57th Dutch Mathematical Olympiad 2018

and the team selection for IMO 2019 United Kingdom

First Round, January 2018

Second Round, March 2018

Final Round, September 2018

BxMO Team Selection Test, March 2019 C

IMO Team Selection Test 1, May 2019 CM

MY

IMO Team Selection Test 2, May 2019

CY

We eat problems
CMY

IMO Team Selection Test 3, May 2019

Junior Mathematical Olympiad, September 2018

for breakfast.
Preferably unsolved ones...

In juli 2011 wordt de internationale wiskunde olympiade

57 Dutch Mathematical
in Nederlandthgehouden: IMO2011
In de opmaat naar IMO2011 wordt op 3 oktober 2008 op

Olympiad 2018
de VU de eerste Junior Wiskunde Olympiade gehouden
voor de 100 beste deelnemers aan de Kangoeroewedstrijd.
International
De JWO wordt een jaarlijks terugkerend evenement. Mathematical
Zie ook: www.wiskundeolympiade.nl/junior Olympiad Am
sponsored by: Centrum Wiskunde & Informatica sterdam 2011
 We thank our sponsors
NEDERLANDSE
WISKUNDE
OLYMPIADE

Contents

1 Introduction
3 First Round, January 2018
7 Second Round, March 2018
12 Final Round, September 2018
19 BxMO Team Selection Test, March 2019
23 IMO Team Selection Test 1, May 2019
27 IMO Team Selection Test 2, May 2019
31 IMO Team Selection Test 3, May 2019
35 Junior Mathematical Olympiad, September 2018

© Stichting Nederlandse Wiskunde Olympiade, 2019 FOUNDATION COMPOSITIO MATHEMATICA

Introduction
The selection process for IMO 2019 started with the first round in January
2018, held at the participating schools. The paper consisted of eight multiple
choice questions and four open questions, to be solved within 2 hours. In
this first round 8924 students from 339 secondary schools participated.

The 1014 best students were invited to the second round, which was held in
March at twelve universities in the country. This round contained five open
questions, and two problems for which the students had to give extensive
solutions and proofs. The contest lasted 2.5 hours.

The 113 best students were invited to the final round. Also some outstanding
participants in the Kangaroo math contest or the Pythagoras Olympiad
were invited. In total about 150 students were invited. They also received
an invitation to some training sessions at the universities, in order to prepare
them for their participation in the final round.

The final round in September contained five problems for which the students
had to give extensive solutions and proofs. They were allowed 3 hours for
this round. After the prizes had been awarded in the beginning of November,
the Dutch Mathematical Olympiad concluded its 57th edition 2018.

The 30 most outstanding candidates of the Dutch Mathematical Olympiad

2018 were invited to an intensive seven-month training programme. The
students met twice for a three-day training camp, three times for a single
day, and finally for a six-day training camp in the beginning of June. Also,
they worked on weekly problem sets under supervision of a personal trainer.

In February a team of four girls was chosen from the training group to
represent the Netherlands at the EGMO in Kyiv, Ukraine, from 7 until 13
April. The team brought home a silver medal, two bronze medals, and a
honourable mention; a very nice achievement. For more information about
the EGMO (including the 2019 paper), see www.egmo.org.

In March a selection test of three and a half hours was held to determine the
ten students participating in the Benelux Mathematical Olympiad (BxMO),
held in Valkenswaard, Netherlands, from 26 until 28 April. The Dutch team
received three silver medals, four bronze medals and a honourable mention.
For more information about the BxMO (including the 2019 paper), see
www.bxmo.org.

1
In May the team for the International Mathematical Olympiad 2019 was
selected by three team selection tests on 29, 30 and 31 May, each lasting
four hours. A seventh, young, promising student was selected to accompany
the team to the IMO as an observer C. The team had a training camp in
Bristol, United Kingdom, from 6 until 14 July.

For younger students the Junior Mathematical Olympiad was held in Oc-
tober 2018 at the VU University Amsterdam. The students invited to
participate in this event were the 100 best students of grade 2 and grade 3
of the popular Kangaroo math contest. The competition consisted of two
one-hour parts, one with eight multiple choice questions and one with eight
open questions. The goal of this Junior Mathematical Olympiad is to scout
talent and to stimulate them to participate in the first round of the Dutch
Mathematical Olympiad.

We are grateful to Jinbi Jin and Raymond van Bommel for the composition
of this booklet and the translation into English of most of the problems and
the solutions.

Dutch delegation
The Dutch team for IMO 2019 in the United Kingdom consists of

• Szabi Buzogány (19 years old) • Matthijs van der Poel (18 years old)
– silver medal at BxMO 2018, – bronze medal at BxMO 2016,
silver medal at BxMO 2019 bronze medal at BxMO 2017
– honourable mention at IMO – observer C at IMO 2016,
2018 silver medal at IMO 2017,
• Jesse Fitié (17 years old) silver medal at IMO 2018
• Jovan Gerbscheid (16 years old) • Richard Wols (16 years old)
– silver medal at BxMO 2018 – bronze medal at BxMO 2018,
– bronze medal at IMO 2018 silver medal at BxMO 2019
• Jippe Hoogeveen (16 years old) – observer C at IMO 2018
– bronze medal at IMO 2018

We bring as observer C the promising young student

• Tjeerd Morsch (17 years old)
– bronze medal at BxMO 2019

The team is coached by

• Quintijn Puite (team leader), Eindhoven University of Technology
• Birgit van Dalen (deputy leader), Leiden University
• Jeroen Huijben (observer B), University of Amsterdam

2
 First Round, January 2018
Problems
A-problems

1. In a classroom there are chairs and stools. On each chair and on each stool
one child is seated. Each chair has 4 legs, each stool has 3 legs and each
child has 2 legs. Together, we have a total of 39 legs.
How many chairs are there in the classroom?
A) 3 B) 4 C) 5 D) 6 E) 9

2. On an island, there are knights and knaves. Knights always speak the truth
and knaves always lie. On the island you meet five people. You know that
four of them are knights and one of them is a knave, but you do not know
who is the knave. They make the following statements about the island
inhabitants:
• A: “All knaves have shoe size 40.”
• B: “All people with shoe size 40 have a goldfish.”
• C: “All people with a goldfish are knaves.”
• D: “I have shoe size 40.”
• E: “I have a goldfish.”
Which of them is the knave?
A) A B) B C) C D) D E) E

3. If you continue the chain of squares and regular pentagons

in the same way, does it connect to itself after going
around? If so, how many pentagons do you need?
A) 9 B) 10 C) 11 D) 12 E) It does not connect.

4. Julian wants to compose a list of integers. He wants the list to be as long

as possible. Each integer on the list must consist of one or more of the
digits 1 to 9. Moreover,
• each of the digits 1 to 9 is used exactly once;
• no integer in the list is divisble by another integer in the list.
What is the maximum number of integers in Julian’s list?
A) 4 B) 5 C) 6 D) 7 E) 8

3
5. Nine people are at a party. While entering, some of them shook hands.
Quintijn is at the party and asks each of the others how many hands they
shook. He gets eight different answers.
How many hands did Quintijn shake?
A) 0 B) 1 C) 2 D) 3 E) 4

6. Birgit is studying positive integers n for which n is divisible by 4, n + 1 is

divisible by 5, and n + 2 is divisible by 6.
How many of such integers n are smaller than 2018?
A) 16 B) 17 C) 18 D) 33 E) 34

7. A frog starts in the point at coordinates (0, 0) in the plane. He can make
three kinds of jumps:
• from (x, y) to (x, y − 5);
• from (x, y) to (x − 2, y + 3);
• from (x, y) to (x + 4, y + 9).
Ahead, there are three juicy snacks that the frog would like to eat: a worm
at (2013, 2018), a beetle at (2018, 2019), and a snail at (2018, 2023).
Which of these snacks can the frog reach?
A) worm and snail B) beetle and snail C) worm and beetle
D) only the beetle E) only the snail

8. Harold draws a trapezium with parallel top and A

bottom sides. The length of the top side is smal- C
ler than the length of the bottom side. The two D
diagonals divide the trapezium into four triangles. B
The area of the upper triangle is called A, of the
lower B, of the left C, and of the right D. An example of such a trapezium
is depicted on the right.
Which of the following equalities holds for any such trapezium?
A) A + C = B + D B) A + D = B + C C) A + B = C + D
D) A : B = D : C E) A : C = D : B

4
 B-problems
The answer to each B-problem is a number.

1. Three years ago, Rosa’s mother was exactly five times as old as Rosa was
at that time. At that moment, Rosa’s mother was just as old as Rosa’s
grandmother was when Rosa’s mother was born. Now, Rosa’s grandmother
is exactly seven times as old as Rosa is.
How old is Rosa’s mother now?

