbers x. Case 2 f (0) = 0

It follows from (3) that f (1) = 0. Setting x = 1 in (1), and using f (1) = 0 yields f (y + f (y)) = 0 ⇒ f (f (y + f (y))) = f (0) ⇒ (y + f (y))f (0) = f (0) (from (2)) ⇒ y + f (y) = 1. (since f (0) = 0) Thus f (y) = 1 − y.

As in solution 1, we check that f (x) = 0 and f (x) = 1 − x each satisfy the given functional equation. 

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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 )) = n2 (a1 + a2 + · · · + an−2 ).

Hence an is divisible by n2 for each integer n ≥ 3. In particular, a2018 is divisible by 20182 .  Comment We did not use the initial condition a1 = 1. Hence the result is true for any integer a1 .

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Solution 2 (Xinyue (Alice) Zhang, year 12, A. B. Paterson College, QLD) We prove an even stronger result, namely a2018 is divisible by 20183 . First we prove by induction that a1 + a2 + · · · + an−1 = n!/2 for all integers n ≥ 2. The base case is true because a1 = 1 =

2! . 2

For the inductive step, assume that a1 + a2 + · · · + an−1 = n!/2 for some 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 a2018 = 2018(a1 + · · · + a2017 ) = 2018(2018 × 2017 × · · · × 3) = 20182 (2017 × 2016 × · · · × 1009 × · · · × 4 × 3) which is obviously a multiple of 20182 . However 2017 × 2016 × · · · × 1009 × · · · × 4 × 3 is divisible by both 2 and 1009, and so it is also divisible by 2018. Therefore a2018 is divisible by 20183 . 

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Solution 3 (Andres Buritica, year 9, Scotch College, VIC) We are given

an = a1 + a2 + · · · + an−1 n

(1)

for any integer n ≥ 2.

Replacing n with n + 1 in (1), we find

⇒

an+1 = a1 + a2 + · · · + an−1 + an n+1 an + an = n an+1 (n + 1)2 = an n

(2)

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.

⇒ ⇒

am (m − 1)2 32 am−1 a3 m2 × × ··· × × × ··· × = am−1 am−2 a2 m−1 m−2 2 2 m (m − 1)! am = a2 4 2 m (m − 1)! am = 2

In particular a2018 = (2018)2 ×

2017! , 2

which is a multiple of 20182 .



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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 C

Q

P

X Y

In circle D, chords QY and QR have equal length. Hence they subtend equal angles at the centre P of this circle. Thus ∠Y P Q = ∠QP R.

(1)

In circle C, chords QX and QR have equal length. Hence they subtend equal angles at the point P on the circumference of this circle. Thus ∠XP Q = ∠QP R. From (1) and (2) we have ∠XP Q = ∠Y P Q. Thus P , X, and Y are collinear.

(2) 

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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 P Y for a second time at X  . It suffices to prove that P RQX  is cyclic as this implies X  = X.

R

Q

P

X Y

As P and Q are centres of circles, we have P R = P Q = P Y and QR = QX  = QY . Hence P QR ≡ P QY (SSS). Thus ∠P RQ = ∠QY P = ∠QY X  = ∠Y X  Q where the last equality is due to QX  = QY . Since ∠P RQ = ∠Y X  Q, it follows that P RQX  is cyclic, as desired.



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Solution 3 (Sharvil Kesarwani, year 11, Merewether High School, NSW) Let Y  be the intersection of ray P X with circle D. It suffices to prove that Y = Y  . Let ∠QY  X = ∠QY  P = α.

D

R C

Q

P

X

α Y

Observe that P Y  = P Q = P R because they are radii of circle D. It follows that ∠P QY  = ∠QY  P = α. Considering the angle sum in P QY yields ∠Y  P Q = 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 C, they subtend equal angles at the point P on the circumference of this circle. Hence ∠QP R = ∠XP Q = ∠Y  P Q = 180◦ − 2α. Since P QR is isosceles with apex P , the angle sum in this triangle yields ∠P RQ = ∠RQP = α. Since P RQX is a cyclic quadrilateral, we have ∠Y  XQ = ∠P RQ = 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 P Q, it follows that Y = Y  . 

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Solution 4 (Alan Offer, AMOC Senior Problems Committee) Comment Let E be the circle in the problem statement centred at Q and passing through R. If we relax the condition that Q is the centre of E by requiring only that the centre of E lies on C, then the conclusion of the problem is still true. Here is a proof of this generalisation of the problem.

D

R C

Q

P X

O

Y

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 E. Since P R = P Y (radii of D), and OR = OY (radii of E), it follows that OP R ≡ OP Y (SSS). Thus ∠Y P O = ∠OP R.

(1)

In circle C, chords OX and OR have equal length. Hence they subtend equal angles at the point P on the circumference of this circle. Thus ∠XP O = ∠OP R. From (1) and (2) we have ∠XP O = ∠Y P O. Thus P , X, and Y are collinear.

(2) 

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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 P Q 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 C. Let D be a circle centred at P and passing through R, and let E be the circle centred at O and passing through R. Suppose that E intersects C again at X and D again at Y . Then P , X, and Y lie on a line. In fact we can say even more! There is a combinatorial symmetry in the problem statement between D and E with respect to C. Specifically, exchanging points P and O also exchanges D and E. Thus if D intersects C again at Q, then O, Q, and Y also lie on a line.

D C

R

P

Q

E

O

X

Y

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7. Solution 1 (Tianyue (Ellen) Zheng, year 12, Smith’s Hill High School, NSW) Since b1 > 1, it follows inductively that bn > 1 for all n ≥ 2. Observe that bn is a perfect square for all n ≥ 2.

