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MATHEMATICS CONTESTS THE AUSTRALIAN SCENE 2015 PART 1: MATHEMATICS CHALLENGE FOR YOUNG AUSTRALIANS KL McAvaney A u s t r a l i a n M at h e mat i c a l O ly m p i a d C omm i t t e e A d e pa r t m e n t o f t h e A u s t r a l i a n M at h e mat i c s T r u s t 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 2 Mathematics Contests The Australian Scene 2015 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. Part 1: Mathematics Challenge for Young Australians 3 ACKNOWLEDGEMENTS The Australian Mathematical Olympiad Committee (AMOC) sincerely thanks all sponsors, teachers, mathematicians and others who have contributed in one way or another to the continued success of its activities. The editors sincerely thank those who have assisted in the compilation of this book, in particular the students who have provided solutions to the 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. 4 Mathematics Contests The Australian Scene 2015 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 see, once again, an Australian-authored question on the paper. This year, it was 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. This year they adopted a specific policy of tackling very hard questions in training, which was daunting for team members at times but seems to have paid off. We congratulate them again on their success. 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 AMOC 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 Norm 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 Part 1: Mathematics Challenge for Young Australians 5 6 Mathematics Contests The Australian Scene 2015 CONTENTS Support for the Australian Mathematical Olympiad Committee Training Program 3 Acknowledgements 4 Preface5 Mathematics Challenge for Young Australians 8 Membership of MCYA Committees 10 Challenge Problems – Middle Primary 12 Challenge Problems – Upper Primary 14 Challenge Problems – Junior 16 Challenge Problems – Intermediate 19 Challenge Solutions – Middle Primary 21 Challenge Solutions – Upper Primary 24 Challenge Solutions – Junior 28 Challenge Solutions – Intermediate 36 Challenge Statistics – Middle Primary 46 Challenge Statistics – Upper Primary 47 Challenge Statistics – Junior 48 Challenge Statistics – Intermediate 49 Australian Intermediate Mathematics Olympiad 50 Australian Intermediate Mathematics Olympiad Solutions 52 Australian Intermediate Mathematics Olympiad Statistics 60 Australian Intermediate Mathematics Olympiad Results 61 Honour Roll 65 Part 1: Mathematics Challenge for Young Australians 7 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 8 Mathematics Contests The Australian Scene 2015 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. Part 1: Mathematics Challenge for Young Australians 9 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 10 Mathematics Contests The Australian Scene 2015 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 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 Part 1: Mathematics Challenge for Young Australians 11 CHALLENGE PROBLEMS – MIDDLE PRIMARY 2015 Challenge Problems - Middle Primary Students may work on each of these four problems in groups of up to three, but must write their solutions individually. MP1 e-Numbers The digits on many electronic devices look like this: When numbers made of these digits are rotated 180◦ ( ), some of them become numbers and some don’t. 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. 12 Mathematics Contests The Australian Scene 2015 1 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. 2 Part 1: Mathematics Challenge for Young Australians 13 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: 14 Mathematics Contests The Australian Scene 2015 3 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. 4 Part 1: Mathematics Challenge for Young Australians 15 CHALLENGE PROBLEMS 2015 Challenge Problems–- JUNIOR 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 4 5 1 2 1 2 3 4 5 6 7 8 9 6 9 • 7 8 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? 16 Mathematics Contests The Australian Scene 2015 5 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 • 3k • Start/Finish 2 km m • Base of Mt KJ 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? 6 Part 1: Mathematics Challenge for Young Australians 17 J6 Tessellating Hexagons Will read in his favourite maths book that this hexagon tessellates the plane. D E C F A B To see if this was true, Will made 16 copies of the hexagon and glued them edge-to-edge onto a piece of paper, with no overlaps and no gaps. a Cut out 16 hexagons from the worksheet and glue them as Will might have done so they entirely cover the dashed rectangle. b Show that your block of 16 hexagons can be translated indefinitely to tessellate the entire plane. c Find all points on the hexagon above that are points of symmetry of your tessellation. d In your tessellation, select four hexagons that form one connected block which will tessellate the plane by translation only. Indicate three such blocks on your tessellation that are not identical. 