30th Iranian Mathematical Olympiad

2012-2013 Young Scholars Club Ministry of Education, I.R.Iran www.ysc.ac.ir

30th

Iranian Mathematical Olympiad

Selected Problems and Their Solutions

1

30th Iranian Mathematical Olympiad

Selected Problems and Their Solutions

This booklet is prepared by Goodarz Mehr, Hesam Rajabzadeh and

Morteza Saghaan.

With special thanks to Ali Khezeli, Atieh Khoshnevis, Mahan Malihi, Omid

Naghshineh Arjmand, Sam Nariman and Erfan Salavati.

Copyright

cYoung Scholars Club 2012-2013. All rights reserved.Ministry of education, Islamic Republic of Iran.

www.ysc.ac.ir

3

Iranian Team Members in the 54th IMO (Santa Marta-Colombia)

From left to right:

Tina Torkaman Aref Sadeghi Alek Bedroya Seyyed Mohammadhossein Seyyedsalehi Mohammad Javad Shabani Zahraei Mahan Tajrobehkar

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Contents

Preface..................................................................................................................... 9

Problems

Second Round......................................................................................................... 13

Third Round........................................................................................................... 14

Team Selection Test................................................................................................ 16

Solutions

Second Round........................................................................................................ 21

Third Round.......................................................................................................... 26

Team Selection Test............................................................................................... 36

7

Preface

The 30th Iranian National Mathematical Olympiad consisted of four rounds. Therst round was held on 24 February 2012 all over the country. The exam consisted of 10

short-answer questions and 15 multiple-choice questions to be solved in 3 hours. In total

more than 25000 students participated in the exam and 1446 of them were validated for

the next round.

The second round was held on 30 April and 1 May 2012. In each day, participants

were given 3 problems to be solved in 4.5 hours. In this round, from a total of 1446

participants, 43 of them were chosen to participate in the third round.

The examination of the third round consisted of ve separate exams, and a nal exam

containing 8 questions, each having it's specied time to be solved in. In the end, 10 peo-

ple were awarded Bronze Medal, 20 people were awarded Silver Medal, and 13 people were

awarded Gold Medal, which the following list represents the names of the Gold Medalists:

Jalal Ahmadi

Mohammad Javad Baferooni

Alek Bedroya

Hossein Hazrati

Mohammad Amin Ketabchi

Hadi Khodabande

Aref Sadeghi

Seyyed Mohammadhossein Seyyedsalehi

Mohammad Javad Shabani Zahraei

Mahan Tajrobehkar

Seyyed Alireza Tavakoli

Tina Torkaman

Mobin Yahyazade Jelodar

The Team Selection Test was held on 6 days, each day having 3 problems to be solved

in 4.5 hours. In the end, the top 6 participants were chosen to participate in the 54th

IMO as members of the Iranian Team.

In this booklet, we present 6 problems of the Second Round, 8 problems of the nal

exam of the Third Round, and 18 problems of the Team Selection Test, together with

their solutions.

It's a pleasure for the authors to oer their grateful appreciation to all the people

who have contributed to the conduction of the 30th Iranian Mathematical Olympiad, in-cluding the National Committee of Mathematics Olympiad, problem proposals, problem

selection teams, exam preparation teams, coordinators, editors, instructors and all who

have shared their knowledge and eort to increase the Mathematics enthusiasm in our

country, and assisted in various ways to the conduction of this scientic event.

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Problems

11

Second Round

1. (Morteza Adl) Consider a circle C1 and a point O on it. Circle C2 with center O,intersects C1 in two points P and Q. C3 is a circle which is externally tangent to C2 atR and internally tangent to C1 at S and suppose that RS passes through Q. Suppose Xand Y are second intersection points of PR and OR with C1. Prove that QX is parallelto SY . (! p.21)

2. (Morteza Saghaan) Suppose n is a natural number. In how many ways can weplace numbers 1; 2; ::::; n around a circle such that each number is a divisor of the sumof it's two adjacent numbers? (! p.21)

3. (Mahyar Sedgaran, Amin Nejatbakhsh) Prove that if t is a natural number thenthere exists a natural number n > 1 such that (n; t) = 1 and none of the numbersn+ t; n2 + t; n3 + t; :::: are perfect powers. (! p.23)

4. (Morteza Saghaan) a) Do there exist 2-element subsets A1; A2 ; A3; ::: of naturalnumbers such that each natural number appears in exactly one of these sets and also for

each natural number n, sum of the elements of An equals 1391 + n?b) Do there exist 2-element subsets A1; A2; A3; ::: of natural numbers such that eachnatural number appears in exactly one of these sets and also for each natural number n,sum of the elements of An equals 1391 + n

2? (! p.23)

5. (Sahand Seifnashri) Consider the second degree polynomial x2 + ax + b withreal coecients. We know that the necessary and sucient condition for this polyno-

mial to have roots in real numbers is that its discriminant, a2 4b, be greater than orequal to zero. Note that the discriminant is also a polynomial with variables a and b.Prove that the same story is not true for polynomials of degree 4: Prove that there doesnot exist a 4 variable polynomial P (a; b; c; d) such that the fourth degree polynomialx4 + ax3 + bx2 + cx+ d can be written as the product of four 1st degree polynomials ifand only if P (a; b; c; d) 0. (All the coecients are real numbers.) (! p.24)

6. (Mehdi E'tesami Fard) The incircle of triangle ABC, is tangent to sides BC;CAand AB at D;E and F respectively. The reection of F with respect to B and thereection of E with respect to C are T and S respectively. Prove that the incenter oftriangle AST is inside or on the incircle of triangle ABC. (! p.24)

13

Third Round

1. (Morteza Saghaan) Let G be a simple undirected graph with vertices v1; v2; :::; vn.We denote the number of acyclic orientations of G with f(G).a) Prove that f(G) f(G v1) + f(G v2) + + f(G vn).b) Let e be an edge of the graph G. Denote by G0 the graph obtained by omitting e andmaking it's two endpoints as one vertex. Prove that f(G) = f(G e) + f(G0).c) Prove that for each > 1, there exists a graph G and an edge e of it such thatf(G)

f(Ge) < . 90 minutes (! p.26)

2. (Ali Khezeli) Suppose S is a convex gure in plane with area 10. Consider a chordof length 3 in S and let A and B be two points on this chord which divide it into threeequal parts. For a variable point X in S fA;Bg, let A0 and B0 be the intersectionpoints of rays AX and BX with the boundary of S. Let S0 be those points of X forwhich AA0 > 13BB

0. Prove that the area of S0 is at least 6.105 minutes (! p.26)

3. (Amirhossein Gorzi) Prove that for each n 2 N there exist natural numbersa1 < a2 < < an such that (a1) > (a2) > > (an). 50 minutes (! p.28)

4. (Hamidreza Ziarati) We have n bags each having 100 coins. All of the bags have10 gram coins except one of them which has 9 gram coins. We have a balance which canshow weights of things that have weight of at most 1 kilogram. At least how many timesshall we use the balance in order to nd the dierent bag? 75 minutes (! p.28)

5. (Mostafa Einollahzade, Erfan Salavati) We call the three variable polynomial Pcyclic if P (x; y; z) = P (y; z; x). Prove that cyclic three variable polynomials P1; P2; P3and P4 exist such that for each cyclic three variable polynomial P , there exists a fourvariable polynomial Q such that

P (x; y; z) = Q(P1(x; y; z); P2(x; y; z); P3(x; y; z); P4(x; y; z)):

60 minutes (! p.29)

6. (Mostafa Einollahzade) a) Prove that a real number a > 0 exists such that for eachnatural number n, there exists a convex n-gon P in plane with lattice points as verticessuch that the area of P is less than an3.b) Prove that there exists b > 0 such that for each natural number n and each n-gon Pin plane with lattice points as vertices, the area of P is not less than bn2.c) Prove that there exist ; c > 0 such that for each natural number n and each n-gonP in plane with lattice points as vertices, the area of P is not less than cn2+.105 minutes (! p.30)

14

7. (Erfan Salavati) The city of Bridge Village has some highways. Highways are

closed curves that have intersections with each other or themselves in 4-way crossroads.Mr.Bridge Lover, mayor of the city, wants to build a bridge on each crossroad in order

to decrease the number of accidents. He wants to build the bridges in such a way that in

each highway, cars pass above and under a bridge alternatively. By knowing the number

of highways determine whether this action is possible or not.

50 minutes (! p.33)

8. (Morteza Saghaan) a) Does there exist an innite subset S of the natural num-bers, such that S 6= N, and such that for each natural number n 62 S, exactly n membersof S are coprime with n?b) Does there exist an innite subset S of the natural numbers, such that for each naturalnumber n 2 S, exactly n members of S are coprime with n? 75 minutes (! p.34)

15

Team Selection Test

1. (Mehdi E'tesami Fard) In acute triangle ABC, let H be the foot of the perpen-dicular from A to BC and also let J and I be the B-excenter of triangle ABH andC-excenter of triangle ACH, respectively. If P is the point of tangency of the incenterof triangle ABC with side BC, prove that points I; J; P and H lie on the same circle.(!p.36)

2. (Ali Khezeli) At most how many subsets of the set f1; 2; :::; ng can be selected insuch a way that if A and B are two selected subsets and A B, then jB Aj 3? (By|X| we mean number of elements of the set X.) (!p.37)

3. (Morteza Saghaan) For nonnegative integers m and n, the sequence a(m;n) ofreal numbers is dened as follows: a(0; 0) is equal to 2, and for each natural number n,a(0; n) = 1 and a(n; 0) = 2. Also for m;n 2 N:

a(m;n) = a(m 1; n) + a(m;n 1)Prove that for each natural number k, all roots of the polynomial Pk(x) =

Pki=0 a(i; 2k+

1 2i)xi are real numbers. (!p.37)

4. (Shayan Dashmiz) Suppose m and n are two nonnegative integers. In the Philoso-pher's Chess, the chessboard is an innite array of same regular hexagon cells.