For any perfect square m2 , all of the factors of m2 , except for m, come in pairs (d, m2 /d) where 1 ≤ d < m. Hence square numbers have an odd number of factors. Thus bn is an odd perfect square for all n ≥ 3. Moreover since bn > 1, it follows that bn ≥ 9 for all n ≥ 3.

Consider any integer n ≥ 3. Then bn = m2 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 < m. Since m2 is odd, each such d is odd. The number of odd positive integers less than m is equal to m−1 . It follows that the number of factors of m2 is 2 at most 2 × m−1 + 1 = m. Therefore bn+1 ≤ bn . 2 However, for bn+1 = bn to occur, all of the odd positive integers less than m must be factors of m2 . In particular m − 2 | m2 . But gcd(m, m − 2) = gcd(m, 2) = 1

⇒

gcd(m2 , m − 2) = 1.

Since m − 2 | m2 , it follows that m − 2 = 1. Thus m = 3, and bn = 9. So from b3 onward, the sequence is strictly decreasing until it reaches the fixed point bn = 9. 

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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 bn+1 ≤ bn with equality if and only if bn = 9. r  i Let bn = p2a i , where p1 < p2 < · · · < pr are the prime factors of bn . i=1

The number of factors of bn is given by

r 

(2ai + 1). Hence bn+1 =

i=1

It follows that bn+1 ≤ bn if and only if r  i=1

pai i ≥

r 

r 

(2ai + 1)2 .

i=1

(2ai + 1).

(1)

i=1

To finish the proof it suffices to show that pa ≥ 2a + 1

(2)

with equality if and only if p = 3 and a = 1. Since pa > 3a whenever p > 3, it suffices to prove (2) just for the case p = 3. 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 + 1 for some a ≥ 1. It follows that 3a+1 ≥ 3(2a + 1) > 2(a + 1) + 1. This concludes the induction and the proof.



Comment Here are two more alternative ways of proving inequality (2). Alternative 1 (Angelo Di Pasquale, Director of Training AMOC) Using the binomial theorem, the inequality follows from     2a 2a 2a 2a · 22 = 8a2 + 1 ≥ (2a + 1)2 . ·2+ p ≥ (1 + 2) ≥ 1 + 2 1



Alternative 2 (Ivan Guo, AMOC Senior Problems Committee) Factoring the difference of perfect ath powers, the inequality follows from pa − 1 ≥ 3a − 1 = (3 − 1)(3a−1 + 3a−2 + · · · + 1) ≥ 2(1 + 1 + · · · + 1) = 2a. 

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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 Si = m1 + m2 + · · · + mi . Note that S1 , S2 , . . . , Sr is a strictly increasing sequence of r ≥ 1010 positive integers, each of which is in the range 1, 2, . . . , 2018. Consider the 1009 pairs (1, 1010), (2, 1011), (3, 1012), . . . , (1009, 2018). By the pigeonhole principle one of the above pairs contains two of the Si . 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 m1 , m2 , . . . , mn . Let Si = m1 +m2 +· · ·+mi for 1 ≤ i ≤ n. If Si ≡ 0 (mod n) for some i, then we are done.

Otherwise, by the pigeonhole principle we have Si ≡ Sj (mod n) for some i < j. It follows that Sj − Si = mi+1 + mi+2 + · · · + mj ≡ 0 (mod n). 

To complete the proof of the given problem, if we have at least 1010 rocks, consider any 1009 of them. By the lemma, a nonempty sub-collection of those rocks has mass equal to a multiple of 1009. Since the total mass of the sub-collection is strictly less than 2018, it is equal to 1009. 

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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 a + 2(b − 1) + M ≤ 2018

(1) (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. 

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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 + 1 rocks 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 + 2 rocks whose total mass is 2k + 2. If all of Amy’s rocks have mass greater than or equal to 2, then the total mass 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 + 1 and mr > 1.

Consider the situation where we have r rocks of masses m1 , m2 , . . . , mr−1 , mr − 1.

Observe that the number of rocks above is greater than or equal to k + 1, and 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 mr − 1 to mr and add 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 + 1 rocks 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.

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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

12

St Joseph's College QLD 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

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

9 10

Ziang (Tommy) Wei

Knox Grammar NSW Queensland Academy for Science, Mathematics and Technology QLD Scotch College VIC

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

Zirui (Harry) Zhang

Christian Brothers College VIC

9

Yufei (Phoebe) Zuo

Tara Anglican School for Girls NSW

12

Jianyi (Jason) Wang

Year

12

11*

*NZ school year

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AUSTRALIAN MATHEMATICAL OLYMPIAD STATISTICS Score Distribution/Problem Number of Students/Score

Problem Number 1

2

3

4

5

6

7

8

0

2

49

59

58

14

25

44

18

1

2

4

10

15

0

4

11

65

2

14

3

1

6

5

23

13

7

3

2

2

0

3

0

0

6

1

4

0

0

0

2

0

1

1

1

5

0

2

0

3

0

1

6

1

6

20

12

1

4

4

2

4

3

7

81

49

50

30

98

65

36

25

Average

6

3.6

3

2.4

6

4.3

3

2.3

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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 BM N C and that the circumcircles of triangles BM H and CN H are tangent to each other. The line through H parallel to BC intersects the circumcircles of the triangles BM H and CN H in the points K and L, respectively. Let F be the intersection point of M K and N L and let J be the incenter of triangle M HN . Prove that F J = F A. Problem 2. Let f (x) and g(x) be given by f (x) = and g(x) = Prove that

1 1 1 1 + + + ··· + x x−2 x−4 x − 2018

1 1 1 1 + + + ··· + . x−1 x−3 x−5 x − 2017 |f (x) − g(x)| > 2

for any non-integer real number x satisfying 0 < x < 2018. Problem 3. A collection of n squares on the plane is called tri-connected if 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.

