1

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

Alber t Einstein

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

A depArtMent Of the AustrAliAn MAtheMAtics trust

MATHEMATICS CONTESTS THE AUSTRALIAN SCENE 2015

PART 1 AND 2A Di Pasquale, N Do and KL McAvaney

2

Published by

AMT Publishing

Australian Mathematics Trust

University of Canberra Locked Bag 1

Canberra GPO ACT 2601

Australia

Tel: 61 2 6201 5137

www.amt.edu.au

AMTT Limited

ACN 083 950 341

Copyright © 2015 by the Australian Mathematics Trust

National Library of Australia Card Number

and ISSN 1323-6490

3

SUPPORT FOR THE AUSTRALIAN MATHEMATICAL OLYMPIAD COMMITTEE TRAINING PROGRAM

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

Trustee

The University of Canberra

Sponsors

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

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

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

Special Thanks

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

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ACKNOWLEDGEMENTS

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

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

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PREFACE

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

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

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

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

Mike Clapper

November 2015

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CONTENTS

Support for the Australian Mathematical Olympiad Committee Training Program 3

Acknowledgements 4

Preface 5

Background Notes on the IMO and AMOC 7

Mathematics Challenge for Young Australians 10

Membership of MCYA Committees 12

Membership of AMOC Committees 14

AMOC Timetable for Selection of the Team to the 2016 IMO 15

Activities of AMOC Senior Problems Committee 16

Challenge Problems 17

Challenge Solutions 26

Challenge Statistics 51

Australian Intermediate Mathematics Olympiad 55

Australian Intermediate Mathematics Olympiad Solutions 57

Australian Intermediate Mathematics Olympiad Statistics 65

Australian Intermediate Mathematics Olympiad Results 66

AMOC Senior Contest 70

AMOC Senior Contest Solutions 71

AMOC Senior Contest Results 79

AMOC Senior Contest Statistics 80

AMOC School of Excellence 81

Australian Mathematical Olympiad 83

Australian Mathematical Olympiad Solutions 85

Australian Mathematical Olympiad Statistics 110

Australian Mathematical Olympiad Results 111

27th Asian Pacific Mathematics Olympiad 113

27th Asian Pacific Mathematics Olympiad Solutions 114

27th Asian Pacific Mathematics Olympiad Results 128

AMOC Selection School 130

IMO Team Preparation School 133

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The Mathematics Ashes 134

The Mathematics Ashes Results 135

IMO Team Leader’s Report 136

International Mathematical Olympiad 140

International Mathematical Olympiad Solutions 142

International Mathematical Olympiad Results 168

Origin of Some Questions 173

Honour Roll 174

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BACKGROUND NOTES ON THE IMO AND AMOC

The Australian Mathematical Olympiad Committee

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

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

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

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

The aims of AMOC include:

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

AMOC schedule from August until July for potential IMO team members

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

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

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

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

The International Mathematical Olympiad

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

9

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

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

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

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

The IMO will be held in Hong Kong in 2016.

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SUMMARY OF AUSTRALIA’S ACHIEVEMENTS AT PREVIOUS IMOS

Year City Gold Silver Bronze HM Rank

1981 Washington 1 23 out of 27 teams

1982 Budapest 1 21 out of 30 teams

1983 Paris 1 2 19 out of 32 teams

1984 Prague 1 2 15 out of 34 teams

1985 Helsinki 1 1 2 11 out of 38 teams

1986 Warsaw 5 15 out of 37 teams

1987 Havana 3 15 out of 42 teams

1988 Canberra 1 1 1 17 out of 49 teams

1989 Braunschweig 2 2 22 out of 50 teams

1990 Beijing 2 4 15 out of 54 teams

1991 Sigtuna 3 2 20 out of 56 teams

1992 Moscow 1 1 2 1 19 out of 56 teams

1993 Istanbul 1 2 3 13 out of 73 teams

1994 Hong Kong 2 3 3 12 out of 69 teams

1995 Toronto 1 4 1 21 out of 73 teams

1996 Mumbai 2 3 23 out of 75 teams

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

1998 Taipei 4 2 13 out of 76 teams

1999 Bucharest 1 1 3 1 15 out of 81 teams

2000 Taejon 1 3 1 16 out of 82 teams

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

2002 Glasgow 1 2 1 1 26 out of 84 teams

2003 Tokyo 2 2 2 26 out of 82 teams

2004 Athens 1 1 2 1 27 out of 85 teams

2005 Merida 6 25 out of 91 teams

2006 Ljubljana 3 2 1 26 out of 90 teams

2007 Hanoi 1 4 1 22 out of 93 teams

2008 Madrid 5 1 19 out of 97 teams

2009 Bremen 2 1 2 1 23 out of 104 teams

2010 Astana 1 3 1 1 15 out of 96 teams

2011 Amsterdam 3 3 25 out of 101 teams

2012 Mar del Plata 2 4 27 out of 100 teams

2013 Santa Marta 1 2 3 15 out of 97 teams

2014 Cape Town1

Perfect Score by Alexander

Gunning3 2 11 out of 101 teams

2015 Chiang Mai 2 4 6 out of 104 teams

12

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 Challenge.

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

Challenge (three weeks during the period March–June)

Enrichment (April–September)

Australian Intermediate Mathematics Olympiad (September).

Challenge

Challenge now consists of four levels. Upper Primary (Years 5–6) and Middle Primary (Years 3–4) present students with four problems each to be attempted over three weeks, students being allowed to work on the problems in groups of up to three participants, but each to 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 weeks, students being allowed to work on the problems with a partner but each must write their solutions individually.

There were 12692 entries (1166 Middle Primary, 3416 Upper Primary, 5006 Junior, 3104 Intermediate) for the Challenge in 2015. The 2015 problems and solutions for the Challenge, together with some statistics, appear later in this book.

Enrichment

This is a six-month program running from April to September, which consists of six 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 12–14 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.

Newton (years 5–6) includes polyominoes, fast arithmetic, polyhedra, pre-algebra concepts, patterns, divisibility and specific problem-solving techniques. There were 1165 entries in 2015.

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 1181 entries in 2015.

Euler (years 7–8) includes primes and composites, least common multiples, highest common factors, arithmetic

12 12

sequences, figurate numbers, congruence, properties of angles and pigeonhole principle. There were 1708 entries in 2015.

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 1150 entries in 2015.

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 818 entries in 2015.

Polya (top 10% year 10) (currently under revision) Topics will include angle chasing, combinatorics, number theory, graph theory and symmetric polynomials. There were 303 entries in 2015.

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 1440 entries for 2015 and 19 perfect scores.

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MEMBERSHIP OF MCYA COMMITTEES

Mathematics Challenge for Young Australians Committee 2015

Director Dr K McAvaney, Victoria

Challenge

Committee Adj Prof M Clapper, Australian Mathematics Trust, ACT Mrs B Denney, NSW Mr A Edwards, Queensland Studies Authority Mr B Henry, Victoria Ms J McIntosh, AMSI, VIC Mrs L Mottershead, New South Wales Ms A Nakos, Temple Christian College, SA Prof M Newman, Australian National University, ACT Dr I Roberts, Northern Territory Ms T Shaw, SCEGGS, NSW Ms K Sims, New South Wales Dr A Storozhev, Attorney General’s Department, ACT Mr S Thornton, South Australia Ms G Vardaro, Wesley College, VIC

Moderators Mr W Akhurst, New South Wales Mr R Blackman, Victoria Ms J Breidahl, Victoria Mr A Canning, Queensland Dr E Casling, Australian Capital Territory Mr B Darcy, South Australia Mr J Dowsey, Victoria Ms P Graham, MacKillop College, TAS Ms J Hartnett, Queensland Ms N Hill, Victoria Dr N Hoffman, Edith Cowan University, WA Ms R Jorgenson, Australian Capital Territory Prof H Lausch, Victoria Mr J Lawson, St Pius X School, NSW Ms K McAsey, Victoria Ms T McNamara, Victoria Mr G Meiklejohn, Department of Education, QLD Mr M O’Connor, AMSI, VIC Mr J Oliver, Northern Territory Mr G Pointer, Marratville High School, SA Dr H Sims, Victoria Mrs M Spandler, New South Wales Ms C Stanley, Queensland Mr P Swain, Ivanhoe Girls’ Grammar School, VIC Dr P Swedosh, The King David School, VIC Mrs A Thomas, New South Wales Ms K Trudgian, Queensland

14 14

Enrichment

Editors Mr G R Ball, University of Sydney, NSW Dr M Evans, International Centre of Excellence for Education in Mathematics, VIC Mr K Hamann, South Australia Mr B Henry, Victoria Dr K McAvaney, Victoria Dr A M Storozhev, Attorney General’s Department, ACT Emeritus Prof P Taylor, Australian Capital Territory Dr O Yevdokimov, University of Southern Queensland

Australian Intermediate Mathematics Olympiad Problems Committee Dr K McAvaney, Victoria (Chair) Adj Prof M Clapper, Australian Mathematics Trust, ACT Mr J Dowsey, University of Melbourne, VIC Dr M Evans, International Centre of Excellence for Education in Mathematics, VIC Mr B Henry, Victoria Assoc Prof H Lausch, Monash University, VIC

14 15

MEMBERSHIP OF AMOC COMMITTEES

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

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

TreasurerDr P Swedosh, The King David School, VIC

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

Chair, ChallengeDr K McAvaney, Deakin University, VIC

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

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

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

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

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AMOC TIMETABLE FOR SELECTION OF THE TEAM TO THE 2016 IMO

August 2015—July 2016

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

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

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

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

Month Activity

August

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

September Australian Intermediate Mathematics Olympiad

December AMOC School of Excellence

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

February Australian Mathematical Olympiad

March Asian Pacific Mathematics Olympiad

April AMOC Selection School

May–June Personal Tutor Scheme for IMO team members

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

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

1. 2015 Australian Mathematical Olympiad

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

2. 2015 Asian Pacific Mathematics Olympiad

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

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

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

4. 2015 AMOC Senior Contest

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

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2015 Challenge Problems - Middle Primary

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

MP1 e-Numbers

The digits on many electronic devices look like this:

When numbers made of these digits are rotated 180 ( ), some of them become numbers and some don’t. Forexample, when 8 is rotated it remains the same, 125 becomes 521, and 14 does not become a number. Except for 0,no number starts with 0.

All rotations referred to below are through 180.

a List all digits which rotate to the same digit.

b List all digits which rotate to a different digit.

c List all numbers from 10 to 50 whose rotations are numbers.

d List all numbers from 50 to 200 that stay the same when rotated.

MP2 Quazy Quilts

Joy is making a rectangular patchwork quilt from square pieces of fabric. All the square pieces are the same size. Shestarts by joining two black squares into a rectangle. Then she adds a border of grey squares.

Next, she adds a border of white squares.

a How many white squares does she need?

Joy continues to add borders of grey and borders of white squares alternately.

b How many grey squares and how many white squares does she need for the next two borders?

c Joy notices a pattern in the number of squares used for each border. Describe this pattern and explain why italways works.

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CHALLENGE PROBLEMS – MIDDLE PRIMARY

19 19

d There are 90 squares in the outside border of the completed quilt. How many borders have been added to Joy’sstarting black rectangle?

MP3 Egg Cartons

Zoe decided to investigate the number of different ways of placing identical eggs in clear rectangular cartons of varioussizes.

For example, there is only one way of placing six eggs in a 2 × 3 carton and only two different ways of placing fiveeggs.

Reflections and rotations of arrangements are not considered different. For example, all of these arrangements are thesame:

a Here are two ways of arranging four eggs in a 2 × 3 carton.

Draw four more different ways.

b Draw six different ways of arranging two eggs in a 2 × 3 carton.

c Draw six different ways of arranging three eggs in a 2 × 3 carton.

d Draw all the different ways of arranging six eggs in a 2 × 4 carton.

MP4 Condates

On many forms, the date has to be written as DD/MM/YY. For example, 25 April 2013 is written 25/04/13. Datessuch as this that use all the digits 0, 1, 2, 3, 4, 5 will be called condates.

a Find two condates in 2015.

b Find the first condate after 2015.

c Find the last condate in any century.

d Explain why no date of the form DD/MM/YY can use all the digits 1, 2, 3, 4, 5, 6.

2

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2015 Challenge Problems - Upper Primary

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

UP1 Rod Shapes

I have a large number of thin straight steel rods of lengths 1 cm, 2 cm, 3 cm and 4 cm. I can place the rods with theirends touching to form polygons. For example, with four rods, two of length 1 cm and two of length 3 cm, I can forma parallelogram which I refer to as (1, 3, 1, 3).

3 cm

1cm

3 cm

1cm

a List all the different equilateral triangles I can make using only three rods at a time.

b Using only three rods at a time, how many different isosceles triangles can I make which are not equilateral triangles?List them all.

c List all the different scalene triangles I can make using only three rods at a time.

d Using only four rods at a time, none of which is 4 cm, how many different quadrilaterals can I make which haveexactly one pair of parallel sides and the other pair of sides equal in length? List them all.

UP2 Egg Cartons

Zoe decided to investigate the number of different ways of placing identical eggs in clear rectangular cartons of varioussizes.

For example, there is only one way of placing six eggs in a 2 × 3 carton and only two different ways of placing fiveeggs.

Reflections and rotations of arrangements are not considered different. For example, all of these arrangements are thesame:

3

CHALLENGE PROBLEMS – UPPER PRIMARY

21 21

a Draw four different ways of arranging two eggs in a 2 × 3 carton.

b Draw six different ways of arranging three eggs in a 2 × 3 carton.

c Draw all the different ways of arranging six eggs in a 2 × 4 carton.

d Zoe discovers that there are 55 ways of arranging three eggs in a 2 × 6 carton. Explain why there must also be 55ways of arranging nine eggs in a 2 × 6 carton.

UP3 Seahorse Swimmers

The coach at the Seahorse Swimming Club has four boys in his Under 10 squad. Their personal best (PB) times for50 metres are:

Mack 55 secondsJack 1 minute 8 secondsZac 1 minute 16 secondsNick 1 minute 3 seconds

a The coach splits the swimmers into two pairs for a practice relay race. He wants the race to be as close as possible.Based on the PB times, who should be in each pair?

b Chloe’s PB time is 1 minute 10 seconds and Sally’s is 53 seconds. The coach arranges the two girls and four boysinto two teams of three with total PB times as close as possible.

i Why can’t the total PB times be the same for the two teams?

ii Who are in each team?

c The swimmers ask the coach what his PB time was when he was 10 years old. The coach replied that if his PBtime was included with all theirs, then the average PB time would be exactly 1 minute. What was the coach’s PBtime?

UP4 Primelandia Money

In Primelandia, the unit of currency is the Tao (T). The value of each Primelandian coin is a prime number of Taos.So the coin with the smallest value is worth 2T. There are coins of every prime value less than 50. All payments inPrimelandia are an exact number of Taos.

a List all combinations of two coins whose sum is 50T.

b What is the smallest payment (without change) which requires at least three coins?

c Suppose you have one of each Primelandian coin with value less than 50T. What is the difference between thelargest even payment that can be made using four coins and the largest even payment that can be made using threecoins?

d A bag contains six different coins. Alice, Bob and Carol take two coins each from the bag and keep them. Theyfind that they have all taken the same amount of money. What is the smallest amount of money that could havebeen in the bag? Explain your reasoning.

4

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2015 Challenge Problems - Junior

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

J1 Quirky Quadrilaterals

Robin and Toni were drawing quadrilaterals on square dot paper with their vertices on the dots but no other dots onthe edges. They decided to use I to represent the number of dots in the interior. Here are two examples:

•

•

•

•

•

•

•

•

•

I = 1

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

I = 6

a Toni said she could draw a rhombus with I = 2 and a kite that is not a rhombus with I = 2. Draw such quadrilateralson square dot paper.

b Toni drew squares with I = 4 and I = 9. Draw such squares on square dot paper.

c Robin drew a square with I = 12. Draw such a square on square dot paper.

d Toni exclaimed with excitement, ‘I can draw rhombuses with any number of interior points’. Explain how this couldbe done.

J2 Indim Integers

Jim is a contestant on a TV game show called Indim. The compere spins a wheel that is divided into nine sectorsnumbered 1 to 9. Jim has nine tiles numbered 1 to 9. When the wheel stops, he removes all tiles whose numbers arefactors or multiples of the spun number.

•

•

34

56

7

2

19

8

1 2 3

4 5 6

7 8 9

Jim then tries to use three of the remaining tiles to form a 3-digit number that is divisible by the number on thewheel. If he can make such a number, he shouts ‘Indim’ and wins a prize.

a Show that if 1, 2, or 5 is spun, then it is impossible for Jim to win.

b How many winning numbers can Jim make if 4 is spun?

c List all the winning numbers if 6 is spun.

d What is the largest possible winning number overall?

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CHALLENGE PROBLEMS – JUNIOR

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J3 Primelandia Money

In Primelandia, the unit of currency is the Tao (T). The value of each Primelandian coin is a prime number of Taos.So the coin with the smallest value is worth 2T. There are coins of every prime value less than 50. All payments inPrimelandia are an exact number of Taos.

a What is the smallest payment (without change) which requires at least three coins?

b A bag contains six different coins. Alice, Bob and Carol take two coins each from the bag and keep them. Theyfind that they have all taken the same amount of money. What is the smallest amount of money that could havebeen in the bag? Explain your reasoning.

c Find five Primelandian coins which, when placed in ascending order, form a sequence with equal gaps of 6T betweentheir values.

d The new King of Primelandia decides to mint coins of prime values greater than 50T. Show that no matter whatcoins are made, there is only one set of five Primelandian coins which, when placed in ascending order, form asequence with equal gaps of 6T between their values.

J4 Condates

On many forms, the date has to be written as DD/MM/YYYY. For example, 25 April 1736 is written 25/04/1736.Dates such as this that use eight consecutive digits (not necessarily in order) will be called condates.

a What is the first condate after the year 2015?

b Why must there be a 0 in every condate in the years 2000 to 2999?

c Why must every condate in the years 2000 to 2999 have 0 as the first digit of the month?

d How many condates are there in the years 2000 to 2999?

J5 Jogging

Theo is training for the annual Mt Killaman Joggin 10 km Torture Track. This ‘fun’ run involves running east on aflat track for 2 km to the base of Mt Killaman Joggin, then straight up its rather steep side a further 3 km to the top.This is the halfway point where you turn around and run back the way you came. This simplified mudmap shouldgive you the general idea.

• •2 km

Start/Finish Base of Mt KJ

3 km

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Top of Mt KJ

Theo can run at 12km/h on the flat. Up the hill he reckons he can average 9km/h and he is good for 15km/h on thedownhill part of the course.

a Calculate Theo’s expected time to complete the course, assuming his estimates for his running speeds are correct.

The day before the race, Theo hears that windy conditions are expected. He knows from experience that this willaffect his speed out on the course. When the wind blows into his face, it will slow him down by 1 km/h, but when it isat his back, he will speed up by the same amount. This applies to his speeds on the flat and on the slope. The windwill either blow directly from the east or directly from the west.

b From which direction should he hope the wind blows if he wishes to minimise his time? Justify your answer.

c Theo would like to finish the race in 50 minutes. He decides to run either the uphill leg faster or the downhill legfaster, but not both. Assuming there is no wind, how much faster would he have to run in each case?

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J6 Tessellating Hexagons

Will read in his favourite maths book that this hexagon tessellates the plane.

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To see if this was true, Will made 16 copies of the hexagon and glued them edge-to-edge onto a piece of paper, withno overlaps and no gaps.

a Cut out 16 hexagons from the worksheet and glue them as Will might have done so they entirely cover the dashedrectangle.

b Show that your block of 16 hexagons can be translated indefinitely to tessellate the entire plane.

c Find all points on the hexagon above that are points of symmetry of your tessellation.

d In your tessellation, select four hexagons that form one connected block which will tessellate the plane by translationonly. Indicate three such blocks on your tessellation that are not identical.

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2015 Challenge Problems - Intermediate

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

I1 Indim Integers

See Junior Problem 2.

I2 Digital Sums

The digital sum of an integer is the sum of all its digits. For example, the digital sum of 259 is 2 + 5 + 9 = 16.

a Find the largest even and largest odd 3-digit multiples of 7 which have a digital sum that is also a multiple of 7.

A digital sum sequence is a sequence of numbers that starts with any integer and has each number after the firstequal to the digital sum of the number before it. For example: 7598, 29, 11, 2. Any digital sum sequence ends in asingle-digit number and this is called the final digital sum or FDS of the first number in the sequence. Thus the FDSof 7598 is 2.

b Find the three largest 3-digit multiples of 7 which have an FDS of 7.

c Find and justify a rule that produces all 3-digit multiples of 7 which have an FDS of 7.

d Find and justify a rule that produces all 3-digit multiples of 8 which have an FDS of 8.

I3 Coin Flips

I have several identical coins placed on a table. I play a game consisting of one or more moves. Each move consists offlipping over some of the coins. The number of coins that are flipped stays the same for each game but may changefrom game to game. A coin showing ‘heads’ is represented by H . A coin showing ‘tails’ is represented by T . So, whena coin is flipped it changes from H to T or from T to H . The same coin may be selected on different moves. In eachgame we start with all coins showing tails.

For example, in one game we might start with five coins and flip two at a time like this:

Start: T T T T TMove 1: T H H T TMove 2: H H T T TMove 3: H H T H H

a Starting with 14 coins and flipping over four coins on each move, explain how to finish with exactly 10 coins headsup in three moves.

b Starting with 154 coins and flipping over 52 coins on each move, explain how to finish with all coins heads up inthree moves.

c Starting with 26 coins and flipping over four coins on each move, what is the minimum number of moves needed tofinish with all coins heads up?

d Starting with 154 coins and flipping over a fixed odd number of coins on each move, explain why I cannot have all154 coins heads up at the end of three moves.

I4 Jogging

See Junior Problem 5.

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CHALLENGE PROBLEMS – INTERMEDIATE

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I5 Folding Fractions

A square piece of paper has side length 1 and is shaded on the front and white on the back. The bottom-right corneris folded to a point P on the top edge as shown, creating triangles PQR, PTS and SUV .

P Q

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a If P is the midpoint of the top edge of the square, find the length of QR.

b If P is the midpoint of the top edge of the square, find the side lengths of triangle SUV .

c Find all positions of P so that the ratio of the sides of triangle SUV is 7:24:25.

I6 Crumbling Cubes

A large cube is made of 1 × 1 × 1 small cubes. Small cubes are removed in a sequence of steps.

The first step consists of marking all small cubes that have at least 2 visible faces and then removing only those smallcubes.

In the second and subsequent steps, the same procedure is applied to what remains of the original large cube after theprevious step.

a Starting with a 10 × 10 × 10 cube, how many small cubes are removed at the first step?

b A different cube loses 200 small cubes at the first step. What are the dimensions of this cube?

c Beginning with a 9 × 9 × 9 cube, what is the surface area of the object that remains after the first step?

d Beginning with a 9 × 9 × 9 cube, how many small cubes are left after the third step?

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2015 Challenge Solutions - Middle Primary

MP1 e-Numbers

a The digits which rotate to the same digit are: 0, 1, 2, 5, 8.

b The digits which rotate to a different digit are 6 and 9.

c Only the digits in Parts a and b can be used to form numbers whose rotations are numbers. So the only numbersfrom 10 to 50 whose rotations are numbers are: 11, 12, 15, 16, 18, 19, 21, 22, 25, 26, 28, 29.

d From Parts a and b, if a number and its rotation are the same, then the first and last digits of the number mustboth be 1, 2, 5, 8, or the first digit is 6 and the last 9 or the first 9 and the last 6. So the numbers from 50 to 200that stay the same when rotated are: 55, 69, 88, 96, 101, 111, 121, 151, 181.

MP2 Quazy Quilts

a The quilt after adding one grey and one white border:

The number of white squares is 18.

b The quilt after adding another grey and another white border:

There are 26 extra grey squares and 34 extra white squares.

c The table shows the number of squares in each border that we have counted so far.

Border 1 2 3 4

Squares 10 18 26 34

The number of squares from one border to the next appears to increase by 8. We now show this rule continues toapply.

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CHALLENGE SOLUTIONS – MIDDLE PRIMARY

29

Except for the corner squares, each square on the previous border corresponds to a square in the new border. Eachof the corner squares C on the previous border corresponds to a corner square C ′ in the new border. In addition,there are two new squares next to each corner.

previous border

new border

previous border

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Thus the number of squares from one border to the next increases by 4 × 2 = 8.

d Alternative i

The first border has 2 + 8 squares.The second border has 2 + 8 + 8 squares.The third border has 2 + 8 + 8 + 8 squares.And so on.

We want 2 + 8 + 8 + · · · = 90.Hence the number of 8s needed is 11.So Joy’s completed quilt has 11 borders.

