XXXIII Brazilian Math Olympiad 2011

(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 Científico e Tecnológico CNPq Instituto do Milênio Avanço Global e Integrado da Matemática Brasileira Comissão Nacional de Olimpíadas 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: obm@impa.br Chair: Carlos Gustavo Tamm de Araújo Moreira, Onofre Campos da Silva Farias Members: Antonio Caminha, Francisco Bruno Holanda, Carlos Yuzo Shine, Cícero 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, Fabrício 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)

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(page 3) 3 Problems 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 a cube with volume ( ) 12 3 5 cm? Problem 3 Wecallanumberpalifitdoesn thaveazerodigitandthesumofthesquares of the digits is a perfect square. For example, 2115522 is pal (because 2 2 +1 2 +1 2 +5 2 +5 2 +2 2 +2 2 = 8 2 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 Inthediagram, O isthecenterofthesquare, 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 cm 2. 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 ADC = CAB. Let M and N be the midpoints of AD and AB. Prove that triangle MNC 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 inthelistandanysumofanyquantityofdistinctnumbersfromthelistwere 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 P 1 P 2 P 3 P 4 P 5 with area 1, there are indices i and j (assume P 6 = P 1 and P 7 = P 2 ) such that: area P i P i+1 P i+2 5 5 10 area P j P j+1 P j+2 5 (Problems)

(page 6) 6 XXXIII Brazilian Math Olympiad 2011 Problem 4 Do there exist 2011 positive integers a 1 < a 2 <... < a 2011 such that gcd(a i,a j ) = a j a i 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 A. Prove that the angle bisectors of BFC and BHC concur at a point on line BC. Problem 6 Let a 1,a 2,...,a 2011 be nonnegative reals with sum 2011 2. Prove that (a n a n+1 ) = (a 1 a 2 )(a 2 a 3 )...(a 2011 a 1 ) 3 3 16. cyc 2.4. Undergraduates Problem 1 For each real number t, let P t (x) = x 3 12x+t and let (t) = max{c R P t (c) = 0} min{c R P t (c) = 0} the difference between the largest and the smallest real roots of P t (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) = min t 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) = x 3 +x 2 4x+1. 6 (Problems)

(page 7) Problems 7 (a) Prove that if r is a root of f(x) then r 2 +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 α β + β γ + γ β Problem 5 If u 1,...,u k R 3, denote by C(u 1,...,u k ) the cone generated by u 1,...,u k : C(u 1,...,u k ) = {a 1 u 1 + +a k u k ;a 1,...,a k [0,+ )}. Let v 1,v 2,v 3,v 4 points randomly and independently chosen from the unit sphere x 2 +y 2 +z 2 = 1. (a) What is the probability that C(v 1,v 2,v 3,v 4 ) = R 3? (b) What is the probability that each of the vectors is needed to generate C(v 1,v 2,v 3,v 4 ), i.e., that C(v 1,v 2,v 3 ) C(v 1,v 2,v 3,v 4 ), C(v 1,v 2,v 4 ) C(v 1,v 2,v 3,v 4 ), C(v 1,v 3,v 4 ) C(v 1,v 2,v 3,v 4 ) and C(v 2,v 3,v 4 ) C(v 1,v 2,v 3,v 4 )? Problem 6 Let(x n ) n 0 beasequenceofintegernumbersthatfulfillsalinearrecursionof order k for a fixed positive integer k, i.e., there exists real constant numbers c 1,c 2,...,c k such that x n+k = k r=1 c rx n+k r, n 0. Suppose k is the minimum positive integer with this property. Prove that c j Z, for all j, 1 j k. 7 (Problems)

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(page 9) 9 Solutions 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 sidelengths is 12 5 2 = 2 5 cm, that is, the red cubes should not have sidelength greater than this length. Cubes with sidelength 2 5 cm are the natural candidates, so we set a new unit u = 2 5 cm. Notice that the bigger cube should have sidelength 6u and the original cube must have sidelength 5u. So we need 6 3 5 3 = 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...1 {{ }. The 2011 fives sum of its digits is 2011. The smallest perfect square greater than 2011 is 45 2 = 2025. Since 2025 2011 = 14 and 14 = 2 (2 2 1 2 )+(3 2 1 2 ), we can exchange two 1s by two 2s and one 1 by one 3. So we obtain the pal number } 11...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 64 4 16 6 = 10. = 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 c If a = d, then x a and x is equal to either b or c. Suppose wlog x = b a. Then we have the following situation: x b d a c b b a 10 (Solutions)

