Mathematics Contests. The Australian Scene 2015

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MATHEMATICS CONTESTS THE AUSTRALIAN SCENE 2015 PART 1 AND 2 A Di Pasquale, N Do and KL McAvaney Le ar n f r om ye ste r d a y, l ive f o r to day, ho pe f o r to m o r r ow. T h e i m p or ta nt th i ng is no t to sto p que stio ning. Alber t Einstein

A u s t r a l i a n M a t h e ma t i c a l O l y m p i a d C omm i t t e e A

department of the

A u s t r a l i a n M at h e mat i c s T r u s t

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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

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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

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Acknowledgements 4 Preface 5 Background Notes on the IMO and AMOC

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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 8

held in Romania. The aims of the IMO include: (1) discovering, encouraging and challenging mathematically gifted school students (2) fostering friendly international relations between students and their teachers (3) sharing information on educational syllabi and practice throughout the world. It was not until the mid-sixties that countries from the western world competed at the IMO. The United States of America first entered in 1975. Australia has entered teams since 1981. Students must be under 20 years of age at the time of the IMO and have not enrolled at a tertiary institution. The Olympiad contest consists of two four-and-a-half hour papers, each with three questions. Australia has achieved varying successes as the following summary of results indicate. HM (Honorable Mention) is awarded for obtaining full marks in at least one question. The IMO will be held in Hong Kong in 2016.

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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

2

11 out of 38 teams

1986

Warsaw

5

15 out of 37 teams

1987

Havana

1988

Canberra

1989

Braunschweig

2

2

22 out of 50 teams

1990

Beijing

2

4

15 out of 54 teams

1991

Sigtuna

1992

Moscow

1

1993

Istanbul

1

1994

1

3 1

15 out of 42 teams 1

1

17 out of 49 teams

3

2

20 out of 56 teams

1

2

1

19 out of 56 teams

2

3

Hong Kong

2

3

3

12 out of 69 teams

1995

Toronto

1

4

1

21 out of 73 teams

1996

Mumbai

2

3

23 out of 75 teams

1997

Mar del Plata

3

1

9 out of 82 teams

1998

Taipei

4

2

13 out of 76 teams

1999

Bucharest

1

1

3

2000

Taejon

1

3

1

16 out of 82 teams

2001

Washington D.C.

1

4

25 out of 83 teams

2002

Glasgow

1

2003

Tokyo

2004

Athens

2005

Merida

2006

Ljubljana

3

2

1

26 out of 90 teams

2007

Hanoi

1

4

1

22 out of 93 teams

2008

Madrid

5

1

2009

Bremen

2

1

2

1

23 out of 104 teams

2010

Astana

1

3

1

1

15 out of 96 teams

2011

Amsterdam

3

3

25 out of 101 teams

2012

Mar del Plata

2

4

27 out of 100 teams

2013

Santa Marta

1

2

3

15 out of 97 teams

2014

Cape Town

1 Perfect Score by Alexander Gunning

3

2

11 out of 101 teams

2015

Chiang Mai

2

4

2

1

13 out of 73 teams

1

15 out of 81 teams

2

1

1

26 out of 84 teams

2

2

2

26 out of 82 teams

1

2

1

27 out of 85 teams

6

10

25 out of 91 teams

19 out of 97 teams

6 out of 104 teams

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

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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

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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

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MEMBERSHIP OF AMOC COMMITTEES Australian Mathematical Olympiad Committee 2015 Chair Prof C Praeger, University of Western AustraliaDeputy Chair Assoc Prof D Hunt, University of New South Wales

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

Treasurer Dr P Swedosh, The King David School, VIC

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

Chair, Challenge Dr K McAvaney, Deakin University, VIC

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

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

State Directors Dr K Dharmadasa, University of Tasmania Dr G Gamble, University of Western Australia Dr Ian Roberts, Northern Territory Dr W Palmer, University of Sydney, NSW Mr D Martin, South Australia Dr V Scharaschkin, University of Queensland Dr P Swedosh, The King David School, VIC Dr Chris Wetherell, Radford College, ACT

Representatives Ms A Nakos, Challenge Committee Prof M Newman, Challenge Committee Mr 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 Outstanding students are identified from AMC results, MCYA, other competitions and recommendations; and eligible students from previous training programs

August

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

July

Short mathematics school for IMO team members 2016 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 2015 Dr A Di Pasquale, University of Melbourne, VIC Dr N Do, Monash University, VIC, (Chair) Dr M Evans, Australian Mathematical Sciences Institute, VIC Dr I Guo, University of Sydney, NSW Assoc Prof D Hunt, University of NSW Dr J Kupka, Monash University, VIC Assoc Prof H Lausch, Monash University, VIC Dr K McAvaney, Deakin University, VIC Dr D Mathews, Monash University, VIC Dr A Offer, Queensland Dr C Rao, NEC Australia, VIC Dr B B Saad, Monash University, VIC Assoc Prof J Simpson, Curtin University of Technology, WA Emeritus Professor P J Taylor, Australian Capital Territory Dr 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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CHALLENGE PROBLEMS PRIMARY 2015 Challenge Problems–- MIDDLE Middle Primary Students may work on each of these four problems in groups of up to three, but must write their solutions individually.

MP1 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. For example, 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. She starts 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 it always works.

1

1

19

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

MP3 Egg Cartons Zoe decided to investigate the number of different ways of placing identical eggs in clear rectangular cartons of various sizes.

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

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

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. Dates such 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.

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CHALLENGE PROBLEMS PRIMARY 2015 Challenge Problems–- UPPER Upper Primary Students may work on each of these four problems in groups of up to three, but must write their solutions individually.

UP1 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 their ends touching to form polygons. For example, with four rods, two of length 1 cm and two of length 3 cm, I can form a parallelogram which I refer to as (1, 3, 1, 3).

1c m

1c m

3 cm

3 cm 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 have exactly 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 various sizes.

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

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

3

2

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 55 ways 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 for 50 metres are: Mack Jack Zac Nick

55 seconds 1 minute 8 seconds 1 minute 16 seconds 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 boys into 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 PB time was included with all theirs, then the average PB time would be exactly 1 minute. What was the coach’s PB time?

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 in Primelandia 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 the largest even payment that can be made using four coins and the largest even payment that can be made using three coins? d A bag contains six different coins. Alice, Bob and Carol take two coins each from the bag and keep them. They find that they have all taken the same amount of money. What is the smallest amount of money that could have been in the bag? Explain your reasoning.

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CHALLENGE PROBLEMS JUNIOR 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 on the 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 quadrilaterals on 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 could be 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 sectors numbered 1 to 9. Jim has nine tiles numbered 1 to 9. When the wheel stops, he removes all tiles whose numbers are factors or multiples of the spun number. • 3

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Jim then tries to use three of the remaining tiles to form a 3-digit number that is divisible by the number on the wheel. 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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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 in Primelandia 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. They find that they have all taken the same amount of money. What is the smallest amount of money that could have been 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 between their values. d The new King of Primelandia decides to mint coins of prime values greater than 50T. Show that no matter what coins are made, there is only one set of five Primelandian coins which, when placed in ascending order, form a sequence 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 a flat 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 should give you the general idea. Top of Mt KJ •

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Theo can run at 12 km/h on the flat. Up the hill he reckons he can average 9 km/h and he is good for 15 km/h on the downhill 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 will affect 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 is at his back, he will speed up by the same amount. This applies to his speeds on the flat and on the slope. The wind will 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 leg faster, 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. D E C

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CHALLENGE PROBLEMS 2015 Challenge Problems–- INTERMEDIATE 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 first equal to the digital sum of the number before it. For example: 7598, 29, 11, 2. Any digital sum sequence ends in a single-digit number and this is called the final digital sum or FDS of the first number in the sequence. Thus the FDS of 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 of flipping over some of the coins. The number of coins that are flipped stays the same for each game but may change from game to game. A coin showing ‘heads’ is represented by H. A coin showing ‘tails’ is represented by T . So, when a coin is flipped it changes from H to T or from T to H. The same coin may be selected on different moves. In each game 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: Move 1: Move 2: Move 3:

T T H H

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a Starting with 14 coins and flipping over four coins on each move, explain how to finish with exactly 10 coins heads up 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 in three moves. c Starting with 26 coins and flipping over four coins on each move, what is the minimum number of moves needed to finish 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 all 154 coins heads up at the end of three moves.

I4 Jogging See Junior Problem 5.

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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 corner is folded to a point P on the top edge as shown, creating triangles P QR, P T S and SU V . T

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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 SU V . c Find all positions of P so that the ratio of the sides of triangle SU V 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 small cubes. In the second and subsequent steps, the same procedure is applied to what remains of the original large cube after the previous 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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CHALLENGE SOLUTIONS MIDDLE PRIMARY 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 numbers from 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 must both 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 200 that 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 Squares

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The number of squares from one border to the next appears to increase by 8. We now show this rule continues to apply. 10

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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 Squares

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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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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 empty positions, 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 month must 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. If the 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. The earliest 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 last condate 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 the month must be 12. So the day must contain two of the digits 3, 4, 5, 6. This is impossible.

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CHALLENGE SOLUTIONS UPPER PRIMARY 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 different length. 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 parallel sides 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 in a 2 × 4 carton.

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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 fill the empty places with 9 brown eggs. Then each placement of 3 white eggs corresponds to a placement of 9 brown eggs. 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 Jack Zac Nick

55 s 1 min 8 s = 68 s 1 min 16 s = 76 s 1 min 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 that exclude Mack. Pair with M Total PB Other pair Total PB

MJ 123 ZN 139

MZ 131 JN 131

MN 118 JZ 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 total PB 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 68 with 63. Each pair then totals 131 as required. Thus Mack is paired with Zac and Jack is paired with Nick.

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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 = 385 seconds. Since 385 is odd and cannot be divided into two equal amounts, the two teams cannot have the same total 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 have the same total PB times. The next closest would be 1 second apart, in which case the times would be 192 s and 193 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 order of their units digits are: 70, 53, 63, 55, 76, 68. The only three that give 3 in the units digit of their total are 70, 55, 68, and these do in fact total 193 s. Thus one team has Chloe, Mack, and Jack with a total PB time of 70 + 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 corresponding teams that exclude Mack. Team with M MJZ MJN MJC MJS MZN MZC MZS MNC MNS MCS

Total PB 199 186 193 176 194 201 184 188 171 178

Other team NCS ZCS ZNS ZNC JCS JNS JNC JZS JZC JZN

Total PB 186 199 192 209 191 184 201 197 214 207

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 Part b, 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 − 37 = 13,

50 − 43 = 7, 50 − 31 = 19,

50 − 41 = 9, 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 3=3 4= 2+2 5=5 6= 3+3

7=7 8= 3+5 9= 2+7 10 = 3 + 7 11 = 11

12 = 5 + 7 13 = 13 14 = 3 + 11 15 = 2 + 13 16 = 3 + 13

17 = 17 18 = 5 + 13 19 = 19 20 = 3 + 17 21 = 2 + 19

22 = 3 + 19 23 = 23 24 = 5 + 19 25 = 2 + 23 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 even prime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallest payment 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 four coins is 47 + 43 + 41 + 37 = 168. If the sum of three primes is even, then one of those primes must be 2. So the largest 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 the sum 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 sum of 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 a multiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to 72T. Sum of six 60 66 72

Sum of pair 20 22 24

All possible pairs 3 + 17, 7 + 13 3 + 19, 5 + 17 5 + 19, 7 + 17, 11 + 13

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

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CHALLENGE SOLUTIONS JUNIOR 2015 Challenge Solutions -–Junior J1 Quirky Quadrilaterals a •

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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 tiles would 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 it cannot 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 if and only if it ends with a 2-digit number that is divisible by 4. The only such 2-digit multiples of 4 that can be made 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 last digit 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 multiple 994 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 8 or 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 but cannot be a winning number from spinning 3 since digit 9 would have been eliminated. So 987 is not a winning number. The next largest number is 986. This cannot be a winning number from spinning 9, 8, or 6. Since 986 is not divisible by 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 divisible by 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 a winning 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 3=3 4= 2+2 5=5 6= 3+3

7=7 8= 3+5 9= 2+7 10 = 3 + 7 11 = 11

12 = 5 + 7 13 = 13 14 = 3 + 11 15 = 2 + 13 16 = 3 + 13

17 = 17 18 = 5 + 13 19 = 19 20 = 3 + 17 21 = 2 + 19

22 = 3 + 19 23 = 23 24 = 5 + 19 25 = 2 + 23 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 even prime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallest payment 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 the sum 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 sum of 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 a multiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to 72T. Sum of six Sum of pair All possible pairs 60 20 3 + 17, 7 + 13 66 22 3 + 19, 5 + 17 72 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 sequence 5, 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 is the only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The difference between any two primes in the sequence is 6, 12, 18, or 24. So no two primes in the sequence can end in the same digit. There are five primes in the sequence so all of 1, 3, 5, 7, 9 must appear as last digits. The only prime whose last 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 is the only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The only prime 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 next four 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 first prime 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 at least 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 five numbers 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 the sequence 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 the sequence 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 that is 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 the original 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 no available 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 is 17/06/2345. 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 year cannot 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. So in 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, 7 without repeating a digit. There are 5 choices for D, then 4 choices for M, and so on. So the number of condates is 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 of condates is 2 × 3 × 2 × 1 = 12. Thus the number of condates in the years 2000 to 2999 is 120 + 12 = 132.

20

37

J5 Jogging 4 a On the flat, Theo will run a total of 4 km at 12 km/h and this will take 12 × 60 = 20 minutes. He will run 3 km 3 uphill at 9 km/h and this will take 9 × 60 = 20 minutes. He will run 3 km downhill at 15 km/h and this will take 3 × 60 = 12 minutes. So the total time taken for Theo to complete the course will be 20 + 20 + 12 = 52 minutes. 15

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

Flat east Uphill Downhill Flat west

2 km 3 km 3 km 2 km

13 km/h 10 km/h 14 km/h 11 km/h

2 So the total time to complete the course would be ( 13 +

3 10

+

3 14

+

Time 2 13 3 10 3 14 2 11

2 11 )

h h h h

× 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 Uphill Downhill Flat west

2 km 3 km 3 km 2 km

11 km/h 8 km/h 16 km/h 13 km/h

2 11 h 3 8 h 3 16 h 2 h 13

2 So the total time to complete the course would be ( 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. 3 3 + 14 hours. If the wind blows from the east, If the wind blows from the west, then his time on the hill will be 10 3 3 then his time on the hill will be 8 + 16 hours. To simplify comparison of these times we divide both by 3 and 1 1 12 multiply both by 2. So we want to compare 15 + 17 = 12 35 with 4 + 8 = 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 3 km uphill at 9 km/h in 20 minutes. So to gain 2 minutes uphill, he will need to run 3 km in 3 18 minutes. Therefore his speed needs to be 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 run 3 3 km in 10 minutes. Therefore his speed needs to be 10 × 60 = 18 km/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 minutes for the hill. 3 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 r3 + 15 = 3 1 1 3 So r = 2 − 5 = 10 . Hence r = 10 km/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 So 3r = 12 − 13 = 16 . Hence r = 18 km/h. This is 3 km/h faster.

21

3 9

+ 3r =

30 60

= 12 .

30 60

= 12 .

3

39

J6 Tessellating Hexagons a The flipped hexagons are shaded. D

E F

D

B

A

E F

C B

A F

A C D

C

E

B F E

A C D

E

B

B D C A

F

B

E

C

D C A

F

A

B

D

E

C

F E

D

B

A

F A

B

D

E

F E

D

B

A

F

C F

A C D

C

E

B F E

A C D

E

B

B D C A

F

B

E

C

D C A

F

A C

B

F E

A

B

D

F E

D

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 = F A, the sum of the three interior angles A, C, D is 360◦ , and the sum 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 is A − BF − E − D − CA − BF − E − D and the top boundary is D − E − F B − AC − D − E − F B − A. So these boundaries are identical. Hence the 16-block can be translated vertically to form an infinite column without gaps and without overlap. The right boundary of this column is DA − F − EB − C − DA − F − EB − C − DA repeated indefinitely. The left boundary of the column is C − 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

39

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 the tessellation 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, F A, has the smallest edge EF attached at one end but not the other. So the midpoints of edges AB, BC, DE, F A are not points of symmetry for the tessellation. To check the midpoint of edge CD, trace a copy of the block and place it exactly over the original block. Then place a pin through the midpoint of edge CD, rotate the traced copy of the block through 180◦ and notice that it again fits exactly over the original block (except for overhang). Thus the midpoint of CD is a point of symmetry for 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

4

41

d The block of hexagons in Part a is reproduced below with hexagons labelled H1 , H2 , H3 , H4 , H1 , H2 , H3 , H4 , and so on as shown. D

E F

D

F

C B

A

H3

A A C D

H4

C

B

H4

E

H3

A C D

B F E E

B

B D C A

H1

F

B

E

C A

H1 A

B

D

E

A

B

D

E

C

H2

F E

D

B

A

F

F E

D

B

A

F

H4

C

H3

A C D

H4

C

B F E

H3

A C D

E

B

H1

F

D C A

C A A C

H2 A

B

C

H2

D

H1

F E B

B

E F

C

H2

D

F

F E

B

F E

D

F E

D

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

24

41

CHALLENGE SOLUTIONS INTERMEDIATE 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 993 984 975 966 957 950 948 941 939 932 923 914

s 21 21 21 21 21 14 21 14 21 14 14 14

7 divides n? no no no yes no no no no no no no no

n 905 894 885 876 867 860 858 851 849 842 833

s 14 21 21 21 21 14 21 14 21 14 14

7 divides n? no no no no no no no no no no 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 and 833 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 994 987 980 973 966 959 952 945 938 931 924 917

s 22 24 17 19 21 23 16 18 20 13 15 17

7 divides s? no no no no yes no no no no no no no

n 910 903 896 889 882 875 868 861 854 847 840 833

s 10 12 23 25 18 20 22 15 17 19 12 14

7 divides s? no no no no no no no no no no no 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 and 833 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, in decreasing order, the first few 3-digit numbers n with digital sum 16 or 25.

