48th Dutch Mathematical Olympiad 2009

and the team selection for IMO 2010 Astana

First Round, January 2009

Final Round, September 2009

BxMO Team Selection Test, March 2010

Benelux Mathematical Olympiad, April 2010 C

IMO Team Selection Tests, June 2010 CM

MY

CY

We eat problems
CMY

Junior Mathematical Olympiad, October 2009 K

for breakfast.
Preferably unsolved ones...

In juli 2011 wordt de internationale wiskunde olympiade

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

Olympiad 2009
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
 WISKUNDE
Hoofdsponsors van de OLYMPIADE
We thank our sponsors
NEDERLANDSE
WISKUNDE 2010
OLYMPIADE
Overige sponsors
en donateurs

Contents

1 Introduction
3 First Round, January 2009
9 Final Round, September 2009
14 BxMO Team Selection Test, March 2010
18 Benelux Mathematical Olympiad, April 2010
24 IMO Team Selection Test 1, June 2010
29 IMO Team Selection Test 2, June 2010 Centraal Bureau voor de Statistiek

33 Junior Mathematical Olympiad, October 2009

© Stichting Nederlandse Wiskunde Olympiade, 2010

Introduction
In 2009 the Dutch Mathematical Olympiad consisted of two rounds. The
first round was held on 30 January 2009 at the participating schools. The
paper consisted of eight multiple choice questions and four open-answer
questions, and students got two hours to work on it. In total 4379 students
of 230 secondary schools participated in this first round.

The best students were invited for the final round. In order to stimulate
young students to participate, we set different thresholds for students from
different grades. Those students from grade 5 (4, ≤ 3) that scored 26 (23,
20) points or more on the first round (out of a maximum of 36 points)
were invited to the final round. Also some outstanding participants in the
Kangaroo math contest or the Pythagoras Olympiad were invited.

For the second consecutive year we organised training sessions at six uni-
versities in the country for the 155 students who had been invited for the
final round. Former Dutch IMO-participants were involved in the training
sessions at each of the universities.

Out of those 155, in total 131 participated in the final round on 18 Septem-
ber 2009 at Eindhoven University of Technology. This final round contained
five problems for which the students had to give extensive solutions and
proofs. They had three hours for the paper. After the prizes had been
awarded in the beginning of November, the Dutch Mathematical Olympiad
concluded its 48th edition 2009. In 2011 we will have our 50th edition.

The 31 most outstanding candidates of the Dutch Mathematical Olympiad

2009 were invited to an intensive seven-month training programme, con-
sisting of weekly problem sets. Also, the students met twice for a three-day
training camp, three times for a day at the university, and finally for a
six-day training camp in the beginning of June.

On 5 March 2010 the first selection test was held. The best ten students
participated in the second Benelux Mathematical Olympiad (BxMO), held
in Amsterdam, the Netherlands.

In June, out of those 10 students and 3 reserve candidates, the team for
the International Mathematical Olympiad 2010 was selected by two team
selection tests on 9 and 12 June 2010. A seventh, young, promising student
was selected to accompany the team to the IMO. The team had a training
camp in Astana from 28 June until 5 July, together with the team from

1
New Zealand.

For younger students we organised the second Junior Mathematical Olym-

piad in October 2009 at the VU University Amsterdam. The students
invited to participate in this event were the 30 best students of grade 1,
grade 2 and grade 3 of the popular Kangaroo math contest. The competi-
tion consisted of two one-hour parts, one with 15 multiple choice questions
and one with 10 open-answer questions. The goal of this Junior Mathe-
matical Olympiad is to scout talent and stimulate them to participate in
the first round of the Dutch Mathematical Olympiad.

The Dutch team for IMO 2010 Kazakhstan consists of

• Guus Berkelmans (16 y.o.)
• Harm Campmans (18 y.o., participated in IMO 2009 as well)
• Madelon de Kemp (17 y.o.)
• David Kok (17 y.o., participated in IMO 2009 as well)
• Daniël Kroes (16 y.o.)
• Merlijn Staps (15 y.o., observer C at IMO 2009)

We bring as observer C the promising young student

• Jeroen Huijben (14 y.o.)

The team is coached by

• Birgit van Dalen (deputy leader), Leiden University
• Johan Konter (team leader), Utrecht University
• Quintijn Puite (observer A), Eindhoven University of Technology
• Sietske Tacoma (observer B), Utrecht University

The Dutch delegation for IMO 2010 Kazakhstan further consists of

• Wim Berkelmans (member AB, observer A), VU University Amster-
dam
• Karst Koymans (observer A), University of Amsterdam
• Aad Loois (observer C)
• Jelle Loois (observer A), ORTEC
• Ronald van Luijk (observer A), Leiden University
• Rozemarijn Schalkx (observer A), Eindhoven University of Technol-
ogy

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.

2
First Round, January 2009
Problems
A-problems

A1. Ella does three tests. During the first one, she answered 60% of
the 25 questions correctly, during the second one, she answered 70%
of the 30 questions correctly, and during the last one, she answered
80% of the 45 questions correctly. Now if we merge these three tests
together to form one of 100 questions, what is the percentage of these
100 questions that Ella answered correctly?
A) 68% B) 70% C) 72% D) 74% E) 76%

A2. How many of the integers from 10 to 99 (10 and 99 included) have the
property that the sum of its digits is equal to the square of an integer?
(An example: The sum of the digits of 27 is equal to 2 + 7 = 9 = 32 .)
A) 13 B) 14 C) 15 D) 16 E) 17

A3. Ronald throws three dice. These dice look just like ordinary dice, but
their faces are numbered differently. The first die has the numbers 1,
1, 2, 2, 3 and 3 on it. The second die has the numbers 2, 2, 4, 4, 6
and 6 on it. And the third die has the numbers 1, 1, 3, 3, 5 and 5 on
it. He then adds up the three numbers he gets from rolling the three
dice. What is the probability that the resulting number is odd?
1 1 1 2 3
A) 4 B) 3 C) 2 D) 3 E) 4

A4. Three distinct numbers from the set

{1, 2, 3, 4, 5, 6, 7, 8, 9} are placed in the three squares
in the top of the figure to the right, after which the + +
numbers are added as described in said figure. We
call Max the highest number that can appear in the
bottom square, and Min the lowest number that
+
can appear there. What is the value of Max − Min?
A) 16 B) 24 C) 25 D) 26 E) 32

3
A5. The lengths of the diagonals of a rhombus have a ratio of 3 : 4. (A
rhombus is an equilateral quadrilateral.) The sum of the lengths of
the diagonals is 56. What is the diameter of this rhombus?
A) 80 B) 96 C) 100 D) 108 E) 160

A6. Wouter is traveling by foot from his home to the fitness center. He
also could have chosen to travel by bike, in which case he would travel
7 times as fast. But he left his bike at home. After walking for 1 km,
continuing to walk would take just as long as walking back to get his
bike, and then travel further by bike. By then, what is the distance
in km to the fitness center?
8 7 6 5 4
A) 7 B) 6 C) 5 D) 4 E) 3

A7. On the sides of an equilateral triangle, we draw

three squares. The sides of these squares that
are parallel to the sides of the triangle are ex-
tended until they intersect. These three intersec-
1 1
tions form another equilateral triangle. Suppose
1
that the length of a side of the original triangle is
equal to 1. What is the length of a side of the large
equilateral triangle?
√ √ √
A) 1 + 2 2 B) 5 − 12 3 C) 3 2
√ √
D) 1 + 2 3 E) 2 6

A8. Consider all four-digit numbers where each of the digit 3, 4, 6 and 7
occurs exactly once. How many of these numbers are divisible by 44?
A) 2 B) 4 C) 6 D) 8 E) 12

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

B1. On a sheet of paper, a grid of 101 by 101 white

squares is drawn. A chain is formed by coloring
squares grey as shown in the figure to the right.
The chain starts in the upper left corner and goes
on until it cannot go on any further. A large piece
of the sheet is torn off. How many squares were col-
ored grey in the original grid of 101 by 101 squares?

4
B2. The integer N consists of 2009 consecutive nines.
A computer calculates N 3 = (99999 . . . 99999)3 . 99999
| .{z
. . 99999}
3 2009×
How many nines does the number N contain in
total?

B3. Using, a wide brush, we paint the diagonals of a

square tile, as in the figure. Exactly half of the
1
area of this tile is covered with paint. Knowing
that as the width of the brush is 1, as indicated
in the figure, what is the length of the side of the
tile?

B4. Determine a triplet of integers (a, b, c) satisfying:

a + b + c = 18
a + b2 + c2 = 756
2

a2 = bc

Solutions
A-problems

A1. C) 72% 60% of 25 is 15; 70% of 30 is 21; and 80% of 45 is 36.

So in total, Ella answered 15 + 21 + 36 = 72 of the 100 questions
correctly.

A2. E) 17 We check how many of these numbers have sum of

digits equal to 1, 2, etc. There is 1 number with sum 1 (being 10);
there are 2 with sum 2 (being 20 and 11); etc.; 9 with sum 9 (being
90, 81, . . . , 18); also, 9 with sum 10 (being 91, 82, . . . , 19); etc.; and
finally, 1 with sum 18 (being 99); see the table below. Then the sum
of digits is a square of an integer (i.e. 1, 4, 9 or 16) in 1+4+9+3 = 17
of the 90 cases.