(page 11) Solutions 11 Let m be the other remainder (that is, m a and m b). Then m cannot be in the same line as a and b. This leaves only one possibility: a b m b m m 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 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 ADC = BAC, triangles ADC and BAC are similar by case SAS. Segments CM and CN are corresponding medians, so CM CN = CA CB and BCN = ACM BCN + NCA = ACM + NCA BCA = NCM. Thus, again by case SAS, triangles CM N and CAB are similar, and therefore CM N is an isosceles triangle with CM = MN. 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 abc = lcm(a,b) lcm(a,c) lcm(b,c) gcd(a,b,c) 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 p x i a i then p min{x i} gcd(a 1,a 2,...,a n ) and p max{x i} lcm(a 1,a 2,...,a n ). Problem 4 The smallest perfect square, apart from 1, is 2 2 = 4. So let a 1,a 2,...,a k be the numbers on the list modulo 4. We cannot have a i = 0; also, there is at most one a i equal to 2 and we cannot have a i = 1 and a j = 3 simultaneously. We claim that among any four distinct numbers a 1,a 2,a 3,a 4 fulfilling the above properties there are three of them whose sum is a multiple of 4. Indeed, there are two equal numbers, say a 1,a 2. We cannot have a 1 = a 2 = 2, so either a 1 = a 2 = 1 or a 1 = a 2 = 3. We can suppose wlog a 1 = a 2 = 1 (otherwise, reverse the signs of all four numbers modulo 4). But since we also cannot have a j = 3 and a 3 = a 4 = 1, one of a 3,a 4, say a 3, is 2. But then a 1 +a 2 +a 3 = 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 16 Since 2 3 = 64 3 = 21+1 3 liesbetween4.52 = 20.25and5 2 andthealtitude of the triangle is 2 3 3 2 = 3, we can cover an square with sidelength 16 with 2 5 16 3 = 60 equilateral triangles. Since 1000 60 = 16, by the pigeon hole 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 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 10 2k 1 + d 10 2k 2 + N 0, N 0 with 2k 2 digits. Then odd(n) = d 10 k 1 +odd(n 0 ) and even(n) = c 10 k 1 +even(n 0 ). So we need to prove that c 10 2k 1 +d 10 2k 2 +N 0 >(c 10 k 1 +even(n 0 )) (d 10 k 1 +odd(n 0 )) c 10 2k 1 +d 10 2k 2 +N 0 > cd 10 2k 2 +d 10 k 1 even(n 0 )+c 10 k 1 odd(n 0 )+odd(n 0 ) even(n 0 ) But this is true, since both odd(n 0 ) and even(n 0 ) are less than 10 k 1 and thus c 10 2k 1 c(d+1) 10 2k 2 > cd 10 2k 2 +c 10 k 1 odd(n 0 ) d 10 2k 2 > d 10 k 1 even(n 0 ) N 0 > odd(n 0 ) even(n 0 ) 3.3. Grades 10 12 Problem 1 Consider the number 55...5 } {{ }. The sum of the squares of its digits is n 5 2 = n times 25n. We can exchange any two fives by one three and one four, so the sum of the squares decreases by 5 2, 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 n 2 k2 n. Choose k such that k 2 n < (k +1) 2. Suppose k 2 < n 2. Then n > 2k2, 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 S i be the set of the stickers that the friend i has, 1 i 33. Since2011 = 33 61 2, considertheexamplewheres i = {k Z 61(i 1) < k 61i} for i = 1,2,...,31, S 32 = {k Z 61 31 < k < 61 32} and S 33 = {k Z 61 32 k 2011}. Notice that S i = 61 for 1 i 31 and S i = 60 for i = 32 and i = 33. Thus S i S j 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 S i S j 2011 1890 = 121. Suppose that S 1 S 2 S 3 > 181. Then one of the sets, say S 1, has more than 181/3 elements, that is, S 1 61. But (S 1 S 2 S 3 )\(S 1 S 2 ) > 181 121 S 3 \(S 1 S 2 ) > 60. But S 3 S 1 = S 3 \S 1 + S 1 S 3 \(S 1 S 2 ) + S 1 > 60+61 = 121, contradiction. Hence S 1 S 2 S 3 181. So, S 1 S 2... S 33 S 1 S 2 S 3 + S 4 S 5 + S 6 S 7 + + S 32 +S 33 181+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, numberthestickersfrom1to2011andlett i bethesetofthestickers that the friend i does not have, 1 i 33. Consider all pairs (x,{i,j}) such that x T i T j. If for every sticker there is a friend that has it, that is, T 1 T 2... T 33 = then for each x there exists k such that x / T k. So each x belongs to at most 32 sets T i and, hence, there exist at most 2010 (32 2) pairs (x,{i,j}). On the other hand, since T i T j m there exists at least m (33 2) pairs (x,{i,j}). Therefore, if T 1 T 2... T 33 = then m ( ) 33 2011 2 ( ) 32 2 m 1890 Thus, if m 1819 there exists an element x that is contained in all sets T i. 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 if k n n 2 n k +1 if k < n and k n n 2 n k if k < n and n mod k = 1 n 2 n k 1 if k < n and n mod k > 1 Problem 3 Let s prove that there exists a triangle P j P j+1 P j+2 with area less than or equal to α = 5 5 2. Suppose that all triangles P j P j+1 P j+2 have area greater 14 (Solutions)