Alternative ii

From Part c we have the following table.

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

Squares 10 18 26 34 42 50 58 66 74 82 90

So Joy’s completed quilt has 11 borders.

MP3 Egg Cartons

a

For each of these arrangements, any rotation or reflection is acceptable.

b

For each of these arrangements, any rotation or reflection is acceptable.

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c

For each of these arrangements, any rotation or reflection is acceptable.

d Each arrangement of six eggs in a 2× 4 carton has two empty positions. By systematically locating the two emptypositions, we see that there are ten different ways to arrange six eggs in a 2 × 4 carton.

For each of these arrangements, any rotation or reflection is acceptable.

MP4 Condates

a Any condate in the year 2015 must be written as DD/MM/15. The remaining digits are 0, 2, 3, 4. So the monthmust be 02, 03, or 04.

If the month is 02, then the day is 34 or 43, which are not allowed. If the month is 03, then the day must be 24. Ifthe month is 04, then the day must be 23. Thus the only condates in 2015 are 24/03/15 and 23/04/15.

b There are no condates in the years 2016, 2017, 2018, 2019 since their last digits are greater than 5.

In 2020, the remaining digits are 1, 3, 4, 5. No two of these form a month.

In 2021, the remaining digits are 0, 3, 4, 5. So the month must start with 0. Then the day must contain two of 3,4, 5, which is impossible.

The year 2022 duplicates 2.

In 2023, the only digits remaining are 0, 1, 4, 5. The day cannot use both 4 and 5, so the month is 04 or 05. Theearliest of these is 04, in which case the day must be 15. So the first condate after 2015 is 15/04/23.

c The latest possible year ends with 54. The latest month in that year is 12. So the latest day is 30. Thus the lastcondate in any century is 30/12/54.

d If DD/MM/YY contains all the digits 1, 2, 3, 4, 5, 6, then no digit can be repeated and no digit is 0. Hence themonth must be 12. So the day must contain two of the digits 3, 4, 5, 6. This is impossible.

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2015 Challenge Solutions - Upper Primary

UP1 Rod Shapes

a There are four equilateral triangles: (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4).

b To form an isosceles triangle that is not equilateral we need two rods of equal length and a third rod with a differentlength. Thus we have only the following choices:(1, 2, 2), (1, 3, 3), (1, 4, 4), (2, 1, 1), (2, 3, 3), (2, 4, 4), (3, 1, 1), (3, 2, 2), (3, 4, 4), (4, 1, 1), (4, 2, 2), (4, 3, 3).

To form a triangle, we must have each side smaller than the sum of the other two sides. This eliminates(2, 1, 1), (3, 1, 1), (4, 1, 1), (4, 2, 2).

So there are only eight isosceles triangles that are not equilateral triangles:(1, 2, 2), (1, 3, 3), (1, 4, 4), (2, 3, 3), (2, 4, 4), (3, 2, 2), (3, 4, 4), (4, 3, 3).

c A scalene triangle has three unequal sides. So there are four possibilities: (1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4).

To form a triangle, we must have each side smaller than the sum of the other two sides. This eliminates(1, 2, 3), (1, 2, 4), (1, 3, 4).

So there is only one scalene triangle: (2, 3, 4).

d If the pair of parallel sides have equal length, then the other pair of sides would be parallel. So the pair of parallelsides must have unequal length.

The possible lengths for the parallel sides are (1, 2), (1, 3), (2, 3).So we have only the following choices for the quadrilateral:(1, 1, 2, 1), (1, 2, 2, 2), (1, 3, 2, 3), (1, 1, 3, 1), (1, 2, 3, 2), (1, 3, 3, 3), (2, 1, 3, 1), (2, 2, 3, 2), (2, 3, 3, 3).

To form a quadrilateral, we must have each side smaller than the sum of the other three sides.This eliminates (1, 1, 3, 1). So there are just eight required quadrilaterals.

UP2 Egg Cartons

a Any four of the following arrangements.

For each of these arrangements, any rotation or reflection is acceptable.

b

For each of these arrangements, any rotation or reflection is acceptable.

c By systematically locating the two empty positions, we see that there are ten different ways to arrange six eggs ina 2 × 4 carton.

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CHALLENGE SOLUTIONS – UPPER PRIMARY

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For each of these arrangements, any rotation or reflection is acceptable.

d Each of Zoe’s arrangements in the 2×6 carton has 3 eggs and 9 empty places. Suppose the 3 eggs are white and fillthe empty places with 9 brown eggs. Then each placement of 3 white eggs corresponds to a placement of 9 browneggs. So the number of ways of arranging 3 eggs equals the number of ways of arranging 9 eggs.

UP3 Seahorse Swimmers

a To compare times, first convert each PB time to seconds.

Mack 55 sJack 1min 8 s = 68 sZac 1min 16 s = 76 sNick 1min 3 s = 63 s

Alternative i

The table shows the total PB times, in seconds, for all pairs that include Mack and for the corresponding pairs thatexclude Mack.

Pair with M MJ MZ MNTotal PB 123 131 118Other pair ZN JN JZTotal PB 139 131 144

Thus the pairs that have the same total PB time are Mack with Zac and Jack with Nick.

Alternative ii

The total PB time for the four swimmers is 55 + 68 + 76 + 63 = 262 seconds. It might be possible that the totalPB time for each pair is 262/2 = 131 seconds. To get the units digit 1 in the total we must pair 55 with 76 and 68with 63. Each pair then totals 131 as required. Thus Mack is paired with Zac and Jack is paired with Nick.

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b i Chloe’s PB time is 70 seconds and Sally’s is 53 seconds. So the total of all six PB times is 262 + 70 + 53 = 385seconds. Since 385 is odd and cannot be divided into two equal amounts, the two teams cannot have the sametotal PB times.

ii Alternative i

The total of all six PB times is 262 + 70 + 53 = 385 seconds. As explained in Part i, two teams cannot havethe same total PB times. The next closest would be 1 second apart, in which case the times would be 192 s and193 s. To see if this is possible, we need to select three PB times and add them to see if they total 192 or 193.It is easier to just add their units digits to see if we get 2 or 3. The swimmers’ PB times in increasing orderof their units digits are: 70, 53, 63, 55, 76, 68. The only three that give 3 in the units digit of their total are70, 55, 68, and these do in fact total 193 s. Thus one team has Chloe, Mack, and Jack with a total PB time of70 + 55 + 68 = 193 s. The other team has Sally, Nick, and Zac with a total PB time of 53 + 63 + 76 = 192 s.

Alternative ii

The table shows the total PB times, in seconds, for all teams that include Mack and for the correspondingteams that exclude Mack.

Team with M Total PB Other team Total PB

MJZ 199 NCS 186MJN 186 ZCS 199

MJC 193 ZNS 192MJS 176 ZNC 209MZN 194 JCS 191

MZC 201 JNS 184MZS 184 JNC 201

MNC 188 JZS 197MNS 171 JZC 214

MCS 178 JZN 207

Thus the two teams that have the closest total PB time are Mack, Jack, Chloe with 193 seconds and Zac, Nick,Sally with 192 seconds.

c If the average of seven PB times is 60 seconds, then the total of their PB times is 7× 60 = 420 seconds. From Partb, the total PB time for the six swimmers is 385 seconds. So the coach’s PB time was 420 − 385 = 35 seconds.

UP4 Primelandia Money

a From 50, we subtract primes from 50 down to 25 and check if the difference is prime:

50− 47 = 3, 50− 43 = 7, 50− 41 = 9,50− 37 = 13, 50− 31 = 19, 50− 29 = 21.

Since 9 and 21 are not prime, the only pairs of primes that total 50 are: (47,3), (43,7), (37,13), (31,19).

b We have the following amounts that can be paid with one or two coins:

2 = 2 7 = 7 12 = 5 + 7 17 = 17 22 = 3 + 193 = 3 8 = 3 + 5 13 = 13 18 = 5 + 13 23 = 234 = 2 + 2 9 = 2 + 7 14 = 3 + 11 19 = 19 24 = 5 + 195 = 5 10 = 3 + 7 15 = 2 + 13 20 = 3 + 17 25 = 2 + 236 = 3 + 3 11 = 11 16 = 3 + 13 21 = 2 + 19 26 = 3 + 23

We now show that 27T requires at least three coins.

Alternative i

From 27, we subtract primes from 27 down to 14 and check if the difference is prime:

27− 23 = 4, 27− 19 = 8, 27− 17 = 10.

None of the differences is prime. Hence 27T is the smallest payment which requires at least three coins.

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Alternative ii

The sum of two odd primes is even. So, if 27 is the sum of two primes, then one of those primes is 2, the only evenprime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallestpayment which requires at least three coins.

c The four largest primes less than 50 are 47, 43, 41, and 37. So the largest even amount that can be made from fourcoins is 47 + 43 + 41 + 37 = 168. If the sum of three primes is even, then one of those primes must be 2. So thelargest even amount that can be made from three coins is 47 + 43 + 2 = 92. The difference is 168 − 92 = 76.

d The bag can’t contain a 2T coin. If it did, then the sum of the pair of coins that includes 2T would be odd and thesum of each other pair of coins would be even.

Since all coins are odd, the sum of all six coins is even. Since the three pairs of coins have the same sum, the sumof all six coins is divisible by 3. The sum of all six coins is at least 3 + 5 + 7 + 11 + 13 + 17 = 56, which is not amultiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to72T.

Sum of six Sum of pair All possible pairs

60 20 3 + 17, 7 + 1366 22 3 + 19, 5 + 1772 24 5 + 19, 7 + 17, 11 + 13

Thus the smallest amount that could have been in the bag is 72T.

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2015 Challenge Solutions - Junior

J1 Quirky Quadrilaterals

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d If I is an odd number we can surround a column of I dots with a rhombus whose vertices are immediately above,below and either side of the middle of the column as the following examples show. Thus one diagonal of therhombus lies on the line through the column of I dots and the other diagonal lies on the line perpendicular to thefirst diagonal and passing through the middle dot.

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CHALLENGE SOLUTIONS – JUNIOR

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If I is an even number we can surround a south-west/north-east diagonal of I dots with a rhombus whose verticesare immediately south-west, north-east, and either side of the middle of the diagonal as the following examplesshow. Thus one diagonal of the rhombus lies on the line through the I dots and the other diagonal lies on the lineperpendicular to the first diagonal and bisecting the distance between the two middle dots.

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These rhombuses are not unique.

J2 Indim Integers

a Since 1 is a factor of every integer, no tile remains if 1 is spun.

If 2 is spun, then all remaining tiles have an odd number. Hence any 3-digit number made from the remaining tileswould be odd and not divisible by 2. So there is no winning number if 2 is spun.

If 5 is spun, then tiles 1 and 5 are removed. No number made from the remaining tiles is a multiple of 5 since itcannot end in 0 or 5. So there is no winning number if 5 is spun.

b The only digits remaining after spinning 4 are 3, 5, 6, 7, 9. A number made from these digits is divisible by 4 ifand only if it ends with a 2-digit number that is divisible by 4. The only such 2-digit multiples of 4 that can bemade are 36, 56, 76, and 96.

Alternative i

So there are 12 winning numbers: 536, 736, 936, 356, 756, 956, 376, 576, 976, 396, 596, 796.

Alternative ii

From each of 36, 56, 76, 96, we form a 3-digit number by choosing one of the remaining three digits as its first digit.So the number of winning numbers is 3 × 4 = 12.

c The only digits remaining after spinning 6 are 4, 5, 7, 8, 9. A 3-digit number is divisible by 6 if and only if its lastdigit is even and the sum of all its digits is divisible by 3. So the only winning numbers if 6 is spun are: 594, 954,894, 984, 498, 948, 798, 978.

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d From Part a, there is no winning number if 1, 2, or 5 is spun.

Alternative i

If 3 is spun, then tile 9 is excluded. So any winning number from spinning 3 must be less than 900.

From Part b, the largest winning number from spinning 4 is 976.

From Part c, the largest winning number from spinning 6 is 984.

The only 3-digit multiples of 7 larger than 984 are 987 and 994. If 7 is spun, then 987 is excluded. The multiple994 is excluded because 9 is repeated.

All 3-different-digit numbers larger than 984 start with 98. So there are no winning numbers larger than 984 if 8or 9 is spun.

So the largest winning number overall is 984.

Alternative ii

The largest number with three different digits is 987. This cannot be a winning number from spinning 9, 8, or 7.Since 987 is not divisible by 6 or 4, it cannot be a winning number from spinning 6 or 4. It is divisible by 3 butcannot be a winning number from spinning 3 since digit 9 would have been eliminated. So 987 is not a winningnumber.

The next largest number is 986. This cannot be a winning number from spinning 9, 8, or 6. Since 986 is not divisibleby 7, 4, or 3, it is not a winning number from spinning 7, 4, or 3. So 986 is not a winning number.

The next largest number is 985. This cannot be a winning number from spinning 9, 8, or 5. Since 985 is not divisibleby 7, 6, 4, or 3, it is not a winning number from spinning 7, 6, 4, or 3. So 985 is not a winning number.

The next largest number is 984. This is divisible by 6 and none of its digits is a factor or multiple of 6. So it is awinning number.

Thus the largest winning number overall is 984.

J3 Primelandia Money

a We have the following amounts that can be paid with one or two coins:

2 = 2 7 = 7 12 = 5 + 7 17 = 17 22 = 3 + 193 = 3 8 = 3 + 5 13 = 13 18 = 5 + 13 23 = 234 = 2 + 2 9 = 2 + 7 14 = 3 + 11 19 = 19 24 = 5 + 195 = 5 10 = 3 + 7 15 = 2 + 13 20 = 3 + 17 25 = 2 + 236 = 3 + 3 11 = 11 16 = 3 + 13 21 = 2 + 19 26 = 3 + 23

We now show that 27T requires at least three coins.

Alternative i

From 27, we subtract primes from 27 down to 14 and check if the difference is prime:

27− 23 = 4, 27− 19 = 8, 27− 17 = 10.

None of the differences is prime. Hence 27T is the smallest payment which requires at least three coins.

Alternative ii

The sum of two odd primes is even. So, if 27 is the sum of two primes, then one of those primes is 2, the only evenprime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallestpayment which requires at least three coins.

b The bag can’t contain a 2T coin. If it did, then the sum of the pair of coins that includes 2T would be odd and thesum of each other pair of coins would be even.

Since all coins are odd, the sum of all six coins is even. Since the three pairs of coins have the same sum, the sumof all six coins is divisible by 3. The sum of all six coins is at least 3 + 5 + 7 + 11 + 13 + 17 = 56, which is not amultiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to72T.

Sum of six Sum of pair All possible pairs

60 20 3 + 17, 7 + 1366 22 3 + 19, 5 + 1772 24 5 + 19, 7 + 17, 11 + 13

Thus the smallest amount that could have been in the bag is 72T.

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c Examining primes less than 50 for pairs that differ by 6 and starting with small primes, we quickly find the sequence5, 11, 17, 23, 29.

d Alternative i

Suppose we have a sequence of five prime numbers in ascending order which are 6 apart. Since 6 is even and 2 isthe only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The differencebetween any two primes in the sequence is 6, 12, 18, or 24. So no two primes in the sequence can end in the samedigit. There are five primes in the sequence so all of 1, 3, 5, 7, 9 must appear as last digits. The only prime whoselast digit is 5 is 5 itself. So the sequence must be 5, 11, 17, 23, 29.

Alternative ii

Suppose we have a sequence of five prime numbers in ascending order which are 6 apart. Since 6 is even and 2 isthe only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The onlyprime ending in 5 is 5 itself. Therefore, if the first prime in the sequence ends in 5, the sequence is 5, 11, 17, 23, 29.

If the first prime ends in 1, then the next four primes end in 7, 3, 9, 5. If the first prime ends in 3, then the nextfour primes end in 9, 5, 1, 7. If the first prime ends in 7, then the next four primes end in 3, 9, 5, 1. If the firstprime ends in 9, then the next four primes end in 5, 1, 7, 3. In each of these four cases, the prime ending in 5 is atleast 1 + 6 = 7, 3 + 6 = 9, 7 + 6 = 13, and 9 + 6 = 15 respectively. This is impossible.

So the only sequence is 5, 11, 17, 23, 29.

Alternative iii

Suppose we have a sequence of five prime numbers in ascending order which are 6 apart. Replace each of the fivenumbers in the sequence with its remainder when divided by 5.

If the first remainder is 0, then the sequence of remainders is 0, 1, 2, 3, 4. If the first remainder is 1, then thesequence of remainders is 1, 2, 3, 4, 0. If the first remainder is 2, then the sequence of remainders is 2, 3, 4, 0,1. If the first remainder is 3, then the sequence of remainders is 3, 4, 0, 1, 2. If the first remainder is 4, then thesequence of remainders is 4, 0, 1, 2, 3.

In all cases there is a remainder of 0, which represents a multiple of 5 in the original sequence. The only prime thatis divisible by 5 is 5 itself. Since there is no positive number that is 6 less than 5, 5 must be the first number in theoriginal sequence. Thus the original sequence is 5, 11, 17, 23, 29.

J4 Condates

a Any condate in a year starting with 20 must be DD/MM/20YY. So the month must be 1M. Then there is noavailable digit for M. So there is no condate in the years 2000 to 2099.

Any condate in a year starting with 21 must be DD/MM/21YY. So the month must be 0M. Hence the day is 3D.Then there is no available digit for D. So there is no condate in the years 2100 to 2199.

There is no condate in a year starting with 22, since 2 is repeated.

Any condate in a year starting with 23 must be DD/MM/23YY. So the month is 0M or 10. If the month is 10,then there is no available digit for the first D in the day. So the month is 0M and the condate is 1D/0M/23YY.The earliest year is 2345 and the earliest month in that year is 06. Hence the first condate after the year 2015 is17/06/2345.

b In any year either 0 or 1 must appear in the month. If 0 is not in the month, then the month is 12. Hence the yearcannot start with 2. So in the years 2000 to 2999, 0 must be in the month for any condate.

c From Part b, 0 must be in the month for any condate from 2000 to 2999. If the 0 is not the first digit of the month,then the month must be 10. Hence the remaining digits are 3, 4, 5, 6, 7, no two of which can represent a day. Soin the years 2000 to 2999, 0 must be the first digit in the month for any condate.

d From Part c, a condate in the years 2000 to 2999 has the form DD/0M/2YYY. So DD is either 1D or 3D.

If the condate is 1D/0M/2YYY, then each of the letters can be replaced by any one of the digits 3, 4, 5, 6, 7without repeating a digit. There are 5 choices for D, then 4 choices for M, and so on. So the number of condatesis 5 × 4 × 3 × 2 × 1 = 120.

If the condate is 3D/0M/2YYY, then D is 1 and the month has 31 days. The remaining available digits are 4, 5,6, 7. So M is 5 or 7. Thus there are 2 choices for M, then 3 choices for the first Y, and so on. So the number ofcondates is 2 × 3 × 2 × 1 = 12.

Thus the number of condates in the years 2000 to 2999 is 120 + 12 = 132.

20

39

J5 Jogging

a On the flat, Theo will run a total of 4 km at 12 km/h and this will take 4

12× 60 = 20 minutes. He will run 3 km

uphill at 9 km/h and this will take 3

9× 60 = 20 minutes. He will run 3 km downhill at 15 km/h and this will take

3

15× 60 = 12 minutes. So the total time taken for Theo to complete the course will be 20 + 20 + 12 = 52 minutes.

b Alternative i

If the wind blows from the west, then Theo’s speed and times for the various sections of the course will be:

Section Distance Speed Time

Flat east 2 km 13km/h 2

13h

Uphill 3 km 10km/h 3

10h

Downhill 3 km 14km/h 3

14h

Flat west 2 km 11km/h 2

11h

So the total time to complete the course would be ( 2

13+ 3

10+ 3

14+ 2

11) × 60 ≈ 51.0 minutes.

If the wind blows from the east, then Theo’s speed and times for the various sections of the course will be:

Section Distance Speed Time

Flat east 2 km 11km/h 2

11h

Uphill 3 km 8 km/h 3

8h

Downhill 3 km 16km/h 3

16h

Flat west 2 km 13km/h 2

13h

So the total time to complete the course would be ( 2

11+ 3

8+ 3

16+ 2

13) × 60 ≈ 53.9 minutes.

So Theo’s time will be less if the wind blows from the west.

Alternative ii

Since Theo runs both ways on the flat, he gains no advantage on the flat from either wind direction.

If the wind blows from the west, then his time on the hill will be 3

10+ 3

14hours. If the wind blows from the east,

then his time on the hill will be 3

8+ 3

16hours. To simplify comparison of these times we divide both by 3 and

multiply both by 2. So we want to compare 1

5+ 1

7= 12

35with 1

4+ 1

8= 12

32. Thus the first time is shorter.

So Theo’s time will be less if the wind blows from the west.

c Alternative i

From Part a, Theo has to gain 2 minutes.

From Part a, he ran 3km uphill at 9 km/h in 20 minutes. So to gain 2 minutes uphill, he will need to run 3 km in18 minutes. Therefore his speed needs to be 3

18× 60 = 10 km/h, which is 1 km/h faster.

From Part a, he ran 3 km downhill at 15 km/h in 12 minutes. So to gain 2 minutes downhill, he will need to run3 km in 10 minutes. Therefore his speed needs to be 3

10× 60 = 18km/h, which is 3 km/h faster.

Alternative ii

Theo wants to complete the course in 50 minutes. From Part a, the flat takes 20 minutes. This leaves 30 minutesfor the hill.

If Theo’s speed uphill is r km/h and downhill is 15 km/h, then the time taken for the hill in hours is 3

r+ 3

15= 30

60= 1

2.

So 3

r= 1

2− 1

5= 3

10. Hence r = 10km/h. This is 1 km/h faster.

If Theo’s speed downhill is r km/h and uphill is 9 km/h, then the time taken for the hill in hours is 3

9+ 3

r= 30

60= 1

2.

So 3

r= 1

2− 1

3= 1

6. Hence r = 18 km/h. This is 3 km/h faster.

21

39 39

J6 Tessellating Hexagons

a The flipped hexagons are shaded.

A B

C

D

E

F A

B

C

DE

F

A

B

C

D E

FAB

C

D

E

F

A B

C

D

E

F A

B

C

DE

F

A

B

C

D E

FAB

C

D

E

F

A B

C

D

E

F A

B

C

DE

F

A

B

C

D E

FAB

C

D

E

F

A B

C

D

E

F A

B

C

DE

F

A

B

C

D E

FAB

C

D

E

F

b Call the block of hexagons in the solution to Part a the 16-block.

Alternative i

We see from the 16-block that AB = ED, BC = FA, the sum of the three interior angles A, C, D is 360, and thesum of the three interior angles B, E, F is also 360.

Indicating a hexagon side with a dash, the bottom boundary of the 16-block isA − BF − E − D − CA − BF − E − D and the top boundary isD − E − FB − AC − D − E − FB − A.So these boundaries are identical. Hence the 16-block can be translated vertically to form an infinite column withoutgaps and without overlap.

The right boundary of this column isDA − F − EB − C − DA − F − EB − C − DA repeated indefinitely.The left boundary of the column isC − BE − F − AD − C − BE − F − AD − C repeated indefinitely.So these boundaries are identical. Hence the column can be translated horizontally to tessellate the plane.

Alternative ii

Make several photocopies of the 16-block. Place them side by side and top to bottom to show how they fit together.

22

41

c There is no point of symmetry for the tessellation inside a hexagon because the hexagon has no point of symmetry.So all points of symmetry for the tessellation must be on the edges of the hexagons. If a point of symmetry for thetessellation is on an edge of a hexagon, then it must be at the midpoint of that edge.

From the block of hexagons in Part a, we see that each of the edges AB, BC, DE, FA, has the smallest edge EFattached at one end but not the other. So the midpoints of edges AB, BC, DE, FA are not points of symmetryfor the tessellation.

To check the midpoint of edge CD, trace a copy of the block and place it exactly over the original block. Thenplace a pin through the midpoint of edge CD, rotate the traced copy of the block through 180 and notice that itagain fits exactly over the original block (except for overhang). Thus the midpoint of CD is a point of symmetryfor the tessellation. Similarly for the midpoint of EF .

So the only points of symmetry for the tessellation are the midpoints of edges CD and EF .

23

41 41

d The block of hexagons in Part a is reproduced below with hexagons labelled H1, H2, H3, H4, H ′

1, H ′

2, H ′

3, H ′

4, andso on as shown.