2. Nanda and Mike both have a note containing the same five-digit number.
Nanda puts the digit 4 in front and the digit 8 at the end of her number to
obtain a seven-digit number. Mike puts one digit in front of his number.
Comparing their new numbers, Nanda’s number turns out to be exactly 6
times as large as Mike’s.
What was their starting number?

3. We consider a square, the circle through the vertices of the square and the
circle touching the four sides of the square (see the left-hand figure). The
ring-shaped area between the two circles is divided into four dark pieces
(inside the square) and four light pieces (outside the square). The area of
the square is 60.

What is the total area of two dark pieces and one light piece together as
depicted in the right-hand figure?

4. Elisa is making so-called dubious dice. Each face of a dubious die contains
one of the numbers 1 to 6, but not all these numbers need to occur and
some may occur more than once. However, from every direction it must
look like a real die. This means that in each corner three different numbers
must meet, no two of which add up to 7 (on a real die such pairs are always
on opposite faces). For example, the numbers 1, 2, and 4 may meet in a
corner, but 1, 2, and 5 may not as 2 + 5 = 7. Of course, a normal die is an
example of a dubious die as well.
Elisa is interested in the sum of the six numbers on a dubious die.
How many possible values are there for this sum?

5
Solutions
A-problems

1. B) 4 5. E) 4

2. C) C 6. E) 34

3. B) 10 7. E) only the snail

4. D) 7 8. E) A : C = D : B

B-problems
1. 33 years

2. 49998

3. 15

4. 19

6
 Second Round, March 2018
Problems
B-problems
The answer to each B-problem is a number.

B1. Anouk, Bart, Celine, and Daan have participated in a math competition.
Each of their scores is a positive integer. The sum of the scores of Bart
and Daan is the same as the sum of the scores of Anouk and Celine. The
sum of the scores of Anouk and Bart is higher than the sum of the scores
of Celine and Daan. Daan’s score is higher than the sum of the scores of
Bart and Celine.
Write down the names of the four students in decreasing order of their
scores.

B2. In the figure, you can see a triangle ABC. A

The angle at B is equal to 63 degrees. Side
AC contains a point D, and side BC con- D
tains a point E. Points B, A, and D lie on 63◦ E C
B
a circle with centre E. Points E and C lie
on a circle with centre D.
What is the angle at point C?

B3. A palindromic number is a positive integer (consisting of one or more digits)

that remains the same when the digits are reversed. For example: 1245421
and 333 are palindromic numbers, but 345 and 100 are not. There is exactly
one palindromic number n with the following property: if you subtract
2018 from n, the result is again a palindromic number.
What number is n?
C

B4. Triangle ABC is isosceles with apex C. The midpoint of

E
AB is point M . On segment CM there is a point D such
|CD| 3 D
that |DM | = 2 . Line BD intersects segment AC in point E.
|CE|
Determine |EA| .
A M B

7
B5. A sawtooth number is a positive integer with the following property: for any
three adjacent digits, the one in the middle is either greater than its two
neighbours or smaller than its two neighbours. For example, the numbers
352723 and 314 are sawtooth numbers, but 3422 and 1243 are not.
How many 8-digit sawtooth numbers exist, for which each of the digits is
equal to 1, 2, or 3?

C-problems For the C-problems not only the answer is important; you also have to
describe the way you solved the problem.

C1. You have n balls that are numbered from 1 to n. You need to distribute
the balls over two boxes. The value of a box is the sum of the numbers of
the balls in that box. Your distribution of balls must obey the following
rules:
• Each box has at least one ball.
• The two boxes do not have the same number of balls.
• The value of the box with the least number of balls must be at least 2
more than the value of the box with the most balls.
Determine for which positive integers n this is possible.
(Prove that for those values of n it is indeed possible, and prove that it is
not possible for other values of n.)

C2. In this problem we consider squares: numbers of the form m2 where m is

an integer.
(a) Does there exist an integer a such that 16 + a, 3 + a, and 16 · 3 + a
are squares?
If so, give such a number a and show that the three numbers are
indeed squares.
If not, prove that such a number a does not exist.
(b) Does there exist an integer a such that 20 + a, 18 + a, and 20 · 18 + a
are squares?
If so, give such a number a and show that the three numbers are
indeed squares.
If not, prove that such a number a does not exist.
(c) Prove that for every odd integer n there exists an integer a such that
2018 + a, n + a, and 2018 · n + a are squares.

8
 Solutions
B-problems

1. A, D, B, C

2. 18 degrees

3. 2442

3
4. 4

5. 110

C-problems

C1. A distribution of the balls that follows all three rules will be called a correct
distribution. The box containing the larger number of balls will be called
the fullest box. It is clear that at least 1 + 2 = 3 balls are needed to obey
the first two rules.
We first consider the case that n is odd.
If n = 3, then the fullest box must have two balls. It therefore has a value
of at least 1 + 2 = 3, while the other box has one ball and hence a value of
at most 3. The third rule is broken so there is no correct distribution.
For n = 5 there is a correct distribution: put balls 1, 2, and 3 in one box
and put balls 4 and 5 in the other box. Since 4 + 5 is at least two more
than 1 + 2 + 3, this is indeed a correct distribution.
If there is a correct distribution for n balls, there is also one for n + 2 balls.
Indeed, we can simply add ball n + 1 to the fullest box and add ball n + 2
to the other box. The value of the fullest box increases by less than the
other box, hence we still follow rule 3.
Since we have a correct distribution for n = 5, we also have a correct
distribution for n = 7. Then, we also find a correct distribution for n = 9,
n = 11, et cetera.
Now we consider the case that n is even.
If n = 4, the fullest box must have at least three balls and hence has a
value of at least 1 + 2 + 3 = 6. The other box has only one ball and has a

9
 value of at most 4. This means that the third rule is broken. There is no
correct distribution.
If n = 6, the fullest box must have at least four balls. It therefore has a
value of at least 1 + 2 + 3 + 4 = 10. The other box has at most two balls
and therefore has a value of at most 5 + 6 = 11. Since 11 < 2 + 10, there is
no correct distribution.
For n = 8, there is a correct distribution: put balls 1 to 5 into one box
and put balls 6 to 8 in the other box. The fullest box then has a value of
1 + 2 + 3 + 4 + 5 = 15, and the other box has a value of 6 + 7 + 8 = 21.
Since 21 ≥ 2 + 15, this is indeed a correct distribution.
Again, we see that a correct distribution of n balls gives a correct distribution
with n + 2 balls by adding ball n + 1 to the fullest box and adding ball
n + 2 to the other box. Since we have a correct distribution for n = 8, we
thus find correct distributions for n = 10, 12, 14, . . ..
We conclude: for n = 1, 2, 3, 4, 6 there is no correct distribution, but for all
other positive integers n a correct distribution does exist.

C2. (a) Yes, such a number a exists. For example, take a = 33. Then we have:
16 + a = 72 , 3 + a = 62 , and 16 · 3 + a = 92 .
A suitable a can be easily found by trying 48 + a = 49, 64, 81.
(b) No, such a number a does not exist. The numbers 20 + a and 18 + a
cannot both be squares since the difference of two squares is never
equal to 2. Indeed, suppose that m2 − n2 = 2. Then we would have
(m + n)(m − n) = 2, where m > n. Since 1 and 2 are the only positive
divisors of 2, we would have m + n = 2 and m − n = 1. But this would
imply that 2m = (m + n) + (m − n) = 3, which is impossible since
2m is even whereas 3 is odd.
(c) Given an odd integer n, we will show that there exist integers a, x, y,
and z such that

2018 + a = x2 , (1)
2
n+a = y , (2)
2018n + a = z2. (3)

If we subtract equation (2) from equation (1), we obtain

2018 − n = x2 − y 2 = (x − y)(x + y).

