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Serbian Mathematical Olympiad 2010

for high school students

Nis, April 67, 2010

Problems and Solutions

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Cover photo: Nis Fortress

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SHORT HISTORY AND SYSTEM

Mathematical competitions in Serbia have been held since 1958. In the first

years only republic competitions within the former Yugoslavia, which Serbia was apart of, were held. The first Federal Mathematical Competition in Yugoslavia washeld in Belgrade in 1960, and since then it was held regularly every year, skippingonly 1999 for a non-mathematical reason. The system has undergone relatively fewchanges. The earliest Federal Competitions were organized for 3rd and 4th gradesof high school only; 2nd grade was added in 1970, and 1st grade in 1974. Since1982, 3rd and 4th grades compose a single category. After the breakdown of theold Yugoslavia in 1991, the entire system was continued in the newly formed FRYugoslavia, later renamed Serbia and Montenegro. The separation of Montenegrofinally made the federal competition senseless as such. Thus, starting with 2007,the federal competition and team selection exam are replaced by a two-day Serbian

Mathematical Olympiad.

Today a mathematical competition season in Serbia consists of four rounds ofincreasing difficulty:

Municipal round, held in early February. The contest consists of 5 problems,each 20 points worth, to be solved in 3 hours. Students who perform wellqualify for the next round (50-60 points are usually enough).

Regional round, held in late February in the same format as the municipalround. Each students score is added to that from the municipal round.Although the number of students qualified for the state round is bounded by

regional quotas, a total score of 110-120 should suffice.

State (republic) round, held in late March in a selected town in the country.There are roughly 200 to 300 participants. The contest consists of 5 problemsin 4 hours.

Serbian Mathematical Olympiad (SMO), held in early April in a selectedplace in the country. The participants are selected through the state round:26 from A category (distribution among grades: 3+5+8+10), 3 from B cate-gory (0+0+1+2), plus those members of the last years olympic team who didnot manage to qualify otherwise. Six most successful contestants are invited

to the olympic team.Since 1998, contests for each grade on the three preliminary rounds are divided

into categories A (specialized schools and classes) and B (others). A student fromcategory B is normally allowed to work the problems for category A instead. Onthe SMO, all participants work on the same problems.

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The Serbian Mathematical Olympiad 2010 for high school students took placein Nis on April 67. There were 29 students from Serbia and 9 guest studentsfrom Republic of Srpska (Bosnia and Herzegovina), the specialized school Kol-mogorov in Moscow, and the host city of Nis taking part on the competition. The

average score on the contest was 12.87 points and all problems were fully solvedby the contestants, except problem 6 on which the maximum achieved score was2. Based on the results of the competition the team of Serbia for the 27-th BalkanMathematical Olympiad and the 51-st International Mathematical Olympiad wasselected:

Teodor von Burg Math High School, Belgrade 36 pointsLuka Milicevic Math High School, Belgrade 35 pointsRade Spegar Math High School, Belgrade 32 pointsMihajlo Cekic Math High School, Belgrade 30 pointsStevan Gajovic Math High School, Belgrade 29 points

Dusan Milijancevic Math High School, Belgrade 26 points

In this booklet we present the problems and full solutions of the Serbian Math-ematical Olympiad and the Balkan Mathematical Olympiad.

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SERBIAN MATHEMATICAL OLYMPIAD

for high school students

Nis , 06.04.2010.

First Day

Some of n towns are connected by two-way airlines. There are m airlines in total.1.For i = 1, 2, . . . , n, let di be the number of airlines going from town i. If 1 di 2010 for each i = 1, 2, . . . , 2010, prove that

n

i=1

d2

i 4022m 2010n.

Find all n for which equality can be attained. (Aleksandar Ilic)

In an acute-angled triangle ABC, M is the midpoint of side BC, and D, E and F2.the feet of the altitudes from A, B and C, respectively. Let H be the orthocenterof ABC, S the midpoint of AH, and G the intersection of F E and AH. If Nis the intersection of the median AM and the circumcircle of BC H, prove thatHM A = GNS. (Marko Djikic)

Let A be an infinite set of positive integers. Find all natural numbers n such that3.for each a A

an + an1 + + a1 + 1 | an! + a(n1)! + + a1! + 1.