A B

C

D

E

F H1

A

B

C

DE

F

H2

A

B

C

D E

F

H4

AB

C

D

E

FH3

A B

C

D

E

F H ′

1

A

B

C

DE

F

H ′

2

A

B

C

D E

F

H ′

4

AB

C

D

E

FH ′

3

A B

C

D

E

F H ′′

1

A

B

C

DE

F

H ′′

2

A

B

C

D E

F

H ′′

4

AB

C

D

E

FH ′′

3

A B

C

D

E

F H ′′′

1

A

B

C

DE

F

H ′′′

2

A

B

C

D E

F

H ′′′

4

AB

C

D

E

FH ′′′

3

Hexagons with the same subscript translate to one another. So there are several blocks of four hexagons thattessellate by translation. Three such blocks are: H1, H2, H3, H4, H1, H ′′

2 , H3, H4, H ′

1, H2, H3, H ′

4.

24

43

2015 Challenge Solutions - Intermediate

I1 Indim Integers

See Junior Problem 2.

I2 Digital Sums

a Alternative i

The table lists 3-digit numbers n in decreasing order whose digit sums s are multiples of 7.

n s 7 divides n? n s 7 divides n?

993 21 no 905 14 no984 21 no 894 21 no975 21 no 885 21 no966 21 yes 876 21 no957 21 no 867 21 no950 14 no 860 14 no948 21 no 858 21 no941 14 no 851 14 no939 21 no 849 21 no932 14 no 842 14 no923 14 no 833 14 yes914 14 no

So the largest even and odd n that are multiples of 7 and have a digital sum that is also a multiple of 7 are 966 and833 respectively.

Alternative ii

The table lists 3-digit numbers n and their digital sums s. The n are multiples of 7 in decreasing order.

n s 7 divides s? n s 7 divides s?

994 22 no 910 10 no987 24 no 903 12 no980 17 no 896 23 no973 19 no 889 25 no966 21 yes 882 18 no959 23 no 875 20 no952 16 no 868 22 no945 18 no 861 15 no938 20 no 854 17 no931 13 no 847 19 no924 15 no 840 12 no917 17 no 833 14 yes

So the largest even and odd n that are multiples of 7 and have a digital sum that is also a multiple of 7 are 966 and833 respectively.

b Alternative i

The largest 3-digit number with a digital sum of 7 is 700. The digital sum s of a 3-digit number is at most 3×9 = 27.So, if the FDS of a 3-digit number n is 7 and n > 700, then the digital sum of n is 16 or 25. The table lists, indecreasing order, the first few 3-digit numbers n with digital sum 16 or 25.

25

CHALLENGE SOLUTIONS – INTERMEDIATE

43 43

s n 7 divides n? s n 7 divides n?

16 970 no 25 997 no961 no 988 no952 yes 979 no943 no 898 no934 no 889 yes925 no 799 no916 no907 no880 no871 no862 no853 no844 no835 no826 yes

So the three largest 3-digit multiples of 7 that have FDS 7 are: 952, 889, 826.

Alternative ii

The table lists, in decreasing order, multiples of 7 and their digital sum sequences (DSS).

n DSS n DSS

994 22, 4 903 12, 3987 24, 6 896 23, 5980 17, 8 889 25, 7973 19, 10, 1 882 18, 9966 21, 3 875 20, 2959 23, 5 868 22, 4952 16, 7 861 15, 6945 18, 9 854 17, 8938 20, 2 847 19, 10, 1931 13, 4 840 12, 3924 15, 6 833 14, 5917 17, 8 826 16, 7910 10, 1

So the three largest 3-digit multiples of 7 that have FDS 7 are: 952, 889, 826.

c Alternative i

The table lists, in decreasing order, the 3-digit integers n that are multiples of 7 together with their FDSs (F).

26

45

n F n F n F n F n F

994 4 896 5 798 6 693 9 595 1987 6 889 7 791 8 686 2 588 3980 8 882 9 784 1 679 4 581 5973 1 875 2 777 3 672 6 574 7966 3 868 4 770 5 665 8 567 9959 5 861 6 763 7 658 1 560 2952 7 854 8 756 9 651 3 553 4945 9 847 1 749 2 644 5 546 6938 2 840 3 742 4 637 7 539 8931 4 833 5 735 6 630 9 532 1924 6 826 7 728 8 623 2 525 3917 8 819 9 721 1 616 4 518 5910 1 812 2 714 3 609 6 511 7903 3 805 4 707 5 602 8 504 9

700 7

497 2 399 3 294 6 196 7490 4 392 5 287 8 189 9483 6 385 7 280 1 182 2476 8 378 9 273 3 175 4469 1 371 2 266 5 168 6462 3 364 4 259 7 161 8455 5 357 6 252 9 154 1448 7 350 8 245 2 147 3441 9 343 1 238 4 140 5434 2 336 3 231 6 133 7427 4 329 5 224 8 126 9420 6 322 7 217 1 119 2413 8 315 9 210 3 112 4406 1 308 2 203 5 105 6

301 4

Thus the only 3-digit multiples of 7 that have FDS 7 are: 133, 196, 259, 322, 385, 448, 511, 574, 637, 700, 763, 826,889, 952. These have a common difference of 63. So the only 3-digit multiples of 7 that have FDS 7 are the integersn = 7 + 63t where t = 2, 3, . . ., 15.

Alternative ii

The three numbers from Part b, 952, 889, 826, have a common difference of 63. This suggests that all 3-digitmultiples of 7 that have FDS 7 have the form 7 + 63m. These numbers are: 133, 196, 259, 322, 385, 448, 511, 574,637, 700, 763, 826, 889, 952. Since 7 divides 7 and 63, all these numbers are multiples of 7. By direct calculation,each has FDS 7. But are there any other 3-digit multiples of 7 that have FDS 7?

Extending the table in the first solution to Part b shows that there are no other n besides those of the form 7+63mthat have digital sum 16 or 25. The following table lists in decreasing order all 3-digit integers with digital sum 7.

n 7 divides n? n 7 divides n? n 7 divides n?

700 yes 340 no 205 no610 no 331 no 160 no601 no 322 yes 151 no520 no 313 no 142 no511 yes 304 no 133 yes502 no 250 no 124 no430 no 241 no 115 no421 no 232 no 106 no412 no 223 no403 no 214 no

So there are no other 3-digit multiples of 7 with FDS 7 besides those of the form 7 + 63m.

27

45 45

Alternative iii

From the table in the first solution to Part b, it appears that integers with the same FDS differ by a multiple of9. This is equivalent to saying they have the same remainder when divided by 9. We now show that this is alwaystrue.

Any positive integer n can be written in the form

n = a + 10b + 100c + · · ·

where a, b, . . . are the digits of n. Thus

n = a + (1 + 9)b + (1 + 99)c + · · ·

= (a + b + c + · · ·) + 9(b + 11c + · · ·)

So n and its digital sum have the same remainder when divided by 9. Therefore all digital sums in the digital sumsequence for n, including its FDS, have the same remainder when divided by 9. Hence, if n is a multiple of 9, thenthe FDS of n is 9, and if n is not a multiple of 9, then the FDS of n is the remainder when n is divided by 9.

So, the FDS of n is 7 if and only if n has the form n = 7 + 9r. Such an n is a multiple of 7 if and only if 7 dividesr. So n is a multiple of 7 and has FDS 7 if and only if n has the form n = 7 + 63t. Hence the only 3-digit multiplesof 7 that have FDS 7 are the integers n = 7 + 63t where t = 2, 3, . . ., 15. These are: 133, 196, 259, 322, 385, 448,511, 574, 637, 700, 763, 826, 889, 952.

d Alternative i

The table lists, in decreasing order, the 3-digit integers n that are multiples of 8 together with their FDSs (F).

n F n F n F n F n F

992 2 896 5 792 9 696 3 592 7984 3 888 6 784 1 688 4 584 8976 4 880 7 776 2 680 5 576 9968 5 872 8 768 3 672 6 568 1960 6 864 9 760 4 664 7 560 2952 7 856 1 752 5 656 8 552 3944 8 848 2 744 6 648 9 544 4936 9 840 3 736 7 640 1 536 5928 1 832 4 728 8 632 2 528 6920 2 824 5 720 9 624 3 520 7912 3 816 6 712 1 616 4 512 8904 4 808 7 704 2 608 5 504 9

800 8 600 6

496 1 392 5 296 8 192 3488 2 384 6 288 9 184 4480 3 376 7 280 1 176 5472 4 368 8 272 2 168 6464 5 360 9 264 3 160 7456 6 352 1 256 4 152 8448 7 344 2 248 5 144 9440 8 336 3 240 6 136 1432 9 328 4 232 7 128 2424 1 320 5 224 8 120 3416 2 312 6 216 9 112 4408 3 304 7 208 1 104 5400 4 200 2

Thus the only 3-digit multiples of 8 that have FDS 8 are: 152, 224, 296, 368, 440, 512, 584, 656, 728, 800, 872,944. These have a common difference of 72. So the only 3-digit multiples of 8 that have FDS 8 are the integersn = 8 + 72t where t = 2, 3, . . ., 13.

Alternative ii

From the third solution to Part c, the FDS of a positive integer n is 8 if and only if n has the form n = 8 + 9r.Such an n is a multiple of 8 if and only if 8 divides r. So n is a multiple of 8 and has FDS 8 if and only if n has theform n = 8 + 72t. Hence the only 3-digit multiples of 8 that have FDS 8 are the integers n = 8 + 72t where t = 2,3, . . ., 13. These are: 152, 224, 296, 368, 440, 512, 584, 656, 728, 800, 872, 944.

28

47

I3 Coin Flips

a Here are three moves that leave exactly 10 coins heads up.

Start: T T T T T T T T T T T T T TMove 1: H H H H T T T T T T T T T TMove 2: H H H H H H H H T T T T T TMove 3: H H H H H H H T H H H T T T

b Since there are 52 coins flipped in each move, the total number of coin flips is 3 × 52 = 156. We start with 154 T sand want to finish with 154 Hs. We can achieve this by flipping 153 coins exactly once and flipping one coin threetimes over the three moves. Here are three moves that leave all coins heads up.

Start:

154︷ ︸︸ ︷

T . . .T

Move 1:

52︷ ︸︸ ︷

H . . .H

102︷ ︸︸ ︷

T . . .T

=

51︷ ︸︸ ︷

H . . .H H

51︷ ︸︸ ︷

T . . .T

51︷ ︸︸ ︷

T . . .T

Move 2:

51︷ ︸︸ ︷

H . . .H T

51︷ ︸︸ ︷

H . . .H

51︷ ︸︸ ︷

T . . .T

Move 3:

51︷ ︸︸ ︷

H . . .H H

51︷ ︸︸ ︷

H . . .H

51︷ ︸︸ ︷

H . . .H

c Since there are 4 coins flipped in each move, the total number of coins flipped in 6 moves is at most 24. So it takesat least 7 moves to finish with all coins heads up. We now show that this can actually be done in 7 moves.

After three moves, each flipping 4 Ts to 4 Hs, we get exactly 12 coins heads up. Then applying the three moves inPart a to the 14 coins that are tails up, gives us exactly 12 + 10 = 22 coins heads up. One more move, flipping 4Ts to 4 Hs, results in all coins heads up. Thus seven moves suffice to get all coins heads up.

So 7 is the minimum number of moves to get all coins heads up.

d Alternative i

Suppose m coins are flipped on each move where m is odd. Two moves give the following results.

Start:

154︷ ︸︸ ︷

T . . .T

Move 1:

m︷ ︸︸ ︷

H . . .H

154−m︷ ︸︸ ︷

T . . .T

Move 2:

m−n︷ ︸︸ ︷

H . . .H

n︷ ︸︸ ︷

T . . .T

m−n︷ ︸︸ ︷

H . . .H

154−2m+n︷ ︸︸ ︷

T . . .T where 0 ≤ n ≤ m andm− n ≤ 154 − m

=

2m−2n︷ ︸︸ ︷

H . . .H

154−2m+2n︷ ︸︸ ︷

T . . .T

To have all 154 coins with heads up after Move 3, we must have m = 154 − 2m + 2n. Thus 3m = 154 + 2n. But3m is odd and 154 + 2n is even. So we cannot have 154 coins heads up after just three moves.

29

47 47

Alternative ii

Three moves give the following results. Note that in a move, an even number of coins may mean 0 coins.

Start:

154︷ ︸︸ ︷

T . . .T

Move 1:

odd︷ ︸︸ ︷

H . . .H

odd︷ ︸︸ ︷

T . . .T

Move 2:

even︷ ︸︸ ︷

H . . .H

odd︷ ︸︸ ︷

T . . .T

even︷ ︸︸ ︷

H . . .H

odd︷ ︸︸ ︷

T . . .T orodd

︷ ︸︸ ︷

H . . .H

even︷ ︸︸ ︷

T . . .T

odd︷ ︸︸ ︷

H . . .H

even︷ ︸︸ ︷

T . . .T givingeven

︷ ︸︸ ︷

H . . .H

even︷ ︸︸ ︷

T . . .T

Move 3:

even︷ ︸︸ ︷

H . . .H

even︷ ︸︸ ︷

T . . .T

odd︷ ︸︸ ︷

H . . .H

odd︷ ︸︸ ︷

T . . .T orodd

︷ ︸︸ ︷

H . . .H

odd︷ ︸︸ ︷

T . . .T

even︷ ︸︸ ︷

H . . .H

even︷ ︸︸ ︷

T . . .T givingodd

︷ ︸︸ ︷

H . . .H

odd︷ ︸︸ ︷

T . . .T

Since 154 is even, we cannot have 154 coins heads up after just three moves.

I4 Jogging

See Junior Problem 5.

I5 Folding Fractions

a Let QR = x. Then PR = 1 − x.

P Q

R

T

V

U

S

1

2

1

2

x

1 − x

1 − x

In PQR Pythagoras’ theorem gives 1

4+ x2 = (1 − x)2 = x2 − 2x + 1.

Hence 2x = 3

4and x = 3

8.

b Alternative i

From Part a, we have:

P Q

R

T

V

U

S

1

2

1

2

3

85

8

a

ba

b

a

b

30

49

Since angles RQP , RPS, PTS, SUV are right angles, triangles RQP , PTS, V US have the same angles as shown.Hence they are similar and we have: V U :US:SV = PT :TS:SP = RQ:QP :PR = 3:4:5.

Hence PS = 5

3PT = 5

3× 1

2= 5

6. So US = UP − SP = 1 − 5

6= 1

6.

Then SV = 5

4US = 5

4× 1

6= 5

24and V U = 3

4US = 3

4× 1

6= 1

8.

Alternative ii

As in Alternative i, V U :US:SV = PT :TS:SP = RQ:QP :PR = 3:4:5. So ST = 4

3PT = 4

3× 1

2= 2

3.

From the side of the square, we have: 1 = UV + V S + ST = 3

5V S + V S + 2

3= 8

5V S + 2

3.

So V S = 1

3× 5

8= 5

24.

From V US, we have: V U = 3

5SV = 3

5× 5

24= 1

8and US = 4

5SV = 4

5× 5

24= 1

6.

Alternative iii

Draw RW perpendicular to TS and let TS = t and V U = r. From Part a, we have:

P Q

R

T

V

U

S

W

1

2

1

2

3

85

8

3

8

t − 3

8

r

r

Now apply Pythagoras’ theorem to the following triangles.

In RWS, RS2 = RW 2 + WS2 = 1 + (t − 3

8)2.

In RPS, RS2 = RP 2 + PS2 = (5

8)2 + PS2.

In PTS, PS2 = PT 2 + TS2 = (1

2)2 + t2.

Hence1 + (t − 3

8)2 = (5

8)2 + (1

2)2 + t2

1 + t2 + 9

64− 3

4t = 25

64+ 1

4+ t2

3

4t = 1 − 1

4+ 9

64− 25

64

= 3

4− 16

64= 3

4− 1

4= 1

2

t = 4

3× 1

2= 2

3

So PS2 = (1

2)2 + (2

3)2 = 1

4+ 4

9= 25

36and PS = 5

6.

Hence US = UP − SP = 1 − 5

6= 1

6.

Since 1 = TS + SV + r = 2

3+ SV + r, SV = 1

3− r.

In V US, Pythagoras’ theorem gives V S2 = V U2 + US2 . Hence

(1

3− r)2 = r2 + (1

6)2

1

9+ r2 − 2

3r = r2 + 1

36

2

3r = 1

9− 1

36= 3

36= 1

12

r = 3

2× 1

12= 1

8

Thus V U = 1

8and SV = 1

3− 1

8= 5

24.

31

49 49

c Since angles RQP , RPS, PTS, SUV are right angles, triangles RQP , PTS, V US have the same angles as shown.Hence they are similar and we have RQ:QP :PR = PT :TS:SP = V U :US:SV = 7:24:25 or 24:7:25.

P Q

R

T

V

U

S

a

ba

b

a

b

Alternative i

If PQ/QR = 7/24, let PQ = 7k. Then QR = 24k and PR = 25k. Since PR = 1 −QR, we have 25k = 1− 24k. So49k = 1 and PQ = 7 × 1

49= 1

7.

If PQ/QR = 24/7, let PQ = 24k. Then QR = 7k and PR = 25k. Since PR = 1 − QR, we have 25k = 1 − 7k. So32k = 1 and PQ = 24 × 1

32= 3

4.

Alternative ii

If PQ/QR = 7/24, then PR/QR = 25/24. Since PR = 1 − QR, we have 24 − 24QR = 25QR, QR = 24/49, andPR = 25/49. From Pythagoras, PQ2 = PR2 − QR2 = 1 − 2QR = 1 − 48/49 = 1/49. So PQ = 1

7.

If PQ/QR = 24/7, then PR/QR = 25/7. Since PR = 1 − QR, we have 7 − 7QR = 25QR, QR = 7/32, andPR = 25/32. From Pythagoras, PQ2 = PR2 − QR2 = 1 − 2QR = 1 − 7/16 = 9/16. So PQ = 3

4.

Alternative iii

Let TP = x. If V U/US = 7/24, let V U = 7k. Then US = 24k and V S = 25k and we have:

P Q

R

T

V

U

S

x

1 − 24k1 − 32k

25k24k

7k

7k

So (1 − 24k)/25k = x/7k = (1 − 32k)/24k.Hence 25x = 7(1 − 24k) and 24x = 7(1 − 32k).Subtracting gives x = 7(32 − 24)k = 56k.Substituting gives 25(56k) = 7(1 − 24k).Hence 1 = 224k and x = 56/224 = 8/32 = 1/4.

32

51

Alternatively, in PTS, Pythagoras’ theorem gives

(1 − 24k)2 = (1 − 32k)2 + (56k)2

(24k)2 − 48k = (32k)2 − 64k + (56k)2

16k = (562 + 322 − 242)k2

1 = (142 + 82 − 62)k = 224k

x = 56/224 = 8/32 = 1/4

If V U/US = 24/7, let V U = 24k. Then US = 7k and V S = 25k. We have:

P Q

R

T

V

U

S

x

1 − 7k1 − 49k

25k7k

24k

24k

So (1 − 7k)/25k = x/24k = (1 − 49k)/7k.Hence 25x = 24(1 − 7k) and 7x = 24(1− 49k).Subtracting gives 18x = 24(49− 7)k = 24(42)k, hence x = 56k.Substituting gives 7(56k) = 24(1− 49k).Hence 49k = 3(1− 49k), 4(49k) = 3, x = 56(3)/4(49) = 6/7.

Alternatively, in PTS, Pythagoras’ theorem gives

(1 − 7k)2 = (1 − 49k)2 + (56k)2

(7k)2 − 14k = (49k)2 − 98k + (56k)2

84k = (562 + 492 − 72)k2

12 = (82 + 72 − 12)7k = (112)7k

3 = (28)7k

x = 6/7

I6 Crumbling Cubes

a The small cubes have at most three faces exposed. The only small cubes that have exactly three faces exposedare the 8 on the corners of the 10 × 10 × 10 cube. The only small cubes that have exactly two faces exposed arethe 8 on each edge of the 10 × 10 × 10 cube that are not on its corners. All other cubes have less than two facesexposed. There are 12 edges on the 10 × 10 × 10 cube, so the number of small cubes removed at the first step is8 + (12 × 8) = 104.

b Alternative i

At the first step the 8 small cubes at the corners are removed. The other 192 cubes come equally from the 12 edgesbut not from the corners. So the number of non-corner small cubes on each edge is 192/12 = 16. Hence each edgein the original cube had a total of 18 small cubes.

Alternative ii

A cube with n small cubes along one edge will lose 8+12(n−2) small cubes at the first step. So 8+12(n−2) = 200,12(n − 2) = 192, n − 2 = 16, n = 18. Thus the original cube was 18 × 18× 18.

33

51 51

Alternative iii

From Part a, a 10 × 10 × 10 cube loses 104 small cubes at the first step. Increasing the cube’s dimension by 1increases the number of lost cubes by 12, one for each edge. Since 200 − 104 = 96 and 96/12 = 8, the cube thatloses 200 small cubes at the first step is 18 × 18× 18.

c Alternative i

At the first step, each of the 6 faces of the 9 × 9× 9 cube are converted to a 7 × 7 single layer of small cubes. Theexposed surface area of this layer is 7 × 7 small faces plus a ring of 4 × 7 small faces. So the surface area of theremaining object is 6 × (49 + 28) = 6 × 77 = 462.

Alternative ii

First remove the middle small cube on one edge of the 9 × 9 × 9 cube. This increases the surface area by 2 smallfaces. Then remove the small cubes either side. This does not change the surface area. Continue until only the twocorner cubes remain. At this stage the surface area has increased by 2 small faces. Repeating this process on all12 edges increases the surface area by 12 × 2 small faces. At this stage all 8 corner cubes have 6 exposed faces. Soremoving the 8 corner cubes reduces the surface area by 8× 6 small faces. The surface area of the original 9× 9× 9cube was 6 × 81 = 486. Hence the surface area of the remaining object is 486 + 24− 48 = 462.

d At the first step, the original 9 × 9 × 9 cube is reduced to a 7 × 7 × 7 cube with a 7 × 7 single layer of small cubesplaced centrally on each face. At this stage the number of small cubes removed is 8 + (12 × 7) = 92.

7 × 7

7×77× 7

At the second step the only small cubes removed are the edge cubes in the 7×7 single layers. This leaves a 7×7×7cube with a 5 × 5 single layer of small cubes placed centrally on each face. So in this step, the number of smallcubes removed is 6(4 + 4 × 5) = 144.

5 × 5

5×55× 5

At the third step the only small cubes removed are the edge cubes in the 5 × 5 single layers and the edge cubes inthe 7 × 7 × 7 cube. So in this step, the number of small cubes removed is 6(4 + 4 × 3) + 8 + (12 × 5) = 164.

Thus the number of small cubes remaining is (9 × 9 × 9) − 92 − 144 − 164 = 329.

34

53

CHALLENGE STATISTICS – MIDDLE PRIMARY

SCORE DISTRIBUTION %/PROBLEM

Score

Challenge Problem

1e-Numbers

2Quazy Quilts

3Egg Cartons

4Condates

Did not attempt 0% 1% 2% 3%

0 7% 6% 10% 10%

1 15% 12% 11% 18%

2 29% 20% 18% 20%

3 32% 29% 33% 24%

4 16% 32% 26% 25%

Mean 2.3 2.7 2.6 2.4

Discrimination Factor 0.5 0.6 0.6 0.6

MEAN SCORE/SCHOOL YEAR/PROBLEM

Year Number of Students

Mean

OverallProblem

1 2 3 4

3 486 8.8 2.1 2.4 2.3 2.1

4 653 10.6 2.6 2.9 2.7 2.6

All Years* 1166 9.8 2.3 2.7 2.6 2.4

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

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.

53 53

CHALLENGE STATISTICS – UPPER PRIMARY

SCORE DISTRIBUTION %/PROBLEM

Score

Challenge Problem

1Rod Shapes

2Egg Cartons

3Seahorse Swimmers

4Primelandia

Money

Did not attempt 1% 0% 1% 3%

0 3% 3% 5% 11%

1 40% 6% 14% 29%

2 24% 20% 20% 26%

3 21% 38% 29% 20%

4 11% 33% 31% 12%

Mean 2.0 2.9 2.7 1.9

Discrimination Factor 0.5 0.4 0.6 0.6

MEAN SCORE/SCHOOL YEAR/PROBLEM

Year Number of Students

Mean

OverallProblem

1 2 3 4

3 486 8.8 2.1 2.4 2.3 2.1

4 653 10.6 2.6 2.9 2.7 2.6

All Years* 1166 9.8 2.3 2.7 2.6 2.4

MEAN SCORE/SCHOOL YEAR/PROBLEM

YEAR Number of Students

Mean

OverallProblem

1 2 3 4

5 1363 8.6 1.8 2.8 2.4 1.7

6 1910 9.8 2.0 3.0 2.8 2.0

7 129 10.9 2.3 3.3 3.2 2.2

All Years* 3416 9.4 2.0 2.9 2.7 1.9

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

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.