The Phoenix piece, which is a special

piece in this kind of chess, moves as fol-

lows:

At rst, the Phoenix selects one of

the six directions and moves m cells inthat direction. Then it turns 60 de-grees clockwise and moves n cells inthat new direction to get to the nal

point.

At most how many cells of the Philoso-

pher's Chess' chessboard exist in such a way

that one cannot start from one of them and

reach another one with a nite number of

movements of the Phoenix piece? (!p.38)

5. (Mahan Malihi) Do there exist natural numbers a; b and c such that a2 + b2 + c2

is divisible by 2013(ab+ bc+ ca)? (!p.39)

6. (Ali Khezeli) Let X;Y;X 0 and Y 0 be four points on a line l with the same order aswritten. Consider arcs !1 and !2 with end points fX;Y g and arcs !01 and !02 with end

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points fX 0; Y 0g on one side of l such that !1 is tangent to !01 and !2 is tangent to !02.Prove that the external common tangent of !1 and !

02 and the external common tangent

of !01 and !2 meet on l. (!p.41)

7. (Mohammad Ali Karami) Nonnegative real numbers p1; p2; :::; pn and q1; q2; :::; qnare given such that

p1 + p2 + :::+ pn = q1 + q2 + :::+ qn:

Among all matrices with nonnegative real entries for which sum of entries of the ith rowis pi and sum of entries of the jth column is qj , nd the maximum value that the traceof the matrix can have. (!p.44)

8. (Yahya Motevassel) Find all innite arithmetic progressions a1; a2; ::: for whichthere exists natural number N > 1 such that for each k 2 N:

a1a2 akjaN+1aN+2 aN+k:(!p.46)

9. (Mohammad Jafari) Find all functions f; g : R+ ! R+ such that f is an increasingfunction and also for each x; y 2 R+:

f(f(x) + 2g(x) + 3f(y)) = g(x) + 2f(x) + 3g(y)g(f(x) + y + g(y)) = 2x g(x) + f(y) + y:(!p.46)

10. (Morteza Saghaan) To each edge of a graph, we have assigned a real number in

a way that for each even walk of this graph, sum of the numbers on edges of the walk,

with minus and plus signs assigned alternatively, is zero. Prove that we can assign a

real number to each vertex so that for each edge, the number assigned to it is the sum

of numbers assigned to it's endpoints. (A walk is a closed path of edges that can have

repeated number of vertices or edges.) (!p.47)

11. (Amirhossein Gorzi) Suppose we have a triangle with side lengths a; b and c suchthat a b c. Prove thatq

a(a+ bpab) +pb(a+ cpac) +

qc(b+ cpbc) a+ b+ c:

(!p.48)

12. (Ali Zamani) Quadrilateral ABCD is inscribed in circle !. Suppose that I1 andI2 are incenters of triangles ACD and ABC and r1 and r2 are inradiuses of trianglesACD and ABC, respectively. We know that r1 = r2. Circle !

0is tangent to sides ABand AD and also is tangent to circle ! at T . Tangents at T and A to circle ! intersect

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each other in K. Prove that I1; I2 and K are collinear. (!p.48)

13. (Ali Zamani) Point P is an arbitrary point inside acute triangle ABC and A1; B1and C1 are reections of P with respect to sides of triangle ABC. Prove that the centroidof triangle A1B1C1 is inside triangle ABC. (!p.50)

14. (Amirhossein Gorzi) We have drawn n rectangles in plane where n is a naturalnumber. Prove that among 4n resulting right angles, at least bp4nc distinct right anglesexist. (!p.50)

15. (Amirhossein Gorzi) a) Does there exist a sequence of natural numbers a1 < a2 < for which there exists a natural number N in a way that for each natural numberm N , am is divisible by exactly d(m) 1 terms of the sequence? (d(n) is the numberof divisors of natural number n.)b) Does there exist a sequence of natural numbers a1 < a2 < for which thereexists a natural number N in a way that for each natural number m N , am is divisibleby exactly d(m) + 1 terms of the sequence? (!p.50)

16. (Amirhossein Gorzi) The function f : Z! Z has the property that for all integersm and n

f(m) + f(n) + f(f(m2 + n2)) = 1:

We know that integers a and b exist such that f(a) f(b) = 3. Prove that integers cand d can be found such that f(c) f(d) = 1. (!p.51)

17. (Mahan Malihi) AD and AH are angle bisector and altitude of vertex A, respec-tively. The perpendicular bisector of AD, intersects semicircles with diameters AB andAC which are drawn outside triangle ABC in X and Y , respectively. Prove that thequadrilateral XYDH is concyclic. (!p.51)

18. (Ali Khezeli) A special kind of parallelogram tile is made up by attaching the

legs of two right isosceles triangles of side length 1. We want to put a number of thesetiles on the oor of an n n room such that the distance from each vertex of each tileto the sides of the room is an integer and also no two tiles overlap. Prove that at least

an area n of the room will not be covered by the tiles. (!p.52)

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Solutions

19

Second Round

1 . Let l be the common tangent of C1 and C3. Then:

_SR = 2\RSl = 2\QSl =

_SQ

_SR =

_RQ = 2\RPQ =

_XQ

)

)_SQ =

_XQ) SQ = XQ: (1)

So it is sucient to prove that QX = XY . Notice that O is the center of C2, so:

OP = OR) \OPR = \ORP )_OX =

_OP +

_XY

)_OQ+

_QX =

_OP +

_XY )

_QX =

_XY : (2)

(1); (2)) SQ = XQ = XY ) SY kQX:

2 . We call an arrangement good if it satises the problem requirements. Consider oneof these arrangements. n should be a divisor of sum of it's two adjacent numbers, but thesum of these two numbers is less than n+ n = 2n. So sum of the two adjacent numbersof n must be n. Thus for a natural number k, 1 k n 1, these two numbers are kand n k. As a result, we see that by omitting n, the condition of the problem is stillsatised, because n nk (mod k) and n k (mod nk), which means k and nkstill divide sum of their two adjacent numbers. So by omitting n, we reach to a goodarrangement of numbers 1; 2; :::; n 1. For n = 1 and n = 2, the answer to the questionis 1; and for n = 3 the answer is 2. Now, by using induction, we prove that the answer

21

to the question is 4 for odd natural numbers n greater than 3, and 2 for even naturalnumbers n greater than 2.

Base case of the induction: n = 4

Suppose that our assumption holds for n = k. A good arrangement of numbers1; 2; :::; k + 1 will become a good arrangement of numbers 1; 2; :::; k by omitting k + 1.

If k+1 is odd, by the induction hypothesis good arrangements of numbers 1; 2; :::; k are

and k+1 could only be placed between numbers 1 and k or between numbers k2 andk2 + 1 which have a sum of k + 1. Thus we have 4 good arrangements in this case.

If k+1 is even, by the induction hypothesis good arrangements of numbers 1; 2; :::; kare

22

and k+1 could only be placed between the two numbers 1 and k in the two bottom-most arrangements. So we have 2 good arrangements in this case.In conclusion, The answer to the problem for n = 1 and n = 2 is 1, for n = 3 and forall other even numbers is 2, and for all other odd numbers is 4.

3 . Let p be a prime factor of t + 1 and let pmjjt + 1. Dene K = m!(pm+1) andn = (t(t + 1) + 1)K . Now, we have nr + t 1 + t (mod pm+1), so if nr + t = bs thens m. On the other hand we have

nr = (t(t+ 1) + 1)Kr =(t(t+ 1) + 1)

Ksrs

= n0s; (Because sjm!jK)so

t = bs nr = bs n0s (n0 + 1)s n0s sn0 2n0 > twhich is a contradiction.

4 . a) Let Ai = fxi; yig. So xi + yi = 1391 + i, but:

13912 + 1391 696 =1391Xi=1

(xi + yi)

= x1 + x2 + + x1391 + y1 + y2 + + y1391 1 + 2 + + 2 1391 = 1391 2783:

23

Contradiction!

b) We construct such Ai's inductively. Let A1 = f1; 1391g. Now, assume thatA1; A2; : : : and Ak are constructed. Let xk+1 be the least positive integer which is not inany of Ai's (1 i k). Obviously xk+1 2k + 1, so the number

yk+1 = 1391 + (k + 1)2 xk+1 1391 + k2

is not in any of Ai's too (because sum of the elements of each Ai, (1 i k) is atmost 1391+ k2 until now). So we can have Ak+1 = fxk+1; yk+1g, and in these Ai's, eachpositive integer has appeared exactly once (because of the method of choosing xk+1 ineach step).

5 . If we put a = c = 0, polynomial x4 + bx2 + d can be written as product of fourlinear terms if and only if quadratic polynomial y2 + by + d has two nonnegative roots.Therefore P (0; b; 0; d) 0 if and only if b 0, d 0 and b24d 0. For a xed b 0 letQb(d) = P (0; b; 0; d). Now, Qb(d) 0 if and only if 0 d b24 and hence by continuityof Qb, Qb(b

2=4) must be zero. This implies that for all b 0, one variable polynomialP (0; b; 0; b

2

4 ) = 0, and hence this polynomial is always zero. This means that polynomial

x4 + bx2 + b2

4 has four real roots for all values of b . Contradiction!