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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.

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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. A

F

M

N J

Z

Y K

L

H

B

C

Since M and N are the midpoints of AB and AC, respectively, it follows that M N  BC  KL. Using this and cyclic BKM H yields ∠N M F = ∠LKF = ∠HKM = ∠HBM = x. Similarly,

∠F N M = ∠F LK = x.

The above angle equalities imply F K = F L and F M = F N . Hence F M · F K = F N · F L. Thus F has equal power with respect to circles BKM H and CHN L. 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 F H is tangent to the two circles at H. Using the alternate segment theorem we deduce ∠F HM = ∠HBM = x and ∠N HF = ∠N CH = x. In particular F H bisects ∠N HM , and so F , J, and H are collinear. Another consequence is that F M HN is cyclic due to ∠N M F = x = ∠N HF. But now we have a 51 Mathematics Contests The Australian Scene 2018 | 30th Asian Pacific Mathematics Olympiad Solutions

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standard configuration where J is the incentre of M N H and F is the intersection of the circumcircle of M N H with the line HJ.1 In this configuration it is known that F M = F J = F N , but here is a proof anyway. We calculate ∠F JM = ∠JHM + ∠HM J = x + ∠JM N = ∠JM F. Thus F M = F J. A similar argument shows that F N = F J. Thus F is the centre of circle M JN . We also calculate N JM = 180◦ − ∠JM N − ∠M N J 1 = 180◦ − (∠HM N + ∠M N H) 2 1 = 180◦ − (180◦ − ∠N HM ) 2 1 = 180◦ − (180◦ − 2x) 2 = 90◦ + x. Also ∠M AN = ∠BAC = 90◦ −x, and so ∠N JM +∠M AN = 180◦ . Thus M JN A is cyclic. Recall F is the centre of circle M JN . Thus F is the centre of circle M JN A, and so F A = F J, as desired. 

1

See configuration B6 in chapter 5 of Problem Solving Tactics published by the AMT.

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Solution 2 (William Hu, year 12, Christ Church Grammar School, WA) As in solution 1, we have KL  BC  M N , and chase out the following equal angles. ∠N M F = ∠HKM = ∠HBM = ∠N CH = ∠N LH = ∠F N M = x

A

F y M

z

x

x

N

J P x

x

K

L

H x

x y

z

B

C

Let ∠CBH = y and ∠HCB = z. From M N  BC we have ∠N M A = x + y and ∠AN M = x + z. Since ∠N M F = ∠F N M = x, it follows that ∠F M A = y

and ∠AN F = z.

Observe that AM N 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 AM N . Also we have ∠N M H  = ∠CBH = y = ∠F M A, and ∠H  N M = ∠HCB = z = ∠AN F . Hence H  and F are isogonal conjugates in AM N . But the orthocentre and circumcentre of any triangle are isogonal conjugates, and so F is the circumcentre of AM N . 2 In order to complete the proof it only remains to show that M JN A is cyclic because then F would be the centre of circle M JN A. 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 ∠N HM = ∠N HP + ∠P HM = ∠N LH + ∠HKM = 2x. Using the same angle calculations near the end of solution 1, we deduce that ∠N JM = 90◦ + x, and ∠M AN = 90◦ − x. Consequently AM JN is cyclic due to ∠N JM + ∠M AN = 180◦ , as desired.  2

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

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2. Solution (Based on the solution by James Bang, year 11, Baulkham Hills High School, NSW) There are two cases: 2n − 1 < x < 2n and 2n < x < 2n + 1. Note that f (2018 − x) = −f (x) and g(2018 − x) = −g(x), that is, a half turn about the point (1009, 0) preserves the graphs of f and g. So it suffices to consider only the case 2n < x < 2n + 1 where n is an integer with 0 ≤ n ≤ 1008. We have the following. 1009 

1  f (x) − g(x) = + x i=1 1009

1 1 − x − 2i x − (2i − 1)



1 1  = + x i=1 (x − (2i − 1))(x − 2i)

(1) (2)

Observe that x − (2i − 1) and x − 2i are both positive if i ≤ n, and they are both 1 negative if i ≥ n + 1. Thus (x−(2i−1))(x−2i) > 0 for each i. Case 1 0 < x < 1

By dropping all terms in (2) with i ≥ 2, we see that f (x) − g(x) >

1 1 + . x (x − 1)(x − 2)

Thus it suffices to show 1 1 + >2 x (x − 1)(x − 2) ⇔ (x − 1)(x − 2) + x > 2x(x − 1)(x − 2) ⇔ −2x3 + 7x2 − 6x + 2 > 0 ⇔ x2 + 2(1 − x)3 > 0.

(3) (4) (5)

Note that (4) follows from (3) because x(x − 1)(x − 2) > 0 for 0 < x < 1.

It remains to observe that (5) is obviously true for 0 < x < 1.

Case 2 2n < x < 2n + 1 where n is an integer with 1 ≤ n ≤ 1008

By dropping all terms in (2) except for those with i = n and i = n + 1, we see that f (x) − g(x) >

1 1 + . (x − (2n − 1))(x − 2n) (x − (2n + 1))(x − (2n + 2))

Let us use the substitution y = x − 2n, so that 0 < y < 1. It suffices to prove that 1 1 + >2 y(y + 1) (y − 1)(y − 2) ⇔ (y − 1)(y − 2) + y(y + 1) > 2y(y + 1)(y − 1)(y − 2) ⇔ −y 4 + 2y 3 + 2y 2 − 3y + 1 > 0 ⇔ y 3 (1 − y) + y(y − 1)2 + (2y − 1)2 > 0.