18 Mathematics Contests The Australian Scene 2015 7 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 T H H H T H T T T T T H T T T H 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. 8 Part 1: Mathematics Challenge for Young Australians 19 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 P Q R S U V 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? 20 Mathematics Contests The Australian Scene 2015 9 CHALLENGE SOLUTIONS PRIMARY 2015 Challenge Solutions–- MIDDLE 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 1 10 2 18 3 26 4 34 The number of squares from one border to the next appears to increase by 8. We now show this rule continues to apply. 10 Part 1: Mathematics Challenge for Young Australians 21 Except for the corner squares, each square on the previous border corresponds to a square in the new border. Each of the corner squares C on the previous border corresponds to a corner square C in the new border. In addition, there are two new squares next to each corner. new border C C previous border C previous border new border C new border previous border previous border C C new border C C 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 1 10 2 18 3 26 4 34 5 42 6 50 7 58 8 66 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. 22 Mathematics Contests The Australian Scene 2015 11 9 74 10 82 11 90 c For each of these arrangements, any rotation or reflection is acceptable. d Each arrangement of six eggs in a 2 × 4 carton has two empty positions. By systematically locating the two 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. 12 Part 1: Mathematics Challenge for Young Australians 23 CHALLENGE SOLUTIONS PRIMARY 2015 Challenge Solutions–- UPPER 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. 24 Mathematics Contests The Australian Scene 2015 13 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. 14 Part 1: Mathematics Challenge for Young Australians 25 b i Chloe’s PB time is 70 seconds and Sally’s is 53 seconds. So the total of all six PB times is 262 + 70 + 53 = 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. 26 Mathematics Contests The Australian Scene 2015 15 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. 16 Part 1: Mathematics Challenge for Young Australians 27 CHALLENGE SOLUTIONS 2015 Challenge Solutions –- JUNIOR Junior J1 Quirky Quadrilaterals a • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Kite Rhombus Another kite b • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • I=4 I=9 c • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • I = 12 d If I is an odd number we can surround a column of I dots with a rhombus whose vertices are immediately above, below and either side of the middle of the column as the following examples show. Thus one diagonal of the rhombus lies on the line through the column of I dots and the other diagonal lies on the line perpendicular to the first diagonal and passing through the middle dot. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • I=1 28 • Mathematics Contests The Australian Scene 2015 I=3 17 I=5 If I is an even number we can surround a south-west/north-east diagonal of I dots with a rhombus whose vertices are immediately south-west, north-east, and either side of the middle of the diagonal as the following examples show. Thus one diagonal of the rhombus lies on the line through the I dots and the other diagonal lies on the line perpendicular to the first diagonal and bisecting the distance between the two middle dots. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • I =2 I =4 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • I=6 These rhombuses are not unique. J2 Indim Integers a Since 1 is a factor of every integer, no tile remains if 1 is spun. If 2 is spun, then all remaining tiles have an odd number. Hence any 3-digit number made from the remaining 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. 18 Part 1: Mathematics Challenge for Young Australians 29 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. 30 Mathematics Contests The Australian Scene 2015 19 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 Part 1: Mathematics Challenge for Young Australians 31 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. 32 Mathematics Contests The Australian Scene 2015 21 3 9 + 3r = 30 60 = 12 . 30 60 = 12 . 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 Part 1: Mathematics Challenge for Young Australians 33 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 . 34 Mathematics Contests The Australian Scene 2015 23 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 Part 1: Mathematics Challenge for Young Australians 35 CHALLENGE SOLUTIONS 2015 Challenge Solutions– -INTERMEDIATE 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. 36 Mathematics Contests The Australian Scene 2015 25 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 Part 1: Mathematics Challenge for Young Australians 37 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. 38 Mathematics Contests The Australian Scene 2015 27 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 Part 1: Mathematics Challenge for Young Australians 39 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. 