2019 AIMO Paper and Solutions W MS

   

2019 Australian Intermediate Mathematics Olympiad Australian Intermediate Mathematics Olympiad 2019 Questions Questions

1.   Gaston Gaston

and Jordon are two two budding chefs who unfortunatel unfortunately y always misread misread the cooking time required requi red for a recipe. recipe. For example, example, if the requir required ed cooking time is written written as 1:32, meani meaning ng 1 hour and 32 minutes, Jordon reads it as 132 minutes while Gaston reads it as 1.32 hours. For one particular recipe, the difference between Jordon’s and Gaston’s misread times is exactly 90 minutes. What is the actual cooking time in minutes? [2 marks]

2.   An

integer integer has 3 digits digits in base 10. When When it is in inter terpre preted ted in base 6 and multipl multiplied ied by 4, the result result is 2 more more than than 3 times times the same same numbe numberr when when it is in inter terpre preted ted in base base 7. What What is the largest integer, in base 10, that has this property? [2 marks]

3.   Let  ABCD  be

a square with side length 24. Let  P  be the point on side  AB  with  AP  = 8, and let  AC   and  DP   D P   intersect at   Q. Determine the area of triangle   CQD . [3 marks]

4.   Jenny Jenny

has a large large bucket bucket of 1000 1000 marbles, marbles, of equal size. size. At least least one is red and the rest rest are green.. She calculates green calculates that if she picks out two marbles marbles at random, random, the probabilit probability y that both are red is the same as the probability probability that they are different different colours. colours. How many many red marbles marbles are in the bucket? [3 marks]

5.   A

beetle starts at corner   A  of a square   ABCD. Each Each minute, the beetle move movess along exactly one side of the square from one corner of the squar squaree to anothe anotherr corner. For example, example, in the first two minutes, the beetle could have one of 4 walks: ABA, ADA, ABC, ADC. How many walks can the beetle have from corner  A  to corner  C  in 10 minutes? [3 marks]

6.   A

leaky boat already has some water on board and water is coming in at a constant rate. Several people are available to operate the manual pumps and, when they do, they all pump at the same rate. Starti Starting ng at a given given time, five people would take take 10 hours to pump the boat dry,, while 12 people would dry would take only 3 hours. How many many people are required to pump it dry in 2 hours from the given time? [4 marks]

PLEASE TURN OVER THE PAGE FOR QUESTIONS 7, 8, 9, AND 10

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7.   A

triangle  P QR  is to be constructed so that the perpendicular bisector of  P  P Q  cuts the side  QR at  N  and the line  P N  splits the angle  QP R  into two angles, not necessarily of integer degrees, in the the rati ratioo 1 : 22. 22. If    QP R  =  p  degrees, where   p  is an integer, find the maximum value of  p  p . [4 marks] the largest number which will divide   a5 + b5 − ab(a3 + b3 ) for   all  primes   a  and   b  greater than 5.

8.  Find

[4 marks]

9.   In

a large botanic garden, there is a long line of 999 regularly spaced positions where plants may grow. At any time, there is at most one plant in each position. In the first year, only the middle position has a plant. In the  n th year, for all  n ≥ 2, there is a plant at position  X  if and only if, in the previous year, the total number of plants in position   X  and either side of   X   is precisely one. Determine, with proof, the total number of plants in the 128th year. [5 marks]

10.   An

even number number of teams teams play in a round round robin tourname tournament nt defined defined as follo follows. ws. Each Each team team playss every other team exactly play exactly once. The matches matches are arranged in rounds. Each Each team plays plays exactly once in each round. We assume there are no draws. (a) Prove that at the end of round 2, the number of teams that have won two matches equals the number of teams that have lost two matches. (b) Find the least number of teams such that at the end of round 3, the number of teams that have won three matches may not equal the number of teams that have lost three matches. (c) Prove that, at the end of round 4, the number of teams that have won 4 matches equals the number of teams that have lost 4 matches, if and only if the number of teams that have won exactly 3 matches equals the number of teams that have lost exactly 3 matches. [5 marks]

Investigation 

Prove that for all even  k ≥ 4, there is a round robin tournament of   k k  teams in which the number of teams that have won  k − 1 matches does not equal the number of teams that have lost  k − 1 matches. [3

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bonus bonus marks marks]

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2019 Australian Intermediate Mathematics Olympiad Australian Intermediate Mathematics Olympiad 2019 Solutions Solutions

1.   Method Method 1

Let the cooking time be h be  h : : m  m,, where h where  h is  is the number of hours and m and  m is  is the number of minutes. Jordon’s time is 100h 100h  + m  +  m   minutes and Gaston’s time is   h  + (m/ m/100) 100) hours, or 60h 60h  + 3m/ m/55 minutes. minu tes. Hence the difference difference of 90 minutes minutes leads to the equation equation 40h 40h + (2 (2m/ m/5) 5) = 90, hence   1 100h 100 h +  + m  m =  = 225. So 0 ≤ 225 − 100 100h h ≤ 60, which implies h implies  h =  = 2. Therefore the actual cooking time in minutes is 2 × 60 + 25 =   145.

