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XXXIIBrazilian Math Olympiad

2010

Editora AOBM

Rio de Janeiro

2011

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Instituto Nacional de Matematica Pura e Aplicada – IMPA

Chair: Cesar Camacho

Sociedade Brasileira de Matematica (Brazilian Mathematical Society)

Chair: Hilario Alencar

Support

Conselho Nacional de Desenvolvimento Cientıfico e Tecnologico – CNPq

Instituto do Milenio Avanco Global e Integrado da Matematica Brasileira

Comissao Nacional de Olimpıadas de Matematica (Mathematical Olympiads National Com-

mittee)

Estrada Dona Castorina, 110 – Jardim Botanico – 22460-320 Rio de Janeiro – RJ

Telefone: (21) 2529-5077 Fax: (21) 2529-5023

web: http://www.obm.org.br

e-mail: obm@impa.br

Chair: Luzinalva Miranda de Amorim, Carlos Yuzo Shine

Members: Antonio Caminha Muniz Neto, Carlos Gustavo Moreira, Edmilson Luis Rodrigues Motta,

Eduardo Wagner, Eduardo Tengan, Elio Mega, Florencio Ferreira Guimaraes Filho, Luciano Guima-

raes Monteiro de Castro, Nicolau Corcao Saldanha, Onofre Campos, Pablo Rodrigo Ganassim, Paulo

Cezar Pinto Carvalho, Ronaldo Alves Garcia, Ralph Costa Teixeira, Yoshiharu Kohayakawa

Junior Members: Alex Correa Abreu, Bernardo Paulo Freitas da Costa, Bruno Holanda, Carlos Au-

gusto David Ribeiro, Carlos Stein Naves de Brito, Cıcero Thiago Magalhaes, Davi Maximo Alexan-

drino Nogueira, Einstein do Nascimento Junior, Emanuel Carneiro, Fabio Dias Moreira, Fabrıcio

Siqueira Benevides, Gabriel Tavares Bujokas, Humberto Naves, Larissa Cavalcante Lima, Marcio

Assad Cohen, Samuel Barbosa Feitosa, Telmo Correa Junior, Tertuliano Franco, Thiago Barros Ro-

drigues Costa, Rodrigo Villard, Yuri Gomes Lima

Executive Secretary: Nelly Carvajal Florez.

Assistant Secretaries: Sonia de Souza Silva de Melo

Typeset with Plain TEX.

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Introduction

1.1. Structure of the Brazilian Math Olympiad

The Brazilian Math Olympiad is a nationwide competition for studentsfrom grade 6 to undergraduates, comprising a total of approximately 400000contestants. Students from grade 6 to 12 have to take three rounds: thefirst round is held in June and consists in multiple choice questions, 20for grades 6 and 7 and 25 for grades 8 to 12. Approximately 10% of thesestudents qualify to the second round in late September, which has two typesof problem: questions in which only the answer, which is an non-negativeinteger less than 10000, is required and problems in which full solutions arerequired. At the same time, undergraduates take the first round, whichconsists in a six-problem test (full solutions required).

Finally, approximately 200 to 400 students in each level go to the finalround, held in late October. Grades 6 and 7 have only one test with fiveproblems; all other students have two tests in two consecutive days, eachone with three problems.

The winners are announced in early December and invited to go to a week-long training camp in late January named Olympic Week. They are in-formed about the selection process of international olympiads like IMO,Cono Sur Olympiad and Iberoamerican Olympiad.

The selection process to both IMO and Cono Sur Olympiad usually consistsin three or four team selection tests and three or four problem sets that thestudents receive. The Cono Sur Olympiad team is usually announced inApril and the IMO team is announced in late April or early May. The ConoSur team goes to a training camp the week before the competition; the IMOteam has a training camp three weeks before IMO.

1

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(Introduction)

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3 Problems

2.1. Grades 6–7

Problem 1

Emerald has several right triangles just like the one in the diagram.

(a) Emerald made the following diagram by coinciding part of the sidesand without superposing triangles. Find the area and the perimeter ofthis diagram.

(b) Following the same rules as above, Emerald constructed the smallestsquare with integer side. Draw a diagram showing how she can do this.

Problem 2

The cells of a 3×3 table were numbered from 1 to 9, each number appearingexactly once. For each row the cell with the greatest number is colored redand the cell with the smallest number is colored green. Let A be the smallestof the numbers in the red cells and B the greatest of the numbers in thegreen cells.

(a) Show a distribution of the numbers in the table such that A−B = 4.

(b) Show a distribution of the numbers in the table such that A−B = −3.

(c) Is it possible that A = 4 and B = 3?

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4 XXXII Brazilian Math Olympiad 2010

Problem 3

Given a solid made of unit cubes, as in figure 1, we can write the numberof unit cubes in each direction, as shown in figure 2.

Emerald Jr made a solid out of unit cubes and drew a figure similar to figure2.

Find a, b, c, d, e, f , x and m.

Problem 4

A positive integer n is clowny if the number obtained by reversing its digitsis greater than n. For example, 2009 is clowny because 9002 is greater than2009; however, 2010 is not clowny because 0102 = 102 is less than 2010 and

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Problems 5

3443 is not clowny because it is equal to the number obtained by reversingits digits. How many four-digit numbers are clowny?

Problem 5

(a) Show a positive integer not greater than 1000 with at least 20 positivedivisors.

(b) Does there exist a positive integer not greater than 11000 with at least200 positive divisors?

2.2. Grades 8–9

Problem 1

See problem 4, Grades 6–7.

Problem 2

Let ABCD be a paralellogram and Γ the circumcircle of the triangle ABD.Lines BC and CD meet Γ at E 6= B and F 6= D respectively. Prove thecircumcenter of the triangle CEF lies on Γ.

Problem 3

Arnold and Bernold play the following game in a m × n board: Arnoldchooses one of its cells and places a knight on it. Then Bernold and Arnoldmove the knight alternately, with the condition that the knight visits a cellat most once. The player who is unable to move the knight loses. Determine,in terms of m and n, which player has the winning strategy.

Remark: the knight always moves two cells in a row or a column and thenone cell in the perpendicular direction.

Problem 4

Let a, b and c be real numbers such that a 6= b and a2(b+ c) = b2(c+ a) =2010. Compute c2(a+ b).