(Milos Milosavljevic)

Time allowed: 270 minutes.Each problem is worth 7 points.

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SERBIAN MATHEMATICAL OLYMPIAD

for high school students

Nis, 07.04.2010.

Second Day

4. Let O be the circumcenter of triangle ABC. A line through O intersects the sidesCA and CB at points D and E respectively, and meets the circumcircle again at

point P = O inside the triangle. A point Q on side AB is such that AQQB

=DP

P E.

Prove that AP Q = 2CAP . (Dusan Djukic)

5. An n n table whose cells are numerated with numbers 1, 2, . . . , n2 in some orderis called Naissus if all products of n numbers written in n scattered cells give thesame residue when divided by n2 + 1. Does there exist a Naissus table for

(a) n = 8;

(b) n = 10?

(n cells are scattered if no two are in the same row or column.) (Marko Djikic)

6. Let a0 and an be different divisors of a natural number m, and a0, a1, a2, . . . , anbe a sequence of natural numbers such that it satisfies

ai+1 = |ai ai1| for 0 < i < n.

If gcd(a0, . . . , an) = 1, show that there is a term of the sequence that is smallerthan

m . (Dusan Djukic)

Time allowed: 270 minutes.Each problem is worth 7 points.

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SOLUTIONS

By the problem conditions, 0

(di

1)(2010

di) holds for each i, i.e. d

2i

1.

2011di 2010. Since ni=1 di = 2m, summing up these inequalities givesn

i=1

d2i 2011 n

i=1

di 2010n = 4022m 2010n,

as desired.

The equality holds if and only if di {1, 2010} for each i {1, 2, . . . , n}.1 Let n = 2k for some k N. Setting an airline between towns i and i + k for

i = 1, . . . , k and no other airlines yields a configuration with di = 1 for all i.

2 Let n = 2k 1 for some k N. We cannot have di = 1 for all i because thesum of all di must be even, so we must have dj = 2010 for some j; hencen 2011. On the other hand, setting an airline between towns 2i and 2i + 1for i = 1006, . . . , k and between 1 and i for 1 i 2010 yields a configurationwith d1 = 2010 and di = 1 for i = 2, . . . , n.

Therefore the equality can be attained if and only if 2 | n, or 2 n and n 2011.

2. Let the line AN meet the circle BH Cagain at point A. Then ABAC is aparallelogram and HCA = HCB +

BC A

= HCB + ABC = 90

, sothe points A , B, N all lie on the cir-cle with diameter HA and thereforeANH = 90.

Since S is the circumcenter of AEF,it follows that SF G = 90 EAF =ACF = ADF; hence the trianglesSF G and SDF are similar and SG SD = SF2 = SN2. This in turnshows that SN G SDN and fi-nally GNS = GDN = HM Nsince the quadrilateral HDMNis cyclic.

A

B CMD

E

F HN

S

A

G

Second solution. Quadrilaterals BDHF and DCEH are cyclic and AF AB =AH AD = AE AC. We apply the inversion Iwith center A and power AF AB.Clearly I(F) = B,I(H) = D,I(E) = C, so Imaps line BC to the circumcircle of

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F HE, i.e. the circle with diameter AH; also, Imaps the circumcircle of BC Hto the circumcircle of F DE which is the nine-point circle of ABC. SinceM AM and Ipreserves line AM, it follows that I(M) = N.Let

I(G) = G and

I(S) = S. Since

I(EF) is the circumcircle of

ABC, S

and I(AH) = AH, points G and S are the second intersection points ofAH withthe circumcircles ofABC and HBC respectively. Thus GNS = GM S =HM A, for G, S are the mirrors of H, A in BC.

3. Consider polynomials P(x) = xn + xn1 + + 1 and Q(x) = xn! + + x1! + 1,and let Q(x) = C(x)P(x) + R(x), where C and R are polynomials with integercoefficients and deg R < deg P. The problem condition claims that P(a) dividesQ(a) and hence also divides R(a) for infinitely many a, but for sufficiently large awe have |R(a)| < |P(a)|, implying R(a) = 0; thus R(x) has infinitely many zeros,so R(x) 0 and P(x) | Q(x).