A3. B) 13 Note that the second die only has even numbers on it,
and that the third die only has odd numbers on it. So essentially
the question is to find the probability that rolling the first die gives
an even number. Since 2 of the 6 numbers on this die are even, this
probability is equal to 26 = 13 .

5
A4. D) 26 Let’s say we put a, b, and c in the
top three squares. Then the result in the bottom a b c
square is a+2b+c. So we can maximize this result
+ +
by making first b, then a and c as large as possible. b+c
a+b
Taking b = 9, a = 8 and c = 7 then yields 33 as
result. In the same way, we can minimize the result +
a+2b+c
by making first b, then a and c as small as possible.
Taking b = 1, a = 2, c = 3 yields 7 as result. The
difference between these numbers is 33 − 7 = 26.

A5. A) 80 Note that the diagonals have lengths

3 4 5 5
7 · 56 = 24 and 7 · 56 = 32. So the halves of diag-
3
4 4
onals have lengths 12 = 3 · 4 and 16 = 4 · 4. So the
rhombus is 4 times larger than the rhombus in the 3
5 5
figure, which consists of four triangles with sides
3,4 and 5. Hence the sides of the original rhombus
have length 4 · 5 = 20, and thus the diameter has
length 4 · 20 = 80.

A6. E) 43 Let x be said distance and let us suppose that he has

spent a quarter of an hour walking by then. Then continuing walking
will take him x quarters of an hour. On the other hand, if he decides
to walk back home to pick up his bike, he’ll first have to spend one
quarter of an hour to get back, and then 1+x 7 quarters of an hour
by bike; since he travels 7 times faster that way. Then we have
x = 1 + 1+x 7 , so 7x = 7 + (1 + x), or 6x = 8. We deduce that
x = 68 = 34 . Taking for ‘quarter of an hour’ any other time unit will
give us the same result.

√
A7. D) 1 + 2 3 In 4ABC, ∠A is half of 60◦ , so
30◦ . Also, ∠C is a right angle, so 4ABC is
a 30◦ -60◦ -90◦ -triangle, where |BC| = 1. So
it’s half of an equilateral triangle with sides
2: |AB| = 2. Now we calculate |AC| √ with the B
|AC| 2 12 =
Theorem
√ of Pythagoras: √ 2 −√
= 2
1
3. So the required length is 3 + 1 + 3.
A 3 C

6
A8. A) 2 Suppose that n, having digits a, b, c and d (so n =
1000a + 100b + 10c + d) is divisible by 44. Then it is also divisible by
11. Since the number m = 1001a + 99b + 11c is also divisible by 11,
so is m − n. Hence m − n = a − b + c − d is a multiple of 11. But this
number is at most the sum of the two highest digits, minus the sum
of the lowest two, so 13 − 7 = 6, and in the same way, we see that this
number is at least −6. So it has to be equal to 0. Thus a + c = b + d,
and since the sum of the digits is 20, we have a + c = b + d = 10.
First suppose that d = 4. Then b = 6 so we get two possibilities for
n, namely 3674 and 7634. But neither of them is divisible by 4, let
alone by 44. Now suppose that d = 6, then we have b = 4, and in this
case we get 3476 and 7436, both of which are divisible by 44. Finally,
note that since n is divisible by 4, d must be even. We deduce that
we have only 2 such numbers that are divisible by 44.
Alternative solution Check all 24 possibilities, or just the 12 even
possibilities, of even only the 6 multiples of 4.

B-problems

B1. 5201 We can subdivide this grid of 1012 squares as follows. In

the upper left corner, we have one (grey) square, then two L-shaped
pieces, one having 3 squares (one of which grey), the other having 5
squares (all of which grey). Then we have two more L-shaped pieces,
one having 7 squares (one of which grey), the other having 9 squares
(all of which grey), etc. Of the last two L-shaped pieces, the first
one has 199 squares (one of which grey), and the second one has 201
squares (all of which grey). We have 50 pairs of L-shapes in total, so
the total number of grey squares is 1+(1+5)+(1+9)+(1+13)+. . .+
(1 + 201) = 1 + (6 + 10 + 14 + . . . + 202) = 1 + 21 · 50 · (6 + 202) = 5201.

Alternative solution In each pair of these L-shapes, there are 4

more grey squares than white squares. So there are 50 · 4 = 200 more
grey squares than white squares in the 1012 − 1 = 10200 squares
contained in the 50 pairs of L-shapes, so we have 5000 white squares
and 5200 grey ones. Since the upper left square is grey, in total, we
have 5201 grey squares.

7
B2. 4017 93 = 729; 993 = 970299; 9993 = 997002999. It seems to
be the case that in general, the third power of a number n consisting
of k consecutive nines takes the following form: first k − 1 nines; then
a 7; then k − 1 zeroes; then a 2; and finally k nines. To prove this, we
write n = 10k − 1. Indeed: (10k − 1)3 = 103k − 3 · 102k + 3 · 10k − 1 =
102k (10k − 3) + (3 · 10k − 1). The number 10k − 3 can be written as
999 . . . 997 with k − 1 nines. Multiplied with 102k this gives a number
that ends in 2k zeroes. Adding 3 · 10k − 1 to this number, the last
k + 1 zeroes are replaced with 2999 . . . 999 with k nines. So in total,
we have (k − 1) + k nines; in our case, k = 2009, so we have 4017
nines.

√
B3. 2 + 2 2 We only need to look at a
quarter of the tile: 4ABC. The area of
4P QR is half of the area of 4ABC. The B Q
1
triangles are similar,
√ so corresponding sides 2 2
have
√ a ratio of 1 : 2, so |QR| : |BC| = 1 : Q
2. Now we calculate |BQ| using the The-
orem of Pythagoras in 4BQQ0 : 2|BQ| q =
2 x P
A
0 2 2 2 1
|BQ | + |BQ| = 1 , so |BQ| = 2 = R
1
√
2 2. Now √ let us write x for |QR|. Then 1
2
√ √ √ 2
we find x + 2 = 2 · x, so x( √ 2 −√1) = 2 C
√
or equivalently, x = √2−12
= 2(2−1 2+1)
=
√ √ √
2+ 2. Hence
√ |BC|
√ = x+ √ 2 = 2+2
√ 2 (or
|BC| = 2 · x = 2 · (2 + 2) = 2 2 + 2).

B4. (a, b, c) = (−12, 6, 24) of (a, b, c) = (−12, 24, 6) (one answer is enough)
We calculate (b + c)2 in two different ways. (b + c)2 = (18 − a)2 =
324 − 36a + a2 and (b + c)2 = b2 + 2bc + c2 = (756 − a2 ) + 2a2 .
So a2 − 36a + 324 = a2 + 756, or −36a = 756 − 324 = 432, so
a = −12. Substituting this in the first and in the last equation, we
obtain the equations b + c = 30 and bc = 144. Trying some divisors
of 144 = 122 , we then should be able to find a solution. Or we can
just substitute c = 30 − b in the last equation, yielding the quadratic
equation b(30 − b) = 144, or equivalently b2 − 30b + 144 = 0. We can
factorize this as (b − 6)(b − 24) = 0 (or we can use the abc-formula)
to see that we have two solutions b = 6 (and c = 24) or b = 24 (and
c = 6).

8
Final Round, September 2009
Problems
For these problems not only the answer is important; you also have to describe the way you
solved the problem.

1. In this problem, we consider integers consisting of 5 digits, of which

the first and last one are nonzero. We say that such an integer is a
palindromic product if it satisfies the following two conditions:
• the integer is a palindrome, (i.e. it doesn’t matter if you read it
from left to right, or the other way around);
• the integer is a product of two positive integers, of which the
first, when read from left to right, is equal to the second, when
read from right to left, like 4831 and 1384.
For example, 20502 is a palindromic product, since 102 · 201 = 20502,
and 20502 itself is a palindrome.
Determine all palindromic products of 5 digits.

2. Consider the sequence of integers 0, 1, 2, 4, 6, 9, 12, . . . obtained by

starting with zero, adding 1, then adding 1 again, then adding 2, and
adding 2 again, then adding 3, and adding 3 again, and so on. If we
call the subsequent terms of this sequence a0 , a1 , a2 , . . . , then we
have a0 = 0, and

a2n−1 = a2n−2 + n, a2n = a2n−1 + n

for all integers n > 1.

Find all integers k > 0 for which ak is the square of an integer.

3. A tennis tournament has at least three participants. Every partici-

pant plays exactly one match against every other participant. More-
over, every participant wins at least one of the matches he plays.
(Draws do not occur in tennis matches.)
Show that there are three participants A, B and C for which the
following holds: A wins against B, B wins against C, and C wins
against A.

