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(page 1) XXXIII Brazilian Math Olympiad 2011 Editora AOBM Rio de Janeiro 2012 () (page 0) Instituto Nacional de Matemática Pura e Aplicada – IMPA Chair: César Camacho Sociedade Brasileira de Matemática (Brazilian Mathematical Society) Chair: Hilário Alencar Support Conselho Nacional de Desenvolvimento Cientı́fico e Tecnológico – CNPq Instituto do Milênio Avanço Global e Integrado da Matemática Brasileira Comissão Nacional de Olimpı́adas de Matemática (Mathematical Olympiads National Committee) Estrada Dona Castorina, 110 – Jardim Botânico – 22460-320 Rio de Janeiro – RJ Telefone: (21) 2529-5077 Fax: (21) 2529-5023 web: http://www.obm.org.br e-mail: [email protected] Chair: Carlos Gustavo Tamm de Araújo Moreira, Onofre Campos da Silva Farias Members: Antonio Caminha, Francisco Bruno Holanda, Carlos Yuzo Shine, Cı́cero Thiago Bernardino Magalhães, Edmilson Luis Rodrigues Motta, Eduardo Tengan, Eduardo Wagner, Emanuel Carneiro, Élio Mega, Fabio Brochero, Luciano Guimarães Monteiro de Castro, Luzinalva Miranda de Amorim, Nicolau Corção Saldanha, Pablo Rodrigo Ganassim, Paulo Cezar Pinto Carvalho, Ralph Costa Teixeira, Samuel Barbosa Feitosa, Yoshiharu Kohayakawa, Yuri Lima Junior Members: Alex Corrêa Abreu, Bernardo Paulo Freitas da Costa, Carlos Augusto David Ribeiro, Carlos Stein Naves de Brito, Davi Máximo Alexandrino Nogueira, Fábio Dias Moreira, Fabrı́cio Siqueira Benevides, Gabriel Tavares Bujokas, Humberto Naves, Larissa Cavalcante Lima, Marcio Assad Cohen, Telmo Correa Júnior, Thiago Barros Rodrigues Costa, Rodrigo Villard Executive Secretary: Nelly Carvajal Flórez Assistant Secretaries: Rosa Morena Freitas Kohn Typeset with Plain TEX. () (page 1) Introduction 1.1. Structure of the Brazilian Math Olympiad The Brazilian Math Olympiad is a nationwide competition for students from grade 6 to undergraduates, comprising a total of approximately 400000 contestants. Students from grade 6 to 12 have to take three rounds: the first round is held in June and consists in multiple choice questions, 20 for grades 6 and 7 and 25 for grades 8 to 12. Approximately 10% of these students qualify to the second round in late September, which has two types of problem: questions in which only the answer, which is an non-negative integer less than 10000, is required and problems in which full solutions are required. At the same time, undergraduates take the first round, which consists in a six-problem test (full solutions required). Finally, approximately 200 to 400 students in each level go to the final round, held in late October. Grades 6 and 7 have only one test with five problems; all other students have two tests in two consecutive days, each one with three problems. The winners are announced in early December and invited to go to a weeklong training camp in late January named Olympic Week. They are informed 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 consists in three or four team selection tests and three or four problem sets that the students receive. The Cono Sur Olympiad team is usually announced in April and the IMO team is announced in late April or early May. The Cono Sur team goes to a training camp the week before the competition; the IMO team has a training camp three weeks before IMO. 1 (Introduction) (page 2) (Introduction) (page 3) Problems 3 2.1. Grades 6–7 Problem 1 Emerald wrote on the blackboard all the integers from 1 to 2011. Then she erased all the even numbers. (a) How many numbers were left on the board? (b) How many of the remaining numbers were written with only the digits 0 and 1? Problem 2 We have a red cube with sidelength 2 cm. What is the minimum number of identical cubes that must be adjoined to the red cube in order to obtain 3 cm? a cube with volume 12 5 Problem 3 We call a number pal if it doesn’t have a zero digit and the sum of the squares of the digits is a perfect square. For example, 2115522 is pal (because 22 + 12 + 12 + 52 + 52 + 22 + 22 = 82 but 304 and 12 are not pal. (a) What is the greatest two-digit pal number? (b) Does there exist a 2011-digit pal number? Problem 4 In the diagram, O is the center of the square, OA = OC = 2, AB = CD = 4, CD is perpendicular to OC, which is perpendicular to OA, which in turn is perpendicular to AB. The square has area 64 cm2 . 3 (Problems) (page 4) 4 XXXIII Brazilian Math Olympiad 2011 (a) Compute the area of trapezoid ABCO. (b) Compute the area of quadrilateral BCDE. Problem 5 Emerald writes the integers from 1 to 9 in a 3 × 3 table, one number in each cell, each number appearing exactly once. Then she computes eight sums: the sums of three numbers on each row, the sums of the three numbers on each column and the sums of the three numbers on both diagonals. (a) Show a table such that exactly three of the eight sums are multiples of 3. (b) Is it possible that none of the eight sums is a multiple of 3? 2.2. Grades 8–9 Problem 1 Emerald writes the integers from 1 to 9 in a 3 × 3 table, one number in each cell, each number appearing exactly once. Then she computes eight sums: the sums of three numbers on each row, the sums of the three number on each column and the sums of the three numbers on both diagonals. Is it possible that none of the eight sums is a multiple of 3? Problem 2 Let ABCD be a convex quadrilateral such that AD = DC, AC = AB and 6 ADC = 6 CAB. Let M and N be the midpoints of AD and AB. Prove that triangle M N C is isosceles. Problem 3 Emerald and Jade play the following game: Emerald writes a list with 2011 positive integers, but does not show it to Jade. Jade’s goal is finding the product of the 2011 numbers in Emerald’s list. In order to do so, she is allowed to ask Emerald the gcd or the lcm of any subset with at least two of the 2011 numbers (as, for instance, “what is the gcd of the first, second, 10th and 2000th numbers from your list?” or “what is the lcm of all the numbers in your list?”). Jade can make as many questions as she wants, but can only obtain her (correct) answers from Emerald after making all her questions (Emerald is generous and also says which answer corresponds to each question). Jade then can use any of the four elementary operations (add, subtract, multiply, divide) with Emerald’s answers. Can Jade make a list of questions that guarantees that she can find the product of the 2011 numbers? 