XXXIII Brazilian Math Olympiad 2011 - OBM

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XXXIIIBrazilian Math Olympiad

2011

Editora AOBM

Rio de Janeiro

2012

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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: [email protected]

Chair: Carlos Gustavo Tamm de Araujo Moreira, Onofre Campos da Silva Farias

Members: Antonio Caminha, Francisco Bruno Holanda, Carlos Yuzo Shine, Cıcero Thiago Ber-

nardino Magalhaes, Edmilson Luis Rodrigues Motta, Eduardo Tengan, Eduardo Wagner, Emanuel

Carneiro, Elio Mega, Fabio Brochero, Luciano Guimaraes Monteiro de Castro, Luzinalva Miranda

de Amorim, Nicolau Corcao Saldanha, Pablo Rodrigo Ganassim, Paulo Cezar Pinto Carvalho, Ralph

Costa Teixeira, Samuel Barbosa Feitosa, Yoshiharu Kohayakawa, Yuri Lima

Junior Members: Alex Correa Abreu, Bernardo Paulo Freitas da Costa, Carlos Augusto David

Ribeiro, Carlos Stein Naves de Brito, Davi Maximo Alexandrino Nogueira, Fabio Dias Moreira,

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

Marcio Assad Cohen, Telmo Correa Junior, Thiago Barros Rodrigues Costa, Rodrigo Villard

Executive Secretary: Nelly Carvajal Florez

Assistant Secretaries: Rosa Morena Freitas Kohn

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 wrote on the blackboard all the integers from 1 to 2011. Then sheerased 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 digits0 and 1?

Problem 2

We have a red cube with sidelength 2 cm. What is the minimum numberof identical cubes that must be adjoined to the red cube in order to obtain

a cube with volume(125

)3cm?

Problem 3

We call a number pal if it doesn’t have a zero digit and the sum of the squaresof the digits is a perfect square. For example, 2115522 is pal (because22 + 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 turnis perpendicular to AB. The square has area 64 cm2.

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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 eachcell, each number appearing exactly once. Then she computes eight sums:the sums of three numbers on each row, the sums of the three numbers oneach 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 of3.

(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 eachcell, each number appearing exactly once. Then she computes eight sums:the sums of three numbers on each row, the sums of the three number oneach column and the sums of the three numbers on both diagonals. Is itpossible 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 and6 ADC = 6 CAB. Let M and N be the midpoints of AD and AB. Provethat triangle MNC is isosceles.

Problem 3

Emerald and Jade play the following game: Emerald writes a list with 2011positive integers, but does not show it to Jade. Jade’s goal is finding theproduct of the 2011 numbers in Emerald’s list. In order to do so, she isallowed to ask Emerald the gcd or the lcm of any subset with at least twoof 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 thenumbers in your list?”). Jade can make as many questions as she wants,but can only obtain her (correct) answers from Emerald after making allher questions (Emerald is generous and also says which answer correspondsto each question). Jade then can use any of the four elementary operations(add, subtract, multiply, divide) with Emerald’s answers. Can Jade make alist of questions that guarantees that she can find the product of the 2011numbers?

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Problems 5

Problem 4

Emerald wrote a list of positive integers. Renan noticed that each numberin the list and any sum of any quantity of distinct numbers from the list weresquare-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 isan 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 numberobtained by writing the digits of odd order of N and even(N) be the k-digitnumber obtained by writing the digits of even order of N . For example,odd(249035) = 405 and even(249035) = 293. Prove that there is no positiveinteger 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 thesquares of the digits is a perfect square. For example, 122 and 34 are palbut 304 and 12 are not pal. Prove that there exists a pal number with ndigits, n > 1.

Problem 2

33 friends are collecting stickers for a 2011-sticker album. A distributionof stickers among the 33 friends is incomplete when there is a sticker thatno friend has. Determine the least m with the following property: everydistribution 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 P1P2P3P4P5 with area 1, there areindices i and j (assume P6 = P1 and P7 = P2) such that:

area△PiPi+1Pi+2 ≤5−

√5

10≤ area△PjPj+1Pj+2

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Problem 4

Do there exist 2011 positive integers a1 < a2 < . . . < a2011 such thatgcd(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 inter-section of BH and AC and E be the intersection of CH and AB. Thecircumcircle of ADE meets the circumcircle of ABC at F 6= A. Prove thatthe angle bisectors of 6 BFC and 6 BHC concur at a point on line BC.