6 . Let I and I 0 be the incenters of triangles ABC and ATS respectively, and ! and !0

be the incircle of triangle ABC and circumcircle of TI 0S respectively.We have

\TDF = \SDE = 90

\EDF = 12\EIF = 90 \A

2

) \TDS = 360 90 90 (90 \A2

)

= 90 +\A2:

On the other hand we have TI 0S = 90+ \A2 ; therefore, TDI0S is cyclic and D is on

!0. We know that, center of !0 is the midpoint of_TS of the circumcircle of triangle ATS,so it is on the line II 0. Hence two circles !; !0 intersect at D, and I 0 is the intersectionof !0 and the line joining the two centers. Thus I 0 is inside !.

24

25

Third Round

1 . a) In each acyclic orientation of G, there is at least one vertex (say vi) such that alledges connected to that are directed toward vi, because if not the orientation will havea cycle which contradicts the fact that the orientation is acyclic. By omiting this vertex

and all edges connected to it, we obtain an acyclic orientation for the graph G vi.The number of acyclic orientations of Gvi is f(Gvi), but for some orientations of

G there exists more than one vertex with the property discussed. Therefore, the assertionis proved.

b) Every acyclic orientation of G e results in at least one acyclic orientation of Gbecause if vi and vj are two endpoints of edge e, then if e cannot be oriented from vito vj , it means that there is a path (in G e) from vj to vi. Similarly, if e cannot beoriented from vj to vi, it means that there is a path (in G e) from vi to vj .We concludethat because of two directed paths from vi to vj and from vj to vi, there exists a directedcycle containing vi and vj in G e, which is a contradiction.So each acyclic orientation of G e results in one or two acyclic orientaions of G.Now, if edge e can be oriented in two directions, then in G e there is no directed patheither form vj to vi or from vi to vj . So if we omit e to get G

0, as stated in the problems

statement, an acyclic orientation of G0 can be found, and the assertion is followed.c) It suces to consider the complete graph with n vertices. In this case f(G) = n!,and according to part (b) we have

f(G e) = n! (n 1)! = (n 1)(n 1)!:

Therefore

f(G)f(Ge) =

n!(n1)(n1)! =

nn1 . For each > 1, there is a natural number n suchthat

n

n 1 = 1 +1

n 1 <

and this completes the proof.

2 . The idea is to remove some neighborhoods of A and B, because near these pointswe cannot bound

AA0BB0 . Let Z be the intersection of the ray AB with the boundary of Sand let ZB0 intersect AA0 in A00. A00 is between A and A0 since S is convex. So, if we letS00 be the set of those points X such that AA00 > 13BB

0, we have S00 S0. Hence, it isenough to prove that the area of S00 is at least 6.By Menelaus's theorem, we have

AA00

A00X XB

0

B0B BZZA

= 1

) AA00

A00X XB

0

B0B= 2:

26

To omit of A00X, we write AA00

A00X in terms of AA00and AX and we treat B0X similarly:

) A00X

AA00=

1

2 XB

0

B0B

) AXAA00

= 1 12(1 BX

BB0)

) AXAA00

=1

2 BXBB0

+1

2:

Let D1 be the set of those points X such that12 BXBB0 + 12 BXBB0 for some constant (for suitable , D1 is a neighborhood of B). If X 62 D1, then

AX

AA00<

BX

BB0

) AA00

BB0>

1

AXBX

:

Let D2 be the set of those points X in the plane such thatAXBX

1

3=

1

3

as desired. So

S D1 D2 S00 S0:Now, we calculate the areas of D1 and D2. If < 3, then D2 is the interior part of theApolonius circle of A and B. Also,

X 2 D1 , 12 BXBB0

+1

2 BX

BB0, BX

BB0 1 then D1 is a neighborhood of B similar to S. Now, we take =32 . We

conclude that the area of D1 is (12)

2times the area of S which is 104 = 2:5. Also, if C and

D are the intersection points of the boundary of D2 with the line AB and C is betweenA and B, we have

AD

DB=

1

2) AD

AD + 1=

1

2) AD = 1

AC

CB=

1

2) AC

1AC =1

2) AC = 1

3:

So the diameter of D2 is 1 +13 =

43 and hence, the area of D1 is (

23)

2 < 1:5. So

TX TSD1D2 2TS TD1 TD2> 10 2:5 1:5 = 6:where TK represents the area of gure K in the plane. Hence, the assertion is proved.

27

3 . First we prove a lemma.

Lemma. Let a > b be two positive integers such that ab > 4. Then there exist somepositive integer l such that b < (2l) = 2l1 < 2l < a.

Proof of lemma. There exists some positive integer m such that 2m1 b < 2m. Soa > 4b 2m+1 and hence for l = m+ 1 we have b < 2l1 < 2l < a, as claimed.

We construct the sequence in the problem inductively. The assertion for n = 1 isobvious. Now, suppose that a1 < a2 < < an is a sequence such that (a1) > (a2) > > (an). Let pN be the greatest prime factor of ai's (pj is the jth prime number),1 i n. Since for every x with prime factors greater than pN , the greatest commondivisor of x and ai is 1 for all 1 i n, we deduce that the sequence a1x < a2x < (a1x).First we claim that there is a positive integer x with prime factors greater that pN suchthat

a1x(a1)(x)

> 4 or equivaletly, x(x) > 4(a1)a1.

To prove this, let xm = pN+1pN+2 pN+m(m 2 N). We have

xm(xm)

=

N+mYi=N+1

pipi 1 =

N+mYi=N+1

1 +

1

pi 1

N+mXi=N+1

1

pi 1 :

The sum

P1i=N+1

1pi1 >

P1i=N+1

1pidiverges by a well-known fact, so we can nd m

such that

xm(xm)

> 4(a1)a1 . We let x = xm. Now, sincea1x

(a1)(x)> 4, according to the

lemma there is some positive integer l such that (a1x) < (2l) < 2l < a1x. So putting

y = 2l leads to the new sequence

y < a1x < a2x < < anx

which has n+ 1 terms and satises the problem's condition.

4 . Obviously, if we are able to nd the dierent bag among n bags with k times usingthe balance, for less than n bags also k times using the balance is enough. Denote byf(k) the maximum number of bags such that the dierent bag among them can be foundwith at most k times using the balance. Our goal is to nd f(k) for all k 2 N.

(i) f(1) = 14. If n = 14, for 1 i 14 we put i 1 coins from the ith bag on thebalance, and if the resulting weight is 910k (91 = 1+2+ +13) grams, then the(k + 1)st bag is dierent, so f(1) 14. To prove the converse inequality, suppose

28

that we have a number of bags and we want to use the balance only once. Let aibe the number of bags for which we take exactly i coins from them. We have

a1 + 2a2 + +mam 100: ()

Note that for i 0; ai 1 because if we take the same number of coins from twodierent bags, we cannot distinguish between them. Now, we want to maximizePm

i=0 ai. Since the coecient of ak in () equals k, it's best for a0 to be maximized,then a1 should be maximized and so on. We conclude that the maximum value willbe 14. So f(1) 14 and consequently f(1) = 14.(ii) f(2) = 60. To nd f(2), again () should be satised. Also, for any integer i 0,we must have ai 14. Here we want to maximize

Pmi=0 ai. The maximum valuewill be gained when a0 = a1 = a2 = a3 = 14 and a4 = 4. Therefore f(2) = 60.

(iii) f(3) = 140. By a similar argument, here () should be satised, and also for eachpositive integer i 0, we must have ai 60, so the maximum value will be gainedwhen a0 = a1 = 60 and a2 = 20; therefore, f(3) = 140.

(iv) Finally, if k > 3, then ai 100 and we must have a0 f(k1), so we get the maxi-mum value of

Pmi=0 ai when a0 = f(k1) and a1 = 100. Hence f(k) = f(k1)+100.

Upshot.

n 2 [1; 14] [15; 60] [61; 140] [100k + 40; 100k + 140]; k 2 N1 2 3 k + 3

5 . LetQ(x; y; z) = P (x; y; z) + P (y; x; z)

R(x; y; z) = P (x; y; z) P (y; x; z):By these denitions it is obvious that Q is a symmetric polynomial and R is anantisymmetric one. Note that R(x; x; z) = 0, so R is divisible by x y. Similarly, y zand z x also divide R. Therefore there is a polynomial S such that

R(x; y; z) = (x y)(y z)(z x)S(x; y; z):And it's obvious that S is a symmetric polnomial. We conclude that

P =1

2Q+

1

2(x y)(y z)(z x)S:

Every symmetric polynomial in three variables can be written as a polynomial of ele-

mentary symmetric polynomials P1 = x+ y+ z, P2 = xy+ yz + zx and P3 = xyz. If weput P4 = (x y)(y z)(z x) (which obviously is cyclic) the proof is complete.

29

6 . We call a convex n-gon A1A2 An with vertices from lattice points good if for eachinteger 1 i n 1, vector !AiAi+1 lies in the rst quadrant of the cartesian plane.Throughout the solution all polygons are considered to have vertices from lattice points.

