Hellenic Mathematical Competitions 2009_booklet

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HELLENIC MATHEMATICAL SOCIETY

2009

Competition Committee

Athens, 2009


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2 Hellenic Mathematical Competitions 2009

() 34

106 79

. 3616532 - 3617784 - Fax: 3641025

e-mail : [email protected]

www.hms.gr

GREEK MATHEMATICAL SOCIETY

34, Panepistimiou (leftheriou Venizelou) Street

GR. 106 79 - Athens - HELLAS

Tel. 3616532 - 3617784 - Fax: 3641025

e-mail : [email protected]

www.hms.gr

Frontispiece:

The image on the front page represents the Discobolus statue (Olympic discus

thrower) which was made by Myron, one of the best sculptors of ancient Greece.

Myron, who lived in 5th-century BC in Athens, was a well-known member of a new

school of Greek art that incorporated motion into free-standing statues. In this

case, Myron has caught a discus thrower at the peak of his backswing, poised for

eternity just before spinning his body in powerful rotations to give the discus even

greater speed at the moment of release. History does not record whether

Discobolus recognized a particular Olympic athlete, but Myron is known to have

produced other statues honouring specific heroes. In any event, it has evolved into

a powerful symbol of the spirit of Olympic athletic competition.

The original Discobolus statue was never recovered; an exact copy of

the statue however is placed at the entrance of Panathinaikon Stadium in Athens,

where the first modern Olympic Games were held in 1896.


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Hellenic Mathematical Competitions 2009 3

Contents

1. Preface . 4

2. 26thHellenic Mathematical Olympiad 2009

A. Juniors .. 5

B. Seniors7

3. Selection examinations 2009

A. Juniors ..13B. Seniors ..17

4. Mediterranean Mathematical Competition 2009 ....... 24

4. 26thBalkan Mathematical Olympiad (Kragujevac) 27

5. 13thJunior Balkan Mathematical Olympiad (Sarajevo) 32


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Preface

On November 3, 2008, more than 12.000 students from all Hellenic High

Schools (Gymnasiums and Lyceums) took part in the first round of the 69stNation-

al Math Competition THALES. The best 1.500 students qualified and took part

in the second round of the 69stNational Math Competition EUCLIDES, held in

January 17, 2009. From this competition about 300 students qualified and took part

in the 26th

Hellenic Math. Olympiad ARCHIMEDES, held in Athens on Febru-

ary 21, 2009. The best 50 students (25 Juniors) qualified to take part in the Selec-

tion Examination for the completion of the Greek teams for the 11thJunior Balkan

Math. Olympiad (JBMO), the 26thBalkan Math. Olympiad (BMO) and the 50th In-

ternational Mathematical Olympiad. This year the Selection Examination was al-

most the same with the 10th

Mediterranean Mathematical Competition 2009 (me-

morial Peter O Halloran).

The Hellenic team for the IMO 2009 consists of the students:

Giechaskiel Ilias Logothetis Fotios Papadimitriou Dimitrios Pappelis Konstantinos Taratoris Evangelos

Zadik Ilias

Twelve days of training were offered to the selected team for the IMO 2009.

Moreover, a training program was offered some Saturdays from November 2008 to

May 2009 to all students wanting to attend.

Athens, July 2009

The competition Committee

of the Hellenic Math. Society


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2266tthh

HHeelllleenniiccMMaatthheemmaattiiccaallOOllyymmppiiaadd22000099ARCHIMEDES

February 21, 2009

A. Juniors

Problem 1

If the number2

2

9 31

7

n

n

is integer, find the possible values ofn .

Solution

We have2 2

2 2 2

9 31 9( 7 ) 32 329 .

7 7 7

n n

n n n

Since is integer, it follows that 2 7n is a divisor of 32 and taking in mind that2 7 8n , we conclude:

2 27 8 , 1 6 , 3 2 1, 9 , 2 5 1, 1, 3 , 3 , 5 , 5n n n .

Alternatively, we can solve the problem by solving the given equation with re-

spect to2n and in the sequel determining the suitable values of for which 2n is

a nonnegative integer.

Problem 2

From the vertex of an equilateral triangle we draw the ray x which in-tersect the side at . On we consider a point such that . Find

the angle .

First solution

Since , point is the circum-

circle of the triangle A. The angle is in-

scribed to the circle , BAC and so:01 3 0

2 .

Second solution

From and = , we have = ,

and the triangle is isosceles. We draw the alti-

tude from , let , .


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Then is also median and bisector of the angle of the triangle . Let

BZ meets AE at K. Then the triangles and

are equal because they have:

= , common side and .