40 Mathematics Contests The Australian Scene 2015 29 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 Part 1: Mathematics Challenge for Young Australians 41 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 = 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 = = = = = 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 = 42 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 = 5 24 . Mathematics Contests The Australian Scene 2015 31 3 36 1 8 = 4 3 1 12 1 2 × 1 2 = 23 . 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 Part 1: Mathematics Challenge for Young Australians 43 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. 44 Mathematics Contests The Australian Scene 2015 33 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 Part 1: Mathematics Challenge for Young Australians 45 CHALLENGE STATISTICS – MIDDLE PRIMARY Mean Score/School Year/Problem Mean Year Number of Students Problem Overall 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 Quazy Quilts 3 Egg Cartons 4 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. 46 Mathematics Contests The Australian Scene 2015 CHALLENGE STATISTICS – UPPER PRIMARY Mean Score/School Year/Problem Mean YEAR Number of Students Problem Overall 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 1 Rod Shapes 2 Egg Cartons 3 Seahorse Swimmers 4 Primelandia Money Did not attempt 1% 0% 1% 3% 0 3% 3% 5% 11% 1 40% 6% 14% 29% 2 24% 20% 20% 26% 3 21% 38% 29% 20% 4 11% 33% 31% 12% Mean 2.0 2.9 2.7 1.9 Discrimination Factor 0.5 0.4 0.6 0.6 Score Please note: The discrimination factor for a particular problem is calculated as follows: (1) The students are ranked in regard to their overall scores. (2) The mean score for the top 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean top score’. (3) The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean bottom score’. (4) The discrimination factor = mean top score – mean bottom score 4 Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is considered to be a good discriminator. Part 1: Mathematics Challenge for Young Australians 47 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 Quirky Quadrilaterals 2 Indim Integers 3 Primelandia Money 4 Condates 5 Jogging 6 Tessellating Hexagons Did not attempt 3% 3% 7% 8% 9% 23% 0 7% 6% 8% 15% 10% 28% 1 8% 17% 24% 14% 18% 20% 2 15% 23% 30% 19% 16% 13% 3 36% 26% 22% 31% 21% 10% 4 31% 26% 10% 13% 27% 5% Mean 2.8 2.5 2.0 2.1 2.4 1.3 Discrimination Factor 0.5 0.6 0.6 0.7 0.7 0.5 Score Please note: The discrimination factor for a particular problem is calculated as follows: (1) The students are ranked in regard to their overall scores. (2) The mean score for the top 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean top score’. (3) The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean bottom score’. (4) The discrimination factor = mean top score – mean bottom score 4 Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is considered to be a good discriminator. 48 Mathematics Contests The Australian Scene 2015 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 1 Indim Integers 2 Digital Sums 3 Coin Flips 4 Jogging 5 Folding Fractions 6 Crumbling Cubes Did not attempt 1% 3% 5% 3% 14% 9% 0 2% 6% 4% 4% 14% 11% 1 7% 11% 6% 9% 21% 7% 2 15% 26% 15% 12% 21% 21% 3 25% 16% 28% 22% 8% 25% 4 50% 38% 42% 49% 22% 27% Mean 3.2 2.7 3.0 3.1 2.0 2.6 Discrimination Factor 0.4 0.6 0.6 0.6 0.7 0.7 Score Please note: The discrimination factor for a particular problem is calculated as follows: (1) The students are ranked in regard to their overall scores. (2) The mean score for the top 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean top score’. (3) The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem including no attempts. Call this mean score the ‘mean bottom score’. (4) The discrimination factor = mean top score – mean bottom score 4 Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is considered to be a good discriminator. Part 1: Mathematics Challenge for Young Australians 49 2015 Australian Intermediate Mathematics Olympiad - Questions AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD 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.) 50 Mathematics Contests The Australian Scene 2015 1 [4 marks] 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] Part 1: Mathematics Challenge for Young Australians 51 AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD 2015 Australian Intermediate Mathematics Olympiad - Solutions 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 Method 1 3 × (1) − 2 × (2): w + d + t = 3 × 329 − 2 × 441 = 987 − 882 = 105. 52 Mathematics Contests The Australian Scene 2015 3 (1) (2) (3) 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 Part 1: Mathematics Challenge for Young Australians 53 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. 