9
 4. Let ABC be an arbitrary triangle. On the perpendicular bisector of
AB, there is a point P inside of triangle ABC. On the sides BC and
CA, triangles BQC and CRA are placed externally. These triangles
satisfy 4BP A ∼ 4BQC ∼ 4CRA. (So Q and A lie on opposite
sides of BC, and R and B lie on opposite sides of AC.)
Show that the points P , Q, C and R form a parallelogram.

5. We number a hundred blank cards on both sides with the numbers 1

to 100. The cards are then stacked in order, with the card with the
number 1 on top.
The order of the cards is changed step by step as follows: at the 1st
step the top card is turned around, and is put back on top of the stack
(nothing changes, of course), at the 2nd step the topmost 2 cards are
turned around, and put back on top of the stack, up to the 100th
step, in which the entire stack of 100 cards is turned around. At the
101st step, again only the top card is turned around, at the 102nd
step, the topmost 2 cards are turned around, and so on.
Show that after a finite number of steps, the cards return to their
original positions.

Solutions

1. Note that the two integers mentioned in the second condition cannot
end in a 0; since otherwise the product would also have to end in
a 0. Thus, the two numbers have an equal number of digits. But
a product of two integers of four or more digits consists of at least
seven digits, hence is too large, and a product of two integers of at
most two digits consists of at most four digits, hence is too small.
So to satisfy the second condition, we have to consider products of
integers consisting of three digits, say abc and cba.
If we work this out, we obtain: abc · cba = (100a + 10b + c)(100c +
10b + a) = 10000ac + 1000b(a + c) + 100(a2 + b2 + c2 ) + 10b(a + c) + ac.
If ac > 9, then this integer consists of more than 5 digits. Hence
ac 6 9. Now if b(a + c) > 9, then the first digit will be greater
than ac, hence it will no longer be equal to the last digit. Hence we
also have b(a + c) 6 9. We can use a similar argument to show that
a2 +b2 +c2 6 9. We deduce from this that 1+b2 +1 6 a2 +b2 +c2 6 9,
and hence that b is equal to one of 0,1,2.

If b = 0, then a2 + c2 6 9, so we have 1 6 a 6 2 and 1 6 c 6 2.

10
 If b = 1, the conditions become a + c 6 9 and a2 + c2 6 8, and we
must also have 1 6 ac 6 9. From the second condition, we obtain
1 6 a 6 2 and 1 6 c 6 2.
If b = 2, the conditions become a + c 6 4 and a2 + c2 6 5. Hence we
have three possibilities for (a, b).
In the above table, we have worked out all the possibilities. It turns
out that there are eight possible palindromic products: 10201, 12321,
14641, 20502, 23632, 26962, 40804 and 44944. 

2. For n > 1, we have a2n = a2n−1 + n = a2n−2 + 2n, so a2n = 0 + 2 +

4 + · · · + 2n = 2(0 + 1 + 2 + · · · + n) = n(n + 1). Since n > 0, we have
n2 < n2 +n < n2 +2n+1 = (n+1)2 . Hence a2n = n(n+1) cannot be a
square; it lies between two subsequent squares. So for the even k > 2,
ak is not a square. From a2n−1 = a2n − n = n(n + 1) − n = n2 , we
deduce that a2n−1 is always a square, hence all odd k are as desired.
Finally, we note that a0 = 0 is a square as well. So ak is a square if
k is odd or k = 0, and it’s not a square otherwise. 

3. We first show that there is a cycle, i.e. m distinct participants

A1 , . . . , Am such that A1 won against A2 , A2 won against A3 , and
so on, up to Am , who won against A1 . Pick an arbitrary participant
B1 . He won against another participant B2 , who won against a cer-
tain B3 , and so on. Since we have only finitely many players, we will
have a repetition in our sequence of participants, i.e., B1 won against
B2 , B2 won against B3 , . . . , Bu won against Bu+1 , where Bu+1 is a
participant that occurred earlier in our sequence, say Bt . Then Bt
up to Bu indeed form a cycle. Note that the length m of a cycle is at
least 3. In fact, we now need to show that there is a cycle of length
3.
Since there exists a cycle, we can take one of minimal length M , say
C1 , . . . , CM . If M > 3, then we consider the match between C1 and
C3 . If C3 won against C1 , then we have obtained a cycle of length
3, which is smaller than M , yielding a contradiction. On the other
hand, if C1 won against C3 , then, by removing C2 , we obtain a cycle
of length M − 1, which is again smaller than M . This is a contra-
diction as well. We deduce that M = 3 and that C1 , C2 , C3 is the
desired cycle of length 3. 

11
 Alternative solution 1: Pick a participant A who has won the least
number of matches, and a participant B who lost to A. Consider
the set X of participants who lost to B. Then the set X does not
contain A. If there is no triple of participants as desired, then A
must have won against all participants in X. But A also won against
B. But then A has won at least one match more than B, yielding a
contradiction. 

4. Note that |AP | = |P B|, and hence that |BQ| = |QC| and |CR| =
|RA|. Since 4BP A ∼ 4BQC, it follows that |P B| |QB|
|AB| = |CB| , and hence
|P B| |AB|
that |QB| = |CB| . Moreover, since Q lies outside triangle ABC and
P lies inside of it, we have

∠QBP = ∠QBC + ∠CBP = ∠P BA + ∠CBP = ∠CBA,

from which follows that 4ABC ∼ 4P BQ. Note that 4ABC is

|AB|
|P B| times larger than 4P BQ. Similarly, it follows that 4ABC ∼
|AB|
4AP R, and note that 4ABC is |P A| times larger than 4AP R. Since
|P A| = |P B|, from which follows that |AB| |AB|
|P A| = |P B| , we deduce that
4P BQ ∼ = 4AP R. Hence |QP | = |RA| = |CR| and |P R| = |BQ| =
|QC|.
So the two pairs of opposite sides of quadrilateral P QCR have equal
length, hence quadrilateral P QCR is a parallelogram. 

12
5. We first consider the effect of two consecutive steps. If before step
2k −1, the stack of cards, from top to bottom, consists of a1 , . . . , a100 ,
the first 2k − 1 cards are reversed in order, so we obtain
a2k−1 , a2k−2 . . . , a2 , a1 , a2k , a2k+1 . . . , a100 . Then, at the 2k th step,
the top 2k cards are reversed in order, yielding
a2k , a1 , a2 , . . . , a2k−2 , a2k−1 , a2k+1 , . . . , a100 . Hence the effect of these
two steps is that the 2k th card is moved up to the top, and the rest
is shifted down. So if we consider the first 100 steps, then first 2 is
put on top, then 4 is put on top of that, then 6, and so on, so we ob-
tain 100, 98, 96, . . . , 4, 2, 1, 3, 5, . . . , 97, 99. We call this a megastep.
In general, a megastep changes the order a1 , a2 , . . . , a99 , a100 into
a100 , a98 , a96 , . . . , a4 , a2 , a1 , a3 , a5 , . . . , a97 , a99 . This megastep is in-
vertible; the order after a megastep can only come from one unique
order. Now we there are only finitely many possible ways (namely
100!) to order the 100 cards, so after applying megasteps for a while,
we’ll obtain an order that has occurred before. Consider the first
megastep that yields an order that has occurred before. If this is not
the starting position, then one megastep before, the position should
have occurred earlier as well; a contradiction. Hence we return to the
starting position after a finite number of (mega)steps. 

Alternative solution: We again consider a megastep. The first

(topmost) card goes to the 51st place; the 51st card goes to the 76th
place, etc. This yields a so-called cycle, which we denote as 1 → 51 →
76 → 13 → · · · → 1. From this notation, we deduce for example that
after two megasteps, the first card moves to the 76th place. If the first
card returns to its original place after k megasteps, we say that the
length of the cycle is k. It’s now clear that the 51st card also returns
to the 51st place after k megasteps. In this way, we can describe the
behavior of the cards in a megastep, with a number of cycles. Now if
we take a common multiple of the lengths of the cycles, all cards will
return to their original places. 

13
BxMO Team Selection Test, March 2010
Problems

1 Let ABCD be a trapezoid with AB k CD, 2|AB| = |CD| and BD ⊥

BC. Let M be the midpoint of CD and let E be the intersection BC
and AD. Let C be the intersection of AM and BD. Let N be the
intersection of OE and AB.
(a) Prove that ABM D is a rhombus.
(b) Prove that the line DN passes through the midpoint of the line
segment BE.

2 Find all functions f : R → R satisfying

f (x)f (y) = f (x + y) + xy

for all x, y ∈ R.

3 Let N be the number of ordered 5-tuples (a1 , a2 , a3 , a4 , a5 ) of positive

integers satisfying
1 1 1 1 1
+ + + + = 1.
a1 a2 a3 a4 a5

Is N even or odd?

4 The two circles Γ1 and Γ2 intersect at P and Q. The common tan-

gent that’s on the same side as P , intersects the circles at A and B,
respectively. Let C be the second intersection with Γ2 of the tan-
gent to Γ1 at P , and let D be the second intersection with Γ1 of the
tangent to Γ2 at Q. Let E be the intersection of AP and BC, and
let F be the intersection of BP and AD. Let M be the image of
P under point reflection with respect to the midpoint of AB. Prove
that AM BEQF is a cyclic hexagon.