Therefore we have: and since it follows that .

Thus quadrilateral is inscrible, and hence:

60

Finally we get

0 90 90 60 30 .

Problem 3

We consider the numbers

1 3 5 5 9 5 5 9 7 2 4 6 5 9 6 5 9 8... a n d ...

4 6 8 5 9 8 6 0 0 5 7 9 5 9 9 60 1 .

Prove that: () , () 15990

.

Solution()To each fraction of A of the form 2 1 , 1, 2 , ..., 2 9 92 2

, cor-

responds a fraction from B of the form 2 , 1, 2, ..., 2992 3

. Since

2 1 20

2 2 2 3

for every

* ,

and so for 1,2,...,299 , by multiplying by parts the above 299 inequalities we

obtain .

()Since 0 , from we get :

2

2 2

1 2 3 1 1 1.

599 600 601 100 599 5990 5990

Problem 4

In the figure we see the paths connecting the square

(point ) with the school of a city (point ). In thesquare there are kpupils starting to go to the school.

They have the possibility to move to the right and up. If


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the pupils are free to choose any allowed path, determine the minimum value of k

so that in any case at least two pupils to follow the same path.

Solution

In the figure they are shown all possible allowed paths starting from the square

and leading to the school.

According to the rule it is clear that in the

nodes1 2 3, , and

1 2 3, , , there is only one

possible path to choose..

Counting easily we find that all possible paths

are 20. Therefore, according to the pigeonhole prin-

ciple, at least two pupils will follow the same path ,

If the number of the pupils is 21k . Hence theminimum value of kis 21.

B. Seniors

Problem 1

Determine the values of the positive integer n for which

9 1

7

n

n

is rational.

Solution

It is enough to prove that there exist*,a b with , 1a b such that:

2

2

9 1

7

n a

n b

. (1)

From this relation we get:

2 2 2 2 2 2

2 2 2 2 2 27 7 ( 9 ) 64 6 479 9 9

a b a b b bnb a b a b a

(2)

Since , 1a b , it follows that 2 2, 1a b and 2 2 29 , 1b a b , and hence

from (2) we get that n is integer, if and only if 2 29 b a is a divisor of 64.

Since ,a b and n are positive integers, it follows that 2 29 8b a , and hence:


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2 29 3 3 8,16, 32, 64b a b a b a . (3)

Moreover, the factors 3 , 3b a b a have sum a multiple of 6 and difference a

multiple of 2 and 3 3b a b a . Therefore from relation (3) the possible casesare the following:

3 , 3 4 , 2 3 , 3 8 , 4 3 , 3 1 6 , 2b a b a b a b a b a b a

, 1,1 , 2, 2 , 7, 3a b a b a b .

The pair , 2 , 2a b is rejected, because g cd 2, 2 2 1 , and therefore

we have the values 1 or 11.n n

Problem 2

A triangle is given and let its circumcenter and1 1 1, , the middles

of its sides , and , respectively. We consider the points2 2 2, , such

that2 1 =

,

2 1

and2 1

, with 0 . Prove that

the lines2 2 2, , are concurrent.

Solution

Let be the orthocenter of the triangle . Then12

and from

2 1

, we find:2

2

.

If2 meets at C (from the similarity of the triangles C and

2C ), we have: 2C C

. It means that2 pass from C which divides

in ratio

2 .

Similarly, we have12

and

2

2

.

Let now C the point of intersection of the lines 2 and . Then we have2

C C

, which means that

2 pass through C which divides in ra-

tio

2 . Similarly, if C is the point of intersection of the lines2

, then

we have that2

pass through C which divides in ratio

2 .


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Since the points C , C , C coincide the lines2 2 2, , are concurrent.

Comments

(1) If 1 , then C coincides with the barycenter of the triangle .

(2)If 2 , then C coincides with the center of the Eulers circle of the trian-

gle . In this case the triangles and 2 2 2 are equal and they have thesame Euler circle.

(3)In any case the triangles and2 2 2 are similar with their sides pa-

rallel. The one triangle is the image of the other with respect to a homothety, and

thus we may have a solution using homotheties.

(4)The problem can be solved by using Analytic Geometry or complex num-

bers.

Problem 3

If the nonnegative real numbers , andy zhave sum 2, prove that:

2 2 2 2 2 2 1.x y y z z x xyz

For which values of , andx y zthe equality is valid?