We denote the least possible value for the area of a convex n-gon and the area of agood convex n-gon by f(n) and g(n), respectively. Furthermore, in such polygons let

ui =!AiAi+1 (1 i n 1). We claim that g(n4 ) f(n) g(n). The rightmostinequality is obvious, since good polygons are a subset of all such polygons. To prove the

leftmost inequality, note that for each convex n-gon P the rightmost, leftmost, topmostand bottommost vertices divide the perimeter of the polygon into at most 4 parts, so at

least one part has at least

n4 veritces and the area of the convexhull of these vertices is

at least g(n4 ) so the claim is proved.

Now, it suces to prove the assertions for g(n). (g(n) = O(n2); :::)

Lemma 1. The area of a good n-gon P with vectors u1; u2; : : : and un1 equals 12P

i

If

Pjjui uj j 16n

32for 1 i n+1, summing these expressions implies the claim.So without loss of generality we can assume thatX

i

jui un+1j 16n32 :

We do not have the assumption of monotonocity of slopes any further. For each i,let Ci = ful : jul un+1j = ig and ki = jCij. Obviously

Pki = n. Elements of Ci lieon two lines parallel to un+1, so for each j n, at most 4 numbers among the valuesjuj ulj (ul 2 Cj) can be equal (except number 0 when uj 2 Ci which is unique).Since these values are all integer, we haveX

ul2Cijuj ulj 4

ki4

2

:

Therefore,

Xln

juj ulj 4Xi

ki4

2

=

1

8

Xi

k2i n

2;

and Xj;l

juj ulj n8

Xi

k2i n2

2;

but

Pki = n and

Piki 16n

32. Now, according to the following lemma we haveP

k2i 32n32, and therefore for suciently large nX

j;l

juj ulj 316n52 n

2

2 1

6n52

and this completes the proof of the claim.

Lemma 2. Let k1; k2; : : : ; kn be nonnegative real numbers such thatP

ki = a andPiki b. Show that

Pk2i a

3

4b .

Proof of lemma. Suppose that values a and b are constants and k1, k2; : : : ; kn arevariables satisfying the conditions for which

Pk2i is minimum. We conclude that ki's,

1 i n, are in decreasing order because if ki < kj for some i < j then we can substituteboth of them with

12(ki + kj) and

Pk2i will have a smaller value.We claim that k1; k2; : : : ; kn must be a block of an arithmetic progression unless theyare zero. Suppose that kj1; kj and kj+1 are nonzero and kj 6= kj1+kj+12 . If we changethese three to kj1 + x; kj 2x; kj+1 + x, it is easy to check that this new sequence alsosatises the conditions. Thus for the dierence of the value

Pk2i for these two sequenceswe have

= ((kj1 + x)2 + (kj 2x)2 + (kj+1 + x)2) (k2j1 + k2j + k2j+1)

31

= 6x2 + 2x(kj1 + kj + kj+1):

Note that the coecient of x is nonzero. Hence we can choose a small value of xhaving a dierent sign from it's coecient for to be less than 0. Therefore, eachthree consecutive nonzero numbers among ki's must form an arithmetic progression. Bymodifying n if necessary, we can assume that the sequence k1; k2; : : : and kn is a decreasingarithmetic progression of nonzero numbers, so ki = rsi for constants r; s 0. We have

a = nr sP ic = r

Pi sP i2 b:HenceX

k2i = nr2 2rs

Xi+ s2

Xi2 = r(nr s

Xi) s(r

Xi s

Xi2);

so Xk2i = ra sc:The values of r and s can be calculated from the linear system of equations above interms of a; b and c:

r =aP

i2 cP inP

i2 (P i)2 ; s = aP

i ncnP

i2 (P i)2 :By Cauchy-Schwarz Inequality we obtain that

Xk2i

(P

ki)2

n=a2

n:

Since r sc is negative we have

aP

i2 cP i anP i+ n2c 0) a(n2(n+1)2 n(n+1)(2n+1)6 ) c(n2 n(n+1)2 ) cn(n+1)2) a(n 2n+13 ) c) n 3 ca + 1 4 ca :

The last inequality follows from the fact that c a (because c =P iki P ki = a).Finally we have

Xk2i

a2

n=

a3

an a

3

4c a

3

4b

as desired.

Claim 3. The area of such good (n+2)-gon is at least 1200n3and therefore f(n) = (n3).

32

To prove this claim it suces to prove that for each set of integer vectors u1; u2; : : :and un+1 with distinct slopes, we haveX

i

Proof of lemma. We proceed by induction on number of edges of graph G. We call acoloring of faces of G satisfying the mentioned property, a good coloring. Suppose thatC is the shortest path in G that lies in just one highway and assume C C1. Note thatbecause of minimality, C cannot cross itself, so is a simple curve, therefore dividing thethe plane into two regions. Now, C C1 is a closed or maybe an empty curve.Suppose that H is the graph consisting of CC1; C2; : : : and Cn. Since the number ofedges of H is less than G, H has a good coloring according to the induction hypothesis.Now, we add curve C to H and change the color of faces inside C. It is very easy tocheck that this is a good coloring of G. The base case is when G is empty. In this casewe have only one region which obviously has a good coloring!

Finally consider a good coloring (by colors black and white) for the graph G andan arbitrary orientation for each highway. We have two types of edges. For some edges

when we are moving through ine (in it's specied direction) our right face is white. We

call such edges, edges of type 1 and others, edges of type 2. Build a bridge in the way of

each edge of type 1.

Note that in each crossroad exactly two of the edges are directed into the crossroad

and only one of those two is of type 1. Hence our construction do not cause any contra-

dictions.

8 . a) We claim that a set cannot exist with this property. Assume to the contrary thatthere exists a set S satisfying the problems condition and let n 2 N , n =2 S. Thusexactly n elements of S are coprime to n. Therefore, there exist innitely many primenumbers which do not belong to S. Let p and q be two of them. Since p =2 S exactly pelements of S are not divisible by p (coprime to p), for each > 1, p 2 S. This is truebecause if p =2 S, there must be p elements of S coprime to p (and therefore coprimeto p), but we have shown that S has exactly p elements of this kind. Finally, the greatestcommon divisor of p and q is 1, so innitely many elements of S are coprime to q =2 S,a contradiction!

b) We construct set the S in the following way:

Suppose a1; a2; a3; : : : are elements of S and at rst all ai's, i 2 N, are 1. In each stepwe multiply ai by some primes.

For n 2 N in step n:(i) We multiply an by some new primes p2n1 and p2n so that an become greater than

an1 (Note that pn is the nth prime number).

(ii) Some elements among fa1; a2; : : : ; an1g are coprime to an. let bn be the numberof such elements, so bn < n < an. Let tn = an bn.(iii) Because of the type of construction, innitely many number of ai's, i > n, arecoprime to an. We leave rst tn and assume that the tnth number is am.

34

(iv) In the sequence am+1; am+2; : : : multiply the rst 2n1terms by p2n1, the next

2n1 terms by p2n and so on.

Therefore at the end of step n:

(i) Exactly an numbers are coprime to an.

(ii) For each sequence (q1; q2; : : : ; qn) of prime numbers such that qi 2 fp2i1; p2ig, thereexist innitely many numbers k such that ak has exactly these prime factors. Sofor each ai there exist innitely many numbers k such that ak has prime factorsdierent from those of ai and consequently is copirme to ai.

35

Team Selection Test

1 . Let J 0 and I 0 be foot of the perpendicular lines from J and I to line BC, respec-tively. Furhtermore, throughout the solution we denote by p(XY Z) and S(XY Z) thesemiperimeter and area of triangle XY Z, respectively.

J 0 is the tangency point of B-excircle of triangle ABH and line BH, so we haveBJ 0 = p(ABH) = AB+BH+AH2 . Also we have BP =

AB+BC+CA2 and hence PJ

0 =BJ 0 BP = AH+CACH2 . II 0 is the radius of C-excircle of triangle ACH, therefore

II 0 =S(ACH)

p(ABH)AH =AH:CH

AC + CH AH :

We claim that PJ 0 = II 0.

PJ 0 = II 0 , AH + CA CH2

=AH:CH

AC + CH AH, 2AH:CH = CA2 (CH AH)2 = CA2 CH2 AH2 + 2CH:AH

, CA2 = CH2 +AH2:Which is true because of the Pythagorean Theorem. By a similar argument we get

PI 0 = JJ 0. Therefore, the right-angled triangles PII 0 and PJJ 0 are congruent, so

\IPI 0 = \PJJ 0 = 90 \JPJ 0 ) \IPJ = 90

HI and HJ are angle bisectors of \AHB and \AHC, respectively, and therefore\IHJ = 12(\AHB+\AHC) = 90. Since \IPJ = \IHJ = 90, we deduce that I, P ,J and H lie on the same circle.

36

2 . We claim that the maximum number of such sets is d2n3 e.We proceed by induction on n. For the base case we must check the claim for n = 1; 2which is obvious. Suppose that the claim is true for n. For n + 2 let G be the familyof those selected subsets of f1; 2; : : : ; n+ 2g. We divide G into four sets with respect totheir intersection with fn+ 1; n+ 2g.