55

CHALLENGE STATISTICS – JUNIOR

SCORE DISTRIBUTION %/PROBLEM

Score

Challenge Problem

1Quirky

Quadrilaterals

2Indim

Integers

3Primelandia

Money

4Condates

5Jogging

6Tessellating Hexagons

Did not attempt 3% 3% 7% 8% 9% 23%

0 7% 6% 8% 15% 10% 28%

1 8% 17% 24% 14% 18% 20%

2 15% 23% 30% 19% 16% 13%

3 36% 26% 22% 31% 21% 10%

4 31% 26% 10% 13% 27% 5%

Mean 2.8 2.5 2.0 2.1 2.4 1.3

Discrimination Factor 0.5 0.6 0.6 0.7 0.7 0.5

MEAN SCORE/SCHOOL YEAR/PROBLEM

Year Number of Students

Mean

OverallProblem

1 2 3 4 5 6

7 2607 11.0 2.6 2.3 1.9 2.0 2.2 1.1

8 2373 13.4 3.0 2.7 2.2 2.3 2.7 1.4

All Years* 5006 12.2 2.8 2.5 2.0 2.1 2.4 1.3

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

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.

55 55

CHALLENGE STATISTICS – INTERMEDIATE

SCORE DISTRIBUTION %/PROBLEM

Score

Challenge Problem

1Indim

Integers

2Digital Sums

3Coin Flips

4Jogging

5Folding

Fractions

6Crumbling

CubesDid not attempt 1% 3% 5% 3% 14% 9%

0 2% 6% 4% 4% 14% 11%

1 7% 11% 6% 9% 21% 7%

2 15% 26% 15% 12% 21% 21%

3 25% 16% 28% 22% 8% 25%

4 50% 38% 42% 49% 22% 27%

Mean 3.2 2.7 3.0 3.1 2.0 2.6

Discrimination Factor 0.4 0.6 0.6 0.6 0.7 0.7

MEAN SCORE/SCHOOL YEAR/PROBLEM

Year Number of Students

Mean

OverallProblem

1 2 3 4 5 6

9 2038 15.2 3.1 2.6 3.0 3.0 2.0 2.5

10 1055 16.6 3.3 2.9 3.2 3.2 2.2 2.7

All Years* 3104 15.7 3.2 2.7 3.0 3.1 2.0 2.6

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

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.

57

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD2015 Australian Intermediate Mathematics Olympiad - Questions

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

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

1. A number written in base a is 123a. The same number written in base b is 146b. What is the minimum value ofa + b? [2 marks]

2. A circle is inscribed in a hexagon ABCDEF so that each side of the hexagon is tangent to the circle. Find theperimeter of the hexagon if AB = 6, CD = 7, and EF = 8. [2 marks]

3. A selection of 3 whatsits, 7 doovers and 1 thingy cost a total of $329. A selection of 4 whatsits, 10 doovers and 1thingy cost a total of $441. What is the total cost, in dollars, of 1 whatsit, 1 doover and 1 thingy? [3 marks]

4. A fraction, expressed in its lowest termsa

b, can also be written in the form

2

n+

1

n2, where n is a positive integer.

If a + b = 1024, what is the value of a? [3 marks]

5. Determine the smallest positive integer y for which there is a positive integer x satisfying the equation213 + 210 + 2x = y2. [3 marks]

6. The large circle has radius 30/√

π. Two circles with diameter 30/√

π lie inside the large circle. Two more circleslie inside the large circle so that the five circles touch each other as shown. Find the shaded area.

[4 marks]

7. Consider a shortest path along the edges of a 7 × 7 square grid from its bottom-left vertex to its top-right vertex.How many such paths have no edge above the grid diagonal that joins these vertices? [4 marks]

8. Determine the number of non-negative integers x that satisfy the equation

⌊

x

44

⌋

=

⌊

x

45

⌋

.

(Note: if r is any real number, then r denotes the largest integer less than or equal to r.) [4 marks]

1

57 57

9. A sequence is formed by the following rules: s1 = a, s2 = b and sn+2 = sn+1 + (−1)nsn for all n ≥ 1.

If a = 3 and b is an integer less than 1000, what is the largest value of b for which 2015 is a member of the sequence?Justify your answer. [5 marks]

10. X is a point inside an equilateral triangle ABC. Y is the foot of the perpendicular from X to AC, Z is the footof the perpendicular from X to AB, and W is the foot of the perpendicular from X to BC.

The ratio of the distances of X from the three sides of the triangle is 1 : 2 : 4 as shown in the diagram.

A

B

C

X

Y

Z

W

1

24

If the area of AZXY is 13 cm2, find the area of ABC. Justify your answer. [5 marks]

Investigation

If XY : XZ : XW = a : b : c, find the ratio of the areas of AZXY and ABC. [2 bonus marks]

2

59

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD SOLUTIONS

2015 Australian Intermediate Mathematics Olympiad - Solutions

1. Method 1

123a = 146b ⇐⇒ a2 + 2a + 3 = b2 + 4b + 6

⇐⇒ (a + 1)2 + 2 = (b + 2)2 + 2

⇐⇒ (a + 1)2 = (b + 2)2

⇐⇒ a + 1 = b + 2 (a and b are positive)

⇐⇒ a = b + 1

Since the minimum value for b is 7, the minimum value for a + b is 8 + 7 = 15.

Method 2

Since the digits in any number are less than the base, b ≥ 7.We also have a > b, otherwise a2 + 2a + 3 < b2 + 4b + 6.

If b = 7 and a = 8, then a2 + 2a + 3 = 83 = b2 + 4b + 6.So the minimum value for a + b is 8 + 7 = 15.

2. Let AB, BC, CD, DE, EF , FA touch the circle at U , V , W , X, Y , Z respectively.

A B

C

D

E

F

U

V

W

X

Y

Z

Since the two tangents from a point to a circle have equal length,UB = BV , V C = CW , WD = DX, XE = EY , Y F = FZ, ZA = AU .

The perimeter of hexagon ABCDEF isAU + UB + BV + V C + CW + WD + DX + XE + EY + Y F + FZ + ZA= AU + UB + UB + CW + CW + WD + WD + EY + EY + Y F + Y F + AU= 2(AU + UB + CW + WD + EY + Y F )= 2(AB + CD + EF ) = 2(6 + 7 + 8) = 2(21) = 42.

3. Preamble

Let the required cost be x. Then, with obvious notation, we have:

3w + 7d + t = 329 (1)

4w + 10d + t = 441 (2)

w + d + t = x (3)

Method 1

3 × (1) − 2 × (2): w + d + t = 3 × 329− 2 × 441 = 987 − 882 = 105.

3

59 59

Method 2

(2) − (1): w + 3d = 112.(1) − (3): 2w + 6d = 329 − x = 2 × 112 = 224.

Then x = 329− 224 = 105.

Method 3

10 × (1) − 7 × (2): w = (203 − 3t)/23 × (2) − 4 × (1): d = (7 + t)/2

Then w + d + t = 210/2− 2t/2 + t = 105.

4. We have2

n+

1

n2=

2n + 1

n2.

Since 2n + 1 and n2 are coprime, a = 2n + 1 and b = n2.So 1024 = a + b = n2 + 2n + 1 = (n + 1)2, hence n + 1 = 32.

This gives a = 2n + 1 = 2× 31 + 1 = 63.

5. Method 1213 + 210 + 2x = y2 ⇐⇒ 210(23 + 1) + 2x = y2

⇐⇒ (25 × 3)2 + 2x = y2

⇐⇒ 2x = y2 − 962

⇐⇒ 2x = (y + 96)(y − 96).

Since y is an integer, both y + 96 and y − 96 must be powers of 2.Let y + 96 = 2m and y − 96 = 2n. Then 2m − 2n = 192 = 26 × 3.

Hence 2m−6 − 2n−6 = 3. So 2m−6 = 4 and 2n−6 = 1.In particular, m = 8. Hence y = 28 − 96 = 256 − 96 = 160.

Method 2

We have y2 = 213 + 210 + 2x = 210(23 + 1 + 2x−10) = 210(9 + 2x−10).

So we want the smallest value of 9 + 2x−10 that is a perfect square.Since 9 + 2x−10 is odd and greater than 9, 9 + 2x−10 ≥ 25.

Since 9 + 214−10 = 25, y = 25 × 5 = 32× 5 = 160.

Comment

Method 1 shows that 213 + 210 + 2x = y2 has only one solution.

6. The centres Y and Y ′ of the two medium circles lie on a diameter of the large circle. By symmetry about thisdiameter, the two smaller circles are congruent. Let X be the centre of the large circle and Z the centre of a smallcircle.

X

Y

Y′

Z

Let R and r be the radii of a medium and small circle respectively. Then ZY = R + r = ZY ′. Since XY = XY ′,triangles XY Z and XY ′Z are congruent. Hence XZ ⊥ XY .

4

61

By Pythagoras, Y Z2 = Y X2 + XZ2. So (R + r)2 = R2 + (2R − r)2.Then R2 + 2Rr + r2 = 5R2 − 4Rr + r2, which simplifies to 3r = 2R.

So the large circle has area π(30/√

π )2 = 900,each medium circle has area π(15/

√π )2 = 225,

and each small circle has area π(10/√

π )2 = 100.Thus the shaded area is 900 − 2 × 225− 2 × 100 = 250.

7. Method 1

Any path from the start vertex O to a vertex A must pass through either the vertex L left of A or the vertex Uunderneath A. So the number of paths from O to A is the sum of the number of paths from O to L and the pathsfrom O to U .

•

••

•

•

AL

U

O

There is only one path from O to any vertex on the bottom line of the grid.

So the number of paths from O to all other vertices can be progressively calculated from the second bottom rowupwards as indicated.

•

•

1 1 1 1 1 1 1

1 2 3 4 5 6 7

2 5 9 14 20 27

5 14 28 48 75

14 42 90 165

42 132 297

132 429

429

Thus the number of required paths is 429.

Method 2

To help understand the problem, consider some smaller grids.

Let p(n) equal the number of required paths on an n × n grid and let p(0) = 1.

5

61 61

Starting with the bottom-left vertex, label the vertices of the diagonal 0, 1, 2, . . . , n.

•

•

•

•

0

1

i

n

Consider all the paths that touch the diagonal at vertex i but not at any of the vertices between vertex 0 andvertex i. Each such path divides into two subpaths.

One subpath is from vertex 0 to vertex i and, except for the first and last edge, lies in the lower triangle of thediagram above. Thus there are p(i − 1) of these subpaths.

The other subpath is from vertex i to vertex n and lies in the upper triangle in the diagram above. Thus there arep(n − i) of these subpaths.

So the number of such paths is p(i − 1) × p(n − i).

Summing these products from i = 1 to i = n gives all required paths. Thus

p(n) = p(n − 1) + p(1)p(n − 2) + p(2)p(n − 3) + · · ·+ p(n − 2)p(1) + p(n − 1)

We have p(1) = 1, p(2) = 2, p(3) = 5. So

p(4) = p(3) + p(1)p(2) + p(3)p(1) + p(3) = 14,p(5) = p(4) + p(1)p(3) + p(2)p(2) + p(3)p(1) + p(4) = 42,p(6) = p(5) + p(1)p(4) + p(2)p(3) + p(3)p(2) + p(1)p(4) + p(5) = 132, andp(7) = p(6) + p(1)p(5) + p(2)p(4) + p(3)p(3) + p(4)p(2) + p(5)p(1) + p(6) = 429.

8. Method 1

Let

⌊

x

44

⌋

=

⌊

x

45

⌋

= n.

Since x is non-negative, n is also non-negative.

If n = 0, then x is any integer from 0 to 44 − 1 = 43: a total of 44 values.

If n = 1, then x is any integer from 45 to 2 × 44 − 1 = 87: a total of 43 values.

If n = 2, then x is any integer from 2 × 45 = 90 to 3 × 44 − 1 = 131: a total of 42 values.

If n = k, then x is any integer from 45k to 44(k + 1) − 1 = 44k + 43: a total of (44k + 43) − (45k − 1) = 44 − kvalues.

Thus, increasing n by 1 decreases the number of values of x by 1. Also the largest value of n is 43, in which casex has only 1 value.

Therefore the number of non-negative integer values of x is 44 + 43 + · · ·+ 1 = 1

2(44× 45) = 990.

Method 2

Let n be a non-negative integer such that

⌊

x

44

⌋

=

⌊

x

45

⌋

= n.

Then

⌊

x

44

⌋

= n ⇐⇒ 44n ≤ x < 44(n + 1) and

⌊

x

45

⌋

= n ⇐⇒ 45n ≤ x < 45(n + 1).

So

⌊

x

44

⌋

=

⌊

x

45

⌋

= n ⇐⇒ 45n ≤ x < 44(n + 1) ⇐⇒ 44n + n ≤ x < 44n + 44.

This is the case if and only if n < 44, and then x can assume exactly 44− n different values.

Therefore the number of non-negative integer values of x is

(44 − 0) + (44 − 1) + · · ·+ (44 − 43) = 44 + 43 + · · ·+ 1 = 1

2(44 × 45) = 990.

6

63

Method 3

Let n be a non-negative integer such that

⌊

x

44

⌋

=

⌊

x

45

⌋

= n.

Then x = 44n + r where 0 ≤ r ≤ 43 and x = 45n + s where 0 ≤ s ≤ 44.

So n = r − s. Therefore 0 ≤ n ≤ 43. Also r = n + s. Therefore n ≤ r ≤ 43.

Therefore the number of non-negative integer values of x is 44 + 43 + · · ·+ 1 = 1

2(44× 45) = 990.

9. Working out the first few terms gives us an idea of how the given sequence develops:

n s2n−1 s2n

1 a b2 b − a 2b − a3 b 3b − a4 2b − a 5b − 2a5 3b − a 8b − 3a6 5b − 2a 13b− 5a7 8b − 3a 21b− 8a

It appears that the coefficients in the even terms form a Fibonacci sequence and, from the 5th term, every oddterm is a repeat of the third term before it.

These observations are true for the entire sequence since, for m ≥ 1, we have:

s2m+2 = s2m+1 + s2m

s2m+3 = s2m+2 − s2m+1 = s2m

s2m+4 = s2m+3 + s2m+2 = s2m+2 + s2m

So, defining F1 = 1, F2 = 2, and Fn = Fn−1 + Fn−2 for n ≥ 3, we have s2n = bFn − aFn−2 for n ≥ 3. Since a = 3and b < 1000, none of the first five terms of the given sequence equal 2015. So we are looking for integer solutionsof bFn − 3Fn−2 = 2015 for n ≥ 3.

s6 = 3b− 3 = 2015, has no solution.s8 = 5b− 6 = 2015, has no solution.s10 = 8b − 9 = 2015 implies b = 253.

For n ≥ 6 we have b = 2015/Fn + 3Fn−2/Fn. Since Fn increases, we have Fn ≥ 13 and Fn−2/Fn < 1 for n ≥ 6.Hence b < 2015/13 + 3 = 158. So the largest value of b is 253.

10. Method 1

We first show that X is uniquely defined for any given equilateral triangle ABC.

Let P be a point outside ABC such that its distances from AC and AB are in the ratio 1:2. By similar triangles,any point on the line AP has the same property. Also any point between AP and AC has the distance ratio lessthan 1:2 and any point between AP and AB has the distance ratio greater than 1:2.

A

B

C

P

Q

1

2

1

4

7

63 63

Let Q be a point outside ABC such that its distances from AC and BC are in the ratio 1:4. By an argumentsimilar to that in the previous paragraph, only the points on CQ have the distance ratio equal to 1:4.

Thus the only point whose distances to AC, AB, and BC are in the ratio 1:2:4 is the point X at which AP andCQ intersect.

Scaling if necessary, we may assume that the actual distances of X to the sides of ABC are 1, 2, 4. Let h be theheight of ABC. Letting | | denote area, we have

|ABC| = 1

2h × AB and

|ABC| = |AXB| + |BXC| + |CXA| = 1

2(2AB + 4BC + AC) = 1

2AB × 7.

So h = 7.

Draw a 7-layer grid of equilateral triangles each of height 1, starting with a single triangle in the top layer, then atrapezium of 3 triangles in the next layer, a trapezium of 5 triangles in the next layer, and so on. The boundaryof the combined figure is ABC and X is one of the grid vertices as shown.

A

B

C

X

Y

Z

W

1

2

4

There are 49 small triangles in ABC and 6.5 small triangles in AZXY . Hence, after rescaling so that the area ofAZXY is 13 cm2, the area of ABC is 13 × 49/6.5 = 98 cm2.

Method 2

Join AX, BX, CX. Since Y AZ = ZBW = 60, the quadrilaterals AZXY and BWXZ are similar. Let XY be1 unit and AY be x. Then BZ = 2x.

A

B

C

X

Yx

2x

1

24

Z

W

By Pythagoras: in AXY , AX =√

1 + x2 and in AXZ, AZ =√

x2 − 3. Hence BW = 2√

x2 − 3.

Since AB = AC, Y C = x +√

x2 − 3.

By Pythagoras: in XY C, XC2 = 1 + (x +√

x2 − 3)2 = 2x2 − 2 + 2x√

x2 − 3and in XWC, WC2 = 2x2 − 18 + 2x

√x2 − 3.

Since BA = BC, 2x +√

x2 − 3 = 2√

x2 − 3 +√

2x2 − 18 + 2x√

x2 − 3.

So 2x −√

x2 − 3 =√

2x2 − 18 + 2x√

x2 − 3.

8

65

Squaring gives 4x2 + x2 − 3 − 4x√

x2 − 3 = 2x2 − 18 + 2x√

x2 − 3, which simplifies to 3x2 + 15 = 6x√

x2 − 3.

Squaring again gives 9x4 +90x2 +225 = 36x4−108x2. So 0 = 3x4−22x2−25 = (3x2−25)(x2 +1), giving x =5√3.

Hence, area AZXY =x

2+√

x2 − 3 =5

2√

3+

4√3

=13

2√

3and

area ABC =

√3

4(2x +

√x2 − 3)2 =

√3

4

( 10√3

+4√3

)2

=49√

3.

Since the area of AZXY is 13 cm2, the area of ABC is (49√3/

13

2√

3) × 13 = 98 cm2.

Method 3

Let DI be the line through X parallel to AC with D on AB and I on BC.Let EG be the line through X parallel to BC with E on AB and G on AC.Let FH be the line through X parallel to AB with F on AC and H on BC.Let J be a point on AB so that HJ is parallel to AC.Triangles XDE, XFG, XHI, BHJ are equilateral, and triangles XDE and BHJ are congruent.

F G

D

E

H

I

J

A

B

C

X

Y

1

24

Z

W

The areas of the various equilateral triangles are proportional to the square of their heights. Let the area ofFXG = 1. Then, denoting area by | |, we have:

|DEX| = 4, |XHI| = 16, |AEG| = 9, |DBI| = 36, |FHC| = 25.

|ABC| = |AEG|+ |FHC|+ |DBI| − |FXG| − |DEX| − |XHI| = 9 + 25 + 36 − 1 − 4 − 16 = 49.

|AZXY | = |AEG| − 1

2(|FXG|+ |DEX|) = 9 − 1

2(1 + 4) = 6.5.

Since the area of AZXY is 13 cm2, the area of ABC is 2 × 49 = 98 cm2.

Method 4

Consider the general case where XY = a, XZ = b, and XW = c.

A

B

C

X

Y

a

bc

Z

W

60120

9

65 65

Projecting AY onto the line through ZX gives AY sin 60 − a cos 60 = b.Hence AY = (a + 2b)/

√3. Similarly, AZ = (b + 2a)/

√3.

Letting | | denote area, we have

|AZXY | = |Y AZ| + |Y XZ|= 1

2(AY )(AZ) sin 60 + 1

2ab sin 120

=√

3

4((AY )(AZ) + ab)

=√

3

12((a + 2b)(b + 2a) + 3ab)

=√

3

12(2a2 + 2b2 + 8ab)

=√

3

6(a2 + b2 + 4ab)

Similarly, |CY XW | =√

3

6(a2 + c2 + 4ac) and |BWXZ| =

√3

6(b2 + c2 + 4bc).

Hence |ABC| =√

3

6(2a2 + 2b2 + 2c2 + 4ab + 4ac + 4bc) =

√3

3(a + b + c)2.

So |ABC|/|AZXY | = 2(a + b + c)2/(a2 + b2 + 4ab).

Letting a = k, b = 2k, c = 4k, and |AZXY | = 13cm2, we have |ABC| = 26(49k2)/(k2 + 4k2 + 8k2) = 98 cm2.

Investigation

Method 4 gives |ABC|/|AZXY | = 2(a + b + c)2/(a2 + b2 + 4ab).

Alternatively, as in Method 3,

|ABC| = |AEG|+ |FHC|+ |DBI| − |FXG| − |DEX| − |XHI|= (a + b)2 + (a + c)2 + (b + c)2 − a2 − b2 − c2 = (a + b + c)2.

Also |AZXY | = |AEG| − 1

2(|FXG|+ |DEX|)

= (a + b)2 − 1

2(a2 + b2)

= 2ab + 1

2(a2 + b2).

So |ABC|/|AZXY | = 2(a + b + c)2/(a2 + b2 + 4ab).

10

67

DISTRIBUTION OF AWARDS/SCHOOL YEAR

Year Number of Students

Number of Awards

Prize High Distinction Distinction Credit Participation

8 341 3 17 39 86 196

9 414 8 45 61 99 201

10 462 11 52 89 139 171

Other 221 4 9 16 41 151

Total 1438 26 123 205 365 719

NUMBER OF CORRECT ANSWERS QUESTIONS 1–8

YearNumber Correct/Question

1 2 3 4 5 6 7 8

8 119 231 282 164 128 79 48 50

9 144 298 347 224 187 138 82 86

10 176 341 377 297 219 208 103 104

Other 66 132 176 81 75 34 33 21

Total 505 1002 1182 766 609 459 266 261

MEAN SCORE/QUESTION/SCHOOL YEAR

Year Number of Students

Mean Score

Overall MeanQuestion

1–8 9 10

8 341 10.1 0.5 0.2 10.9

9 414 11.8 0.9 0.5 13.2

10 462 13.0 1.1 0.6 14.6

Other 221 8.6 0.4 0.2 9.3

All Years 1438 11.3 0.8 0.4 12.5

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD STATISTICS

67 67

Name School Year Score

Prize

Matthew Cheah Penleigh and Essendon Grammar School, VIC 10 35

Puhua Cheng Raffles Institution, Singapore 8 35Ariel Pratama Junaidi Anglo-Chinese School, Singapore 10 35

Evgeni Kayryakov Childrens Academy 21st Century, Bulgaria 7 35

Wei Khor Jun Raffles Institution, Singapore 8 35

Jack Liu Brighton Grammar, VIC 9 35

Jerry Mao Caulfield Grammar School, Wheelers Hill,VIC 9 35

Liao Meng Anglo-Chinese School, Singapore 10 35

Nguyen Hoai Nam Anglo-Chinese School, Singapore 10 35

Aloysius Ng Yangyi Raffles Institution, Singapore 7 35

Kohsuke Sato Christ Church Grammar School, WA 10 35

Yuelin Shen Scotch College, WA 10 35

Chen Tan Xu Raffles Institution, Singapore 7 35

Kit Victor Loh Wai Raffles Institution, Singapore 8 35

Jianzhi Wang Raffles Institution, Singapore 9 35

Zhe Xin Raffles Institution, Singapore 9 35

Austin Zhang Sydney Grammar School, NSW 10 35

Yu Zhiqiu Anglo-Chinese School, Singapore 10 35

Lin Zien Anglo-Chinese School, Singapore 9 35

Bobby Dey James Ruse Agricultural High School, NSW 10 34

Goh Ethan Raffles Institution, Singapore 7 34

Yulong Guo Hwa Chong Institution, Singapore 9 34Edwin Winata Hartanto Anglo-Chinese School, Singapore 10 34

Hristo Papazov Childrens Academy 21st Century, Bulgaria 10 34

Zhang Yansheng Chung Cheng High School, Singapore 9 34

Guowen Zhang St Joseph's College, QLD 9 34

HIGH DISTINCTION

Ivan Ganev Childrens Academy 21st Century, Bulgaria 10 33

Theodore Leebrant Anglo-Chinese School, Singapore 9 33

Yu Peng Ng Hwa Chong Institution, Singapore 8 33

Cheng Shi Hwa Chong Institution, Singapore 8 33

Kean Tan Wee Raffles Institution, Singapore 7 33

Sharvil Kesarwani Merewether High School, NSW 8 32Hong Rui Benjamin Lee Hwa Chong Institution, Singapore 10 32