(6) (7) (8)

Note that (7) follows from (6) because y(y + 1)(y − 1)(y − 2) > 0 for 0 < y < 1.

It remains to observe that (8) is obviously true for 0 < y < 1.



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Comment 1 A shortcut can be made in the preceding solution as follows. 1009 

 1 = f (x) − g(x) − x+1 i=0 ⇒

f (x) − g(x) >

1009  i=0

1 1 − x − 2i x − (2i − 1)



1 (x − (2i − 1))(x − 2i)

Having the above inequality removes the need to treat 0 < x < 1 as a separate case. Comment 2 Here is an alternative proof of inequality (6) from the preceding solution.

⇔ ⇔ ⇔

1 1 + y(y + 1) (y − 1)(y − 2) 1 1 1 1 − + − y y+1 y−2 y−1 1 1 1 1 − + − y y−1 y−2 y+1 1 1 1 + − y(1 − y) y − 2 y + 1

>2 >2 >2 > 2.

However since y and 1 − y are both positive, the GM–HM inequality yields  2 1 1 2 · ≥ = 4. y 1−y y + (1 − y) Also for 0 < y < 1 we have 1 > −1 and y−2

−

1 > −1. y+1

Putting it all together yields 1 1 1 + − > 4 − 1 − 1 = 2, y(1 − y) y − 2 y + 1 

as desired.

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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 total number of edges of G is equal to 3n . This implies that n is even. 2 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 triconnected 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 a A-pieces and b B-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

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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)

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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 + 1 for 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 + 5 for 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 = 5 and n = 17.

Case 5 n = 5

From (4), we require r + s = 4. And r ≡ s (mod 3) from (1). Thus (r, s) = (2, 2). But this violates (2). So there are no solutions in this case. Case 6 n = 17 From (4), we require r + s = 10. And r ≡ s (mod 3) from (1). Thus (r, s) = (5, 5), (2, 8), or (8, 2). But these violate (2). So there are no solutions in this case either. Having covered all cases, the proof is complete.



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5. Solution (Based on the presentation by William Hu, year 12, Christ Church Grammar 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) = a0 + a1 x + · · · + an xn for some integers a0 , a1 , a2 , . . . , an where an =  0. Let S be the set of P -good real numbers. Clearly Z ⊆ S. Also the problem statement informs us that if r, s ∈ S then rs ∈ S. Consider the polynomial

Q(x) = 2n P (x) − P (2x) =

n−1  i=0

(2n − 2i )ai xi .

If r ∈ S, then since 2 ∈ S, we have 2r ∈ S. Thus Q(r) = 2r P (r) − P (2r) ∈ Z. So we have shown the following. Each real number that is P -good is also Q-good.

(1)

We claim that a1 = a2 = · · · = an−1 = 0. Since 2n − 2i = 0 for i < n, this is the same as saying that Q is a constant polynomial. Suppose, for the sake of contradiction, that deg(Q) = m where 1 ≤ m ≤ n−1. Since P and Q are polynomials of positive degree, both are strictly monotonic continuous functions for all sufficiently large real numbers u. But any strictly monotonic continuous real function is also a bijection. So P attains each integer value between P (u) and P (u + 1) inclusive exactly once. Hence there are exactly 1 + |P (u + 1) − P (u)| P -good real numbers in the interval [u, u + 1]. We similarly reason that there are exactly 1 + |Q(u + 1) − Q(u)| Q-good real numbers in the interval [u, u + 1]. But from (1), each P -good number in [u, u + 1] is also Q-good. Hence 1 + |Q(u + 1) − Q(u)| ≥ 1 + |P (u + 1) − P (u)|,

(2)

for all sufficiently large u. Let P1 (x) = P (x+1)−P (x) and Q1 (x) = Q(x+1)−Q(x). Note that deg(P1 ) = n − 1 > m − 1 = deg(Q1 ). Hence |P1 (x)| > |Q1 (x)| for all sufficiently large x, which contradicts (2). So we have shown that a1 = a2 = · · · = an−1 = 0. Hence P (x) = an xn + a0 . Let r = 1 1 . Since P (r) = ±1 + a0 ∈ Z, we have r ∈ S. Thus also r2 ∈ S. So

P (r2 ) =

|an | n 1 + an

a0 ∈ Z. Hence an = ±1, which concludes the proof.



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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 Rank Country

Number of Contestants

Total Score

Gold Awards

Silver Awards

Bronze Awards

Honourable Mentions

1

Republic of Korea

10

320

1

2

4

3

2

USA

10

306

1

2

4

3

3

Japan

10

254

1

2

4

3

4

Singapore

10

231

1

2

4

3

5

Canada

10

220

1

2

4

3

6

Russia

10

210

1

2

4

3

7

Taiwan

10

208

1

2

4

3

8

Islamic Republic of Iran

10

186

1

2

4

3

9

Thailand

10

185

1

2

4

3

10

Indonesia

10

173

1

2

4

3

11

Peru

10

166

1

2

4

3

12

Australia

10

165

0

3

4

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

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Rank Country

Number of Contestants

Total Score

Gold Awards

Silver Awards

Bronze Awards

Honourable Mentions

25

Turkmenistan

10

85

0

0

2

8

26

Tajikistan

10

75

0

0

0

10

27

Bolivia

5

66

0

2

1

1

28

Colombia

9

60

0

0

2

3

29

Syria

10

60

0

0

2

4

30

Kyrgyzstan

8

54

0

0

1

5

31

Pakistan

10

47

0

0

1

1

32

Sri Lanka

6

40

0

0

0

5

33

El Salvador

4

28

0

0

1

1

34

Nicaragua

10

24

0

0

0

3

35

Trinidad and Tobago

10

24

0

0

0

1

36

Panama

3

21

0

0

0

3

37

Cambodia

5

20

0

0

0

1

38

Costa Rica

10

12

0

0

0

1

39

Guatemala

2

7

0

0

0

0

352

4716

12

43

109

124

Total

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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