(page 15) Solutions 15 than α. Let diagonals P 1 P 4 and P 3 P 5 meet at Q. Since Q P 3 P 5, areap 1 P 2 Q max(areap 1 P 2 P 5,areaP 1 P 2 P 3 ) < α, so areap 1 P 2 P 4 = 1 areap 1 P 4 P 5 areap 2 P 3 P 4 > 1 2α. Thus We also have P 1Q 1 2α, Therefore P 1 Q = areap 1P 2 Q < α P 1 P 4 areap 1 P 2 P 4 1 2α P 4 Q = areap 1P 3 P 5 1 2α 1 α < P 1Q P 1 P 4 < α contradiction. areap 3 P 4 P 5. Since areap 3 P 4 P 5 < α and areap 1 P 3 P 5 > P 1 Q P 4 Q > 1 2α α P 1Q > 1 2α P 1 P 4 1 α 1 2α = 5α2 5α+1<0 5 5 10 The proof of the other inequality is analogous. Problem 4 <α< 5+ 5, 10 The answer is yes and you can construct an example in several ways. The main observation is that gcd(a i,a j ) = a j a i a j a i a i. In fact, if gcd(a i,a j ) = a j a i then a j a i a i and, conversely, if a j a i a i then a j a i a i + (a j a i ) a j a i a j, so a j a i gcd(a i,a j ). But gcd(a i,a j ) a i and gcd(a i,a j ) a j implies gcd(a i,a j ) a j a i, so gcd(a i,a j ) = a j a i. 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 (x 1,x 2,...,x k 1 ) with k 1 terms such that gcd(x i,x j ) = x j x i, construct a new sequence adding x 0 to every term and putting x 0 at its beginning: (x 0,x 1 +x 0,x 2 +x 0,...,x k 1 +x 0 ). All we need to do is to find x 0. By the previous observation, we need x j x i x i +x 0 and x i x 0. We already have that x j x i x i, so a good choice is x 0 = lcm(x 1,x 2,...,x k 1 ), because by definition x i x 0 and, since x i x 0 and x j x i x i, x j x i x 0, so x j x i x i +x 0. 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 EFB = 180 FEA = 180 FDA = FDC and FBE = FBA = FCA = FCD, so triangles BEF and CDF are similar. Thus BF FC = BE CD = BHcos EBH CHcos DCH = BHcos(90 BAC) CHcos(90 BAC) = BH CH and the result follows. Problem 6 In what follows, indices are taken modulo 2011 and E = cyc (a n a n+1 ). Lemma. If E is maximum, for every i {1,2,...,2011}, one of the numbers a i 1,a i,a i+1 is zero. Proof. Suppose, by means of contradiction, that E is maximum and there exists a i such that a i 1,a i,a i+1 are all nonzero (that is, a i 1 a i a i+1 > 0). Define A = {a i a i > 0} and B = {a i a i 1 a i a i+1 > 0}. Then B A and 16 (Solutions)