10
We therefore choose x and y such that x + y = 2018 − n and x − y = 1.
Addition of these two equations gives us 2x = 2019 − n, subtraction
gives us 2y = 2017 − n. We therefore choose
2019 − n 2017 − n
x= and y= .
2 2
Since n is odd, these are integers. Indeed, we now have x − y = 1 and
x + y = 2018 − n, and hence that

2018 − n = x2 − y 2 . (4)

We let a = y 2 − n. Then equation (2) certainly holds. Equation (1)

holds as well, since 2018 + a = 2018 + y 2 − n = x2 (the second equation
follows from (4)).
Finally, we show that 2018n + a is a square. We successively obtain

2018n + a = 2018n + y 2 − n
(2017 − n)2
= 2017n +
4
4 · 2017n + 20172 − 2 · 2017n + n2
=
4
20172 + 2 · 2017n + n2
=
4
2
(2017 + n)2

2017 + n
= = .
4 2

Since n is odd, the number 2017+n 2 is an integer. By choosing z =

2017+n
2 , we have found integers a, x, y, and z for which equations (1),
(2), and (3) are true.

11
 Final Round, September 2018
Problems

1. We call a positive integer a shuffle number if the following hold:

(1) All digits are nonzero.
(2) The number is divisible by 11.
(3) The number is divisible by 12. If you put the digits in any other order,
you again have a number that is divisible by 12.
How many 10-digit shuffle numbers are there?

2. The numbers 1 to 15 are each coloured blue or red. Determine all possible
colourings that satisfy the following rules:

• The number 15 is red.

• If numbers x and y have different colours and x + y ≤ 15, then x + y
is blue.
• If numbers x and y have different colours and x · y ≤ 15, then x · y is
red.

3. Determine all triples (x, y, z) consisting of three distinct real numbers, that
satisfy the following system of equations:

x2 + y 2 = −x + 3y + z,
2 2
y +z = x + 3y − z,
2 2
x +z = 2x + 2y − z.

12
4. In triangle ABC, ∠A is smaller than ∠C. Point D lies on the (extended)
line BC (with B between C and D) such that |BD| = |AB|. Point E lies
on the bisector of ∠ABC such that ∠BAE = ∠ACB. Line segment BE
intersects line segment AC in point F . Point G lies on line segment AD
such that EG and BC are parallel.
Prove that |AG| = |BF |.

C
E
F

5. At a quiz show there are three doors. Behind exactly one of the doors, a
prize is hidden. You may ask the quizmaster whether the prize is behind
the left-hand door. You may also ask whether the prize is behind the
right-hand door. You may ask each of these two questions multiple times,
in any order that you like. Each time, the quizmaster will answer ‘yes’
or ‘no’. The quizmaster is allowed to lie at most 10 times. You have to
announce in advance how many questions you will be asking (but which
questions you will ask may depend on the answers of the quizmaster).
What is the smallest number you can announce, such that you can still
determine with absolute certainty the door behind which the prize is hidden?

13
 Solutions

1. First observe that a shuffle number can only contain the digits 2, 4, 6, and
8. Indeed, if we place the digits in any order, we obtain an even number
(since it is divisible by 12) because of property (3). Since the last digit of
an even number is also even, the digit must be 2, 4, 6, or 8 since it cannot
be 0 because of property (1). Since we can put any of the digits in the last
position, this holds for each digit of a shuffle number.
Next, observe that a shuffle number can only contain the digits 4 and 8.
Indeed, suppose that we had a shuffle number containing the digit 2. We
could reorder the digits so that the last digit is 2. The last two digits would
then be 22, 42, 62, or 82. But then the number would not be divisible by
4 (and hence also not divisible by 12), contradicting property (3). In the
same way we see that a shuffle number cannot contain digit 6 because a
number ending in digits 26, 46, 66, or 86 is not divisible by 4.
A shuffle number is divisible by 3 (since it is divisible by 12), hence
the sum of its digits is divisible by 3 as well. If our 10-digit number
has k eights and 10 − k fours, then the sum of its digits is equal to
8k + 4(10 − k) = 40 + 4k = 36 + 4(k + 1). This is divisible by 3 if and only
if k + 1 is divisible by 3. That is, if k = 2, k = 5, or k = 8. We see that a
shuffle number must have 2 eights and 8 fours, or 5 eights and 5 fours, or 8
eights and 2 fours. Note that each of those numbers satisfy (1) and (3). It
remains to examine which of these numbers are divisible by 11 (prop. (2)).
For this we use the 11-criterion: a number is divisible by 11 if the alternating
sum of the digits is divisible by 11. By alternating sum we mean that
instead of adding them, we alternately add and subtract. Since in our
case all digits are equal to 4 or to 8, the alternating sum must even be
divisible by 4 × 11 = 44. Since the alternating sum cannot be greater than
8 + 8 + 8 + 8 + 8 − 4 − 4 − 4 − 4 − 4 = 20 (and cannot be smaller than −20),
it must be equal to 0. In other words: the sum of the five digits in the odd
positions must equal the sum of the five digits in the even positions. This
means that the number of digits 8 in the odd positions must be equal to
the number of digits 8 in the even positions. We examine the three cases
that we found before:

• Suppose exactly 2 digits are eights. The only requirement is now that
there is exactly 1 eight in the odd positions and exactly 1 eight in the
even positions. There are 5 × 5 = 25 ways to do this.
• Suppose exactly 5 digits are eights. Since we cannot divide the eights
into two equal groups, there are no solutions in this case.

14
 • Suppose exactly 8 digits are eights. The only requirement is now that
we have 4 eights in the odd positions and 4 eights in the even positions.
In other words: 1 odd position must be a four and 1 even position
must be a four. Again we find 5 × 5 = 25 possibilities.

We conclude that the total number of 10-digit shuffle numbers is 25+25 = 50.

2. We first consider the case that 1 is red. Then all numbers from 1 to 15
are red. Indeed, suppose that some number k is blue. Then 1 and k have
different colours, hence by the third rule the number 1 · k = k must be
red. But this contradicts the assumption that it was blue. Observe that
colouring all numbers red indeed gives a correct colouring.
Now we consider the case that 1 is blue. Observe that when two numbers
sum to 15, those numbers must have the same colour by the second rule.
We get the following pairs of numbers of the same colour: 1 and 14, 2 and
13, 3 and 12, 4 and 11, 5 and 10, 6 and 9, 7 and 8.
The number 2 is blue. Indeed, suppose that 2 is red. From the second rule,
we derive that 3 = 1 + 2 is blue. Repeatedly applying the same rule we find
that 5 = 3 + 2 is blue, that 7 = 5 + 2 is blue, and finally that 15 is blue.
Since 15 is in fact not blue, 2 cannot be red.
The number 7 is blue. Indeed, suppose that 7 is red. It then follows from
the second rule that 8 = 1 + 7 is blue. However, 7 and 8 must have the
same colour, so this cannot be the case.
The number 4 is blue. Indeed, suppose that 4 is red. It then follows from
the second rule that 11 = 4 + 7 is blue. But 4 and 11 have the same colour,
so this cannot be the case.
Recall, that 3 and 12 have the same colour, and so do 6 and 9. In fact,
all four numbers must have the same colour. Indeed, otherwise 9 = 3 + 6
would be blue by the second rule and also 12 = 3 + 9 would be blue by the
second rule.
So far, we know that 15 is red, that 1, 2, 4, 7, 8, 11, 13, 14 are blue, that
3, 6, 9, 12 have the same colour, and that 5, 10 have the same colour. The
numbers 3 and 5 cannot both be red, since if 3 is red, 5 = 2 + 3 is blue by
the second rule. The three remaining colour combinations for the numbers
3 and 5 result in the following three colourings:
(1) Only 15 is red.
(2) Only 5, 10, 15 (the numbers divisible by 5) are red.
(3) Only 3, 6, 9, 12, 15 (the numbers divisible by 3) are red.

15
 It is easy to check that all three colourings are indeed correct. We write this
out for the third colouring, the other two can be checked in a similar way.
That the sum of a red and a blue number is always blue follows from the
fact that the sum of a number divisible by 3 and a number not divisible by
3 is itself not divisible by 3. That the product of a blue and a red number
is always red follows from the fact that the product of a number divisible
by 3 and a number not divisible by 3 is itself divisible by 3.
We conclude that there are 4 correct colourings in total: the colouring in
which all numbers are red, and colourings (1), (2), and (3).

3. If we subtract the second equation from the first, we obtain

x2 − z 2 = −x + z − (x − z),

which can be rewritten as

(x − z)(x + z) = −2(x − z),

and then simplified further to get

(x − z)(x + z + 2) = 0.

Since x and z must be different, x − z is nonzero and we can conclude that

x + z + 2 = 0, hence z = −x − 2. If we subtract the third equation from
the second, we obtain

y 2 − x2 = x + 3y − (2x + 2y),

which can be rewritten as

(y − x)(y + x) = 1(y − x),

and then simplified further to get

(y − x)(y + x − 1) = 0.