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD RESULTS

69

Name School Year Score

Chenxu Li Raffles Institution, Singapore 8 32

William Li Barker College, NSW 9 32

Han Yang Hwa Chong Institution, Singapore 9 32

Stanley Zhu Melbourne Grammar School, VIC 9 32

Anand Bharadwaj Trinity Grammar School, VIC 9 31

William Hu Christ Church Grammar School, WA 9 31

Xianyi Huang Baulkham Hills High School, NSW 10 31

Wanzhang Jing James Ruse Agricultural High School, NSW 10 31

Yuhao Li Hwa Chong Institution, Singapore 9 31

Steven Lim Hurlstone Agricultural High School, NSW 9 31

John Min Baulkham Hills High School, NSW 9 31

Elliott Murphy Canberra Grammar School, ACT 10 31

Longxuan Sun Hwa Chong Institution, Singapore 8 31

Boyan Wang Hwa Chong Institution, Singapore 9 31

Sean Zammit Barker College, NSW 10 31

Atul Barman James Ruse Agricultural High School, NSW 10 30

Atanas Dinev Childrens Academy 21st Century, Bulgaria 9 30

Bill Hu James Ruse Agricultural High School, NSW 10 30

Phillip Huynh Brisbane State High School, QLD 10 30

Wei Ci William Kin Hwa Chong Institution, Singapore 10 30

Yang Lee Ker Raffles Institution, Singapore 9 30

Forbes Mailler Canberra Grammar School, ACT 9 30

Moses Mayer Surya Institute, Indonesia 9 30

Zlatina Mileva Childrens Academy 21st Century, Bulgaria 8 30

Kirill Saulov Brisbane Grammar School, QLD 10 30

Yuxuan Seah Raffles Institution, Singapore 8 30

Kieran Shivakumaarun Sydney Boys High School, NSW 10 30

Liang Tan Xue Raffles Institution, Singapore 10 30

Nicholas Tanvis Anglo-Chinese School, Singapore 9 30

An Aloysius Wang Hwa Chong Institution, Singapore 10 30

William Wang Queensland Academy for Science, Mathematics and Technology, QLD 10 30

Joshua Welling Melrose High School, ACT 9 30

Seung Hoon Woo Hwa Chong Institution, Singapore 10 30

Chen Yanbing Methodist Girls' School, Singapore 10 30

Guangxuan Zhang Raffles Institution, Singapore 10 30

Chi Zhang Yu Raffles Institution, Singapore 7 30

Keer Chen Presbyterian Ladies’ College, NSW 10 29

Linus Cooper James Ruse Agricultural High School, NSW 9 29

Liam Coy Sydney Grammar School, NSW 7 29

69 69

Name School Year Score

Rong Dai Xiang Raffles Institution, Singapore 8 29

Hong Pei Goh Hwa Chong Institution, Singapore 10 29

Tianjie Huang Hwa Chong Institution, Singapore 9 29

Ricky Huang James Ruse Agricultural High School, NSW 10 29

Tianjie Huang Hwa Chong Institution, Singapore 9 29

Yu Jiahuan Raffles Girls' School, Singapore 9 29

Tony Jiang Scotch College, VIC 10 29

Charles Li Camberwell Grammar School, Vic 9 29

Steven Liu James Ruse Agricultural High School, NSW 10 29

Hilton Nguyen Sydney Technical High School, NSW 9 29

James Nguyen Baulkham Hills High School, NSW 10 29

Trung Nguyen Penleigh and Essendon Grammar School, VIC 10 29

James Phillips Canberra Grammar School, ACT 9 29

Ryan Stocks Radford College, ACT 9 29

Hadyn Tang Trinity Grammar School, VIC 6 29Stanve Avrilium Widjaja Surya Institute, Indonesia 7 29

Wang Yihe Anglo-Chinese School, Singapore 9 29

Sun Yue Raffles Girls' School, Singapore 9 29

Wang Beini Raffles Girls' School, Singapore 10 28

Chwa Channe Raffles Girls' School, Singapore 8 28

Keiran Hamley All Saints Anglican School, QLD 10 28

Lee Shi Hao Anglo-Chinese School, Singapore 9 28

Zhu Jiexiu Raffles Girls' School, Singapore 9 28

Jodie Lee Seymour College, SA 10 28

Yu Hsin Lee Hwa Chong Institution, Singapore 9 28

Phillip Liang James Ruse Agricultural High School, NSW 9 28

Anthony Ma Shore School, NSW 10 28

Dzaki Muhammad Surya Institute, Indonesia 9 28

Daniel Qin Scotch College, VIC 10 28

Sang Ta Randwick Boys High School, NSW 8 28

Ruiqian Tong Presbyterian Ladies’ College, VIC 9 28

Hu Xing Yi Methodist Girls' School, Singapore 10 28

Zhao Yiyang Methodist Girls' School, Singapore 10 28

Claire Yung Lyneham High School, ACT 10 28

Xuan Ang Ben Raffles Institution, Singapore 9 27

Hantian Chen James Ruse Agricultural High School, NSW 10 27

Jasmine Jiawei Chen Pymble Ladies’ College, NSW 10 27

Harry Dinh James Ruse Agricultural High School, NSW 10 27

Tan Jian Yee Chung Cheng High School, Singapore 10 27

70

Name School Year Score

Daniel Jones All Saints Anglican Senior School, QLD 10 27

Winfred Kong Hwa Chong Institution, Singapore 10 27

Adrian Law James Ruse Agricultural High School, NSW 10 27

Jason Leung James Ruse Agricultural High School, NSW 7 27Sabrina Natashya Liandra Surya Institute, Indonesia 9 27

Angela (Yunyun) Ran The Mac.Robertson Girls’ High School, VIC 9 27

Yi Shen Xin Raffles Institution, Singapore 7 27

Peter Tong Yarra Valley Grammar, VIC 9 27

Jordan Truong Sydney Technical High School, NSW 10 27

Andrew Virgona Concord High School, NSW 9 27

Tommy Wei Scotch College, VIC 9 27

Tianyi Xu Sydney Boys High School, NSW 9 27

Wu Zhen Anglo-Chinese School, Singapore 10 27

Gordon Zhuang Sydney Boys High School, NSW 9 27

Zhou Zihan Raffles Girls' School, Singapore 8 27

Amit Ben-Harim McKinnon Secondary College, VIC 9 26

Hu Chen The King's School, NSW 10 26

Merry Xiao Die Chu North Sydney Girls' High School, NSW 9 26

Li Haocheng Anglo-Chinese School, Singapore 9 26

William Hu Rossmoyne Senior High School, WA 10 26

Laeeque Jamdar Baulkham Hills High School, NSW 9 26

Yasiru Jayasoora James Ruse Agricultural High School, NSW 7 26

Arun Jha Perth Modern School, WA 10 26

Tony Li Sydney Boys High School, NSW 10 26

Zefeng Jeff Li Glen Waverley Secondary College, VIC 8 26

Adrian Lo Newington College, NSW 7 26

Lionel Maizels Norwood Secondary College, VIC 8 26

Marcus Rees Taroona High School, TAS 8 26

Elva Ren Presbyterian Ladies College, VIC 10 26

Aidan Smith All Saints' College, WA 8 26

Jacob Smith All Saints' College, WA 9 26

Keane Teo Hwa Chong Institution, Singapore 10 26

Jeffrey Wang Shore School, NSW 10 26

Xinlu Xu Presbyterian Ladies’ College, Sydney, NSW 10 26

Jason (Yi) Yang James Ruse Agricultural High School, NSW 8 26

Shukai Zhang Hwa Chong Institution, Singapore 10 26

Yanjun Zhang Hwa Chong Institution, Singapore 9 26

Kevin Zhu James Ruse Agricultural High School, NSW 10 26

Jonathan Zuk Elwood College, VIC 8 26

71

AMOC SENIOR CONTESTTHE 2015 AMOC SENIOR CONTEST

Tuesday, 11 August 2015

Time allowed: 4 hours

No calculators are to be used.

Each question is worth seven points.

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

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

because 9 = 2 + 3 + 4.

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

4-addy and 5-addy?

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

and 6-addy?

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

am+1 =1a1 + 2a2 + 3a3 + · · ·+mam

amfor m ≥ 1.

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

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

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

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

Determine the minimum possible number of students in the group.

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

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

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

Prove that BF bisects the angle AFC.

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

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

⌊a

p

⌋+

⌊2a

p

⌋+

⌊3a

p

⌋+ · · ·+

⌊pa

p

⌋= 100.

c© 2015 Australian Mathematics Trust

72

THE 2015 AMOC SENIOR CONTEST

Solutions and cumulative marking scheme

c© 2015 Australian Mathematics Trust

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

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

9 = 2 + 3 + 4.

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

and 5-addy?

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

6-addy?

Solution (Angelo Di Pasquale)

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

by 3 and greater than 3. 1

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

that are divisible by 5 and greater than 10.

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

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

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

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

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

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

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

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

are necessarily odd. 5

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

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

1

AMOC SENIOR CONTEST SOLUTIONS

73

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

am+1 =1a1 + 2a2 + 3a3 + · · ·+mam

amfor m ≥ 1.

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

Solution 1 (Norman Do)

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

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

of m shifted by 1.

amam+1 = 1a1 + 2a2 + 3a3 + · · ·+mam

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

Subtracting the first equation from the second yields

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

for all m ≥ 1. 2

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

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

= 2k + (2k − 2) + · · ·+ 2

= 2 [k + (k − 1) + · · ·+ 1]

= k(k + 1). 3

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

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

= (2k − 1) + (2k − 3) + · · ·+ 3

= k2 − 1. 4

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

am =

m2

4 , if m is even,

m2+34 , if m is odd.

5

If m is odd, then

am+1 − am =(m+ 1)2

4− m2 + 3

4=

2m− 2

4,

and if m is even, then

am+1 − am =(m+ 1)2 + 3

4− m2

4=

2m+ 4

4. 6

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

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

2

74

Solution 2 (Angelo Di Pasquale)

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

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

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

Consider the following sequence of equalities.

2n∑i=1

iai =

n∑i=1

(2i− 1) a2i−1 +

n∑i=1

2i a2i

=

n∑i=1

(2i− 1) (i2 − i+ 1) +

n∑i=1

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

=

n∑i=1

4i3 − 3i2 + 3i− 1

=

n∑i=1

3i3 + (i− 1)3 3

= 3

n∑i=1

i3 +

n−1∑i=1

i3

=3n2(n+ 1)2

4+

(n− 1)2n2

4

(since

n∑i=1

i3 =n2(n+ 1)2

4

)

= n2(n2 + n+ 1)

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

It follows that

a2n+1 =1a1 + 2a2 + 3a3 + · · ·+ 2na2n

a2n= n2 + n+ 1. 4

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

state that∑2n

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

2n+1∑i=1

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

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

= (n+ 1)2a2n+1

It follows that

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

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

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

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

1. 7

3

75

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

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

pair of problems, exactly two students solved them both.

Determine the minimum possible number of students in the group.

Solution (Norman Do)

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

example.

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

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

Student 3 solved problems 1, 4, 5.

Student 4 solved problems 2, 4, 5.

Student 5 solved problems 3 and 4.

Student 6 solved problems 3 and 5. 2

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

solved 2 problems. Therefore, a students solved(42

)= 6 pairs of problems, b students

solved(32

)= 3 pairs of problems and c students solved

(22

)= 1 pair of problems. Since we

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

that a+ b+ c ≤ 5.

There are(52

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

two students solved them both, so we must have

6a+ 3b+ c = 20. 4

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

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

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

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

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

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

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

4

76

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

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

line AD meet the side CD at F .

Prove that BF bisects the angle AFC.

Solution 1 (Alan Offer)

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

A B

CD

E

F

H

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

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

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

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

pairs of equal ratios

AF

AD=

AH

AF

AE

AD=

AH

AC

AB

AE=

AC

AB. 3

Putting these together, we have

AF 2 = AD ·AH = AC ·AE = AB2. 6

So triangle BAF is isosceles and we have

∠AFB = ∠ABF = 90 − ∠CBF = ∠CFB,

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

have proven that BF bisects the angle AFC. 7

Solution 2 (Angelo Di Pasquale)

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

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

the circumcircle of triangle BEC, we have

AB2 = AE ·AC. 1

5

77

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

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

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

respect to the circumcircle of triangle EFC, we have

AF 2 = AE ·AC. 4

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

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

Solution 3 (Ivan Guo)

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

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

inversion f with centre A and radius AF . 2

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

to AB through F . Thus,

f(circle AEFH) = line CD. 4

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

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

Hence, AB = AF . 6

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

Solution 4 (Chaitanya Rao)

Consider the following chain of equalities.

DF 2 = AD ·DH (ADF ∼ FDH) 1

= AD · (AH −AD)

= AD ·(AB2

BC−AD

)(ABC ∼ HAB) 3

= AB2 −AD2 (AD = BC) 4

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

So triangle BAF is isosceles and we have

∠AFB = ∠ABF = 90 − ∠CBF = ∠CFB,

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

have proven that BF bisects the angle AFC. 7

6

78

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

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

p

⌋+

⌊2a

p

⌋+

⌊3a

p

⌋+ · · ·+

⌊pa

p

⌋= 100.

Solution 1 (Norman Do)

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

We divide the problem into the following two cases.

The number a is divisible by p.

If we write a = kp, the equation becomes

a(p+ 1)

2= 100 ⇒ kp(p+ 1) = 200.

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

there are no such primes p. 1

The number a is not divisible by p.

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

p+

2a

p+

3a

p+ · · ·+ pa

p

)−

(a

p

+

2a

p

+

3a

p

+ · · ·+

pa

p

)= 100.

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

a(p+ 1)

2−

[a

p

+

2a

p

+

3a

p

+ · · ·+

pa

p

]= 100. 3

We will prove that the sequence of numbersap

,2ap

,3ap

, . . . ,

pap

is a rearrange-

ment of the sequence of numbers 0p ,

1p ,

2p , . . . ,

p−1p . 4

Observe that if k is a positive integer, thenkap

is one of the numbers 0

p ,1p ,

2p , . . . ,

p−1p .

So it suffices to show that no two of the numbersap

,2ap

,3ap

, . . . ,

pap

are equal.

Suppose for the sake of contradiction thatiap

=

jap

, where 1 ≤ i < j ≤ p. Then

jap − ia

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

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

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

sequence of numbersap

,2ap

,3ap

, . . . ,

pap

is a rearrangement of the sequence

of numbers 0p ,

1p ,

2p , . . . ,

p−1p . 5

We may now write the equation as

a(p+ 1)

2− p− 1

2= 100 ⇒ (a− 1)(p+ 1) = 198. 6

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

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

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

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

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

7

79

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

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

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

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

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

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

ar

p− 1 +

as

p− 1 <

⌊ar

p

⌋+

⌊as

p

⌋<

ar

p+

as

p⇒ a− 2 <

⌊ar

p

⌋+

⌊as

p

⌋< a.

But since⌊arp

⌋+

⌊asp

⌋is an integer, we conclude that

⌊ar

p

⌋+

⌊as

p

⌋= a− 1. 3

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

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

we obtain (p− 1

2

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

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

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

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

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

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

8

80

Name School Year Score

Prize

Yong See Foo Nossal High School VIC 11 35

Kevin Xian James Ruse Agricultural High School NSW 11 35

Wilson Zhipu Zhao Killara High School NSW 11 35

Ilia Kucherov Westall Secondary College VIC 11 34

Seyoon Ragavan Knox Grammar School NSW 11 34

High Distinction

Jongmin Lim Killara High School NSW 11 32

Matthew Cheah Penleigh and Essendon Grammar School VIC 10 29

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

Distinction

Alexander Barber Scotch College VIC 11 28

Michelle Chen Methodist Ladies’ College VIC 11 28

Thomas Baker Scotch College VIC 11 27

Linus Cooper James Ruse Agricultural High School NSW 9 27

Steven Lim Hurlstone Agricultural High School NSW 9 27

Eric Sheng Newington College NSW 11 27

William Song Scotch College VIC 11 27

Jack Liu Brighton Grammar VIC 9 26

Leo Li Christ Church Grammar School WA 11 25

Bobby Dey James Ruse Agricultural High School NSW 10 24

Michael Robertson Dickson College ACT 11 24

Charles Li Camberwell Grammar School VIC 9 22

Isabel Longbottom Rossmoyne Senior High School WA 10 22

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

AMOC SENIOR CONTEST RESULTS

81

SCORE DISTRIBUTION/PROBLEM

Problem Number

Number of Students/ScoreMean

0 1 2 3 4 5 6 7

1 1 0 0 0 1 5 46 31 6.2

2 16 9 8 1 2 8 15 25 4.1

3 25 4 11 2 4 5 8 25 3.5

4 54 13 2 0 0 0 0 15 1.5

5 65 2 1 0 0 4 6 6 1.2

AMOC SENIOR CONTEST STATISTICS

82

AMOC SCHOOL OF EXCELLENCE2014 AMOC School of Excellence

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

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

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

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

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

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

Angelo Di PasqualeDirector of Training, AMOC

1

83

Name School Year

Seniors

Thomas Baker Scotch College VIC 10

Matthew Cheah Penleigh and Essendon Grammar School VIC 9

Yong See Foo Nossal High School VIC 10

Alexander Gunning Glen Waverley Secondary College VIC 11

Leo Li Christ Church Grammar School WA 10

Allen Lu Sydney Grammar School NSW 11

Seyoon Ragavan Knox Grammar School NSW 10

Kevin Shen St Kentigern College NZ 11*

Yang Song James Ruse Agricultural High School NSW 11

Kevin Xian James Ruse Agricultural High School NSW 10

Jeremy Yip Trinity Grammar School VIC 11

Henry Yoo Perth Modern School WA 11

Juniors

Adam Bardrick Whitefriars College VIC 8

Rachel Hauenschild Kenmore State High School QLD 9

William Hu Christ Church Grammar School WA 8

Shivasankaran Jayabalan Rossmoyne Senior High School WA 8

Tony Jiang Scotch College VIC 9

Sharvil Kesarwani Merewether High School NSW 7

Ilia Kucherov Westall Secondary College VIC 10

Adrian Law James Ruse Agricultural High School NSW 9

Charles Li Camberwell Grammar School VIC 8

Jack Liu Brighton Grammar School VIC 8

Isabel Longbottom Rossmoyne Senior High School WA 9

Hilton Nguyen Sydney Technical High School NSW 8

Madeline Nurcombe Cannon Hill Anglican College QLD 10

Zoe Schwerkolt Fintona Girls' School VIC 10

Katrina Shen James Ruse Agricultural High School NSW 7

Eric Sheng Newington College NSW 10

Wen Zhang St Joseph's College QLD 8

PARTICIPANTS AT THE 2014 AMOC SCHOOL OF EXCELLENCE

* Equivalent to year 10 in Australia.

84

AUSTRALIAN MATHEMATICAL OLYMPIADTHE 2015 AUSTRALIAN MATHEMATICALOLYMPIAD

DAY 1

Tuesday, 10 February 2015

Time allowed: 4 hours

No calculators are to be used.

Each question is worth seven points.

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

an+1 = 2an − an−1 + 2, for n ≥ 2.

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

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

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

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

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

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

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

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

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

are friends.

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

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

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

through P parallel to AC at the point D.

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

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

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

c© 2015 Australian Mathematics Trust

85

THE 2015 AUSTRALIAN MATHEMATICALOLYMPIAD

DAY 2

Wednesday, 11 February 2015

Time allowed: 4 hours

No calculators are to be used.

Each question is worth seven points.

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

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

and CZ at X and Y , respectively.

Prove that the quadrilateral BXYD is cyclic.

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

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

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

Prove that there exist infinitely many positive integers n such that

(p(n+ 1)− p(n)

) (p(n)− p(n− 1)

)> 0.

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

such that no two are parallel.

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

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

apple.

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

c© 2015 Australian Mathematics Trust

86

AUSTRALIAN MATHEMATICAL OLYMPIAD SOLUTIONS

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

an+1 = 2an − an−1 + 2, for n ≥ 2.

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

First we prove the following formula by induction.

an = n2 + 3, for n ≥ 1.

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

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

ak+1 = 2ak − ak−1 + 2 (given)

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

= k2 + 2k + 4

= (k + 1)2 + 3.

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

Using the formula, we calculate

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

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

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

= (m2 +m+ 3)2 + 3

= am2+m+3.

Hence amam+1 is a term of the sequence.

17

87

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

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

an+1 − an = an − an−1 + 2, for n ≥ 2.

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

a2 − a1 = 3

a3 − a2 = 5

a4 − a3 = 7

...

am − am−1 = 2m− 1

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

am − a1 = 3 + 5 + · · ·+ 2m− 1.

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

am − a1 =(2m+ 2)(m− 1)

2= m2 − 1.

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

It is now a simple matter to calculate

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

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

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

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

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

a = 3, d = 2 and k = m− 2.

18

1

1

88

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

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

am+2 = 2am+1 − am + 2

= 2(am+1 + 1)− am

am+3 = 2am+2 − am+1 + 2

= 2((2am+1 + 1)− am)− am+1 + 2

= 3(am+1 + 2)− 2am

Similarly, am+4 = 4(am+1 + 3)− 3am.

We prove the following general formula by induction.

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

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

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

am+r+1 = 2am+r − am+r−1 + 2

= 2(r(am+1 + r − 1)− (r − 1)am)

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

= (r + 1)(am+1 + r)− ram.

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

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

am+am = am(am+1 + am − 1)− (am − 1)am

= amam+1.

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

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

19

89

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

Answer: 1000

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

Lemma The function f has the following two properties.

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

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

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

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

f(n+ 1) = n+ 1− s(n+ 1)

= n+ 1− (s(n)− 9k + 1)

= n− s(n) + 9k

= f(n) + 9k

> f(n).

From the lemma it follows that

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

< f(20) = f(21) = · · · = f(29)

...

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

< f(10010).

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

20

90

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

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

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

s(n) = ak + ak−1 + · · ·+ a1 + a0

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

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

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

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

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

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

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

11(b− e) + (c− f) = 0. (2)

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

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

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

21

91

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

Answer: 45

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

First we exhibit a friendless subset T of size 45.

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

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

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

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

22

92

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

Consider the following nine lines of numbers.

11

21, 22, 12

31, 32, 33, 23, 13

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

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

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

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

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

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

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

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

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

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

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

23

93

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

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

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

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

B C

A

O

P

D

Q

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

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

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

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

24

2

2

94

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

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

B C

A

O

P

D

Q

E

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

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

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

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

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

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95

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

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

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

B C

A

O

P

D

X

Y

(a) Using directed angles, we have

∠(OY,OD) = ∠(BO,BA) (OD ‖ BA)

= ∠(AB,AO) (OB = OA)

= ∠(OD,OP ). (BA ‖ OD)

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

∠(Y D,OD) = ∠(OD,PD)

= ∠(OD,OP ) + ∠(OP, PD)

= ∠(BA,OA) + ∠(OA,AC)

(OD ‖ BA and PD ‖ AC)

= ∠(BA,AC).

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

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

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

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96

we have

2s =OY

sin∠(Y D,OD)

=2r

sin∠(BA,AC)

≥ 2r.

Therefore s ≥ r, as required.

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

sin(BA,AC) = 1.

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

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97

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

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

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

BA

C

D

X

Z

Y

xx

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

∠BDY = 90 − x. (1)

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

∠Y XB = 90 + x. (2)

Adding (1) and (2) yields

∠BDY + ∠Y XB = 180,

and so BXYD is a cyclic quadrilateral.

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98

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

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

It follows that

∠AYD = ∠ACD (ADY C cyclic)

= 90 − ∠DAC (angle sum CAD)

= 90 − ∠BAC

= ∠CBA. (angle sum ABC)

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

BA

C

D

X

Z

Y

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99

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

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

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

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

∠BAC = 90 − 2x (angle sum ABC)

⇒ ∠ZAY = 45 − x (AY bisects ∠BAC)

⇒ ∠Y ZA = 45 + x (angle sum AZY )

⇒ ∠ZDY = 45 + x (Y D = Y Z)

⇒ ∠DY Z = 90 − 2x (angle sum Y DZ)

⇒ ∠DYX = 180 − 2x (AX ⊥ CZ)

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

BA

C

D

X

Z

Y

2x

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100

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

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

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

We now have

∠CXH = ∠ZHX (BC ‖ ZH)

= ∠ZHY

= ∠ZDY. (HDZY cyclic)

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

BA

C

D

X

Z

Y

H

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101

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

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

∠ACY = 90 − ∠XAC = ∠CXY.

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

AC2 = AY · AX. (1)

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

∠ACD = 90 − ∠BAC = ∠CBD.

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

AC2 = AD · AB. (2)

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

AY · AX = AD · AB,

and so BXYD is cyclic.

BA

C

D

X

Z

Y

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102

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

Answer: 1008

Let f(x) = p(x)− q(x), where

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

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

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

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

f(0) = (−1)× (−3)× · · · × (−2015)

− (−2)× (−4)× · · · × (−2014)

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

> 0.