25

4

43

s 16

n 970 961 952 943 934 925 916 907 880 871 862 853 844 835 826

7 divides n? no no yes no no no no no no no no no no no yes

s 25

n 997 988 979 898 889 799

7 divides n? no no no no yes no

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 994 987 980 973 966 959 952 945 938 931 924 917 910

DSS 22, 4 24, 6 17, 8 19, 10, 1 21, 3 23, 5 16, 7 18, 9 20, 2 13, 4 15, 6 17, 8 10, 1

n 903 896 889 882 875 868 861 854 847 840 833 826

DSS 12, 3 23, 5 25, 7 18, 9 20, 2 22, 4 15, 6 17, 8 19, 10, 1 12, 3 14, 5 16, 7

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

43

n 994 987 980 973 966 959 952 945 938 931 924 917 910 903

F 4 6 8 1 3 5 7 9 2 4 6 8 1 3

n 896 889 882 875 868 861 854 847 840 833 826 819 812 805

F 5 7 9 2 4 6 8 1 3 5 7 9 2 4

497 490 483 476 469 462 455 448 441 434 427 420 413 406

2 4 6 8 1 3 5 7 9 2 4 6 8 1

399 392 385 378 371 364 357 350 343 336 329 322 315 308 301

3 5 7 9 2 4 6 8 1 3 5 7 9 2 4

n 798 791 784 777 770 763 756 749 742 735 728 721 714 707 700 294 287 280 273 266 259 252 245 238 231 224 217 210 203

F 6 8 1 3 5 7 9 2 4 6 8 1 3 5 7 6 8 1 3 5 7 9 2 4 6 8 1 3 5

n 693 686 679 672 665 658 651 644 637 630 623 616 609 602

F 9 2 4 6 8 1 3 5 7 9 2 4 6 8

196 189 182 175 168 161 154 147 140 133 126 119 112 105

7 9 2 4 6 8 1 3 5 7 9 2 4 6

n 595 588 581 574 567 560 553 546 539 532 525 518 511 504

F 1 3 5 7 9 2 4 6 8 1 3 5 7 9

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 integers n = 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-digit multiples 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 + 63m that have digital sum 16 or 25. The following table lists in decreasing order all 3-digit integers with digital sum 7. n 700 610 601 520 511 502 430 421 412 403

7 divides n? yes no no no yes no no no no no

n 340 331 322 313 304 250 241 232 223 214

7 divides n? no no yes no no no no no no no

n 205 160 151 142 133 124 115 106

7 divides n? no no no no yes no no no

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

27

4

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 of 9. This is equivalent to saying they have the same remainder when divided by 9. We now show that this is always true. 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 sum sequence for n, including its FDS, have the same remainder when divided by 9. Hence, if n is a multiple of 9, then the 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 divides r. 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 multiples of 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 992 984 976 968 960 952 944 936 928 920 912 904

F 2 3 4 5 6 7 8 9 1 2 3 4

496 488 480 472 464 456 448 440 432 424 416 408 400

1 2 3 4 5 6 7 8 9 1 2 3 4

n 896 888 880 872 864 856 848 840 832 824 816 808 800 392 384 376 368 360 352 344 336 328 320 312 304

F 5 6 7 8 9 1 2 3 4 5 6 7 8 5 6 7 8 9 1 2 3 4 5 6 7

n 792 784 776 768 760 752 744 736 728 720 712 704

F 9 1 2 3 4 5 6 7 8 9 1 2

296 288 280 272 264 256 248 240 232 224 216 208 200

8 9 1 2 3 4 5 6 7 8 9 1 2

n 696 688 680 672 664 656 648 640 632 624 616 608 600 192 184 176 168 160 152 144 136 128 120 112 104

F 3 4 5 6 7 8 9 1 2 3 4 5 6 3 4 5 6 7 8 9 1 2 3 4 5

n 592 584 576 568 560 552 544 536 528 520 512 504

F 7 8 9 1 2 3 4 5 6 7 8 9

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 integers n = 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 the form 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

45

I3 Coin Flips a Here are three moves that leave exactly 10 coins heads up. Start: Move 1: Move 2: Move 3:

T T T T T T T T T T T HHHHT T T T T T T HHHHHHHHT T T HHHHHHHT HHH

T T T T

T T T T

T 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 s and want to finish with 154 Hs. We can achieve this by flipping 153 coins exactly once and flipping one coin three times over the three moves. Here are three moves that leave all coins heads up. 154

Start:

   T . . .T 52

102

      Move 1: H . . . H T . . . T 51

51

=

51

         H . . .H H T . . .T T . . .T 51

51

51

         Move 2: H . . . H T H . . . H T . . . T 51

51

51

         Move 3: H . . . H H H . . . H 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 takes at 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 in Part a to the 14 coins that are tails up, gives us exactly 12 + 10 = 22 coins heads up. One more move, flipping 4 Ts 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. 154

Start:

   T . . .T m

154−m

      Move 1: H . . . H T . . . T m−n

n

m−n

154−2m+n

            Move 2: H . . . H T . . . T H . . . H T . . . T

=

2m−2n

154−2m+2n

   H . . .H

   T . . .T

where 0 ≤ n ≤ m and m − n ≤ 154 − m

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

29

4

47

Alternative ii Three moves give the following results. Note that in a move, an even number of coins may mean 0 coins. 154

Start:

   T . . .T odd

odd

      Move 1: H . . . H T . . . T even

odd

even

odd

            Move 2: H . . . H T . . . T H . . . H T . . . T or odd

even

odd

even

            H . . . H T . . . T H . . . H T . . . T giving even even       H . . .H T . . .T even

even

odd

odd

            Move 3: H . . . H T . . . T H . . . H T . . . T or odd

odd

even

even

            H . . . H T . . . T H . . . H T . . . T giving odd

odd

      H . . .H 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 P R = 1 − x. T

1 2

1 2

P

Q

x

1−x

R

S

1−x

U V

In P QR Pythagoras’ theorem gives Hence 2x = 34 and x = 38 .

1 4

+ x2 = (1 − x)2 = x2 − 2x + 1.

b Alternative i From Part a, we have: 1 2

T

1 2

P

a

Q

b 5 8

3 8

a R

b S U

b a V

30

47

Since angles RQP , RP S, P T S, SU V are right angles, triangles RQP , P T S, V U S have the same angles as shown. Hence they are similar and we have: V U :U S:SV = P T :T S:SP = RQ:QP :P R = 3:4:5. Hence P S = 35 P T = Then SV = 54 U S =

5 1 5 3 × 2 = 6 . So U S = U P − SP 5 5 × 16 = 24 and V U = 34 U S = 34 4

5 1 6 = 6. 1 = 18 . 6

=1− ×

Alternative ii As in Alternative i, V U :U S:SV = P T :T S:SP = RQ:QP :P R = 3:4:5. So ST = 43 P T = From the side of the square, we have: 1 = U V + V S + ST = 53 V S + V S + 5 So V S = 13 × 58 = 24 . From V U S, we have: V U = 35 SV =

3 5

×

5 24

=

1 8

and U S = 45 SV =

4 5

×

2 3

5 24

4 3

×

1 2

= 23 .

= 85 V S + 23 . = 16 .

Alternative iii Draw RW perpendicular to T S and let T S = t and V U = r. From Part a, we have: T

1 2

1 2

P

3 8

Q

3 8

5 8

W

t−

R

3 8

S U

r V r

Now apply Pythagoras’ theorem to the following triangles. In RW S, RS 2 = RW 2 + W S 2 = 1 + (t − 38 )2 . In RP S, RS 2 = RP 2 + P S 2 = ( 58 )2 + P S 2 . In P T S, P S 2 = P T 2 + T S 2 = ( 12 )2 + t2 . Hence

1 + (t − 38 )2 9 1 + t2 + 64 − 34 t 3 t 4 t

( 58 )2 + ( 12 )2 + t2 25 + 14 + t2 64 9 1 − 14 + 64 − 25 64 3 16 3 1 − = − 4 64 4 4 = 4 1 2 3 × 2 = 3

= = = = =

1 2

5 So P S 2 = ( 12 )2 + ( 23 )2 = 14 + 49 = 25 36 and P S = 6 . 5 1 Hence U S = U P − SP = 1 − 6 = 6 . Since 1 = T S + SV + r = 23 + SV + r, SV = 31 − r.

In V U S, Pythagoras’ theorem gives V S 2 = V U 2 + U S 2 . Hence

1 9

Thus V U =

1 8

and SV =

1 3

−

1 8

=

( 31 − r)2 + r 2 − 23 r 2 3r r

= = = =

r 2 + ( 16 )2 1 r 2 + 36 1 1 9 − 36 = 3 1 2 × 12 =

3 36 1 8

=

1 12

5 24 .

31

4

49

c Since angles RQP , RP S, P T S, SU V are right angles, triangles RQP , P T S, V U S have the same angles as shown. Hence they are similar and we have RQ:QP :P R = P T :T S:SP = V U :U S:SV = 7:24:25 or 24:7:25. T

Q

P

a

b

a R

b S

b a

U

V

Alternative i If P Q/QR = 7/24, let P Q = 7k. Then QR = 24k and P R = 25k. Since P R = 1 − QR, we have 25k = 1 − 24k. So 1 49k = 1 and P Q = 7 × 49 = 17 . If P Q/QR = 24/7, let P Q = 24k. Then QR = 7k and P R = 25k. Since P R = 1 − QR, we have 25k = 1 − 7k. So 1 32k = 1 and P Q = 24 × 32 = 34 . Alternative ii If P Q/QR = 7/24, then P R/QR = 25/24. Since P R = 1 − QR, we have 24 − 24QR = 25QR, QR = 24/49, and P R = 25/49. From Pythagoras, P Q2 = P R2 − QR2 = 1 − 2QR = 1 − 48/49 = 1/49. So P Q = 17 . If P Q/QR = 24/7, then P R/QR = 25/7. Since P R = 1 − QR, we have 7 − 7QR = 25QR, QR = 7/32, and P R = 25/32. From Pythagoras, P Q2 = P R2 − QR2 = 1 − 2QR = 1 − 7/16 = 9/16. So P Q = 34 . Alternative iii Let T P = x. If V U/U S = 7/24, let V U = 7k. Then U S = 24k and V S = 25k and we have: T

P

x

1 − 32k

Q

1 − 24k R

S

24k

25k

U

7k

V

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

49

Alternatively, in P T S, Pythagoras’ theorem gives (1 − 24k)2 (24k)2 − 48k 16k 1 x

= = = = =

(1 − 32k)2 + (56k)2 (32k)2 − 64k + (56k)2 (562 + 322 − 242 )k 2 (142 + 82 − 62 )k = 224k 56/224 = 8/32 = 1/4

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

P

x

1 − 49k

Q

1 − 7k R

S

7k

25k

U

24k

V

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 P T S, Pythagoras’ theorem gives (1 − 7k)2 (7k)2 − 14k 84k 12 3 x

= = = = = =

(1 − 49k)2 + (56k)2 (49k)2 − 98k + (56k)2 (562 + 492 − 72 )k 2 (82 + 72 − 12 )7k = (112)7k (28)7k 6/7

I6 Crumbling Cubes a The small cubes have at most three faces exposed. The only small cubes that have exactly three faces exposed are the 8 on the corners of the 10 × 10 × 10 cube. The only small cubes that have exactly two faces exposed are the 8 on each edge of the 10 × 10 × 10 cube that are not on its corners. All other cubes have less than two faces exposed. There are 12 edges on the 10 × 10 × 10 cube, so the number of small cubes removed at the first step is 8 + (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 edges but not from the corners. So the number of non-corner small cubes on each edge is 192/12 = 16. Hence each edge in 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

5

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 1 increases the number of lost cubes by 12, one for each edge. Since 200 − 104 = 96 and 96/12 = 8, the cube that loses 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. The exposed surface area of this layer is 7 × 7 small faces plus a ring of 4 × 7 small faces. So the surface area of the remaining 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 small faces. Then remove the small cubes either side. This does not change the surface area. Continue until only the two corner cubes remain. At this stage the surface area has increased by 2 small faces. Repeating this process on all 12 edges increases the surface area by 12 × 2 small faces. At this stage all 8 corner cubes have 6 exposed faces. So removing the 8 corner cubes reduces the surface area by 8 × 6 small faces. The surface area of the original 9 × 9 × 9 cube 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 cubes placed centrally on each face. At this stage the number of small cubes removed is 8 + (12 × 7) = 92.

7×7

7×

7×

7

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 × 7 cube with a 5 × 5 single layer of small cubes placed centrally on each face. So in this step, the number of small cubes removed is 6(4 + 4 × 5) = 144.

5×5

5×

5×

5

5

At the third step the only small cubes removed are the edge cubes in the 5 × 5 single layers and the edge cubes in the 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

51

CHALLENGE STATISTICS – MIDDLE PRIMARY MEAN SCORE/SCHOOL YEAR/PROBLEM Mean Year

Number of Students

Overall

Problem 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.

SCORE DISTRIBUTION %/PROBLEM Challenge Problem Score

1 e-Numbers

2

3

4

Quazy Quilts

Egg Cartons

Condates

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

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.

5

53

CHALLENGE STATISTICS – UPPER PRIMARY MEAN SCORE/SCHOOL YEAR/PROBLEM Mean YEAR

Number of Students

Overall

Problem 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.

SCORE DISTRIBUTION %/PROBLEM Challenge Problem 3

4

Egg Cartons

Seahorse Swimmers

Primelandia Money

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

Score

1

2

Rod Shapes Did not attempt

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

CHALLENGE STATISTICS – JUNIOR MEAN SCORE/SCHOOL YEAR/PROBLEM Mean Year

Number of Students

Problem

Overall

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.

SCORE DISTRIBUTION %/PROBLEM Challenge Problem 1

2

3

Quirky Quadrilaterals

Indim Integers

Did not attempt

3%

0

Score

6

Primelandia Money

4

5

Condates

Jogging

Tessellating Hexagons

3%

7%

8%

9%

23%

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

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.

5

55

CHALLENGE STATISTICS – INTERMEDIATE MEAN SCORE/SCHOOL YEAR/PROBLEM Mean Year

Number of Students

Problem

Overall

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.

SCORE DISTRIBUTION %/PROBLEM Challenge Problem 5

6

Jogging

Folding Fractions

Crumbling Cubes

5%

3%

14%

9%

6%

4%

4%

14%

11%

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

Score

1 Indim Integers

2

3

4

Digital Sums

Coin Flips

Did not attempt

1%

3%

0

2%

1

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

2015 Australian INTERMEDIATE Intermediate Mathematics Olympiad OLYMPIAD - Questions AUSTRALIAN MATHEMATICS 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 of a + 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 the perimeter 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 1 thingy cost a total of $441. What is the total cost, in dollars, of 1 whatsit, 1 doover and 1 thingy? [3 marks] a 2 1 , can also be written in the form + 2 , where n is a positive integer. b n n If a + b = 1024, what is the value of a? [3 marks]

4. A fraction, expressed in its lowest terms

5. Determine the smallest positive integer y for which there is a positive integer x satisfying the equation 213 + 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 circles lie 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 x = . 44 45 (Note: if r is any real number, then r denotes the largest integer less than or equal to r.)

1

[4 marks]

5

57

9. A sequence is formed by the following rules: s1 = a, s2 = b and sn+2 = sn+1 + (−1)n sn 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 foot of 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. B

W Z

4

2 X 1

A

Y

C

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

[2 bonus marks]

57

2015 AustralianINTERMEDIATE Intermediate Mathematics OlympiadOLYMPIAD - Solutions AUSTRALIAN MATHEMATICS 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 , F A touch the circle at U , V , W , X, Y , Z respectively. E X D Y

W

F C Z V A

U

B

Since the two tangents from a point to a circle have equal length, U B = BV , V C = CW , W D = DX, XE = EY , Y F = F Z, ZA = AU . The perimeter of hexagon ABCDEF is AU + U B + BV + V C + CW + W D + DX + XE + EY + Y F + F Z + ZA = AU + U B + U B + CW + CW + W D + W D + EY + EY + Y F + Y F + AU = 2(AU + U B + CW + W D + 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 4w + 10d + t = 441 w+d+t = x

(1) (2) (3)

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

3

5

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)/2 3 × (2) − 4 × (1): d = (7 + t)/2

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

2 1 2n + 1 + = . n n2 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.