5 For any non-negative integer n, we say that a permutation

(a0 , a1 , . . . , an ) of {0, 1, . . . , n} is quadratic if k + ak is a square for
k = 0, 1, . . . , n. Show that for any non-negative integer n, there exists
a quadratic permutation of {0, 1, . . . , n}.

14
Solutions

1 From 2|AB| = |CD| and AB k CD, we deduce that AB is a mid-

segment in triangle CDE. Hence A is the midpoint of DE. Since
∠DBE = 90◦ , by Thales’ Theorem, A is the center of the circle pass-
ing through D, B and E. Hence |AD| = |AE| = |AB|. We already
know that |AB| = |DM | = |M C|. In a similar way, using Thales’
Theorem, we obtain |BM | = |CM | = |DM |. Hence all sides of the
quadrilateral ABM D have the same length; ABM D is a rhombus
(a). The intersection of the diagonals of a rhombus is the midpoint
of both diagonals, hence O is the midpoint of BD. Since A is the
midpoint of DE, it follows that N is the centroid of triangle BDE.
Hence DN passes through the midpoint of BE (b). 

2 First, note that the function given by f (x) = 0 for all x ∈ R, does
not satisfy the functional equation. Hence there exists an x0 ∈ R for
which f (x0 ) 6= 0. Substituting x = x0 and y = 0 gives f (x0 )f (0) =
f (x0 ). Since f (x0 ) 6= 0, we obtain f (0) = 1. Now substituting x = 1
and y = −1 gives f (1)f (−1) = f (0) − 1 = 1 − 1 = 0. Hence f (1) = 0
or f (−1) = 0.
We now distinguish the two cases. First, suppose that f (1) = 0, and
substitute x = 1. We then get 0 = f (1)f (y) = f (1 + y) + y for all
y ∈ R, so f (1 + y) = −y for all y ∈ R. Substituting y = t − 1 now
gives f (t) = −t + 1 for all t ∈ R. This is our first candidate solution.
Now suppose that f (−1) = 0 and substitute x = −1. We obtain
0 = f (−1)f (y) = f (−1 + y) − y for all y ∈ R, hence f (−1 + y) = y
for all y ∈ R. Substituting y = t + 1, we obtain f (t) = t + 1 for all
t ∈ R. This is our second candidate solution.
Since we have covered all the cases, these are the only two possible
solutions. Let’s check them.
If f (t) = −t + 1 for all t ∈ R, then for all x, y ∈ R:

f (x)f (y) = (−x + 1)(−y + 1) = (−x − y + 1) + xy = f (x + y) + xy.

If f (t) = t + 1 for all t ∈ R, then for all x, y ∈ R:

f (x)f (y) = (x + 1)(y + 1) = (x + y + 1) + xy = f (x + y) + xy.

Hence both candidate solutions are indeed solutions of the functional

equation. 

15
3 Solution I. Consider an unordered 5-tuple that satisfies the given
equation, and suppose that it consists of the distinct integers b1 , . . . ,
bk , where bi occurs exactly ti times. Hence t1 +· · ·+tk = 5. Note that
5!
this unordered 5-tuple can be ordered in t1 !···t k!
ways. This is odd
if and only if three factors 2 occur in the denominator, which holds
if and only if 4! or 5! occurs in the denominator. Hence the only
unordered 5-tuples that can be ordered in a odd number of ways, are
those that consist of at least 4 numbers that are the same.
Such a 5-tuple has the form (a, a, a, a, b), where b can be equal to a,
and where a, b are positive integers. We obtain the equation a4 + 1b = 1,
or, equivalently, 4b + a = ab. We deduce that b has to be a divisor
of a and that a is a divisor of 4b. Hence there are three possibilities:
a = b, a = 2b and a = 4b. In the first case, we obtain 5b = b2 ,
hence b = 5, yielding the solution (5, 5, 5, 5, 5). In the second case,
we obtain 6b = 2b2 , hence b = 3, yielding the solution (6, 6, 6, 6, 3).
In the third and final case, we obtain 8b = 4b2 , so b = 2, yielding the
solution (8, 8, 8, 8, 2). Hence there are three unordered 5-tuples which
can be ordered in an odd number of ways. Hence N is odd. 
Solution II. Let (a1 , a2 , a3 , a4 , a5 ) be an ordered 5-tuple that satisfies
the given equation. Then also the 5-tuple (a2 , a1 , a3 , a4 , a5 ). If a1 6=
a2 , this is a different ordered 5-tuple. Hence there is an even number
of solutions (a1 , a2 , a3 , a4 , a5 ) with a1 6= a2 . So we may only consider
the solutions with a1 = a2 . In the same way, we show that there is
an even number of solutions (a1 , a1 , a3 , a4 , a5 ) with a3 6= a4 . Hence
we may only consider the solutions with a1 = a2 , a3 = a4 . For
each solution (a1 , a1 , a3 , a3 , a5 ) with a1 6= a3 , there is yet another
solution (a3 , a3 , a1 , a1 , a5 ), hence there is an even number of solutions
(a1 , a1 , a3 , a3 , a5 ) with a1 6= a3 . Thus we may only consider the
solution of the form (a, a, a, a, b), where b can be equal to a.
We obtain the equation a4 + 1b = 1, or equivalently, 4b + a = ab. We
can rewrite this last equation as (a − 4)(b − 1) = 4. Since b is a
positive integer, we have b − 1 ≥ 0. Hence b − 1 is a positive divisor
of 4, namely 1, 2 or 4. This yields the three solutions (a, b) = (8, 2),
(a, b) = (6, 3) and (a, b) = (5, 5). Since the number of solutions that
we covered previously, is even, we deduce that N is odd. 

16
4 We express all relevant angles in terms of α = ∠BAP and β =
∠P BA. The midpoint of AB is, by definition, also the midpoint of
P M , hence the diagonals of quadrilateral divide each other into equal
segments. Hence AP BM is a parallelogram and we have ∠AM B =
∠AP B. Note that (using the sum of the angles of a triangle) ∠AP B+
α + β = 180◦ . Hence 180◦ − ∠AM B = α + β.
By the inscribed angle theorem, applied on the tangent AB to Γ1 ,
we have α = ∠ADP . Then, by the inscribed angle theorem, applied
on the chord AP of Γ1 , it follows that ∠ADP = ∠AQP . Hence
α = ∠AQP . In a similar way, we deduce that β = ∠P CB = ∠P QB.
Hence ∠AQB = ∠AQP + ∠P QB = α + β = 180◦ − ∠AM B, so
AQBM is a cyclic quadrilateral.
Now let S be the intersection of DP and AB. Using the inscribed an-
gle theorem, we deduce that ∠SP B = ∠P CB = ∠P BS = ∠P BA =
β. Hence ∠DP F = ∠SP B = β, by opposite angles. Now we apply
the exterior angle theorem to 4DF P , to obtain ∠AF B = ∠AF P =
∠F DP + ∠DP F = α + β = ∠AQB. Hence AF QB is a cyclic quadri-
lateral. Hence F lies on the circumcircle of the cyclic quadrilateral
AQBM . Similarly, we see that E also lies on that circle. Hence
AM BEQF is a cyclic hexagon. 

5 We use induction on n. For n = 0 the permutation (0) is quadratic,

since 0 + 0 is a square.
Now let l ≥ 0 and assume that for every n ≤ l, there exists a quadratic
permutation (induction hypothesis). Consider n = l + 1. Let m be
such that m2 is the least square greater than or equal to l + 1. Now
we have l ≥ (m − 1)2 = m2 − 2m + 1, so 2(l + 1) ≥ 2m2 − 4m + 4 =
m2 + (m − 2)2 ≥ m2 . We deduce that there exists an integer p with
0 ≤ p ≤ l + 1 such that (l + 1) + p = m2 .
Now define a permutation as follows. If p ≥ 1, then we take
(a0 , a1 , . . . , ap−1 ) a quadratic permutation of {0, 1, . . . , p − 1}; which
exists, by our induction hypothesis, since p − 1 ≤ l. For p ≤ i ≤ l + 1,
we define ai = m2 − i. (Note that for p = 0, we now also have define
a permutation of (a0 , a1 , . . . , al+1 ).)
Now the sequence (ap , ap+1 , . . . , al+1 ) is exactly (l +1, l, . . . , p), which
implies that, together with the initial part, we have used all val-
ues from 0 up to l + 1. Also, ai + i is a square for all i. Hence
(a0 , a1 , . . . , al+1 ) is a quadratic permutation of {0, 1, . . . , l + 1}. 

17
Benelux Mathematical Olympiad, April 2010
Problems

1. A finite set of integers is called bad if its elements add up to 2010. A fi-
nite set of integers is a Benelux-set if none of its subsets is bad. Deter-
mine the smallest integer n such that the set {502, 503, 504, . . . , 2009}
can be partitioned into n Benelux-sets.
(A partition of a set S into n subsets is a collection of n pairwise
disjoint subsets of S, the union of which equals S.)

2. Find all polynomials p(x) with real coefficients such that

p(a+b−2c)+p(b+c−2a)+p(c+a−2b) = 3p(a−b)+3p(b−c)+3p(c−a)

for all a, b, c ∈ R.