Since x and y must be different, y − x is nonzero and we can conclude that

y + x − 1 = 0, hence y = 1 − x.
If we now substitute y = 1 − x and z = −x − 2 in the first equation, we get

x2 + (1 − x)2 = −x + 3(1 − x) + (−x − 2),

which we can rewrite as

2x2 − 2x + 1 = −5x + 1.

16
 This is a quadratic equation 2x2 + 3x = 0, whose roots are x = 0 and
x = − 32 . In both cases we can deduce the values of y and z from the formulas
y = 1 − x and z = −x − 2. This gives the solutions (x, y, z) = (− 32 , 25 , − 21 )
and (x, y, z) = (0, 1, −2).
For these solutions we know that they satisfy the first equation. Since the
difference between the first and the second equation gives zero on both
sides (since we chose z = −x − 2), the second equation is satisfied as well.
Because of the choice y = 1 − x, also the third equation is satisfied. We
conclude that both solutions indeed satisfy all three equations.

4. Because BE is the bisector of ∠ABC, it follows that ∠ABE = ∠CBF .

It was given that ∠F CB = ∠EAB. Since the angles in a triangle sum
to 180◦ , both in triangle ABE and in triangle CBF , we must also have
∠BF C = ∠BEA. Since ∠BF C and ∠AF E are opposite angles, we see
that ∠AF E = ∠F EA, and hence that triangle AEF is isosceles.

C
E
F

B
A
G

We will show that 4ABF and 4EGA are congruent.

We see that ∠EGA = ∠CDA, since EG and BC are parallel. In the
isosceles triangle ABD, we see that ∠CDA = 12 (180◦ − ∠ABD), which is
equal to 12 ∠ABC (because of the straight angle). This in turn, is equal
to ∠ABF , because BF is the bisector of ∠ABC. Altogether, we find that
∠EGA = ∠ABF .
Triangle ABD is isosceles, so ∠BAG = ∠CDA. As we just observed, this
last angle is equal to ∠ABF . We have also seen that ∠EGA = ∠ABF .
Because the three angles in triangle GEA sum to 180◦ , we get
∠GEA = 180◦ −∠EGA−∠EAB−∠BAG = 180◦ −∠ABF −∠F CB−∠ABF.
Since 2 · ∠ABF = ∠ABC (since BF is the bisector) we thus find that
∠GEA = 180◦ − ∠F CB − ∠ABC = ∠BAC (because the angles in triangle
ABC sum to 180◦ ), hence ∠GEA = ∠BAF .

17
 Using the fact that |AE| = |AF |, which follows from the fact that 4AEF
is isosceles, we can now conclude that 4ABF and 4EGA are congruent.
This immediately implies |AG| = |BF | as required.

5. The smallest number of questions that suffices is 32. First we will show a
strategy to locate the prize in only 32 questions.
Start by asking the quizmaster if the prize is behind the left-hand door.
Repeat this until you are sure whether the prize is there or not. You
can only be 100% sure when you have received the same answer 11 times,
because the quizmaster can lie a maximum of 10 times. After doing this,
suppose that the quizmaster has lied n times. This means that you have
thus far asked 11 + n questions and the quizmaster is entitled to 10 − n
more lies.
Now ask the quizmaster whether the prize is behind the right-hand door,
and keep asking this until you are 100% sure of the true answer. Since the
quizmaster can lie only 10 − n more times, you need at most 2(10 − n) + 1 =
20 − 2n + 1 questions for this. In total you have now asked 32 − n questions.
We conclude that with this strategy, 32 questions always suffice.
Having only 31 questions, it is not always possible to locate the prize. We
will show how the quizmaster can make sure of that. At the beginning,
as long as you have asked about both doors at most 10 times, he always
answers ‘no’. To keep it simple, we assume that the left-hand door is the
first door that you query for the eleventh time (the other case is completely
analogous). We consider the situation that the prize is not behind the left-
hand door (which may happen). In this case, we show that the quizmaster
can make sure that after 31 questions, you cannot tell whether the prize is
behind the middle door or behind the right-hand door.
So far, the quizmaster has been speaking the truth about the left-hand
door, and he will continue to do so: when asked about the left-hand door,
he will always answer ‘no’. For the right-hand door, the quizmaster will
keep saying ‘no’ up to and including the tenth time he is asked about this
door. The next ten times he is asked about this door, he will answer ‘yes’.
Since you ask at most 20 questions about the right-hand door (since you
already queried the left-hand door at least 11 times), the quizmaster needs
to lie at most 10 times. Whether the prize is behind the middle door or
behind the right-hand door, in both cases the quizmaster gives the same 31
answers. Therefore, you cannot determine the door behind which the prize
is located.

18
 BxMO Team Selection Test, March 2019
Problems

1. Prove that for each positive integer n there are at most two pairs (a, b) of
positive integers with following two properties:
(i) a2 + b = n,
(ii) a+b is a power of two, i.e. there is an integer k ≥ 0 such that a+b = 2k .

2. Let ABC be a triangle and let I be the incentre of this triangle. The line
through I perpendicular to AI intersects the circumcircle of 4ABC in
points P and Q, where P lies on the same side of AI as B. Let S be the
second intersection point of the circumcircles of 4BIP and 4CIQ. Prove
that SI is the angle bisector of ∠P SQ.

3. Let x and y be positive real numbers.

1. Prove: if x3 − y 3 ≥ 4x, then x2 > 2y.

2. Prove: if x5 − y 3 ≥ 2x, then x3 ≥ 2y.

4. Do there exist a positive integer k and a non-constant sequence a1 , a2 , a3 , . . .

of positive integers such that an = gcd(an+k , an+k+1 ) for all positive
integers n?

5. In a country, there are 2018 cities, some of which are connected by roads.
Each city is connected to at least three other cities. It is possible to travel
from any city to any other city using one or more roads. For each pair of
cities, consider the shortest route between these two cities. What is the
greatest number of roads that can be on such a shortest route?

19
 Solutions

1. Suppose there are three or more of such pairs for one particular n. Then
by the pigeonhole principle, there are at least two such pairs, say (a, b) and
(c, d), with a ≡ c mod 2. Write a + b = 2k and c + d = 2` where k and `
are positive integers. Without loss of generality, assume that ` ≥ k. We
see that
2` − 2k = (c + d) − (a + b)
= (c + n − c2 ) − (a + n − a2 ) = c − a − c2 + a2
= (a + c − 1)(a − c).
As ` ≥ k, we have that 2k is a divisor of 2` − 2k . Because a ≡ c mod 2,
the integer a + c − 1 is odd, hence 2k is a divisor of a − c. As 2` − 2k ≥ 0
and a + c − 1 > 0, we have a − c ≥ 0. On the other hand, we have
a − c < a + b = 2k , hence 0 ≤ a − c < 2k , while 2k | a − c. We conclude
that a − c = 0, hence a = c. Now we have b = n − a2 = n − c2 = d, hence
the two pairs (a, b) and (c, d) are identical, which gives a contradiction.
Therefore, there at most two pairs (a, b) having the two properties. 

2. Write ∠CAB = 2α, ∠ABC = 2β, and ∠BCA = 2γ. As the angles in
triangle 4AIB add up to 180◦ , we have
∠BIA = 180◦ − ∠IAB − ∠ABI = 180◦ − α − β = 90◦ + γ,
hence
∠BIP = ∠BIA − ∠AIP = 90◦ + γ − 90◦ = γ = ∠BCI.
 Moreover, as BP QC is a cyclic quadrilateral, we have

180◦ − ∠BP I = 180◦ − ∠BP Q = ∠BCQ = ∠BCI + ∠ICQ.

If we now consider the sum of the angles in triangle BP I, and use both
previous results, we find

∠P BI = 180◦ − ∠BP I − ∠BIP = ∠BCI + ∠ICQ − ∠BCI = ∠ICQ.

Using the fact that BSIP and CQIS are cyclic quadrilaterals, we get

∠P SI = ∠P BI = ∠ICQ = ∠ISQ,

hence SI is the angle bisector of angle P SQ. 

3. 1. We have x3 − 4x ≥ y 3 > 0, hence x(x2 − 4) > 0. As x is positive, this

yields x2 − 4 > 0, hence x2 > 4. That means that x > 2. Moreover,
we have x3 − y 3 ≥ 4x > 0, hence x > y. The combination of these
two results (which is allowed because x and y are both positive) gives
x2 = x · x > 2 · y = 2y.
2. We have (x4 − 4)2 ≥ 0. Expanding yields x8 − 8x4 + 16 ≥ 0. Because x
is positive, we can multiply this with x without changing the inequality
sign, hence we have x9 ≥ 8x5 − 16x. The inequality in the assumption
gives x5 − 2x ≥ y 3 . If we combine this with the preceding inequality,
we get x9 ≥ 8(x5 − 2x) ≥ 8y 3 . It follows that x3 ≥ 2y. 