Problem 5

Let O be the intersection point of the diagonals of the cyclic quadrilateralABCD. The circumcircles of triangles AOB and COD meet lines BC andAD again at M , N , P and Q. Prove that the quadrilateral MNPQ isinscribed in a circle with center O.

Problem 6

The sidelenghts and area of a triangle are all integer numbers. Find theminimum value of its area.

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6 XXXII Brazilian Math Olympiad 2010

2.3. Grades 10–12

Problem 1

Find all functions f from real numbers to real numbers such that

f(a+ b) = f(ab)

for all irrationals a, b.

Problem 2

Let P (n) be a polynomial with real coefficients. Prove that there existintegers n and k such that k has at most n digits and at least P (n) divisors.

Problem 3

What is the maximum area of the shadow cast by a unit cube? We un-derstand “area of the shadow” of something as the area of its orthogonalprojection in a given plane.

Problem 4

Let ABCD be a quadrilateral with 6 ABC 6= 90◦. Let M and N be themidpoints of AD and CD, respectively. Prove that the lines perpendicu-lar to BC passing through M and perpendicular to AB passing throughN and BD are concurrent if and only if the diagonals BD and AC areperpendicular.

Problem 5

Find all values of n such that there exists a set S of n points in plane,no three of them collinear, with the following property: one can color thepoint such that if three of such points have the same color or three differentcolors then they do not determine an obtuse angle. The number of colors isunlimited.

Problem 6

Find all positive integers a, b such that

3a = 2b2 + 1.

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Problems 7

2.4. Undergraduates

Problem 1

Compute∫ π/4

0

x

(sinx+ cosx) cosxdx.

Problem 2

See problem 3, Grades 10–12.

Problem 3

Let n be an integer and n1 be one of its divisors. Let A be a n×n symmetricmatrix defined by ai,i = 4, ai,i+1 = ai+1,i = −1 for all i such that 1 ≤ i ≤ n−1and i+1 is not a multiple of n1, ai,i+n1

= ai+n1,i = −1 and ai,j = 0 otherwise.

Prove that A has an inverse and that all of the entries in the inverse arepositive.

Problem 4

Define the polynomials(

xj

)

= x(x−1)(x−2)...(x−j+1)j! for j positive integer and

(

x0

)

= 1.

(a) Prove that all non-nil polynomials can be written uniquely as a linearcombination of such polynomials

(

xn

)

.

(b) Let c(n, k) be the coefficient of(

xk

)

in xn (as described in the previousitem). Compute

c(n, k) + c(n, k + 1)

c(n+ 1, k + 1).

Problem 5

For each finite subset F of the space R3, define Vr(F ) as the union of theopen spheres with center on each point of F and radius r. Prove that, for0 < r < R,

vol(VR(F )) ≤ R3

r3vol(Vr(F )).

Problem 6

Prove that if 102n+8 · 10n+1 has a prime factor of the form 60k+7, k andn both positive integers, then n and k are both even.

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9 Solutions

3.1. Grades 6–7

Problem 1

The triangle is the 3–4–5 triangle.

(a) Since four triangles were used, the area is 4· 3·42 = 24 cm2. The perimeteris 2 · (5 + 4 + (4− 3) + 3) = 26 cm.

(b) Since the area of each triangle is 3·42 = 6 cm2, the side of the square

must be a multiple of 6. Moreover, since we are covering the sides ofthe square with the sides of the triangle, the side of the square must beequal to 3x+4y+5z, x, y, z ≥ 0. Though 6 = 2 · 3, it’s not hard to seethat it’s not possible to cover a square with side 6 cm with triangles.So the smallest square has side 12 cm and can be made of twelve 3× 4rectangles.

Problem 2

(a) For example,

7 4 18 5 29 6 3

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10 XXXII Brazilian Math Olympiad 2010

Notice that A = min{7, 8, 9} = 7 and B = max{1, 2, 3} = 3, so A−B =4.

(b) For example,

1 2 34 5 76 8 9

Notice that A = min{3, 7, 9} = 3 and B = max{1, 4, 6} = 6, so A−B =−3.

(c) No, it’s not possible. Since B = 3 is the greatest of the numbers ingreen cells, which are the smallest ones in their respective rows, thenumbers 1, 2 and 3 must be on different rows. Now A = 4 means that4 is the greatest number of a row, but it could only happen if two fothe numbers 1, 2, 3 are on the same row as 4, which cannot happen.

Problem 3

Consider the layer 3 × 3 × 1 to the left, with numbers 2, 2, 1 on one sideand a, 3, 2. The number of unit cubes in this layer is 2 + 2+ 1 = a+3+ 2,so a = 0. Similarly, considering the middle horizontal layer, 2 + x + 2 =3 + 3 + 1 ⇐⇒ x = 3. Analogously, 1 + 2 + 1 = m + 2 + 2 ⇐⇒ m = 0,a + 1 + d = 1 + 3 + m ⇐⇒ 0 + 1 + d = 1 + 3 + 0 ⇐⇒ d = 3. Now,2 + x + 2 = 1 + b + c ⇐⇒ b + c = 6. Since b, c ≤ 3, b = c = 3. Finally,3+b+e = 3+3+2 ⇐⇒ 3+e = 5 ⇐⇒ e = 2 and 1+c+f = 2+1+2 ⇐⇒3 + f = 4 ⇐⇒ f = 1.

Problem 4

Let (abcd) be a four-digit number a, b, c, d being its digits. So a number isclowny if and only if (dcba) > (abcd). This means that either d > a or d = aand c > b. Notice that we cannot have both d = a and b = c because itwould imply (dcba) = (abcd). There are 9·8

2 = 36 choices in the first case(recall that d > a > 0) and 9· 10·92 = 405 choices in the second case (9 choicesfor a = d and 10·9

2 for c > b ≥ 0). So there are 36 + 405 = 441 four-digitclowny numbers.

Problem 5

(a) For example, 900 = 22 · 32 · 52, which has (2 + 1) · (2 + 1) · (2 + 1) = 27positive divisors.