54 Mathematics Contests The Australian Scene 2015 5 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 Part 1: Mathematics Challenge for Young Australians 55 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 56 Mathematics Contests The Australian Scene 2015 C 7 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 Part 1: Mathematics Challenge for Young Australians 57 √ √ √ 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 58 Mathematics Contests The Australian Scene 2015 c b 120 X ◦ a Y C 9 ◦ ◦ 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 Part 1: Mathematics Challenge for Young Australians 59 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 60 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 Mathematics Contests The Australian Scene 2015 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 Chenxu Li Raffles Institution, Singapore 8 32 William Li Barker College, NSW 9 32 Part 1: Mathematics Challenge for Young Australians 61 NAME SCHOOL YEAR SCORE 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 Rong Dai Xiang Raffles Institution, Singapore 8 29 Hong Pei Goh Hwa Chong Institution, Singapore 10 29 62 Mathematics Contests The Australian Scene 2015 NAME SCHOOL YEAR SCORE 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 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 Part 1: Mathematics Challenge for Young Australians 63 NAME SCHOOL YEAR SCORE 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 64 Mathematics Contests The Australian Scene 2015 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. Mathematics Challenge for Young Australians Problems Committee for Challenge Dr K McAvaney Mr B Henry Prof P J O’Halloran Dr R A Bryce Adj Prof M Clapper Ms L Corcoran Ms B Denney Mr J Dowsey Mr A R Edwards Dr M Evans Assoc Prof H Lausch Ms J McIntosh Mrs L Mottershead Miss A Nakos Dr M Newman Ms F Peel Dr I Roberts Ms T Shaw Ms K Sims Dr A Storozhev Prof P Taylor Mrs A Thomas Dr S Thornton Miss G Vardaro Victoria, (Director) Member Victoria (Director) Member University of Canberra, ACT Australian National University, ACT Australian Mathematics Trust, ACT Australian Capital Territory New South Wales University of Melbourne, VIC Department of Education, QLD Scotch College, VIC Monash University, VIC AMSI, VIC New South Wales Temple Christian College, SA Australian National University, ACT St Peter’s College, SA Northern Territory SCEGGS, NSW New South Wales Attorney General’s Department, ACT Australian Mathematics Trust, ACT New South Wales South Australia Wesley College, VIC 9 years; 2006–2015; 1 year 2005–2006 17 years; 1990–2006; 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 University of Toronto, Canada Rose Hulman Institute of Technology, USA Department of Education, Slovakia Institute for Educational Research, Slovakia Canada University of Waterloo, Canada St Petersburg State University, Russia University College, Ireland University of Alberta, Canada Academy of Science, China University of Birmingham, United Kingdom Hong Kong University of Waterloo, Canada India New Caledonia Austria United States of America Colombia United Kingdom Bulgaria Latvia University of Rostock, Germany 1991, 1993, 1993 1993 1992 1992, 1994 1994 1995, 1995 1996 1997 1997 1998 1999 1999, 2000 2000 2001 2001, 2002 2003 Visiting members Prof E Barbeau Prof G Berzsenyi Dr L Burjan Dr V Burjan Mrs A Ferguson Prof B Ferguson Dr D Fomin Prof F Holland Dr A Liu Prof Q Zhonghu Dr A Gardiner Prof P H Cheung Prof R Dunkley Dr S Shirali Mr M Starck Dr R Geretschlager Dr A Soifer Prof M Falk de Losada Mr H Groves Prof J Tabov Prof A Andzans Prof Dr H-D Gronau 2004, 2008 2002 2005 2006, 2009 2013 2010 Part 1: Mathematics Challenge for Young Australians 65 Prof J Webb Mr A Parris Dr A McBride Prof P Vaderlind Prof A Jobbings Assoc Prof D Wells University of Cape Town, South Africa Lynwood High School, New Zealand University of Strathclyde, United Kingdom Stockholm University, Sweden United Kingdom 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 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 66 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 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 Mathematics Contests The Australian Scene 2015 2003, 2011 2004 2007 2009, 2012 2014 2015 Moderators for Challenge continued 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 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 Prof P O’Halloran Mr G Ball Dr M Evans Mr K Hamann Assoc Prof H Lausch Dr A Storozhev Victoria (Chairman) University of Canberra, ACT (Director) University of Sydney, NSW Scotch College, VIC South Australia Monash University, VIC 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 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 Part 1: Mathematics Challenge for Young Australians 67 Prof P O’Halloran Dr W Pender Mr R Vardas University of Canberra, ACT Sydney Grammar School, NSW 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 Mrs K Sims Chapman Primary School, ACT Mrs A Thomas New South Wales Australian Intermediate Mathematics Olympiad Committee Dr K McAvaney Adj Prof M Clapper Mr J Dowsey Dr M Evans Mr B Henry Assoc Prof H Lausch Mr R Longmuir 68 Victoria (Chair) Australian Mathematics Trust, ACT University of Melbourne, VIC AMSI, VIC Victoria (Chair) Member Monash University, VIC China Mathematics Contests The Australian Scene 2015 9 years; 2007–2015 2 years; 2014–2015 17 years; 1999–2015 17 years; 1999–2015 8 years; 1999–2006 9 years; 2007–2015 17 years; 1999–2015 2 years; 1999–2000