4 (Problems) (page 5) Problems 5 Problem 4 Emerald wrote a list of positive integers. Renan noticed that each number in the list and any sum of any quantity of distinct numbers from the list were square-free (that is, not divisible by any perfect square except, of course, 1). What is the maximum quantity of numbers that Emerald’s list can have? Problem 5 Consider 1000 points inside a square with√sidelength 16. Prove that there is an equilateral triangle with sidelength 2 3 that covers at least 16 of those points. Problem 6 For each positive integer N with 2k digits, let odd(N ) be the k-digit number obtained by writing the digits of odd order of N and even(N ) be the k-digit number obtained by writing the digits of even order of N . For example, odd(249035) = 405 and even(249035) = 293. Prove that there is no positive integer N with 2k digits such that N = odd(N ) · even(N ). 2.3. Grades 10–12 Problem 1 We call a number pal if it doesn’t have a zero digit and the sum of the squares of the digits is a perfect square. For example, 122 and 34 are pal but 304 and 12 are not pal. Prove that there exists a pal number with n digits, n > 1. Problem 2 33 friends are collecting stickers for a 2011-sticker album. A distribution of stickers among the 33 friends is incomplete when there is a sticker that no friend has. Determine the least m with the following property: every distribution of stickers among the 33 friends such that, for any two friends, there are at least m stickers both don’t have, is incomplete. Problem 3 Prove that, for all convex pentagons P1 P2 P3 P4 P5 with area 1, there are indices i and j (assume P6 = P1 and P7 = P2 ) such that: √ 5− 5 ≤ area △Pj Pj+1 Pj+2 area △Pi Pi+1 Pi+2 ≤ 10 5 (Problems) (page 6) 6 XXXIII Brazilian Math Olympiad 2011 Problem 4 Do there exist 2011 positive integers a1 < a2 < . . . < a2011 such that gcd(ai , aj ) = aj − ai for any i, j such that 1 ≤ i < j ≤ 2011? Problem 5 Let ABC be an acute triangle and H is orthocenter. Let D be the intersection of BH and AC and E be the intersection of CH and AB. The circumcircle of ADE meets the circumcircle of ABC at F 6= A. Prove that the angle bisectors of 6 BF C and 6 BHC concur at a point on line BC. Problem 6 Let a1 , a2 , . . . , a2011 be nonnegative reals with sum 2011 2 . Prove that √ Y 3 3 . (an − an+1 ) = |(a1 − a2 )(a2 − a3 ) . . . (a2011 − a1 )| ≤ cyc 16 2.4. Undergraduates Problem 1 For each real number t, let Pt (x) = x3 − 12x + t and let ∆(t) = max{c ∈ R | Pt (c) = 0} − min{c ∈ R | Pt (c) = 0} the difference between the largest and the smallest real roots of Pt (x). Determine the range of values that ∆(t) can assume as t varies. Problem 2 Consider a regular n-gon inscribed in the unit circle. Compute the sum of the areas of all triangles determined by the vertices of the n-gon. Problem 3 Let n be a positive integer and A a subset of Z/(n), the set of the integers modulo n, define f (A) = mint∈Z/(n) |A ∩ (A + t)|, where A + t = {x + t, x ∈ A} ⊂ Z/(n). Define g(n) = max{f (A); A ⊂ Z/(n), |A| = ⌊n/2⌋}. (a) Prove that g(n) ≤ ⌈n/4⌉ − 1, ∀n ≥ 1. (b) Prove that g(n) = ⌈n/4⌉ − 1 for infinite values of n ≥ 1. Problem 4 Consider the polynomial f (x) = x3 + x2 − 4x + 1. 6 (Problems) (page 7) Problems 7 (a) Prove that if r is a root of f (x) then r2 + r − 3 is also a root of f (x). (b) Let α, β, γ be the three roots of f (x), in some order. Determine all possible values of γ α β + + β γ β Problem 5 If u1 , . . . , uk ∈ R3 , denote by C(u1 , . . . , uk ) the cone generated by u1 , . . . , uk : C(u1 , . . . , uk ) = {a1 u1 + · · · + ak uk ; a1 , . . . , ak ∈ [0, +∞)}. Let v1 , v2 , v3 , v4 points randomly and independently chosen from the unit sphere x2 + y 2 + z 2 = 1. (a) What is the probability that C(v1 , v2 , v3 , v4 ) = R3 ? (b) What is the probability that each of the vectors is needed to generate C(v1 , v2 , v3 , v4 ), i.e., that C(v1 , v2 , v3 ) 6= C(v1 , v2 , v3 , v4 ), C(v1 , v2 , v4 ) 6= C(v1 , v2 , v3 , v4 ), C(v1 , v3 , v4 ) 6= C(v1 , v2 , v3 , v4 ) and C(v2 , v3 , v4 ) 6= C(v1 , v2 , v3 , v4 )? Problem 6 Let (xn )n≥0 be a sequence of integer numbers that fulfills a linear recursion of order k for a fixed positive integer i.e., there exists real constant numbers Pk, k c1 , c2 , . . . , ck such that xn+k = r=1 cr xn+k−r , ∀n ≥ 0. Suppose k is the minimum positive integer with this property. Prove that cj ∈ Z, for all j, 1 ≤ j ≤ k. 7 (Problems) (page 8) (Problems) (page 9) Solutions 9 3.1. Grades 6–7 Problem 1 (a) The erased numbers were 2 = 2 · 1, 4 = 2 · 2, . . ., 2010 = 2 · 1005. So 2011 − 1005 = 1006 numbers were left on the board. (b) We can list the numbers: they are 1, 11, 101, 111, 1001, 1011, 1101, 1111, a total of 8. OR we can argue that the number is of the form (abc1), where a, b, c are digits equal to either 0 or 1. Notice that the units digit must be 1. Problem 2 The bigger cube has sidelength 12 5 cm, so the difference between the side2 lengths is 12 5 − 2 = 5 cm, that is, the red cubes should not have sidelength greater than this length. Cubes with sidelength 52 cm are the natural candidates, so we set a new unit u = 25 cm. Notice that the bigger cube should have sidelength 6 u and the original cube must have sidelength 5 u. So we need 63 − 53 = 91 red cubes. Problem 3 (a) First notice that 86 is pal. Then it’s not hard to check by hand that every number from 87 to 99 is not pal. (b) The answer is yes. First consider the 2011-digit number |11 {z . . . 