G0 = fA 2 Gjn+ 1 =2 A;n+ 2 =2 AgG1 = fA 2 Gjn+ 1 2 A;n+ 2 =2 AgG2 = fA 2 Gjn+ 1 2 A;n+ 2 2 AgG3 = fA 2 Gjn+ 1 =2 A;n+ 2 2 AgFor i = 0; 1; 2 let Fi be the family of A fn+ 1; n+ 2g for each A 2 Gi. We claim thatfor each B f1; 2; : : : ; ng, B is in at most one of the sets F0;F1 and F2, Because ifB 2 Fi;Fj , where i < j, it means that there exist Bi and Bj in G such that

Bi fn+ 1; n+ 2g = B; Bj fn+ 1; n+ 2g = B:

Now, Bi Bj but jBj Bij = jBj j jBij = j i < 3 which is a contradiction, sojG0j + jG1j + jG2j 2n. The number of elements of G3 is at most d2n3 e according to theinduction hypothesis, so

jGj 2n + d2n3 e = d2n+2

3 e:To construct an example for this upper bound, for i = 0; 1; 2 let Si be the family ofsubsets A f1; 2; : : : ; ng such that jAj i (mod 3). We have

jS0j+ jS1j+ jS2j = 2n:

Since these sets are disjoint, one of them has at least d2n3 e elements which obviouslysatises the problem condition.

3 . Let Qk(x) =Pk

i=0 a(i; 2k 2i)xi. According to the recurrence relation for a(m;n),we get

Pk(x) = xPk1(x) +Qk(x)

Qk(x) = xQk1(x) + Pk1(x):

So Qk(x) = Pk(x)xPk1(x) and therefore, Pk(x)xPk1(x) = x(Pk1(x)xPk1(x))+Pk1(x). Finally, we get

Pk(x) = (2x+ 1)Pk1(x) x2Pk1(x):

Hence we get a recurrence relation for Pk+1(x) where k 2, P1(x) = 3x + 1 andP2(x) = 5x

2 + 5x+ 1.

37

Claim. For each positive integer k 2 all of the roots of Pk(x) and Pk1(x) are realand distinct. Furthermore, if a1 < a2 < < ak1 and b1 < b2 < < bk are roots ofPk1 and Pk, respectively, we have

b1 < a1 < b2 < a2 < < ak1 < bk:

Proof. We proceed by induction. For the base case k = 2, 13 is the only root of P1 andP2(

13 ) < 0 so

13 lies between the two roots of P2(x).Suppose that Pk1(x) and Pk(x) satisfy the induction hypothesis. We know Pk+1(x) =

(2x+ 1)Pk(x) x2Pk1(x). Now, we consider the signs of Pk and Pk1 on dierent realnumbers. First suppose that k is even.

x 1 b1 a1 b2 a2 bk1 ak1 bk +1Pk + 0 0 + 0 0 +

x2Pk1 0 + 0 0 + +

Since Pk+1(x) = (2x + 1)Pk(x) x2Pk1(x), for the sign of Pk+1 in respective bi's wehave

x 1 b1 a1 b2 a2 bk1 ak1 bk +1Pk+1 + + +

According to the change of signs of Pk+1(x) and by the Mean Value Theorem, foreach 1 i k 1, Pk+1 has a root between bi and bi+1. Also, it has one root less thanb1 and one root greater than bk, which completes the proof for even values of k. Thecase where k is an odd number is similar, the only dierence being the signs of b1 and 1.

Note that

(i) 0 is not a root of any of the Pk's. Because Pk(0) = a(0; 2k + 1) = 1, Pk1(x) andx2Pk1(x) have the same sign.

(ii) The coecient of xk in Pk(x) equals a(2k; 1) > 0, so Pk(+1) > 0 and the sign ofPk(1) is related to the parity of k.

4 . By considering centers of the hexagons, we get the following equivalent problem:

"Two vertices and of the triangular lattice are called equivalent if is equalto sum of nitely many vectors from the set:

A = f(n~im~j);(n~k +m~i);(n~j +m~k)g

Where

~i = (1; 0);~j = (12 ;p32 ) and

~k = (12 ;p32 ) according to the normal Cartesiancoordinates.

38

What is the maximum number of nonequivalent vertices?"

Note that

~k =~i+~j, thus we can determine vectors of set A using only ~i and ~j:

A = f(n~im~j);((n+m)~i+ n~j);(m~i+ (m+ n)~j)g:

Since the rst vector n~im~j is the dierence of the two others, two vertices and are equivalent i their dierence can be written as a linear combination of vectors(n+m)~i+ n~j and m~i+ (m+ n)~j with integer coecients.

Note that for integers a; b; x and y the space generated by the vectors fa~i+b~j; x~i+y~jg(using integer coecients) equals the space generated by vectors f(ax)~i+(by)~j; x~i+y~jg. Therefore, we can change former vectors to get simpler vectors to work with. Notethat in this change the value of ay bx (determinant of the matrix formed by vectors(a; b) and (x; y) as rows) is invariant, because ay bx = (a x)y (b y)x. Now,in a similar way to Euclidean Algorithm, if a x we change fa~i + b~j; x~i + y~jg byf(a x)~i + (b y)~j; x~i + y~jg. Repeating this process several times leads to vectors ofthe form s~i+ r~j and 0~i+ t~j. It means that we can suppose the coecient of ~i for one ofthem is zero. hence, if we want to go from one vertex to another, the dierence of the

coecients of

~i must be a multiple of s, since t~j does not aect the coecient of ~i. As aresult, for two equivalet vertices, we can rst use vector s~i + r~j several times to matchtheir rst coordinate. Then, we must use t~j to match the second coordinate. Hence wehave st nonequivalent vertices. But

st = det

s r0 t

which is invariant during the process.

det

s r0 t

= det

m+ n mn m+ n

= m2 +mn+ n2

So the maximum number of nonequivalent vertices is m2 +mn+ n2.

5 .

Lemma 1. Let A 2 (mod 3) be a positive integer. Then there exists a prime numbr psuch that p 2 (mod 3) and pjjA where is an odd integer.

Proof of lemma. Assume to the contrary that there is not such a prime number p.Therefore, if A = p11 p

22 pkk is the prime factorization of A, we have two cases for

pi's, 1 i k, to consider:

Case 1. pi 1 (mod 3) ) pii 1 (mod 3).

39

Case 2. pi 2 (mod 3) and 2ji ) pii (p2i )i2 1 (mod 3).

As a result, we must have A 1 (mod 3) which is a contradiction.

Lemma 2. Let p 2 (mod 3) be a prime number. Show that f03; 13; : : : ; (p 1)3g is acomplete residue system modulo p.

Proof of lemma. Obviously i3 03 (mod p) i i 0 (mod p). Suppose that p - i; j.Our goal is to show that i3 j3 (mod p) i i j (mod p). One part of the proof isobvious. To prove the other part, suppose that p = 3t + 2. Then by Fermat's LittleTheorem, we have i3t+1 j3t+1 1 (mod p). Hence, we have

i3ti i3t+1 j3t+1 (j3)tj i3tj (mod p):

Since (i; p) = 1, we get i j (mod p).

Now we are ready to solve the main problem. We claim that there is no such triple.

Assume to the contrary that

a2 + b2 + c2 = 2013k(ab+ bc+ ca)

for some positive integer k.First, without loss of generality we can suppose that a, b and c have no common factor,because if (a; b; c) = d > 1, we can divide them by d to get a new triple with no commonfactor. We have (a+ b+ c)2 = (2013k + 2)(ab+ bc+ ca). 2013k + 2 2 (mod 3), so bylemma 1 there is some prime number p 2 (mod 3) such that p2n+1jj2013k+2 (n 0).

p2n+1jj2013k + 2 ) p2n+1j(a+ b+ c)2 ) p2n+2j(a+ b+ c)2

) p2n+2j(2013k + 2)(ab+ bc+ ca)) pjab+ bc+ ca:As a result, pja+ b+ c and pjab+ bc+ ca. Hence

0 ab+ bc+ ca ab+ c(a+ b) ab+ c(c) (mod p) ) ab c2 (mod p)

) c3 abc (mod p):By a similar argument, a3 b3 abc (mod p), so by lemma 2 we deduce a b c(mod p) and since pja+b+c and 3 - p, we nd that p divides a, b and c, which contradictsour assumption that (a; b; c) = 1.

40

6 . Without loss of generality, we can suppose that !1 is shorter than !2. Let l0be the

common external tangent of !01 and !2 and let it intersects l at O. Let OA be tangentto !1 at A, B := !1 \!01, C := !01 \ l0, D := l0 \!2, E := !2 \!02 and let OF be tangentto !02 at F . We must prove that A, O and F are collinear.Let the common internal tangent of !1 and !

01 meet l at O

0, so

O0X:O0Y = O0B2 = O0X 0:O0Y 0:

Thus the power of O0 with respect to !2 and !02 is equal to this value. Hence the circleC1 with center O

0and radius O0B passes through E and is orthogonal to all arcs !1, !2,

!01 and !02. We also have OA2 = OX OY = OD2 and OF 2 = OX 0 OY 0 = OC2, Sothe circle C2 with center O and radius OA is orthogonal to both !1 and !2 and also thecircle C3 with center O and radius OF is orthogonal to both !

01 and !

02.

We use the following classic lemma without proving it.

Lemma 1. Let circle ! be orthogonal to circles C1 and C2 and intersect them in fA1; B1gand fA2; B2g, respectively. Let O;O1 and O2 be the centers of !;C1 and C2, respectively.Suppose that the right angles \OA1O1 and \OA2O2 have dierent orientations; i.e. oneof them is clockwise and the other one is not. Then the pairs fA;A0g and fB;B0g aredirect anti-homologous points and pairs fA;B0g and fA0; Bg are inverse anti-homologouspoints for C1 and C2.

Using the lemma for circle C2, we deduce that A and D are inverse anti-homologouspoints for !1 and !2 (we do not check the conditions here for brevity). Using the lemmaagain for circle C1, we get that B and E are also inverse anti-homologous points for !1and !2. So, the quadrilateral ADEB is cyclic.