We also have

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

> 0,

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

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

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

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

f(0) > 0

f(2) < 0

f(4) > 0

f(6) < 0

...

f(2014) < 0

f(2016) > 0

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

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

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103

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

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

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

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

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

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

2+ 1 = 2015−x

2.3 Taking into account the minus sign at

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

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

Hence we have the following inequalities.

f(1) > 0

f(3) < 0

f(5) > 0

...

f(2013) > 0

f(2015) < 0

f(2016) > 0

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

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

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

y = p(x)

y = q(x)

2013 2014 2015 2016

y

x. . .1 2 3 4 5 6 7

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

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3

3

104

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

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

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

(p(n+ 1)− p(n)

)(p(n)− p(n− 1)

)> 0.

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

(p(n+ 1)− p(n)

)(p(n)− p(n− 1)

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

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

n

p(n)

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

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

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

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

2< q and p(q − 1) ≤ q−1

2< q.

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

4Also known as ‘proof by contradiction’.

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4

4

105

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

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

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

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

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

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

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

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

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

But 34m = 81m ≡ 1m (mod 5).

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

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

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

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

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

Hence p(2q + 1) ≤ 2q+13

< q, which is a contradiction.

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

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

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

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

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

3< q.

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

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

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

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

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

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

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

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

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

5Equivalent to year 11 in Australia.

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5

5

106

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

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

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

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

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

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

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

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

p(q2k − 1) < p(q2

k

) < p(q2k

+ 1). (1)

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

b+ 1) = 2 for all integers

1 ≤ a < b.

Proof Let d = gcd(q2a+1, q2

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

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

q2a

+ 1 | q2a+1 − 1

q2a+1 − 1 | q2a+2 − 1

q2a+2 − 1 | q2a+3 − 1

...

q2b−1 − 1 | q2b − 1

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

yields d | 2. Since q2a+ 1 and q2

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

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

p(q2k − 1) < q and p(q2

k

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

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

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

). Hence

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

For the inductive step, suppose we know that

p(q2j − 1) < q and p(q2

j

+ 1) < q,

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

j − 1)(q2j+ 1), we have

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

j+1+1) < q.

This completes the induction.

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

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

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

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

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

39

6

6

109

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

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

Answer: n⌊n−12

⌋

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

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

⌊n−12

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

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

⌊n−12

⌋.

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

n. Next translate

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

⌊n−12

⌋base-specified isosceles triangles.

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

⌊n−12

⌋pairs b, c such that

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

⌊n−12

⌋, as desired.

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

−→

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110

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

⌋are possible. The

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

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

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111

AUSTRALIAN MATHEMATICAL OLYMPIAD STATISTICS

SCORE DISTRIBUTION/PROBLEM

Number of Students/Score

Problem Number

1 2 3 4 5 6 7 8

0 10 14 29 80 17 60 47 86

1 7 2 1 8 0 1 30 8

2 1 7 18 0 6 3 1 1

3 2 10 4 0 5 4 4 4

4 13 10 11 1 0 5 0 0

5 6 9 9 0 0 1 0 3

6 17 26 11 8 1 7 0 0

7 50 28 23 9 77 25 24 4

Average Mark 5.2 4.6 3.4 1.2 5.4 2.5 2.0 0.6

112

AUSTRALIAN MATHEMATICAL OLYMPIAD RESULTS

Name School Year

Perfect Score and Gold

Alexander Gunning Glen Waverley Secondary College VIC 12

Seyoon Ragavan Knox Grammar School NSW 11

Gold

Jeremy Yip Trinity Grammar School VIC 12

Yong See Foo Nossal High School VIC 11

Ilia Kucherov Westall Secondary College VIC 11

Yang Song James Ruse Agricultural High School NSW 12

Allen Lu Sydney Grammar School NSW 12

Alan Guo Penleigh and Essendon Grammar School VIC 12

Thomas Baker Scotch College VIC 11

Richard Gong Sydney Grammar School NSW 10

Henry Yoo Perth Modern School WA 12

Silver

Matthew Cheah Penleigh and Essendon Grammar School VIC 10

Kevin Xian James Ruse Agricultural High School NSW 11

Kevin Shen Saint Kentigern College NZ 12 NZ

George Han Westlake Boys High School NZ 13 NZ

Wilson Zhao Killara High School NSW 11

Michelle Chen Methodist Ladies' College VIC 11

Leo Li Christ Church Grammar School WA 11

Jack Liu Brighton Grammar School VIC 9

Charles Li Camberwell Grammar School VIC 9

Anthony Pisani St Paul's Anglican Grammar School VIC 8

Ivan Zelich Anglican Church Grammar School QLD 12

Michael Robertson Dickson College ACT 11

Edward Chen Palmerston North Boys High School NZ 11 NZ

William Hu Christ Church Grammar School WA 9

Martin Luk King's College NZ 13 NZ

113

Name School Year

Bronze

Jerry Mao Caulfield Grammar School VIC 9

Zoe Schwerkolt Fintona Girls' School VIC 11

Harish Suresh James Ruse Agricultural High School NSW 11

Justin Wu James Ruse Agricultural High School NSW 11

Austin Zhang Sydney Grammar School NSW 10

Isabel Longbottom Rossmoyne Senior High School WA 10

Anand Bharadwaj Trinity Grammar School VIC 9

Devin He Christ Church Grammar School WA 11

Miles Yee-Cheng Lee Auckland International College NZ 12 NZ

Bobby Dey James Ruse Agricultural High School NSW 10

Tony Jiang Scotch College VIC 10

Nitin Niranjan All Saints Anglican School QLD 12

William Song Scotch College VIC 11

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

Prince Michael Balanay Botany Downs Secondary College NZ 13 NZ

Michael Chen Scotch College VIC 12

William Wang King's College NZ 12 NZ

Xuzhi Zhang Auckland Grammar School NZ 13 NZ

Keiran Lewellen Home Schooled NZ 10 NZ

Eryuan Sheng Newington College NSW 11

Linus Cooper James Ruse Agricultural High School NSW 9

David Steketee Hale School WA 12

Alexander Barber Scotch College VIC 11

Matthew Jones All Saints Anglican School QLD 12

Madeline Nurcombe Cannon Hill Anglican College QLD 11

Steven Lim Hurlstone Agricultural High School NSW 9

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27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD

XXVII Asian Pacific Mathematics Olympiad

March, 2015

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

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

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

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

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

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

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

ai + 2.

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

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

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

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

(i) an+2 is divisible by an;

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

115

27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD SOLUTIONS

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

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

A

B C

X

Y

D

ZW

V

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

∠(WZ, V Z) = ∠(Y Z,DZ)

= ∠(Y C,DC) (CDZY cyclic)

= ∠(AC,BC).

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

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

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

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

43

1

2

1

2

116

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

Answer: No

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

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

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

and sof(2n)

f(2n+1)=

f(2n+3)

f(2n+4). (1)

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

(22(2n+5)

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

and sof(2n+1)

f(2n+2)=

f(2n+3)

f(2n+4). (2)

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

f(2n)

f(2n+1)=

f(2n+1)

f(2n+2). (3)

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

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

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

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

a · ar = ar5

⇒ a = r4. (4)

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

a · ar2 = ar7

⇒ a = r5. (5)

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

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117

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

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

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

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

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

Next we have

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

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

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118

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

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

(f(2a)f(2c)

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

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

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

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

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

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

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

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

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

In what follows we use (4) repeatedly.

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

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

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

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

Thus f(49) = f(420).

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

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119

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

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

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

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

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

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

f(a2)f(b) = f(b2)f(a) ⇒ f(a2)

f(a)=

f(b2)

f(b).

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

f(a2) = kf(a), for all a ∈ S. (1)

Substituting this into the functional equation yields

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

k, for all a, b ∈ S with a = b. (2)

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

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

k=

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

k2=

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

k.

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

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

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120

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

Answer: n = 60

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

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

Therefore,pi+1

qi+1

=2pi + qi

qior

pi+1

qi+1

=pi

pi + 2qi. (1)

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

qn= 2014 we cannot have pi and qi both

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

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

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

qi+1= 2pi+qi

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

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

pi odd ⇒ pi+1 = pi +qi2

and qi+1 =qi2

for i < n. (3)

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

qi+1= pi

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

Hence

pi even ⇒ pi+1 =pi2

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

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

pi + qi = 2015 for i ≤ 2015. (5)

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

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

qi2,qi2

)

≡(pi2,qi2

)(mod 2015). (6)

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

A simple induction on (6) yields,

(pn, qn) ≡(p02n

,q02n

)(mod 2015). (7)

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

qn ≡ 1 (mod 2015).

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121

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

2n ≡ 1 (mod 2015).

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

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

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

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122

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

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

integer i, we have ai =ai+1 − 1

2or ai =

2ai+1

1− ai+1

. Since ai > 0 we deduce that

ai =

ai+1 − 1

2if ai+1 > 1

2ai+1

1− ai+1

if ai+1 < 1.

(1)

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

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

ai =

u− v

2vif u > v

2u

v − uif u < v.

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

(ui+1, vi+1) =

(ui − vi, 2vi) if ui > vi

(2ui, vi − ui) if ui < vi,(2)

then an−i =ui

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

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

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

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

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

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

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

Next we prove by induction that

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

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

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

≡ (−2i+1, 2i+1) (mod 2015)

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123

and

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

≡ (−2i+1, 2i+1) (mod 2015).

Either way, this completes the inductive step.

We are given that a0 =un

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

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

2n ≡ 1 (mod 2015).

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

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

2, 2011

4, 2007

8, 1999

16, 1983

32, 1951

64, 1887

128, 1759

256, 1503

512, 9911024

, 198233

, 194966

, 1883132

, 1751264

, 1487528

, 9591056

,191897

, 1821194

, 1627388

, 1239776

, 4631552

, 9261089

, 1852163

, 1689326

, 1363652

, 7111304

, 1422593

, 8291186

, 1658357

, 1301714

, 5871428

, 1174841

, 3331682

,6661349

, 1332683

, 6491366

, 1298717

, 5811434

, 1162853

, 3091706

, 6181397

, 1236779

, 4571558

, 9141101

, 1828187

, 1641374

, 1267748

, 5191496

, 1038977

, 611954

,1221893

, 2441771

, 4881527

, 9761039

, 195263

, 1889126

, 1763252

, 1511504

, 10071008

, 20141.

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

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

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124

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

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

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

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

θ

θ1

θ2

α

β

θα

β

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

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

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

O

B

R

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

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125

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

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

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

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126

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

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

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

O

B′

R′

R

B

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

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127

5. Solution (APMO Problem Selection Committee)

There are two families of answers.

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

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

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

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

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

sn = (n− 1)an + hn−1,

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

an+1 = (n+ 1)an + nan−1 + δn (1)

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

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

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

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

Proof For n ≥ 3 we have

an = nan−1 + (n− 1)an−2 + δn−1

> nan−1 + 3. (2)

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

an = nan−1 + (n− 1)an−2 + δn−1

< nan−1 + (an−1 − 3) + δn−1

< (n+ 1)an−1. (3)

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

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

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

< (n2 + 5n+ 5)an.

Also for n ≥ 4,

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

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

= (n2 + 5n+ 3)an.

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

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

n+2an.

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128

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

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

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

n+2an, as desired.

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

n+2an.

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

an. By lemma 2, there

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

m+3am+1.

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

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

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

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

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

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

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

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

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

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

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129

TOP 10 AUSTRALIAN SCORES

Name School Year Total Award

Jeremy Yip Trinity Grammar School VIC 12 28 Gold

Alexander Gunning Glen Waverley Secondary College VIC 12 28 Silver

Yang Song James Ruse Agricultural High School NSW 12 23 Silver

Henry Yoo Perth Modern School WA 12 20 Bronze

Thomas Baker Scotch College VIC 11 20 Bronze

Allen Lu Sydney Grammar School NSW 12 20 Bronze

Ilia Kucherov Westall Secondary College VIC 11 19 Bronze

Seyoon Ragavan Knox Grammar School NSW 11 17

Yong See Foo Nossal High School VIC 11 16

Alan Guo Penleigh and Essendon Grammar School VIC 12 14

COUNTRY SCORES

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

1 USA 10 298 1 2 4 3

2 Korea 10 279 1 2 4 3

3 Russia 10 266 1 2 4 3

4 Singapore 10 259 1 2 4 3

5 Japan 10 256 1 2 4 3

6 Canada 10 237 1 2 4 3

7 Thailand 10 228 1 2 4 3

8 Taiwan 10 222 1 2 4 3

9 Australia 10 205 1 2 4 3

10 Brazil 10 202 1 2 4 3

11 Peru 10 185 0 3 4 3

12 Mexico 10 169 1 1 5 3

13 Hong Kong 10 167 0 3 4 3

14 Kazakhstan 10 163 0 2 5 3

15 Indonesia 10 161 0 3 4 3

16 Malaysia 10 134 0 3 3 3

17 India 10 127 0 1 5 3

27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD RESULTS

130

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

17 Tajikistan 10 127 0 0 6 4

19 Bangladesh 10 122 0 0 7 1

20 Philippines 10 105 0 3 0 2

21 Turkmenistan 10 99 0 0 4 3

22 Saudi Arabia 10 94 0 0 4 1

23 New Zealand 10 86 0 2 1 1

24 Argentina 10 73 0 0 1 3

24 Colombia 10 73 0 1 3 1

26 Syria 6 52 0 0 1 5

27 Sri Lanka 7 48 0 0 1 4

28 El Salvador 7 47 0 0 2 2

29 Trinidad and Tobago 10 31 0 0 1 0

30 Ecuador 10 27 0 0 1 1

31 Costa Rica 5 20 0 0 0 1

32 Panama 2 12 0 0 0 0

33 Cambodia 2 9 0 0 0 1

Total 299 4583 11 42 102 81

131

AMOC SELECTION SCHOOL2015 IMO Team Selection School

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

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

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

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

Angelo Di PasqualeDirector of Training, AMOC

2015 Australian IMO team

Name Year School State

Alexander Gunning 12 Glen Waverley Secondary College VIC

Ilia Kucherov 11 Westall Secondary College VIC

Seyoon Ragavan 11 Knox Grammar School NSW

Yang Song 12 James Ruse Agricultural High School NSW

Kevin Xian 11 James Ruse Agricultural High School NSW

Jeremy Yip 12 Trinity Grammar School VIC

Reserve

Yong See Foo 11 Nossal High School VIC

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2015 AUSTRALIAN IMO TEAM

Name School Year

Alexander Gunning Glen Waverley Secondary College VIC 12

Ilia Kucherov Westall Secondary College VIC 11

Seyoon Ragavan Knox Grammar School NSW 11

Yang Song James Ruse Agricultural High School NSW 12

Kevin Xian James Ruse Agricultural High School NSW 11

Jeremy Yip Trinity Grammar School VIC 12

Reserve

Yong See Foo Nossal High School VIC 11

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

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PARTICIPANTS AT THE 2015 IMO SELECTION SCHOOL

Name School Year

Seniors

Thomas Baker Scotch College VIC 11

Matthew Cheah Penleigh and Essendon Grammar School VIC 10

Michelle Chen Methodist Ladies' College VIC 11

Yong See Foo Nossal High School VIC 11

Alexander Gunning Glen Waverley Secondary College VIC 12

Alan Guo Penleigh and Essendon Grammar School VIC 12

Ilia Kucherov Westall Secondary College VIC 11

Leo Li Christ Church Grammar School WA 11

Allen Lu Sydney Grammar School NSW 12

Seyoon Ragavan Knox Grammar School NSW 11

Kevin Xian James Ruse Agricultural High School NSW 11

Jeremy Yip Trinity Grammar School VIC 12

Henry Yoo Perth Modern School WA 12

Wilson Zhao Killara High School NSW 11

Juniors

Bobby Dey James Ruse Agricultural High School NSW 10

Rachel Hauenschild Kenmore State High School QLD 10

William Hu Christ Church Grammar School WA 9

Tony Jiang Scotch College VIC 10

Charles Li Camberwell Grammar School VIC 9

Jack Liu Brighton Grammar School VIC 9

Isabel Longbottom Rossmoyne Senior High School WA 10

Hilton Nguyen Sydney Technical High School NSW 9

Tommy Wei Scotch College VIC 9

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

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IMO TEAM PREPARATION SCHOOLIMO Team Preparation School

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

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

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

Angelo Di PasqualeIMO Team Leader

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THE MATHEMATICS ASHESThe 2015 Mathematical Ashes: AUS v UK

Exam

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

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

Prove that BR ‖ AC.

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

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

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

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

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

How many ways are there to define such a rule?

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

Results

Q1 Q2 Q3 Σ

AUS 1 7 7 7 21

AUS 2 7 0 7 14

AUS 3 7 7 0 14

AUS 4 7 5 0 12

AUS 5 3 6 1 10

AUS 6 6 7 0 13

Σ 37 32 15 84

Q1 Q2 Q3 Σ

UNK 1 7 7 5 19

UNK 2 7 2 0 9

UNK 3 5 0 0 5

UNK 4 3 7 7 17

UNK 5 7 6 7 20

UNK 6 7 7 0 14

Σ 36 29 19 84

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THE MATHEMATICS ASHES RESULTS

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

Code Name Q1 Q2 Q3 Total

AUS1 Alexander Gunning 7 7 7 21

AUS2 Ilia Kucherov 7 0 7 14

AUS3 Seyoon Ragavan 7 7 0 14

AUS4 Yang Song 7 5 0 12

AUS5 Kevin Xian 3 6 1 10

AUS6 Jeremy Yip 6 7 0 13

TOTAL 37 32 15 84

UNITED KINGDOM

Code Name Q1 Q2 Q3 Total

UNK1 Joe Benton 7 7 5 19

UNK2 Lawrence Hollom 7 2 0 9

UNK3 Samuel Kittle 5 0 0 5

UNK4 Warren Li 3 7 7 17

UNK5 Neel Nanda 7 6 7 20

UNK6 Harvey Yau 7 7 0 14

TOTAL 36 29 19 84

AUSTRALIA

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

Thailand

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

6

12

3

45

6

1

2

3

4

5

6

139

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

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

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

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

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

2017 Brazil

2018 Romania

2019 United Kingdom

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

www.imo-official.org

Angelo Di PasqualeIMO Team Leader, Australia

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

7

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

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INTERNATIONAL MATHEMATICAL OLYMPIAD

Language: English

Day: 1

S

A B S C S AC = BC S

A B C S P S PA = PB = PC

n 3 n

n 3 n

(a, b, c)

ab− c, bc− a, ca− b

2

2 2n n

ABC AB > AC Γ H

F A M BC Q

Γ ∠HQA = 90 K Γ ∠HKQ = 90

A B C K Q Γ

KQH FKM

IMO Papers

Day: 1

Friday, July 10, 2015

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

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

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

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

ab− c, bc− a, ca− b

is a power of 2.

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

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

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

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

8

IMO Papers

Day: 1

Friday, July 10, 2015

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

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

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

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

ab− c, bc− a, ca− b

is a power of 2.

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

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

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

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

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142

Language: English

Day: 2

ABC Ω O Γ A

BC D E B D E C

BC F G Γ Ω A F

B C G Ω K

BDF AB L

CGE CA

FK GL X X

AO

R f : R → R

f(x+ f(x+ y)

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

x y

a1, a2, . . .

1 aj 2015 j 1

k + ak = ℓ+ aℓ 1 k < ℓ

b N

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ 10072

m n n > m N

Day: 2

Saturday, July 11, 2015

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

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

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

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

for all real numbers x and y.

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

(i) 1 ≤ aj ≤ 2015 for all j ≥ 1;

(ii) k + ak = + a for all 1 ≤ k < .

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

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ ≤ 10072

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

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

9

Day: 2

Saturday, July 11, 2015

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

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

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

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

for all real numbers x and y.

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

(i) 1 ≤ aj ≤ 2015 for all j ≥ 1;

(ii) k + ak = + a for all 1 ≤ k < .

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

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ ≤ 10072

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

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

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INTERNATIONAL MATHEMATICAL OLYMPIAD SOLUTIONS

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

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

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

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

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

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

O

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

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

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

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

⌊n−12

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

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

⌋.

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

(n2

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

(n2

).

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

⌋≥

(n2

), which simplifies to⌊

n−12

⌋≥ n−1

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

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

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

⌊n−12

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

n2

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

total indegree being equal to the total outdegree.

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

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

O

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

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

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

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

We divide our solution into five cases as follows.

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

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

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

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

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

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

b2 − c = 2x (1)

b(c− 1) = 2y (2)

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

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

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

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

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

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

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

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

ab− c ≡ 0 (mod ac− b)

⇔ a(ac)− c ≡ 0 (mod ac− b)

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

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

ac− b ≤ 2(a+ 1).

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

(c− 3)a < 2. (4)

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

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

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

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

3a− b ≡ 0 (mod 3b− a)

⇔ 3(3b)− b ≡ 0 (mod 3b− a)

⇔ 8 ≡ 0 (mod 3b− a). (b is odd)

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

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

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

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

ab− c = b(3b− 8)− 3

= (3b+ 1)(b− 3)

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

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

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

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

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

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

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

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

ac− b ≡ 0 (mod bc− a) (5)

⇔ (bc)c− b ≡ 0 (mod bc− a)

⇔ 2kn(c2 − 1) ≡ 0 (mod 2k(nc−m))

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

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

nc−m ≤ 2(c+ 1)

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

⇔ 4k(n2 − 2n) ≤ n

m< 1. (m > n)

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

(c− 1)(c+ 1) ≡ 0 (mod c−m)

⇔ (m− 1)(m+ 1) ≡ 0 (mod c−m).

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

c−m ≤ 2(m+ 1)

⇔ 4km−m ≤ 2m+ 2 (c=4km− 1)

⇔ 4km < 3m+ 2

⇒ k = 1.

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

a(2a− 1)− 2 ≡ 0 (mod 3a− 2)

⇔ 3a(6a− 3) ≡ 18 (mod 3a− 2)

⇔ 2 ≡ 18 (mod 3a− 2).

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

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

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

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

ab− c = 2γ and ac− b = 2β.

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

ab ≤ 2c and ac ≤ 2b. (7)

Multiplying these inequalities together yields

a2bc ≤ 4bc.

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

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

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

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

Hence a > b > c ≥ 2.

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

bc− a | ca− b | ab− c.

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

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

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

ca− b | 2(a− 1)b

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

⇔ ca− b | 2(b− c)

⇒ ca− b ≤ 2(b− c). (b = c)

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

ca− b | 2(a+ 1)b

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

⇔ ca− b | 2(b+ c)

⇒ ca− b ≤ 2(b+ c).

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

ca ≤ 3b+ 2c.

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

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

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

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

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

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

ca− b | ab− c

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

⇒ 3b− 2 | 3((2b− 1)b− 2)− 2b(3b− 2)

⇒ 3b− 2 | b− 6.

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

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

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

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

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

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

A

B C

H

F

Q

EA′

Next we show that CHBA′ is a parallelogram.

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

∠BCA′ = ∠EBC (A′B = CE)

= ∠EAC (ECAB cyclic)

= 90 − ∠ACB (AE ⊥ BC)

= ∠CBH. (BH ⊥ AC)

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

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

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

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A

B C

H

F

Q

EA′

K

M

R

S

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

SF · SM = SH2 = SK2.

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

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

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

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

KA′ about K.

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

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

A

B C

H

F

M

Q

K

A′

S′

M ′

F ′

S

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

Since S ′F ′Q ∼ S ′QM ′ (AA), we have

S ′F ′

S ′Q=

S ′Q

S ′M ′ .

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

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

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

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Solution 3 (IMO Problem Selection Committee)

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

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

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

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

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

∠KFM = ∠TKM.

We have

∠KFM = ∠TKM

⇔ 90 + ∠KFA = ∠TKQ+ 90 + ∠HKM

⇔ ∠KFA = ∠Q′HA′ + ∠HKM. (1)

Thus it suffices to establish (1).

Let J be the midpoint of HQ′. Observe that triangles KHE and AHQ′ are similarwith F and J being the midpoints of corresponding sides. Hence ∠KFA = ∠HJA.

A

B C

H

FM

Q

K

EA′

Q′

J

T

Observe also that triangles KHA′ and QHQ′ are similar with M and J being themidpoints of corresponding sides. Hence ∠HKM = ∠JQH.

Thus our task is reduced to proving

∠HJA = ∠Q′HA′ + ∠JQH.

Let us draw a new diagram that will help us focus on the task at hand.

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A

H

Q

A′

Q′

J

O

Note that QAQ′A′ is a rectangle. Let O be its centre. We also know that H lieson side A′Q and that J is the midpoint of Q′H. Thus J and O both lie on themid-parallel of QA′ and Q′A. Hence

∠HJA = ∠HJO + ∠OJA

= ∠Q′HA+ ∠Q′AJ. (A′Q ‖ JO ‖ Q′A)

Thus it suffices to prove that ∠JQH = ∠Q′AJ . However, this is an immediateconsequence of the fact that JO a line of reflective symmetry of the rectangle.