Solution

We will use the well-known inequality 2 22 , which is valid for all

, . The equality holds for . Thus we have


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2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2

2 2 2

2

12 2 2 2

2

12 2 2 2

2

12 (1)

2

12

2

12

2

12

x y y z z x xyz x y y z z x xyz

xy xy yz yz zx zx xyz

xy x y yz y z yz y z xyz

xy yz zx x y z xyz yzx zxy xyz

xy yz zx x y z xyz x y z

xy yz zx x

2 2 , (since 2).y z x y z

Till now we have shown that

2 2 2 2 2 2 2 2 21

2x y y z z x xyz xy yz zx x y z

, (2)

and the equality holds, as we see from (1), when :

or , 0 or , 0 or , 0x y z x y z y z x z x y .

Since 2x y z , equality holds when:

2 2 2

, , , , 1,1, 0 1,0,1 0,1,13 3 3

x y z

. (3)

In the sequel we will use the known inequality2

,2

, ,

putting 2 2 22 2 2 ,xy yz zx x y z . Thus we have

2 2 2 2 2 2

22 2 2

4

1 12 2 2

2 4

1 2 2 2 11. (4)

4 2 16

xy yz zx x y z xy yz zx x y z

xy yz zx x y zx y z

From (2) and (4) we obtain the inequality2 2 2 2 2 2 1.x y y z z x xyz

The equality holds when in inequality (4) we have:2 2 22 2 2xy yz zx x y z ,

which in consideration with (3) gives:

, , 1 , 1 , 0 1 , 0 , 1 0 , 1 , 1x y z .


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Problem 4

Let 1 2 3 4 5 6, , , , ,z z z z z zbe six pairwise different complex numbers which their

images1 2 3, , ,

4 5 6, , are consecutive points of the circle with center

O(0,0) and radius > 0r . If w is a solution of the equation 2 1 0z z and2

1 3 5 0z w z w z () ,

2

2 4 6 0z w z w z ()

Prove that: ()the triangle1 3 5

is equilateral,

() 1 2 2 3 3 4 4 5 5 6 6 1z z z z z z z z z z z z 1 4 2 5 3 63 3 3z z z z z z .Solution

() Since w is a root of the equation 01zz2 , we have 01ww2 .

Multiplying both parts by w , :

0www 23 11www0

23

1w3 .

From the last equation we find 1w . Substituting in relation (I)

1ww2 , we find: 0zwz)w1(z 531 0zwzwzz 5311 5113 zzw)zz( .

Hence

5113 zzw)zz( 5113 zzwzz 5113 zzzz ().

Substituting in relation () 1ww 2 , we find:

0z)1w(zwz 52

32

1 0zzwzwz 532

32

1 352

31 zzw)zz( .

Hence we have

352

31 zzw)zz( 352

31 zzwzz 3513 zzzz ().

From () and () we obtain the equalities:

155331 zzzzzz ,

that is the triangle531 is equilateral..

()Similarly, using relation ()) we prove that the triangle 642 is equilat-

eral. From a known proposition of Euclidean Geometry we have

that 416121 , and then using measures of complex numbers we

have:


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411621 zzzzzz . (1)

Similarly, from the equality634323 using measures of com-

plex numbers we get:

634332 zzzzzz . (2)

Also, from equality 256545 we find

526554 zzzzzz . (3)

Summing up by parts the relations (1), (2) and (3) and using the equalities

526341 zzzzzz

we find:

166554433221 zzzzzzzzzzzz

635241 zz3zz3zz3 .


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SELECTION EXAMINATION 2009April 11, 2009

A. Juniors

Problem 1

A pupil has 7 pieces of paper. He chooses some of them and cuts each of them

into seven pieces. In the sequel, he chooses some of the pieces and cuts each of

them into seven pieces. He continues this procedure many times with the pieces he

has in hands every time. Is it possible to have some time 2009 pieces of paper?

Solution

Let he choose at the beginning 1 from the seven pieces and each of them into

seven pieces. Then he will have totally1 1 17 7 7 6 pieces of paper. Sup-

pose that in the next step he chooses 2 pieces of paper and cuts each of them into

seven pieces. Then he will have totally1 2 2 1 27 6 7 7 6( ) pieces of

paper. If he continues this procedure times, then he will have totally

1 27 6( )

pieces of paper. Therefore we are looking for the value of

satisfying the equation

1 2 1 27 6( ) 2009 6( ) 2002 ,

which is absurd, because 2002 is not divided by 6. Hence it is not possible for him

to have some time 2009 pieces of paper.

Problem 2

Let ABCD be a convex quadrilateral inscribed in a circle ( , )O R . With centers

the vertices of the quadrilateral and radiusR we draw circles ,AC A R , ,BC B R ,

,CC C R , ,DC D R . Circles AC and BC meet at K , circles BC and CC

meet at L , circlesC

C andD

C meet at and the circlesD

C ,A

C meet at N.