4. We have

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

5. Method 1

213 + 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 this diameter, the two smaller circles are congruent. Let X be the centre of the large circle and Z the centre of a small circle.

Y X Z

Y

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

59

By Pythagoras, Y Z 2 = Y X 2 + XZ 2 . So (R + r)2 = R2 + (2R − r)2 . Then R2 + 2Rr + r 2 = 5R2 − 4Rr + r 2 , which simplifies to 3r = 2R. √ 2 So the large circle has area π(30/ √ π ) = 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 U underneath A. So the number of paths from O to A is the sum of the number of paths from O to L and the paths from O to U . •

L•

•A • 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 row upwards as indicated. •

429

132 429

•

42

132 297

14

42

90

165

5

14

28

48

75

2

5

9

14

20

27

1

2

3

4

5

6

7

1

1

1

1

1

1

1

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

6

61

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

n

i• 1• 0

•

Consider all the paths that touch the diagonal at vertex i but not at any of the vertices between vertex 0 and vertex 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 the diagram 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 are p(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, and p(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     x x Let = = n. 44 45 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 − k values. Thus, increasing n by 1 decreases the number of values of x by 1. Also the largest value of n is 43, in which case x has only 1 value. Therefore the number of non-negative integer values of x is 44 + 43 + · · · + 1 = 21 (44 × 45) = 990.

Method 2 

   x x Let n be a non-negative integer such that = = n. 44 45     x x = n ⇐⇒ 44n ≤ x < 44(n + 1) and = n ⇐⇒ 45n ≤ x < 45(n + 1). Then 44 45     x x = = n ⇐⇒ 45n ≤ x < 44(n + 1) ⇐⇒ 44n + n ≤ x < 44n + 44. So 44 45 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 = 21 (44 × 45) = 990.

6

61

Method 3    x x = = n. 44 45 Then x = 44n + r where 0 ≤ r ≤ 43 and x = 45n + s where 0 ≤ s ≤ 44. Let n be a non-negative integer such that



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 = 21 (44 × 45) = 990.

9. Working out the first few terms gives us an idea of how the given sequence develops: n 1 2 3 4 5 6 7

s2n−1 a b−a b 2b − a 3b − a 5b − 2a 8b − 3a

s2n b 2b − a 3b − a 5b − 2a 8b − 3a 13b − 5a 21b − 8a

It appears that the coefficients in the even terms form a Fibonacci sequence and, from the 5th term, every odd term 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+3 s2m+4

= = =

s2m+1 + s2m s2m+2 − s2m+1 s2m+3 + s2m+2

= =

s2m 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 = 3 and b < 1000, none of the first five terms of the given sequence equal 2015. So we are looking for integer solutions of 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 less than 1:2 and any point between AP and AB has the distance ratio greater than 1:2. B

2

4

P Q 1

1 A

C

7

6

63

Let Q be a point outside ABC such that its distances from AC and BC are in the ratio 1:4. By an argument similar 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 and CQ 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 the height of ABC. Letting | | denote area, we have |ABC| = 21 h × AB and |ABC| = |AXB| + |BXC| + |CXA| = 12 (2AB + 4BC + AC) = 12 AB × 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 a trapezium of 3 triangles in the next layer, a trapezium of 5 triangles in the next layer, and so on. The boundary of the combined figure is ABC and X is one of the grid vertices as shown. B

W 4

Z

2 X 1 Y

A

C

There are 49 small triangles in ABC and 6.5 small triangles in AZXY . Hence, after rescaling so that the area of AZXY 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 BW XZ are similar. Let XY be 1 unit and AY be x. Then BZ = 2x. B

2x W Z

4

2

X 1

A

x

Y

C

√ √ √ 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, XC 2 = 1 +√(x + x2 − 3)2 = 2x2 − 2 + 2x x2 − 3 and in XW C, W C 2 = 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

63

√ √ √ Squaring gives 4x2 + x2 − 3 − 4x x2 − 3 = 2x2 − 18 + 2x x2 − 3, which simplifies to 3x2 + 15 = 6x x2 − 3.

5 Squaring again gives 9x4 + 90x2 + 225 = 36x4 − 108x2 . So 0 = 3x4 − 22x2 − 25 = (3x2 − 25)(x2 + 1), giving x = √ . 3 x √ 2 4 13 5 Hence, area AZXY = + x − 3 = √ + √ = √ and 2 3 2 3 √ 2 3 √ √ 4 2 49 3 3 10 2 2 √ +√ (2x + x − 3) = =√ . area ABC = 4 4 3 3 3 13 49 Since the area of AZXY is 13 cm2 , the area of ABC is ( √ / √ ) × 13 = 98 cm2 . 3 2 3 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 F H 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, XF G, XHI, BHJ are equilateral, and triangles XDE and BHJ are congruent. B

J

H

E Z

W 4

2

X I

D 1 A

F Y G

C

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

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

|ABC| = |AEG| + |F HC| + |DBI| − |F XG| − |DEX| − |XHI| = 9 + 25 + 36 − 1 − 4 − 16 = 49. |AZXY | = |AEG| − 12 (|F XG| + |DEX|) = 9 − 12 (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. B

W Z 60◦ A

c b 120

X ◦

a

Y

C

9

6

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 |

= = = = = =

Similarly, |CY XW | = Hence |ABC| =

√

3 (2a2 6

√ 3 2 6 (a

|Y AZ| + |Y XZ| 1 1 ◦ ◦ 2 (AY )(AZ) sin 60 + 2 ab sin 120 √ 3 4 ((AY )(AZ) + ab) √ 3 12 ((a + 2b)(b + 2a) + √ 3 2 2 12 (2a + 2b + 8ab) √ 3 2 (a + b2 + 4ab) 6

+ c2 + 4ac) and |BW XZ| =

+ 2b2 + 2c2 + 4ab + 4ac + 4bc) =

√ 3 2 6 (b

√ 3 (a 3

3ab)

+ c2 + 4bc).

+ b + c)2 .

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

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

Investigation Method 4 gives |ABC|/|AZXY | = 2(a + b + c)2 /(a2 + b2 + 4ab). Alternatively, as in Method 3,

|ABC| = |AEG| + |F HC| + |DBI| − |F XG| − |DEX| − |XHI| = (a + b)2 + (a + c)2 + (b + c)2 − a2 − b2 − c2 = (a + b + c)2 .

Also |AZXY | = |AEG| − 12 (|F XG| + |DEX|) = (a + b)2 − 12 (a2 + b2 ) = 2ab + 12 (a2 + b2 ).

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

10

65

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD STATISTICS DISTRIBUTION OF AWARDS/SCHOOL YEAR Number of Awards

Year

Number of Students

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 Year

Number 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 Mean Score Year

Number of Students

Question

Overall Mean

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

6

67

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD RESULTS Name

School

Year

Score

Prize Matthew Cheah

Penleigh and Essendon Grammar School, VIC

10

35

Puhua Cheng

Raffles Institution, Singapore

8

35

Ariel 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

34

Edwin 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

32

Hong Rui Benjamin Lee

Hwa Chong Institution, Singapore

10

32

67

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

6

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

29

Stanve 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

69

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

27

Sabrina 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

7

THE 2015SENIOR AMOC SENIOR CONTEST AMOC 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 am

for 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 AF C. 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 2a 3a pa + + + ··· + = 100. p p p p

c 2015 Australian Mathematics Trust 

71

THE 2015 AMOC SENIOR CONTEST Solutions and cumulative marking scheme c 2015 Australian  Mathematics Trust AMOC SENIOR

CONTEST SOLUTIONS

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 68 4 2 = 34. (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

72

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

1a1 + 2a2 + 3a3 + · · · + mam am

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

1

Subtracting the first equation from the second yields am+1 am+2 − am am+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

= k 2 − 1.

4

Using a1 = 1 and a2 = 1, we obtain the formula   m2 , if m is even, 4 am = 2 +3 m  4 , if m is odd. If m is odd, then

am+1 − am =

2m − 2 (m + 1)2 m2 + 3 − = , 4 4 4

am+1 − am =

2m + 4 (m + 1)2 + 3 m2 − = . 4 4 4

5

and if m is even, then 6

In particular, it follows that a2 < a3 < a4 < · · · . Since a2000 = 1 000 000, the largest integer n such that an < 1 000 000 is 1999. 7 2

73

Solution 2 (Angelo Di Pasquale) We will prove by induction that a2k−1 = k 2 − k + 1 and a2k = k 2 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 

iai =

i=1

= =

n  i=1 n  i=1 n  i=1

=

n 

(2i − 1) a2i−1 +

2i a2i

i=1

(2i − 1) (i2 − i + 1) +

n 

(2i) (i2 )

(by the inductive assumption)

i=1

4i3 − 3i2 + 3i − 1 3i3 + (i − 1)3

i=1 n 

=3 =

n 

3

i +

i=1 3n2 (n

2

+ 4

n−1 

3

i3

i=1 1)2

+

(n − 1)2 n2 4

2

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

n   n2 (n + 1)2  since i3 = 4 i=1

(by the inductive assumption)

It follows that a2n+1 =

1a1 + 2a2 + 3a3 + · · · + 2na2n = n2 + n + 1. a2n

4

Now consider the following sequence of equalities, which uses the facts derived above that  2 2 2 state that 2n i=1 iai = n (n + n + 1) and a2n+1 = n + n + 1. 2n+1 

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

i=1

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

It follows that a2n+2 =

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

5

So we have shown that the two formulas a2k−1 = k 2 − k + 1 and a2k = k 2 hold for k = 1, 2, . . . , n + 1. This completes the induction and the rest of the proof follows Solution 1. 7

3 74

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 4 solved problems 2, 4, 5.

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

Student 5 solved problems 3 and 4.

Student 3 solved problems 1, 4, 5.

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.

4

75

7

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 AF C.

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

E D

F

C

H Since F D is the altitude of the right-angled triangle AF H, we have AF H ∼ 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 AH = AD AF

AE AH = AD AC

AB AC = . AE AB

3

Putting these together, we have AF 2 = AD · AH = AC · AE = AB 2 .

6

So triangle BAF is isosceles and we have ∠AF B = ∠ABF = 90◦ − ∠CBF = ∠CF B, where the last equality uses the angle sum in triangle BCF . Since ∠AF B = ∠CF B, we have proven that BF bisects the angle AF C. 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 AB 2 = AE · AC. 5

76

1

Next, by the alternate segment theorem and the fact that AB  CD, we have ∠EF A = ∠EAB = ∠ECF . Hence, by the alternate segment theorem again, the circumcircle of triangle EF C is tangent to the line AF at F . Considering the power of the point A with respect to the circumcircle of triangle EF C, we have AF 2 = AE · AC.

4

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

6

Hence, ∠AF B = ∠ABF = ∠CF B, 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 AEF H must map to a line parallel to AB through F . Thus, f (circle AEF H) = 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 ∠BF C = ∠ABF = ∠AF B.

7

Solution 4 (Chaitanya Rao) Consider the following chain of equalities. DF 2 = AD · DH = AD · (AH − AD)   AB 2 − AD = AD · BC = AB 2 − AD2

(ADF ∼ F DH)

1

(ABC ∼ HAB)

3

(AD = BC)

4

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

6

So triangle BAF is isosceles and we have ∠AF B = ∠ABF = 90◦ − ∠CBF = ∠CF B, where the last equality uses the angle sum in triangle BCF . Since ∠AF B = ∠CF B, we have proven that BF bisects the angle AF C. 7

6

77

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         2a 3a pa a + + + ··· + = 100. p p p p 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) = 100 2

⇒

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 2a 3a pa a 2a 3a pa + + + ··· + − + + + ··· + = 100. p p p p p p p p Summing the terms of the arithmetic progression on the left-hand side yields         a 2a 3a pa a(p + 1) − + + + ··· + = 100. 3 2 p p p p      3a   pa  We will prove that the sequence of numbers ap , 2a is a rearrangep , p ,..., p p−1 0 1 2 ment of the sequence of numbers p , p , p , . . . , p . 4  ka  p−1 0 1 2 Observe that if k is a positive integer, then p is one of the numbers p , p , p , . . . , p .      3a    So it suffices to show that no two of the numbers ap , 2a , p , . . . , pa are equal. p p  ia   ja  Suppose for the sake of contradiction that p = p , where 1 ≤ i < j ≤ p. Then

a(j−i) − ia must be an integer. It follows that either a is divisible by p or j − i is p = p 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      3a   pa  sequence of numbers ap , 2a is a rearrangement of the sequence p , p ,..., p p−1 0 1 2 of numbers p , p , p , . . . , p . 5 ja p

We may now write the equation as a(p + 1) p − 1 − = 100 2 2

⇒

(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

78

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 as ar as ar as ar as −1+ −1< + < + ⇒ a−2< + < a. p p p p p p p p     as But since ar is an integer, we conclude that p + p 

   as ar + = a − 1. p p

The numbers {1, 2, . . . , p − 1} can be partitioned into

p−1 2

3 pairs whose sum is p.

4

Using the equation above for each such pair and substituting into the original equation, we obtain   p−1 (a − 1) + a = 100 ⇒ (a − 1)(p + 1) = 198. 6 2 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

79

AMOC SENIOR CONTEST RESULTS 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

80

AMOC SENIOR CONTEST STATISTICS SCORE DISTRIBUTION/PROBLEM Number of Students/Score

Problem Number

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

81

Mean

SCHOOL OF EXCELLENCE 2014 AMOCAMOC 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 AIMO and the AMOC Senior Contest. This year AMOC went ahead with a new initiative in an attempt to widen the net of identification. 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 top students from among schools that may not normally enter students in the AIMO. This turned out to be quite successful and prompted us to increase the number of invitations we 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 junior students, 16 of whom were attending for the first time. There were 12 students making up the senior group. The program covered the four major areas of number theory, geometry, combinatorics and algebra. Each day would start at 8am with lectures or an exam and go until 12 noon or 1pm. After a one-hour lunch break they would have a lecture at 2pm. At 4pm participants would usually have free time, followed by dinner at 6pm. Finally, each evening would round out with a problem session, topic review, or exam review from 7pm until 9pm. Another new initiative we tried was to invite two of our more experienced senior students to give a lecture. The rationale behind this is that teaching a subject is highly beneficial to the teacher because it can really solidify the foundations of the teacher’s own understanding. 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 was assigned the senior transformation geometry lecture. I did some trial runs with them prior to the lectures so as to trouble shoot any problems, as well as to give them some practice for the real thing. Overall it was successful, and I will likely try it again in the future. Many thanks to Andrew Elvey Price, Ivan Guo, Victor Khou, and Sampson Wong, who served 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 assisted in lecturing and marking. Angelo Di Pasquale Director of Training, AMOC

1 82

PARTICIPANTS AT THE 2014 AMOC SCHOOL OF EXCELLENCE 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

* Equivalent to year 10 in Australia.

83

THE 2015 AUSTRALIAN MATHEMATICAL OLYMPIAD AUSTRALIAN MATHEMATICAL OLYMPIAD 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 am am+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 

84

OLYMPIAD 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 BXY D 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 

85

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 = 2 because 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 2 = 2(k + 3) − ((k − 1) + 3) + 2 (inductive assumption) = k 2 + 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 am am+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 am am+1 is a term of the sequence.

17

86



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 we go 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 on the LHS, and we have am − a1 = 3 + 5 + · · · + 2m − 1. Using the standard formula11 for summing an arithmetic progression, we find (2m + 2)(m − 1) 2 2 = m − 1.

am − a1 =

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

It is now a simple matter to calculate

am am+1 = (m2 + 3)((m + 1)2 + 3) = m4 + 2m3 + 7m2 + 6m + 12 = (m2 + m + 3)2 + 3. Hence am am+1 = an , where n = m2 + m + 3. Therefore, am am+1 is a term of the sequence. 

1 1

The standard formula for the sum a + (a + d) + (a + 2d) + · · · + (a + kd) is a = 3, d = 2 and k = m − 2.

18 87

(2a+kd)(k+1) . 2

In our case

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 in terms 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 = am am+1 . Hence am am+1 = ak , where k = m + am . Therefore, am am+1 is a term of the sequence.  Comment This proof shows that the conclusion of the problem remains true for any 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 88

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 different values 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 easily follows 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 the right-hand end of n are 9s, but the (k + 1)th digit from the right-hand end of n is not a 9. In going from n to n + 1, the rightmost k digits all change from 9 to 0, and the (k + 1)th digit from the right-hand end of n increases by 1. Hence s(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 required number of different values of f in the range from 1 up to 9999 is simply equal to the number of multiples of 10 from 10 up to 10000. The number of these is 10000 ÷ 10 = 1000. 