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Solution 4 (Jacob Tsimerman, Leader of the 2015 Canadian IMO team)

Points A′, E and Q′ are defined as in solution 3. Furthermore, as in solution 3, wehave M is the midpoint of A′H; points A′, M , H and Q are collinear; points Q′, Hand K are collinear; and F is the midpoint of HE.

Since the three chords AE, QA′ and KQ′ are concurrent we have

HA ·HE = HQ ·HA′ = HK ·HQ′ = r2,

for some positive real number r. Let f be the inversion with centre H and radiusr, followed by the reflection1 in H. Let A′′ and Q′′ be the reflections of H in A andQ, respectively. Then f has the effect of exchanging the following pairs of points.

A ↔ E A′′ ↔ F Q ↔ A′ Q′′ ↔ M K ↔ Q′

A

B C

H

FM

Q

K

EA′

Q′

A′′

Q′′

Hence circle KQH is transformed into line Q′A′, and circle FKM is transformedinto circleA′′Q′Q′′. So it suffices to show thatQ′A′ is tangent to circleA′′Q′Q′′. SinceQ′A′ ‖ AQ ‖ A′′Q′′, it suffices to show that A′′Q′Q′′ is isosceles with Q′A′′ = Q′Q′′.

Observe Q′A ‖ A′Q because AQA′Q′ is a rectangle. So Q′A ⊥ A′′Q′′. But A is thecentre of circle A′′HQ′′ because it is the midpoint of A′′H and ∠A′′Q′′H = 90. ThusQ′A is the perpendicular bisector of A′′Q′′. From this it follows that Q′A′′ = A′Q′′,as desired.

1Reflection in a point is the same as a 180 rotation about that point.

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156

Comment Solutions 1 and 3 first establish that Q, H, M and A′ are collinear, andthat M and F are the respective midpoints of A′H and EH. After this the pointsB and C are no longer relevant to the solution. The crux of matter boils down tothe following.

Let AQA′Q′ be a rectangle inscribed in a circle Γ. Let H be any point onthe line A′Q. Let E and K be the respective second points of intersectionof the lines AH and Q′H with Γ. Let M and F be the midpoints of A′Hand EH, respectively. Then circles KQH and KMF are tangent to eachother.

A

H

FM

Q

K

EA′

Q′

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Here is an even more minimal way of looking at the same thing.

Let A′EKQ be a cyclic quadrilateral. Let H be a point on the line A′Qsuch that ∠HEA′ = ∠QKH = 90. Let M and F be the midpoints ofA′H and EH, respectively. Then circles KQH and KMF are tangent toeach other. (*)

H

FM

Q

K

EA′

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Solution 5 (Panupong Pasupat, one of the Coordinators at the 2015 IMO)

It is sufficient to prove statement (*) found in the comments on the previous page.

Let T be a point on the tangent to circle KQH at K, such that T and Q lie on thesame side of the line KH. Let W be the midpoint of QH. Since ∠QKH = 90, itfollows that W is the centre of circle KQH. Hence WH = WK, and we may let

∠WKH = ∠KHW = α.

From the angle sum in KQH, we have ∠HKQ = 90−α. Since A′EKQ is cyclic,we have ∠AEK = 90 + α, and so ∠KEH = α. Thus by the alternate segmenttheorem, circle HEK is tangent to line A′Q at H. Then since MH = ME, itfollows by symmetry that MH and ME are the common tangents from M to circleHEK.

H

FM

Q

K

EA′

T

W

The configuration where MH and ME are common tangents to the circumcircle ofHEK, is a standard one. In particular it yields an alternative characterisation ofthe symmedian.2 Thus we may let

∠HKM = ∠FKE = β.

From the exterior angle sum in EKF , we have

∠KFH = α + β = ∠WKM. (1)

Since WK is the radius of circle KQH, we have TK ⊥ KW . Thus adding 90 toboth sides of (1) yields

∠KFM = ∠TKM.

Hence from the alternate segment theorem, TK is tangent to circle FKM at K.Thus circles FKM and KQH are tangent at K.

2For more details see the section Alternative characterisations of symmedian found in chapter 5 of ProblemSolving Tactics published by the AMT.

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2

159

4. Solution 1 (Kevin Xian, year 11, James Ruse Agricultural High School, NSW.Kevin was a Silver medallist with the 2015 Australian IMO team.)

A

B CD E

F

G

K

L

X

O

Observe that AF = AG (radius of Γ), and OF = OG (radius of Ω). So F andG are symmetric in AO. Thus lines FK and GL intersect on AO if and only if∠KFA = ∠AGL. We have,

∠KFA = ∠FKB − ∠FAB (exterior angle AFK)

= ∠FDB − ∠FGB (BDKF and AFBG cyclic)

= ∠FGE − ∠FGB (FDEG cyclic)

= ∠BGE

= ∠CEG− ∠CBG (exterior angle GEB)

= ∠CLG− ∠CAG (ECGL and ABCG cyclic)

= ∠AGL. (exterior angle GLA)

Comment A careful analysis of this solution shows that the result is still true ifwe only assume that the centre of Γ lies on the line AO.

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Solution 2 (Found independently by Ilia Kucherov, year 11, Westall SecondaryCollege, VIC, and Jeremy Yip, year 12, Trinity Grammar School, VIC. Ilia andJeremy were Silver medallists with the 2015 Australian IMO team.)

As in solution 1 it suffices to show that ∠KFA = ∠AGL.

Let the lines DF and EG intersect Ω for a second time at points P and Q, respec-tively. Then

∠PDE = ∠FGQ (FDEG cyclic)

= ∠FPQ. (FQPG cyclic)

Thus BC ‖ QP . Since BQPC is cyclic, it follows that BQPC is an isoscelestrapezium with BQ = PC. Hence ∠BGQ = ∠PFC, that is,

∠BGE = ∠DFC. (1)

A

B CD E

F

G

PQ

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A

B CD E

F

G

K

L

X

O

Since ABCG is cyclic, we have ∠LAG = ∠CAG = ∠CBG = ∠EBG. Since ECGLis cyclic, ∠CLG = ∠CEG, and so ∠GLA = ∠GEB. Thus GAL ∼ GBE (AA).Hence

∠AGL = ∠BGE. (2)

Similarly∠KFA = ∠DFC. (3)

Combining (1), (2) and (3), it follows that

∠KFA = ∠AGL.

Comment There are two pairs of similar triangles associated with circles ECGLand Ω. They are GAL ∼ GBE and GLE ∼ GAB. This is a standardconfiguration which can help fast track the route to a solution.3

3For more details see the section Similar Switch found in chapter 5 of Problem Solving Tactics publishedby the AMT.

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162

5. Solution 1 (Seyoon Ragavan, year 11, Knox Grammar School, NSW. Seyoon wasa Gold medallist with the 2015 Australian IMO team.)

We show that the only answers are: f(x) = x for all x ∈ R and f(x) = 2− x for allx ∈ R.

We are asked to find all functions f : R → R such that for all x, y ∈ R,

f(x+ f(x+ y)) + f(xy) = x+ f(x+ y) + yf(x). (1)

Set y = 1 in (1) to find that for all x ∈ R,

x+ f(x+ 1) is a fixed point of f . (2)

With (2) in mind, set x = 0 and y = z + f(z + 1) in (1) to find that for all z ∈ R,

f(0) = f(0)(z + f(z + 1)). (3)

Case 1 f(0) = 0

Equation (3) implies z + f(z + 1) = 1 for all z ∈ R. Putting z = x− 1, this may berearranged to f(x) = 2− x for all x ∈ R. We verify this is a solution.

LHS = 2− (x+ 2− (x+ y)) + 2− xy = 2 + y − xy

RHS = x+ 2− (x+ y) + y(2− x) = 2 + y − xy = LHS

Case 2 f(0) = 0

Set x = 0 in (1) to find that for all y ∈ R,

f(f(y)) = f(y). (4)

Set y = 0 in (1) to find that for all x ∈ R,

x+ f(x) is a fixed point of f . (5)

Let S denote the set of fixed points of f . Suppose that u ∈ S. Then (5) with x = utells us that 2u ∈ S. And (2) with x = u− 1 tells us that 2u− 1 ∈ S. Hence

u ∈ S ⇒ 2u, 2u− 1 ∈ S. (6)

Since 0 ∈ S, applying (6) tells us that −1 ∈ S. Applying (6) again tells us that−2,−3 ∈ S. Continuing inductively, we find that all negative integers are in S.

On the other hand, if x is any positive integer, choose an integer y such that y < 0,x+ y < 0, 2x+ y < 0 and xy < 0 (any y < −2x will do). Using these values in (1)yields f(x) = x. Hence

Z ⊆ S. (7)

Since f(1) = 1, we may set x = 1 in (1) to find that for all y ∈ R,

f(1 + f(y + 1)) + f(y) = y + 1 + f(y + 1). (8)

Suppose that u, u + 1 ∈ S. Then setting y = u in (8) tells us that u + 2 ∈ S. Alsosetting y = u− 1 in (8) tells us that u− 1 ∈ S. An immediate corollary of this is

u, u+ 1 ∈ S ⇒ u+ n ∈ S for any n ∈ Z. (9)

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Let y ∈ R. Then from (2) with x = y−1 we have y+ f(y)−1 ∈ S, and from (5) wehave y+ f(y) ∈ S. Hence with u = y+ f(y)− 1 in (7), we find that for each n ∈ Z,

y + f(y) + n ∈ S. (10)

Using the change of variables y = x+m and n = −m, we have for each m ∈ Z andx ∈ R,

x+ f(x+m) ∈ S. (11)

If we set y = m in (1) and use (11), then for each m ∈ Z and x ∈ R we have,

f(mx) = mf(x). (12)

If we replace y with f(y) in (10) and remember that f(f(y)) = f(y) from (4), wefind

2f(y) + n ∈ S. (13)

Let y ∈ R and let y = 2x. Then applying (12) and (13), we are able to deduce thatf(y) + 1 = f(2x) + 1 = 2f(x) + 1 ∈ S. Hence for all y ∈ R we have

f(y) + 1 ∈ S. (14)

Finally, put x = 1 in (1). Then using f(1 + f(1 + y)) = 1+ f(1 + y) from (14), andf(1) = 1 from (7), we deduce that f(y) = y for all y ∈ R. This is easily seen to bea solution.

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Solution 2 (Based on the presentation by Alex Gunning, year 12, Glen WaverleySecondary College, VIC. Alex was a Gold medallist with the 2015 Australian IMOteam.)

We are asked to find all functions f : R → R such that for all x, y ∈ R,

f(x+ f(x+ y)) + f(xy) = x+ f(x+ y) + yf(x). (1)

The case f(0) = 0 is handled as in solution 1.

It remains to deal with the case f(0) = 0. All equations that follow will hold truefor all z ∈ R.Replacing x with y and y with x in (1) yields

f(y + f(x+ y)) + f(xy) = y + f(x+ y) + xf(y). (2)

Computing the difference (1)− (2), we have for all x, y ∈ R,

f(x+ f(x+ y))− f(y + f(x+ y)) = x− y + yf(x)− xf(y). (3)

Putting y = −x in (3) yields

f(x)− f(−x) = 2x− xf(x)− xf(−x). (4)

Suppose that f(x) = x. Then (4) can be rearranged to yield

(x− 1)(f(−x) + x)) = 0.

If x = 1, then f(−x) = −x. If x = 1, then putting (x, y) = (−1, 1) in (1), yieldsf(−1) = −1. Hence if S is the set of fixed points of f , we have shown that

x ∈ S ⇒ −x ∈ S. (5)

Putting (x, y) = (z, 0) in (1), we find,

z + f(z) ∈ S. (6)

Putting (x, y) = (0, z) in (1), we find,

f(z) ∈ S. (7)

Putting (x, y) = (f(z), 0) in (1) and using (7), we find,

2f(z) ∈ S. (8)

Put (x, y) = (z + f(z),−f(z)) in (3). The LHS of (3) is f(z + 2f(z)). We alsohave x = z + f(z) ∈ S from (6). And y = −f(z) ∈ S from (7) and (5). Thusyf(x) − xf(y) = yx − xy = 0, and so the RHS of (3) is just x − y = z + 2f(z).Hence

z + 2f(z) ∈ S. (9)

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Put (x, y) = (z,−z − f(z)) in (3). Note that y = −z − f(z) ∈ S from (6) and (5).And x+ y = −f(z) ∈ S from (7) and (5). Thus (3) simplifies to

f(2x+ y)− f(x+ 2y) = x− y + yf(x)− xy.

However, x+ 2y = −z − 2f(z) ∈ S from (9) and (5). So (3) simplifies further to

f(2x+ y) = 2x+ y + yf(x)− xy.

Writing x and y in term of z, and tidying up yields

f(z − f(z)) = z − f(z) + z2 − f(z)2. (10)

Put (x, y) = (z−f(z), f(z)) in (3). The LHS of (3) equals f(z)−f(2f(z)) = −f(z)from (8). Hence

−f(z) = x− y + yf(x)− xf(y)

= z − 2f(z) + yf(x)− xy (since y = f(z) ∈ S)

⇒ f(z)− z = f(z)(f(x)− x) (y = f(z))

= f(z)(z2 − f(z)2). (by (10) as x = z − f(z))

This can be written in the form

(f(z)− z)(f(z)2 + zf(z) + 1) = 0. (11)

Solving for f(z) in (11), we find f(z) ∈z, −z+

√z2−4

2, −z−

√z2−4

2

. Since f(z) ∈ R,

we have f(z) = z for all |z| < 2. Hence (−2, 2) ⊆ S. Finally, from (8) we see thatz ∈ S implies 2z ∈ S. This allows us to deduce that (−2k, 2k) ⊆ S for all positiveintegers k. Hence S = R, and f(x) = x for all x ∈ R.

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6. This was the hardest problem of the 2015 IMO. Only 11 of the 577 contestants wereable to solve this problem completely.

The authors of this problem were Ross Atkins and Ivan Guo of Australia. Ross andIvan were Bronze medallists with the 2003 Australian IMO Team and Ivan was aGold medallist with the 2004 Australian IMO Team. The problem was inspired by anotation for juggling (Ross is also a juggler) in which each ai represents the airtimeof a ball thrown at time i, and b is the total number of balls.

Solution 1 (Alex Gunning, year 12, Glen Waverley Secondary College, VIC. Alexwas a Gold medallist with the 2015 Australian IMO team.)

Let S be the set of positive integers which are not of the form n + an for somepositive integer n. Note that S is nonempty because 1 ∈ S. Let s1 < s2 < · · · bethe elements of S listed in increasing order.

Lemma |S| ≤ 2015.

Proof Assume that |S| ≥ 2016. Choose n so that an+n ≥ s2016. Since an ≤ 2015,this implies that s1, s2, . . . , s2016 ∈ 1, 2, . . . , n + 2015. However, the n numbers1 + a1, 2 + a2, . . . , n + an are not equal to any si and are also members of the set1, 2, . . . , n+ 2015. Hence 1, 2, . . . , n+ 2015 contains at least n+ 2016 differentnumbers, contradiction.

We claim that if b = |S| and if N is larger than all members of S, then the inequalityposed in the problem statement is true.

Let n be any integer satisfying n ≥ N . We shall find bounds for∑n

j=1(j + aj) andhence also for

∑nj=1(aj − b). In what follows, let L be the following list of n + b

distinct positive integers.

1 + a1, 2 + a2, . . . , n+ an, s1, s2, . . . , sb

For the lower bound, since the n+ b numbers in L are distinct, their sum is greaterthan or equal to

∑n+bj=1 j. Hence we have

n∑j=1

(j + aj) +b∑

j=1

sj ≥n+b∑j=1

j

⇒n∑

j=1

aj ≥n+b∑

j=n+1

j −b∑

j=1

sj

=b(2n+ b+ 1)

2− s

⇒n∑

j=1

(aj − b) ≥ b2 + b

2− s, (1)

where s =∑b

j=1 sj.

For the upper bound, observe that s1, s2, . . . , sb are b members belonging to the setT = 1, 2, . . . , n + 1. The remaining n + 1− b members of T must be of the formj + aj where j ≤ n, and so are in L. The sum of these n+ 1− b numbers is exactly

n+1∑j=1

j −b∑

j=1

sb.

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All together there are exactly n numbers of the form j + aj in L and so far we haveaccounted for n+ 1− b of them.

Consider the remaining b−1 numbers of the form j+aj which are in L. When listedin decreasing order, they can be no larger than n+2015, n+2014, . . . , n+2015−b+2,respectively. Hence their sum is at most

∑b−1j=1(n+ 2016− j). Thus

n∑j=1

(j + aj) ≤n+1∑j=1

j −b∑

j=1

sb +b−1∑j=1

(n+ 2016− j)

⇒n∑

j=1

aj ≤ n+ 1 +(b− 1)(2n+ 4032− b)

2− s

⇒n∑

j=1

(aj − b) ≤ 4033b− b2 − 4030

2− s. (2)

Summarising (1) and (2), we have established the following bounds for any n ≥ N .

4033b− b2 − 4030

2− s ≤

n∑j=1

(aj − b) ≤ b2 + b

2− s. (3)

Now let m,n be any two integers satisfying n > m ≥ N . Since also m ≥ N , (3) isalso satisfied if n is replaced by m. Thus

∣∣∣∣∣n∑

j=m+1

(aj − b)

∣∣∣∣∣ =∣∣∣∣∣

n∑j=1

(aj − b)−m∑j=1

(aj − b)

∣∣∣∣∣

≤∣∣∣∣(b2 + b

2− s

)−

(4033b− b2 − 4030

2− s

)∣∣∣∣

= (b− 1)(2015− b)

≤((b− 1) + (2015− b)

2

)2

(AM–GM)

= 10072.

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168

Solution 2 (A juggling interpretation solution by the authors Ross Atkins andIvan Guo)

Suppose you are juggling several balls using only one hand. At the ith second, if aball lands in your hand, it is thrown up immediately. If no ball lands, you insteadreach for an unused ball (that is, a ball that has not been thrown yet) and throw itup. In both cases, a ball is thrown so that it will stay in the air for ai seconds. Thecondition ai + i = aj + j ensures that no two balls land at the same time.

Let b be the total number of balls used. If b > 2015, then eventually a ball muststay in the air for more than 2015 seconds, contradicting ai ≤ 2015. So b is finiteand bounded by 2015.

Select N so that all b balls have been introduced by the Nth second. For all i ≥ N ,denote by Ti the total remaining airtime of the current balls, immediately after theith throw is made. (That is, we calculate the remaining airtime for each currentball, and add these values together.) Consider what happens during the next second.The airtime of each of the b balls is reduced by 1. At the same time a ball is thrown,increasing its airtime by ai+1. Thus we have the equality Ti+1 − Ti = ai+1 − b. Thisgives a nice representation of the required sum,

n∑i=m+1

(ai − b) = Tn − Tm.

To complete the problem, it suffices to identify the maximal and minimal possiblevalues of the total remaining airtime Ti. Since no two balls can land at the sametime, the minimal value is 1 + 2 + · · · + b. On the other hand, the maximal valueis 1 + 2015 + 2014 + · · · + (2015 − b + 2). (Note that there must be a ball with aremaining airtime of 1 since something must be caught and thrown every second.)Taking the difference between these two sums, we find that

|Tn − Tm| ≤(4032− b)(b− 1)

2− (b+ 2)(b− 1)

2

= (2015− b)(b− 1)

≤ 10072 (GM ≤ AM)

as required.

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169

INTERNATIONAL MATHEMATICAL OLYMPIAD RESULTS

Mark Q1 Q2 Q3 Q4 Q5 Q6

0 93 256 408 91 153 521

1 89 151 122 36 255 11

2 5 77 12 61 34 15

3 21 27 1 18 90 6

4 72 8 3 11 8 3

5 12 13 0 1 4 3

6 20 14 1 8 3 7

7 265 31 30 351 30 11

Total 577 577 577 577 577 577

Mean 4.3 1.4 0.7 4.8 1.5 0.4

Name Q1 Q2 Q3 Q4 Q5 Q6 Score Award

Alex Gunning 7 6 7 7 2 7 36 Gold

Ilia Kucherov 7 2 0 7 3 0 19 Silver

Seyoon Ragavan 7 7 1 7 7 0 29 Gold

Yang Song 7 2 1 7 3 0 20 Silver

Kevin Xian 7 3 1 7 3 0 21 Silver

Jeremy Yip 7 6 1 7 2 0 23 Silver

Totals 42 26 11 42 20 7 148

Australian average 7.0 4.3 1.8 7.0 3.3 1.2 24.7

IMO average 4.3 1.4 0.7 4.8 1.5 0.4 13.0

The medal cuts were set at 26 for Gold, 19 for Silver and 14 for Bronze.

AUSTRALIAN SCORES AT THE IMO

MARK DISTRIBUTION BY QUESTION

170

Rank Country Total

1 United States of America 185

2 China 181

3 South Korea 161

4 North Korea 156

5 Vietnam 151

6 Australia 148

7 Iran 145

8 Russia 141

9 Canada 140

10 Singapore 139

11 Ukraine 135

12 Thailand 134

13 Romania 132

14 France 120

15 Croatia 119

16 Peru 118

17 Poland 117

18 Taiwan 115

19 Mexico 114

20 Hungary 113

20 Turkey 113

22 Brazil 109

22 Japan 109

22 United Kingdom 109

25 Kazakhstan 105

26 Armenia 104

27 Germany 102

28 Hong Kong 101

29 Bulgaria 100

29 Indonesia 100

29 Italy 100

29 Serbia 100

SOME COUNTRY TOTALS

171

DISTRIBUTION OF AWARDS AT THE 2015 IMO

Country Total Gold Silver Bronze HM

Albania 37 0 0 0 3

Algeria 60 0 1 1 2

Argentina 70 0 0 1 4

Armenia 104 0 1 5 0

Australia 148 2 4 0 0

Austria 63 0 0 3 1

Azerbaijan 73 0 0 2 4

Bangladesh 97 0 1 4 1

Belarus 84 0 0 3 3

Belgium 67 0 1 0 3

Bolivia 5 0 0 0 0

Bosnia and Herzegovina 76 0 0 2 4

Botswana 1 0 0 0 0

Brazil 109 0 3 3 0

Bulgaria 100 0 2 1 2

Cambodia 24 0 0 0 2

Canada 140 2 0 4 0

Chile 12 0 0 0 1

China 181 4 2 0 0

Colombia 72 0 0 4 0

Costa Rica 53 0 0 2 2

Croatia 119 1 3 1 0

Cuba 15 0 0 1 0

Cyprus 58 0 1 0 2

Czech Republic 74 0 0 3 3

Denmark 52 0 0 2 1

Ecuador 27 0 0 0 2

El Salvador 14 0 0 0 0

Estonia 51 0 0 1 3

Finland 26 0 0 0 1

France 120 0 3 3 0

Georgia 80 0 1 3 1Germany 102 0 2 3 0

Ghana 5 0 0 0 0

Greece 71 0 1 2 2

172

Country Total Gold Silver Bronze HMHong Kong 101 0 2 3 1

Hungary 113 0 3 3 0

Iceland 41 0 0 0 3

India 86 0 1 2 3

Indonesia 100 0 2 4 0

Iran 145 3 2 1 0

Ireland 37 0 0 0 3

Israel 83 1 0 2 2

Italy 100 1 2 0 0

Japan 109 0 3 3 0

Kazakhstan 105 1 1 2 2

Kosovo 24 0 0 0 1

Kyrgyzstan 17 0 0 0 0

Latvia 36 0 0 0 3

Liechtenstein 18 0 0 1 0

Lithuania 54 0 0 1 1

Luxembourg 12 0 0 0 1

Macau 88 0 1 2 3

Macedonia (FYR) 45 0 0 1 1

Malaysia 66 0 0 3 1

Mexico 114 1 2 3 0

Moldova 85 0 1 2 3

Mongolia 74 0 0 2 4

Montenegro 19 0 0 1 0

Morocco 27 0 0 0 1

Netherlands 76 0 0 3 1

New Zealand 72 0 0 2 4

Nicaragua 26 0 0 0 3

Nigeria 22 0 0 0 2

North Korea 156 3 3 0 0

Norway 54 0 1 0 2

Pakistan 25 0 0 1 0

Panama 9 0 0 0 0

Paraguay 53 0 0 3 0

Peru 118 2 2 1 0

Philippines 87 0 2 2 1

Poland 117 1 1 4 0

173

Country Total Gold Silver Bronze HM

Portugal 70 0 0 3 1

Puerto Rico 18 0 0 1 0

Romania 132 1 4 1 0

Russia 141 0 6 0 0

Saudi Arabia 81 0 1 3 2

Serbia 100 1 1 2 2

Singapore 139 1 4 1 0

Slovakia 97 0 2 3 0

Slovenia 46 0 0 1 1

South Africa 68 0 0 1 2

South Korea 161 3 1 2 0

Spain 47 0 0 1 2

Sri Lanka 51 0 0 0 4

Sweden 63 0 0 2 2

Switzerland 74 0 0 3 2

Syria 69 0 1 1 3

Taiwan 115 0 4 1 1

Tajikistan 57 0 1 1 2

Tanzania 0 0 0 0 0

Thailand 134 2 3 1 0

Trinidad and Tobago 26 0 1 0 0

Tunisia 41 0 0 1 2

Turkey 113 0 5 0 0

Turkmenistan 64 0 0 2 2

Uganda 6 0 0 0 0

Ukraine 135 2 3 1 0

United Kingdom 109 0 4 1 1

United States of America 185 5 1 0 0

Uruguay 16 0 0 0 1

Uzbekistan 64 0 0 3 2

Venezuela 13 0 0 0 1

Vietnam 151 2 3 1 0

Total (104 teams, 577 contestants) 39 100 143 126

174

ORIGIN OF SOME QUESTIONS

Senior ContestQuestion 1 was submitted by Angelo Di Pasquale.Questions 2, 3 and 5 were submitted by Norman Do.Question 4 was submitted by Alan Offer.