(Points , , ,K L M N are the second common points of the corresponding circles,

given that all of them pass through point O ). Prove that the quadrilateral KLMNis

parallelogram

.

First solution


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The line segment AB connects the centers of the circlesA

C andB

C , and there-

fore it is the perpendicular bisector of the common chord OK. Since the circles

AC and BC have the same radius, the quadrilateral AOBKis rhombus. Thus point

1K is the middle ofAB .

Similarly, we can show that1L is the middle ofBC, 1M is the middle of CD

and1

N is the middle ofAD .

From the trianglesOKL , OLM, OMNand ONKwe conclude that:1 1KL K L ,

1 1LM L M , 1 1MN M N and 1 1NK N K (because the line segments 1 1K L , 1 1L M ,

1 1M N and 1 1N K connect the middles of the sides of a triangle).

Hence the quadrilateralsKLMNand1 1 1 1K L M N have their sides parallel. But we

know that the middles of the sides of a quadrilateral define a parallelogram. So the

quadrilateral KLMNis parallelogram.

Second solution

Since the line segmentKO is the common chord of the equal circlesAC and

BC , it is the perpendicular bisector of the line of the centers AB and vise versa.

Hence the quadrilateralKAOB is rhombus, and so


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AK OB . (1)

Similarly the quadrilateral LCOB is rhombus and

CL OB . (2)

From (1) and (2) KACL ,

KL AC . (3)

Working similarly we can prove that the quadrilateral MNACis parallelogram

and that

NM AC . (4)

From (3) and (4) it follows that KLMNis parallelogram .

Problem 3

Let , , are positive integers such that the number

2 3

2 3

is rational. Prove that the number2 2 2

is integer.

Solution. First of all, it is easy to see that:

1 2 3 42 3 2 3 , with 1 2 3 4, , ,

1 3 and 2 4 .

In fact, we can write the first relation in the form

1 3 4 22 ( ) 3 ,

and if1 3 0 , then 4 2

1 3

3

2

, absurd. Hence 1 3 and 2 4 .

The converse is clear.

Let now 2 3

2 3

. Then and , that is

2

.

Hence we have:


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2 2 2 2 2 2 2

2 2

2

( ) ( )( ),

and so .

Problem 4

Determine positive integers , ,x y zwhich satisfy the system

1.

x y z xy yz zx

xyz

and have the least possible sum.

First solution

We write the system in the form

xy yz zx x y z (1)

1xyz . (2)

Subtracting the two equations by parts we find

1

1 0

xyz xy yz zx x y z

xyz xy yz zx x y z

1 1 1 1 0

1 1 0

xy z x z y z z

z xy x y

1 1 1 0

1 or 1 or 1.

z x y

x y z

For 1,x from (1) and (2) we have 1yz , which have the solutions

1

, , , 0y z a aa

,

and so, the solutions of the system are

1

, , 1, , , 0.x y z a aa

Similarly, considering 1y or 1z we find the solutions

1 1

, , ,1, or , , , ,1 , 0.x y z a x y z a aa a

Since for each 0a we have

11 1 2 3x y z a

a .


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Equality holds for 1a , it follows that between the solutions of the system, the

, , 1,1,1x y z is that having the least possible sumx y z .

Second solution

Let ( , , )x y z is the solution of the system with the least possible sum. Then,

from the inequality of arithmetic geometric mean we have

3 33

x y zxyz x y z

,

while equality holds for x y z .

Hence the least possible value of the sum x y z , between the solution of the

given system is 3 and it happens for x y z .

For y z , from the equation 1, , , 0xyz x y z , it follows that

1x y z , which satisfies also the equation x y z xy yz zx .

B. Seniors

Problem 1

If a is an even positive integer and 1 ... 1n na a a , *n , is a per-

fect square, prove that a is a multiple of 8.

Solution

Since a is an even positive integer, it follows that is odd. Therefore A will

be a perfect square of an odd integer, that is

2 22 1 4 4 1 4 1 1,

where is a positive integer. Since one of the two integers and 1 is even, we

have

4 1 1 8 1, where is appositive integer

1 1

1 1

1 ... 8 ... 1 8

8 ... 1 8 , since 8, ... 1 1.

n n n

n n

a a a a a a

a a a a a a

Problem 2


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Let the triangle ABC has barycenter G and circumcenter O . The perpendicular

bisectors of GA , GB and GC intersect at the points1 1 1A ,B ,C . Prove that O is the

barycentre of the triangle1 1 1

A B C .

Solution

Let F,E,D be the middles of the sides AB,AC,BC , respectively.