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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

School

State

Reserve Vacant in 2018

2018 Australian IMO team Name

Year

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

Hadyn Tang

9

Trinity Grammar School

VIC

Guowen Zhang

12

St Joseph’s College, Gregory Terrace

QLD

10

James Ruse Agricultural High School

NSW

NSW

Reserve Yasiru Jayasooriya

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Participants at the 2018 AMOC Selection School Name

m/f

Year

School

State

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

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

Ethan Tan

m

12

Cranbrook School

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

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

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

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

Genevieve Conway

f

8

Methodist Ladies’ College

Eva Ge

f

9

James Ruse Agricultural High School

NSW

Mikaela Gray

f

9

Brisbane State High School

QLD

Claire Huang

f

10

Radford College

ACT

Anagha Kanive-Hariharan

f

10

James Ruse Agricultural High School

NSW

Linda Lu

f

11

The Mac.Robertson Girls’ High School

VIC

Ying Tan

f

10

Presbyterian Ladies’ College

VIC

Jianyi (Jason) Wang

m

10

Queensland Academy for Science, Mathematics & Technology

QLD

Christine Ye

f

8

Lauriston Girls' School

VIC

Elizabeth Yevdokimov

f

9

St Ursula’s College

QLD

Zirui (Harry) Zhang

m

9

Christian Brothers College

VIC

Senior

WA

WA NSW

Intermediate

VIC

NSW

Junior

Mathematics Contests The Australian Scene 2018 | AMOC Selection School

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The 7th EGMO,

The 7th European Girls’ Mathematical Olympiad (EGMO) was held 9–15 April 2018, The 7th European Girls’ in the city of Florence, Italy. This was the first time Italy has hosted an Olympiad and in the city of Florence, Ital it was the biggest EGMO to date with 8 new countries participating in 2018: Australia, it was the biggest EGMO t LEADER’S Austria, Bolivia, Canada,EGMO Germany, TEAM Greece, Mongolia, and Peru.REPORT For this 7th EGMO, a Bolivia, Canada, G Austria, total of 195 contestants from 52 countries participated (137 contestants from 36 official total of 195 contestants fro 1 European countries)Girls’ . Peru had the youngest team withwas an average of 2018, 14 years andcity ofcountries)1 . Peru The 7th European Mathematical Olympiad (EGMO) held 9–15age April inEuropean the 10Florence, months,Italy. Latvia had the oldest team with an average age of 18 years and 7 months, This was the first time Italy has hosted an Olympiad and it was the biggest EGMOLatvia to 10 months, had the and Lichtenstein, a countryparticipating of just 40000, smallest team of one. Canada, Germany, date with 8 new countries in had 2018:the Australia, Austria, Bolivia, Greece, and Lichtenstein, a country Mongolia, and Peru. For this 7thmay EGMO, a total of 195 contestants from 52 countries participated Each participating country send a team of up to four students, a Team Leader Each participating coun 1 (137a contestants from 36 official countries) . Leaders form what is called the and Deputy Team Leader. At European the EGMO, the Team and a Deputy Team Leader Jury, in youngest 2018 was team chaired byanRoberto For and decisions such Latvia as approving Peruwhich had the with averageDvornicich. age of 14 years 10 months, had the oldest Jury, which in 2018 was cha each paper as marking schemes, all7Leaders and motions are carried by team with as anwell average age of 18 years and months,may and vote, Lichtenstein, a country of just 40000, had the each paper as well as marki team of one. a smallest simple majority of those voting. For decisions such as determining medal cut-offs and majority of those v a simple coordination appeals, however, European teams may vote. and Each participating country mayonly sendLeaders a team of of official up to four students, a Team Leader a Deputy appeals, howev coordination Participating countries are invited to submit up to six proposed problems to be in received Team Leader. At the EGMO, the Team Leaders form what is called the Jury, which 2018 was chairedcountries a Participating bybythe Problem SelectionFor Committee, which 2018 was chaired Gobbino. Roberto Dvornicich. decisions such as in approving each paperby as Massimo well as marking schemes, all by the Problem Selection C Leaders may vote, and the motions are problems carried byand a simple majority of problems those voting. For decisions such as This committee selects contest has alternative prepared if acommittee This selects the determining andknown. coordination appeals,problems however,were only presented Leaders oftoofficial European teams problem turnsmedal out tocut-offs be already The selected the Jury problem turns out to be alre inmay theirvote. first meeting on Tuesday 10 April. With no objections from the Jury, problems for first meeting on Tue in their Participating are invited submit up to six proposed problems to be both Day 1 andcountries Day 2 papers were to approved, including translations into the 33 received languages both by Daythe 1 and Day 2 pape Problemby Selection Committee, which in 2018 was chaired by Massimo Gobbino. This committee selects required the contestants. The Jury, Observers, and any others who have knowledge required by the contestants problems and has alternative prepared a problem turns oututmost to be already ofthe thecontest problems and solutions before the problems examinations, are iftrusted to do their of the problems known. The selected problems were presented to the Jury in their first meeting on Tuesday 10 April. and solutio to ensure that no contestant has any information, direct or indirect, about any proposed to ensureincluding that no contestan With no objections from the Jury, problems for both Day 1 and Day 2 papers were approved, problem. problem. translations into the 33 languages required by the contestants. The Jury, Observers, and any others The six problems on the EGMO contest papers may be described as follows. The six their problems on the who have knowledge of the problems and solutions before the examinations, are trusted to do

utmost to ensure that no contestant has any information, direct or indirect, about any proposed 1. A lovely geometry problem about a point lying on a circle, proprosed by Velina 1. A lovely geometry pr problem.