1}. The 2011 fives sum of its digits is 2011. The smallest perfect square greater than 2011 is 452 = 2025. Since 2025 − 2011 = 14 and 14 = 2 · (22 − 12 ) + (32 − 12 ), we can exchange two 1s by two 2s and one 1 by one 3. So we obtain the pal number |11 {z . . . 1} 223. 2008 fives Problem 4 (a) The trapezoid OABC has area ′ ′ ′ AB+OC 2 ′ · OA = 4+2 2 · 2 = 6. (b) Let A , B , C and D be the reflections of A, B, C and D across O, respectively. Because O is the center of the square, B ′ and D′ lie on the sides of the square. So the square is divided into four congruent 9 (Solutions) (page 10) 10 XXXIII Brazilian Math Olympiad 2011 (non-convex) polygons, each with area 16 − 6 = 10. 64 4 = 16. Then BCDE has area Problem 5 (a) For instance, 1 2 3 4 5 6 8 9 7 The trick is to only adjust the last row. The usual order 7, 8, 9 yields all sums to be multiple of 3, so it’s just a matter of rearranging them. (b) No, it’s not possible. First, notice that the sum of three numbers x, y, z is a multiple of 3 iff x ≡ y ≡ z (mod 3) or x, y, z are 0, 1, 2 mod 3 in some order. Let a, b, c, d be the numbers in the corner modulo 3. So two of them are equal. We can suppose wlog that they are either a = b or a = d. Also, let x be the number in the central cell modulo 3. a b x c d If a = d, then x 6= a and x is equal to either b or c. Suppose wlog x = b 6= a. Then we have the following situation: a b b c a 10 (Solutions) (page 11) Solutions 11 Let m be the other remainder (that is, m 6= a and m 6= b). Then m cannot be in the same line as a and b. This leaves only one possibility: a b m b mm a But the remaining a will necessarily yield a line with all three remainders. Now if a = b, then both c and d are different from a (otherwise, we reduce the problem to the previous case). If d 6= c, a, c, d are the three distinct remainders, and we have no possibility for x. So c = d. a a x c c But this prevents the other remainder m to appear in the middle row, leaving only two cells for three numbers, which is not possible. So, in both cases, one of the sums is a multiple of 3. 3.2. Grades 8–9 Problem 1 See problem 5.b, grades 6–7. Problem 2 Since AD = CD, AB = AC and 6 ADC = 6 BAC, triangles ADC and BAC are similar by case SAS. Segments CM and CN are corresponding CA 6 6 BCN + 6 N CA = 6 medians, so CM CN = CB and BCN = ACM ⇐⇒ 6 ACM + 6 N CA ⇐⇒ 6 BCA = 6 N CM . Thus, again by case SAS, triangles CM N and CAB are similar, and therefore CM N is an isosceles triangle with CM = M N . 11 (Solutions) (page 12) 12 XXXIII Brazilian Math Olympiad 2011 Problem 3 She can obtain the product of any two numbers a and b by asking gcd(a, b) and lcm(a, b), since lcm(a, b) · gcd(a, b) = ab. The identity lcm(a, b) · lcm(a, c) · lcm(b, c) · gcd(a, b, c) abc = lcm(a, b, c) essentially finishes the proof, since the 2011 numbers can be divided into a set of three numbers and 1004 sets of two numbers. It remains to prove the above identity. But this follows from the facts that max(x, y)+max(x, z)+max(y, z)+min(x, y, z)−max(x, y, z) = x+y+z, and if pxi k ai then pmin{xi } k gcd(a1 , a2 , . . . , an ) and pmax{xi } k lcm(a1 , a2 , . . . , an ). Problem 4 The smallest perfect square, apart from 1, is 22 = 4. So let a1 , a2 , . . . , ak be the numbers on the list modulo 4. We cannot have ai = 0; also, there is at most one ai equal to 2 and we cannot have ai = 1 and aj = 3 simultaneously. We claim that among any four distinct numbers a1 , a2 , a3 , a4 fulfilling the above properties there are three of them whose sum is a multiple of 4. Indeed, there are two equal numbers, say a1 , a2 . We cannot have a1 = a2 = 2, so either a1 = a2 = 1 or a1 = a2 = 3. We can suppose wlog a1 = a2 = 1 (otherwise, reverse the signs of all four numbers modulo 4). But since we also cannot have aj = 3 and a3 = a4 = 1, one of a3 , a4 , say a3 , is 2. But then a1 + a2 + a3 = 1 + 1 + 2 = 4. So the quantity of numbers is at most 3. 5, 13 and 17 is an example of a list with three numbers. Problem 5 2 √ = Since 216 3 64 1 2 2 3 = 21+ 3 lies between 4.5 = 20.25 and 5 and the altitude √ √ of thetriangle is 2 3·2 3 = 3, we can cover ansquare with sidelength 16 with 16 = 16, by the pigeon hole 2 · 5 · 3 = 60 equilateral triangles. Since 1000 60 principle there is an equilateral triangle that covers at least 17 points. Problem 6 We will prove by induction that odd(N ) · even(N ) < N for all positive integers N with 2k digits. If N = 10a + b, a, b ∈ {0, 1, 2, . . . , 9}, a 6= 0, N = 10a + b > a · b + b ≥ a · b = even(N ) · odd(N ). Now suppose that N has 2k > 2 digits and that the claim is true for all numbers with 2k − 2 digits. Let c and d be the two leftmost digits of N , 12 (Solutions) (page 13) Solutions 13 so that N = c · 102k−1 + d · 102k−2 + N0 , N0 with 2k − 2 digits. Then odd(N ) = d · 10k−1 + odd(N0 ) and even(N ) = c · 10k−1 + even(N0 ). So we need to prove that c · 102k−1 + d · 102k−2 + N0 > (c · 10k−1 + even(N0 )) · (d · 10k−1 + odd(N0 )) ⇐⇒ c · 102k−1 + d · 102k−2 + N0 > cd · 102k−2 + d · 10k−1 · even(N0 ) + c · 10k−1 · odd(N0 ) + odd(N0 ) · even(N0 ) But this is true, since both odd(N0 ) and even(N0 ) are less than 10k−1 and thus c · 102k−1 ≥ c(d + 1) · 102k−2 > cd · 102k−2 + c · 10k−1 · odd(N0 ) d · 102k−2 > d · 10k−1 · even(N0 ) N0 > odd(N0 ) · even(N0 ) 3.3. Grades 10–12 Problem 1 Consider the number 55 . . . 