20 89

Solution 2 (Seyoon Ragavan, year 11, Knox Grammar School, NSW) Suppose that the decimal representation of n is ak ak−1 . . . a1 a0 where ak = 0. Then we have the following.

⇒

n = 10k ak + 10k−1 ak−1 + · · · + 101 a1 + 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 if n = 10000 then n − s(n) = 9999. So it only remains to deal with positive integers that are less than 9999. If k ≤ 3, then n − s(n) = 999a3 + 99a2 + 9a1 . It follows that the nifty numbers less than 999 are precisely those numbers of the form 999a3 + 99a2 + 9a1 , where a1 , 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 contradiction is 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 nifty number. However, we exclude the combination a1 = a2 = a3 = 0 because although 0 is nifty, it is not positive. In summary, there are 999 nifty numbers from the case k ≤ 3, and one nifty number from the case k ≥ 4. This yields a total of 1000 nifty numbers. 

21 90

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 a friendless subset T as follows. Draw a 9 × 9 square grid. If the number 10a + b is in T 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 the first diagram below. No two numbers in T are friends because their smaller digits are both odd and hence cannot differ by 1. Since there are 45 shaded squares, we have shown that |T | = 45 is possible. 1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 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 the second diagram above. Observe that the numbers in each pair are friends. Hence T can have at most one number from each pair. Since there are 9 unpaired numbers in the top row and 36 pairs, this shows that T has at most 45 numbers. 

22 91

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 S that 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 integers we can choose from the ith line without choosing neighbours is i. This shows that T 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 without choosing neighbours. There is only one way to do this, namely take every second number starting from the left of each line. It is still necessary to verify that T contains no friends because even some nonadjacent 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 integers in T , the difference between their smaller digits is even, and thus cannot be equal to 1. Hence T contains no friends.  Comment This proof shows that there is exactly one subset of S of maximal size that contains no friends.

23 92

4. Comment All solutions that were dependent on how the diagram was drawn received a penalty deduction of 1 point. The easiest way to avoid diagram dependence was 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 parallel to n.22 Solution 1 (Yang Song, year 12, James Ruse Agricultural High School, NSW) Let Q be the intersection of lines OC and P D. Since AC  P Q and A is the midpoint of OP , it follows that C is the midpoint of OQ. P

D A

O B

C

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 a circle, 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 of Ω. 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 a diameter of Γ. 

22 For

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

24 93

Solution 2 (Ilia Kucherov, year 11, Westall Secondary College, VIC) Let E be the intersection of lines AC and OD. Since AC  P D and A is the midpoint of OP , it follows that E is the midpoint of OD. P

D

A

E

O B

C Q

(a) As in solution 1, we use directed angles. Since OE  BA, we have ∠(OE, EC) = ∠(BA, AC), which is fixed because A lies on Γ. This implies that E lies on a fixed circle, ω say, through O and C. Since E is the midpoint of OD, the point D is the image of E under the dilation centred at O and of enlargement factor 2. The image of ω under the dilation 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 2s . 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, as required.  (b) From the preceding analysis we have s = r if and only if OC is a diameter of ω. 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 a diameter of Γ. 

25 94

Solution 3 (Kevin Xian, year 11, James Ruse Agricultural High School, NSW) The motivation for this solution comes from considering a couple of special positions for A. If A is diametrically opposite C, then D = O. If A is diametrically opposite B, then this gives a second position for D. With this in mind, let X be the point on Γ that is diametrically opposite B, and let Y be the point such that X is the midpoint of OY . P

D

A

Y X O B

C

(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 symmetric in the line OD. It follows that ∠(Y D, OD) = ∠(OD, P D) = ∠(OD, OP ) + ∠(OP, P D) = ∠(BA, OA) + ∠(OA, AC) (OD  BA and P D  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, Ω say, that passes through O and Y . Let s be the radius of Ω. Then by the extended sine rule applied to ODY

26 95

we have OY sin ∠(Y D, OD) 2r = sin ∠(BA, AC) ≥ 2r.

2s =



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 diameter of Γ. 

27 96

5. This was the easiest problem of the competition. Of the 106 students who sat the AMO, 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 bisector at 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. C

X Y

x A

x D

Z

B

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 BXY D is a cyclic quadrilateral.

28 97

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 ∠AY D = ∠ACD = 90◦ − ∠DAC = 90◦ − ∠BAC = ∠CBA.

(ADY C cyclic) (angle sum CAD) (angle sum ABC)

Hence ∠AY D = ∠XBD, and so BXY D is cyclic. C

X Y

A

D

Z

B



29 98

Solution 3 (Wen Zhang, year 9, St Joseph’s College, QLD) We are given ACZ is isosceles with AC = AZ. Hence by symmetry, the angle bisector at A is both the median and the altitude from A. Thus Y is the midpoint of CZ. Since CDZ is right-angled at D, the point Y is the circumcentre of CDZ. Hence we have Y C = Y D = Y Z. Let ∠CBA = ∠XBD = 2x. We now compute some other angles in the diagram. ⇒ ⇒ ⇒ ⇒ ⇒

∠BAC = 90◦ − 2x ∠ZAY = 45◦ − x ∠Y ZA = 45◦ + x ∠ZDY = 45◦ + x ∠DY Z = 90◦ − 2x ∠DY X = 180◦ − 2x

(angle sum ABC) (AY bisects ∠BAC) (angle sum AZY ) (Y D = Y Z) (angle sum Y DZ) (AX ⊥ CZ)

Hence ∠XBD + ∠DY X = 180◦ , and so BXY D is cyclic. C

X Y

2x A

D

Z

B



30 99

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 that H is the orthocentre of CAZ. Hence ZH ⊥ AC. But since AC ⊥ BC, we have ZH  BC. Observe also that HDZY is cyclic because ∠ZDH = 90◦ = ∠HY Z.

We now have

∠CXH = ∠ZHX = ∠ZHY = ∠ZDY.

(BC  ZH) (HDZY cyclic)

Hence ∠CXY = ∠BDY , and so BXY D is cyclic. C

X Y H

A

D

Z

B



31 100

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 CAY and 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 AC 2 = 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 AC 2 = AD · AB.

(2)

Combining (1) and (2), it follows that AY · AX = AD · AB, and so BXY D is cyclic. C

X Y

A

D

Z

B



32 101

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 that f (x) is a polynomial of degree 1008 because p(x) has degree 1008 while q(x) has degree 1007. Observe that p(x) is the product of an even number of brackets and q(x) is the product 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 to 3 > 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 number of negative brackets in the product for q(x) by 1. This has the effect of changing the sign of f (x). In general, each time we increase x by 2 up to a maximum of x = 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 intermediate 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 solutions to f (x) = 0. Finally, since f (x) is a polynomial of degree 1008, the equation f (x) = 0 has at most 1008 roots. Hence we may conclude that f (x) = 0 has exactly 1008 distinct real solutions.  33 102

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 deduce f (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 terms to the right of ∗ are negative. Hence the total number of bracketed terms that are negative is equal to 2014−(x+1) + 1 = 2015−x .33 Taking into account the minus sign at 2 2 the front, it follows that for each x ∈ {1, 3, 5, . . . , 2015}, we have f (x) > 0 for x ≡ 1 (mod 4)

and

Hence we have the following inequalities.

f (x) < 0 for x ≡ 3 (mod 4).

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 intermediate 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 different 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 has at most 1008 roots. Thus we may conclude that f (x) = 0 has exactly 1008 distinct real 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

y = p(x) y = q(x) ... 1

3 3

2

3

4

5

6

7

2013 2014 2015 2016

x

This 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 corresponds to putting the ∗ to the right of the last term (x − 2014).

34 103

7. There are many different ways to solve this problem. Over 10 different methods of solution 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 adjacent numbers cannot be divisible by the same prime. Hence there is a positive integer N such 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 that the graph of p(n) versus n alternates between peaks and valleys once n > N . This can be schematically visualised as follows. p(n)

n Consider the number 2m where m > 1 is large enough so that 2m > N . Note that p(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 for every n > N , it follows that there is a valley at every even number and a peak at every odd number once we pass N . Consider the number 3m . Note that p(3m ) = 3. However, from the preceding paragraph, there is a peak at 3m because it is odd and greater than N . Hence p(3m − 1) = 2 and p(3m + 1) = 2. This is possible if and only if 3m − 1 and 3m + 1 are powers of 2. But the only powers of 2 that differ by 2 are 21 and 22 . This implies m = 1, which is a contradiction.  Comment 1 Some contestants found a second way to deduce that the graph of p(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 peak at q. This is because p(q) = q, while p(q + 1) ≤ q+1 < q and p(q − 1) ≤ q−1 < q. 2 2

Comment 2 All solutions used the method of indirect proof.44 They all established that the graph of p(n) versus n (for n > N ) has peaks at odd values of n and valleys

4 4

Also known as ‘proof by contradiction’.

35 104

at even values of n. We present some of the variations that contestants used to derive 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. Hence 32m+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+1 3



< 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 them is 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) ≤

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

2q+1 3

< q. 

Variation 5 (Kevin Shen, year 1255 , 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−3 is 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 that 2q−1 (q − 1) + 1 is prime. 

5 5

Equivalent to year 11 in Australia.

36 105

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 we may choose j > 3 such that qj > N and such that the list q1 , q2 , . . . , qj contains an odd number of primes of the form of 4k + 3. Consider the number m = q1 q2 · · · qj . Note that p(m) = qj . Furthermore, there is a peak 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 prime that is greater than pj . Thus p(m − 1) > p(m), which contradicts that there is a peak at m. 

37 106

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 k

k

k

p(q 2 − 1) < p(q 2 ) < p(q 2 + 1). Lemma 1 1 ≤ a < b.

a

(1) b

For any odd prime q, we have gcd(q 2 + 1, q 2 + 1) = 2 for all integers a

b

Proof Let d = gcd(q 2 + 1, q 2 + 1). Using the difference of perfect squares factorisation we know x − 1 | x2 − 1 and x + 1 | x2 − 1. From this we deduce the following chain of divisibility. a

a+1

q

2a+1

2a+2

q

2a+2

q2 + 1 | q2

q2 a

b−1

−1|q −1|q .. .

2a+3

−1

−1 −1

b

− 1 | q2 − 1

b

b

b

It follows that q 2 + 1 | q 2 − 1, and so d | q 2 − 1. Combining this with d | q 2 + 1 a b yields d | 2. Since q 2 + 1 and q 2 + 1 are even, we conclude that d = 2.  Lemma 2

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

k

p(q 2 − 1) < q

and p(q 2 + 1) < q

for k = 1, 2, . . . .

Proof We proceed by induction under the assumption that (1) is false.   . Hence For the base case k = 1, since q is odd we have q 2 − 1 = 2(q − 1) q+1 2 2 2 p(q − 1) < q. However, since (1) is false for k = 2, we also have p(q + 1) < q. For the inductive step, suppose we know that j

p(q 2 − 1) < q

j

and p(q 2 + 1) < q, j+1

j

j

for some positive integer j. Then since q 2 − 1 = (q 2 − 1)(q 2 + 1), we have j+1 j+1 p(q 2 −1) < q. However, since (1) is false for k = j+1, we also have p(q 2 +1) < q. This completes the induction.  To complete the proof of the problem, let us focus on the following list of numbers. L = p(q 2 + 1), p(q 4 + 1), p(q 8 + 1), . . . From lemma 1, q 2 + 1, q 4 + 1, q 8 + 1, . . . all have pairwise greatest common divisor equal to 2. Hence at most one of the numbers in L is equal to 2, while the others are distinct odd primes. This means that L contains infinitely many different prime numbers. This is a contradiction because lemma 2 implies that L contains only finitely many different numbers. 

38 107

Comment Jeremy’s proof is rather impressive because it establishes a much stronger result than what was required. The AMO problem only required a proof that 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 does occur infinitely often. Jeremy’s proof establishes that p(n − 1) < p(n) < p(n + 1) definitely occurs infinitely often.66

66 Erd¨ os

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

39 108

8. This was the most difficult problem of the 2015 AMO. Just four contestants managed to solve it completely. Solution (Seyoon Ragavan, year 11, Knox Grammar School, NSW)   Answer: n n−1 2

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 with base BC, and isosceles with base AC. In this way an equilateral triangle counts as three isosceles triangles depending on which side is considered the base. Thus the number of apples Maryam receives is equal to the number of base-specified isosceles triangles. Let S denote the set of lines that Maryam draws. Consider a line a ∈ S. We determine how many ways a can be the base of an isosceles triangle. Observe that if a, b, c and a, d, e both form isosceles triangles with base a, then since no two lines are parallel,   the pairs {b, c} and {d, e} are either equal or disjoint. Thus there are at most n−1 pairs {b, c} from S such that a, b, c forms an isosceles triangle with base 2 a. Since there are n possibilities  fora, we conclude that the number of base-specified isosceles triangle is at most n n−1 . 2

It remains to show that this is attainable. Take n equally spaced lines passing ◦ through the origin so that the angle between consecutive lines is 180 . Next translate n each of the lines so that no three of them are concurrent. We claim this configuration  n−1  of lines results in n 2 base-specified isosceles triangles.

For any line x, let x denote the line after it has been translated. Consider any triangle bounded by lines a , b , c . It is isosceles with base a if and only if before the translations, the lines b and c were symmetric in a. There are n ways of choosing  n−1 a. Once a is chosen, then by our construction there are 2 pairs b, c such that they are symmetric in a. Indeed, if n is odd, all the other lines are involved in a symmetric pair about a. If n is even then exactly one line is not involved in a symmetric pair about a because it is perpendicular to a. Thus  the total number of base specified isosceles triangles for this construction is n n−1 , as desired.  2

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

−→

40 109

  are possible. The Comment Other constructions that achieve the bound n n−1 2 following one is by Andrew Elvey Price, Deputy Leader of the 2015 Australian IMO team. If n is odd, Maryam can achieve this by drawing the lines to form the sides of a regular polygon with n sides. If n is even, Maryam can achieve this by drawing the lines to form all but one of the sides of a regular polygon with n + 1 sides. 

41 110

AUSTRALIAN MATHEMATICAL OLYMPIAD STATISTICS SCORE DISTRIBUTION/PROBLEM Problem Number

Number of Students/Score

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

5.2

4.6

3.4

1.2

5.4

2.5

2.0

0.6

Average Mark



111

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

William Hu

Christ Church Grammar School WA

Martin Luk

King's College NZ

11 NZ 9 13 NZ

112

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

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

Michael Chen

Scotch College VIC

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

12 NZ

13 NZ 12

113

27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD

March, 2015 Time allowed: 4 hours No calculators are to be used Each problem is worth 7 points Problem 1. Let ABC be a triangle, and let D be a point on side BC. A line through D intersects side AB at X and ray AC at Y . The circumcircle of triangle BXD intersects the circumcircle ω of triangle ABC again at point Z = B. The lines ZD and ZY intersect ω again at V and W , respectively. Prove that AB = V W . Problem 2. Let S = {2, 3, 4, . . .} denote the set of integers that are greater than or equal to 2. Does there exist a function f : S → S such that f (a)f (b) = f (a2 b2 ) for all a, b ∈ S with a = b?

Problem 3. A sequence of real numbers a0 , a1 , . . . is said to be good if the following three 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, no two of which are parallel. Of the 2n lines, n are colored blue, the other n are colored red. 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 that there exists a circle that intersects B in exactly 2n − 1 points, and also intersects R in exactly 2n − 1 points. Problem 5. Determine all sequences a0 , a1 , a2 , . . . of positive integers with a0 ≥ 2015 such 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+1 a0 .

114

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 CDY are concurrent at point Z.11 Hence CDZY is cyclic. A

V X

D

B

W

Z

C

Y

Let ∠(m, n) denote the directed angle between any two lines m and n.22 We have ∠(W Z, V Z) = ∠(Y Z, DZ) = ∠(Y C, DC) (CDZY cyclic) = ∠(AC, BC). Therefore V W and AB subtend equal or supplementary angles in ω. It follows that V W = AB.  Comment All solutions that were dependent on how the diagram was drawn received a penalty deduction of 1 point. The easiest way to avoid diagram dependence was to use directed angles as in the solution presented above.

11

The point Z is also the Miquel point of the four lines AB, AY , BC and XY . It is the common point of the circumcircles of ABC, AXY , BDX and CDY . See the sections entitled Pivot theorem and Four lines and four circles found in chapter 5 of Problem Solving Tactics published by the AMT. 22 This is the angle by which one may rotate m anticlockwise to obtain a line parallel to n. For more details, see the section Directed angles in chapter 17 of Problem Solving Tactics published by the AMT.

43 115

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 so

Similarly, and so

f (2n+3 ) f (2n ) = . f (2n+1 ) f (2n+4 )

(1)

  f (2n+1 )f (2n+4 ) = f 22(2n+5) = f (2n+2 )f (2n+3 ), f (2n+3 ) f (2n+1 ) = . f (2n+2 ) f (2n+4 )

(2)

Combining (1) and (2), we find that f (2n+1 ) f (2n ) = . 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 a and 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 such functions. 