 

1

Method Metho d 2 

Let   m   be the number of minutes Let  minutes as read by Jordon. Then Gaston Gaston reads this as  as   m/ m/100 100 hours = 60m/ 60m/100 100 minutes. Hence the difference of 90 minutes leads to the equation 90 = m =  m − 3m/ m/55 = 2m/ 5, which gives m gives  m =  = 225.   1 m/5, Therefore the actual cooking time in minutes is 2 × 60 + 25 =   145.

 

1

2.  Let

the three-digit number be  abc  abc.. Since 4 × abc6  = 3 × abc7  + 2, we have 4(36a + 6b 4(36a 6b + c  + c)) = 3(49a 3(49a + 7b 7b +  + c  c)) + 2 144a 144 a + 24b 24b + 4c 4c  = 147a 147a + 21b 21b + 3c 3c + 2 3b + c  + c =  = 3a + 2

 

Hence 3 divides c divides  c − 2. Since c Since  c 0 and   P QR exists. QR  exists. So the maximum integer value of  p of  p  is   172.  

8.   Expanding,

1

rearranging, and factorising gives:

a5 + b5 − a4 b − b4 a = a(a4 − b4 ) − b(a4 − b4 ) = (a − b)( )(a a4 − b4 ) = (a − b)2 (a2 + b2 )( )(a a + b) 1

Let   b  be an Let  any y prime prime greater greater than 5. Choose Choose   a  to be some prime greater than 5 but not equal Therefore   b  cannot to  to   b. Then  Then   b  does not divide  a.  a . Hence  Hence   b   divides neither  neither   a − b   nor nor   a4 − b4 . Therefore 4 4 4 4 divide (a (a − b)(a )(a − b ). So, if a prime  prime   p  divides (a (a − b)( )(a a − b ) for all primes  a   and and   b  greater than 5, then p then  p  = b. So So   p  is 2, 3, or 5. As  As   a   and  and   b  are both odd, both   a + b   and and   a2 + b2 are even and 4 divides (a (a − b)2 . Henc Hencee 16 divide div idess the given given expres expressio sion, n, but we can do better than this. Each Each of   a   and   b  has the form − 4m + 1 or 4m 4m 1. If they have the same form, then 16 divides (a ( a − b)2 . If they have different forms, then 4 divides a divides  a + b. Either way, 32 divides the given expression. Since 3 divides neither   a   nor nor   b, each has the form 3m 3 m + 1 or 3m 3m − 1. If they have have the same same   1 form, then 3 divides a divides  a − b. If they have different forms, then 3 divides  a + b. Since 5 divides neither   a   nor nor   b, each has the form 5m 5 m ± 1 or 5m 5m ± 2. If both have have the the form 2 2 5m ± 1 or both have the form 5m 5 m ± 2, then 5 divides a divides  a − b . If one has the form 5m 5m ± 1 and 4 2 2 4 the other has the form 5m 5m ± 2, then 5 divides a divides  a + b . Hence 5 divides a divides  a − b . So the largest guaranteed divisor of the given expression is at least 32 × 3 × 5 = 480.

 

1

Substituting  b = 7 and a Substituting b and  a = 13 in the last factorisation gives (36)(218)(20) = 2 5 × 32 × 5 × 109. 6 Hence neither 2 nor 52 is a guaranteed factor of the given expression. Substituting   b  = 11 and   a  = 13 in the last factorisation gives (4)(290)(24) = 26 × 3 × 5 × 29. Substituting  Hence 32 is not a guaranteed factor of the given expression. So the largest guaranteed divisor is   480.