3. On a line l there are three different points A, B and P in that order.

Let a be the line through A perpendicular to l, and let b be the line
through B perpendicular to l. A line through P , not coinciding with l,
intersects a in Q and b in R. The line through A perpendicular to BQ
intersects BQ in L and BR in T . The line through B perpendicular
to AR intersects AR in K and AQ in S.

(a) Prove that P , T , S are collinear.

(b) Prove that P , K, L are collinear.

4. Find all quadruples (a, b, p, n) of positive integers, such that p is a

prime and
a3 + b3 = pn .

18
Solutions

1. As 502 + 1508 = 2010, the set S = {502, 503, . . . , 2009} is not a

Benelux-set, so n = 1 does not work. We will prove that n = 2 does
work, i.e. that S can be partitioned into 2 Benelux-sets.
Define the following subsets of S:

A = {502, 503, . . . , 670},

B = {671, 672, . . . , 1005},
C = {1006, 1007, . . . , 1339},
D = {1340, 1341, . . . , 1508},
E = {1509, 1510, . . . , 2009}.

We will show that A ∪ C ∪ E and B ∪ D are both Benelux-sets.

Note that there does not exist a bad subset of S of one element, since
that element would have to be 2010. Also, there does not exist a bad
subset of S of more than three elements, since the sum of four or more
elements would be at least 502 + 503 + 504 + 505 = 2014 > 2010. So
any possible bad subset of S contains two or three elements.
Consider a bad subset of two elements a and b. As a, b ≥ 502 and
a + b = 2010, we have a, b ≤ 2010 − 502 = 1508. Furthermore, exactly
one of a and b is smaller than 1005 and one is larger than 1005. So
one of them, say a, is an element of A∪B, and the other is an element
of C ∪ D. Suppose a ∈ A, then b ≥ 2010 − 670 = 1340, so b ∈ D.
On the other hand, suppose a ∈ B, then b ≤ 2010 − 671 = 1339, so
b ∈ C. Hence {a, b} cannot be a subset of A ∪ C ∪ E, nor of B ∪ D.
Now consider a bad subset of three elements a, b and c. As a, b, c ≥
502, a + b + c = 2010, and the three elements are pairwise distinct,
we have a, b, c ≤ 2010 − 502 − 503 = 1005. So a, b, c ∈ A ∪ B. At least
one of the elements, say a, is smaller than 2010
3 = 670, and at least
one of the elements, say b, is larger than 670. So a ∈ A and b ∈ B.
We conclude that {a, b, c} cannot be a subset of A ∪ C ∪ E, nor of
B ∪ D.
This proves that A ∪ C ∪ E and B ∪ D are Benelux-sets, and therefore
the smallest n for which S can be partitioned into n Benelux-sets is
n = 2. 
Remark. Observe that A ∪ C ∪ E1 and B ∪ D ∪ E2 are also Benelux-
sets, where {E1 , E2 } is any partition of E.)

19
2. For a = b = c, we have 3p(0) = 9p(0), hence p(0) = 0. Now set
b = c = 0, then we have

p(a) + p(−2a) + p(a) = 3p(a) + 3p(−a)

for all a ∈ R. So we find a polynomial equation

p(−2x) = p(x) + 3p(−x). (1)

Note that the zero polynomial is a solution to this equation. Now

suppose that p is not the zero polynomial, and let n ≥ 0 be the
degree of p. Let an 6= 0 be the coefficient of xn in p(x). At the
left-hand side of (1), the coefficient of xn is (−2)n · an , while at the
right-hand side the coefficient of xn is an + 3 · (−1)n · an . Hence
(−2)n = 1 + 3 · (−1)n . For n even, we find 2n = 4, so n = 2, and
for n odd, we find −2n = −2, so n = 1. As we already know that
p(0) = 0, we must have p(x) = a2 x2 + a1 x, where a1 and a2 are real
numbers (possibly zero).
The polynomial p(x) = x is a solution to our problem, as

(a+b−2c)+(b+c−2a)+(c+a−2b) = 0 = 3(a−b)+3(b−c)+3(c−a)

for all a, b, c ∈ R. Also, p(x) = x2 is a solution, since

(a + b − 2c)2 + (b + c − 2a)2 + (c + a − 2b)2

= 6(a2 + b2 + c2 ) − 6(ab + bc + ca)
= 3(a − b)2 + 3(b − c)2 + 3(c − a)2

for all a, b, c ∈ R.
Now note that if p(x) is a solution to our problem, then so is λp(x)
for all λ ∈ R. Also, if p(x) and q(x) are both solutions, then so is
p(x) + q(x). We conclude that for all real numbers a2 and a1 the
polynomial a2 x2 + a1 x is a solution. Since we have already shown
that there can be no other solutions, these are the only solutions. 

20
3. Solution 1.
(a) Since P , R and Q are collinear, we have 4P AQ ∼ 4P BR,
hence
|AQ| |AP |
= .
|BR| |BP |
Conversely, P , T and S are collinear if it holds that
|AS| |AP |
= .
|BT | |BP |
So it suffices to prove
|BT | |AS|
= .
|BR| |AQ|

Since ∠ABT = 90◦ = ∠ALB and ∠T AB = ∠BAL, we have

4ABT ∼ 4ALB. And since ∠ALB = 90◦ = ∠QAB and
∠LBA = ∠ABQ, we have 4ALB ∼ 4QAB. Hence 4ABT ∼
4QAB, so
|BT | |AB|
= .
|BA| |AQ|
Similarly, we have 4ABR ∼ 4AKB ∼ 4SAB, so
|BR| |AB|
= .
|BA| |AS|

Combining both results, we get

|BT | |BT |/|BA| |AB|/|AQ| |AS|
= = = ,
|BR| |BR|/|BA| |AB|/|AS| |AQ|
which had to be proved.
(b) Let the line P K intersect BR in B1 and AQ in A1 and let the
line P L intersect BR in B2 and AQ in A2 . Consider the points
A1 , A and S on the line AQ, and the points B1 , B and T on the
line BR. As AQ k BR and the three lines A1 B1 , AB and ST
are concurrent (in P ), we have

A1 A : AS = B1 B : BT,

where all lengths are directed. Similarly, as A1 B1 , AR and SB

are concurrent (in K), we have

A1 A : AS = B1 R : RB.

21
 This gives
BB1 RB1 RB + BB1 BB1 BB1
= = =1+ =1− ,
BT RB RB RB BR
so
1
BB1 = 1 1 .
BT + BR

Similary, using the lines A2 B2 , AB and QR (concurrent in P )

and the lines A2 B2 , AT and QB (concurrent in L), we find

B2 B : BR = A2 A : AQ = B2 T : T B.

This gives
BB2 T B2 T B + BB2 BB2 BB2
= = =1+ =1− ,
BR TB TB TB BT
so
1
BB2 = 1 1 .
BR + BT
We conclude that B1 = B2 , which implies that P , K and L are
collinear.


Solution 2. As ∠AKB = ∠ALB = 90◦ , the points K and L belong

to the circle with diameter AB. Since ∠QAB = ∠ABR = 90◦ , the
lines AQ and BR are tangents to this circle.
Apply Pascal’s theorem to the points A, A, K, L, B and B, all on the
same circle. This yields that the intersection Q of the tangent in A
and the line BL, the intersection R of the tangent in B and the line
AK, and the intersection of KL and AB are collinear. So KL passes
through the intersection of AB and QR, which is point P . Hence P ,
K and L are collinear. This proves part b.
Now apply Pascal’s theorem to the points A, A, L, K, B and B.
This yields that the intersection S of the tangent in A and the line
BK, the intersection T of the tangent in B and the line AL, and the
intersection P of KL and AB are collinear. This proves part a. 