Ivanova, Bulgaria.

The six problems on the EGMO contest papers may be described as follows.

Ivanova, Bulgaria.

2. A theory problem integers as a product ofby numbers of 2. theA number 1. A number lovely problemabout aboutexpressing a point lying on a circle, proposed Velina Ivanova, Bulgaria.theory prob  geometry  1   form 1 + , proposed by Mihail Baluna, Romania. 1 k 2. A number theory problem about expressing integers as a product of numbers of the form 1 + k , propose proposed by Mihail Baluna, Romania. 3. A fun combinatorics problem involving a Jury choosing and then rearranging the

3. A fun 3. order A fun of combinatorics involving Jury choosing and then rearranging the order of combinatorics p contestants problem in a queue. It wasaproposed by Hungary. order of contestants in contestants in a queue. It was proposed by Hungary.

4. A involvingbalanced balanced configurations configurations of of dominoes dominoes on an an n × 4. A combinatorics problem involving It 4. nAboard. combinatorics probl was proposed by Merlijn by Staps, the Netherlands. board. It was proposed Merlijn Staps, the Netherlands. 5. A classical circle geometry problem, proposed by Dominika Regiec, Poland.

board. It was propose

5. A classical circle geometry problem, proposed by Dominika Regiec, Poland.

5. A classical 6. A tricky algebra problem involving the closeness of pairs of elements in a set of positive integers.circle geom 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 prob

Like positive for the International 1–3Netherlands. appeared on the Day 1 Paper and integers. It Mathematical was proposed Olympiad, by MerlijnProblems Staps, the positive integers. It w problems 4–6 appeared on the Day 2 Paper. These papers were held on Wednesday 11 April and Like for12the International Olympiad, Problems appeared onpaper the Thursday April respectively,Mathematical with the contestants allowed four and1–3 a half hours per Like to forwork the Internation individually the problems4–6 andappeared write their problem was scored out of a Day 1 Paper on and problems onattempted the Day 2proofs. Paper.Each These papers were held Day 1 Paper and problems of seven points. onmaximum Wednesday 11 April and Thursday 12 April respectively, with the contestants allowed

on Wednesday 11 April and four and a half paper to work individually on the writefour their Prior to the firsthours day ofper competition, the opening ceremony wasproblems held in theand beautiful Teatro andVerdi a half hours per p attempted proofs. Each problem wasand scored out of a maximum of seven included points. speeches and hosted by Alessandra Caraceni Massimo Gobbino. The occasion from the attempted proofs. Each pro 1

Vice Mayor of Florence and various sponsors for the event as well as several jokes about a lack of a timetable for the week. parade teams wasofcomplete with beautifully-designed background This is a new record for theThe EGMO, withofthe number participating countries increasing every 1 year. This is a new record for the EG slides each country accompaniment of 16 theofunique instrumental In 2012 for for the first edition and therean were only 19 countries, them official Europeanmusic ones. of Lame Da Barba. In 2012 the first edition the Finally, it was declared that the 2018 EGMO was officially open. After the opening ceremonyfor there 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 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

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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

8.8%

17.5%

25.5%

51.8%

Proportion from official European teams

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

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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

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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

17

39

52

45

Total (52 teams, 195 contestants)

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

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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 M Q 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) < f (x) + f (y). (Pairs (x1 , y1 ) and (x2 , y2 ) are different if x1 = x2 or y1 = y2 .) Problem 3. The n contestants of an EGMO are named C1 , . . . , Cn . After the competition 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

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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. (a) Prove that for every real number t such that 0 < t < 12 there exists a positive integer n with the following property: for every set S of n positive integers there exist two different elements x and y of S, and a non-negative integer m (i.e. m ≥ 0), such that |x − my| ≤ ty. (b) Determine whether for every real number t such that 0 < t < positive integers such that |x − my| > ty

1 2

there exists an infinite set S of

for every pair of different elements x and y of S and every positive integer m (i.e. m > 0).

Language: English

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EUROPEAN GIRLS’ MATHEMATICAL OLYMPIAD SOLUTIONS Solutions to the 2018 European 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.) C

M

A

B

Q N O

P

Consider triangles QN P and QM C. Vertically opposite angles N QP and M QC are equal. We also have that CM is perpendicular to AB since ABC is isosceles and M is the midpoint of AB. Given that N P is also perpendicular to AB, CM and N P are parallel. Therefore ∠N P Q = ∠M CQ and ∠P N Q = ∠CM Q. Hence triangles QN P and QM C are similar. Since QP = 2QC, it follows that N P = 2M C. Furthermore, N P is parallel to M C. −−→ The point N is therefore a translation of P by the fixed vector 2M C. 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, 2M C = 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.