Let, also 11CB , 11CA , 11BA be the perpendicular bisectors of the line segments

GA , GB and GC , respectively. Then the points 11 B,A and 1C are the circumcen-

ters of the triangles GBC, GAC and GAB, respectively. Hence DA1 , EB1 and

FC1 are the perpendicular bisectors of the sides BC , AC and AB , respectively,

and therefore they will pass through the circumcenter O of the triangle ABC .

Next, we will show that DA1 , EB1 and FC1 are the medians of the triangle

111 CBA . Let the extension of DA1 , meets 11CB at N . We will prove that N is the

middle of the line segment11CB .

From the inscrible quadrilateral 1AMEB (o90EM ), we have M A E


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1 MB E . Also, from the inscrible quadrilateral DOEC ( o90ED ), we get

NOEDCE . Therefore the triangles ADC and NOB1 are similar, and so

1NB AD

=NO CD

. (1)

From the inscrible quadrilateral 1AMFC (o90FM ), we have

xFCMFAM 1 and similarly from DOFB (o90FD ), we obtain

that yNOFDBF . From the above equalities the triangles ADB and NOC1 are

similar and therefore:

1NC AD

NO BD . (2)

From (1) and (2) we get11 NCNB . In a similar way we prove that EB1 , FC1

are the other two medians of the triangle 111 CBA .

Problem 3

Find all triples of real numbers (x, y, z) which are greater than 3 and satisfy the

equality:2 2 2

( 2) ( 4) ( 6)36

2 4 6

x y z

y z z x x y

.

Solution

Since , ,x y zare greater than 3, it follows that 2,y z 4, 6z x x y are

positive. Thus, fromCauchy-Schwarz inequality we get:

2 2 2

2( 2) ( 4) ( 6) ( 2) ( 4) ( 6) ( 12)2 4 6

x y zy z x z x y x y z

y z x z x y

2 2 2 2( 2) ( 4) ( 6) 1 ( 12)

2 4 6 2 ( 6)

x y z x y z

y z x z x y x y z

.

From the hypothesis of the problem it follows that2

( 12) 72( 6)

x y zx y z

, (1)

where as the equality holds when:

2 4 6

2 4 6

x y z

y z x z x y

( ) 2 1

( ) 4( 1)

( ) 6( 1)

y z x

x z y

x y z

. (2)


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Moreover, we observe that:2 2 2( 12) ( 12)

,( 6) ( 12) 18 18

x y z x y z

x y z x y z

where we have put 12x y z . Since we have

2

22 24 18 72 4 18 4 18 0 36 0,18

it follows that2( 12)

72( 6)

x y z

x y z

. (3)

Equality holds when:

12 36 24x y z x y z (4)

From relations (1) and (3) follows that:2( 12)

72( 6)

x y z

x y z

,

and thus equations (2) are (4) are valid, and hence we have the system

(2 1)( ) 12( 1)1

24

x y z

x y z

.

For 1 , from relations (2) we get the system :

4

8 , , 10,8,6

12

y z x

x z y x y z

x y z

.

Therefore the unique solution of the problem is the triple , , 10,8,6x y z ,

taking in mind that it satisfies the equation 24x y z .

Problem 4

In the plane are given different points such that any three of them are not

collinear. We color these points red, green and black. In the sequel we consider allline segments with ends these points and we correspond to each of them an

algebraic value according to the following rules:

1) If at least one of the ends of the line segment is black, then it has algebraic

value 0.

2) If both ends of the line segment have the same colour, red or green then it has

the algebraic value 1.


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3) If ends of the line segment have different colour red or green, then it has the

algebraic value -1.

Determine the least possible value of the sum of algebraic values of the all line

segments.

Solution

From the three rules for the determination of the algebraic value of the line

segments we have the following table:

Let now that we have red, green and black points. Then it is clear

that .

The red points determine2

line segments having their ends red and so

they have algebraic values 1. The green points determine2

line segments

with both ends green and therefore with algebraic values 1. The number of the line

segments having ends with different colors, red or green, and so with algebraic

value -1, are . All the other line segments have algebraic value 0, because they

have at least one of their ends black.

The sum of the algebraic values of all existing line segments is:

2 2

! !

( 2)!2! ( 2)!2!

( 1) ( 1) 22 2 2

2 2

22 2 2

2

2 2

(because )

2

- --

2 2

2

2 2 2

.


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From the last expression of S we conclude that:2

(1)

If is even (let 2 ), then relation (1)becomes: .

Equality in the last relation holds, if and only if2

and 0 .

For example, for 4 , we have the following result:

If is odd ( 2 1 ), then relation (1): 2 1 12 2

.