Lemma. Let 1, 2, . . . , n N. Polynomial P(x) divides Q(x) = xn

+xn1 + + 1 if and only if{0, 1, 2, . . . , n} is a complete residue systemmodulo n + 1.

Proof. Let ri be the residue of i modulo n + 1. Suppose that P(x) | Q(x).Since P(x) divides xn+1 1 which divides xi xri , it also divides Q1(x) =xr1 + + xrn + 1. Since deg Q1 deg P, it follows that Q1 = c P for someconstant c, and this is obviously only possible if c = 1 and {r1, . . . , rn} ={1, . . . , n}. The other direction is trivial.

We conclude from the lemma that the desired numbers n are those for which

{0, 1!, . . . , n!

}is a complete residue system modulo n + 1.

Ifn > 3 and n +1 is a composite number, then n + 1 | n!, so this case is impossible.If n + 1 = p > 3 is a prime, then by Wilsons theorem (p 1)! 1 (mod p), so(p 2)! 1 = 1! (mod p), again impossible. The only remaining possibilities aren = 1 and n = 2, and it is directly verified that these values satisfiy the conditionsof the problem.

Second solution. We shall prove a stronger statement: If A = an + + a + 1divides an! + + a1! + 1 for some integer a > 1, then n {1, 2}.Suppose that A divides a1 + + an + 1 for some positive integers 1, . . . , n,and let ri be the remainder of i upon division by n + 1. Like in the first solution,

A divides C = ar1 + + arn + 1 = cnan + + c1a + c0, where c0, . . . , cn arenonnegative integers with the sum n + 1.

Of all numbers of the form B = bnan+ +b1a+b0 that are divisible by A (bi N0),

consider any one with the minimum possible value of the sum b0 + + bn. Eachbi is less than a, otherwise substituting (bi, bi+1) with (bi n, bi+1 + 1) yields a

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representation of B with a smaller sum b0 + + bn (where bn+1 = b0). ThereforeB < a A, which implies B = kA for some k {1, . . . , n}, so bi = k by uniquenessof base a representation. It follows that the minimum possible value of b0 + + bnequals n + 1 and is attained only when bi = 1 for all i.

Since c0 + + cn = n + 1, it follows from above that ci = 1 for all i. Thus{r1, . . . , rn} = {1, . . . , n} and we can proceed like in the first solution to obtainn {1, 2}.

4. The given configuration is only possible ifABC is acute-angled. Denote P AD =, QP A = , BC A = . Then AP B = 2 and DAP +EBP = AP B ACB = , so P BE = and BP Q = 2 . Since AP D = BP E =90 , we also have ADP = 90 + and BE P = 90 + .The sine theorem in triangles AP D and P BE gives us

DP

P E =

DP

P A P A

P B P B

P E =

sin cos

sin( )cos( ) P A

P B =

sin2

sin(2 2) P A

P B .

On the other hand,AQ

QB=

AQ

AP AP

BP

BP

QB=

sin

sin(2 ) AP

P B.

NowAQ

QB=

DP

P Eimplies f(2) = f(),

where f(x) =sin x

sin(2 x) . However, fis strictly increasing on (0, 2) because

f(x) = sin(2 2x)(sin(2 x))2 > 0, so it imme-

diately follows that = 2. A B

C

O

D

E

P

Q

2

5. Suppose that there exists a Naissus table 8 8 and that the product of any 8scattered numbers gives a remainder r modulo 82 + 1 = 65 = 5 13. Since thesenumbers can be chosen so as to contain a multiple of 13, r is divisible by 13. Onthe other hand, the table can be partitioned into 8 pairwise disjoint groups of 8scattered cells each, and at least one of these groups does not contain any multipleof 13, so the product over this group of scattered cells is not divisible by 13, a

contradiction. Hence there is no Naissus table for n = 8.

For n = 10, n2 + 1 = 101 is a prime, and there is a primitive root g modulo 101.Then the table (ai,j)0i,j9 such that ai,j g10i+j (mod 101) contains each num-ber from {1, . . . , 100} exactly once. Now let a0,p(0), . . . , a9,p(9) be scattered num-bers; note that p(0), . . . , p(9) is a permutation of 0, . . . , 9, so p(0)+ +p(9) = 45.