Problem 6

Let a1, a2, . . . , a2011 be nonnegative reals with sum 20112 . Prove that

∣∣∣∣∣

∏

cyc

(an − an+1)

∣∣∣∣∣= |(a1 − a2)(a2 − a3) . . . (a2011 − a1)| ≤

3√3

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). De-termine 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 ofthe 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 integersmodulo 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.

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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 allpossible values of

α

β+

β

γ+

γ

β

Problem 5

If u1, . . . , uk ∈ R3, denote by C(u1, . . . , uk) the cone generated by u1, . . . , uk:

C(u1, . . . , uk) = {a1u1 + · · ·+ akuk; a1, . . . , ak ∈ [0,+∞)}.

Let v1, v2, v3, v4 points randomly and independently chosen from the unitsphere x2 + y2 + z2 = 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 generateC(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 k, i.e., there exists real constant numbersc1, c2, . . . , ck such that xn+k =

∑kr=1 crxn+k−r, ∀n ≥ 0. Suppose k is the

minimum positive integer with this property. Prove that cj ∈ Z, for all j,1 ≤ j ≤ k.

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(Problems)

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9 Solutions

3.1. Grades 6–7

Problem 1

(a) The erased numbers were 2 = 2 · 1, 4 = 2 · 2, . . ., 2010 = 2 · 1005. So2011− 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 aredigits equal to either 0 or 1. Notice that the units digit must be 1.

Problem 2

The bigger cube has sidelength 125 cm, so the difference between the side-

lengths is 125 − 2 = 2

5 cm, that is, the red cubes should not have sidelengthgreater than this length. Cubes with sidelength 2

5 cm are the natural can-didates, so we set a new unit u = 2

5 cm. Notice that the bigger cube shouldhave sidelength 6u and the original cube must have sidelength 5u. So weneed 63 − 53 = 91 red cubes.

Problem 3

(a) First notice that 86 is pal. Then it’s not hard to check by hand thatevery number from 87 to 99 is not pal.

(b) The answer is yes. First consider the 2011-digit number 11 . . . 1︸︷︷︸

2011 fives

. The

sum of its digits is 2011. The smallest perfect square greater than 2011is 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 obtainthe pal number 11 . . . 1

︸︷︷︸

2008 fives

223.

Problem 4

(a) The trapezoid OABC has area AB+OC2 ·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 onthe sides of the square. So the square is divided into four congruent

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(Solutions)

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(non-convex) polygons, each with area 644 = 16. Then BCDE has area

16− 6 = 10.

Problem 5

(a) For instance,

1 2 34 5 68 9 7

The trick is to only adjust the last row. The usual order 7, 8, 9 yieldsall 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, zis a multiple of 3 iff x ≡ y ≡ z (mod 3) or x, y, z are 0, 1, 2 mod 3in some order. Let a, b, c, d be the numbers in the corner modulo 3. Sotwo of them are equal. We can suppose wlog that they are either a = bor a = d. Also, let x be the number in the central cell modulo 3.

a bx

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 bb

c a

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Solutions 11

Let m be the other remainder (that is, m 6= a and m 6= b). Then m cannotbe in the same line as a and b. This leaves only one possibility:

a bm bm 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 reducethe problem to the previous case). If d 6= c, a, c, d are the three distinctremainders, and we have no possibility for x. So c = d.

a ax

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 andBAC are similar by case SAS. Segments CM and CN are correspondingmedians, so CM

CN = CACB and 6 BCN = 6 ACM ⇐⇒ 6 BCN + 6 NCA =

6 ACM + 6 NCA ⇐⇒ 6 BCA = 6 NCM . Thus, again by case SAS,triangles CMN and CAB are similar, and therefore CMN is an isoscelestriangle with CM = MN .

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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 aset of three numbers and 1004 sets of two numbers.

It remains to prove the above identity. But this follows from the facts thatmax(x, y)+max(x, z)+max(y, z)+min(x, y, z)−max(x, y, z) = x+y+z, andif pxi ‖ ai then pmin{xi} ‖ gcd(a1, a2, . . . , an) and pmax{xi} ‖ lcm(a1, a2, . . . , an).

Problem 4

The smallest perfect square, apart from 1, is 22 = 4. So let a1, a2, . . . , ak bethe numbers on the list modulo 4. We cannot have ai = 0; also, there is atmost 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 theabove 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 wealso cannot have aj = 3 and a3 = a4 = 1, one of a3, a4, say a3, is 2. Butthen a1 + a2 + a3 = 1 + 1 + 2 = 4.