(b) No, there doesn’t. Let n be a number with at least 200 divisors. Ifthe i-th divisor is d, then the i-th to last divisor is n

d . Let m be the100th divisor. So m ≥ 100 and n

m > m ⇐⇒ n > m2 = 10000. Close

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Solutions 11

enough, but how do we fix this? First notice that if 97, 99 and 100are all divisors of n, then n ≥ lcm(97, 99, 100) > 11000. So the keyobservation is considering the 98th, 99th and 100th divisors. Let k, ℓand m be such divisors. Notice that if m ≥ 105 we are done becausethen n > m2 = 11025 > 11000. So 98 ≤ k < ℓ < m ≤ 104. Butgcd(x, y) ≤ |x − y| implies that n ≥ lcm(k, ℓ,m) ≥ k·ℓ·m

(ℓ−k)(m−ℓ)(m−k) ≥98·99·100

(104−98)·32 > 11000. Here we used the fact that if x+y ≤ 2t then xy ≤ t2

applied to x = ℓ− k, y = m− ℓ and t = 3.

3.2. Grades 8–9

Problem 1

See problem 4, Grades 6–7.

Problem 2

Consider all angles oriented and modulo 180◦. Let O be the center of thecircle. In the cyclic pentagonal ABEDF , 6 EBF = 6 EDF = 6 EDC =6 CED + 6 DCE = 6 BED + 6 DCE = 6 BAD + 6 DCE = 26 DCE. Thismenas that if M is the midpoint of the arc FE that does not contain A,6 FME = 6 EBF = 26 DCE = 26 FCE. Since ME = MF , M is thecircumcenter of triangle CEF .

Problem 3

Suppose, without loss of generality, m ≤ n. If m = 2, Arnold has winningstrategy if ond only if n is not a multiple of 4; para m ≥ 3, Arnold haswinning strategy if and only if m and n are both odd.

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12 XXXII Brazilian Math Olympiad 2010

Suppose m = 2. If 4 does not divide n, Arnold can win placing the knighton the first column if n = 4k + 1 and on the second column if n = 4k + r,r = 2 or r = 3. The knight should always be moved two columns ahead oneach move, allowing exactly 2k moves.

Now consider n = 4k. Divide the table into 2 × 4 subtables, and pair thecells in such a way that it’s always possible to move the knight between cellsfrom each pair:

1 2 3 43 4 1 2

Since the whole table is divided into pairs, Bernold can always move, nomatter where Arnold places the knight: it suffices to motve the knight tothe other cell in the pair. When Arnold plays, he will move the knight to acell from another pair, and Bernold can repeat this strategy. So if n = 4kBernold has winning strategy.

This finishes the case m = 2. The case m ≥ 3 follows in a similar fashion,dividing the table into smaller subtables:

1 2 3 43 6 1 52 5 4 6

1 2 3 4 5 63 4 1 7 8 92 7 8 9 6 5

1 2 34 A 12 3 4

1 2 3 4 53 4 6 A 72 1 7 5 6

Those tables prove that Bernold has winning strategy for tables 3×n, n evenand Arnold has winning strategy for tables 3 × n, n odd: divide the tableinto one 3×3 or 3×5 subtable and several 3×4 subtables; it suffices to placethe knight on the cell marked with an A and then use the aforementionedBernold’s strategy.

The case m = 4 can be verified using several 4×2 subtables if n is even andone 4× 3 subtable and several 4× 2 subtables if n is odd. This also provesthat if either Arnold or Bernold has winning strategy for a m×n table thenhe also has winning strategy for a (m + 4) × n table, m ≥ 3. So is sufficesto solve the problem for m ∈ {3, 4, 5, 6}.The case m = 5 can be solved using the following subtables:

1 2 3 4 5 63 4 1 7 8 92 10 11 9 6 512 13 14 15 7 810 11 12 13 14 15

1 2 3 4 53 4 5 6 72 1 10 8 911 A 12 7 612 10 11 9 8

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Solutions 13

Notice that Bernold has winning strategy for a 5× 4 table and Arnold haswinning strategy for a 5×3 table. Thus, in the case 5×n, n even, we merge5 × 4 subtables if n is a multiple of 4 and one 5 × 6 subtable and 5 × 4subtables if n = 4k + 2; if n is odd, we merge several 5 × 4 subtables to a5× 3 or 5× 5 subtable, if n = 4k + 3 or n = 4k + 1, respectively.

The case 6× n follows directly from the case 3× n if n is even (merge two3 × n tables) and merging 6 × 4 subtables to a 6 × 3 or 6 × 5 subtable, ifn = 4k + 3 or n = 4k + 1, respectively. All the cases are covered.

Problem 4

Since a 6= b, a2(b + c) = b2(c + a) ⇐⇒ a2b + a2c − b2c − ab2 = 0 ⇐⇒ab(a−b)+c(a−b)(a+b) = 0 ⇐⇒ ab+ca+bc = 0. So (a−c)(ab+bc+ca) =0 ⇐⇒ a2b+ abc+ a2c = abc+ bc2 + ac2 ⇐⇒ c2(a+ b) = a2(b+ c) = 2010.

Problem 5

Consider angles oriented modulo 180◦. Since 6 PCO = 6 BCA = 6 BDA =6 ODN are inscribed in the circumcircle of the triangle OCD, OP = ON .Analogously, OM = OQ.

Now, 6 QOP = 6 CPO = 6 CDO = 6 CDB = 6 CAB = 6 OAB = 6 OQB =6 OQP , so OP = OQ, and analogously, OM = ON .

Since OM = ON = OP = OQ, the result follows.

Problem 6

The 3–4–5 triangle has area 3·42 = 6. We will prove that no other trian-

gle with integer sidelengths and area has smaller area. Let a, b, c be the

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14 XXXII Brazilian Math Olympiad 2010

sidelenghts. Then its area is S =√

s(s− a)(s− b)(s− c), where s = a+b+c2 .

Since the area is also an integer, a+ b+ c is even, and s, s− a, s− b, s− care all integers.

Now, notice that the triangle cannot be equilateral, since equilateral tri-angles with an integer side have irrational area. So, at least two of thethree integer numbers s − a, s − b, s − c are distinct and, since s =(s − a) + (s − b) + (s − c) ≥ 1 + 1 + 2 = 4, S ≥

√4 · 2 · 2 · 1 =

√8, so

S ≥ 3.

If S is odd, s ≥ 5 and s−a, s−b, s−c are all odd, so S ≥√5 · 3 · 1 · 1 =

√15,

so S ≥ 5. The only relevant case is S = 5. But this would imply two of s,s− a, s− b, s− c being equal to 5, which is impossible.