4. Such a k and a sequence do not exist. We prove this by contradiction,

so suppose they do exist. Note that an | an+k and an | an+k+1 for all
n ≥ 1. Using simple induction, it follows that an | an+`k and an | an+`k+`
for all ` ≥ 0. We will prove by induction to m that an | an+mk+(m+1)
for all 0 ≤ m ≤ k − 1. For m = k − 1, this follows from an | an+k·k =
an+(k−1)k+k . Now suppose that for a certain m with 1 ≤ m ≤ k −
1 we have that an | an+mk+(m+1) . We also know that an | an+mk+m .
Therefore, as an+(m−1)k+m = gcd(an+mk+m , an+mk+m+1 ), we also have
an | an+(m−1)k+m . This finishes the induction argument. Substituting
m = 0 yields an | an+1 .
Because an | an+1 , we also have gcd(an , an+1 ) = an for all n. Hence,
an = an−k for all n ≥ k + 1. Now we have an−k | an−k+1 | an−k+2 | . . . |
an = an−k . Because these are all positive integers, an−k , an−k+1 , . . . , an
must all be equal. This must be true for all n ≥ k + 1, hence the sequence
is constant, which gives a contradiction. 

21
5. The greatest number of roads that can be in such a shortest route, is
1511. We first describe a country for which this number is attained. Divide
the cities in 504 groups: two groups of five cities (A0 , B0 , C0 , D0 , E0 ) and
(A503 , B503 , C503 , D503 , E503 ), and groups of four cities (Ai , Bi , Ci , Di ) for
1 ≤ i ≤ 502. For all i with 0 ≤ i ≤ 503: connect Ai to Bi , connect Ai to
Ci , connect Bi to Di , and connect Ci to Di . Moreover, connect E0 to A0 ,
B0 , and C0 . Connect E503 to B503 , C503 , and D503 . For each 1 ≤ i ≤ 502:
connect Bi to Ci . Finally, for each 0 ≤ i ≤ 502: connect Di to Ai+1 . Now
every city is connect two exactly three other cities.
If we now want to travel from A0 to D503 , then we have to travel through
all Ai and Di , because the only connections between the different groups
are there, and each group is only connected to the previous and the next
group. Moreover, for 0 ≤ i ≤ 503, the city Ai is not connected to Di , which
causes the route within group i to go through either Bi or Ci . At least
three of the four or five cities of each group must lie on the route. In total,
this route visits at least 3 · 504 = 1512 cities and has at least 1511 roads.
We will now prove that the shortest route between two cities cannot have
more than 1511 roads. Consider such a shortest route which visits the cities
A0 , A1 , . . . , Ak consecutively. Each of the cities Ai has one more neighbour
besides Ai−1 and Ai+1 , which we will call Bi (note that the cities B0 , B1 ,
. . . , Bk do not have to be distinct). Moreover, A0 and Ak are connected
to a third city, say C0 6= B0 and Ck 6= Bk respectively. If one of the cities
Bi or Ci equals one of the cities Aj , then we could have found a shorter
route by going directly from Ai to Aj (or vice versa), which would be a
contradiction. Hence, the cities Bi and Ci are not equal to any of the cities
Aj .
If one of the cities Bi is connected to four cities Aj , say Bi is connected to
Ai , Am , An , and Ap with i < m < n < p, then we could shorten the route
by going from Ai to Bi and then to Ap . This makes us skip at least two
cities of the original route (Am and An ) and in their place we only visit Bi ,
hence this new route is shorter, which would be a contradiction. Therefore,
within the cities Bi with 3 ≤ i ≤ k − 3 there are at least k−5 3 distinct cities.
For B0 and C0 , the route can only be shortened if one of these cities equals
Bi for certain i ≥ 3. That would be a contradiction, hence B0 and C0 are
two distinct cities, not equal to Bi for i ≥ 3. In the same way, Bk and
Ck are two distinct cities, not equal to Bi for i ≤ k − 3. Altogether, we
have k + 1 cities on the route itself and at least k−5 3 + 2 + 2 other cities.
Therefore, k−53 + k + 5 ≤ 2018, hence 4k − 5 + 15 ≤ 3 · 2018 = 6054, hence
4k ≤ 6044, hence k ≤ 1511.
We conclude that the greatest number of roads occurring in a shortest route
is 1511. 

22
 IMO Team Selection Test 1, May 2019
Problems

1. Let P (x) be a quadratic polynomial with two distinct real roots. For all
real numbers a and b satisfying |a|, |b| ≥ 2017, we have P (a2 + b2 ) ≥ P (2ab).
Show that at least one of the roots of P is negative.

2. Write Sn for the set {1, 2, . . . , n}. Determine all positive integers n for
which there exist functions f : Sn → Sn and g : Sn → Sn such that for
every x exactly one of the equalities f (g(x)) = x and g(f (x)) = x holds.

3. Let n be a positive integer. Determine the maximum value of gcd(a, b) +

gcd(b, c) + gcd(c, a) for positive integers a, b, c such that a + b + c = 5n.

4. We are given a triangle ABC. On edge AC there are points D and E such
that the order of points on this line is A, E, D, C. The line through E
parallel to BC intersects the circumcircle of 4ABD in a point F , with E
and F lying on opposite sides of AB. The line through E parallel to AB
intersects the circumcircle of 4BCD in a point G, with E and G lying on
opposite sides of BC. Prove that DEF G is a cyclic quadrilateral.

23
 Solutions
1. Write P (x) = c(x − d)(x − e), where d and e are the distinct roots of P ,
so d =6 e. We also have c 6= 0, since otherwise P is not quadratic. We
distinguish two cases. First suppose that c > 0. Substituting b = −a =
−2017 in P (a2 + b2 ) ≥ P (2ab), we obtain P (2a2 ) ≥ P (−2a2 ), and therefore

c(2a2 − d)(2a2 − e) ≥ c(−2a2 − d)(−2a2 − 2).

Dividing both sides by the positive real number c, expanding, and then
canceling the terms 4a4 and de, we get

−(d + e) · 2a2 ≥ (d + e) · 2a2 ,

or equivalently,
4a2 (d + e) ≤ 0.
Dividing by 4a2 = 4 · 20172 , we see that d + e ≤ 0. As the roots d and e
are distinct, one of them must be negative.
Now consider the remaining case c < 0. Then the graph of P opens up
to the bottom, and therefore is descending on the right of the apex of the
parabola. Choose any a 6= b with a, b ≥ 2017 and a, b both to the right
side of the apex. Then by AM-GM we have a2 + b2 > 2ab. Since a, b are
positive, at least 1, and to the right of the apex, we have 2ab > a, so 2ab
(and therefore a2 + b2 as well) also lies to the right of the apex. Since P
is descending on the right side of the apex, we have P (a2 + b2 ) < P (2ab),
contrary to the given condition. We deduce that the case c < 0 is impossible
and therefore that c > 0, from which we conclude that at least one of the
roots of P is negative. 

2. We first show that if n = 2m for some positive integer m, such functions

exist. Define
( (
x if 1 ≤ x ≤ m, x + m if 1 ≤ x ≤ m,
f (x) = g(x) =
x − m if m + 1 ≤ x ≤ 2m, x if m + 1 ≤ x ≤ 2m.

First note that all values lie in Sn , so these indeed define functions Sn →
Sn . The range of f is equal to {1, 2, . . . , m} and that of g is equal to
{m + 1, m + 2, . . . , 2m}. So f (g(x)) 6= x if x ≥ m + 1 and g(f (x)) 6= x if
x ≤ m. Moreover, for x ≤ m we have f (g(x)) = f (x + m) = x + m − m = x,
and for x ≥ m + 1 we have g(f (x)) = g(x − m) = x − m + m = x. Therefore
for every x ∈ Sn , exactly one of f (g(x)) = x and g(f (x)) = x holds.