Australian Mathematical OlympiadQuestions 1, 2, 5 and 6 were submitted by Norman Do.Question 3 was submitted by Andrei Storozhev.Questions 4 and 7 were submitted by Angelo Di Pasquale.Question 8 was submitted by Andrew Elvey Price.

Asian Pacific Mathematical Olympiad 2015Question 2 was composed by Angelo Di Pasquale and submitted by the AMOC Senior Problems Com-mittee.

International Mathematical Olympiad 2015Question 6 was composed by Ross Atkins and Ivan Guo, and submitted by the AMOC Senior ProblemsCommittee. Ivan provided the following background information on the problem.

The original idea for this problem came about while Ross was reading the paper PositroidVarieties: Juggling and Geometry by Knutson, Lam and Speyer, in which the “excitationnumber” of a periodic juggling sequence was discovered. It seemed obvious that this wassimilar to some specific elementary result that could be proven using elementary methods.We had some difficulties in phrasing the problem in a concise self-contained way. Intuitively,each term ai in the sequence corresponds to throwing a ball at the ith second with an air timeof ai. The inequality condition ensures that no two balls land simultaneously.

The first formulation of the problem was to show that the long-term average of the sequenceconverges to an integer b, which is the total number of balls. However, the usage of limits wasinappropriate for an olympiad problem. We then came up with three more versions whichinvolved bounding the partial sums. Eventually we settled on the most difficult version, withthe explicit bound of |

∑(ai − b)| ≤ 10072. Interestingly, the term ai − b can be interpreted

as the change in the “total air time” on the ith second, while 10072 is the difference betweenmaximal and minimal possible “total air times”, after the introduction of all b balls. The finalwording may be a little difficult for students who are unfamiliar with the construct: “thereexists an N such that for all m > n > N”.

It is possible to solve the problem combinatorially without invoking any physical interpreta-tions, juggling or otherwise. Furthermore, as demonstrated by some at the IMO, the problemcan also be tackled using purely algebraic approaches. Overall, we are very happy with theproblem and we hope everyone enjoyed it.

It is worth noting that one of the authors of the paper that inspired this problem was Thomas Lam, amember of the 1997 Australian IMO team and recipient of an IMO gold medal.

1

ORIGIN OF SOME QUESTIONS

175

HONOUR ROLL

Because of changing titles and affiliations, the most senior title achieved and later affiliations are generally used, except for the Interim committee, where they are listed as they were at the time.

Interim Committee 1979–1980Mr P J O’Halloran Canberra College of Advanced Education, ACT, ChairProf A L Blakers University of Western AustraliaDr J M Gani Australian Mathematical Society, ACT,Prof B H Neumann Australian National University, ACT,Prof G E Wall University of Sydney, NSWMr J L Williams University of Sydney, NSW

Mathematics Challenge for Young Australians

Problems Committee for ChallengeDr K McAvaney Victoria, (Director) 9 years; 2006–2015; Member 1 year 2005–2006Mr B Henry Victoria (Director) 17 years; 1990–2006; Member 9 years 2007–2015Prof P J O’Halloran University of Canberra, ACT 5 years; 1990–1994Dr R A Bryce Australian National University, ACT 23 years; 1990–2012Adj Prof M Clapper Australian Mathematics Trust, ACT 3 years; 2013–2015Ms L Corcoran Australian Capital Territory 3 years; 1990–1992Ms B Denney New South Wales 6 years; 2010–2015Mr J Dowsey University of Melbourne, VIC 8 years; 1995–2002Mr A R Edwards Department of Education, QLD 26 years; 1990–2015Dr M Evans Scotch College, VIC 6 years; 1990–1995Assoc Prof H Lausch Monash University, VIC 24 years; 1990–2013Ms J McIntosh AMSI, VIC 14 years; 2002–2015Mrs L Mottershead New South Wales 24 years; 1992–2015Miss A Nakos Temple Christian College, SA 23 years; 1993–2015Dr M Newman Australian National University, ACT 26 years; 1990–2015Ms F Peel St Peter’s College, SA 2 years; 1999, 2000Dr I Roberts Northern Territory 3 years; 2013–2015Ms T Shaw SCEGGS, NSW 3 years; 2013–2015Ms K Sims New South Wales 17 years; 1999–2015Dr A Storozhev Attorney General’s Department, ACT 22 years; 1994–2015Prof P Taylor Australian Mathematics Trust, ACT 20 years; 1995–2014Mrs A Thomas New South Wales 18 years; 1990–2007Dr S Thornton South Australia 18 years; 1998–2015 Miss G Vardaro Wesley College, VIC 22 years: 1993–2006, 2008–2015Visiting membersProf E Barbeau University of Toronto, Canada 1991, 2004, 2008Prof G Berzsenyi Rose Hulman Institute of Technology, USA 1993, 2002Dr L Burjan Department of Education, Slovakia 1993Dr V Burjan Institute for Educational Research, Slovakia 1993Mrs A Ferguson Canada 1992Prof B Ferguson University of Waterloo, Canada 1992, 2005Dr D Fomin St Petersburg State University, Russia 1994Prof F Holland University College, Ireland 1994Dr A Liu University of Alberta, Canada 1995, 2006, 2009Prof Q Zhonghu Academy of Science, China 1995

176

Dr A Gardiner University of Birmingham, United Kingdom 1996Prof P H Cheung Hong Kong 1997Prof R Dunkley University of Waterloo, Canada 1997Dr S Shirali India 1998Mr M Starck New Caledonia 1999Dr R Geretschlager Austria 1999, 2013Dr A Soifer United States of America 2000Prof M Falk de Losada Colombia 2000Mr H Groves United Kingdom 2001 Prof J Tabov Bulgaria 2001, 2010Prof A Andzans Latvia 2002 Prof Dr H-D Gronau University of Rostock, Germany 2003 Prof J Webb University of Cape Town, South Africa 2003, 2011Mr A Parris Lynwood High School, New Zealand 2004Dr A McBride University of Strathclyde, United Kingdom 2007Prof P Vaderlind Stockholm University, Sweden 2009, 2012Prof A Jobbings United Kingdom 2014 Assoc Prof D Wells United States of America 2015

Moderators for ChallengeMr W Akhurst New South WalesMs N Andrews ACER, Camberwell, VICProf E Barbeau University of Toronto, CanadaMr R Blackman VictoriaMs J Breidahl St Paul’s Woodleigh, VICMs S Brink Glen Iris, VICProf J C Burns Australian Defence Force Academy, ACTMr A. Canning Queensland Mrs F Cannon New South WalesMr J Carty ACT Department of Education, ACTDr E Casling Australian Capital TerritoryMr B Darcy South AustraliaMs B Denney New South WalesMr J Dowsey VictoriaBr K Friel Trinity Catholic College, NSWDr D Fomin St Petersburg University, RussiaMrs P Forster Penrhos College, WAMr T Freiberg QueenslandMr W Galvin University of Newcastle, NSWMr M Gardner North Virginia, USAMs P Graham TasmaniaMr B Harridge University of Melbourne, VICMs J Hartnett Queensland Mr G Harvey Australian Capital TerritoryMs I Hill South AustraliaMs N Hill VictoriaDr N Hoffman Edith Cowan University, WAProf F Holland University College, IrelandMr D Jones Coff’s Harbour High School, NSWMs R Jorgenson Australian Capital TerritoryAssoc Prof H Lausch VictoriaMr J Lawson St Pius X School, NSWMr R Longmuir ChinaMs K McAsey Victoria

177

Moderators for Challenge continuedDr K McAvaney VictoriaMs J McIntosh AMSI, VICMs N McKinnon VictoriaMs T McNamara VictoriaMr G Meiklejohn Queensland School Curriculum Council, QLDMr M O’Connor AMSI, VICMr J Oliver Northern TerritoryMr S Palmer New South WalesDr W Palmer University of Sydney, NSWMr G Pointer South AustraliaProf H Reiter University of North Carolina, USAMr M Richardson Yarraville Primary School, VICMr G Samson Nedlands Primary School, WAMr J Sattler Parramatta High School, NSWMr A Saunder VictoriaMr W Scott Seven Hills West Public School, NSWMr R Shaw Hale School, WAMs T Shaw New South WalesDr B Sims University of Newcastle, NSWDr H Sims VictoriaMs K Sims New South WalesProf J Smit The NetherlandsMrs M Spandler New South WalesMr G Spyker Curtin University, WAMs C Stanley Queensland Dr E Strzelecki Monash University, VICMr P Swain Ivanhoe Girls Grammar School, VICDr P Swedosh The King David School, VICProf J Tabov Academy of Sciences, BulgariaMrs A Thomas New South WalesMs K Trudgian QueenslandProf J Webb University of Capetown, South AfricaMs J Vincent Melbourne Girls Grammar School, VIC

Mathematics Enrichment Development

Enrichment Committee — Development Team (1992–1995)Mr B Henry Victoria (Chairman)Prof P O’Halloran University of Canberra, ACT (Director)Mr G Ball University of Sydney, NSWDr M Evans Scotch College, VICMr K Hamann South AustraliaAssoc Prof H Lausch Monash University, VICDr A Storozhev Australian Mathematics Trust, ACT Polya Development Team (1992–1995)Mr G Ball University of Sydney, NSW (Editor)Mr K Hamann South Australia (Editor)Prof J Burns Australian Defence Force Academy, ACTMr J Carty Merici College, ACTDr H Gastineau-Hill University of Sydney, NSWMr B Henry VictoriaAssoc Prof H Lausch Monash University, VICProf P O’Halloran University of Canberra, ACT

178

Dr A Storozhev Australian Mathematics Trust, ACTEuler Development Team (1992–1995)Dr M Evans Scotch College, VIC (Editor)Mr B Henry Victoria (Editor)Mr L Doolan Melbourne Grammar School, VICMr K Hamann South AustraliaAssoc Prof H Lausch Monash University, VICProf P O’Halloran University of Canberra, ACTMrs A Thomas Meriden School, NSWGauss Development Team (1993–1995)Dr M Evans Scotch College, VIC (Editor)Mr B Henry Victoria (Editor)Mr W Atkins University of Canberra, ACTMr G Ball University of Sydney, NSWProf J Burns Australian Defence Force Academy, ACTMr L Doolan Melbourne Grammar School, VICMr A Edwards Mildura High School, VICMr N Gale Hornby High School, New ZealandDr N Hoffman Edith Cowan University, WAProf P O’Halloran University of Canberra, ACTDr W Pender Sydney Grammar School, NSWMr R Vardas Dulwich Hill High School, NSWNoether Development Team (1994–1995)Dr M Evans Scotch College, VIC (Editor)Dr A Storozhev Australian Mathematics Trust, ACT (Editor)Mr B Henry VictoriaDr D Fomin St Petersburg University, RussiaMr G Harvey New South WalesNewton Development Team (2001–2002)Mr B Henry Victoria (Editor)Mr J Dowsey University of Melbourne, VICMrs L Mottershead New South WalesMs G Vardaro Annesley College, SAMs A Nakos Temple Christian College, SAMrs A Thomas New South WalesDirichlet Development Team (2001–2003)Mr B Henry Victoria (Editor)Mr A Edwards Ormiston College, QLDMs A Nakos Temple Christian College, SAMrs L Mottershead New South Wales

Australian Mathematical Olympiad CommitteeThe Australian Mathematical Olympiad Committee was founded at a meeting of the Australian Academy of Science at its meeting of 2–3 April 1980.

* denotes Executive Position

Chair*Prof B H Neumann Australian National University, ACT 7 years; 1980–1986Prof G B Preston Monash University, VIC 10 years; 1986–1995Prof A P Street University of Queensland 6 years; 1996–2001Prof C Praeger University of Western Australia 14 years; 2002–2015Deputy Chair*Prof P J O’Halloran University of Canberra, ACT 15 years; 1980–1994Prof A P Street University of Queensland 1 year; 1995Prof C Praeger, University of Western Australia 6 years; 1996–2001

179

Assoc Prof D Hunt University of New South Wales 14 years; 2002–2015Executive Director*Prof P J O’Halloran University of Canberra, ACT 15 years; 1980–1994Prof P J Taylor University of Canberra, ACT 18 years; 1994–2012Adj Prof M G Clapper University of Canberra, ACT 3 years; 2013–2015

SecretaryProf J C Burns Australian Defence Force Academy, ACT 9 years; 1980–1988Vacant 4 years; 1989–1992Mrs K Doolan Victorian Chamber of Mines, VIC 6 years; 1993–1998

Treasurer*Prof J C Burns Australian Defence Force Academy, ACT 8 years; 1981–1988Prof P J O’Halloran University of Canberra, ACT 2 years; 1989–1990Ms J Downes CPA 5 years; 1991–1995Dr P Edwards Monash University, VIC 8 years; 1995–2002Prof M Newman Australian National University, ACT 6 years; 2003–2008Dr P Swedosh The King David School, VIC 7 years; 2009–2015

Director of Mathematics Challenge for Young Australians*Mr J B Henry Deakin University, VIC 17 years; 1990–2006Dr K McAvaney Deakin University, VIC 10 years; 2006–2015

Chair, Senior Problems CommitteeProf B C Rennie James Cook University, QLD 1 year; 1980Mr J L Williams University of Sydney, NSW 6 years; 1981–1986Assoc Prof H Lausch Monash University, VIC 27 years; 1987–2013Dr N Do Monash University, VIC 2 years; 2014–2015

Director of Training*Mr J L Williams University of Sydney, NSW 7 years; 1980–1986Mr G Ball University of Sydney, NSW 3 years; 1987–1989Dr D Paget University of Tasmania 6 years; 1990–1995Dr M Evans Scotch College, VIC 3 months; 1995Assoc Prof D Hunt University of New South Wales 5 years; 1996–2000Dr A Di Pasquale University of Melbourne, VIC 15 years; 2001–2015

Team LeaderMr J L Williams University of Sydney, NSW 5 years; 1981–1985Assoc Prof D Hunt University of New South Wales 9 years; 1986, 1989, 1990, 1996–2001Dr E Strzelecki Monash University, VIC 2 years; 1987, 1988Dr D Paget University of Tasmania 5 years; 1991–1995Dr A Di Pasquale University of Melbourne, VIC 13 years; 2002–2010, 2012–2015Dr I Guo University of New South Wales 1 year; 2011

Deputy Team LeaderProf G Szekeres University of New South Wales 2 years; 1981–1982Mr G Ball University of Sydney, NSW 7 years; 1983–1989Dr D Paget University of Tasmania 1 year; 1990Dr J Graham University of Sydney, NSW 3 years; 1991–1993Dr M Evans Scotch College, VIC 3 years; 1994–1996Dr A Di Pasquale University of Melbourne, VIC 5 years; 1997–2001Dr D Mathews University of Melbourne, VIC 3 years; 2002–2004Dr N Do University of Melbourne, VIC 4 years; 2005–2008Dr I Guo University of New South Wales 4 years; 2009–10, 2012–2013

180

Mr G White University of Sydney, NSW 1 year; 2011Mr A Elvey Price Melbourne University, VIC 2 years; 2014–2015

State DirectorsAustralian Capital TerritoryProf M Newman Australian National University 1 year; 1980Mr D Thorpe ACT Department of Education 2 years; 1981–1982Dr R A Bryce Australian National University 7 years; 1983–1989Mr R Welsh Canberra Grammar School 1 year; 1990Mrs J Kain Canberra Grammar School 5 years; 1991–1995Mr J Carty ACT Department of Education 17 years; 1995–2011Mr J Hassall Burgmann Anglican School 2 years; 2012–2013Dr C Wetherell Radford College 2 years; 2014–2015New South WalesDr M Hirschhorn University of New South Wales 1 year; 1980Mr G Ball University of Sydney, NSW 16 years; 1981–1996Dr W Palmer University of Sydney, NSW 19 years; 1997–2015Northern TerritoryDr I Roberts Charles Darwin University 2 years; 2014–2015QueenslandDr N H Williams University of Queensland 21 years; 1980–2000Dr G Carter Queensland University of Technology 10 years; 2001–2010Dr V Scharaschkin University of Queensland 4 years; 2011–2014Dr A Offer Queensland 1 year; 2015South Australia/Northern TerritoryMr K Hamann SA Department of Education 19 years; 1980–1982, 1991–2005, 2013Mr V Treilibs SA Department of Education 8 years; 1983–1990Dr M Peake Adelaide 8 years; 2006–2013Dr D Martin Adelaide 2 years; 2014–2015TasmaniaMr J Kelly Tasmanian Department of Education 8 years; 1980–1987Dr D Paget University of Tasmania 8 years; 1988–1995Mr W Evers St Michael’s Collegiate School 9 years; 1995–2003Dr K Dharmadasa University of Tasmania 12 years; 2004–2015VictoriaDr D Holton University of Melbourne 3 years; 1980–1982Mr B Harridge Melbourne High School 1 year; 1982Ms J Downes CPA 6 years; 1983–1988Mr L Doolan Melbourne Grammar School 9 years; 1989–1998Dr P Swedosh The King David School 18 years; 1998–2015Western AustraliaDr N Hoffman WA Department of Education 3 years; 1980–1982Assoc Prof P Schultz University of Western Australia 14 years; 1983–1988, 1991–1994, 1996–1999Assoc Prof W Bloom Murdoch University 2 years; 1989–1990Dr E Stoyanova WA Department of Education 7 years; 1995, 2000–2005Dr G Gamble University of Western Australia 10 years; 2006–2015

EditorProf P J O’Halloran University of Canberra, ACT 1 year; 1983Dr A W Plank University of Southern Queensland 11 years; 1984–1994Dr A Storozhev Australian Mathematics Trust, ACT 15 years; 1994–2008Editorial ConsultantDr O Yevdokimov University of Southern Queensland 7 years; 2009–2015

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Other Members of AMOC (showing organisations represented where applicable)Mr W J Atkins Australian Mathematics Foundation 18 years; 1995–2012Dr S Britton University of Sydney, NSW 8 years; 1990–1998Prof G Brown Australian Academy of Science, ACT 10 years; 1980, 1986–1994Dr R A Bryce Australian Mathematical Society, ACT 10 years; 1991–1998 Mathematics Challenge for Young Australians 13 years; 1999–2012Mr G Cristofani Department of Education and Training 2 years; 1993–1994Ms L Davis IBM Australia 4 years; 1991–1994Dr W Franzsen Australian Catholic University, ACT 9 years; 1990–1998Dr J Gani Australian Mathematical Society, ACT 1980Assoc Prof T Gagen ANU AAMT Summer School 6 years; 1993–1998Ms P Gould Department of Education and Training 2 years; 1995–1996Prof G M Kelly University of Sydney, NSW 6 years; 1982–1987Prof R B Mitchell University of Canberra, ACT 5 years; 1991–1995Ms Anna Nakos Mathematics Challenge for Young Australians 13 years; 2003–2015Mr S Neal Department of Education and Training 4 years; 1990–1993Prof M Newman Australian National University, ACT 15 years; 1986–1998 Mathematics Challenge for Young Australians 10 years; 1999–2002, (Treasurer during the interim) 2009–2014Prof R B Potts University of Adelaide, SA 1 year; 1980Mr H Reeves Australian Association of Maths Teachers 11 years; 1988–1998 Australian Mathematics Foundation 2014–2015Mr N Reid IBM Australia 3 years; 1988–1990Mr R Smith Telecom Australia 5 years; 1990–1994Prof P J Taylor Australian Mathematics Foundation 6 years; 1990–1994, 2013Prof N S Trudinger Australian Mathematical Society, ACT 3 years; 1986–1988Assoc Prof I F Vivian University of Canberra, ACT 1 year; 1990Dr M W White IBM Australia 9 years; 1980–1988

Associate Membership (inaugurated in 2000)Ms S Britton 16 years; 2000–2015Dr M Evans 16 years; 2000–2015Dr W Franzsen 16 years; 2000–2015Prof T Gagen 16 years; 2000–2015Mr H Reeves 16 years; 2000–2015Mr G Ball 16 years; 2000–2015

AMOC Senior Problems CommitteeCurrent membersDr N Do Monash University, VIC (Chair) 2 years; 2014–2015 (member) 11 years; 2003–2013M Clapper Australian Mathematics Trust 3 years; 2013-2015Dr A Di Pasquale University of Melbourne, VIC 15 years; 2001–2015Dr M Evans Australian Mathematical Sciences Institute, VIC 26 years; 1990–2015Dr I Guo University of Sydney, NSW 8 years; 2008–2015Dr J Kupka Monash University, VIC 13 years; 2003–2015Dr K McAvaney Deakin University, VIC 20 years; 1996–2015Dr D Mathews Monash University, VIC 15 years; 2001–2015Dr A Offer Queensland 4 years; 2012–2015Dr C Rao IBM Australia 16 years; 2000–2015Dr B B Saad Monash University, VIC 22 years; 1994–2015Dr J Simpson Curtin University, WA 17 years; 1999–2015Dr I Wanless Monash University, VIC 16 years; 2000–2015

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Previous membersMr G Ball University of Sydney, NSW 16 years; 1982–1997Mr M Brazil LaTrobe University, VIC 5 years; 1990–1994Dr M S Brooks University of Canberra, ACT 8 years; 1983–1990Dr G Carter Queensland University of Technology 10 years; 2001–2010Dr J Graham University of Sydney, NSW 1 year; 1992Dr M Herzberg Telecom Australia 1 year; 1990Assoc Prof D Hunt University of New South Wales 29 years; 1986–2014Dr L Kovacs Australian National University, ACT 5 years; 1981–1985Assoc Prof H Lausch Monash University, VIC (Chair) 27 years; 1987–2013 (member) 2 years; 2014-2015Dr D Paget University of Tasmania 7 years; 1989–1995Prof P Schultz University of Western Australia 8 years; 1993–2000 Dr L Stoyanov University of Western Australia 5 years; 2001–2005Dr E Strzelecki Monash University, VIC 5 years; 1986–1990Dr E Szekeres University of New South Wales 7 years; 1981–1987Prof G Szekeres University of New South Wales 7 years; 1981–1987Em Prof P J Taylor Australian Capital Territory 1 year; 2013Dr N H Williams University of Queensland 20 years; 1981–2000

Mathematics School of ExcellenceDr S Britton University of Sydney, NSW (Coordinator) 2 years; 1990–1991Mr L Doolan Melbourne Grammar, VIC (Coordinator) 6 years; 1992, 1993–1997Mr W Franzsen Australian Catholic University, ACT (Coordinator) 2 years; 1990–1991Dr D Paget University of Tasmania (Director) 5 years; 1990–1994Dr M Evans Scotch College, VIC 1 year; 1995Assoc Prof D Hunt University of New South Wales (Director) 4 years; 1996–1999Dr A Di Pasquale University of Melbourne, VIC (Director) 16 years; 2000–2015

International Mathematical Olympiad Selection SchoolMr J L Williams University of Sydney, NSW (Director) 2 years; 1982–1983Mr G Ball University of Sydney, NSW (Director) 6 years; 1984–1989Mr L Doolan Melbourne Grammar, VIC (Coordinator) 3 years; 1989–1991Dr S Britton University of Sydney, NSW (Coordinator) 7 years; 1992–1998Mr W Franzsen Australian Catholic University, ACT (Coordinator) 8 years; 1992–1996, 1999–2001Dr D Paget University of Tasmania (Director) 6 years; 1990–1995Assoc Prof D Hunt University of New South Wales (Director) 5 years; 1996–2000Dr A Di Pasquale University of Melbourne, VIC (Director) 15 years; 2001–2015