(page 17) Solutions 17 B. Let a k = minb and consider a k 1 and a k+1. We have the following cases: a k < a k 1 and a k < a k+1. Let { 0, if ai a = 0 or i = k i = a i + a k A 1, if a i > 0 and i k That is, we make a k be zero and distribute it among the remaining nonzero terms. So a i a i+1 remains unchanged if a i,a i+1 A and k / {i,i+1}, or a i,a i+1 / A; increases from a i a i+1 = max{a i,a i+1 } to max{a i,a i+1 } + a k A 1 if a i / A or a i+1 / A, but not both; increases from a k±1 a k = a k±1 a k to a k±1 + a k A 1 if k {i,i+1}. a k 1 < a k < a k+1. This means that a k 1 / B, and a k B, a k 1 > 0, that is, a k 1 A \ B, which means a k 2 = 0. In this case, we enchange (a k 1,a k ) for (a k 1,a k ) = (a k 1 + a k,0). Then a i a i+1 remains unchanged for i / {k 2,k 1,k}; for i = k 2 increases from a k 2 a k 1 = a k 1 to a k 2 a k 1 = a k 1 + a k ; for i = k 1 increases from a k 1 a k = a k a k 1 to a k 1 a k = a k 1 + a k ; for i = k increases from a k a k+1 = a k+1 a k to a k a k+1 = a k+1. a k 1 > a k > a k+1. Analogous to the previous case. a k > a k 1 and a k > a k+1. This means a k 1,a k+1 A\B, that is, a k 2 = a k+2 = 0. In this case, exchange (a k 1,a k,a k+1 ) for (a k 1,a k,a k+1 ) = (a k 1 + a k /2,0,a k+1 + a k /2). All differences a i a i+1 remain unchanged except if i {k 2,k 1,k,k + 1}. The only change is (a k 2 a k 1 )(a k 1 a k )(a k a k+1 )(a k+1 a k+2 ) = a k 1 (a k a k 1 )(a k a k+1 )a k+1 to (a k 2 a k 1 )(a k 1 a k )(a k a k+1 )(a k+1 a k+2) = (a k 1 + a k /2) 2 (a k+1 +a k /2) 2. But (a k 1 +a k /2) 2 (a k+1 +a k /2) 2 = (a k 1 (a k 1 +a k )+a 2 k/4)(a k+1 (a k+1 +a k )+a 2 k/4) > a k 1 (a k +a k 1 )(a k +a k+1 )a k+1 > a k 1 (a k a k 1 )(a k a k+1 )a k+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,a k,0), we obtain the product (a k 1 a k )(a k a k+1 ) = a 2 k ; for a group (0,a k,a k+1,0), the obtain (a k 1 a k )(a k a k+1 )(a k+1 a k+2 ) = a k a k+1 a k+1 a k. 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. IfE ismaximumthenthereisexactlyonegroupwithtwononzero 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 aretheoneswithoddindices, thatis, a 1,a 3,...,a 2011. Inthiscase, weobtain the product a 1 a 2011 a 1 a 2011 a 2 3a 2 5...a 2 2009, and we can optimizer it locally. Let a 1 + a 2011 = s and suppose wlog a 1 > a 2011. Let α,β be positive real numbers to be determined. By AM GM, a 1 a 2011 (a 1 a 2011 ) = 1 αβ (αa 1)(βa 2011 )(a 1 a 2011 ) 1 αβ = 1 αβ So we choose α and β such that ( αa1 +βa 2011 +(a 1 a 2011 ) ( (α+1)a1 +(β 1)a 2011 we obtain s in the end, that is, α+1 = β 1 β α = 2; the equality can occur, that is, αa 1 = βa 2011 = a 1 a 2011 a 2011 = (1 α)a 1 and a 1 = (β+1)a 2011, that is, 1 = (1 α)(β+1) αβ = α β = 2. Thus α and β are the roots of the quadratic t 2 2t 2 = 0. Hence α = 3 1 and β = 1+ 3, and a 1 a 2011 (a 1 a 2011 ) 1 ( ) 3 (α+1)a1 +(β 1)a 2011 αβ 3 ( ) 3 = 1 3(a1 +a 2011 ) 3 = 2 3 18 s3 Now we optimize the rest. If a 3 +a 5 + +a 2009 = 2011 2 s, ( ) 2008 ( a 2 3a 2 5...a 2 a3 +a 5 + +a 2011 2009 2 s 2009 = 1004 1004 3 3 ) 3 ) 3 ) 2008 18 (Solutions)