22
4. Let (a, b, p, n) be a solution. Note that we can write the given equation
as
(a + b)(a2 − ab + b2 ) = pn .
As a and b are positive integers, we have a + b ≥ 2, so p | a + b.
Furthermore, a2 − ab + b2 = (a − b)2 + ab, so either a = b = 1 or
a2 −ab+b2 ≥ 2. Assume that the latter is the case. Then p is a divisor
of both a+b and a2 −ab+b2 , hence also of (a+b)2 −(a2 −ab+b2 ) = 3ab.
This means that p either is equal to 3 or is a divisor of ab. Since p
is a divisor of a + b, we have p | a ⇔ p | b, hence either p = 3, or
p | a and p | b. If p | a and p | b, then we can write a = pa0 , b = pb0
with a0 and b0 positive integers, and we have (a0 )3 + (b0 )3 = pn−3 , so
(a0 , b0 , p, n − 3) then is another solution (note that (a0 )3 + (b0 )3 is a
positive integer greater than 1, so n − 3 is positive).
Now assume that (a0 , b0 , p0 , n0 ) is a solution such that p - a. From the
reasoning above it follows that either a0 = b0 = 1, or p0 = 3. After
all, if we do not have a0 = b0 = 1 and we have p0 6= 3, then p | a.
Also, given an arbitrary solution (a, b, p, n), we can divide everything
by p repeatedly until there are no factors p left in a.
Suppose a0 = b0 = 1. Then the solution is (1, 1, 2, 1).
Suppose p0 = 3. Assume that 32 | (a20 − a0 b0 + b20 ). As 32 | (a0 + b0 )2 ,
we then have 32 | (a0 + b0 )2 − (a20 − a0 b0 + b20 ) = 3a0 b0 , so 3 | a0 b0 .
But 3 - a0 by assumption, and 3 | a0 + b0 , so 3 - b0 , which contradicts
3 | a0 b0 . We conclude that 32 - (a20 − a0 b0 + b20 ). As both a0 + b0 and
a20 − a0 b0 + b20 must be powers of 3, we have a20 − a0 b0 + b20 = 3. Hence
(a0 − b0 )2 + a0 b0 = 3. We must have (a0 − b0 )2 = 0 or (a0 − b0 )2 = 1.
The former does not give a solution; the latter gives a0 = 2 and b0 = 1
or a0 = 1 and b0 = 2.
So all solutions with p - a are (1, 1, 2, 1), (2, 1, 3, 2) and (1, 2, 3, 2).
From the above it follows that all other solutions are of the form
(pk0 a0 , pk0 b0 , p0 , n0 + 3k), where (a0 , b0 , p0 , n0 ) is one of these three
solutions. Hence we find three families of solutions:
• (2k , 2k , 2, 3k + 1) with k ∈ Z≥0 ,
• (2 · 3k , 3k , 3, 3k + 2) with k ∈ Z≥0 ,
• (3k , 2 · 3k , 3, 3k + 2) with k ∈ Z≥0 .

It is easy to check that all these quadruples are indeed solutions. 

23
IMO Team Selection Test 1, June 2010
Problems

1 Let ABC be an acute triangle such that ∠BAC = 45◦ . Let D a

point on AB such that CD ⊥ AB. Let P be an internal point of the
segment CD. Prove that AP ⊥ BC if and only if |AP | = |BC|.

2 Let A and B be positive integers. Define the arithmetic sequence

a0 , a1 , a2 , . . . by an = An + B. Suppose that there exists an n ≥ 0
2
such that an is a square. Let M be a positive integer such that √ M
is the smallest square in the sequence. Prove that M < A + B.

3 Let n ≥ 2 be a positive integer and p a prime such that n | p − 1 and

p | n3 − 1. Show that 4p − 3 is a square.

4 Let ABCD be a cyclic quadrilateral satisfying ∠ABD = ∠DBC. Let

E be the intersection of the diagonals AC and BD. Let M be the
midpoint of AE, and N be the midpoint of DC. Show that M BCN
is a cyclic quadrilateral.

5 Find all triples (x, y, z) of real (but not necessarily positive) numbers
satisfying
3(x2 + y 2 + z 2 ) = 1,
x2 y 2 + y 2 z 2 + z 2 x2 = xyz(x + y + z)3 .

Solutions

1 Let E be the intersection of AP and BC. Note that ∠DCA = 90◦ −

∠CAD = 90◦ − ∠CAB = 45◦ , so 4ACD is isosceles: |AD| = |CD|.
Now suppose that |AP | = |BC|. Since ∠ADP = 90◦ = ∠CDB, we
have 4ADP ∼ = 4CDB, by (SSR). Hence ∠AP D = ∠CBD, from
which follows that
∠CEA = ∠CEP = 180◦ − ∠EP C − ∠P CE
= 180◦ − ∠AP D − ∠DCB = 180◦ − ∠CBD − ∠DCB
= ∠BDC = 90◦ ,

24
 hence AP ⊥ BC.
Conversely, suppose that AP ⊥ BC, or equivalently, ∠CEP = 90◦ .
Then we have

∠AP D = ∠EP C = 90◦ − ∠P CE = 90◦ − ∠DCB = ∠CBD.

Since we also have ∠ADP = 90◦ = ∠CDB, from (SAA), it follows

that 4ADP ∼ = 4CDB. Hence |AP | = |BC|. 

√
2 If M ≤ A, then automatically, M < A + B, so we’re done in this
case.
Hence suppose that M > A. Let k be such that ak = M 2 . So
Ak + B = M 2 . Since 0 < M − A < M , the square (M − A)2 is smaller
than M 2 . We have (M − A)2 = M 2 − 2M A + A2 = M 2 − A(2M − A).
So if k − (2M − A) ≥ 0, then

ak−(2M −A) = A(k − (2M − A)) + B = (Ak + B) − 2M A + A2

= M 2 − 2M A + A2 = (M − A)2 ,

and in that case, M 2 is not the smallest square in the sequence,

contrary to our assumption. Hence k − (2M − A) < 0. From this, it
follows that A(k−(2M −A))+B < B, or equivalently,√(M −A)2 < B.
Since M −√A is positive, we deduce that M − A < B, hence that
M < A + B. 

3 Solution I. From n | p − 1, we deduce that n < p. Since p is prime,

p divides one of the two factors of n3 − 1 = (n − 1)(n2 + n + 1), but p
is too large to be a divisor of n − 1 > 0. Hence p | n2 + n + 1. Since
n | p − 1, we can write p as kn + 1, where k is a positive integer. Since
kn + 1 | n2 + n + 1, we have

kn + 1 | k(n2 + n + 1) − (n + 1)(kn + 1) = k − n − 1.

We now distinguish three cases.

Case 1: k > n + 1. In this case, k − n − 1 is positive, and we must
have kn + 1 ≤ k − n − 1. It follows that (k + 1)n ≤ k − 2. But the
left hand side is clearly larger than the right hand side, yielding a
contradiction.
Case 2: k < n + 1. In this case, k − n − 1 is negative, and we must
have kn + 1 ≤ n + 1 − k. It follows that (k − 1)n ≤ −k. But the

25
 left hand side is non-negative, and the right hand side is negative,
yielding a contradiction.
Case 3: k = n + 1. Now we have p = (n + 1)n + 1 = n2 + n + 1.
Hence 4p − 3 = 4n2 + 4n + 1 = (2n + 1)2 is a square.
We deduce that in the only possible case, 4p − 3 is a square. 

Solution II. As in the first solution, we show that p | n2 + n + 1.

If p = n2 + n + 1, then we have 4p − 3 = 4n2 + 4n + 1 = (2n + 1)2 ,
which is a square. Now suppose that p 6= n2 + n + 1. Then there is
an integer m > 1 such that pm = n2 + n + 1. Reducing modulo n, we
get pm ≡ 1 mod n. From n | p − 1, we deduce that p ≡ 1 mod n,
hence m ≡ 1 mod n. So p and m are at least n + 1. However,
(n + 1)2 = n2 + 2n + 1 > n2 + n + 1 = pm, yielding a contradiction.


4 Solution I. Since ABCD is a cyclic quadrilateral, we have ∠BDC =

∠BAC = ∠BAE. Also, from the given properties, it follows that
∠CBD = ∠EBA. Hence we have, by (aa), 4DCB ∼ 4AEB. We
now show that this implies that 4N CB ∼ 4M EB.
First of all, we have ∠N CB = ∠DCB = ∠AEB = ∠M EB. Also,
|N C| 2|N C| |DC| |CB|
we have |M E| = 2|M E| = |AE| = |EB| . By (sas), we deduce that
indeed 4N CB ∼ 4M EB. Hence ∠BM C = ∠BM E = ∠BN C, and
it follows that M BCN is a cyclic quadrilateral. 

Solution II. As in the first solution, we show that 4DCB ∼ 4AEB.

Now note that the lines BM and BN are median in these two similar
triangles. Hence the corresponding angles ∠BM E and ∠BN C are
equal. Hence ∠BM C = ∠BM E = ∠BN C, from which we deduce
that M BCN is a cyclic quadrilateral. 

5 We first show that for reals x, y and z satisfying the first condition,
we have
x2 y 2 + y 2 z 2 + z 2 x2 ≥ xyz(x + y + z)3 . (2)
So the solutions that we are looking for, are the cases in which equality
holds in 2.
2 2
Let a, b and c be reals. Then (a−b)2 ≥ 0, so a +b
2 ≥ ab with equality
if and only if a = b. We repeat this for the pairs (b, c) and (c, a), and
after summing, it follows that
a2 + b2 + c2 ≥ ab + bc + ca

26
with equality if and only if a = b = c.
We apply this inequality to the triple (xy, yz, zx), and obtain

x2 y 2 + y 2 z 2 + z 2 x2 ≥ xy 2 z + yz 2 x + zx2 y = xyz(x + y + z)

with equality if and only if xy = yz = zx. We also apply it to (x, y, z)

to obtain x2 + y 2 + z 2 ≥ xy + yz + zx. Hence from the first condition,
it follows that

1 = 3(x2 + y 2 + z 2 ) = (x2 + y 2 + z 2 ) + 2(x2 + y 2 + z 2 )

≥ x2 + y 2 + z 2 + 2xy + 2yz + 2zx = (x + y + z)2 ,

with equality if and only if x = y = z. Now we’re going to combine

these two inequalities,

x2 y 2 + y 2 z 2 + z 2 x2 ≥ xyz(x + y + z) (3)

and
1 ≥ (x + y + z)2 (4)
2 2
To this end, we first multiply (4) by the non-negative factor x y +
y 2 z 2 + z 2 x2 ; we obtain

(x2 y 2 + y 2 z 2 + z 2 x2 ) · 1 ≥ (x2 y 2 + y 2 z 2 + z 2 x2 ) · (x + y + z)2

with equality if and only if x = y = z or x2 y 2 + y 2 z 2 + z 2 x2 = 0.