24
 Now suppose that n is odd, say equal to 2m + 1, and suppose that f and
g are functions satisfying the given conditions. Without loss of generality,
we assume that f (g(x)) = x for (at least) m + 1 elements of Sn , say for
x = x1 , . . . , xm+1 . Suppose that for some i, j with i ≤ i, j ≤ m + 1 we
have g(xi ) = xj . Then f (xj ) = f (g(xi )) = xi , so g(f (xj )) = g(xi ) = xj .
But now we have both f (g(x)) = x and g(f (x)) = x for x = xj , which
gives a contradiction. So we see that for all i with 1 ≤ i ≤ m + 1 the
number g(xi ) is not equal to any of the xj with 1 ≤ j ≤ m + 1. Since
there are m numbers in Sn not equal to any of the xj with 1 ≤ j ≤ m + 1,
and since there are m + 1 possible values of i, it follows that two of the
g(xi ) must be equal, say g(xk ) = g(xl ) with 1 ≤ k < l ≤ m + 1. But then
xk = f (g(xk )) = f (g(xl )) = xl , which gives a contradiction. Therefore no
such f and g can exist if n is odd.
Hence n satisfies the given conditions if and only of n is even. 

3. Write G = gcd(a, b) + gcd(b, c) + gcd(c, a). Without loss of generality we

assume that a ≤ b ≤ c. Then we have gcd(a, b) ≤ a, gcd(b, c) ≤ b, and
gcd(c, a) ≤ a. Hence

G ≤ a + b + a ≤ a + b + c ≤ 5n.

If 3 | n, then we can achieve G = 5n via a = b = c = 53 n. Since all three

gcd’s equal 53 n, we see that G = 3 · 53 n = 5n.
So suppose that 3 - n. Then a, b, and c cannot all be equal to each other.
We distinguish a number of cases.
Case 1a. b = c and b ≤ 2n. Since in this case we have a = 6 b, it follows that
gcd(a, b) = gcd(a, b − a) ≤ b − a. Hence G ≤ (b − a) + b + a = 2b ≤ 4n.
Case 1b. b = c and b > 2n. We have a + 2b = 5n, so a = 5n − 2b, from which we
deduce that G ≤ a + b + a = 10n − 4b + b = 10n − 3b < 10n − 6n = 4n.
Case 2a. b 6= c and c−a ≥ n. Now we have G ≤ a+b+a = (a+b+c)−(c−a) =
5n − (c − a) ≤ 5n − n = 4n.
6 c and c − a < n. Since a ≤ b ≤ c we now also have c − b < n.
Case 2b. b =
Moreover, we have b 6= c and therefore a 6= c. So gcd(c, a) = gcd(c −
a, a) ≤ c − a < n and gcd(b, c) = gcd(b, c − b) ≤ c − b < n. Also note
that a ≤ 53 n. Therefore G ≤ 53 n + n + n < 4n.
So in all cases we have G ≤ 4n. We can achieve G = 4n via a = n and
b = c = 2n, since then we have gcd(a, b) = gcd(c, a) = n and gcd(b, c) = 2n.
Therefore the maximum value of G is 5n if 3 | n and 4n if 3 - n. 

25
4. Let G0 be the intersection of the line BF
and the circumcircle 4BCD. We show that
DEF G0 is a cyclic quadrilateral and that G =
G0 .
Since EF k BC we have 180◦ − ∠DEF =
∠AEF = ∠ACB, and since BDCG0 is a cyc-
lic quadrilateral we have ∠ABC = ∠DCB =
∠DG0 B = ∠DG0 F , therefore 180◦ −∠DEF =
∠DG0 F . Therefore DEF G0 is a cyclic quad-
rilateral. It follows that we have ∠DEG0 =
∠DF G0 = ∠DF B. Since ADBF is a cyclic quadrilateral we have ∠DF B =
∠DAB, so ∠DEG0 = ∠DAB, and therefore EG0 k AB. As G0 lies on the
circumcircle of 4BCD, we deduce that G = G0 . 

26
 IMO Team Selection Test 2, May 2019
Problems

1. In each of the different grades of a high school there are an odd number of
pupils. Each pupil has a best friend (who possibly is in a different grade).
Everyone is the best friend of their best friend. In the upcoming school
trip, every pupil goes to either Rome or Paris. Show that the pupils can be
distributed over the two destinations in such a way that
(i) every student goes to the same destination as their best friend;
(ii) for each grade the absolute difference between the number of pupils
that are going to Rome and that of those who are going to Paris is
equal to 1.

2. Determine all 4-tuples (a, b, c, d) of positive real numbers satisfying a + b +

c + d = 1 and
 2 2  2 2
a b c d
max , · max , = (min(a + b, c + d))4 .
b a d c

3. Let ABC be an acute triangle with circumcentre O. Point Q lies on the

circumcircle of 4BOC and OQ is a diameter of this circle. Point M lies
on CQ and point N lies on the interior of the line segment BC in such a
way that AN CM is a parallelogram. Show that the circumcircle of 4BOC
and the lines AQ and N M are concurrent.

4. Find all functions f : Z → Z satisfying

• f (p) > 0 for all prime numbers p,

f (p)
• p | f (x) + f (p) − x for all x ∈ Z and all prime numbers p.

27
 Solutions
1. We call a pupil a singleton if their best friend is in a different grade.
First give all singletons a destination as follows. Start by choosing any
singleton A1 and send them to Paris together with their best friend A2 .
Suppose the destinations for pupils A2m+1 , A2m+2 , . . . , A2k , with m ≤ 0
and k ≥ 1 are chosen. If possible, first choose in the grade of A2k a
singleton A2k+1 who doesn’t have a destination yet, and send them to the
city different from the destination of A2k , together with their best friend
A2k+2 . If possible, then choose in the grade of A2m+1 a singleton A2m
who doesn’t have a destination yet, and send them to the city different
from the destination of A2m+1 , together with their best friend A2m−1 .
Continue this process until pupils A2m+1 , A2m+2 , . . . , A2k with m ≤ 0 and
k ≥ 1 have destinations assigned to them and there are no singletons in the
grades of A2m+1 and A2k who don’t have a destination yet. These pupils
A2m+1 , A2m+2 , . . . , A2k are by construction such that A2i−1 and A2i are
best friends and A2i and A2i+1 have different destinations.
In every grade except that of A2m+1 and A2k the same number of pupils
is assigned to each destination. In the grades of A2m+1 and A2k (which
are different: otherwise no other singletons are in this grade, and all other
pupils form pairs of best friends, making the number of pupils in that class
even; contradiction!) one destination is assigned one more pupil than the
other, and these grade no longer have any singletons without destination.
Repeat this process until every singleton has a destination. All remaining
pupils now are in the same grade as their best friends, so every grade has
an even number of pupils without destination and therefore an odd number
of pupils with destination. So every grade contains a pupil on an end of
exactly one of the sequences of pupils (which then must be the last time a
pupil of this grade occurs in such a sequence). Therefore for each grade,
one destination is assigned exactly one more pupil of that grade than the
other.
Now consider any grade, and suppose that the number of singletons assigned
to Paris is one more than that assigned to Rome. Give the pupils in this
grade whose best friends are also in this grade a destination as follows. If
the number of pairs of best friends is odd, first send one pair to Rome;
otherwise do nothing. From the remaining even number of pairs of best
friends, send half of them to Paris and half of them to Rome.
Now note that in the first step, if the number of pairs was odd, then Rome
is assigned one more pupil than Paris. Therefore in any grade, the absolute
difference of the number of students going to Paris and the number going
to Rome is equal to 1. 

28
2. We assume that a + b ≤ c + d. As we can simultaneously interchange a, b
and c, d without changing the problem, this assumption will not give any
loss of generality. We also assume that a ≤ b and c ≤ d; no generality is
lost here as we can interchange a and b (resp. c and d) without changing
2 2
the problem. Now we have a3 ≤ b3 , so ab ≤ ba , and analogously we have
2 2
c d
d ≤ c . The given equation is now equivalent to

b2 d2
a · c = (a + b)4 .
2 2
As ba ≥ b and dc ≥ d, the left hand side is at least bd. We now show that
bd ≤ (a + b)4 , from which follows that equality must hold everywhere.
Since a ≤ b we have b ≥ 12 (a + b), and analogously we have d ≥ 12 (c + d).
Since c+d ≥ a+b, it follows that d ≥ 21 (a+b). Moreover, from a+b+c+d = 1
and a + b ≤ c + d we deduce that a + b ≤ 12 , so

bd ≤ 14 (a + b)2 ≤ (a + b)2 (a + b)2 = (a + b)4 .

Therefore, each inequality used in the above must actually be an equality.

2
Since we have used ba ≥ b it follows that a = b, and analogously that c = d,
and since we have used a + b ≤ 12 it follows that a + b = 12 . Therefore
a = b = c = d = 14 .
So the only possible solutions is (a, b, c, d) = ( 14 , 14 , 14 , 14 ). Substitution of
1
this candidate solution gives 16 on both sides of the equation, and therefore
actually is a solution. 