5}. The sum of the squares of its digits is n · 52 = | {z n times 25n. We can exchange any two fives by one three and one four, so the sum of the squares decreases by 52 , until we run out of fives. So we can get any sum from 25 · ⌈n/2⌉ and 25 · n. So it suffices to show that there is an integer k such that n2 ≤ k 2 ≤ n. Choose k such that k 2 ≤ n < (k + 1)2 . Suppose k 2 < n2 . Then n > 2k 2 , and (k + 1)2 > n > 2k 2 =⇒ (k + 1)2 ≥ 2k 2 + 2 ⇐⇒ k 2 − 2k + 1 ≤ 0 ⇐⇒ (k − 1)2 ≤ 0, which is false except for k = 1, or 2 < n < 4, that is, n = 3. But the statement of the problem itself gives an example with n digits: 122. Problem 2 Number the stickers from 1 to 2011 and let Si be the set of the stickers that the friend i has, 1 ≤ i ≤ 33. Since 2011 = 33·61−2, consider the example where Si = {k ∈ Z | 61(i−1) < k ≤ 61i} for i = 1, 2, . . . , 31, S32 = {k ∈ Z | 61 · 31 < k < 61 · 32} and S33 = {k ∈ Z | 61 · 32 ≤ k ≤ 2011}. Notice that |Si | = 61 for 1 ≤ i ≤ 31 and |Si | = 60 for i = 32 and i = 33. Thus |Si ∪ Sj | ≤ 2 · 61, and therefore m > 2011 − 2 · 61 = 1889. 13 (Solutions) (page 14) 14 XXXIII Brazilian Math Olympiad 2011 Now we prove that the minimum value of m is, in fact, m = 1890. First, notice that if m = 1890 then |Si ∪ Sj | ≤ 2011 − 1890 = 121. Suppose that |S1 ∪ S2 ∪ S3 | > 181. Then one of the sets, say S1 , has more than 181/3 elements, that is, |S1 | ≥ 61. But |(S1 ∪ S2 ∪ S3 ) \ (S1 ∪ S2 )| > 181 − 121 ⇐⇒ |S3 \ (S1 ∪ S2 )| > 60. But |S3 ∪ S1 | = |S3 \ S1 | + |S1 | ≥ |S3 \ (S1 ∪ S2 )| + |S1 | > 60 + 61 = 121, contradiction. Hence |S1 ∪ S2 ∪ S3 | ≤ 181. So, |S1 ∪S2 ∪. . .∪S33 | ≤ |S1 ∪S2 ∪S3 |+|S4 ∪S5 |+|S6 ∪S7 |+· · ·+|S32 +S33 | ≤ 181 + 15 · 121 = 1996, and there exists sixteen stickers that none of the 33 friends have. Another solution: We will prove that m = 1890 in another way. The example for m = 1889 is the same from the previous solution. Again, number the stickers from 1 to 2011 and let Ti be the set of the stickers that the friend i does not have, 1 ≤ i ≤ 33. Consider all pairs (x, {i, j}) such that x ∈ Ti ∩ Tj . If for every sticker there is a friend that has it, that is, T1 ∩ T2 ∩ . . . ∩ T33 = ∅ then for each x there exists k such that x ∈ / Tk . So eachx belongs to at most 32 sets Ti and, hence, there exist at most 2010 · 32 2 pairs (x,{i, j}). On the other hand, since |Ti ∩ Tj | ≥ m there exists at least m · 33 2 pairs (x, {i, j}). Therefore, if T1 ∩ T2 ∩ . . . ∩ T33 = ∅ then m· 33 2 ≤ 2011 · 32 2 ⇐⇒ m ≤ 1890 Thus, if m ≥ 1819 there exists an element x that is contained in all sets Ti . Remark: One can prove in a similar fashion that if the sticker album has n stickers and there are k friends, the minimum value of m is n−1 n − 2 n + 1 k n − 2 nk n − 2 nk − 1 if if if if k k k k ≥n < n and k | n < n and n mod k = 1 < n and n mod k > 1 Problem 3 Let’s prove that√ there exists a triangle Pj Pj+1 Pj+2 with area less than or equal to α = 5−2 5 . Suppose that all triangles Pj Pj+1 Pj+2 have area greater 14 (Solutions) (page 15) Solutions 15 than α. Let diagonals P1 P4 and P3 P5 meet at Q. Since Q ∈ P3 P5 , area P1 P2 Q ≤ max(area P1 P2 P5 , area P1 P2 P3 ) < α, so area P1 P2 P4 = 1 − area P1 P4 P5 − area P2 P3 P4 > 1 − 2α. Thus area P1 P2 Q α P1 Q = < P1 P4 area P1 P2 P4 1 − 2α We also have 1 − 2α, P1 Q P4 Q = area P1 P3 P5 area P3 P4 P5 . Since area P3 P4 P5 < α and area P1 P3 P5 > P1 Q 1 − 2α P1 Q 1 − 2α > ⇐⇒ > P4 Q α P1 P4 1−α Therefore √ √ 1 − 2α P1 Q α 5− 5 5+ 5 2 < < =⇒ 5α − 5α + 1 < 0 ⇐⇒ <α< , 1−α P1 P4 1 − 2α 10 10 contradiction. The proof of the other inequality is analogous. Problem 4 The answer is yes and you can construct an example in several ways. The main observation is that gcd(ai , aj ) = aj − ai ⇐⇒ aj − ai | ai . In fact, if gcd(ai , aj ) = aj − ai then aj − ai | ai and, conversely, if aj − ai | ai then aj − ai | ai + (aj − ai ) ⇐⇒ aj − ai | aj , so aj − ai | gcd(ai , aj ). But gcd(ai , aj ) | ai and gcd(ai , aj ) | aj implies gcd(ai , aj ) | aj − ai , so gcd(ai , aj ) = aj − ai . Once this fact is established, one can construct the sequence inductively as follows: first consider the two-term sequence (1, 2). Now, given a sequence 15 (Solutions) (page 16) 16 XXXIII Brazilian Math Olympiad 2011 (x1 , x2 , . . . , xk−1 ) with k − 1 terms such that gcd(xi , xj ) = xj − xi , construct a new sequence adding x0 to every term and putting x0 at its beginning: (x0 , x1 + x0 , x2 + x0 , . . . , xk−1 + x0 ). All we need to do is to find x0 . By the previous observation, we need xj − xi | xi + x0 and xi | x0 . We already have that xj − xi | xi , so a good choice is x0 = lcm(x1 , x2 , . . . , xk−1 ), because by definition xi | x0 and, since xi | x0 and xj − xi | xi , xj − xi | x0 , so xj − xi | xi + x0 . So we obtained a new sequence with k terms and the result follows by induction. Problem 5 By the angle bisector theorem, it suffices to prove that BF FC = BH HC . We have 6 EF B = 180◦ − 6 F EA = 180◦ − 6 F DA = 6 F DC and 6 F BE = 6 F BA = 6 F CA = 6 F CD, so triangles BEF and CDF are similar. Thus BF BE BH cos 6 EBH BH cos(90◦ − 6 BAC) BH = = = = FC CD CH cos 6 DCH CH cos(90◦ − 6 BAC) CH and the result follows. Problem 6 Q In what follows, indices are taken modulo 2011 and E = cyc (an − an+1 ). Lemma. If E is maximum, for every i ∈ {1, 2, . . . , 2011}, one of the numbers ai−1 , ai , ai+1 is zero. Proof. Suppose, by means of contradiction, that E is maximum and there exists ai such that ai−1 , ai , ai+1 are all nonzero (that is, ai−1 ai ai+1 > 0). Define A = {ai | ai > 0} and B = {ai | ai−1 ai ai+1 > 0}. Then B ⊂ A and 16 (Solutions) (page 17) Solutions 17 B 6= ∅. Let ak = min B and consider ak−1 and ak+1 . We have the following cases: • ak < ak−1 and ak < ak+1 . Let 0, ′ ai = a + a k , i |A|−1 if ai = 0 or i = k if