(page 19) Solutions 19 Now we finish the problem. Let γ be a positive real number to be determined. E = a 1 a 2011 a 1 a 2011 a 2 3a 2 5...a 2 2009 ( 3 2011 ) 2008 18 s3 2 s 3 = 1004 18γ 3(γs)3 ( 2011 ) 3 3γs+2008 2 s 2011 1004 18γ 3 2011 ( ) 2011 3 2011+(3γ 2)s = 18γ 3 2011 We choose γ = 2/3, so ( 2011 a 1 a 2011 a 1 a 2011 a 2 3a 2 5...a 2 2009 3 3 16 2 s 1004 ) 2008 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 a 3 = a 5 = = a 2009 = 1, a 1 = 3+ 3 4 and a 2011 = 3 3 4 or it is a cyclic permutation (or we reverse the order of the variables). 3.4. Undergraduates Problem 1 Let Q(x) = x 3 12x. Then Q (x) = 3x 2 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 P t(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 2 2,48] = [36,48]. So 36 ( (t)) 2 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 ABC is R2 2 (sin2 A + sin2 B + sin2 C). Notice that if B > 90 then sin2 B < 0. So the sum is equal to the sum of the areas of triangles OA i A j with a plus sign or a minus sign, depending on the third vertex A k of the triangle A i A j A k : if A k lies on the major arc A i A j then we have a plus sign; else we have a minus sign (it won t matter if A i A j is a diameter, because in that case the area of OA i A j is zero). Therefore, if A i A j subtend an minor arc of k 2π n, 1 k n/2, the area of the triangle OA i A j 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. Considering that there are n arcs with length k 2π n, if θ = 2π n the required sum is S = n n/2 (n 2k)sinkθ = n2 n/2 n/2 sinkθ n ksinkθ 2 2 k=1 Consider the sums S 1 (θ) = n/2 k=1 sinkθ and S 2(θ) = n/2 k=1 coskθ = S 2(θ) = n/2 n2 k=1 ksinkθ. So we want to compute 2 S 1(θ)+n S 2(θ). But S 2 (θ)+is 1 (θ) = n/2 k=1 coskθ +isinkθ = n/2 k=1 k=1 ω k = ω ω n/2 1 ω 1 = ω n/2 /2+1/2ω n/2 /2 ω n/2 /2 ω 1/2 ω ( ( ) 1/2 ( )) ( n/2 +1)θ ( n/2 +1)θ = cos +isin 2 2 k=1 sin( n/2 θ/2), sin(θ/2) 20 (Solutions)

(page 21) Solutions 21 where ω = cosθ +isinθ. So Finally, sin S 1 = ( ) ( n/2 +1)θ 2 sin sin(θ/2) ( ) n/2 θ 2 = cos(θ/2) cos(( n/2 +1/2)θ) 2sin(θ/2) = 1 cos(( n/2 +1/2)θ) cot(θ/2) 2 2sin(θ/2) ( ) ( ) cos ( n/2 +1)θ 2 sin n/2 θ 2 S 2 = = sin(θ/2)+sin(( n/2 +1/2)θ) sin(θ/2) 2sin(θ/2) = 1 2 + sin(( n/2 +1/2)θ) 2sin(θ/2) S 2(θ) ( n (( 2 + 1 = 2) cos n ) ( 2 + 1 2) θ sin θ ) 2 1 2 cos( (( θ 2) sin n ) ) 2 + 1 2 θ 2sin 2 (θ/2) = n/2 cos(( n/2 + 1/2)θ) 1 sin( n/2 θ) 2 sin(θ/2) 4 sin 2 (θ/2) So the required sum is n 2 ( 1 2 2 cos(( n/2 +1/2)θ) cot(θ/2) 2sin(θ/2) ( n/2 n 2 ) + cos(( n/2 + 1/2)θ) sin(θ/2) 1 4 ) sin( n/2 θ) sin 2 (θ/2) If n is even, n/2 = n/2 and substituting θ = 2π n the sum simplifies to n 2 4 cot π n Ifnisodd, n/2 = (n 1)/2andsubstitutingθ = 2π n thesumalsosimplifies to n 2 4 cot π n 21 (Solutions)

(page 22) 22 XXXIII Brazilian Math Olympiad 2011 So the answer is n 2 4 cot π n Problem 3 (a) Let A = {a 1,a 2,...,a n/2 }. Consider the sum n 1 A (A+t) t=0 Now each element a A appears in a set A+t i A = n/2 times: choose t i = a a i for each i = 1,2,..., n/2. So n 1 ( n 2 A (A+t) = 2 ) t=0 and the average of A (A+t) is 1 ( n ) 2 n n 2 4. Since A (A + 0) = A > n 4 is above average, there is a t such that A (A+t) is below average, so f(a) A (A+t) < n n 4 = f(a) 1. 4 So g(n) n 4 1. (b) Let p 3 (mod 4) ne a prime, and set We have to show that A = (F p ) 2 (non-zero quadratic residues modulo p) (A+t) A p 4 1 for all t F p Since this is clear for t = 0, we henceforth assum that t 0. From now on, all equalities are in F p, that is, taken modulo p. 22 (Solutions)