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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.) C

M

A

B

Q N O

P

As in solution 1 (with comments), N P is parallel to CM and hence CO (C, M , O are collinear with M the midpoint of CO), triangles QN P and QM C are similar, and N P = 2M C = OC. Since N P and CO are equal and parallel, OCN P is a parallelogram. Since adjacent sides OC and OP are equal, OCN P 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. 

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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 M Q (if a = 0), is y=

2b + 1 1 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. 

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2. Solution 1 (Based on the partial solution by Grace He, year 10, Methodist Ladies’ College, VIC. Grace was a Silver medallist with the 2018 Australian EGMO team.) We have the set A=



1 1 + : k = 1, 2, 3, . . . k



=



 k+1 : k = 1, 2, 3, . . . . k

For every integer x ≥ 2, x−1

k+1 2 3 x x = × × ··· × = . 1 2 x − 1 k=1 k 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 =

Similarly,

3 2

× 2n+1 , the product of n + 2 elements of A, f (3 × 2n ) = n + 2.

and since 33× 2n =

33 × 2n > 2n+5 ⇒ f (33 × 2n ) ≥ n + 6

33 × 2n+5 , 32

the product of n+ 6 elements of A, f (33 ×2n ) = n + 6.

Lastly, f (11) ≥ 5 since 11 cannot be written as the product of four or fewer elements of A: 24 > 11, 23 × 32 > 11, and any other product of at most four elements of A does not exceed 23 × 34 = 10 23 < 11. Since 11 = 11 × 54 × 2 × 2 × 2, we have that 10 f (11) = 5. Putting these results together, f (33 × 2n ) = n + 6 < n + 2 + 5 = f (3 × 2n ) + f (11) for all integers n ≥ 0. Hence, there are infinitely many pairs, x = 3 × 2n , y = 11, with x ≥ 2, y ≥ 2 and f (xy) < f (x) + f (y). 

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Solution 2 (2018 EGMO Problem Selection Committee) First we show that f (xy) < f (x) + f (y) for (x, y) = (7, 7). We have that f (7) ≥ 4 since 7 cannot be written as the product of three or fewer elements of A: 23 > 7, and any other product of at most three elements of A does not exceed 22 × 32 = 6 < 7. Also f (49) ≤ 7 since 49 = 2 × 2 × 2 × 2 × 2 × 32 × 49 . Hence 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) < f (x) + f (y). This implies that there exists an M large enough such that whenever a > M or b > M holds 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) < f (x) + f (y) and let n > 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) < f (x) + f (y).

Therefore there exist infinitely many pairs (x, y) with x ≥ 2, y ≥ 2, and f (xy) < f (x) + f (y). 

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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) (a) We prove by induction that each contestant Ci can move at most 2n−i − 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 Cn . Once she has passed Cn there is no way 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. Assume that each contestant Cj , for j > i, can move at most 2n−j − 1 times and consider how many times contestant Ci can pass Cj where j > i. If Cj moves at most 2n−j − 1 times then, by passing Ci every move, she can pass Ci at most 2n−j − 1 times. Since Cj may have started ahead of Ci , we have that Ci can pass Cj at most 2n−j times. Therefore we have that Ci can pass Cn at most 20 times, Cn−1 at most 21 times, Cn−2 at most 22 times, . . ., Ci+1 at most 2n−i−1 times. Each time Ci moves, she must pass at least one Cj where j > i, and in the extreme case, she passes all of C1 , C2 , . . . , Ci−1 and one Cj . Therefore, the maximum number of times Ci can move is 1 + 2 + 22 + · · · + 2n−i−1 = 2n−i − 1.  (b) What remains is to show how the Jury can arrange the queue initially, and what moves they can make, to achieve this maximal number of moves for each contestant. That is, 20 − 1 moves for Cn , 21 − 1 moves for Cn−1 , 22 − 1 moves for Cn−2 , . . ., 2n−1 − 1 moves for C1 and so a total number of moves (i.e. euros) of (20 − 1) + (21 − 1) + (22 − 1) + · · · + (2n−1 − 1) = 2n − 1 − n. We prove by induction that by lining the contestants up in reverse order, Cn , Cn−1 , . . . , C1 , the jury can collect 2n − 1 − n euros before the 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

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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

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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 Ci will move them ahead of exactly i of {C1 , C2 , . . . , Ck } 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 collected. 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 and i < j (we will call this a reverse pair), we assign a weight of 2i−1 . We define the total weight of an arrangement as the sum of the weights for each reverse pair. In this way, the maximum possible total weight of an arrangement occurs when Ci is behind exactly all contestants with a larger number than herself, and so in front of all contestants with a smaller number than herself. That is, for the arrangement Cn , Cn−1 , Cn−2 , . . . , C2 , C1 . This maximum total weight of an arrangement is (n − 1)20 + (n − 2)21 + (n − 3)22 + · · · + (2)2n−3 + (1)2n−2 = 2n − n − 1. Now consider the total weight of an arrangement before and after Ci moves 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 < i. This gives a reduction of at least 2i−1 in the total weight, and an addition of at most 20 + 21 + 22 + · · · + 2i−2 = 2i−1 − 1. Hence the total weight of an arrangement must decrease when any contestant Ci moves forward in the queue. Since the maximum total weight of an arrangement is 2n − n − 1, and this total weight decreases every move, at most 2n − n − 1 are possible. 

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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 First consider the case where n is a multiple of 3. Since k ≥ 1, we have that D ≥ 2n . 3 The following diagram shows a balanced configuration for n = 3 with k = 1 and 2×3 = 2 dominoes on the board. 3

If n is a multiple of 3, we can obtain a balanced configuration with k = 1 and n3 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 configuration is 2n . 3 If n is not a multiple of 3, then k is a multiple of 3. Hence k ≥ 3 and D ≥ 2n.