Similarly, by using the lemma twice for circles C1 and C3, we get that pairs fB;Egand fC;Fg are inverse anti-homologous points for !01 and !02. So the quadrilateral BEFCis also cyclic.

According to the lemma for circle C1, points B and E are direct anti-homologouspoints for !01 and !2, and so are C and D. Hence, the quadrilateral BCDE is also cyclic.By the last three paragraphs, we deduce that all points A;B; : : : ; F lie on the cir-cumcircle of triangle BCD, which we call it C4 (checking distinctness of the points is notmentioned here). Now, circles C2; C3 and C4 have a line of symmetry, so their intersec-tions and centers also have a line of symmetry. It follows that O;A and F are collinearand the problem is solved.

Second Solution. To prove that A;B; : : : ; F lie on a circle, we can use the followinglemma.

Lemma 2. If a cricle ! is tangent to circles C1 and C2, then the points of tangency areanti-homologous points for C1 and C2. If none of C1 and C2 contains the other and ! istangent to them both internally or both externally, then the points of tangency are direct

anti-homologous points.

41

Note that these lemmas remain true if one of the circles is a line (considered as a

circle with innite radius).

If we suppose that interior of l0 (considered as a circle) is the half-plane not containingX, then according to lemma 2 for circle !01, points B and C are direct anti-homologousfor circles !1 and l

0. By applying lemma 1 for circle C2, points A and D are also anti-homologous points for !1 and l

0. Using the fact that the arcs are all in one side of l, itis easily seen that A and D are also direct anti-homologous points, so ABCD is a cyclicquadrilateral.

Similarly, by considering anti-homologous points for pairs of circles f!2; !01g, fl0; !02gwe conclude that quadrilaterals BCDE and CDEF are also cyclic, so points A, B, C,D, E and F lie on a circle which we call it C4.

The rest of the proof is the same as the rst solution.

Third Solution. Dene points A;B; : : : ; F and circles C1, C2 and C3 similar to the rstsolution. We can suppose that X is between X 0 and O, without loss of generality. Thus,C1 and C2 are disjoint and C3 contains both of them. According to lemma 1 and thefact that !1 lies entirely on one side of l, points A and B are direct anti-homologouspoints for C2 and C1. Similarly, B and C are inverse anti-homologous points for C1 andC3 and points C and D are direct anti-homologous points for C3 and C2. Pairs of pointsfD;Eg and fE;Fg are also anti-homologous in a similar manner as are pairs fA;Bg andfB;Cg, respectively.Our motivation is that two anti-homologous points are related to each other by an

inversion depending only on the two circles. So the problem says that the composition of

six particular inversions is the identity. With this intuition in mind, we move all points

to C1. Let A0, C 0, D0 and F 0 be points on C1 such that pairs fA;A0g and fD;D0g are

42

direct homologous points for circles C1 and C2, and pairs fC;C 0g and fF; F 0g are inversehomologous points for circles C1 and C3, respectively.

We have

O0D0 k OD k OC k O0C 0;so, A0D0 passes through O0. The lines A0B and D0E pass through the direct homotheticcenter of C1 and C2, which lies on l. The lines BC

0and EF 0 pass through the inversehomothetic center of C1 and C3, which also lies on l. So, by Pascal's theorem for thecyclic hexagon A0BC 0D0EF 0, we conclude that A0F 0 also passes through O0. Hence, wesimilarly have

OA k O0A0 k O0F 0 k OF;so O;A and F are collinear and we are done.

Comment 1. If we extend the arcs to perfect circles, then the other external tangent

of !1 and !02 and the other external tangent of !2 and !

01 also meet on l.

Comment 2. If we let !1 and !2 be in dierent sides of l, then the assertion is no longerright. Instead, the corresponding internal common tangents meet on l. Also an assertionlike that of the rst comment holds.

Comment 3. We can solve the problem using hyperbolic geometry. If we let the half-

plane be the Poincar model, then the arcs and the common tangents are equidistant

curves; i.e. the locus of points with a constant distance from a given line. This solution

gives interesting generalizations of the problem. For example, see the following gures.

43

7 . First Solution. We use the notation A for matrices and Aij for the entry of row iand column j.First, observe that Aii pi, since the sum of the ith row is pi and the entries arenonnegative. Similarly, Aii qi. Thus, Aii min(pi; qi) and it follows that trace(A)min(p1; q1) + min(p2; q2) + ::: + min(pn; qn). It remains to prove there is a matrix withthe property that Aii = min(pi; qi) for every 1 i n.We proceed by induction on n. The case n = 1 is obvious, since p1 = q1. Assumethe assertion holds for n = 1; 2; : : : ; k and consider the case n = k + 1. We can supposeq1 p1 without loss of generality. Let A11 = q1 and Ai1 = 0 for 2 i n. We claim thatthere exist nonnegative real numbers i := A1i for 2 i n such that these conditionshold:

nPi=2

i = p1 q1;i qi min(pi; qi) for every 2 i n: (1)To verify our claim, rst we prove a lemma.

Lemma. Let a; b1; b2; : : : ; bk be nonnegative real numbers such that a b1+b2+ +bk.Then there exist nonnegative real numbers a1; a2; : : : ; ak such that a1 + a2 + + ak = aand ai bi for every 1 i k.

44

Proof of lemma. We use induction on k. For k = 1 the lemma is obvious. If bk > a,we can set a1 = a2 = = ak1 = 0 and ak = a, so suppose that bk a. Let ak := bk.Now, we must have a1 + a2 + + ak1 = a bk. but a bk b1 + b2 + + bk bk =b1 + b2 + + bk1 and the assertion follows by the induction hypothesis.

We have

nXi=2

(qi min(pi; qi)) =nXi=1

(qi min(pi; qi)) =nXi=1

(pi min(pi; qi))

p1 min(p1; q1) = p1 q1:

Hence, by the lemma we can nd nonnegative real numbers 2; 3; : : : ; n such thatequation 1 holds. Let A1i = i for 2 i n. The remaining entries of A should satisfy

Ai2 +Ai3 + +Ain = pi for 2 i n,

A2i +A3i + +Ani = qi i for 1 i n.

We have qi i min(pi; qi) by equation 1, so min(pi; qi i) = min(pi; qi). Hence,the rest of the matrix can be lled using the induction hypothesis.

Second Solution. We can change the order of p1; p2; : : : ; pn and q1; q2; : : : ; qn arbitrarilyby changing corresponding rows and columns of the matrix. So we can suppose p1 q1; p2 q2; : : : ; pk qk and pk+1 qk+1; : : : ; pn qn. Let Aii = min(pi; qi) for 1 i n. Also let Aij = 0 if i 6= j and either i k or j > k. The remaining entriesfAij : i > k; j kg should satisty:

Ai1 +Ai2 + +Aik = pi qi for i > k,

A(k+1)i + +Ani = qi pi for i k.

i.e. we should nd a matrix with nonnegative entries for the given row sums and column

sums. We can construct such a matrix using induction on the number of rows as follows.

By the lemma above, we can nd An1; An2; : : : ; Ank such that their sum is pn qnand Ani qi pi for each i k. Then, we delete the n'th row and replace qi pi byqi pi Ani for each i k. We can proceed by using induction and the desired matrixwill be constructed.

Comment. This problem originally arises from probability theory. Suppose

Pi pi = 1and we want to have two random variables X;Y such that P(X = i) = pi and P(Y =i) = qi for 1 i n. The problem wants the maximum possible value of P(X = Y ) overall possible joint distributions of X and Y .

45

8 . We denote the problem's condition by . Let d be the common dierence and a+ dbe the initial term of this arithmetic proression, where a and d are integers such thata + nd > 0 for each n 2 N (Note that a may be negative but a + d; a + 2d; : : : are allpositvie). Hence, d is positive and d > a.

Putting k = N +m for nonnegative integer m in gives usa1a2 aN+mjaN+1aN+2 a2N+m;so by omitting equal terms:

a1a2 aN jak+1ak+2 ak+Nfor each k N (m 0). We denote this relation by . Let n0 > N be a natural numbercongruent to 1 modulo P = a1a2 aN . Then, by we have:

P j(a+ (n0 + 1)d)(a+ (n0 + 2)d) (a+ (n0 +N)d):Since P jn0 + 1, we get

P ja(a+ d) (a+ (N 1)d):Since P = (a+ d)(a+ 2d) (a+Nd), we get a+Ndja. Now, if a > 0, a+Nd a, soNd 0. Because N > 1 this implies d = 0, so an = a for all natural numbers n. Anyconstant sequence is an answer.

If a = 0, an = a + nd = nd, and this is another sequence satisfying the problem'scondition.

If a < 0, a + Nd jaj = a, so Nd 2a. Since N 2 and d > a we have2a < Nd 2a which is a contradiction.Finally, an = nd for d > 0 and constant sequences, are the only sequences satisfying .