If S is even, the only relevant case is S = 4. But then all of s, s−a, s−b, s−care powers of two. So if s > 4 then s ≥ 8 and s−a, s− b, s− c would be, insome order, 1, 1, 2, which is not possible because s = (s−a)+(s−b)+(s−c).If s = 4, the only possibility would be s− a, s− b, s− c being 1, 1, 2, whichdoes not work either.

3.3. Grades 10–12

Problem 1

Let f(0) = k. Plugging a =√2 and b = −

√2 one obtain f

(√2 + (−

√2))

=

f(√

2(−√2))

⇐⇒ f(−2) = k.

Let α ∈ R\Q. Since the quadratic equation x2−αx−2 = 0 has discriminant∆ = α2+8 > 0, its roots have sum α ∈ R \Q and product −2 ∈ Q, at leastone of its roots m is irrational; the other root, n = −2/m, is also irrational.Thus we can plug m and n, obtaining f(m + n) = f(mn) ⇐⇒ f(α) =f(−2) = k.

Now let q ∈ Q. The quadratic equation x2 − qx−√2 = 0 has discriminant

∆ = q2 + 4√2 > 0, sum of the roots q ∈ Q and product of the roots

−√2 ∈ R \ Q, so one of its roots r is irrational; the other root, s = q − r,

is also irrational. we can plug r and s, obtaining f(r + s) = f(rs) ⇐⇒f(q) = f

(

−√2)

= k.

So all the functions are the constant functions f(x) = k, k ∈ R.

Problem 2

Let d be the degree of P , and consider d+1 distinct primes p1, p2, . . . , pd+1.Let A = p1p2 . . . pd+1 and kN = AN . If A has r digits, then kN has at mostrN digits. On the other hand, kN has (N +1)d+1 positive divisors. Since P

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Solutions 15

has degree d, P (rN) < (N + 1)d+1 for all sufficiently large N , which solvesthe problem.

Problem 3

Let ABCD and EFGH be two opposite faces, AE, BF , CG and DH beingedges of the cube, and let X ′ be the orthogonal projection of point X ontothe plane. Notice that {A,G}, {B,H}, {C,E} and {D,F} are pairs ofopposite vertices. Suppose, without loss of generality, that A′ lies on theboundary of the projection of the cube. Then, considering the symmetryof the cube around the center of the cube, its symmetric point G′ lies onthe boundary as well. Two of the three neighboring vertices of A are goingto be neighbors of A′ in the projection (unless, say, face AEHD projectsonto a line; but in this case we consider a degenerate vertex inside this line).Suppose without loss that these neighbors are B′ and D′. So E ′ is insidethe projection. Again by symmetry H ′ and F ′ lie on the boundary of theprojection and C ′ lies inside the projection. Finally, since

−→AE =

−−→BF =−−→

CG =−−→DH, the projection of the cube is A′D′H ′G′F ′B′.

The faces ABCD, BCGF and CDHG project onto the parallelograms (orline segments) A′B′C ′D, B′C ′G′F and C ′D′H ′G′. Draw diagonals B′D′,B′G′ and D′G′. The area of the projection is then twice the area of thetriangle B′D′G′, which is at most the area of triangle BDG. This triangle

is equilateral with side√2, so the desired maximum is 2

(√2)

2√3

4 =√3.

Problem 4

Consider a homothety with center on D that takes M to A and N to C.So the perpendicular lines are mapped to the altitudes of the triangle ABCrelative to A and C, and the intersection P of the perpendicular lines is

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16 XXXII Brazilian Math Olympiad 2010

mapped to the orthocenter H of triangle ABC.

Notice that the condition that the perpendicular lines and BD are concur-rent is equivalent to B, P and D being collinear. But the homothety impliesthat D, P and H are collinear, so the three lines are concurrent if and onlyif BH and BD coincide, that is, BD ⊥ AC, since BH ⊥ AC.

Problem 5

We start with the following

Lemma. Every set with 5 or more points in the plane, no three of themcollinear, has three points that determine an obtuse angle. Besides, everyset with 4 points, no three of them determining an obtuse angle, is uniquelydefined by three of its points (that is, the position of one point can bedetermined from the other three points).

Proof. First, notice that the points should be vertices of a convex polygon.Otherwise, one of the points, say P , is inside a triangle with vertices on otherthree points A, B, C from the set, and since 6 APB+ 6 BPC+ 6 CPA = 360◦,of of the three angles 6 APB, 6 BPC, 6 CPA is greater than 90◦. Finally,since the sum of the internal angles of a convex n-gon is (n−2) ·180◦, one ofthe internal angles is greater than or equal to (n−2)·180◦

n =(

1− 2n

)

180◦ > 90◦

for n > 4. This proves the first part of the lemma.

The second part follows from the fact that if we have four points then allinternal angles of the quadrilateral with vertices on the four points must beequal to 90◦, that is, the quadrilateral is a rectangle.

Because of the lemma, there are at most four points with either the samecolor or three different colors in S. Let’s prove then that S cannot have

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Solutions 17

more than 12 points. If S has more than 12 points, since there are at mostfour colors and more than 3 · 4 = 12 points then there are four points withthe same color. Besides, since there are at most four points with the samecolor, there are also four points with four different colors.

Now we prove that S cannot have simultaneously four points with the samecolor and four points with four different colors. Suppose, by means of con-tradiction, the contrary. Let P = {A,B,C,D} be a set of points from Swith the same color and E, F and G points from S with three differentcolors and a different color from the points from P. So {A,E, F,G} and{B,E, F,G} are both sets of points with four different colors. But thelemma implies that E, F and G determine the position of both A and B,that is A = B, contradiction.

So S has at most 12 points. Consider the following examples with 12 points:consider the vertices of three unit squares whose centers are the vertices ofan equilateral triangle with a sufficiently large side. Color the vertices ofeach square with a single color, so we have three different colors, one foreach square. The angles determine by three points of the same color are allright angles, and the angles determined by three points with three differentcolor can be arbitrarily close to 60◦.

It is easy to obtain set S with less than 12 points deleting points from theexamples given above. So the answer is n ≤ 12.