The following diagram shows a balanced configuration for n = 4 with k = 3 and 2 × 4 = 8 dominoes on the board.

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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

For n ≡ 1(mod 3) with n ≥ 7, n−1 Block B’s are placed along a main diagonal of the 3 (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 Block A’s are placed along a main diagonal of the (n − 7) × (n − 7) sub-board 3 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 the board in this construction is 4 × n−1 + 2 × n−4 + 4 = 2n. 3 3

... ...

For n ≡ 2(mod 3) with n ≥ 5, n−2 Block B’s are placed along a main diagonal of the 3 (n − 2) × (n − 2) sub-board, as shown in the diagram below. Each of these dominoes contribute 2 to the value of each of the first n − 2 rows and n − 2 columns. Then n−2 3 Block A’s are placed along a main diagonal of the (n − 5) × (n − 5) sub-board as 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

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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 of dominoes placed on the board in this construction is 4 × n−2 + 2 × n−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 is therefore 2n if n is a multiple of 3, and 2n otherwise.  3 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

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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 Australian 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 . C

Γ Ω Q

V

O1

O2 P

A

U

B

M We first show that V , U , and M are collinear. Let O1 and O2 be the centres of Ω and Γ respectively, and let V U intersect Γ at M  . We have ∠O1 U V = ∠O1 V U = ∠O2 V M  = ∠O2 M  V and hence O1 U  O2 M  . Since O1 U ⊥ AB (AB is tangent to Ω at U ), O2 M  ⊥ AB and hence M  is the midpoint of the arc AB (that does not contain C) and M  = M . (Indeed, the dilation with centre V that sends Ω to Γ sends U to the point of Γ where the tangent of Γ is parallel to AB and this point is M ). Triangles M AV and M U A are similar since ∠AM V = ∠U M A and ∠M V A = ∠M CA = ∠M CB = ∠M AB = ∠M AU . It follows that M U × M V = M A2 = M B2. Computing the power of M with respect to Ω we have that M P × M Q = M U × M V = M B2. Now we have that triangles M BP and M QB are similar as ∠BM P = ∠QM B and MP B = M . In particular, we then have ∠M BP = ∠M QB. We already have that MB MQ ∠M CB = ∠M CA = ∠M BA. Putting these together, we can conclude ∠ABP = ∠M BP − ∠M BA = ∠M QB − ∠M CB = ∠QBC,

as required.



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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 M A = M B. 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 M C and Ω, and since the only fixed point on the line segment M C is its intersection with the inversion circle, P and Q are swapped. This implies that M P × M Q = M B 2 . 

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6. Solution (part a) (2018 EGMO Problem Selection Committee) Let t be a real number such that 0 < t < 12 . We wish to find a positive integer n, such that for every set S of n integers, denoted s1 < s2 < s3 < · · · < sn , there exist two elements si and sj such that |sj − msi | ≤ tsi for a non-negative integer m. In other words sj is within tsi of a multiple of si . tsi si

0

sj

2si

3si

4si

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 si , namely 1 × si (m = 1). Hence the inequality is satisfied.

Otherwise, si+1 − si > tsi ⇒ si+1 > (1 + t)si for all 1 ≤ i ≤ n − 1. That is,

sn > (1 + t)sn−1 > (1 + t)2 sn−2 > (1 + t)3 sn−3 > · · · > (1 + t)n−1 s1 . This gives that sn > (1 + t)n−1 s1 ⇒ s1
tsi for every pair of different elements si and sj of S and for every 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 ts1 < (s1 − 1)/2. This is equivalent to s1 > 1/(1 − 2t).

Having chosen s1 , certain numbers are excluded from being in S. S cannot 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

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ts1
ts1 and (m + 1)s1 − ((m + 1/2)s1 − 1/2) = (s1 − 1)/2 > ts1 and these half-multiples are even further from other multiples of s1 .

ts1
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 s2 (m − t) > s1 ⇒ ms2 − s1 > ts2 ⇒ |s1 − ms2 | > ts2 .

So in particular, we will choose s2 to be whichever of the half-multiples (5/2)s1 ±1/2 is odd and note that since s1 > 1, s2 = (5/2)s1 ± 1/2 = 2s1 + (s1 ± 1)/2 > 2s1 .

Now this excludes even more numbers from S. Again we can be sure that the halfmultiples 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 s1 s2 · · · sk with sk+1 > 2si for all 1 ≤ i ≤ k. In this way, sk+1 is an odd half-multiple of each of s1 , s2 , . . . , sk and so not excluded by any of these elements. Furthermore, sk+1 > 2si ⇒ sk+1 (m − t) > si ⇒ msk+1 − si > tsk+1 ⇒ |si − msk+1 | > tsk+1 and so for all 1 ≤ i ≤ k, si is not excluded by sk+1 .

Specifically, we choose sk+1 to be an odd integer given by sk+1 = 5s1 s2 · · · sk /2 ± 1/2.

Note that since each of s1 , s2 , . . . , sk is greater than 1, sk+1 > 5si /2 ± 1/2 = 2si + (si ± 1)/2 > 2si for all 1 ≤ i ≤ k. Also note that the si sequence is increasing.

Therefore, we have found such an infinite set S of integers s1 < s2 < s3 < · · · , each chosen to be odd and given by s1 > 1/(1 − 2t) and sk+1 = 5s1 s2 · · · sk /2 ± 1/2.  22 Mathematics Contests The Australian Scene 2018 | European Girls' Mathematical Olympiad Solutions

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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

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