9 . f is an increasing function so f(x) + x is strictly increasing and therefore injective.Let A = f(x) + 2g(x) + 3f(y) and B = f(y) + 2g(y) + 3f(x). We have

f(A) +A = 3(f(x) + f(y) + g(x) + g(y));

f(B) +B = 3(f(y) + f(x) + g(y) + g(x)):

Since f(A) +A = f(B) +B and f(x) + x is injective, we deduce that A = B:

f(x) + 2g(x) + 3f(y) = f(y) + 2g(y) + 3f(x):

This implies that f(x) g(x) = f(y) g(y) for all x; y > 0. Thus, f(x) g(x) equalssome contant value c. Putting x = y in the second equation gives g() = 3x+ c > c, so gtakes any value greater than c and consequently, f takes any value greater than 2c. Bysubstituting g(x) = f(x) c in the rst equation we get

f(3(f(x) + f(y)) 2c) = 3(f(x) + f(y)):

46

Therefore, f(x) = x 2c for large values of x. Thus, we can nd values of x and y suchthat f(x) = x 2c and f(y) = y 2c. Therefore, from the second equation we have

g(f(x) + y + g(y)) = g(x 2c+ y + y 3c) = x+ 2y 8c

2x g(x) + f(y) + y = 2x x+ 3c+ y 2c+ y = x+ 2y + c:Hence, c = 0 and f(x) = g(x) for all x 2 R+. Furthermore, we have f(x) = g(x) = xfor large values of x. Now, let y be suciently large such that f(y) = y and let x be anarbitrary positive real number in the second equation.

g(f(x) + y + g(y)) = g(f(x) + 2y) = f(x) + 2y

2x g(x) + f(y) + y = 2x x+ y + y = x+ 2y:Therefore, f(x) = x for all x 2 R+, so we get f(x) = g(x) = x for all x 2 R+ and it'seasy to check that this is indeed a solution.

10 . Denote by G the graph in the problem. We proceed by induction on the number ofedges of the complement graph G.The base case is when G is a complete graph.Let v1; v2; vi and vj be some four vertices of the complete graph G. By the problem'scondition for the 4-cycle v1vivjv2v1 we get

N(viv1)N(viv2) = N(vjv1)N(vjv2)

Where by N(vv0) we mean the number assigned to the edge vv0. Thus, for each i, thenumber N(viv1) N(viv2) is constant. We denote by K this constant value. Now, letni be the number we want to assign to the vertex vi. We must have

n1 n2 = N(v1vi)N(v2vi) = Kn1 + n2 = N(v1v2):

By solving this system of equations we get the numbers due to the vertices v1 and v2.Now, for each i let ni = N(viv1)n1. We claim that these numbers satisfy the assertionfor complete graphs. For this, we must show that for every i and j, ni + nj = N(vivj).For j = 2 this is equivalent to the second relation in the above system.

For other values of j, consider 4-cycle v1vivjv2v1. we have

N(v1v2) +N(vivj) = N(v1vi) +N(v2vj):

Thus,

n1 + n2 +N(vivj) = n1 + ni + nj + n2 ) N(vivj) = ni + nj :This nishes the proof of the base case.

47

Suppose that G does not contain edge vivj .We claim that we can add edge vivj and assign a number to it such that the assignednumber satises the problem's condition for G [ fvivjg and therefore, according to theinduction hypothesis the statement holds for G [ fvivjg.If the edge vivj is not included in any even walk, we can choose an arbitrary numberfor N(vivj), and if the edge vivj is included in some even walk, the assigned number isdetermined uniquely.

Furthermore, if the edge vivj is included in two even walks C1 and C2, C1 [C2 fvivjgis also an even walk which satises the problem's condition. Therefore, the assigned

numbers to the edge vivj by C1 and C2 are equal.

11 . Let a = x2, b = y2 and c = z2 for a y z 0. We havex2 y2 + z2 (y + z)2 ) x y + z;so x, y and z are sides of a triangle.Set Ax =

py2 + z2 yz, Ay =

pz2 + x2 zx and Az =

px2 + y2 xy. Then,

Ax Ay , y2 + z2 yz x2 + z2 xz , (x y)(x+ y z) 0Ay Az , x2 + z2 xz x2 + y2 xy , (y z)(y + z x) 0:Hence, Ax Ay Az.Therefore,

xAz + yAy yAz + xAy , (Az Ay)(x y) 0yAy + zAx zAy + yAx , (Ay Ax)(y z) 0:For the main problem we have

2(xAz + yAy + zAx) = (xAz + yAy) + (yAy + zAx) + xAz + zAx (yAz + xAy) + (zAy + yAx) + xAz + zAx = (y + z)Ax + (x+ z)Ay + (x+ y)Az:

By Cauchy-Schwarz Inequality we have

(x+ y)Az = (x+ y)px2 + y2 xy =

p(x+ y)(x3 + y3) x2 + y2;and similar inequalities hold for (x+ z)Ay and (y + z)Ax. Finally, summing these threeinequlities implies the assersion.

12 .

Lemma. Let ABD be a triangle with circumcircle !. Suppose that !0 is a circle tangentto sides AB and AD at E and F , respectively, and tangent to ! at T . M and N are

midpoints of shorter arcs

_AB and

_AD, respectively. Then tangents to ! at A and T , andline MN are concurrent.

48

Proof of lemma.

\NAF = 12

_ND =

1

2

_NA = \ATN;

so triangles TAN and AFN are similar, so NFNA =NANT , and consequently NA

2 = NF:NT .In a similar way we can show that MA2 = ME:MT . Thus,

MA

AN

2=ME:MT

NF:NT: ()

T is the external homethetic center of ! and !0. This homothety maps E and F to Mand N , respectively, so EF jjMN and by Thales' Theorem MEMF = MTNT . This and ()imply

MA

NA

2=ME

NF:MT

NT=

MT

NT

2) MA

NA=MT

NT:

Therefore,M is the harmonic conjugate ofN with respect to A and T , and hence tangentsto ! at A and T meet on the line MN .

We keep using the notations introduced in the lemma. We showed that K lies on theline MN and now we want to show that K lies on the line I1I2. We have

KM

KN=AM: sin(\MAK)AN: sin(\NAK) =

AM

AN:sin(\MCA)sin(\NCA) =

49

MI2NI1

:sin(\MCA)

r2:

r1sin(\NCA) =

MI2NI1

:CI1CI2

:

As a result,

MKKN :

NI1I1C

: CI2I2M = 1 and by Menelaus' Theorem, K, I1 and I2 are collinear.

13 . Let A1, B1 and C1 be reections of P with re-spect to BC, AC and AB, respectively. Moreover,we denote by A2 and A

0the midpoints of B1C1and PA1, respectively. Points B2, C2, B

0and C 0

are dened similarly. Note that A2C0 k PB1 and

A2B0 k PC1, so A2C 0 ? AC and A2B0 ? AB.Therefore, A2 is the orthocenter of triangle AB

0C 0.

\AB0C 0 \AB0P = 90:Similarly, \AC 0B0 90. Furthermore, \B0AC 0 = \BAC < 90 because triangle

ABC is acute. Therefore, triangle AB0C 0 is acute, so its orthocenter A2 lies inside itand therefore, inside triangle ABC. By similar arguments points B2 and C2 lie insidetrianlgle ABC. Hence, the centroid of triangle A2B2C2 lies inside triangle ABC, but thecentroid of triangle A2B2C2 coincides with that of triangle A1B1C1, and this completesthe proof.

14 . First suppose that all rectangles have sides parallel to coordinate axis. There arefour types of right angles: p, q, x and y.Suppose that there are a number of angles p and b number of angles y. Since eachrectangle is determined by these two angles, we get ab n, and by AM GM inequalitywe get

a+b2

2 ab n, so a + b 2pn. Similarly, if there are c number of angles qand d number of angles x, we can deduce that c+ d 2pn. Thus we have at least 4pnright angles.

If there are dierent directions for rectangles, we denote by n1; n2; : : : and nk the numberof rectangles in each direction. Since n =

Pi ni, we get

4pn 4pn1 + 4pn2 + + 4pnk Number of angles:

15 . a) It suces to nd a sequence an such that a1 - an for all natural numbers n > 1 andijj if and only if aijaj for all natural numbers i; j 2. Since (2m1; 2n1) = 2(m;n)1,the sequence dened by a1 = 2 and an = 2

n 1 for n > 1 has this property.

50

b) We prove a stronger assertion.

For each function f : N ! N such that f(n) n for n > M , where M;n 2 N, thereexists a sequence a1 < a2 < a3 < such that an is divisible by exactly f(n) terms ofthe sequence for suciently large positive integers n.Let a1; a2; : : : ; aM be a strictly increasing sequence of prime numbers. Then, for k > M ,set ak = [a1; a2; : : : ; af(k)1]qk where qk is a new prime number specically for ak. Obvi-ously, we can choose qk large enough so that ak became greater than ak1.By this construction ak is divisible by a1; a2; : : : ; af(k)1 and itself which are f(k) num-bers. Furthermore, ak is not divisible by aj for some other j, because each aj has itsspecic prime number qj which has not appeared in the denition of ak.

16 . Let A = ff(x) f(y)jx; y 2 Zg. By the problem's condition we know 3 2 A, andour goal is to show that 1 2 A.Obviously, A is closed under multiplication by 1 ((f(x) f(y)) = f(y) f(x)). Weclaim that A is closed under summation.Suppose that ; 2 A, so there are some integers m1, m2, n1 and n2 such that =f(m1) f(m2) and = f(n1) f(n2). We have

f(m1) + f(n1) + f(f(m21 + n

21)) = 1

f(m2) + f(n2) + f(f(m22 + n

22)) = 1:

Subtracting these two equalities gives

+ = f(f(m22 + n22)) f(f(m21 + n21)) 2 A

It is a well-known fact that any subset of Z closed under summation and multiplicationby 1 is kZ for some positive integer k. Since 3 2 A, kj3, so k = 1 or k = 3. If k = 3,then for each x; y 2 Z, f(x)f(y) is divisible by 3 and thus all numbers in the form f(x)have the same remainder modulo 3. We denote this common remainder by r. Hence,

1 f(m) + f(n) + f(f(m2 + n2)) r + r + r 3r 0 (mod 3);

which is a contradiction. Therefore, k = 1, so A = Z. Thus, 1 2 A, so there exist integersc and d such that f(c) f(d) = 1.