Problem 6

The only solutions are (1, 1), (2, 2) and (5, 11).

If a is even and greater than 2, the equation can be rewritten as (3a/2 −1) · (3a/2 + 1) = 2b2. but gcd(3a/2 − 1, 3a/2 + 1) = gcd(3a/2 − 1, 2) = 2, so3a/2 + 1 = 4u2 and 3a/2 − 1 = 2v2 or 3a/2 + 1 = 2u2 and 3a/2 − 1 = 4v2.

In the former case, 3a/2 = (2v − 1)(2v + 1), and since gcd(2v − 1, 2v + 1) =gcd(2v − 1, 2) = 1, 2v − 1 = 1 ⇐⇒ v = 1 and a/2 = 1 ⇐⇒ a = 2 andthus b = 2.

In the latter case, 3a/2 = 4v2 + 1 =⇒ 0 ≡ v2 + 1 (mod 3) ⇐⇒ v2 ≡ −1(mod 3), which is impossible.

If a is odd, the equation is equivalent to 3 · (3(a−1)/2)2 − 2b2 = 1. Letc = 3(a−1)/2. Let’s find the solutions to 3c2 − 2b2 = 1 (∗). Since

(√3 +

√2)(

√3−

√2) = 1 =⇒ (

√3 +

√2)2k+1(

√3−

√2)2k+1 = 1

and

(√3 +

√2)2k+1 = ck

√3 + bk

√2 y (

√3−

√2)2k+1 = ck

√3− bk

√2 (∗∗)

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18 XXXII Brazilian Math Olympiad 2010

(ck, bk) are solutions to (∗), for integers k ≥ 0. Now suppose that (α, β) isa different solution. Then there exists an integer k such that

(√3 +

√2)2k−1 < α

√3 + β

√2 < (

√3 +

√2)2k+1

⇐⇒√3 +

√2 <

α√3 + β

√2

(5 + 2√6)k−1

< 9√3 + 11

√2

⇐⇒√3 +

√2 < (α

√3 + β

√2) · (5− 2

√6)k−1 < 9

√3 + 11

√2

It’s not hard to prove by induction that (α√3 + β

√2) · (5 − 2

√6)k−1 =

θ√3+φ

√2, θ and φ both integers. Moreover, θ

√3+φ

√2 > 1 > 1

θ√3+φ

√2=

θ√3− φ

√2 > 0, so θ

√3 > φ

√2 > 0. Hence (θ, φ) is a solution to (∗), with

1 < θ < 9. But it can be verified that there are no solutions with 1 < θ < 9,contradiction.

Now suppose that c > 9, that is, k > 1. By (∗∗) and the binomial theorem,

c =

k∑

m=0

(

2k + 1

2m+ 1

)

·3m ·2k−m = (2k+1) ·2k +(

2k + 1

3

)

·3 ·2k−1+ · · · (∗∗∗)

Thus 3 divides 2k + 1. Let 3t be the largest power of 3 that divides 2k + 1.Since

(

2k+13

)

· 3 · 2k−1 = (2k + 1) · k · (2k − 1) · 22k−1, the largest power of3 that divides the second term of the right hand side of (∗ ∗ ∗) is also 3t

and, letting 3s be the largest power of 3 that divides 2m+1, the m-th term(

2k+12m+1

)

· 3m · 2k−m = 2k+12m+1

(

2k2m

)

· 3m · 2k−m has at least t+m− s ≥ t+ 3s+12 − s

factors 3 (t from 2k + 1, m from 3m, minus s from 2m + 1). Since 3s =(1 + 2)s ≥ 1 + 2s ⇐⇒ 3s+1

2 − s ≥ 1 for s ≥ 0, with equality if and only ifs = 0 or s = 1, the m-th term has more than t + 1 factors 3 except whenm = 3s+1

2 . If s = 0 we have m = 1 ⇐⇒ 2m + 1 = 3, a known solution; ifs = 1 we have m = 2 which has no factors 3. Thus, since k > 1, all termsstarting from the third one have at least t+ 2 factors 3, and

c = (2k + 1) · 2k + (2k + 1) · k · (2k − 1) · 2k−1 + 3t+2 ·N⇐⇒ 3(a−1)/2 = (2k + 1) · 2k−1 · [(2k + 1)(k − 1) + 3] + 3t+2 ·N

Notice that since 3 divides (2k+ 1) then it also divides (2k+ 1)(k− 1) + 3;moreover, 9 divides (2k + 1)(k − 1), and then the largest power of 3 thatdivides (2k + 1)(k − 1) + 3 is 3. Hence the largest power of 3 that divides(2k+1) · 2k−1 · [(2k+1)(k− 1)+3] is t+1 (t factors from 2k+1 and 1 from(2k + 1)(k − 1) + 3). Finally,

3(a−1)/2 = c = 3t+1(1 + 3N),

which is impossible because N > 0.

So there are no more solutions.

18 (Solutions)

(page 19)

Solutions 19

3.4. Undergraduates

Problem 1

First notice the identity sinx + cosx =√2(

cosx cos π4 + sinx sin π

4

)

=√2 cos

(

π4 − x

)

. So

I =

∫ π/4

0

x

(sinx+ cosx) cosxdx =

∫ π/4

0

x√2 cos

(

π4 − x

)

cosxdx

Now consider the substitution y = π4 − x. So dx = −dy and

∫ π/4

0

x√2 cos

(

π4 − x

)

cosxdx =

∫ 0

π/4

−π4 − y√

2 cos y cos(

π4 − y

)dy

=

∫ π/4

0

π4 − y√

2 cos y cos(

π4 − y

)dy =

∫ π/4

0

π4 − x√

2 cos(

π4 − x

)

cosxdx

Summing the two integrals, we find

2I =

∫ π/4

0

x√2 cos

(

π4 − x

)

cosxdx+

∫ π/4

0

π4 − x√

2 cos(

π4 − x

)

cosxdx

=

∫ π/4

0

π4 − x+ x√

2 cos(

π4 − x

)

cosxdx =

∫ π/4

0

π4√

2 cos(

π4 − x

)

cosxdx

So I = π8

∫ π/4

01

(sinx+cosx) cosxdx.