Since S is integer we conclude that:1

2

.

We check now when equality holds in the last relation. We observe that the

case odd ( 2 1 ) comes from the case even ( 2 ) by adding one

more point. The point we add in the case even ( 2 ) can be blank, red or

green, and so we have the following cases:

Case 1

Let the point we add is blank. Then the new produced line segments have alge-

braic values 0 and the equality in this case holds when:2

1

and 1 .

The sum of all algebraic values is:2

1

.


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Case 2

Let the point we add is red. We had red and green points:1

2

. Now

with the new point we can create - 12

line segments having algebraic value 1 and

- 1

2

line segments having algebraic value -1. Equality in this case holds when

1

2

,

1

2

and 0 .

The sum of all algebraic values remains: 1

2

.

Case 3

If the point we add is green, in a similar way we conclude that the equality

holds when 1

2

,

1

2

and 0 Again:

1

2

.

From all the above we conclude that the least possible value of is2

.


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MMeeddiitteerrrraanneeaannMMaatthheemmaattiiccaallCCoommppeettiittiioonn22000099

Problem 1

Determine all integers 1n for which there exists n real numbers

1 2, ,...,

nx x in the closed interval [-4; 2] such that the following three condi-

tions are fulfilled:

- the sum of these real numbers is at least n :

- the sum of their squares is at most 4n :

- the sum of their fourth powers is at least 34n :

Solution

Since the data of the problem concern n real numbers1 2, ,...,

nx x in the closed

interval 4,2 , we consider the polynomial

2

4 2 1P x x x x ,

which in 4,2 satisfies the relation

2

4 2 1 0P x x x x . (1)

Adding by parts the inequalities coming from (1) for1 2, ,...,

nx x x , and taking

in mind the conditions of the problem, we find:

4 211 1 1

0 ... 11 18 8 34 11 4 18 8 0. (2)n n n

n i i i

i i i

P x P x x x x n n n n n

Hence, since 0,iP x for all 1,2,...,i n , from relation (2) we have:

0,iP x for all 1,2,...,i n , which means that 4,1,2 ,ix for all

1, 2,...,i n . We suppose that from the integers1 2, ,...,

nx x , a are equal to

4 , b are equal to 1 and c are equal to 2. Then we have a b c n and from the data of the problems we have the inequalities

4 2 5

16 4 4 4

256 16 34 222 33 18

a b c a b c c a

a b c a b c b a

a b c a b c a b c

.

By multiplying both parts of the first inequality with 18 and the second with

33 and summing the produced inequalities by parts we get the inequality


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33 18 132 90 222b c a a a ,which in combination with the inequality 222 33 18a b c gives:

33 18 222b c a ,which is valid, if and only if 4b a , 5c a , that is

10a b c a or 10a n .

Therefore the numbers1 2, ,...,

nx x x there exist, if and only if, n is a multiply

of n . For 10 ,n m where m is a positive integer a possible solution arises

by taking m times the number -4, 4m times the number 1 and 5m times

the number 2.

Problem 2


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Problem 3

Solution

Problem 4

Let x, y, zbe positive real numbers. Prove that

2 2cyclic cyclic

xy x

2x zxy x y

.

Solution

Given inequality is equivalent to:

1 1 1 1 1 1x y y z z x z x y

1 1 1 2 2 2y x z y x z x y z

.

New we can take substitutionx y z

a, b, c abc 1y z x

, so our inequality

becomes:


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1 1 1 1 1 1

1 1 1 2 a 2 b 2 c1 a 1 b 1 c

a b c

.

After some computations and also using abc 1 , this becomes equivalent to:

2 2

3 a b c 3 ab bc ca ab bc ca a b c

ab bc ca ab bc ca a b c a b c

12 4 a b c ab bc ca.

9 2 ab bc ca 4 a b c

Now, we take substitution a b c S , ab bc ca P and inequality

becomes:

3 3 2 2 22 2

3S 3P SP 12 4S P P 4S 3P S PS 6P 27S 27P 15PS

9 2P 4S P PS S

( )

It is not difficult to prove that 2S 3P and also S 3, P 3 (by AM GM

and abc 1 ). Therefore:

3 2 2 2 2

3 2 2 2

4S 4S S 12PS , PS S PS 3PS , 3P S 3 3 S 27S ,

P P P 3 P 9P, 6 P 6P P 6 3P 18P.

By summing these inequalities we found ( ) to be true so our proof is

finished. Equality is obviously achieved when

2S 3P 9 a b c 1 x y z .