ai > 0 and i 6= k That is, we make ak be zero and distribute it among the remaining nonzero terms. So |ai − ai+1 | remains unchanged if ai , ai+1 ∈ A and k ∈ / {i, i + 1}, or ai , ai+1 ∈ / A; increases from |ai − ai+1 | = max{ai , ai+1 } to max{ai , ai+1 } + ak / A or ai+1 ∈ / A, but not both; increases from |ak±1 − ak | = |A|−1 if ai ∈ ak if k ∈ {i, i + 1}. ak±1 − ak to ak±1 + |A|−1 • • • ak−1 < ak < ak+1 . This means that ak−1 ∈ / B, and ak ∈ B, ak−1 > 0, that is, ak−1 ∈ A \ B, which means ak−2 = 0. In this case, we enchange (ak−1 , ak ) for (a′k−1 , a′k ) = (ak−1 + ak , 0). Then |ai − ai+1 | remains unchanged for i ∈ / {k − 2, k − 1, k}; for i = k − 2 increases from |ak−2 − ak−1 | = ak−1 to |ak−2 − a′k−1 | = ak−1 + ak ; for i = k − 1 increases from |ak−1 − ak | = ak − ak−1 to |a′k−1 − a′k | = ak−1 + ak ; for i = k increases from |ak − ak+1 | = ak+1 − ak to |a′k − ak+1 | = ak+1 . ak−1 > ak > ak+1 . Analogous to the previous case. ak > ak−1 and ak > ak+1 . This means ak−1 , ak+1 ∈ A\B, that is, ak−2 = ak+2 = 0. In this case, exchange (ak−1 , ak , ak+1 ) for (a′k−1 , a′k , a′k+1 ) = (ak−1 + ak /2, 0, ak+1 + ak /2). All differences |ai − ai+1 | remain unchanged except if i ∈ {k − 2, k − 1, k, k + 1}. The only change is |(ak−2 − ak−1 )(ak−1 − ak )(ak − ak+1 )(ak+1 − ak+2 )| = ak−1 (ak − ak−1 )(ak − ak+1 )ak+1 to |(ak−2 − a′k−1 )(a′k−1 − a′k )(a′k − a′k+1 )(a′k+1 − ak+2 )| = (ak−1 + ak /2)2 (ak+1 + ak /2)2 . But (ak−1 + ak /2)2 (ak+1 + ak /2)2 = (ak−1 (ak−1 + ak ) + a2k /4)(ak+1 (ak+1 + ak ) + a2k /4) > ak−1 (ak + ak−1 )(ak + ak+1 )ak+1 > ak−1 (ak − ak−1 )(ak − ak+1 )ak+1 Since we covered all cases, the lemma holds. Now we only have groups with one or two consecutive nonzero variables. For a group (0, ak , 0), we obtain the product |(ak−1 − ak )(ak − ak+1 )| = a2k ; for a group (0, ak , ak+1 , 0), the obtain |(ak−1 − ak )(ak − ak+1 )(ak+1 − ak+2 )| = ak ak+1 |ak+1 − ak |. Notice that the groups can be interchanged, such that we can suppose wlog that all groups with two nonzero variables are contiguous. 17 (Solutions) (page 18) 18 XXXIII Brazilian Math Olympiad 2011 Lemma. If E is maximum then there is exactly one group with two nonzero variables. Suppose, that there are at least two groups of nonzero variables (0, a, b, 0) and (0, c, d, 0). By the above remark, we can suppose wlog that the groups are consecutive, that is, it’s (0, a, b, 0, c, d, 0). Exchange these variables for (0, a + b/2, 0, (b + c)/2, 0, d + c/2, 0). The product abcd|(a − b)(c − d)| is exchanged for (a + b/2)2 ((b + c)/2)2 (d + c/2)2 . But we already know that (a+b/2)2 > a|a−b|, (d+c/2)2 > d|c−d| and, by AM–GM, ((b+c)/2)2 ≥ bc. Multiplying everything yields the lemma. Combining the two lemmas, we can suppose wlog that the nonzero variables are the ones with odd indices, that is, a1 , a3 , . . . , a2011 . In this case, we obtain the product a1 a2011 |a1 − a2011 |a23 a25 . . . a22009 , and we can optimizer it locally. Let a1 + a2011 = s and suppose wlog a1 > a2011 . Let α, β be positive real numbers to be determined. By AM–GM, 1 (αa1 )(βa2011 )(a1 − a2011 ) a1 a2011 (a1 − a2011 ) = αβ 3 αa1 + βa2011 + (a1 − a2011 ) 1 ≤ αβ 3 3 1 (α + 1)a1 + (β − 1)a2011 = αβ 3 So we choose α and β such that • • we obtain s in the end, that is, α + 1 = β − 1 ⇐⇒ β − α = 2; the equality can occur, that is, αa1 = βa2011 = a1 − a2011 ⇐⇒ a2011 = (1 − α)a1 and a1 = (β + 1)a2011 , that is, 1 = (1 − α)(β + 1) ⇐⇒ −αβ = α − β = −2. Thus√−α and β are the √roots of the quadratic t2 − 2t − 2 = 0. Hence α = 3 − 1 and β = 1 + 3, and 3 (α + 1)a1 + (β − 1)a2011 1 a1 a2011 (a1 − a2011 ) ≤ αβ 3 !3 √ √ 1 3(a1 + a2011 ) 3 3 = = s 2 3 18 Now we optimize the rest. If a3 + a5 + · · · + a2009 = 2011 2 − s, 2008 2011 2008 a3 + a5 + · · · + a2009 2 −s a23 a25 . . . a22009 ≤ = 1004 1004 18 (Solutions) (page 19) Solutions 19 Now we finish the problem. Let γ be a positive real number to be determined. E = a1 a2011 |a1 − a2011 |a23 a25 . . . a22009 √ √ 2008 2008 2011 2011 3 3 3 3 2 −s 2 −s = ≤ s · (γs) 18 1004 18γ 3 1004 !2011 2011 √ −s 2 3γs + 2008 · 1004 3 ≤ 18γ 3 2011 √ 2011 3 2011 + (3γ − 2)s = 18γ 3 2011 We choose γ = 2/3, so a1 a2011 |a1 − a2011 |a23 a25 . . . a22009 √ 3 3 ≤ 16 Remark: The latter part of the problem can be solved with Calculus, but we decided to give an elementary solution. Also, the √equality occurs iff √ 3+ 3 a3 = a5 = · · · = a2009 = 1, a1 = 4 and a2011 = 3−4 3 or it is a cyclic permutation (or we reverse the order of the variables). 3.4. Undergraduates Problem 1 Let Q(x) = x3 − 12x. Then Q′ (x) = 3x2 − 12 has roots −2 and 2, and Q has a local minimum at (2, −16) and a local maximum at (−2, 16). So Q(x) = −t has three (not necessarily distinct) real roots for −16 ≤ t ≤ 16 and one real root for t < −16 and t > 16, which means that ∆(t) = 0 for t < −16 or t > 16. So from now on we consider only t ∈ [−16, 16]. Let u ≤ v ≤ w be the roots. Since Pt′ (x) = 0 ⇐⇒ x = −2 or x = 2 we have u ≤ −2 ≤ v ≤ 2 ≤ w. In particular, −2 ≤ v ≤ 2, and v can assume any value in this interval: if t = −16, v = −2 and if t = 16, v = 2. But we know that u + v + w = 0 and uv + vw + uw = −12, so u + w = −v and uw = −12 − v(u + w) = v 2 − 12, so (∆(t))2 = (w − u)2 = (w + u)2 − 4uw = 48 − 3v 2 , which lies in the range [48 − 3 · 22 , 48] = [36, 48]. So √ 2 36 ≤ (∆(t)) ≤ 48 ⇐⇒ 6 ≤ ∆(t) ≤ 4 3. √ So the range of ∆(t) is {0} ∪ [6, 4 3]. 