Q
3. Let T be the intersection of the cir-
cumcircle of 4BOC with AQ. We
show that T lies on N M . Write α = C
∠BAC. Then we have ∠BOC =
M
2α by the inscribed angle theorem.
N
Since |OB| = |OC| and ∠OCQ = O
90◦ = ∠OBQ by Thales’s theorem, B
we have 4OBQ ∼ = 4OCQ. Hence
A

we have ∠COQ = ∠BOQ = α.

As AN CM is a parallelogram, we see that ∠AM C = 180◦ − ∠M CN =
∠QCN = ∠QCB = ∠QOB = α. Hence we have ∠CN A = ∠AM C = α.
Now we see that ∠QT B = ∠QOB = α = ∠CN A, from which follows that
∠AT B = 180◦ − ∠QT B = 180◦ − ∠CN A = ∠AN B. Hence AT N B is a
cyclic quadrilateral. Also note that AT CM is a cyclic quadrilateral since
∠AM C = α = ∠COQ = ∠CT Q = 180◦ − ∠AT C.

29
 As AT CM is a cyclic quadrilateral, we have ∠AT M = ∠ACM = 180◦ −
∠QCB − ∠BCA, using supplementary angles at C. Recall that ∠QCB =
∠QOB = α = ∠CAB, so using the sum of the angles in triangle ABC,
we obtain ∠AT M = 180◦ − ∠CAB − ∠BCA = ∠ABC. Since AT N B is a
cyclic quadrilateral, we see that ∠ABC = ∠ABN = 180◦ − ∠AT N , so we
have ∠AT M = 180◦ − ∠AT N . Therefore M , T , and N are collinear, as we
have set out to prove. 

4. Suppose that f is a function satisfying both relations.

f (p)
Substituting x = p gives p | 2f (p) − p and as p is prime, we have
p | 2f (p). So either p = 2 or p | f (p).

f (p)
Substituting x = 0 gives p | f (0) + f (p) − 0 and as p is prime, we
have p | f (0) + f (p). Since p | f (p) if p 6= 2, it follows that p | f (0) if p 6= 2.
So f (0) is divisible by infinitely many prime numbers and therefore must
be equal to 0. From 2 | f (0) + f (2) we now see that 2 | f (2), so p | f (p) for
all prime numbers p.
Now the second of the given relations translates to f (x)f (p) ≡ p mod p for
all integers x and prime numbers p. It follows that p | f (x) if and only if
p | x. Applying this observation to the case in which x is a prime number
q 6= p, then we see that f (q) is not divisible by any prime number p 6= q.
As f (q) > 0, we deduce that f (q) is a power of q. Fermat’s Little Theorem
states that for all prime numbers p and integers n we have np ≡ n mod p,
t
so we also have np ≡ n mod p for all non-negative integers t. As f (p) is
t
of the form p with t a non-negative integer for all prime numbers p, we
deduce from the second given relation that f (x) ≡ x mod p for all integers
x and prime numbers p. Thus p | f (x) − x for all integers x and all prime
numbers p.
Therefore, for any fixed x ∈ Z the integer f (x) − x is divisible by infinitely
many prime numbers and therefore equal to 0. Hence f (x) must be equal
to x for all integers x.
Now suppose that f (x) = x for all integers x. Then f (p) > 0 for all prime
numbers p and

f (p)
f (x) + f (p) − x = (x + p)p − x ≡ xp − x ≡ 0 mod p

for all integers x and all prime numbers p by Fermat’s Little Theorem.
It follows that f (x) = x is the unique function Z → Z satisfying both
relations. 

30
 IMO Team Selection Test 3, May 2019
Problems

1. In a cyclic quadrilateral ABCD, the intersection of the diagonals is called

E. A line through E, not equal to AC or BD, intersects AB in P and
BC in Q. The circle that is tangent to P Q in E and also goes through D,
intersects the circumcircle of ABCD another time in the point R. Prove
that B, P , R, and Q lie on a circle.

2. Let n be a positive integer. Prove that n2 + n + 1 cannot be written as

the
√ product of two positive integers of which the difference is smaller than
2 n.

3. Find all functions f : Z → Z satisfying the following two conditions:

(i) for all integers x we have f (f (x)) = x;
(ii) for all integers x and y such that x + y is odd, we have f (x) + f (y) ≥
x + y.

4. There are 300 participants to a mathematics competition. After the com-

petition some of the contestants play some games of chess. Each two
contestants play at most one game against each other. There are no three
contestants, such that each of them plays against each other. Determine
the maximum value of n for which it is possible to satisfy the following
conditions at the same time: each contestant plays at most n games of
chess, and for each m with 1 ≤ m ≤ n, there is a contestant playing exactly
m games of chess.

31
 Solutions

1. We consider the configuration in which P lies on the interior of AB and

Q does not lie on the interior of BC; other configurations can be solved
analogously.

Because DBCR is cyclic, we have ∠QCR = 180◦ − ∠BCR = ∠BDR.

As EQ is tangent to the circle through E, D, and R we have ∠BDR =
∠EDR = ∠QER, hence ∠QCR = ∠QER. This implies that ERQC is a
cyclic quadrilateral. Analogously, EP AR is also cyclic.
We now get ∠P RQ = ∠P RE + ∠ERQ = ∠P AE + 180◦ − ∠ECQ =
∠BAE + ∠ECB. Because of the sum of the angles in 4ABC this equals
180◦ − ∠ABC = 180◦ − ∠P BQ. Hence ∠P RQ = 180◦ − ∠P BQ, and hence
BP RQ is a cyclic quadrilateral. 

2. Suppose that a and b are √ positive integers satisfying ab = n2 + n + 1. We

will prove that |a − b| ≥ 2 n. Note that (a − b)2 ≥ 0. Adding 4ab to both
sides yields
(a + b)2 = (a − b)2 + 4ab ≥ 4ab = 4n2 + 4n + 4 > 4n2 + 4n + 1 = (2n + 1)2 .
As (a + b)2 is a square, we have (a + b)2 ≥ (2n + 2)2 . Then it follows that
(a − b)2 = (a + b)2 − 4ab ≥ (2n + 2)2 − 4(n2 + n + 1) = 4n
√
hence |a − b| ≥ 2 n. 

32
3. The function f (x) = x for all x, satisfies the conditions. Now suppose that
f (x) = x does not hold for all x. As of (i), there is both a value of x for
which f (x) > x and a value of x for which f (x) < x. Let a ∈ Z be such
that f (a) < a and consider an arbitrary x with x 6≡ a mod 2. Then we
have f (x) + f (a) ≥ x + a, hence f (x) ≥ x + a − f (a) > x. Write w = f (x),
then we have f (w) = f (f (x)) = x < f (x) = w. If w 6≡ a mod 2, then we
would have f (w) > w, which gives a contradiction. Hence f (x) = w ≡ a
mod 2 and also f (x) 6≡ x mod 2.
Let x1 and x2 both be of different parity from a. Then they both do
not have the same parity as their function values, hence x1 + f (x2 ) and
x2 + f (x1 ) are odd. By substituting x = x1 , y = f (x2 ), and also x = x2 ,
y = f (x1 ), we find

f (x1 )+x2 = f (x1 )+f (f (x2 )) ≥ x1 +f (x2 ) = f (f (x1 ))+f (x2 ) ≥ f (x1 )+x2 .

Hence, equality must hold everywhere and f (x1 ) + x2 = x1 + f (x2 ). Fix

x1 and write c = f (x1 ) − x1 , then we have f (x2 ) = x2 + c for all x2 not
having the same parity as a. Moreover, we know that c must be odd. If
z = x2 + c, then z can take all possible values not having the same parity
as x2 (and thus having the same parity as a). We have

f (z) = f (x2 + c) = f (f (x2 )) = x2 = z − c.

Now write d = c if a is odd and d = −c if a is even, then the function

satisfies (
x + d if x is even,
f (x) =
x − d if x is odd.
We check functions of this shape for any arbitrary odd d. We see that f (x)
and x never have the same parity, hence f (f (x)) = x + d − d = x, which
means (i) is satisfied. Moreover, if x + y is odd, then x and y have different
parities, hence f (x) + f (y) = x + y + d − d = x + y ≥ x + y, which means
(ii) is satisfied as well.
We conclude that the functions of this form together with the function
f (x) = x for all x, are the solutions. 