Next, we multiply (3) by the non-negative factor (x + y + z)2 , from
which follows that

(x2 y 2 + y 2 z 2 + z 2 x2 ) · (x + y + z)2 ≥ xyz(x + y + z) · (x + y + z)2

with equality if and only if xy = yz = zx or x + y + z = 0. Combining

these yields

(x2 y 2 + y 2 z 2 + z 2 x2 ) · 1 ≥ (x2 y 2 + y 2 z 2 + z 2 x2 ) · (x + y + z)2

≥ xyz(x + y + z) · (x + y + z)2 ,

with equality if and only if

x = y = z or x2 y 2 + y 2 z 2 + z 2 x2 = 0

and (xy = yz = zx or x + y + z = 0) .

From the second condition, it follows that indeed, equality must hold.
Hence (x, y, z) must satisfy the conditions above. Furthermore, we

27
still have the first condition, which says that 3(x2 + y 2 + z 2 ) = 1:

3(x2 + y 2 + z 2 ) = 1,
x=y=z or x2 y 2 + y 2 z 2 + z 2 x2 = 0,
xy = yz = zx or x + y + z = 0.

We distinguish two cases. First, suppose that x = y = z. Then

we also have xy = yz = xz. Using the first condition, we obtain
1 = 3(x2 + y 2 + z 2 ) = 9x2 = (3x)2 , hence x = y = z = ± 31 .
Next, suppose that x2 y 2 + y 2 z 2 + z 2 x2 = 0; then each term must be
zero, hence xy = yz = zx = 0. From xy = 0, we deduce without loss
of generality that x = 0, and also, from yz = 0, we deduce without
loss of generality
√ that y = 0. In this way, √ we find the√solutions
(0, 0, ± 13 3), and also the solutions (0, ± 31 3, 0) and (± 13 3, 0, 0).
Remark. Once you’ve shown (3) and (4), you can also obtain (2)
by multiplying (4) by xyz(x + y + z), but you’ll need to show that
xyz(x + y + z) is non-negative. From the second condition, we deduce
that xyz(x + y + z)3 is a sum of squares, hence non-negative. If
x + y + z 6= 0, it follows that xyz(x + y + z) ≥ 0, and otherwise,
we have xyz(x + y + z) = 0. Hence indeed, in both cases we have
xyz(x + y + z) ≥ 0.
Using this method, the equality case becomes:

3(x2 + y 2 + z 2 ) = 1,
xy = yz = zx,
x=y=z or xyz(x + y + z) = 0.

28
IMO Team Selection Test 2, June 2010
Problems

1. Consider sequences a1 , a2 , a3 , . . . of positive integers. Determine the

smallest possible value of a2010 if
(i) an < an+1 for all n ≥ 1,
(ii) ai + al > aj + ak for all quadruples (i, j, k, l) which satisfy 1 ≤
i < j ≤ k < l.

2. Find all functions f : R → R which satisfy

f (x) = max (2xy − f (y))

y∈R

for all x ∈ R.
(In general the expression a = max g(s) means: a ≥ g(s) for all s ∈ S
s∈S
and furthermore there is an s ∈ S for which a = g(s).)

1

b be positive integers such that M (a, b) = a −
3. (a) Let a and b +
b b + a3 is an integer. Proof that M (a, b) is a square.
(b) Find nonzero integers a and b such that M (a, b) is a positive
integer, but not a square.

4. Let ABCD be a square with circumcircle Γ1 . Let P be a point on

the arch AC that also contains B. A circle Γ2 touches Γ1 in P and
also touches the diagonal AC in Q. Let R be a point on Γ2 such that
the line DR touches Γ2 . Proof that |DR| = |DA|.

5. The polynomial A(x) = x2 + ax + b with integer coefficients has the

following property: for each prime p there is an integer k such that
A(k) and A(k + 1) are both divisible by p. Proof that there is an
integer m such that A(m) = A(m + 1) = 0.

29
Solutions

1. By induction we prove that an − a1 ≥ 2n−1 − 1 for all n ≥ 2. For

n = 2 this reduces to a2 − a1 ≥ 1 and this follows from condition (i).
Let m ≥ 2 and suppose that am − a1 ≥ 2m−1 − 1 (IH). We apply
condition (ii) with i = 1, j = k = m and l = m + 1. This yields
a1 + am+1 > 2am . So
IH
am+1 − a1 > 2am − 2a1 ≥ 2(2m−1 − 1) = 2m − 2,

and since am+1 is a positive integer, this yields am+1 − a2 ≥ 2m − 1.

This completes the induction.
We now know that for n ≥ 1:

an ≥ 2n−1 − 1 + a1 ≥ 2n−1

and in particular a2010 ≥ 22009 .

On the other hand we prove that a2010 is possible by showing that the
sequence given by an = 2n−1 satisfies the conditions. This sequence
consists of positive integers and is strictly increasing (condition (i)).
Let (i, j, k, l) be a quadruple that satisfies 1 ≤ i < j ≤ k < l. It is
true that

aj + ak = 2j−1 + 2k−1 ≤ 2k−1 + 2k−1 = 2k ≤ 2l−1 < 2i−1 + 2l−1 ,

and thus the sequence also satisfies condition (ii).

We conclude that the smallest possible value of a2010 is equal to 22009 .


2. For all x ∈ R it is true that

f (x) = max (2xy − f (y)) ≥ 2x2 − f (x),

y∈R

so 2f (x) ≥ 2x2 , that is f (x) ≥ x2 .

Because (x − y)2 ≥ 0, it is true that x2 ≥ 2xy − y 2 for all x, y ∈
R. Because we already have showed that f (y) ≥ y 2 , we know that
2xy − f (y) ≤ 2xy − y 2 ≤ x2 and consequently

f (x) = max (2xy − f (y)) ≤ x2 .

y∈R

We conclude f (x) = x2 .

30
 We check if this function satisfies. Let x ∈ R be given. Because
(x − y)2 ≥ 0, it is true that x2 ≥ 2xy − y 2 for all y ∈ R with equality
iff x = y, therefore
x2 = max (2xy − y 2 ).
y∈R

3. (a) Because a + b2 is an integer, − 1b + 3b a is also an integer. We may

−a+3b2
write this as ab . We know that ab is a divisor of 3b2 − a. In
particular b is a divisor of 3b2 − a and therefore b | a. However,
this means that b2 is a divisor of ab and consequently it is also
a divisor of 3b2 − a, which yields b2 | a. We now write a = mb2
with m a positive integer. Then is is true that mb3 is a divisor
of 3b2 − mb2 , so mb is a divisor of 3 − m. This yields that m is
a divisor of 3 (that is m = 1 or m = 3) and that b is a divisor of
3 − m.
First suppose that m = 3. Then we have a = 3b2 . Filling this in
yields M (3b2 , b) = 3b2 − 1b + b2 + 1b = 4b2 and that is the square
of 2b.
Now suppose that m = 1. From b | 3 − m follows b = 1 or b = 2.
In the first case we have a = 1 and in the second case a = 4.
Filling in the first possibility yields M (1, 1) = 1 − 1 + 1 + 3 =
4, which is a square. Filling in the second possibility yields
M (4, 2) = 4 − 21 + 4 + 32 = 0, which also is a square.
We conclude that M (a, b) is a square in all cases.
(b) Take a = 4 and b = −2. Then we have M (4, −2) = 7, which is
a positive integer, but not a square.
After doing part (a) this answer is not hard to find. You know
that a has to be equal to mb2 , but now for some m ∈ Z, and that
m has to be a divisor of 3. Further more m = 3 will not work,
because it yields a square. You can try the other possibilities of
m yourself.


4. Let M be the intersection of AC and BD (that is the midpoint of

Γ1 ) and let N be the midpoint of Γ2 . First we prove that P , Q and
D are collinear.
If P = B, then Q = M and it is trivial. If not, let S be the in-
tersection of P Q and BD. We will first prove that S = D. Notice

31
 that M , N and P are collinear and that QN is parallel to DB. This
yields because of F-angles that ∠P SB = ∠P QN = ∠N P Q, because
|N P | = |N Q|. By Z-angles we see that ∠P M B = ∠M N Q and be-
cause of the exterior angle theorem inside triangle P QN that is equal
to ∠P QN + ∠N P Q = 2∠P SB. So ∠P M B = ∠P SB, together with
the inscribed angle theorem this yields that S lies on Γ1 . So S = D,
from which follows that P , Q and D are collinear.
Because ∠DP B = ∠DM Q = 90◦ and ∠P DB = ∠QDM , it is true
|DP |
that 4DP B ∼ 4DM Q. This yields |DB| = |DM |
|DQ| . Since |DB| =
2|DM |, this is equivalent to |DP ||DQ| = 2|DM |2 .√The power theo-
rem on Γ2 yields |DP ||DQ| = |DR|2 , so |DR| = 2|DM | = |DA|.