44

116

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 so f (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 such function exists. 

45 117

Solution 3 (Jeremy Yip, year 12, Trinity Grammar School, VIC) Suppose such a function exists. Let a < b < c be positive integers. We work with the quantity f (2a )f (2b )f (2c ) in two different ways.  a  f (2 )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)

 f (2a )f (2b ) f (2c ) = f (22a+2b )f (2c ) (since b > a) = f (24a+4b+2c )

(if c = 2a + 2b)

(2)

(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) (a, b, c) = (1, 2, 4) (a, b, c) = (1, 3, 5) (a, b, c) = (1, 4, 7)

⇒ f (49 ) = f (410 ) = f (411 ) ⇒ f (410 ) = f (412 ) = f (413 ) ⇒ f (413 ) = f (415 ) = f (417 ) ⇒ 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 that no such function exists. 

46 118

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 (a4 b4 c4 ) = f (a2 )f (b2 c2 ) = f (a2 )f (b)f (c). Furthermore, since ac > b and c > a, we have f (a4 b4 c4 ) = f (b2 )f (a2 c2 ) = f (b2 )f (a)f (c). Comparing these two equations, we find that for all a, b ∈ S, 2

2

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

⇒

f (b2 ) f (a2 ) = . f (a) 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 (a)f (a4 ) f (a)f (a5 ) f (a)f (a)f (a2 ) = . = k k2 k

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

Putting this into the functional equation yields k 2 = k. But 0, 1 ∈ S and hence there is no solution. 

47 119

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 nonnegative integer i let ai = pqii where pi and qi are positive integers with gcd(pi , qi ) = 1. Therefore, pi+1 2pi + qi pi+1 pi = or = . (1) qi+1 qi qi+1 pi + 2qi If each of pi and qi are odd, then so are 2pi + qi , qi , pi , and pi + 2qi . Thus when the RHSs of (1) are reduced to lowest terms, the numerators and denominators are still odd. 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 pqnn = 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) because that implies pi+1 and qi+1 are both odd, which contradicts (2). So we must have i+1 i the first option in (1), namely, pqi+1 = 2piq+q . From (2), qi is even, and so we have i gcd(2pi + qi , qi ) = gcd(2pi , qi ) = 2. Hence pi odd

⇒

pi+1 = pi +

qi 2

and qi+1 =

qi 2

for i < n.

(3)

On the other hand, a similar argument shows that if pi is even for some i < n, then pi i+1 we must take the second option in (1), namely pqi+1 = pi +2q , and gcd(pi , pi +2qi ) = 1. i Hence pi even

⇒

pi+1 =

pi 2

and qi+1 =

pi + qi 2

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. If pi is odd, we may combine (3) and (5) to find  qi qi  (pi+1 , qi+1 ) = pi + , 2 2 p q  i i ≡ , (mod 2015). 2 2

(5)

(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 ) ≡

p 2

0 n,

q0  2n

(mod 2015).

We require qn = 1. However, from (5), this is true if and only if qn ≡ 1 (mod 2015). 48 120

(7)

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 see that 5 | 2n − 1 ⇔ 4 | n 13 | 2n − 1 ⇔ 12 | n 31 | 2n − 1 ⇔ 5 | n. Since 60 = lcm(4, 12, 5) we conclude that 2015 | 2n − 1 if and only if 60 | n. The smallest such positive integer is n = 60. 

49 121

Solution 2 (Henry Yoo, year 12, Perth Modern School, WA) Clearly all members of the sequence are positive rational numbers. For each positive ai+1 − 1 2ai+1 or ai = . Since ai > 0 we deduce that integer i, we have ai = 2 1 − ai+1  ai+1 − 1   if ai+1 > 1  2 ai = (1)  2ai+1   if ai+1 < 1. 1 − ai+1

Thus ai is uniquely determined from ai+1 . Hence we may start from an = 2014 and simply run the sequence backwards until we reach a positive integer. From (1), if ai+1 = uv , then  u−v    2v ai =  2u   v−u

if u > v if u < v.

Consequently, if we define the sequence of pairs of integers (u0 , v0 ), (u1 , v1 ), . . . by (u0 , v0 ) = (2014, 1) and  (ui − vi , 2vi ) if ui > vi (2) (ui+1 , vi+1 ) = (2ui , vi − ui ) if ui < vi , then an−i =

ui vi

for 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, and so d is odd. If ui > vi , then from (2) we have d | ui − vi and d | 2vi . This easily implies d | ui and d | vi . If ui < vi , we similarly conclude from (2) that d | ui and d | 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 ) i+1

≡ (−2 50 122

,2

i+1

)

(mod 2015) (mod 2015)

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 = uvnn is 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 from an = 2014 until a positive integer is reached, the result would be 2014 , 1 1918 , 97 666 , 1349 122 , 1893

2013 , 2 1821 , 194 1332 , 683 244 , 1771

2011 , 4 1627 , 388 649 , 1366 488 , 1527

2007 , 8 1239 , 776 1298 , 717 976 , 1039

1999 , 16 463 , 1552 581 , 1434 1952 , 63

1983 , 32 926 , 1089 1162 , 853 1889 , 126

1951 , 64 1852 , 163 309 , 1706 1763 , 252

1887 , 128 1689 , 326 618 , 1397 1511 , 504

1759 , 256 1363 , 652 1236 , 779 1007 , 1008

1503 , 512 711 , 1304 457 , 1558 2014 . 1

991 , 1024 1422 , 593 914 , 1101

1982 , 33 829 , 1186 1828 , 187

1949 , 66 1658 , 357 1641 , 374

1883 , 132 1301 , 714 1267 , 748

1751 , 264 587 , 1428 519 , 1496

1487 , 528 1174 , 841 1038 , 977

959 , 1056 333 , 1682 61 , 1954

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 yields a good sequence is a0 = 2014.

51 123

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 r be 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 defined by θ are shaded. Also we orient our configuration so that the x-axis is the angle bisector of r and b that passes through the two shaded regions. Let O be the intersection of lines r and b. Let  be another line in the configuration. Four options arise that need to be considered.

β

θ2 β θ α

θ α θ1

In the first two diagrams,  does not pass through O. Observe that , r and b enclose a triangle which is either completely shaded or completely unshaded. If the triangle is 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 the triangle 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 > θ and θ + β > θ, it would follow that  could not be red or blue. Hence  lies in the shaded 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 and b in the right-hand shaded region. We claim that Γ has the required property. It suffices to show that every line in the configuration, apart from r and b, intersects Γ in two distinct points.

B

O

R

Let R and B be the points of tangency of Γ with r and b, respectively, and let T be the union of the segments OB and OR. 52 124

If  ( = r, b) is any line of the configuration, then the part of  lying in the right shaded region is an infinite ray  . Let F be the figure enclosed by T and the minor arc RB of Γ. Then  passes into the interior of F , and so intersects the boundary of F at least twice. Since  cannot intersect T twice, it must intersect the minor arc RB. Finally,  cannot be tangent to Γ because that would imply that  intersects both OB and OR (not at O). It follows that  intersects Γ at two distinct points, as desired. 

53 125

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 , . . . , 2n be the lines listed in order of increasing gradient. Then there is a k such that lines k and k+1 are oppositely coloured. By rotating our coordinate system and cyclicly relabelling our lines we can ensure that 1 , 2 , . . . , 2n are listed in order of increasing gradient, 1 and 2n are oppositely coloured, and no line is vertical. Without loss of generality 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. Let R = 1 ∩  and B = 2n ∩ . Since the lines 2 , 3 , . . . , 2n−1 have gradients lying in between those of 1 and 2n , we can move  far enough to the right so that all the intersection points of 2 , 3 , . . . , 2n−1 with  lie between R and B. Let Γ be the excircle of ORB opposite A. We claim that Γ has the required properties. To prove this, it suffices to show that each line i (2 ≤ i ≤ 2n − 1) intersects Γ twice. B

B

O R

R

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

54 126

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 therefore a1 ≥ a30 ≥ 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 that an ≥ 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) an . n+2

55 127

Proof Using the recurrence an+3 = (n + 3)an+2 + (n + 2)an+1 + δn+2 and writing an+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 = Lemma 3

For n ≥ 1, we have an+1 =

(n+1)(n+3) an , n+2

as desired.



(n+1)(n+3) an . n+2

an . By lemma 2, there Proof Suppose there exists n ≥ 1 such that an+1 = (n+1)(n+3) n+2 is a greatest integer 1 ≤ m ≤ 3 with this property. Then am+2 = (m+2)(m+4) am+1 . m+3 If δm+1 = 0, then am+1 = δm+1 = 0.

(m+1)(m+3) am , m+2

which contradicts our choice of m. Thus

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. Combining the two divisibility conditions, we obtain am | (m + 4)δm+1 .

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

an for n ≥ 1. Substituting n = 1 yields 3 | a1 . Letting a1 = 3c, So, an+1 = (n+1)(n+3) n+2 we have by induction that an = n!(n + 2)c for n ≥ 1. Since |s2 − 2a1 | = 1, we then get a0 = c ± 1, yielding the two families of solutions. Finally, since (n+2)n! = n!+(n+1)!, it follows that sn+1 = c(n+2)!+(−1)n (c−a0 ). Hence both families of solutions satisfy the given conditions. 

56 128

27TH ASIAN PACIFIC MATHEMATICS OLYMPIAD RESULTS 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

129

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

299

4583

11

42

102

81

Total

130

AMOC SELECTION SCHOOL 2015 IMO Team Selection School The 2015 IMO Selection School was held 5–14 April at Robert Menzies College, Macquarie University, Sydney. The main qualifying exams are the AMO and the APMO from which 25 students are selected for the school. The routine is similar to that for the December School of Excellence; however, there is the added interest of the actual selection of the Australian IMO team. This year the IMO would be held in Chiang Mai, Thailand. The students are divided into a junior group and a senior group. This year there were 10 juniors and 15 seniors. It is from the seniors that the team of six for the IMO plus one reserve team member is selected. The AMO, the APMO and the final three senior exams at the school are the official selection criteria. My thanks go to Andrew Elvey Price, Ivan Guo, Victor Khou, and Konrad Pilch, who assisted 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, Jonathan Zheng, and Damon Zhong, all of whom came in to give lectures or help with the marking of exams.

Angelo Di Pasquale Director of Training, AMOC

2015 Australian IMO team Name Alexander Gunning Ilia Kucherov Seyoon Ragavan Yang Song Kevin Xian Jeremy Yip Reserve Yong See Foo

Year 12 11 11 12 11 12 11

School State Glen Waverley Secondary College VIC Westall Secondary College VIC Knox Grammar School NSW James Ruse Agricultural High School NSW James Ruse Agricultural High School NSW Trinity Grammar School VIC Nossal High School

3 131

VIC



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

Nossal High School VIC

11

Reserve Yong See Foo

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

132

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

133

TEAM School PREPARATION SCHOOL IMO Team IMO Preparation The pre-IMO July school is always a great reality check when it comes to our perception of the team’s ability. This is of course because we train with the UK team. Our joint training 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, free time in the afternoon while their papers were being assessed, a short debrief of the exam late in the afternoon followed by going out to dinner each evening. The results of training showed that both teams were quite strong, with the UK having the edge. The final exam also doubles as the annual Mathematics Ashes contest. Australia won the Ashes in its inaugural year, lost them the next year, and have not been able to win them back since. There have been a few close calls, and even a tie in 2011. In another heart-breaking nail biter, the UK again retained the Ashes after both teams tied on 84 points apiece. Angelo Di Pasquale IMO Team Leader

1 134

THE MATHEMATICS ASHES The 2015 Mathematical Ashes: AUS v UK Exam 1. Does there exist a 2015 × 2015 array of distinct positive integers such that the sums of 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 triangle ABC with AB > BC. The angle bisector of ∠ABC intersects Ω at M = B. Let Γ be the circle with diameter BM . The angle bisectors of ∠AOB and ∠BOC intersect Γ at points P and Q, respectively. The point R is chosen on the line P Q so that BR = M R. 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 every real number x, there is exactly one card in the deck that has x written on it. Now two players draw disjoint sets A and B of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions: 1. The winner only depends on the relative order of the 200 cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner. 2. If we write the elements of both sets in increasing order as A = {a1 , a2 , . . . , a100 } and 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 and B 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 B such that A beats B according to one rule, but B beats A according to the other.) Results

AUS AUS AUS AUS AUS AUS Σ

1 2 3 4 5 6

Q1 7 7 7 7 3 6 37

Q2 7 0 7 5 6 7 32

Q3 7 7 0 0 1 0 15

Σ 21 14 14 12 10 13 84

UNK UNK UNK UNK UNK UNK Σ

2 135

1 2 3 4 5 6

Q1 7 7 5 3 7 7 36

Q2 7 2 0 7 6 7 29

Q3 5 0 0 7 7 0 19

Σ 19 9 5 17 20 14 84

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:

AUSTRALIA 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

37

32

15

84

Q1

Q2

Q3

Total

TOTAL

UNITED KINGDOM Code

Name

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

36

29

19

84

TOTAL

136

The 56th International Mathematical Olympiad, Chiang Mai, IMO TEAM LEADER’S REPORT 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 students from 104 countries participating. Of these, 52 were girls. Each participating country may send a team of up to six students, a Team Leader and a Deputy Team Leader. At the IMO the Team Leaders, as an international collective, form what is called the Jury. This Jury was chaired by Soontorn Oraintara. The first major task facing the Jury is to set the two competition papers. During this period the Leaders and their observers are trusted to keep all information about the contest problems completely confidential. The local Problem Selection Committee had already shortlisted 29 problems from 155 problem proposals submitted by 53 of the participating countries from around the world. During the Jury meetings one of the shortlisted problems had to be discarded from consideration due to being too similar to material already in the public domain. Eventually, the Jury finalised the exam questions and then made translations 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 the night before that exam was scheduled. This probably resulted in a harder than intended second 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 the Netherlands. It concerns finite sets of points in the plane in which the perpendicular bisector of any pair of points in such a set also contains another point of the set. 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 certain two 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 sequence of integers. Although it does not seem so at first sight, the problem is much more combinatorial 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 Saturday 11 July. Each paper had three problems. The contestants worked individually. They were allowed four and a half hours per paper to write their attempted proofs. Each problem was scored out of a maximum of seven points. For many years now there has been an opening ceremony prior to the first day of competition. HRH Crown Princess Sirindhorn presided over the opening ceremony. Following the formal speeches there was the parade of the teams and the 2015 IMO was declared open. 5 137

After the exams the Leaders and their Deputies spent about two days assessing the work of the students from their own countries, guided by marking schemes, which had been discussed earlier. A local team of markers called Coordinators also assessed the papers. They too were guided by the marking schemes but are allowed some flexibility if, for example, a Leader brought something to their attention in a contestant’s exam script that is not covered by the marking scheme. The Team Leader and Coordinators have to agree on scores for each student of the Leader’s country in order to finalise scores. Any disagreements that cannot be resolved in this way are ultimately referred to the Jury. The IMO paper turned out to be quite difficult. While the easier problems 1 and 4 were quite accessible, the other four problems 2, 3, 5 and 6 were found to be the most difficult combination of medium and difficult problems ever seen at the IMO. There were only around 30 complete solutions to each of problems 2, 3 and 5. Problem 6 was very difficult, 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.11 Consequently, there were 282 (=48.9%) medals awarded. The medal distributions22 were 39 (=6.8%) gold, 100 (=17.3%) silver and 143 (=24.8%) bronze. These awards were presented at the closing ceremony. Of those who did not get a medal, a further 126 contestants received an honourable mention for solving at least one question perfectly. Alex Song of Canada was the sole contestant who achieved the most excellent feat of a perfect score of 42. He now leads the IMO hall of fame, being the most decorated contestant in IMO history. He is the only person to have won five IMO gold medals.33 He was given a standing ovation during the presentation of medals at the closing ceremony. Congratulations to the Australian IMO team on an absolutely spectacular performance this year. They smashed our record rank44 to come 6th, and they also smashed our record medal haul, bringing home two Gold and four Silver medals.55 This is the first time that each team member has achieved Silver or better. The team finished ahead of many of the traditionally stronger teams. In particular, they finished ahead of Russia, whom we would have considered as untouchable. Congratulations to Gold medallist Alexander Gunning, year 12, Glen Waverley Secondary College. He is now the most decorated Australian at the IMO, being the only Australian to have won three Gold medals at the IMO. On each of these occasions he also finished in the top 10 in individual rankings.66 He is now equal 17th on the IMOs all-time hall of fame. 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. 11

This was the lowest ever cut for gold, and the equal lowest ever cut for silver. (This was indicative of the difficulty of the exam, not the standard of the contestants.) 2 2 The total number of medals must be approved by the Jury and should not normally exceed half the total number of contestants. The numbers of gold, silver and bronze medals must be approximately in the ratio 1:2:3. 3 3 In his six appearances at the IMO, Alex Song won a bronze medal in 2010, and followed up with gold medals in 2011, 2012, 2013, 2014 and 2015. 4 4 The ranking of countries is not officially part of the IMO general regulations. However, countries are ranked each year on the IMO’s official website according to the sum of the individual student scores from each country. 5 5 Australia’s best performance prior to this was the dream team of 1997. They came 9th, with a medal tally of two Gold, three Silver and one Bronze. 6 6 In his four appearances at the IMO, Alexander won a bronze medal in 2012, and followed up with gold medals in 2013 (8th), 2014 (1st) and 2015 (4th).