19 (Solutions) (page 20) 20 XXXIII Brazilian Math Olympiad 2011 Problem 2 First consider a triangle ABC and its circumcenter O. Then the area of 2 ABC is R2 (sin 26 A + sin 26 B + sin 26 C). Notice that if 6 B > 90◦ then sin 26 B < 0. So the sum is equal to the sum of the areas of triangles OAi Aj with a plus sign or a minus sign, depending on the third vertex Ak of the triangle Ai Aj Ak : if Ak lies on the major arc Ai Aj then we have a plus sign; else we have a minus sign (it won’t matter if Ai Aj is a diameter, because in that case the area of OAi Aj is zero). Therefore, if Ai Aj subtend an minor arc of k · 2π n , 1 ≤ k ≤ ⌊n/2⌋, the area of the triangle OAi Aj appears with a minus sign k − 1 times and with a plus sign n − (k − 1) − 2 = n − k − 1 times. So it contributes with the sum n − k − 1 − (k − 1) = n − 2k times. 2π Considering that there are n arcs with length k · 2π n , if θ = n the required sum is ⌊n/2⌋ ⌊n/2⌋ ⌊n/2⌋ X n2 X n X (n − 2k) sin kθ = sin kθ − n k sin kθ S= 2 2 k=1 k=1 k=1 P⌊n/2⌋ P⌊n/2⌋ Consider the sums S1 (θ) = k=1 sin kθ and S2 (θ) = k=1 cos kθ =⇒ P⌊n/2⌋ n2 ′ S2 (θ) = − k=1 k sin kθ. So we want to compute 2 S1 (θ) + n · S2′ (θ). But S2 (θ) + iS1 (θ) = ⌊n/2⌋ X k=1 cos kθ + i sin kθ = ⌊n/2⌋ X k=1 ωk = ω · ω ⌊n/2⌋ − 1 ω−1 ω ⌊n/2⌋/2 − ω −⌊n/2⌋/2 = ω ⌊n/2⌋/2+1/2 ω 1/2 − ω −1/2 sin(⌊n/2⌋θ/2) (⌊n/2⌋ + 1)θ (⌊n/2⌋ + 1)θ + i sin · , = cos 2 2 sin(θ/2) 20 (Solutions) (page 21) Solutions 21 where ω = cos θ + i sin θ. So sin (⌊n/2⌋+1)θ 2 sin ⌊n/2⌋θ 2 cos(θ/2) − cos((⌊n/2⌋ + 1/2)θ) sin(θ/2) 2 sin(θ/2) cos((⌊n/2⌋ + 1/2)θ) 1 = cot(θ/2) − 2 2 sin(θ/2) (⌊n/2⌋+1)θ ⌊n/2⌋θ cos sin 2 2 sin(θ/2) + sin((⌊n/2⌋ + 1/2)θ) S2 = = sin(θ/2) 2 sin(θ/2) 1 sin((⌊n/2⌋ + 1/2)θ) = + 2 2 sin(θ/2) S1 = = Finally, S2′ (θ) n θ sin 2θ − 12 cos 2θ sin 2 sin2 (θ/2) ⌊n/2⌋ cos((⌊n/2⌋ + 1/2)θ) 1 sin(⌊n/2⌋θ) = − 2 sin(θ/2) 4 sin2 (θ/2) = 2 + 1 2 cos n 2 + 1 2 n 2 + 1 2 θ So the required sum is n2 1 cos((⌊n/2⌋ + 1/2)θ) + cot(θ/2) − 2 2 2 sin(θ/2) ⌊n/2⌋ cos((⌊n/2⌋ + 1/2)θ) 1 sin(⌊n/2⌋θ) n − 2 sin(θ/2) 4 sin2 (θ/2) If n is even, ⌊n/2⌋ = n/2 and substituting θ = 2π n the sum simplifies to n2 π cot 4 n If n is odd, ⌊n/2⌋ = (n−1)/2 and substituting θ = to π n2 cot 4 n 2π n the sum also simplifies 21 (Solutions) (page 22) 22 XXXIII Brazilian Math Olympiad 2011 So the answer is n2 π cot 4 n Problem 3 (a) Let A = {a1 , a2 , . . . , a⌊n/2⌋ }. Consider the sum n−1 X t=0 |A ∩ (A + t)| Now each element a ∈ A appears in a set A + ti |A| = ⌊n/2⌋ times: choose ti = a − ai for each i = 1, 2, . . . , ⌊n/2⌋. So n−1 X t=0 |A ∩ (A + t)| = j n k2 2 and the average of |A ∩ (A + t)| is 1 j n k2 n ≤ . n 2 4 Since |A ∩ (A + 0)| = |A| > n4 is above average, there is a t such that |A ∩ (A + t)| is below average, so lnm n − 1. f (A) ≤ |A ∩ (A + t)| < =⇒ f (A) ≤ 4 4 So g(n) ≤ n 4 (b) Let p ≡ 3 − 1. (mod 4) ne a prime, and set A = (Fp× )2 (non-zero quadratic residues modulo p) We have to show that |(A + t) ∩ A| ≥ lpm 4 − 1 for all t ∈ Fp Since this is clear for t = 0, we henceforth assum that t 6= 0. From now on, all equalities are in Fp , that is, taken modulo p. 22 (Solutions) (page 23) Solutions 23 We have that x2 ∈ A ∩ (A + t) if and only if there exists y 2 ∈ A such that x2 = y 2 + t ⇐⇒ (x − y)(x + y) = t x − y = u for some u ∈ Fp× ⇐⇒ x + y = u−1 t u + u−1 t x = 2 ⇐⇒ for some u ∈ Fp× −1 u t−u y = 2 Therefore |A ∩ (A + t)| equals the number of elements of Fp of the form x2 = u2 + 2t + u−2 t2 , 4 u ∈ Fp× , with x 6= 0 and y 6= 0, that is, u + u−1 t 6= 0 ⇐⇒ u2 6= ±t −1 u t − u 6= 0 Notice that since p ≡ 3 (mod 4) ±t ±1 t t = =± p p p p and therfore there is exactly one quadratic residue in {−t, t}. On the other hand, it u2 6= v 2 , u2 + 2t + u−2 t2 v 2 + 2t + v −2 t2 t2 t2 = ⇐⇒ u2 − v 2 = 2 − 2 ⇐⇒ u2 v 2 = t2 4 4 u v Hence, as u2 runs over the non-zero quadratic residues, with exception of ±t, we obtain each x2 ∈ A ∩ (A + t) twice (observe that the case u2 = v 2 is excluded since u2 · u2 = t2 ⇐⇒ u2 = ±t). Therefore |A ∩ (A + t)| = as required. p−1 2 −1 p − 3 lpm − 1, = = 2 4 4 23 (Solutions) (page 24) 24 XXXIII Brazilian Math Olympiad 2011 Problem 4 (a) First notice that, since r is a root, then r3 + r2 − 4r + 1 = 0 ⇐⇒ r2 + r − 3 = 1 − 1r . So we need to prove that 3 2 1 1 1 +1=0 + 1− −4 1− 1− r r r 3 3 1 2 1 4 ⇐⇒ 1 − + 2 − 3 + 1 − + 2 − 4 + + 1 = 0 r r r r r r 1 4 1 ⇐⇒ − 1 − + 2 − 3 = 0 r r r ⇐⇒ r3 + r2 − 4r + 1 = 0, which is true. (b) Iterating 1− 1r , we get 1− 1−1 1 = r 1 1−r . It’s not hard to see that r, 1− 1r = 1 and 1−r are all distinct. In fact, if r = 1 − and r3 = −1, so r2 − 4r = 0, which is not true. r−1 r So there are two possible ways of computing 1 • (α, β, γ) = r, r−1 r , 1−r : α β + β γ 1 r then r2 − r + 1 = 0, + αγ : γ r2 (r − 1)2 1 α β + + = − − β γ α r−1 r r(r − 1) r3 − (r − 1)3 − 1 3r(r − 1) = = =3 r(r − 1) r(r − 1) • 1 (α, β, γ) = r, 1−r , r−1 r : r−1 γ r α β + 2 + + = −r(r − 1) − β γ α (r − 1)2 r 6 5 4 3 −r + 3r − 3r + r − 3r2 + 3r − 1 = r4 − 2r3 + r2 Since −r6 + 3r5 − 3r4 + r3 − 3r2 + 3r − 1 = (r3 + r2 − 4r + 1) · (−r3 + 4r2 − 11r + 29) − 80r2 + 130r − 30 = −80r2 + 130r − 30 and r4 − 2r3 + r2 = (r3 + r2 − 4r − 1) · (r − 3) + 8r2 − 13r + 3 = 8r2 − 13r + 3. So α β γ −80r2 + 130r − 30 + + = = −10. β γ α 8r2 − 13r + 3 24 (Solutions) (page 25) Solutions 25 1 Another way to solve this problem is realizing that if (α, β, γ) = r, r−1 r , 1−r then the other sum is actually αβ + βγ + αγ . By Vieta’s formula, σ1 = α+β+γ = −1, σ2 = αβ + βγ + γα = −4 and σ3 = αβγ = −1. α β γ β γ α α2 β + α2 γ + β 2 α + β 2 γ + γ 2 α + γ 2 β + + + + + = β γ α α β γ αβγ (−1) · (−4) − 3(−1) σ1 σ2 − 3σ3 = = −7, = σ3 −1 so γ α β + + = −7 − 3 = −10. α β γ Problem 5 (a) The probability that the cone of the four vectors is proper is probability that the cone is all of R3 is 81 . 