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9.  With

1 representing a plant and 0 representing a vacant position, the following list of binary sequences seque nces illustrate illustrate the plant patterns in the first 9 years. Vacant acant p positio ositions ns beyond the last plant in either direction are ignored. Year 1  

2

 

3

 

4 5

6 7 8 9

  1   11   100

Plant Pattern 1 111 10001 1110111 100000001 1110000011 0001000100 1011101110 00 00 00 00 00

1 01 111 00 01

 

1

Define the  length  of   of a binary sequence to be the number of digits it contains. For or n  n =  = 1, 2, 3, we make two observations. (1) In year 2n the plant pattern is 11101110 . . . 1110111 . . . 01110111 and has length 2n+1 − 1. (2) In year 2n + 1 the plant pattern is 100 . . . 0 . . . 001 and has length 2 n+1 + 1.   1 We use induction to prove these statements are true for  al l  n  n

≥

1.

First assume statements (1) and (2) are true for  n  equal to some positive integer k integer  k.. k

k

 

1 k

In year 2 , the length of the plant pattern either side of the centre 0 is 2 − 1. In year 2 + 1 there are precisely 2k − 1 vacant positions between each plant and the centre vacant position. So by assumption, in the year 2 k + 2k = 2 k+1 the plant pattern will be two copies of the plant pattern for year 2k separ separated ated by precisely precisely one 0. Thus Thus in year year 2 k+1 the plant pattern will be k +1 1110 . . . 1110111 . . . 0111 and have length 2(2 − 1) + 1 = 2k+2 − 1. The Then, n, in year 2k+1 + 1, the plant pattern will be 100 . . . 0 . . . 001 and have length 2k+1 + 1.   1 Thus statements (1) and (2) are true for  n =  n  = k  k +  + 1, and therefore true for all  n

≥

1.

Since 128 = 27 , the plant pattern for year 128 is 11101110 . . . 1110111 . . . 01110111 and has length 28 − 1. If we place a 0 at the right-hand right-hand end of this sequence, sequence, we will have have 28 digits of  which exactly 28 /4 will be 0. So the number of plants in year 128 will be 2 8 − 28 /4 = 28 − 26 =   1 26 (4 − 1) =   192.

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10.   (a)

Method Metho d 1

After two two rounds, each team has two wins, wins, or two two losses, or one win and one loss. Since the total number of wins equals the total number of losses and the last category of teams accounts for an equal number of wins and losses, the number of teams with two wins must equal the number of teams with two losses.   1 Method Metho d 2 

At the end of round 2, let  r i  equal the number of teams that have won   i   matches. Then   r1  also equals the number of teams that have lost one match and   r0  equals the number of teams that havee lost two matches. hav matches. Since the total number number of wins equals equals the total number number of losses, losses, we have 2r2  +  r1  =  r 1  + 2 r0 . Hence   r2  =  r 0 .   1 Method Metho d 3 

If a team scores 1 if it wins and −1 if it loses, then the total of all team scores after each round is zero. Hence the total of all team scores after any number of rounds is zero. After two rounds a team may have   −2, 0 or 2 points. Hence the number of teams with 2 points must equal the number of teams with   −2 points.   1 Comment 

The statement statement in part (a) is not necessarily necessarily true if dra drawn wn matches are allowed. allowed. For example, with 4 teams, one team may win two matches and the other 2 matches may be drawn. (b) If there are three rounds, then there are at least 4 teams. In the following rounds there are 4 teams denoted 1, 2, 3, 4, and   a → b  indicates team   a  defeated team   b. Round 1: Round 2: Round 3:

1 → 2, 1 → 3, 1 → 4,

3→4 4→2 2→3

 

1

Thus one team Thus team wins wins 3 match matches es and no team loses 3 match matches. es. So the lea least st number number of teams teams such that at the end of round 3 the number of teams that have won three matches does not   1 necessarily equal the number of teams that have lost three matches is   4. (c) At the end of round 4, let   ri  equal the number of teams that have won   i  matches and let   si equal the number of teams that have lost   i  matches. Since the total number of wins equals the   1 total number of losses, we have 4 r4  + 3 r3  + 2 r2  +  r 1  = 4s4  + 3 s3  + 2 s2  +  s1 . Since a team wins   i  matches if and only if it loses 4 − i   matches, we have   r1   =   s3 ,   r2   =   s2 , r3   =   s1 , and 4r4  + 3r3  + 2r2  +  s 3   = 4s4  + 3s3  + 2r2  +  r 3 . He Henc ncee 4r4  + 2r3   = 4s4  + 2s3 . So   1 r4  =  s 4  if and only if   r3  =  s 3 .

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Investigation  Method 1

For  k  teams, we generalise the round in part (b) above in which team 1 defeats all other teams and every every team has at least one win. In each column column except the first, the winning and losing losing teams cycle through the sequence 2, 3, 4,  . . . ,  k . For a tidier table, we let  k  = 2n.   1 bonus mark mark

R d 1: R d 2: R d 3: .. .