Let f(x) = ln(sinx + cosx) and g(x) = ln(cosx). Notice that f ′(x) =cosx−sinxsinx+cosx = f ′(x) = cos2 x−sinx cosx

(sinx+cosx) cosx and g′(x) = − sinxcosx = − sin2 x+sinx cosx

(sinx+cosx) cosx . So

f ′(x)− g′(x) = cos2 x−sinx cosx+sin2 x+sinx cosx(sinx+cosx) cosx = 1

(sinx+cosx) cosx . Hence

I =π

8

∫ π/4

0

1

(sinx+ cosx) cosxdx =

π

8

(

f(π

4

)

− g(π

4

)

− (f(0)− g(0)))

=π

8

(

ln(√2)− ln

(√2

2

)

− ln 1 + ln 1

)

=π ln 2

8

Problem 2

See problem 3, grades 10–12.

19

(Solutions)

(page 20)

20 XXXII Brazilian Math Olympiad 2010

Problem 3

Let’s find the inverse of A/4 = I −X, where all entries in X are either 0 or1/4. We will use the series

(I −X)−1 = I +X +X2 +X3 + · · ·

First let’s prove that this series converges. It suffices to show that themaximum M such that ‖Xw‖ ≤ M‖w‖ for all column vectors w of X is lessthan 1, so the sum of the entries always decrease by a factor smaller than 1if you multiply a vector by X2; then we sum the series as (I +X)(I +X2 +X4 + · · ·). Notice that every row of X has at most four nonzero entries, allof which are equal to 1/4. So if w = (a1, a2, . . . , an) then every entry of Xw

is of the formar1+ar2+···+ars

4 , s ≤ 4. By the Cauchy-Schwarz inequality, its

square is at most s16 (a

2r1+a2r2 + · · ·+a2rs) ≤

a2r1+a2r2+···+a2rs4 , with equality if and

only if ar1 = ar2 = · · · = ars and s = 4. Summing over all the rows, the sumof squares of the coordinates of Xw is at most a21 + a22 + · · · + a2n = ‖w‖2,because all columns of X have at most four nonzero entries. But equalitywould only happen if all entries ai are equal and s = 4 always, which doesnot happen for, say, the first row. So M < 1 and the series converges.

Consider the graph whose vertices are the numbers v1, v2, . . . , vn and weconnect vi and vj if and only if the entry xij in X is 1/4. By the defini-tion of the matrix A, this graph has a lattice-like configuration: it can besplit in several paths v1v2 . . . vn1−1, vkn1

vkn1+1 . . . v(k+1)n1−1, 1 ≤ k ≤ nn1

− 1,vrvr+n1

vr+2n1. . . vr+n−n1

, 1 ≤ r ≤ n1. It is clear that this graph is connected.Since the entry mij in Xk is nonzero if and only if there exists a circuitfrom i to j with k edges, for all i, j there is k such that the correspondingentry mij in Xk is nonzero. This proves that all entries in the inverse of Ais positive.

Problem 4

(a) Induct on the degree n of the polynomial P (x) =∑n

k=0 akxk. It is

immediate if n = 0. Since the degree of(

xk

)

, k < n, is k, the coefficient

on(

xn

)

on P (x) is an ·n!. Now consider the polynomial P (x)−an ·n!(

xn

)

.

This is a polynomial with degree less than n, so P (x) − an · n!(

xn

)

can

be uniquely represented as a linear combination of the polynomials(

xk

)

,and the result follows.

20

(Solutions)

(page 21)

Solutions 21

(b) We have xn =∑n

k=0 c(n, k)(

xk

)

, so

xn+1 =

n∑

k=0

c(n, k)

(

x

k

)

x

=

n∑

k=0

c(n, k)

(

x

k

)

(x+ 1)−n∑

k=0

c(n, k)

(

x

k

)

=

n∑

k=0

c(n, k)(k + 1)

(

x+ 1

k + 1

)

−n∑

k=0

c(n, k)

(

x

k

)

=

n∑

k=0

c(n, k)(k + 1)

((

x

k + 1

)

+

(

x

k

))

−n∑

k=0

c(n, k)

(

x

k

)

=

n∑

k=0

(c(n, k − 1)k + (k + 1− 1)c(n, k))

(

x

k

)

=

n∑

k=0

k(c(n, k − 1) + c(n, k))

(

x

k

)

in which we consider c(n,−1) = 0.

Thus c(n + 1, k) = k(c(n, k − 1) + c(n, k)) ⇐⇒ c(n,k)+c(n,k−1)c(n+1,k) = 1

k , andc(n,k+1)+c(n,k)

c(n+1,k+1) = 1k+1 .

Problem 5

Let F = {P1, P2, . . . , Pn} and let αij be the perpendicular plane bisector ofPi and Pj for i 6= j. Those planes define n convex regions R1, R2, . . . , Rn,where Ri is the intersection of the half-spaces determined by αij that containPi. Finally, let Ar(i) be the intersection of Ri with the sphere with centerPi and radius r. Thus, since the (disjoint) union of the regions Ri is thewhole space, Vr(F ) is the disjoint union of Ar(1), Ar(2), . . ., Ar(n).

Now, for each point Pi, apply a homothety with center Pi and ratio r/R < 1.It is clear that the image of AR(i) is contained in Ar(i), since a sphere withradius R is taken to a sphere with radius r and the planes αij are taken to

planes α′ij closer to Pi. Thus vol(Ar(i)) ≥

(

rR

)3vol(Ar(i)) and the result

follows by summing up these inequalities for i = 1, 2, . . . , n.

Problem 6

Let p = 60k+7 be such a prime. Then 102n+8 ·10n+1 ≡ 0 (mod p) ⇐⇒(10n − 1)2 ≡ −10n+1 (mod p). Now suppose n is odd. Then (10n − 1)2 ≡

21

(Solutions)

(page 22)

22 XXXII Brazilian Math Olympiad 2010

−1 · (10(n+1)/2)2 (mod p), and −1 is a quadratic residue. But by the Eulercriterion,

(−1p

)

= (−1)(p−1)/2 = −1, a contradiction. So n is even and,

moreover,(−10

p

)

= 1 ⇐⇒(

2p

)(

5p

)

= −1 ⇐⇒(

2p

)

= −(

5p

)

.