7 8 so the Construct a vector u12 that is normal to the plane spanned by v1 and v2 oriented so that v3 · u12 > 0. Then the half-space {w | w · u12 ≥ 0} contains the cone generated by {v1 , v2 , v3 }. It follows that if v4 · u12 > 0, the cone generated by all four vectors will be contained in the same half-space. So to keep the cone from being proper, we must assume that u4 · u12 < 0. Similarly, find u13 orthogonal to v1 and v3 with v2 ·u13 > 0 and u23 orthogonal to v2 and v3 with v1 · u13 > 0. The cone is proper – contained in a half space – if and only if at least one of the three values v4 · uij > 0. I further claim that if all three of those dot products are negative, then the cone covers all of space. If v1 , v2 and v3 are fixed, the three signs of the dot products v4 · uij are not independent. But if we average over all choices of the vectors, then −v1 occurs exactly as often as +v1 and so on. We conclude that on average the three dot products in question are negative with probability 81 . (b) Given v1 , v2 and v3 then v4 lies in the interior of the cone generated by those three if and only if v4 · uij > 0 for all three such dot products. So there is a 18 chance that v4 lies in C(v1 , v2 , v3 ). Similarly, there is a 81 chance that v2 lies in C(v1 , v3 , v4 ). These two events are disjoint: only one vector can be in the interior of a triangle of the other three. So the probability we seek is the union of four disjoint events, each of probability 18 , which gives a probability of 21 . 25 (Solutions) (page 26) 26 XXXIII Brazilian Math Olympiad 2011 Problem 6 (i) Let (an ) be the sequence of integers obtained by shifting an by i positions, (i) i.e., an = an+i . Then (1) (k) (a(0) n ), (an ), . . . , (an ) is a basis for the k-dimensional C-vector space of sequences (bn ) satisfying bn+k = ck−1 bn+k−1 + · · · + c0 bn (n ≥ 0) (0) (∗) (k−1) In fact, any non-trivial C-linear relation among (an ), . . ., (an imply that k is not minimal. ) would Now let f (x) = xk − ck−1 xk−1 − · · · − c0 and α1 , . . . , αk be the roots of f (x) (listed with multiplicity). First, observe that all coefficients ci are rational since they are solutions to the linear system with integer coefficients ak c0 a0 a1 . . . ak−1 a2 . . . ak c1 ak+1 a1 . = . . . .. . .. .. .. . .. .. ck−1 ak−1 ak . . . a2k−2 a2k−1 (i) (the matrix is non-singular since the (an )’s are linear independent.) Therefore the αi ’s are algebraic numbers. Define tn = α1n + α2n + · · · + αkn (n ≥ 0) Then all tn ∈ Q (they are symmetric expression on the roots of f (x) ∈ Q[x]) and the tn satisfy (∗), hence there are ri such that (1) (k−1) tn = r0 a(0) n + r1 an + · · · + rk−1 an (i) Clearly ri ∈ Q since tn , an ∈ Q for all n. To sum up, if d > 0 is an integer such that dri ∈ Z, i = 0, 1, . . . , k − 1, then dtn ∈ Z 26 (Solutions) (page 27) Solutions 27 for all n. Next, we show that this implies that the αi are algebraic integers, which in turn implies that all the ci are integers, finishing the problem. By Newton’s identities, we may write the elementary symmetric polynomials in αn as polynomials with rational coefficients of tn = α1n + · · · + αkn t2n = α12n + · · · + αk2n .. . tkn = α1kn + · · · + αkkn Therefore the minimal polynomials of αin over Q have coefficients with bounded denominators, independent of n (they depend only on d and k). Hence there exists an integer ∆ > 0 such that ∆ · αin are algebraic integers for all n. But that implies that the ring Z[αi ] is contained in the finitely generated Z-module ∆1 OQ(αi ) , where OQ(αi ) is the ring of algebraic integers in the field Q(αi ). Therefore Z[αi ] itself is finitely generated as a Z-module (use the fact that Z is noetherian or that OQ(αi ) is a free Z-module, together with the structure theorem of finitely generated modules over a PID). Hence αi is an algebraic integer, as required. Alternatively, suppose that αi is not an algebraic integer, so that it has prime ideal factorization (αi ) = pe1i . . . pess with ei < 0 for some i, say e1 . Then nes 1 (∆αin ) = (∆) · pne 1 . . . ps would have a negative p1 -exponent for n sufficiently large, contradicting the fact that ∆ · αin is an algebraic integer for all n ≥ 0. 27 (Solutions) (page 28) (Solutions) (page 29) Winners in 2011 29 4.1. Grades 6–7 Gold medals Pedro Henrique Sacramento de Oliveira Rogerio Aristida Guimarães Junior Mateus Siqueira Thimóteo William Hideki Kondo Bruna Malvar Castello Branco Nathan Bonetti Teodoro Silver medals Mariana Miwa Okuma Miyashiro Lucas dos Anjos Dantas Teixeira Maria Júlia Costa Medeiros Mateus Pereira Carolina Carvalho Silva Laura Mello D’Urso Vianna Henrique Gontijo Chiari Nicolas Wolaniuk do Amaral Carvalho Lucas Diniz Gonçalves Villas Boas Bronze medals Leonardo de Matos Fellippetti Mariano Lúcia Verônica Copque Aguiar de Souza Adriana Mayumi Shiguihara Daniel Akira Hasimoto Rodrigo Gonçalves Correa César Ricardo Silva Filippi Marina Maciel Ansanelli Henrique Bittencourt Netto Monteiro Julia Perdigão Saltiel Jonathan Pereira Maria Lucas Tokio Kawahara Sandra Ayumi Nihama João Guilherme Madeira Araújo Andrey Jhen Shan Chen 29 (Winners in 2011 (page 30) 30 XXXIII Brazilian Math Olympiad 2011 Bruno Brasil Meinhart Daniel Quintão de Moraes Diene Xie Honorable mention Felipe Reyel Feitosa Henrique Corato Zanarella Alı́cia Fortes Machado André Yuji Hisatsuga Bernardo Puetter Schaeffer Bruno Teixeira Gomes Eduardo Reis Cavalcante de Farias Bruno Vinicius da Silva Alves Adriano Henrique de C. A. e Silva Fernando Seiji B. dos Santos Alba Clara Vasconcelos Leopoldo Feitosa Bruno Kenzo Ozaki Eduardo Lennert Rammé Iara Rohn Kombrink Davies Victor Alves Benevides Samuel Sena Galvão Vitor Thomaz da Cruz Francisco Bruno Dias Ribeira da Silva Bryan Diniz Borck Jonathan Aires Pinheiro Nicolas Meira Sinott Lopes Rafael Tchen Yin Hang Wei João Alberto Moreira Serôdio Loc Dominguez Vinı́cius Soares de Abreu Silva Breno Maia Baptista Luı́sa Höller Lee Brendon Diniz Borck Eduardo Emilio Costa Trunci Bernardo Gabriele Collaço Lucas Hideki Takeuchi Okamura Plinio Melo Guimarães Valério Rodrigo Vieira Casanova Monteiro Victor