6 138

And congratulations to our four Silver medallists: Ilia Kucherov, year 11, Westall Secondary 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 Grammar School. Three members of this year’s team are eligible for selection to the 2016 IMO team. So while it is unlikely we will be able to repeat this year’s stellar performance, the outlook seems promising. Congratulations also to Ross Atkins and Ivan Guo, who were IMO medallists with the Australian team when they were students.77 They were the authors of the juggling-inspired IMO problem number six. In fact Ross is a proficient juggler. The 2015 IMO was organised by: The Institute for the Promotion of Teaching Science and Technology, Chiang Mai University, The Mathematical Association of Thailand under the Patronage of His Majesty the King, and The Promotion of Academic Olympiad and Development of Science Education Foundation. The 2016 IMO is scheduled to be held July 6-16 in Hong Kong. Venues for future IMOs have been secured up to 2019 as follows. 2017 2018 2019

Brazil Romania United Kingdom

Much of the statistical information found in this report can also be found at the official website of the IMO. www.imo-official.org

Angelo Di Pasquale IMO Team Leader, Australia

7 7

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

7 139

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

84 140

INTERNATIONAL MATHEMATICAL OLYMPIAD IMO Papers

Language: English Day: 1 Day: 1

Friday, July 10, 2015

IMO Papers

Problem 1. We say that a finite set S of points in the plane is balanced if, for any two different points A and B in S, there is a point C in S such that AC = BC. We say that Day:P 1in S is centre-free if for any three different points A, B and C in S, there is no point S such that P A = P B = P C. Friday, July 10, 2015 (a) Show that for all integers n ≥ 3, there exists a balanced set consisting of n points. S set S of points in the plane is balanced if, for any two Problem 1. We say that a finite (b) Determine integers n C≥ 3 for balanced set that conA B points S Aalland Sis which = S We say different B in S, there a pointthere CAC in exists S BC sucha that AC =centre-free BC. sisting of n points. A B C S P S P A = P B = S is centre-free if for any three different points A, B and C in S, there is no point PP Cin S such that P A = P B = P C. n�3 n Problem 2. Determine all triples (a, b, c) of positive integers such that each of the (a) Show that for all integers n ≥ 3, there exists a balanced set consisting of n points.n n� 3 numbers c, which bc −there a, ca − b a balanced centre-free set con(b) Determine all integers n ≥ab3−for exists sisting is a power of of 2. n points.

(a, b, c) (A power of 2 is an integer of the form 2n , where n is a non-negative integer.) − c, bc ca positive −b Problem 2. Determine allabtriples (a,−b,a,c) of integers such that each of the Problem 3. Let ABC be an acute triangle with AB > AC. Let Γ be its circumcircle, numbers 2 H its orthocentre, and F the foot abof − the c, altitude bc − a, from ca −A. b Let M be the midpoint of BC. n ◦ n 2 2 Let Q be the point on Γ such that ∠HQA = 90 , and let K be the point on Γ such that is a power of ◦2. ∠HKQ = 90 . Assume that the points A, B, C, K and Q are all different, and lie on Γ (A power of ABC 2 is an integer of the form 2n , where is a non-negative integer.) AB >nAC Γ H in this order. F A M BC Q Prove that the circumcircles ◦ of triangles KQH and F KM are tangent to each ◦other. Γ = 90be K triangle with ABΓ > AC. Let∠HKQ = circumcircle, 90 Problem 3. ∠HQA Let ABC an acute Γ be its A B C K Q Γ 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 F =KM 90◦ , and let K be the point on Γ such that KQH ∠HKQ = 90◦ . Assume that the points A, B, C, K and Q are all different, and lie on Γ in this order.

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

Language: English

Time: 4 hours and 30 minutes Each problem is worth 7 points 141

Language: English Day: Day: 2 Saturday, July 11, 2015

Day: 2

Problem 4. Triangle ABC has circumcircle Ω and circumcentre O. A circle Γ with centre A intersects the segment BC at points D and E, such thatSaturday, B, D, E July and C all 11,are 2015 different and lie on line BC in this order. Let F and G be the points of intersection of Γ and Ω, such that A, F , B, C and G lie this order.OLet K be the second point ABC Ω on Ω in A Problem 4. Triangle ABC has circumcircle Ω and circumcentre O. Γ A circle Γ with of intersection ofBC the circumcircle triangle BDFB and the AB. Let L be the D BCofat E D such E segment C B, D, centre A intersects the segment points D and E, that E and C are all second of intersection of the circumcircle of triangle CGE and the segment CA. BC point A F different and lie on lineFBC inGthis order. Let F and G be theΓpointsΩof intersection of B C Suppose G Ω K that the lines F K and GL are different and intersect at the point X. Prove that Γ and Ω, such that A, F , B, C and G lie on Ω in this order. Let K be the second point AB L of triangle BDF and the segment AB. Let L be the X lies on the line AO.circumcircle ofBDF intersection of the CGE point of intersection CA of the circumcircle of triangle CGE and the segment CA. second Problem 5. Let R denote the set of real numbers. Determine all functions f :XR → R X point Suppose that theF K lines F GL K and GL are different and intersect at the X. Prove that satisfying the equation AO X lies on the line AO. f (x + f (x + y)) + f (xy) = x + f (x + y) + yf (x) → R f: R → R Problem 5.R Let R denote the set of real numbers. Determine fall: R functions satisfying equation  x and y.  for all realthe numbers f x + f (x + y) + f (xy) = x + f (x + y) + yf (x) f (x + f (xa+, ay)), .+ = x +satisfies f (x + y) + following yf (x) Problem 6. x The ysequence . . fof(xy) integers the conditions: 1

2

for all real numbers x and y. a1 ,all a2 ,j. .≥ . 1; (i) 1 ≤ aj ≤ 2015 for

The jsequence a1 , a2 , . . . of integers satisfies the following conditions: 1 1Problem  aj k+2015 (ii) a6. k = + a for all 1 ≤ k < . ℓ+ a≤ 1 all k< ℓ 1; k +(i) ak �=≤ ℓ 2015 for j≥ ajthere Prove1that exist two positive integers b and N such that b N  (ii) k + ak = + a for all 1 ≤ k < n .     n     (a − b) ≤ 10072 j     2 (aj − b)b and 1007  j=m+1 Prove that there exist two positive integers N such that   j=m+1   n   for all integers m and n satisfying n > m ≥ N .  m n n > m  N (aj − b) ≤ 10072   j=m+1

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

Language: English

Language: English

Time: 4 hours and 30 minutes Each problem is worth 7 points 142

Time: 4 hours and 30 minutes

2

INTERNATIONAL MATHEMATICAL OLYMPIAD SOLUTIONS 1. Solution (All members of the 2015 Australian IMO team solved this problem using similar methods. Here we present the solution by Yang Song, year 12, James Ruse Agricultural High School, NSW. Yang was Silver medallist with the 2015 Australian IMO 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 and B be any two vertices of P. Since n is odd, one side of the line AB contains an odd number of vertices of P. Thus if we enumerate the vertices of P in order from 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 collection of k unit equilateral triangles, all of which have a common vertex, O say, and exactly one pair of them has a second common vertex. Note that apart from O, 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 construction there 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 is odd. Note that the construction used in the solution to part (a) is centre-free. We shall 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 cannot be centre-free  n−1  because AX = AY = AZ. Hence for a given point A, there are at most 2 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−1 . 2 On the other hand, since S is balanced, for each pair of points X, Y ∈ S, there is at leastone  point A such that A → {X, Y }. Since the number ofnpairs  n {X, Y } is 2 , the total number of instances of A → {X, Y } is at least 2 .     ≥ n2 , which simplifies to If we combine our estimates, we obtain n n−1 2  n−1 ≥ n−1 . This final inequality is impossible if n is even.  2 2 58

143

Comment 1 A nice graph theoretical interpretation of the solution to part (b) was given by Kevin Xian, year 11, James Ruse Agricultural High School, NSW. Kevin was a Silver medallist with the 2015 Australian IMO team. Let S  denote the set of unordered pairs of elements of S. Form a directed bipartite graph G as follows. The vertex set of G is S ∪ S  . The directed edges of G are simply all the instances of A → {X, Y} asper the solution to part (b) above. Then n−1 the  total outdegree of G is at most n 2 , while the total indegree of G is at least n . The inequality found in the solution to part (b) is simply a consequence of the 2 total indegree being equal to the total outdegree. Comment 2 Alternative constructions for n odd in part (a) were found by Yang Song, year 12, James Ruse Agricultural High School, NSW, and Jeremy Yip, year 12, Trinity Grammar School, VIC. Yang and Jeremy were Silver medallists with the 2015 Australian IMO team. For n = 2k + 1, let S be the set of vertices of a collection of k unit equilateral triangles, all of which have exactly a common vertex, O say, and no two of which have any other common vertices besides O. Note that apart from O, all of the vertices 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 open problem. From the foregoing, we see that for n odd there are two infinite families of balanced sets, and for n even there is one infinite family. It is unknown if there are any other infinite families which do not fit into this scheme. Note however, there are sporadic constructions which do not fit into either of these families. Some of these are shown below. (All drawn segments are of unit length.)

59 144

2. Solution 1 (Jeremy Yip, year 12, Trinity Grammar School, VIC. Jeremy was a Silver 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 permutations. 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 b(c − 1) = 2y

(1) (2)

for some non-negative integers x and y. Equation (2) implies that b = 2p and c − 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 are 2 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 is odd. Case 2 Without loss of generality we may suppose that a > b > c, where a, b, c are 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 is divisible by 4, while the other is even but not divisible by 4. Since ac − b is a power of two, it follows that either ac − b | 2(a − 1) or ac − b | 2(a + 1). Either way, we have ac − b ≤ 2(a + 1). Since a > b, we have ac − a < 2(a + 1), which can be rearranged as (c − 3)a < 2. 60 145

(4)

Observe that a ≥ 5, b ≥ 3 and c ≥ 1. This is because a, b, c are all odd and a > 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 they sum 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). Hence 8b − 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 of two. 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 impossible for 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. Then a = 2k m and b = 2k n 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 of the same parity because otherwise mc − n would be an odd number that is greater than 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, the smaller divides the larger. Therefore,

⇔

ac − b ≡ 0 (mod bc − a) (bc)c − b ≡ 0 (mod bc − a)

⇔ 2k n(c2 − 1) ≡ 0 (mod 2k (nc − m)) ⇔ (c − 1)(c + 1) ≡ 0 (mod nc − m). (n is odd) 61 146

(5)

(6)

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

k

⇔ n(4 mn − 1) − m ≤ 2 · 4k mn (c = ab − 1 = 4k mn − 1) n ⇔ 4k (n2 − 2n) ≤ 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 and m + 1 are consecutive even positive integers, we deduce that c − m ≤ 2(m + 1)

⇔ 4k m − m ≤ 2m + 2

⇔ ⇒

4k m < 3m + 2 k = 1.

(c=4k m − 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 = 2α A, b = 2β B and c = 2γ C, where α ≥ β ≥ γ ≥ 1, and A, B and C are odd positive integers. It follows that ab − c = 2γ (2α+β−γ AB − C) and ac − b = 2β (2α+γ−β AC − B). Since α ≥ β ≥ γ ≥ 1, we have that 2α+β−γ AB − C and 2α+γ−β AC − 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 a2 bc ≤ 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 all different. So this case does not occur, and the proof is complete.  62 147

Solution 2 (Alex Gunning, year 12, Glen Waverley Secondary College, VIC. Alex was 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 in solution 1. Without loss of generality we may assume that a > b > c. If c = 1, then we would require both a − b and b − a to be powers of two. But this is impossible because their sum is zero. Hence a > b > c ≥ 2.

It easily follows that ab − c ≥ ca − b ≥ bc − a. Since a smaller power of two always divides 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 must have ca − b | b. Thus ca ≤ 2b < 2a, which yields the contradiction c < 2. Henceforth we 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 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). (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. Thus a = 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 because it 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 + 2 are two powers of two that differ by 8. The only such powers of two are 8 and 16. This quickly yields a = 7 and implies (a, b, c) = (7, 5, 3). 63 148

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. This implies (a, b, c) = (11, 6, 2). 

64 149

3. Solution 1 (Alex Gunning, year 12, Glen Waverley Secondary College, VIC. Alex was 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 of intersection 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 since QA ⊥ QH, it follows that QHA is a straight line. Thus Q lies on the line A H. A Q

H

B

F

A

C

E

Next we show that CHBA is a parallelogram. Since AA is a diameter of Γ we have AE ⊥ A E. However AE ⊥ BC. Thus BC  A E. Since A EBC is cyclic, it follows that A ECB is an isosceles trapezium with CE = A B. Thus ∠BCA = ∠EBC = ∠EAC = 90◦ − ∠ACB = ∠CBH.

(A B = CE) (ECAB cyclic) (AE ⊥ BC) (BH ⊥ AC)

Thus BH  A C. A similar argument shows that CH  A B, and so CHBA is a parallelogram. Finally, since the diagonals of any parallelogram bisect each other, and M is the midpoint of BC, it follows that M is also the midpoint of A H. Thus M lies on the line 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 perpendicular to A Q. Observe that RH is tangent to circles KQH, F HM and EHA . This is because each of these circles has a diameter which lies on the line A Q. Applying the radical axis theorem to the three circles Γ, KQH and EHA , we deduce that lines RH, A E and QK are concurrent. Hence R lies on QK.

65 150

A Q

K

H

S B

M

A

F

C

E

R

Recall that M is the midpoint of A H and that BC  A E. Thus the midpoint, S say, of RH lies on the line BC. Since HK ⊥ KR we have that S is the centre of circle RKH, and so SH = SK. But since SH is tangent to circle KQH at H and SH = SK, it follows that SK is tangent to circle KQH at K. Considering the power of point S with respect to circle F HM , we have SF · SM = SH 2 = SK 2 . It follows that SK is tangent to circle F KM at K. Since circle KQH is also tangent to SK at K, we conclude that circles KQH and F KM are tangent to each other at K. 

66 151

Solution 2 (Andrew Elvey Price, Deputy Leader of the 2015 Australian IMO team) Point A is defined as in solution 1. Furthermore, as in solution 1, we establish that M 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 have KAA ∼ KQH. Hence there is a spiral symmetry, f say, centred at K, such that f (A ) = A and f (H) = Q. Note that f is the composition of a 90◦ rotation KA about K with a dilation of factor KA  about K. Let M  and F  be the respective images of M and F under f . Thus M  F  ⊥ M F . But since AH ⊥ M F , it follows that M  F   AH.

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

F S

K

H M B

F

S

C

A

Let S be the preimage of S  under f . (Note that S happens to lie on the line M F , as shown in the diagram, because S  lies on line M  F  . But we will not need this fact.) Since S  F  Q ∼ S  QM  (AA), we have S Q S F  = . S Q S M  It follows that S  F  · S  M  = S  Q2 = S  K 2 .

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

Applying the inverse of f , we have SK is tangent to circle F KM at K. But since S is the preimage of S  under f , we also have ∠S  KS = 90◦ , and so SK is tangent to circle KQH. Hence circles F KM and KQH are tangent at K.  67 152

Solution 3 (IMO Problem Selection Committee) Points A and E are defined as in solution 1. Moreover, as in solution 1, we establish that M is the midpoint of A H, and A , M , H and Q are collinear. Note that since M F  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 lie on the same side of the line KH. By the alternate segment theorem we have ∠T KQ = ∠KHQ. By the alternate segment theorem, it is sufficient to prove that ∠KF M = ∠T KM. We have ∠KF M = ∠T KM ⇔ 90 + ∠KF A = ∠T KQ + 90◦ + ∠HKM ⇔ ∠KF A = ∠Q HA + ∠HKM. ◦

(1)

Thus it suffices to establish (1). Let J be the midpoint of HQ . Observe that triangles KHE and AHQ are similar with F and J being the midpoints of corresponding sides. Hence ∠KF A = ∠HJA.

A

Q

H

J

T

K

Q B

M

A

F

C

E

Observe also that triangles KHA and QHQ are similar with M and J being the midpoints of corresponding sides. Hence ∠HKM = ∠JQH. Thus our task is reduced to proving ∠HJA = ∠Q HA + ∠JQH. Let us draw a new diagram that will help us focus on the task at hand. 68 153

A

Q

O J

H

Q

A

Note that QAQ A is a rectangle. Let O be its centre. We also know that H lies on side A Q and that J is the midpoint of Q H. Thus J and O both lie on the mid-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 immediate consequence of the fact that JO a line of reflective symmetry of the rectangle. 