By the quadratic reciprocity lemma,(

5p

)(

p5

)

= (−1)5−12 · p−1

2 = 1 ⇐⇒(

5p

)

=(

p5

)

=(

25

)

= −1. So(

2p

)

= 1 ⇐⇒ (−1)p2−18 = 1, so p2−1

8 = (30k+3)(15k+2)is even. Since 30k + 3 is odd, 15k + 2 is even, that is, k is even.

Notice that we did not need the factor 3, that is, the problem holds forp = 20k + 7.

22

(Solutions)

(page 23)

23 Winners in 2010

4.1. Grades 6–7

Gold medals

Ana Emılia Hernandes Dib

Pedro Henrique Alencar Costa

Ryunosuke Watanabe Tagami

Helena Veronique Rios

Italo Lesione de Paiva Rocha

Jose Henrique Carvalho

Silver medals

Juliana Bacelar de Freitas

Daniel Lima Braga

Hermes Lins e Nascimento

Laıs Monteiro Pinto

Lucca Morais de Arruda Siaudzionis

Leandro Alves Cordeiro

Henrique Gontijo Chiari

Andre Akinaga Benites

Gabriel Diniz Vieira e Sousa

Rafael Seiji Uezu Higa

Adriana de Sousa Figueiredo

Gustavo Figueiredo Serra

Bronze medals

Matheus Uchoa Constante

Kristian Holanda Nogueira

Fabio Itikama

Loic Dominguez

Jiang Zhi

Ricardo Ken Wang Tsuzuki

Ana Caroline Obana da Cruz

Ana Paula Lopes Schuch

Jose Marcio Machado de Brito

Lucas Bastos Germano

Victoria Moreira Reis Cogo

23

(Winners in 2010)

(page 24)

24 XXXII Brazilian Math Olympiad 2010

Thiago Araujo Oliveira

Gabriel Toneatti Vercelli

Nathan Bonetti Teodoro

Jefferson Daxian Hong

Cristobal Sciutto Rodriguez

Aruana Almeida Correa

Cynthia Lacroix Herkenhoff

Honorable mention

Kaıque Maestrini Sacchi

Igor de Lacerda

Rafael Reple Geromee

Leonardo de Matos Felippetti Mariano

Gabriel Passamani Correa

Daniel de Almeida Souza

Diego Teixeira Nogueira Fidalgo

Natan Novelli Tu

Ricardo Borsari Brinati

Rafael Neves Vieira

Juliano Pecica Negri

Gustavo Rodrigues Machado

Zoltan Flamarion Glueck Carvalho

Gabriel Ribeiro Barbosa

Pedro Henrique Rocha de Freitas

Pedro Henrique Sacramento de Oliveira

Guilherme Goulart Kowalczuk

Pedro de Vasconcellos Oporto

Aryssa Victoria Shitara

Ives Vaz Caldeira Lopes

Marcos Vinıcius de Oliveira Soares

Jessica Carolina Zilio

Joao Pedro Graca Melo Vieira

Henrique Medici Pontieri

Gabriel Caino Castilho Rodrigues

Tamara P. de A. Moraes

Karine Quaresma Lima

Natalia Brasileiro Lins Barbosa

Lucki Li

Heloısa Antunes de Medeiros

24 (Winners in 2010)

(page 25)

Winners in 2010 25

Iuri Grangeiro Carvalho

Lara Sampaio Pinheiro de Freitas

Maria Julia Costa Medeiros

Kevin Korpasch

Sofıa Leite Correia Lima

Joao Baptista de Paula e Silva

Bernardo Puetter Schaeffer

Julia Bertelli

Rafael Purim de Azevedo

Pedro Henrique da Silva Dias

Marcelo Bandeira de Melo Boavista

Gabriel Branco Frizzo

Maria Eduarda Muller Eyng

Henrique Martınez Rocamora

Felipe Roz Barscevicius

Joao Vitor Vaz Oliveira

Mateus Siqueira Thimoteo

Ebenezer Pinto Bandeira Neto

Maria Clara Vasconcelos Andrade

Rafael Beck

Arthur Monteiro Dos Santos

Julia Wotzasek Pereira

Gabriel Oliveira Rigo

Leonardo Galante Barco

Bruno Scatolini

Lucas Pereira Galvao de Barros

Vıtor Ossamu Rodrigues Okamura

4.2. Grades 8–9

Gold medals

Rafael Rodrigues Rocha de Melo

Vinıcius Canto Costa

Henrique Vieira G. Vaz

Fellipe Sebastiam da Silva P. Pereira

Roberto Tadeu Abrantes de Araujo

Pedro Victor Falci de Rezende

25

(Winners in 2010)

(page 26)

26 XXXII Brazilian Math Olympiad 2010

Silver medals

Alessandro A. de Oliveira Pacanowski

Lincoln de Queiroz Vieira

Tadeu Pires de Matos Belford Neto

Vitor Ramos de Paula

Francisco Markan Nobre de Souza Filho

Jair Gomes Soares Junior

Breno Soares da Costa Vieira

Gabriel Jose Moreira da Costa Silva

Pedro Morais de Arruda Siaudzionis

Gabriel Sena Galvao

Fabio da Silva Soares

Michel Rozenberg Zelazny

Bruno Eidi Nishimoto

Franco Matheus de Alencar Severo

Aime Parente de Sousa

Bronze medals

Marcos Paulo Nunes de Lima Silva

Gabriel Nogueira Coelho de Togni de Souza

Rafael Tedeschi Eugenio Pontes Barone

Murilo Corato Zanarella

Rodrigo Sanches Angelo

Alexandre Perozim de Faveri

Luıze Mello D’urso Vianna

Maria Clara Cardoso

Liara Guinsberg

Lucas Cawai Juliao Pereira

Luis Guilherme Gomes Aguiar

Carlos Adriano Vieira

Daniel Santana Rocha

Raphael Mendes de Oliveira

Samuel Brasil de Albuquerque

Gustavo Souto Henriques Campelo

Honorable mention

Lucas de Moura Herlin

Vitor Dias Gomes Barrios Marin

Joao Pedro Sedeu Godoi

26

(Winners in 2010)

(page 27)