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2266tthh

BBaallkkaannMMaatthheemmaattiiccaallOOllyymmppiiaaddKragujevac, Serbia, April 30, 2009

Problem 1

Solve in the set of positive integers the equation23 5x y z .

Solution

Working with respect modulo 3 in both sides of the equation we get that2 ( 1) (mod3) yz .

When y is even, we have2 1(mod3), z (impossible). Hence

12 1y y , 1 0,1,2,...y (1)

Working with respect modulo 4 in both sides of the equation we get2 ( 1) 1 (mod 4). x yz

Ifxis odd, then the above equation becomes 2 2(mod 4)z , (impossible).

Hence we have

12x x , 1 1,2,...x (2)

Using (1) and (2) in the given equation, we obtain

1 1 1 1 12 2 1 2 123 5 5 (3 )(3 )x y y x xz z z (3)

We have that (3, ) 1z , (otherwise 3|z, which is absurd according to (3)) and so

1 1 1 1(3 ,3 ) 2 3 ,2 2 3 , 2x x x xz z z z .

Hence 1 1(3 ,3 ) 1 or 2x xz z and since zis even (otherwise (3) is absurd), we

have

1 1(3 ,3 ) 1 x xz z .

Moreover, from (3), since 1 12 15 , 3 0 y x z , we get 13 0 x z and

since 13 1 x z , we finally have

1 1 12 13 1 , 3 5x x yz z .1 12 12 3 5 1x y (4)

Now we distinguish the cases:

If1 22x x , then (4) can be written as

2 12 9 5 25 1 x y and considering

both sides mod24 we find2 1 22 9 5 1 1(mod 24) 9 3(mod12) x y x ,


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which is absurd, because the left half side gives always remainder 9, when divided

by 12.

If1 22 1 x x , then (4) can be written as

2 12 3 9 5 25 1.x y (5)

Therefore, for2

0x we find1

0y , and finallyy= 1,x= 2 andz= 2.

If2

1x , then considering both sides mod9 we obtain 15 7 1(mod9)y ,

which is valid, only when1 1 21(mod3) 3 1y y y , (since

37 1(mod9) ).

If2

0y , then equation (5) has not solutions, where as for2

1y , substituting

1y into (5) we have mod7 in both sides

2 2 2 2

2 2

2

3 16 2 5 4 1(mod 7) 6 2 20 64 1(mod 7)

6 2 ( 1) 1 1(mod 7)

6 2 0(mod 7), absurd.

x y x y

x y

x

Hence the given equation has the unique solution (x, y ,z)=(2, 1, 2).

Problem 2

Line Nis parallel to the side BCof the triangleABC, where ,Nare

points of the sides , ,AB AC respectively. Lines BNand CMmeet at pointP

The circumcircles of the triangles BMPand CNPintersect at two different

pointsP Q . Prove that : BAQ CAP .

Solution

We put , and .BAQ PAN QAP x From the inscribed quadrilaterals

BMPQ and QPNCwe get:

BQ BPQ QCN and QNC QPC MBQ ,

and therefore the quadrilaterals AMQCand ABQNare inscrible. From the hypo-

thesis of the problem we observe that at point Pare concurrent the Cevian lines

passing through the vertices of the triangles ABCand QMN.

Using trigonometric form of Cevas theorem we have the relations:

sin sin sin1

sinsin sin

BAP ACP CBP

PBAPAC PCB , (1)


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sin sin sin1

sin sinsin

MNP NQP QMP

PNQ PMNPQM . (2)

From the inscribed quadrilaterals , ,MBQP PQCN ABQN , MQCand the

relationMN BC , we obtain the angle equalities

, , , ,

, , , ,

BAP x PAC QMP PBQ QAN x ACP NQP

CBP NMP PCB MNP PBA PQM PNQ PCQ MAQ

and therefore by dividing by parts (1) and (2) we have:

sin sin

1 sin sin sinsin sin

xsin x x

x

cos cos 2 cos cos 2 cos 2 cos 2x x x x x x ,

from which , given that 0 180x BAC , it follows that: .

Second solution (D. Papadimitriou)

It is enough to prove that ,AQ APare isogonal conjugates of the sides ,AB AC.

If APmeets sideBCat T, then from the theorem of Ceva we have:

1BT CN AM

TC NA MB . (3)


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Since MN BC , it follows that :

AM NA

MB CN , (4)

and hence from (3) and (4), it follows that BT TC , that is point Tis the middle

of the side BC. Hence, it is enough to show that Q is a symmedian of the trian-

gle BC. Equivalently, it is enough to show that point Q satisfies the equality:

,QK AB

QL AC (5)

where QK AB and QL AC .