1 → 2, 1 → 3, 1 → 4,

3 → 2n, 4 → 2, 5 → 3,

4 → 2n − 1,     5 → 2n,   6 → 2,

Rd  n − 1: Rd  n : Rd  n  + 1: .. .

  n + 1 1 → n, 1 → n + 1,   n + 2 1 → n + 2,   n + 3

Rd 2n − 3: Rd 2n − 2:

1 → 2n − 2, 1 → 2n − 1,

2n − 1 → 2n − 3, 2n → 2n − 2,

2n → 2n − 4,   . . . ,   n − 2 → n − 1 2 → 2n − 3,   . . . ,   n − 1 → n

Rd 2n − 1:

1 → 2n,

2 → 2n − 1,

3 → 2n − 2,

n − 1,   n,     → n + 1,

→

→

n + 2 n + 3 n + 4

. . .  ,   n + 1 . . .  ,   n + 2 . . .  ,   n + 3

n − 2,   . . .  , n − 1,   . . .  ,   . . .  , → n,

→

→

 

n + 2 → n + 3 → n  + 4

→

2n − 1 → 2n 2n → 2 2→3

. . .  ,   n

→

n + 1

It is straightforward to check that each team plays exactly once in each round and each team playss every other team exactly play exactly once in the tournamen tournament. t. From Column 1, team 1 wins all its matches. matc hes. From column 2, each of teams 2 to   k  win at lea least st one match. match. Thus Thus one team wins k − 1 matches and no team loses  k − 1 matches.   2 bonus marks marks Method 2 

Denote the teams by 1, 2, 3,   . . .   ,   k . Represen Representt each each team by a unique unique appropriately appropriately labelled ve vert rtex ex in a grap graph. h. Dr Draw aw an ar arro row w fr from om   a   to   b   to indicate that team   a   defeated defeated team   b. 1 bonus mark

Place vertices 2, 3,  . . .  ,  k  clockwise evenly spaced in a circle with vertex 1 at its centre. Draw an arrow from 1 to  i , then  i +1 to  i − 1,  i + 2 ttoo  i − 2, and so on until all vertices are exhausted. Note that the first arrow is perpendicular to all the other arrows. This represents a legitimate round in the tournament, as shown here for  k  = 6 and   i  = 2. 2 •

6

•

•

3

•

1

•

5

•

4

Now rotate the set of arrows so there is an arrow from 1 to  j . This represents another legitimate round that has no match in common with the previous round. Rotating the set of arrows so that  i  = 2, 3, . . . , k  gives all  k −1 rounds of a legitimate tournament. In this tournament, team 1 wins  k − 1 matches and each team wins at least one match so none loses  k − 1 matches. The procedure for   k  = 6 is illustrated here:

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6

2

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2

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4 2 bonus marks

© 2019 Australian Mathematics Trust

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Marking Scheme

Marking Scheme

1.   Establishing

a relevant equation. Correct answer (145).

 

1

 

a relevant equation. Correct answer (552).

1

2.   Establishing

 

1

 

relevant diagram and construction. Establishing relevant equations. Correct answer (216).

1

3.   A

 

1

 

1

 

an expression for one probability. Establishing an expression for the other probability. Correct answer (667).

1

4.   Establishing

 

1  

1

 

1

5.  Establishing

the destination after an odd number of minutes. Establishing the number of walks in 9 minutes.

 

1

 

1

 

Correct answer (512). 6.   Establishing relevant notation. Establishing a relevant equation. Eliminating a variable. Correct answer (17).

1  

1  

1

 

1

 

1

7.   A

relevant diagram and construction. Establishing an expression for a relevant angle. Finding a tight upper bound for   p. Finding a second tight upper bound and correct answer (172). Maximum of 2 marks if only one bound found.

 

1  

1

 

1  

useful factorisation.   Establishing factors 32 and 3.   Establishing factor 5.   Eliminating other factors and calculating correct answer (480).

1

8.   A

9.   Establishing

several small cases. Observing a possible general pattern. Initiating induction. Proving the inductive step. Correct conclusion and answer (192).

1 1 1  

 

1  

1

 

1  

1  

proof for (a).   A valid tournament for (b). Correct conclusion and answer (4) for (b). Establishing useful notation and a useful equation for (c). Correct conclusion for (c).

1

1

10.  Correct

1  

1  

1  

 

1 1

Investigation :

A useful approach.

 

bonus 1  

Substantial progress. A correct tournament.

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bonus 1  

bonus 1

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