Winners in 2010 27

Suzane Eberhart Ribeiro da Silva

Icaro Sampaio Viana

Pedro Henrique Bortolozo Maria

Fabio Kenji Arai

Guilherme de Oliveira Rodrigues

Alexandre Mendonca Cardoso

Leyberson Pereira Assuncao

Rubens Martins Bezerra Farias

Joao Vıtor Fernandes Paiva

Bruno Almeida Costa

Daniel Lima Santanelli

Marılia Nascimento Monteiro

Igor Albuquerque Araujo

Josue Knorst

Ricardo Vieira Marques

Julio Cesar de Barros

Thomas Akio Ikeda Valvassori

Gabriel Fazoli Domingos

Henrique Luan Gomes Pereira Braga

Beatriz Yumi Ota

Kiane Sassaki Menezes

Eric Gripa Marques

Samuel Kuo Chen Shao

Pedro Henrique Jagosenit Vilaca

Caio de Souza Camara

Lucas David Noveline

Lucas Rebelo Vieira da Silva

Elias Brito Oliveira

Guilherme Ryu Odaguiri Kobori

Mariana Souza de Araujo

Francisco Claudio Coelho

Murilo Leao Pereira

Jadi Diniz Guimaraes de Queiroz

Caio Lima Albuquerque

Carolina Lima Guimaraes

27

(Winners in 2010)

(page 28)

28 XXXII Brazilian Math Olympiad 2010

4.3. Grades 10–12

Gold medals

Gustavo Lisboa Empinotti

Marcelo Tadeu de Sa Oliveira Sales

Joao Lucas Camelo Sa

Hanon Guy Lima Rossi

Maria Clara Mendes Silva

Silver medals

Matheus Secco Torres da Silva

Lucas Lourenco Hernandes

Deborah Barbosa Alves

Henrique Gasparini Fiuza do Nascimento

Luiz Filipe Martins Ramos

Andre Macieira Braga Costa

Thiago Saksanian Hallak

Victor Juca Martins

Caıque Porto Lira

Gustavo Haddad Francisco e Sampaio Braga

Alvaro Lopes Pedroso

Andre Amaral de Sousa

Bronze medals

Marcos Massayuki Kawakami

Carlos Henrique de Andrade Silva

Rafael Kazuhiro Miyazaki

Andre Saraiva Nobre dos Santos

Daniel Eiti Nishida Kawai

Lucas de Freitas Smaira

Cassio dos Santos Sousa

Alessandro Macedo de Araujo

Breno Vieira da Silva Passos

Iago Dalmaso Brasil Dias

Isabella Amorim Goncalez

Daniel dos Santos Bossle

Davi Coelho Amorim

Lucas Mestres Mendes

Vinıcius Gomes Pereira

28(Winners in 2010)

(page 29)

Winners in 2010 29

Renan Pablo da Cruz

Jonas Rocha Lima Amaro

Iuri Rezende Souza

Honorable mention

Matheus Araujo Marins

Felipe Vieira de Paula

Rafael Farias Marinheiro

Elvis Falcao de Araujo

Pablo Almeida Gomes

Paulo Gabriel Ramos Monteiro

Victor de Oliveira Bitaraes

Daniel Caueh Dunaiski Figueira Leal

Raphael Julio Barcelos

Fernando Fonseca Andrade Oliveira

Felipe Mendes dos Santos

Felipe Abella Cavalcante Mendonca de Souza

Francisco Raul Lobo Rodrigues

Gabriel Leite de Carvalho

Andre Austregesilo Scussel

Victorio Takahashi Chu

Victor Jose Tiburtius Franco

Matheus Cavalcante Lima

Cleberton de Santana Oliveira

Mauro Brito Junior

Gabriel Jose Guimaraes Barbosa

Lucas Colucci Cavalcante de Souza

Sarah Villanova Borges

Ivan Tadeu Ferreira Antunes Filho

Dalton Felipe de Menezes

Thiago de Paula Vasconcelos

Jardiel Freitas Cunha

Ana Beatriz Prudencio de A. Reboucas

Rafael Sussumu Yamaguti Miada

Davi Sampaio de Alencar

Bruno Ferri de Moraes

29

(Winners in 2010)

(page 30)

30 XXXII Brazilian Math Olympiad 2010

4.4. Undergraduates

Gold medals

Rafael Tupynamba Dutra

Renan Henrique Finder

Regis Prado Barbosa

Ramon Moreira Nunes

Thomas Yoiti Sasaki Hoshina

Silver medals

Guilherme Rodrigues Nogueira de Souza

Jorge Henrique Craveiro de Andrade

Rafael Assato Ando

Gabriel Luıs Mello Dalalio

Charles Barbosa de Macedo Brito

Leonardo Ribeiro de Castro Carvalho

Marcelo Matheus Gauy

Leandro Farias Maia

Bronze medals

Adenilson Arcajo de Moura Junior

Paulo Andre Carvalho de Melo

Joas Elias dos Santos Rocha

Guilherme Lourenco Mejia

Reinan Ribeiro Souza Santos

Rafael Alves da Ponte

Davi Lopes Alves de Medeiros

Luca Mattos Moller

Renato Reboucas de Medeiros

Danilo Furlan Kaio

Rafael Endlich Pimentel

Paulo Sergio de Castro Moreira

Honorable mention

Carlos Coelho Lechner

Thiago Ribeiro Ramos

Hugo Fonseca Araujo

Alysson Espındola de Sa Silveira

Jordan Freitas Piva

30

(Winners in 2010)

(page 31)

Winners in 2010 31

Erik Fernando de Amorim

Daniel Ungaretti Borges

Antonio Deromir Neves Silva Junior

Rafael Parpinel Cavina

Isaque Santa Brigida Pimentel

Mateus Oliveira de Figueiredo

Davi Dos Santos Lima

Bruno da Silva Santos

Francisco Osman Pontes Neto

Breno Vieira de Aguiar

Ricardo Turolla Bortolotti

Guilherme Philippe Figueiredo

Daniel de Barros Soares

Hudson do Nascimento Lima

Eduardo Fischer

Luty Rodrigues Ribeiro

Jose Leandro Pinheiro

Caio Ishizaka Costa

Gabriel Caser Brito

Leonardo Donisete da Silva

Alan Anderson da Silva Pereira

Diego Andres de Barros Lima Barbosa

Renato Dias Costa

Ivan Guilhon Mitoso Rocha

Willy George do Amaral Petrenko

Leonardo Borges Avelino

Jose Armando Barbosa Filho

31

(Winners in 2010)