5. Let p be a prime and let k be such that A(k) and A(k + 1) are both
divisible by p. The difference between A(k) and A(k + 1) is also
divisible by p and this is equal to

A(k + 1) − A(k) = (k + 1)2 + a(k + 1) + b − (k 2 + ak + b) = 2k + 1 + a,

so 2k ≡ −1 − a mod p. Because A(k) is divisible by p, the integer

4A(k) is also divisible by p, so we have modulo p that

4A(k) ≡ 4k 2 + 4ak + 4b ≡ (−1 − a)2 + 2(−1 − a)a + 4b

≡ −a2 + 4b + 1 mod p.

The right hand side is independent of k. We now see that for each
prime p the number −a2 + 4b + 1 has to be divisible by p. This is
only possible if −a2 + 4b + 1 = 0. So we have a2 = 4b + 1. We now
see that a has to be odd and we write a = 2c + 1 with c ∈ Z. Then
it is true that 4b + 1 = a2 = 4c2 + 4c + 1, so b = c2 + c. Therefore,
the polynomial can be written as A(x) = x2 + (2c + 1)x + (c2 + c)
for an integer c. We can factor this as A(x) = (x + c)(x + c + 1),
consequently x = −c and x = −c − 1 are zeroes of the polynomial.
These are both integer points. We conclude that m = −c − 1 satisfies
A(m) = A(m + 1) = 0. 

32
Junior Mathematical Olympiad, October 2009
Problems
Part 1

1. The station hall of Nijmegen is tiled in a repeating

pattern with white, speckled and black tiles (see
figure). Which part is speckled?
1 1 1 1 1
A) 10 B) 9 C) 8 D) 6 E) 4

2. There are 120 chairs in a row. A number of chairs is already occupied

in such a way that if someone comes in, this person can only sit on a
chair next to someone else. What is the smallest amount of occupied
chairs such that this may happen?
A) 40 B) 41 C) 59 D) 60 E) 80

3. Anne has got a lot of tiles like in the figure. She

can lay down four of these tiles such that the white
lines form a closed circuit. What is the smallest
number of tiles with which she can lay down a
larger circuit?
A) 6 B) 8 C) 9 D) 10 E) 12

4. Bram has got coins of 5 cent, 10 cent, 20 cent and 50 cent, of each
type at least one. In total he has got 9 coins and together they are
worth 2,10 euro. How many coins of 20 cent does Bram have?
A) 1 B) 2 C) 3 D) 4 E) 5

5. Exactly in the middle of a square with side length

7 a grey square is drawn. The area of the white ?
part of the large square is three times as large as
the area of the grey square. What is the width of 7
the white ring?
A) 1 B) 1 13 C) 1 12 D) 1 23 E) 1 34

33
 6. Alice, Bas, Chris, Daan and Eva know each other very well. Each
of them always tells the truth or always lies. Chris says: “Alice is
honest”, on which Eve replies: “Chris lies! ” Chris says: “Bas is a
liar.” Eva claims: “Daan is honest.” Which two persons could both
be telling the truth?
A) Alice and Bas B) Bas and Chris C) Chris and Daan
D) Daan and Eva E) Eva and Alice

7. Jan is looking for a triangle which has got circumference 21 and in-
teger side lengths. Moreover, for each pair of sides the length of one
of them has to divide the length of the other. How many different
triangle with these properties exist?
A) 0 B) 1 C) 3 D) 5 E) 6

8. How many triangles are in this figure?

A) 20 B) 25 C) 30 D) 35 E) 40

9. Vincent wants to tile a 1,30 by 1,70 meter terrace with 50 by 20

centimeter tiles. Because it is not possible to exactly tile the terrace
he has to let some pieces of some tiles outside of the terrace. How
many tiles does Vincent need to cover his terrace?
A) 22 B) 23 C) 24 D) 27 E) 28

10. A large bag contains a number of balls: more then 100, but less than
150. If you divide the balls (evenly) over 7 children, then 2 balls will
remain. If you divide the balls over 6 children, 3 balls will remain.
How many balls will remain if you divide the balls over 5 children?
A) 0 B) 1 C) 2 D) 3 E) 4

11. A pyramid with a square base is cut into two pieces

by a straight cut. Which form cannot be the cut?
A) triangle B) square C) pentagon
D) hexagon E) All four forms are possible.

34
12. Mister Jansen departs every morning at 8:00 to his work place by
car. If he (constantly) drives 40 km/h, he would arrive three minutes
late at his work place. If he drives 60 km/h, he would arrive three
minutes too early. How fast does he need to drive to be exactly on
time?
A) 48 B) 49 C) 50 D) 51 E) 52

13. The 2 × 2 × 2-cube in the figure has got four trans-

parent cubes and four nontransparent cubes. You
cannot look through the cube when looking from
the front, side and upper face. A 3 × 3 × 3-cube
has got 27 cubes. What is the smallest amount of
cubes you have to make nontransparent such that
you can’t look through the cube from the front,
side and upper face?
A) 9 B) 10 C) 12 D) 13 E) 14

14. Sir Tuinder has got a ‘half-open’ fence around his garden made of
shelves. The shelves are alternating at the front and the rear side,
exactly in front of the middle of the gap between the two shelves on
the other side. Here you can see the fence from above. The width of
the shelves is 21 cm and the thickness is 3 cm. Between the front- and
the rear side is a distance of 2 cm. How many cm has the distance
between two shelves next to each other be such that people you can’t
look into the garden from the outside?
A) 7 B) 8 C) 9 D) 10 E) 11

21 ?
3
2

15. Rectangle ABCD has got area 1. The points P , D R C

Q, R and S are the middles of the sides and T T
is the middle of RS. What is the area of triangle S Q
AQT ?
1 3 5 9 17
A) 4 B) 8 C) 16 D) 32 E) 64 A P B

35
Part 2
The answer to each problem is a number.

1. The numbers on the faces of this cube are six con- 11

secutive integers. If we add up the numbers on
two opposing faces, we will always get the same 15
answer. What is the sum of all six numbers? 14

2. The entrance tickets of a amusement park normally cost 38 euro, but

when it is a rainy day, you can go in for half of that price. Last week
there 800 tickets were sold, together costing 19057 euro. How many
tickets were sold for half the price?

3. In a 4 × 4 × 4-cube each small cube has got 3, 4,

5 or 6 neighbors (that touch in a face to the small
cube). What is the average number of neighbors
that a small cube has got?

4. Jaap has got eight integers. Of each integer the first and last dig-
its are 1 and the other digits are 0. The eight integers have got
2, 3, 5, 9, 17, 33, 65 and 129 digits. If Jaap multiplies these eight digits
and adds up the digit of the result, which number will he get?

5. A rectangular table has got three rows and two

columns. On how many ways can you fill in the
numbers 1 to 6 in the six positions in the table
such that in each row the first number is greater
than the second?

36
 6. A square with area 54 is divided into four equal
squares. The upper left square is colored grey;
the lower right square is again divides into four
equal squares, and so on. The pattern continues
infinitely long. What is the area of the grey region?

7. Start with a number of two digits (the first digit is nonzero) and do
the following: multiply the two digits and add the same two digits
to the product. In this way 27 will yield 14 + 2 + 7 = 23. For how
many two digit numbers the result is equal the the number, you began
with?

8. A line segment is drawn in between the points (36,22)

(1, 1) and (36, 22). How many points with inte-

ger coordinates lie on this line segment, the end
(1,1)
points included?

9. Anneke can drive from A to B on the high way or via a shorter road.
On the high way Anneke can drive 120 km/h and on the shorter road
she can drive 60 km/h. The high way is 8 km longer, but the trip
takes 8 minutes less. How many kilometers long is the high way from
A to B?

32
10. In the figure you see a square. The square is di-
vided into four rectangles. Of three rectangles the
circumference is given and written inside the rect- 36 42
angle. What is the area of the fourth rectangle?

37
Solutions
Part 1
1
1. C) 8 6. D) Daan and Eva 11. D) hexagon

2. A) 40 7. C) 3 12. A) 48

3. E) 12 8. D) 35 13. A) 9

4. A) 1 9. B) 23 14. C) 9

5. E) 1 34 10. A) 0 15. C) 5
16

Part 2

1. 81 5. 90 8. 8

2. 597 6. 18 9. 32

3. 4 12 7. 9 10. 42

4. 111
| .{z
. . 111}
256 ones

38
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Contents

1 Introduction
3 First Round, January 2009
9 Final Round, September 2009
14 BxMO Team Selection Test, March 2010
18 Benelux Mathematical Olympiad, April 2010
24 IMO Team Selection Test 1, June 2010
29 IMO Team Selection Test 2, June 2010 Centraal Bureau voor de Statistiek

33 Junior Mathematical Olympiad, October 2009

© Stichting Nederlandse Wiskunde Olympiade, 2010