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The product of numbers ai,p(i) is congruent to gp(0)g10+p(1) g90+p(9) = g450+45 =

g495 no matter which scattered cells are chosen, so this is an example of a Naissustable.

6.Let p and q be two smallest (different) terms of the sequence. Assume that ak = pand al = q (k < l) and that (without loss of generality) am / {p, q} for k < m < l.If l > k + 2 then ak+3 = ak+2 + ak+1 (otherwise ak+3 = |ak+2 ak+1| = p) andsimilarly al3 = al2 + al1. Now if am is the maximal term between ak and al,we have k + 2 < m < l 2, so am+2 = am am+1 = am1 and am+1 = am am1 = am2. Thus the sequence obtained by excluding terms am, am+1, am+2 willstill satisfy the imposed conditions. Substituting 5-tuples (u,v,u + v,u,v) in thesequence with pairs (u, v) can be repeated until there is at most one term remainingbetween p and q. However, if we go backward now, we will have to insert either por q in between. This leads to the concludion that in the original sequence therecould be at most one term between ak = p and al = q; hence either l = k + 1 or

l = k + 2 (in the latter case, ak+1 = p + q).Denote vk = (1, 0), vl = (0, 1) and, for each i, vi+2 = ivi +

ivi+1 if ai+2 = iai +

iai+1 (i, i {1, 1}). A simple induction shows that if vi = (xi, yi) then ai =

xip+yiak+1. Since xi+1yi+2xi+2yi+1 = xi+1(iyi+iyi+1)(ixi+ixi+1)yi+1 =i(xiyi+1 xi+1yi), it follows that for each i we have xiyi+1 xi+1yi {1, 1},so (xi, yi) = 1. Suppose that xi < 0 or yi < 0 for some i > k (at least one ofxi, yi must be positive because so is ai), and let j be the smallest such i. Sincexjyj1 xj1yj {1, 1} (i xjyj < 0), it follows that vj1 {(0, 1), (1, 0)},so keeping in mind that either vj = vj1 vj2 or vj = vj2 vj1 we candeduce that vj2 {(1, 0), (0, 1)}. (Indeed, xj2, yj2 0, so if we assume e.g.vj1 = (0, 1), relation xj2yj1

xj1yj2

{1, 1

}will imply xj2 = 1, which

together with vj = (vj1 vj2) and xjyj < 0 implies yj2 = 0.) It follows thatvj {(1, 1), (1, 1)} and aj = |p q| < max{p, q}, contradicting the choice of pand q.

This shows that, if p, q are two smallest terms of the sequence, then each term isof the form ai = xip + yiq with xi, yi N and (xi, yi) = 1. Note that p and qmust be coprime. Now let m = da0 = ean. There exist x0, y0, xn, yn N such that(x0, y0) = (xn, yn) = 1 and a0 = x0p + y0q, an = xnp + ynq, so m = dx0p + dy0q =exnp + eynq and (dx0, dy0) = (exn, eyn). Consequently, p | dy0 eyn, so dy0 > por eyn > p, which finally implies m > pq and therefore min(p, q) qi1(q 1) + 1 > 23 (2qi + 2). The other three casesare similarly shown to be impossible.

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Mathematical Competitions in Serbiahttp://srb.imomath.com/

The IMO Compendium Olympiad Archive Mathematical Society of Serbiahttp://www.imocompendium.com/ http://www.dms.org.rs/

The IMO Compendium

This book attempts to gather all the problems appearing on the IMO, as wellas the so-called short-lists from 35 years, a total of 864 problems, all of whichare solved, often in more than one way. The book also contains 1036 problemsfrom various long-lists over the years, for a grand total of 1900 problems.In short, The IMO Compendium is an invaluable resource, not only for high-school students preparing for mathematical competitions, but for anyone wholoves and appreciates math.

Publisher: Springer (2006); Hardcover, 746 pages; Language: English; ISBN: 0387242996

For information on how to order, visit http://www.imocompendium.com/