33
4. We will prove that the maximum value of n is 200. We will first give an
example with n = 200. Consider players A1 , A2 , . . . , A200 and B1 , B2 ,
. . . , B100 . These are 300 players in total. Let Bi play a game against the
players Aj with 1 ≤ j ≤ i + 100, and assume no other contestants play a
game against each other. Then Bi has exactly 100 + i opponents, so for
101 ≤ m ≤ 200 there is a contestant playing exactly m games of chess.
Moreover, for j > 100, the player Aj is playing against the contestants Bi
with i ≥ j − 100; these are 100 − (j − 101) = 201 − j players. Because
j can vary from 101 up to and including 200, the number of opponents
varies from 100 up to and including 1. Hence, for 1 ≤ m ≤ 100 there is
also a contestant playing exactly m games of chess. Lastly, for j ≤ 100,
contestant Aj plays against all Bi ; these are 100 opponents. Hence, there
is no contestant playing more than 200 games of chess. We see that this
example meets the requirements for n = 200.
Now we will prove that n > 200 cannot be realised. We prove this by
contradiction, so assume that n > 200. Then in any case there is a
contestant A playing exactly 201 games of chess, against players B1 , B2 ,
. . . , B201 . In total, there are 300 contestants, so besides them and A,
there are another 300 − 1 − 201 = 98 contestants, who we will call C1 , . . . ,
C98 . If a contestant Bi is playing a game against another contestant Bj ,
then together with A they form a triple that plays three games among
each other; this is not allowed. Hence Bi does not play against any of the
other Bj . Therefore, they play at most 1 + 98 = 99 games. This means
that the contestants who play against exactly m other contestants with
100 ≤ m ≤ 200, must all be among the contestants Ci . However, these are
only 98 distinct contestants, which gives a contradiction. We conclude that
n > 200 is impossible and that n = 200 is the maximum.

34
 Junior Mathematical Olympiad, September 2018

Problems
Part 1

1. When writing down the date 12 August 2018 using eight digits, each digit
occurs exactly twice: 12-08-2018. There are more dates in 2018 having the
same property. How many dates in 2018 have this property, including 12
August?
A) 5 B) 6 C) 7 D) 8 E) 9

2. In the crossword puzzle on the right, a b

each square has to be filled with
one of the digits 1 to 9. A digit
c
may occur multiple times. For the
2-digit numbers formed in the rows
and columns we are given the following four hints:
Across Down
a. An odd number a. A square
c. A square b. An odd number
The puzzle has more than one solu-
tion, but the digit in the top-left
corner is always the same. Which
digit is in the top-left corner?
A) 1 B) 2 C) 3 D) 4 E) 6

3. Sophie likes to wear red and blue T-shirts. She decided to wear either a
red or a blue T-shirt each day, starting from 1 January 2019. She does not
want to say which colour she will be wearing on 1 and 2 January. From 3
January on, she will choose the colour of her T-shirt each day according to
the following rule: she chooses red if she wore two different colours the last
two days, and she chooses blue if she wore the same colour the last two
days.
By following this rule, she will wear a blue T-shirt on her birthday, 14
January. Is it possible to determine with certainty the colours of the T-
shirts she will be wearing on 28 and 29 January?

35
 A) 28th: red, 29th: blue D) 28th: could be either 29th: blue
B) 28th: blue 29th: blue E) 28th: blue 29th: could be either
C) 28th: blue 29th: red

4. We have an isosceles triangle with two angles of 45 de-

grees; the long side has length 1. Within the triangle we
put two squares of the same size. We can do this in two
ways depicted in the two figures.
What is the total area of the two squares in the top
figure minus the total area of the two squares in the
bottom figure?
1 1 1 1 1
A) 72 B) 48 C) 36 D) 24 E) 18

5. On a glass table, we arrange 100 dice tightly in a 10 by

10 square. This is done in such a way that if two dice are
touching each other along a face, these faces have the
same number of pips. Both the top and bottom faces of
the 100 dice are visible. Altogether, on the front, rear,
left, and right side there are 40 dice faces visible. We add the pips on all of
these 240 visible faces.
What is the largest outcome that we can get?
In the figure on the right, you can see a net representing a dice.
A) 840 B) 880 C) 920 D) 1240 E) 1440

6. A large square is subdivided into 16 squares. We put crosses in two of these

squares (see the figure). This can be done in different ways. Sometimes
two such ways only differ by a rotation, for example the left two squares
below. In this case, we consider the two ways as the same one and only
count it once.

Pay attention: the right two squares are not considered to be the same. You
could obtain the right square from the middle one by reflection, but not by
using a rotation!
In how many different ways can you put the two crosses?
A) 21 B) 30 C) 32 D) 34 E) 36

36
7. Using 27 small cubes, each coloured black or white, we build a 3×3×3 cube.
This large cube has six views: a front, rear, left, right, top, and bottom
view. In the figure, five of the views of the large cube are depicted.

Which could be the sixth view of the large cube?

A) B) C) D) E)

8. How many distinct pairs of digits a and b are there such that 5a68 × 865b
is divisible by 824?
Pay attention: we are counting pairs, so for example if for a = 0 the values
b = 0, b = 1, and b = 2 are all valid, then these are counted as three
different pairs.
A) 10 B) 11 C) 15 D) 19 E) 21

Part 2

1. Anne, Bert, Christiaan, Dirk, and Eveline are particpating in a chess

tournament. They find out that their average age is exactly 28 years.
Exactly one year later Anne, Bert, Christiaan, and Dirk participate together
with Freek in the tournament. This time, their average age is exactly 30
years.
How many years older is Freek compared to Eveline?

x x
2. What is the smallest positive integer x for which the outcome of 2 + 3 +
x x x
4 + 5 + 6 is an integer?

37
3. For each of the two fractions 2018 2054
2011 and 2019 we subtract the same integer a
from both the numerator and the denominator. The two fractions we get,
are equal.
What integer is a?

4. The sides of the square ABCD have length 10. The points P and Q lie on
the line connecting the midpoints of AD and BC. If we connect P with
A and C, and also Q with A and C, then the square is divided into three
parts having equal area.
What is the length of P Q? (The figure is not drawn to scale.)

D C

P
Q

A B

5. Jan has written the number A in two different ways as a fraction. Unfortu-
nately, two numbers have become invisible due to spilled ink spots:

+3 15
A= = .
12 26 −

Lia has found Jan’s note and wants to find all possibilities for the number
A. She knows that the numbers underneath the ink spots must be positive
integers, but the number A does not have to be an integer. Because she
does not know which numbers are hidden below the ink spots, she searches
for all combinations for which the equality holds. In this way, she finds
multiple possible values for A. The largest one she calls Amax and the
smallest one she calls Amin .
Determine Amax − Amin .

38
6. Tim draws five straight lines in a plane. The lines are continuing indefinitely,
and there are no three lines going through the same point. For each
intersection point of two lines he gets one sweet, and for each set of two or
more parallel lines he gets one sweet as well. For instance, in the example
below, he gets 8 sweets because firstly there are 7 intersection points, and
secondly there is 1 set of three parallel lines, which is worth another sweet.

What are all possible numbers of sweets that Tim can get?

7. Six equilateral and equally sized triangles are glued to form a parallelogram
ABCD, see the figure.

D C
Y
X

A B

The length of AC is 10. What is the length of XY ?

Attention: the figure is not drawn to scale.

8. In a class room there are a number of students. They find out that for each
triplet of students, the following two statements are both true:
• Two of them never wrote a report together.
• Two of them did once write a report together.
What is the maximum possible number of students in the class room?

39
Answers

Part 1

1. B) 6

2. E) 6

3. D) 28th: could be either, 29th: blue

1
4. A) 72

5. C) 920

6. C) 32

7. D)

8. D) 19

Part 2
57
1. 5 years 5. 4 (= 14 14 )

2. 20 6. 1, 5, 8, 10

5 10
3. 2009 7. 6 (= 12 )

20
4. 3 (= 6 23 ) 8. 5

40
 We thank our sponsors
NEDERLANDSE
WISKUNDE
OLYMPIADE

Contents

1 Introduction
3 First Round, January 2018
7 Second Round, March 2018
12 Final Round, September 2018
19 BxMO Team Selection Test, March 2019
23 IMO Team Selection Test 1, May 2019
27 IMO Team Selection Test 2, May 2019
31 IMO Team Selection Test 3, May 2019
35 Junior Mathematical Olympiad, September 2018

© Stichting Nederlandse Wiskunde Olympiade, 2019 FOUNDATION COMPOSITIO MATHEMATICA