69 154

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, we have M is the midpoint of A H; points A , M , H and Q are collinear; points Q , H and 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 radius 1 1 in H. Let A and Q be the reflections of H in A and r, followed by the reflection Q, respectively. Then f has the effect of exchanging the following pairs of points. A↔E

A ↔ F

Q ↔ A

Q ↔ M

K ↔ Q

A

A Q

Q K

H Q B

M A

F

C

E

Hence circle KQH is transformed into line Q A , and circle F KM is transformed into circle A Q Q . So it suffices to show that Q A is tangent to circle A Q Q . Since Q 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 the centre of circle A HQ because it is the midpoint of A H and ∠A Q H = 90◦ . Thus Q A is the perpendicular bisector of A Q . From this it follows that Q A = A Q , as desired. 

1

1

Reflection in a point is the same as a 180◦ rotation about that point.

70 155

Comment Solutions 1 and 3 first establish that Q, H, M and A are collinear, and that M and F are the respective midpoints of A H and EH. After this the points B and C are no longer relevant to the solution. The crux of matter boils down to the following. Let AQA Q be a rectangle inscribed in a circle Γ. Let H be any point on the line A Q. Let E and K be the respective second points of intersection of the lines AH and Q H with Γ. Let M and F be the midpoints of A H and EH, respectively. Then circles KQH and KM F are tangent to each other.

A

Q

K

H Q M

A

F

E

71 156

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 Q such that ∠HEA = ∠QKH = 90◦ . Let M and F be the midpoints of A H and EH, respectively. Then circles KQH and KM F are tangent to each other. (*)

Q

K

H

M

A

F

E

72 157

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 the same side of the line KH. Let W be the midpoint of QH. Since ∠QKH = 90◦ , it follows that W is the centre of circle KQH. Hence W H = W K, and we may let ∠W KH = ∠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 segment theorem, circle HEK is tangent to line A Q at H. Then since M H = M E, it follows by symmetry that M H and M E are the common tangents from M to circle HEK.

Q

T

W K

H

M F

A

E

The configuration where M H and M E are common tangents to the circumcircle of HEK, is a standard one. In particular it yields an alternative characterisation of the symmedian.22 Thus we may let ∠HKM = ∠F KE = β. From the exterior angle sum in EKF , we have

∠KF H = α + β = ∠W KM.

(1)

Since W K is the radius of circle KQH, we have T K ⊥ KW . Thus adding 90◦ to both sides of (1) yields ∠KF M = ∠T KM. Hence from the alternate segment theorem, T K is tangent to circle F KM at K. Thus circles F KM and KQH are tangent at K.  2 2

For more details see the section Alternative characterisations of symmedian found in chapter 5 of Problem Solving Tactics published by the AMT.

73 158

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

X K L

F

G

O

B

D

E

C

Observe that AF = AG (radius of Γ), and OF = OG (radius of Ω). So F and G are symmetric in AO. Thus lines F K and GL intersect on AO if and only if ∠KF A = ∠AGL. We have, ∠KF A = ∠F KB − ∠F AB = ∠F DB − ∠F GB = ∠F GE − ∠F GB = ∠BGE = ∠CEG − ∠CBG = ∠CLG − ∠CAG = ∠AGL.

(exterior angle AF K) (BDKF and AF BG cyclic) (F DEG cyclic) (exterior angle GEB) (ECGL and ABCG cyclic) (exterior angle GLA)



Comment A careful analysis of this solution shows that the result is still true if we only assume that the centre of Γ lies on the line AO.

74 159

Solution 2 (Found independently by Ilia Kucherov, year 11, Westall Secondary College, VIC, and Jeremy Yip, year 12, Trinity Grammar School, VIC. Ilia and Jeremy were Silver medallists with the 2015 Australian IMO team.) As in solution 1 it suffices to show that ∠KF A = ∠AGL. Let the lines DF and EG intersect Ω for a second time at points P and Q, respectively. Then ∠P DE = ∠F GQ (F DEG cyclic) = ∠F P Q. (F QP G cyclic) Thus BC  QP . Since BQP C is cyclic, it follows that BQP C is an isosceles trapezium with BQ = P C. Hence ∠BGQ = ∠P F C, that is, ∠BGE = ∠DF C.

(1)

A

F G

B

D

E

Q

C

P

75 160

A

X K L

F

G

O

B

D

E

C

Since ABCG is cyclic, we have ∠LAG = ∠CAG = ∠CBG = ∠EBG. Since ECGL is cyclic, ∠CLG = ∠CEG, and so ∠GLA = ∠GEB. Thus GAL ∼ GBE (AA). Hence ∠AGL = ∠BGE. (2) Similarly

∠KF A = ∠DF C.

(3)

Combining (1), (2) and (3), it follows that ∠KF A = ∠AGL.



Comment There are two pairs of similar triangles associated with circles ECGL and Ω. They are GAL ∼ GBE and GLE ∼ GAB. This is a standard configuration which can help fast track the route to a solution.33

3 3

For more details see the section Similar Switch found in chapter 5 of Problem Solving Tactics published by the AMT.

76 161

5. Solution 1 (Seyoon Ragavan, year 11, Knox Grammar School, NSW. Seyoon was a 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 all x ∈ 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 be rearranged 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 = u tells 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. Also setting y = u − 1 in (8) tells us that u − 1 ∈ S. An immediate corollary of this is u, u + 1 ∈ S ⇒ u + n ∈ S 77 162

for any n ∈ Z.

(9)

Let y ∈ R. Then from (2) with x = y − 1 we have y + f (y) − 1 ∈ S, and from (5) we have 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 and x ∈ 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), we find 2f (y) + n ∈ S. (13) Let y ∈ R and let y = 2x. Then applying (12) and (13), we are able to deduce that f (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), and f (1) = 1 from (7), we deduce that f (y) = y for all y ∈ R. This is easily seen to be a solution. 

78 163

Solution 2 (Based on the presentation by Alex Gunning, year 12, Glen Waverley Secondary College, VIC. Alex was a Gold medallist with the 2015 Australian IMO team.) 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 true for 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), yields f (−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 also have x = z + f (z) ∈ S from (6). And y = −f (z) ∈ S from (7) and (5). Thus yf (x) − xf (y) = yx − xy = 0, and so the RHS of (3) is just x − y = z + 2f (z). Hence z + 2f (z) ∈ S. (9) 79 164

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) + z 2 − 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 ⇒ f (z) − z = f (z)(f (x) − x) = f (z)(z 2 − f (z)2 ).

(since y = f (z) ∈ S) (y = f (z)) (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)

  √ √ 2 2 Solving for f (z) in (11), we find f (z) ∈ z, −z+ 2 z −4 , −z− 2 z −4 . Since f (z) ∈ R, we have f (z) = z for all |z| < 2. Hence (−2, 2) ⊆ S. Finally, from (8) we see that z ∈ S implies 2z ∈ S. This allows us to deduce that (−2k , 2k ) ⊆ S for all positive integers k. Hence S = R, and f (x) = x for all x ∈ R. 

165

80

6. This was the hardest problem of the 2015 IMO. Only 11 of the 577 contestants were able to solve this problem completely. The authors of this problem were Ross Atkins and Ivan Guo of Australia. Ross and Ivan were Bronze medallists with the 2003 Australian IMO Team and Ivan was a Gold medallist with the 2004 Australian IMO Team. The problem was inspired by a notation for juggling (Ross is also a juggler) in which each ai represents the airtime of 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. Alex was 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 some positive integer n. Note that S is nonempty because 1 ∈ S. Let s1 < s2 < · · · be the 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 numbers 1 + a1 , 2 + a2 , . . . , n + an are not equal to any si and are also members of the set {1, 2, . . . , n + 2015}. Hence {1, 2, . . . , n + 2015} contains at least n + 2016 different numbers, contradiction.  We claim that if b = |S| and if N is larger than all members of S, then the inequality posed in the problem statement is true. n Let n be any integer satisfying n ≥ N . We shall find bounds for j=1 (j + aj ) and  hence 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, n+bsince the n + b numbers in L are distinct, their sum is greater than or equal to j=1 j. Hence we have n 

b 

(j + aj ) +

j=1

⇒

j=1

n  j=1

sj ≥ aj ≥

n+b 

j

j=1

n+b 

j=n+1

j−

b 

sj

j=1

b(2n + b + 1) −s 2 n  b2 + b − s, (aj − b) ≥ 2 j=1 =

⇒ where s =

b

j=1

(1)

sj .

For the upper bound, observe that s1 , s2 , . . . , sb are b members belonging to the set T = {1, 2, . . . , n + 1}. The remaining n + 1 − b members of T must be of the form j + aj where j ≤ n, and so are in L. The sum of these n + 1 − b numbers is exactly n+1  j=1

j−

81 166

b  j=1

sb .

All together there are exactly n numbers of the form j + aj in L and so far we have accounted for n + 1 − b of them.

Consider the remaining b−1 numbers of the form j +aj which are in L. When listed in decreasing order, they can be no larger than b−1n+2015, n+2014, . . . , n+2015−b+2, respectively. Hence their sum is at most j=1 (n + 2016 − j). Thus n  j=1

(j + aj ) ≤

⇒ ⇒

n  j=1

n  j=1

n+1  j=1

j−

aj ≤ n + 1 +

(aj − b) ≤

b 

sb +

j=1

b−1  j=1

(n + 2016 − j)

(b − 1)(2n + 4032 − b) −s 2

4033b − b2 − 4030 − s. 2

(2)

Summarising (1) and (2), we have established the following bounds for any n ≥ N . n

 4033b − b2 − 4030 b2 + b −s≤ − s. (aj − b) ≤ 2 2 j=1

(3)

Now let m, n be any two integers satisfying n > m ≥ N . Since also m ≥ N , (3) is also satisfied if n is replaced by m. Thus    n  n m           (aj − b) =  (aj − b) − (aj − b)      j=m+1

j=1

j=1

 2      b +b 4033b − b2 − 4030  −s − − s  ≤ 2 2 = (b − 1)(2015 − b) 2  (b − 1) + (2015 − b) ≤ 2

(AM–GM) 

= 10072 .

82 167

Solution 2 (A juggling interpretation solution by the authors Ross Atkins and Ivan Guo) Suppose you are juggling several balls using only one hand. At the ith second, if a ball lands in your hand, it is thrown up immediately. If no ball lands, you instead reach for an unused ball (that is, a ball that has not been thrown yet) and throw it up. In both cases, a ball is thrown so that it will stay in the air for ai seconds. The condition 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 must stay in the air for more than 2015 seconds, contradicting ai ≤ 2015. So b is finite and bounded by 2015.

Select N so that all b balls have been introduced by the N th second. For all i ≥ N , denote by Ti the total remaining airtime of the current balls, immediately after the ith throw is made. (That is, we calculate the remaining airtime for each current ball, 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. This gives 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 possible values of the total remaining airtime Ti . Since no two balls can land at the same time, the minimal value is 1 + 2 + · · · + b. On the other hand, the maximal value is 1 + 2015 + 2014 + · · · + (2015 − b + 2). (Note that there must be a ball with a remaining airtime of 1 since something must be caught and thrown every second.) Taking the difference between these two sums, we find that (4032 − b)(b − 1) (b + 2)(b − 1) − 2 2 = (2015 − b)(b − 1) ≤ 10072 (GM ≤ AM)

|Tn − Tm | ≤



as required.

83 168

INTERNATIONAL MATHEMATICAL OLYMPIAD RESULTS MARK DISTRIBUTION BY QUESTION 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

AUSTRALIAN SCORES AT THE IMO Name

Q1

Q2

Q3

Q4

Q5

Q6

Score

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.

169

Award

SOME COUNTRY TOTALS 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

170

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

1

Germany

102

0

2

3

0

Ghana

5

0

0

0

0

Greece

71

0

1

2

2

171

Country

Total

Gold

Silver

Bronze

HM

Hong 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

172

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

39

100

143

126

Total (104 teams, 577 contestants)

173

ORIGIN OF SOME QUESTIONS

ORIGIN OF SOME QUESTIONS

Senior Contest Question 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 Olympiad Questions 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 2015 Question 2 was composed by Angelo Di Pasquale and submitted by the AMOC Senior Problems Committee. International Mathematical Olympiad 2015 Question 6 was composed by Ross Atkins and Ivan Guo, and submitted by the AMOC Senior Problems Committee. Ivan provided the following background information on the problem. The original idea for this problem came about while Ross was reading the paper Positroid Varieties: Juggling and Geometry by Knutson, Lam and Speyer, in which the “excitation number” of a periodic juggling sequence was discovered. It seemed obvious that this was similar 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 time of 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 sequence converges to an integer b, which is the total number of balls. However, the usage of limits was inappropriate for an olympiad problem. We then came up with three more versions which involved bounding the partial sums. Eventually we settled on the most difficult version, with  the 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 between maximal and minimal possible “total air times”, after the introduction of all b balls. The final wording may be a little difficult for students who are unfamiliar with the construct: “there exists an N such that for all m > n > N ”. It is possible to solve the problem combinatorially without invoking any physical interpretations, juggling or otherwise. Furthermore, as demonstrated by some at the IMO, the problem can also be tackled using purely algebraic approaches. Overall, we are very happy with the problem 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, a member of the 1997 Australian IMO team and recipient of an IMO gold medal.

1

174

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 1979–1980 Mr P J O’Halloran Prof A L Blakers Dr J M Gani Prof B H Neumann Prof G E Wall Mr J L Williams

Canberra College of Advanced Education, ACT, Chair University of Western Australia Australian Mathematical Society, ACT, Australian National University, ACT, University of Sydney, NSW University of Sydney, NSW

Mathematics Challenge for Young Australians Problems Committee for Challenge Dr K McAvaney Victoria, (Director) Mr B Henry Victoria (Director) Prof P J O’Halloran University of Canberra, ACT Dr R A Bryce Australian National University, ACT Adj Prof M Clapper Australian Mathematics Trust, ACT Ms L Corcoran Australian Capital Territory Ms B Denney New South Wales Mr J Dowsey University of Melbourne, VIC Mr A R Edwards Department of Education, QLD Dr M Evans Scotch College, VIC Assoc Prof H Lausch Monash University, VIC Ms J McIntosh AMSI, VIC Mrs L Mottershead New South Wales Miss A Nakos Temple Christian College, SA Dr M Newman Australian National University, ACT Ms F Peel St Peter’s College, SA Dr I Roberts Northern Territory Ms T Shaw SCEGGS, NSW Ms K Sims New South Wales Dr A Storozhev Attorney General’s Department, ACT Prof P Taylor Australian Mathematics Trust, ACT Mrs A Thomas New South Wales Dr S Thornton South Australia Miss G Vardaro Wesley College, VIC Visiting members Prof E Barbeau University of Toronto, Canada Prof G Berzsenyi Rose Hulman Institute of Technology, USA Dr L Burjan Department of Education, Slovakia Dr V Burjan Institute for Educational Research, Slovakia Mrs A Ferguson Canada Prof B Ferguson University of Waterloo, Canada Dr D Fomin St Petersburg State University, Russia Prof F Holland University College, Ireland Dr A Liu University of Alberta, Canada Prof Q Zhonghu Academy of Science, China 175

9 years; 2006–2015; Member 1 year 2005–2006 17 years; 1990–2006; Member 9 years 2007–2015 5 years; 1990–1994 23 years; 1990–2012 3 years; 2013–2015 3 years; 1990–1992 6 years; 2010–2015 8 years; 1995–2002 26 years; 1990–2015 6 years; 1990–1995 24 years; 1990–2013 14 years; 2002–2015 24 years; 1992–2015 23 years; 1993–2015 26 years; 1990–2015 2 years; 1999, 2000 3 years; 2013–2015 3 years; 2013–2015 17 years; 1999–2015 22 years; 1994–2015 20 years; 1995–2014 18 years; 1990–2007 18 years; 1998–2015 22 years: 1993–2006, 2008–2015 1991, 1993, 1993 1993 1992 1992, 1994 1994 1995, 1995

2004, 2008 2002

2005

2006, 2009

Dr A Gardiner University of Birmingham, United Kingdom Prof P H Cheung Hong Kong Prof R Dunkley University of Waterloo, Canada Dr S Shirali India Mr M Starck New Caledonia Dr R Geretschlager Austria Dr A Soifer United States of America Prof M Falk de Losada Colombia Mr H Groves United Kingdom Prof J Tabov Bulgaria Prof A Andzans Latvia Prof Dr H-D Gronau University of Rostock, Germany Prof J Webb University of Cape Town, South Africa Mr A Parris Lynwood High School, New Zealand Dr A McBride University of Strathclyde, United Kingdom Prof P Vaderlind Stockholm University, Sweden Prof A Jobbings United Kingdom Assoc Prof D Wells United States of America

Moderators for Challenge Mr W Akhurst Ms N Andrews Prof E Barbeau Mr R Blackman Ms J Breidahl Ms S Brink Prof J C Burns Mr A. Canning Mrs F Cannon Mr J Carty Dr E Casling Mr B Darcy Ms B Denney Mr J Dowsey Br K Friel Dr D Fomin Mrs P Forster Mr T Freiberg Mr W Galvin Mr M Gardner Ms P Graham Mr B Harridge Ms J Hartnett Mr G Harvey Ms I Hill Ms N Hill Dr N Hoffman Prof F Holland Mr D Jones Ms R Jorgenson Assoc Prof H Lausch Mr J Lawson Mr R Longmuir Ms K McAsey