17 . LetM , N and P be the midpoints of segments AB, AC and AD, respectively. SinceM and N are the centers of circles with diameters AB and AC, respectively, and AD isthe bisector of \BAC, we have

MP

NP=BD

CD=AB

AC=AM

AN=MX

NY:

Hence, \PXM +\PY N = 180, or \PXM = \PY N . However, \PXM < \BXA =90, and \PY N < \AY C = 90; therefore, the rst case cannot hold, so \PXM =

51

\PY N (1). Furthermore, \MPX = \NPY (2). By (1) and (2) we deduce thattriangles PXM and PY N are similar, So

\PMX = \PNY ) 180 \B + 180 2\ABX = \C + 2\ACY

) \ABX + \ACY = 180 \B \C + 90 \ABX + 90 \ACY) \ABX + \ACY = \BAC + \XAB + \CAY (3):Since \AHB = \AHC = 90, H lies on the circumcircles of triangles AXB and AY C,so \ABX = \AHX (4) and \ACY = \AHY (5). Thus,

\XHY = \XHA+ \AHY = \ABX + \ACY = \Y AX By 3; 4; 5:

Since XY is the perpendicular bisector of segment AD, we get \XAY = \XDY .Therefore, \Y HX = \Y DX, so the quadrilateral XHDY is cyclic.

18 . Suppose that the room in the problem is the square [0; n] [0; n] in the Cartesianplane. Furthermore, throughout the solution we use these denitions:

A lattice point: A point in the plane with integer coordinates. A lattice square: A square with lattice points as vertices and side lengths one. A lattice edge: A side of a lattice square. A lattice triangle: A right isosceles triangle of side lenght 1 and lattice points asvertices.

52

A small triangle: A right isosceles triangle of side lenghtp22 and a lattice edge as

its hypotenuse.

The boundary of each tile contains two lattice edges and two diameters of two lattice

squares, so we have two types of tiles. By these denitions the following lemma is obvious.

Lemma. Suppose that e is a lattice edge of the lattice square S. Furthermore, assumethat e is not included in the boundary of any tile that lies in the same side of e as S. Bythese conditions, there exist a lattice triangle and a small triangle containing e which iscontained in S that are not covered by any tiles. Consider a graph with lattice edges as it's vertices. Two vertices are connected if and

only if their corresponding edges are on the boundary of the same tile. In this graph the

degree of each vertex is at most two. Furthermore, two connected vertices correspond to

parallel edges, and if the degree of a vertex corresponding to a horizontal (vertical) edge

is 2, one of the edges connected to this edge in the graph is above (right of) it and the

other lies below (left of) it. Hence, each connected component of the graph is a path

containing parallel lattice edges, and the order of vertices in this path coincides with

the order of one of the coordinates of the corresponding edges (x-coordinate for verticaledges and y-coordinate for horizontal edges).Thus, lattice edges on the line y = n lie on dierent paths. We denote these paths byU1; U2; : : : ; Un. Note that some of Ui's might contain only one vertex. The bottommostlattice edges of these paths are dierent. Let SUi be the lattice square below the bot-tommost edge of Ui. If SUi lies above the line y = 0, according to the lemma one of thelattice triangles in SUi is not covered. We denote this triangle by TUi. Furthermore,the upper small triangle in SUi is not covered. We denote this small triangle by QUi. Inthe same manner, we denote by D1; D2; : : : and Dn the paths containing lattice edges onthe line y = 0, by L1; L2; : : : and Ln the paths containing the edges on the line x = 0,and nally by R1; R2; : : : and Rn the paths containing lattice edges on the line x = n.Triangles TDi, TLi, TRi, QDi, QLi and QRi are dened similarly. We have three cases:

Case1. Suppose that there exists a path like P containing one lattice edge of the linesx = 0 and x = n. In this case all the tiles of paths U1; U2; : : : and Un are abovethe tiles of the path P and tiles of the paths D1; D2; : : : and Dn lie below thetiles of the path P . Now, 2n paths D1; D2; : : : and Dn, and U1; U2; : : : and Unare all distinct, so the triangles TU1; TU2; : : : and TUn, and TD1; TD2; : : : andTDn can be dened and are disjoint. Since these triangles are not covered, andthe area of each of them is

12 , the sum of the area of these triangles is n. Hence,the assertion is proved in this case.

Case2. There exists a path P containing one lattice edge of the lines y = 0 and y = n.This case is similar to case 1.

53

An example for case 2.

Case3. There are not any paths of cases 1 or 2, so small triangles QDi, QUi, QLi andQRi, for 1 i n, are all dened and are disjoint. Since these 4n triangles arenot covered by any tiles and the area of each is

14 , the sum of the areas of them

is n, so the proof is complete.

An example for case 3.

54

The Holy Month of Ramadan

and

Muslim Mathematicians

This year the IMO and the month of Ramadan are coincident.The holy month of

Ramadan, which is called as the month of mercy and forgiveness by Allah, is the ninth

month of the lunar months. According to the Islamic regulations, Muslims should fast

every day of this month from sunrise to sunset and refrain from eating and drinking,

except for those who are ill or traveling.

O, believers! Prescribed to you is the fast, as it was prescribed to those

before you, that you may guard against evil. (2-183)

For a certain number of days (you have to fast) but if any one of you be

sick, or on a journey, then (he shall fast a like) number of other days,

and those who are not able to do it, (for being too old, or permanently

sick) can be redeemed by feeding a poor. He that does good of his own

accord, it is better for him. And it is better for you to fast, if you could

knew it. (2-184)

The base of the lunar months is the rotation of the moon around the Earth. Each

lunar month begins when the new moon is sighted. Every twelve lunar months forms

a lunar year, which is almost equal to 354 or 355 days. Islamic regulations like the

law of fasting and also the important Islamic ceremonies like the Hadj and the religious

fetes are based on the lunar months, which explains why determining the lunar month

duration has been so important among the Muslims. The need for sighting the moon

crescent along with the need for orientation specially for determining the direction of

Kiblah (which is the Ka'ba in the city of Mecca) has caused the progress of Astronomy

in the Islamic countries.

"The prediction of the visibility of the moon crescent has been one ofthe most serious problems Muslims have faced, because the moon in its

rotation around the Earth is located near the sun from the view point

of an observer on Earth, at the beginning and the ending of the lunar

month. Therefore the crescent sighting is only possible in a short period

of time near sunset. Important factors in sighting the crescent are the

location of the moon and the sun relative to the horizon of the observer,

the pause of the moon after the sunset, the location of the moon in

its orbit relative to the Earth, the brightness of the crescent and the

weather. Muslim astronomers knew that determining the probability of

sighting the crescent requires solving complicated spherical astronomy

problems, so they could obtain the relative positions of the sun and the

moon on the horizon of the observer. The method used was in a way

with which the observer, using his calculations, knew the location whe-

55

re he was meant to search for the thin and dim crescent; therefore

observers tried to sight the moon crescent in the sunset of the 29th dayof the lunarmonth in a serene and an unhindered horizon, and they

would repeat their sightings the night after in case their sightings were

unsuccessful. One of the simplest criteria in sighting the crescent has

been brought up by Kharazmi (Mohammad ibn Musa al-Khwarizm),

the Iranian mathematician and astronomer. The basis of this criterion

was the calculation of the time dierence between the moonset and the

sunset at the local horizon. If this dierence reaches 12 degrees (equal

to 48 minutes of time) the moon crescent will be visible. As time passed,

Muslim astronomers obtained more complicated and precise criteria for

the prediction of the visibility of the crescent."Translation Resource: Astronomy in Iran and the World of Islam(In

Persian), Ali Akbar Velayati, pp 237-241.

56

The Holy Month of Ramadan and Muslim

Mathematicians The holy month of Ramadan, which is called as the month of mercy and

forgiveness by Allah, is the ninth month of the lunar months. According to the Islamic regulations, Muslims should fast every day of this month from sunrise to sunset and refrain from eating and drinking, except for those who are ill or traveling.

O, believers! Prescribed to you is the fast, as it was prescribed to those before you, that you may guard against evil. (2-183) for a certain number of days (you have to fast) but if any one of you be sick,

or on a journey, then (he shall fast a like) number of other days, and those who are not able to do it, (for being too old, or permanently sick) can be redeemed by feeding a poor. He that does good of his own accord, it is better for him. And it is better for you to fast, if you could knew it. (2-184)

The base of the lunar months is the rotation of the moon around the

Earth. Each lunar month begins when the new moon is sighted. Every twelve lunar months forms a lunar year, which is almost equal to 354 or 355 days. Islamic regulations like the law of fasting and also the important Islamic ceremonies like the Hadj and the religious fetes are based on the lunar months, which explains why determining the lunar month duration has been so important among the Muslims. The need for sighting the moon crescent along with the need for orientation specially for determining the direction of Kiblah (which is the Kaba in the city of Makah) has caused the progress of Astronomy in the Islamic countries.

Read more about the role of Muslim astronomers and mathematicians in finding and using the lunar months inside the booklet on pages 55 and 56.