M. K. Tsuda 30 (Winners in 2011 (page 31) Winners in 2011 31 Arthur Henrique Craveiro Costa Pedro Orii Antonácio Gabriel Moura Braúna Victória Santos Duarte Ramos Ítalo Rennan Lima Amanda Barbosa Schirmbeck Thiago Ferreira Teixeira Gabriel Dante Cawamura Seppelfelt Lucas Siqueira Aragão Milena Delarete Drummond Marques Rodrigo Moutinho Faria Daniel Lopes de Castro João Vitor Vaz Oliveira Matheus Bevilacqua 4.2. Grades 8–9 Gold medals Alessandro A. P. de Oliveira Pacanowski Gabriel Fazoli Domingos Daniel Santana Rocha Vitor Dias Gomes Barrios Marin Luı́ze Mello DUrso Vianna Silver medals Daniel Lima Braga Fábio da Silva Soares João Pedro Sedeu Godoi Murilo Corato Zanarella Bruno Eidi Nishimoto Mariana Teatini Ribeiro Samuel Brasil de Albuquerque Lucas Mioranci Mateus Bezrutchka Ana Karoline Borges Carneiro Ana Emı́lia Hernandes Dib Bronze medals Pedro Henrique Alencar Costa 31 (Winners in 2011 (page 32) 32 XXXIII Brazilian Math Olympiad 2011 Pedro Augusto Brasileiro Lins Barbosa Gabriel Mayrink Verdun Leonardo Santos Matiello Matheus Cariús Castro Lucca Morais de Arruda Siaudzionis Luiz Cláudio Sampaio Ramos Matheus Carioca Sampaio José Wanderclesson Nobre Damasceno Filho Suzane Eberhart Ribeiro da Silva Estevão Waldow Erika Rizzo Aquino Pedro Jorge Luz Alves Cronemberger Alexandre Mendonça Cardoso Ricardo Ken Wang Tsuzuki Honorable mention Leonardo Alves Ramalho Ana Paula Lopes Schuch Flávia Nakazato Hokama Lucas Bastos Germano Helena Veronique Rios Isabelle Ferreira de Oliveira Rafael Wilton Barboza Coracini Eduardo Serpa Giovana Sachett Maia Paulo Henrique Omena de Freitas Amanda Vidotto Cerqueira Bruno Cicone de Almeida Gabriel Picanço Costa Guilherme Anitele Silva Mateus Arraes Feitosa Borges Rodrigo Zanette de Magalhães Luis Eduardo de Sousa Lima Gabriel Vidigal de Paula Santos Bruno Almeida Costa João Baptista de Paula e Silva Gabriel Ribeiro Barbosa Kevin Eiji Inashita Dimas Macedo de Albuquerque 32 (Winners in 2011 (page 33) Winners in 2011 33 Mauricio Najjar da Silveira Juliano Petry Pesarico Bruna Caroline Pimentel Gonçalves Gustavo Torres da Silva Artur Corassa Martins Italo Lesione de Paiva Rocha Nathan Antonio de Azevedo Milagres Juliana Amoedo Amoedo Plácido Victória Moreira Reis Cogo Leandro Alves Cordeiro Rômulo Gabriel Lima da Costa Bruno Vasconcelos Silva Alexandro Vı́tor Serafim de Carvalho Cristhian Mafalda Douglas Matos Gomes Gabriel Diniz Vieira e Sousa Enrico Pascucci Löffel Ricardo Vidal Mota Peixoto 4.3. Grades 10–12 Gold medals João Lucas Camelo Sá Henrique Gasparini Fiuza do Nascimento Rafael Kazuhiro Miyazaki André Macieira Braga Costa Rodrigo Sanches Angelo Maria Clara Mendes Silva Silver medals Victor de Oliveira Bitaraes Tadeu Pires de Matos Belfort Neto Rafael Rodrigues Rocha de Melo Gustavo Haddad Francisco e Sampaio Braga Daniel Eiti Nishida Kawai Henrique Vieira G. Vaz Carlos Henrique de Andrade Silva Victor Hugo Corrêa Rodrigues Franco Matheus de Alencar Severo 33 (Winners in 2011 (page 34) 34 XXXIII Brazilian Math Olympiad 2011 Gabriel Ilharco Magalhães Bronze medals Lucas Lourenço Hernandes Ivan Tadeu Ferreira Antunes Filho Kayo de França Gurgel Michel Rozenberg Zelazny Alexandre Perozim de Faveri Davi Coelho Amorim Marcos Massayuki Kawakami Daniel dos Santos Bossle Gabriel Militão Vinhas Lopes Mateus Henrique Ramos de Souza Victor Venturi Ramon Silva de Lima Gabriel José Moreira da Costa Silva Otávio Augusto de Oliveira Mendes Marcelo Luiz Gonçalves Honorable mention Artur Dubeux Duarte Natan Lima Viana Bruno Silva Mucciaccia Juliana Lemes Arai Matheus Henrique Alves Moura Felipe Sampaio Lima Davi Sampaio de Alencar Pedro Morais de Arruda Siaudzionis Luiz Castelo Branco Cavalcante Glauber de Lima Guarinello Victor Oliveira Reis José Ney Alves Feitosa Neto Andre Bandeira Pinheiro Fernando Lima Saraiva Filho Rafael Tedeschi Eugênio Pontes Barone Vinı́cius Canto Costa Lincoln de Queiroz Vieira Thiago Poeiras Silva André Amaral de Souza 34 (Winners in 2011 (page 35) Winners in 2011 35 Carlos Alexandre Silva dos Santos Felipe Viana Sousa Liara Guinsberg Otavio Araújo de Aguiar Rodolfo Rodrigues da Costa Caı́que Porto Lira Kelvin Azevedo Santos Eric Tada de Souza Marcelo Cargnelutti Rossato Marina Pessoa Mota 4.4. Undergraduates Gold medals Renan Henrique Finder Rafael Tupynambá Dutra Régis Prado Barbosa Guilherme Rodrigues Nogueira de Souza Davi Lopes Alves de Medeiros Silver medals Gabriel Luis Mello Dalalio Felipe Gonçalves Assis Darcy Gabriel Augusto de Camargo Cunha Hugo Fonseca Araújo Matheus Secco Torres da Silva Erik Fernando de Amorim Lucas Colucci Cavalcante de Souza Bronze medals José Leandro Pinheiro Reinan Ribeiro Souza Santos Daniel de Barros Soares Rafael Endlich Pimentel Carlos Henrique Melo Souza Ivan Guilhon Mitoso Rocha Thiago Ribeiro Ramos Lucas de Freitas Smaira Paulo Sérgio de Castro Moreira 35 (Winners in 2011 (page 36) 36 XXXIII Brazilian Math Olympiad 2011 Alexandre Azevedo César Ricardo Turolla Bortolotti Robério Soares Nunes Marcelo Matheus Gauy Charles Barbosa de Macedo Brito Honorable mention Renato Dias Costa Felipe Vincent Yannik Romero Pereira Rafael Alves da Ponte Carlos Coelho Lechner João Fernando Doriguello Diniz Luiz Filipe Martins Ramos Guilherme de Sena Brandine Bruno de Nadai Sarnaglia Iuri Rezende Souza Pedro Veras Bezerra da Silva Douglas Machado dos Santos Leandro Farias Maia Willy George do Amaral Petrenko Cassio Henrique Vieira Morais Michel Faleiros Martins José Armando Barbosa Filho Alysson Espı́ndola de Sá Silveira Fernando Nascimento Coelho Gabriel Caser Brito Fernando Fonseca Andrade Oliveira Breno Vieira de Aguiar Thales Graça Athanásio Tiago Fonseca Gabriel Queiroz de Brito Melo Rafael Pereira de Paula de Lucas Simon 36 (Winners in 2011