New South Wales ACER, Camberwell, VIC University of Toronto, Canada Victoria St Paul’s Woodleigh, VIC Glen Iris, VIC Australian Defence Force Academy, ACT Queensland New South Wales ACT Department of Education, ACT Australian Capital Territory South Australia New South Wales Victoria Trinity Catholic College, NSW St Petersburg University, Russia Penrhos College, WA Queensland University of Newcastle, NSW North Virginia, USA Tasmania University of Melbourne, VIC Queensland Australian Capital Territory South Australia Victoria Edith Cowan University, WA University College, Ireland Coff’s Harbour High School, NSW Australian Capital Territory Victoria St Pius X School, NSW China Victoria

176

1996 1997 1997 1998 1999 1999, 2000 2000 2001 2001, 2002 2003 2003, 2004 2007 2009, 2014 2015

2013

2010

2011

2012

Moderators for Challenge continued Dr K McAvaney Ms J McIntosh Ms N McKinnon Ms T McNamara Mr G Meiklejohn Mr M O’Connor Mr J Oliver Mr S Palmer Dr W Palmer Mr G Pointer Prof H Reiter Mr M Richardson Mr G Samson Mr J Sattler Mr A Saunder Mr W Scott Mr R Shaw Ms T Shaw Dr B Sims Dr H Sims Ms K Sims Prof J Smit Mrs M Spandler Mr G Spyker Ms C Stanley Dr E Strzelecki Mr P Swain Dr P Swedosh Prof J Tabov Mrs A Thomas Ms K Trudgian Prof J Webb Ms J Vincent

Victoria AMSI, VIC Victoria Victoria Queensland School Curriculum Council, QLD AMSI, VIC Northern Territory New South Wales University of Sydney, NSW South Australia University of North Carolina, USA Yarraville Primary School, VIC Nedlands Primary School, WA Parramatta High School, NSW Victoria Seven Hills West Public School, NSW Hale School, WA New South Wales University of Newcastle, NSW Victoria New South Wales The Netherlands New South Wales Curtin University, WA Queensland Monash University, VIC Ivanhoe Girls Grammar School, VIC The King David School, VIC Academy of Sciences, Bulgaria New South Wales Queensland University of Capetown, South Africa 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, NSW Dr M Evans Scotch College, VIC Mr K Hamann South Australia Assoc Prof H Lausch Monash University, VIC Dr 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, ACT Mr J Carty Merici College, ACT Dr H Gastineau-Hill University of Sydney, NSW Mr B Henry Victoria Assoc Prof H Lausch Monash University, VIC Prof P O’Halloran University of Canberra, ACT 177

Dr A Storozhev Australian Mathematics Trust, ACT Euler Development Team (1992–1995) Dr M Evans Scotch College, VIC (Editor) Mr B Henry Victoria (Editor) Mr L Doolan Melbourne Grammar School, VIC Mr K Hamann South Australia Assoc Prof H Lausch Monash University, VIC Prof P O’Halloran University of Canberra, ACT Mrs A Thomas Meriden School, NSW Gauss Development Team (1993–1995) Dr M Evans Scotch College, VIC (Editor) Mr B Henry Victoria (Editor) Mr W Atkins University of Canberra, ACT Mr G Ball University of Sydney, NSW Prof J Burns Australian Defence Force Academy, ACT Mr L Doolan Melbourne Grammar School, VIC Mr A Edwards Mildura High School, VIC Mr N Gale Hornby High School, New Zealand Dr N Hoffman Edith Cowan University, WA Prof P O’Halloran University of Canberra, ACT Dr W Pender Sydney Grammar School, NSW Mr R Vardas Dulwich Hill High School, NSW Noether Development Team (1994–1995) Dr M Evans Scotch College, VIC (Editor) Dr A Storozhev Australian Mathematics Trust, ACT (Editor) Mr B Henry Victoria Dr D Fomin St Petersburg University, Russia Mr G Harvey New South Wales Newton Development Team (2001–2002) Mr B Henry Victoria (Editor) Mr J Dowsey University of Melbourne, VIC Mrs L Mottershead New South Wales Ms G Vardaro Annesley College, SA Ms A Nakos Temple Christian College, SA Mrs A Thomas New South Wales Dirichlet Development Team (2001–2003) Mr B Henry Victoria (Editor) Mr A Edwards Ormiston College, QLD Ms A Nakos Temple Christian College, SA Mrs L Mottershead New South Wales

Australian Mathematical Olympiad Committee The Australian Mathematical Olympiad Committee was founded at a meeting of the Australian Academy of Science at its meeting of 2–3 April 1980. * denotes Executive Position Chair* Prof B H Neumann Prof G B Preston Prof A P Street Prof C Praeger Deputy Chair* Prof P J O’Halloran Prof A P Street Prof C Praeger,

Australian National University, ACT Monash University, VIC University of Queensland University of Western Australia

7 years; 1980–1986 10 years; 1986–1995 6 years; 1996–2001 14 years; 2002–2015

University of Canberra, ACT University of Queensland University of Western Australia

15 years; 1980–1994 1 year; 1995 6 years; 1996–2001

178

Assoc Prof D Hunt Executive Director* Prof P J O’Halloran Prof P J Taylor Adj Prof M G Clapper

University of New South Wales

14 years; 2002–2015

University of Canberra, ACT University of Canberra, ACT University of Canberra, ACT

15 years; 1980–1994 18 years; 1994–2012 3 years; 2013–2015

Australian Defence Force Academy, ACT Victorian Chamber of Mines, VIC

9 years; 1980–1988 4 years; 1989–1992 6 years; 1993–1998

Australian Defence Force Academy, ACT University of Canberra, ACT CPA Monash University, VIC Australian National University, ACT The King David School, VIC

8 2 5 8 6 7

Secretary Prof J C Burns Vacant Mrs K Doolan

Treasurer* Prof J C Burns Prof P J O’Halloran Ms J Downes Dr P Edwards Prof M Newman Dr P Swedosh

years; years; years; years; years; years;

1981–1988 1989–1990 1991–1995 1995–2002 2003–2008 2009–2015

Director of Mathematics Challenge for Young Australians* Mr J B Henry Dr K McAvaney

Deakin University, VIC Deakin University, VIC

17 years; 1990–2006 10 years; 2006–2015

Chair, Senior Problems Committee Prof B C Rennie Mr J L Williams Assoc Prof H Lausch Dr N Do

James Cook University, QLD University of Sydney, NSW Monash University, VIC Monash University, VIC

1 year; 1980 6 years; 1981–1986 27 years; 1987–2013 2 years; 2014–2015

Director of Training* Mr J L Williams Mr G Ball Dr D Paget Dr M Evans Assoc Prof D Hunt Dr A Di Pasquale

University of Sydney, NSW University of Sydney, NSW University of Tasmania Scotch College, VIC University of New South Wales University of Melbourne, VIC

7 years; 1980–1986 3 years; 1987–1989 6 years; 1990–1995 3 months; 1995 5 years; 1996–2000 15 years; 2001–2015

University of Sydney, NSW University of New South Wales

5 years; 1981–1985 9 years; 1986, 1989, 1990, 1996–

Monash University, VIC University of Tasmania University of Melbourne, VIC University of New South Wales

2 years; 1987, 1988 5 years; 1991–1995 13 years; 2002–2010, 2012–2015 1 year; 2011

Team Leader Mr J L Williams Assoc Prof D Hunt 2001 Dr E Strzelecki Dr D Paget Dr A Di Pasquale Dr I Guo

Deputy Team Leader Prof G Szekeres Mr G Ball Dr D Paget Dr J Graham Dr M Evans Dr A Di Pasquale Dr D Mathews Dr N Do Dr I Guo

University of New South Wales University of Sydney, NSW University of Tasmania University of Sydney, NSW Scotch College, VIC University of Melbourne, VIC University of Melbourne, VIC University of Melbourne, VIC University of New South Wales 179

2 7 1 3 3 5 3 4 4

years; 1981–1982 years; 1983–1989 year; 1990 years; 1991–1993 years; 1994–1996 years; 1997–2001 years; 2002–2004 years; 2005–2008 years; 2009–10, 2012–2013

Mr G White Mr A Elvey Price

University of Sydney, NSW Melbourne University, VIC

1 year; 2011 2 years; 2014–2015

State Directors Australian Capital Territory Prof M Newman Australian National University Mr D Thorpe ACT Department of Education Dr R A Bryce Australian National University Mr R Welsh Canberra Grammar School Mrs J Kain Canberra Grammar School Mr J Carty ACT Department of Education Mr J Hassall Burgmann Anglican School Dr C Wetherell Radford College New South Wales Dr M Hirschhorn University of New South Wales Mr G Ball University of Sydney, NSW Dr W Palmer University of Sydney, NSW Northern Territory Dr I Roberts Charles Darwin University Queensland Dr N H Williams University of Queensland Dr G Carter Queensland University of Technology Dr V Scharaschkin University of Queensland Dr A Offer Queensland South Australia/Northern Territory Mr K Hamann SA Department of Education 2013 Mr V Treilibs SA Department of Education Dr M Peake Adelaide Dr D Martin Adelaide Tasmania Mr J Kelly Tasmanian Department of Education Dr D Paget University of Tasmania Mr W Evers St Michael’s Collegiate School Dr K Dharmadasa University of Tasmania Victoria Dr D Holton University of Melbourne Mr B Harridge Melbourne High School Ms J Downes CPA Mr L Doolan Melbourne Grammar School Dr P Swedosh The King David School Western Australia Dr N Hoffman WA Department of Education Assoc Prof P Schultz University of Western Australia 1996–1999 Assoc Prof W Bloom Murdoch University Dr E Stoyanova WA Department of Education Dr G Gamble University of Western Australia

1 year; 1980 2 years; 1981–1982 7 years; 1983–1989 1 year; 1990 5 years; 1991–1995 17 years; 1995–2011 2 years; 2012–2013 2 years; 2014–2015 1 year; 1980 16 years; 1981–1996 19 years; 1997–2015 2 years; 2014–2015 21 years; 1980–2000 10 years; 2001–2010 4 years; 2011–2014 1 year; 2015 19 years; 1980–1982, 1991–2005, 8 years; 1983–1990 8 years; 2006–2013 2 years; 2014–2015 8 years; 1980–1987 8 years; 1988–1995 9 years; 1995–2003 12 years; 2004–2015 3 years; 1980–1982 1 year; 1982 6 years; 1983–1988 9 years; 1989–1998 18 years; 1998–2015 3 years; 1980–1982 14 years; 1983–1988, 1991–1994, 2 years; 1989–1990 7 years; 1995, 2000–2005 10 years; 2006–2015

Editor Prof P J O’Halloran Dr A W Plank Dr A Storozhev Editorial Consultant Dr O Yevdokimov

University of Canberra, ACT University of Southern Queensland Australian Mathematics Trust, ACT

1 year; 1983 11 years; 1984–1994 15 years; 1994–2008

University of Southern Queensland

7 years; 2009–2015

180

Other Members of AMOC (showing organisations represented where applicable) Mr W J Atkins Dr S Britton Prof G Brown Dr R A Bryce Mr G Cristofani Ms L Davis Dr W Franzsen Dr J Gani Assoc Prof T Gagen Ms P Gould Prof G M Kelly Prof R B Mitchell Ms Anna Nakos Mr S Neal Prof M Newman Prof R B Potts Mr H Reeves Mr N Reid Mr R Smith Prof P J Taylor Prof N S Trudinger Assoc Prof I F Vivian Dr M W White

Australian Mathematics Foundation University of Sydney, NSW Australian Academy of Science, ACT Australian Mathematical Society, ACT Mathematics Challenge for Young Australians Department of Education and Training IBM Australia Australian Catholic University, ACT Australian Mathematical Society, ACT ANU AAMT Summer School Department of Education and Training University of Sydney, NSW University of Canberra, ACT Mathematics Challenge for Young Australians Department of Education and Training Australian National University, ACT Mathematics Challenge for Young Australians (Treasurer during the interim) University of Adelaide, SA Australian Association of Maths Teachers Australian Mathematics Foundation IBM Australia Telecom Australia Australian Mathematics Foundation Australian Mathematical Society, ACT University of Canberra, ACT IBM Australia

18 years; 1995–2012 8 years; 1990–1998 10 years; 1980, 1986–1994 10 years; 1991–1998 13 years; 1999–2012 2 years; 1993–1994 4 years; 1991–1994 9 years; 1990–1998 1980 6 years; 1993–1998 2 years; 1995–1996 6 years; 1982–1987 5 years; 1991–1995 13 years; 2003–2015 4 years; 1990–1993 15 years; 1986–1998 10 years; 1999–2002, 2009–2014 1 year; 1980 11 years; 1988–1998 2014–2015 3 years; 1988–1990 5 years; 1990–1994 6 years; 1990–1994, 2013 3 years; 1986–1988 1 year; 1990 9 years; 1980–1988

Associate Membership (inaugurated in 2000) Ms S Britton Dr M Evans Dr W Franzsen Prof T Gagen Mr H Reeves Mr G Ball



16 16 16 16 16 16

years; years; years; years; years; years;

2000–2015 2000–2015 2000–2015 2000–2015 2000–2015 2000–2015

AMOC Senior Problems Committee Current members Dr N Do M Clapper Dr A Di Pasquale Dr M Evans Dr I Guo Dr J Kupka Dr K McAvaney Dr D Mathews Dr A Offer Dr C Rao Dr B B Saad Dr J Simpson Dr I Wanless

Monash University, VIC (Chair) (member) Australian Mathematics Trust University of Melbourne, VIC Australian Mathematical Sciences Institute, VIC University of Sydney, NSW Monash University, VIC Deakin University, VIC Monash University, VIC Queensland IBM Australia Monash University, VIC Curtin University, WA Monash University, VIC

181

2 years; 2014–2015 11 years; 2003–2013 3 years; 2013-2015 15 years; 2001–2015 26 years; 1990–2015 8 years; 2008–2015 13 years; 2003–2015 20 years; 1996–2015 15 years; 2001–2015 4 years; 2012–2015 16 years; 2000–2015 22 years; 1994–2015 17 years; 1999–2015 16 years; 2000–2015

Previous members Mr G Ball Mr M Brazil Dr M S Brooks Dr G Carter Dr J Graham Dr M Herzberg Assoc Prof D Hunt Dr L Kovacs Assoc Prof H Lausch Dr D Paget Prof P Schultz Dr L Stoyanov Dr E Strzelecki Dr E Szekeres Prof G Szekeres Em Prof P J Taylor Dr N H Williams

University of Sydney, NSW LaTrobe University, VIC University of Canberra, ACT Queensland University of Technology University of Sydney, NSW Telecom Australia University of New South Wales Australian National University, ACT Monash University, VIC (Chair) (member) University of Tasmania University of Western Australia University of Western Australia Monash University, VIC University of New South Wales University of New South Wales Australian Capital Territory University of Queensland

16 years; 1982–1997 5 years; 1990–1994 8 years; 1983–1990 10 years; 2001–2010 1 year; 1992 1 year; 1990 29 years; 1986–2014 5 years; 1981–1985 27 years; 1987–2013 2 years; 2014-2015 7 years; 1989–1995 8 years; 1993–2000 5 years; 2001–2005 5 years; 1986–1990 7 years; 1981–1987 7 years; 1981–1987 1 year; 2013 20 years; 1981–2000

Mathematics School of Excellence Dr S Britton Mr L Doolan Mr W Franzsen Dr D Paget Dr M Evans Assoc Prof D Hunt Dr A Di Pasquale

University of Sydney, NSW (Coordinator) Melbourne Grammar, VIC (Coordinator) Australian Catholic University, ACT (Coordinator) University of Tasmania (Director) Scotch College, VIC University of New South Wales (Director) University of Melbourne, VIC (Director)

2 years; 1990–1991 6 years; 1992, 1993–1997 2 years; 1990–1991 5 years; 1990–1994 1 year; 1995 4 years; 1996–1999 16 years; 2000–2015

International Mathematical Olympiad Selection School Mr J L Williams Mr G Ball Mr L Doolan Dr S Britton Mr W Franzsen Dr D Paget Assoc Prof D Hunt Dr A Di Pasquale

University of Sydney, NSW (Director) University of Sydney, NSW (Director) Melbourne Grammar, VIC (Coordinator) University of Sydney, NSW (Coordinator) Australian Catholic University, ACT (Coordinator) University of Tasmania (Director) University of New South Wales (Director) University of Melbourne, VIC (Director)

182

2 years; 1982–1983 6 years; 1984–1989 3 years; 1989–1991 7 years; 1992–1998 8 years; 1992–1996, 1999–2001 6 years; 1990–1995 5 years; 1996–2000 15 years; 2001–2015