But the triangles ,QBM QNC are congruent, because they have:

andMQB MPB NPC NQC QMB QPB QCN ,

from the inscribed quadrilaterals andNPQC MBQP. Thus we have:

, (since ).QK MB AB

MN BCQL NC AC

Problem 3

912 rectangle is partitioned into unit squares. The centres of all the unit

squares, except for the four corner squares and the eight squares sharing a common

side with one of them, are coloured red. Is it possible to label these red centres

1 2 96, ,....,C C C in such a way that the following two conditions are both fulfilled:

(i) the distances 1 2 2 3 95 96 96 1, ....,, ,C C C C C C C C are all equal to 13 ,

(ii)the closed broken line 1 2 96 1....C C C C has a centre of symmetry.

Solution

Such a broken line does not exists. To show this, color the red point squares in

a chess pattern (black and white), so that every two red points at distance 1 lie in

squares of different color. It is easy to see that any two red points at distance 13

lie on squares of different colors, so black and white alternate along the broken

line. Also, the center of symmetry of the line must coincide with that of the set of

points, and thus with that of the rectangle.

Consider now the points 2,2 and 8,11 (as usual the point ,i j is the

center of the unit square in the i -th row and the j -th column). The line can be di-

vided in two parts one leading from A to B , and the other from B to A. If they


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are symmetric to each other, each of them must consist of 96:2=48 edges. So an

even number of edges connect A to B, hence A and B must lie in squares of same

color, which is not true.

So, each pert is symmetric to itself (since the symmetrical of the pert leading

from A to B, can only be the other part, case dismissed in the above, or itself; and

same for the part leading from B to A), and each part contains an odd number of

edges. Since the edges can be divided in symmetric pairs, each part must contain

some edge symmetric to itself. Only two such edges are possible: one joining 4,5

and 6,8 , and the other joining 6,5 and 4,8 .

Consider now the point 2,2 . It can only be joined to 5,4 and 4,5 , so the

line must include these two edges. A similar consideration for the points 8,2 ,

8,11 2,11 shows that the line must include the edges 4,5 2,2 5,4

8,2 6,5 4,8 2,11 5,9 8,11 6,8 4,5 . But this is a closed broken

line that does not contain all the points, a contradiction.

Problem 4

Let * 1,2,3,... be the set of positive integers. Find all functions

* *:f such that

2 2 2 22 2 ,f f m f n m n for all *,m n .Solution

First we prove that f is injective. In fact, for any fixed n , if 1 2f m f m ,

then: 2 2 2 22 2 2 21 1 2 22 2 2m n f f m f n f f m f n m n , whence2 2

1 2m m and 1 2m m .

Since f is injective we have:

2 2 2 2 2 2 2 22 2 2 2f m f n f p f q m n p q (1)

Putting 1f a , we find for 1m n , 23 3f a . Then from (1) we get

2 2 2 2 2

2 2 2 2 25 2 3 2 3 3 3 27f a f a f a f a f a .

Since the solution of the equation 2 22 27x y in positive integers is ,x y

= 3,3 and , 5,1x y , it follows that 2 1f a and 25 5f a .Also, from (1) we have


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2 2 2 2

2 2 2 22 4 2 2 5 24f a f a f a f a .

Since the unique solution in positive integers of the equation 2 2 12x y is

, 4, 2x y , it follows that 22 2f a and 24 4f a .Using (1) again we can have

2 2 2 2

2 2 2 24 2 3 2 1f k a f k a f k a f ka ,

as it arises easily by the identity 2 2 22

4 2 1 2 3k k k k , and therefore

2 ,f ka k by induction on k. Then 2 1a a f and thus 1a . Hence

,f k k for any *k . Finally it is easy to verify that ,f k k is a solution of

the problem.

1133tthhJJuunniioorrBBaallkkaannMMaatthheemmaattiiccaallOOllyymmppiiaaddSarajevo, Bosnia and Herzegovina, 25-30 June 2009

Problem 1.

Let ABCDEbe a convex pentagon such that DEBCCDAB and k a circlewith center on side AE that touches the sides CDBCAB ,, and DE at points

RQP ,, and S (different from vertices of the pentagon) respectively. Prove thatlines PSand Eare parallel.

Problem 2.

Solve in non-negative integers the equation

2932 cba .

Problem 3.

Let zyx ,, be real numbers such that

1,,0 zyx and zyxxyz 111 .

Show that at least one of the numbers xzzyyx 1,1,1 is greater than or

equal to41 .

Problem 4.

Each one of 2009 distinct points in the plane is colored in blue or red, so that on

every blue-centered unit circle there are exactly two red points. Find the greatest

possible number of blue points.

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