So the quantity of numbers is at most 3. 5, 13 and 17 is an example of alist with three numbers.

Problem 5

Since(

162√3

)2

= 643 = 21+ 1

3 lies between 4.52 = 20.25 and 52 and the altitude

of the triangle is 2√3·√3

2 = 3, we can cover an square with sidelength 16 with2 · 5 ·

⌈163

⌉= 60 equilateral triangles. Since

⌊100060

⌋= 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 positiveintegers 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 allnumbers with 2k − 2 digits. Let c and d be the two leftmost digits of N ,

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Solutions 13

so that N = c · 102k−1 + d · 102k−2 + N0, N0 with 2k − 2 digits. Thenodd(N) = d · 10k−1 + odd(N0) and even(N) = c · 10k−1 + even(N0). So weneed 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 andthus

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︸︷︷︸

n times

. The sum of the squares of its digits is n ·52 =

25n. We can exchange any two fives by one three and one four, so the sumof the squares decreases by 52, until we run out of fives. So we can get anysum from 25 · ⌈n/2⌉ and 25 ·n. So it suffices to show that there is an integerk such that n

2 ≤ k2 ≤ n. Choose k such that k2 ≤ n < (k + 1)2. Supposek2 < n

2 . Then n > 2k2, and (k+1)2 > n > 2k2 =⇒ (k+1)2 ≥ 2k2+2 ⇐⇒k2 − 2k + 1 ≤ 0 ⇐⇒ (k − 1)2 ≤ 0, which is false except for k = 1, or2 < n < 4, that is, n = 3. But the statement of the problem itself gives anexample with n digits: 122.

Problem 2

Number the stickers from 1 to 2011 and let Si be the set of the stickers thatthe 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} andS33 = {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.

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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 181/3elements, 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 33friends have.

Another solution: We will prove that m = 1890 in another way. Theexample 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 stickersthat the friend i does not have, 1 ≤ i ≤ 33.

Consider all pairs (x, {i, j}) such that x ∈ Ti ∩ Tj . If for every sticker thereis a friend that has it, that is, T1 ∩ T2 ∩ . . . ∩ T33 = ∅ then for each x thereexists k such that x /∈ Tk. So each x belongs to at most 32 sets Ti and,hence, there exist at most 2010 ·

(322

)pairs (x, {i, j}). On the other hand,

since |Ti ∩ Tj | ≥ m there exists at least m ·(332

)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 hasn stickers and there are k friends, the minimum value of m is

n− 1 if k ≥ nn− 2

⌊nk

⌋+ 1 if k < n and k | n

n− 2⌊nk

⌋if k < n and n mod k = 1

n− 2⌊nk

⌋− 1 if k < n and n mod k > 1

Problem 3

Let’s prove that there exists a triangle PjPj+1Pj+2 with area less than or

equal to α = 5−√5

2 . Suppose that all triangles PjPj+1Pj+2 have area greater

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Solutions 15

than α.

Let diagonals P1P4 and P3P5 meet at Q. Since Q ∈ P3P5, areaP1P2Q ≤max(areaP1P2P5, areaP1P2P3) < α, so areaP1P2P4 = 1 − areaP1P4P5 −areaP2P3P4 > 1− 2α. Thus

P1Q

P1P4=

areaP1P2Q

areaP1P2P4<

α

1− 2α

We also have P1QP4Q

= areaP1P3P5

areaP3P4P5. Since areaP3P4P5 < α and areaP1P3P5 >

1− 2α,P1Q

P4Q>

1− 2α

α⇐⇒ P1Q

P1P4>

1− 2α

1− α

Therefore

1−2α

1−α<

P1Q

P1P4<

α

1−2α=⇒ 5α2−5α+1< 0 ⇐⇒ 5−

√5

10< α <

5+√5

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. Themain 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 | aithen 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, sogcd(ai, aj) = aj − ai.

Once this fact is established, one can construct the sequence inductively asfollows: first consider the two-term sequence (1, 2). Now, given a sequence

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(x1, x2, . . . , xk−1) with k−1 terms such that gcd(xi, xj) = xj −xi, constructa 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 theprevious observation, we need xj −xi | xi +x0 and xi | x0. We already havethat xj − xi | xi, so a good choice is x0 = lcm(x1, x2, . . . , xk−1), becauseby definition xi | x0 and, since xi | x0 and xj − xi | xi, xj − xi | x0, soxj−xi | xi+x0. So we obtained a new sequence with k terms and the resultfollows by induction.

Problem 5

By the angle bisector theorem, it suffices to prove that BFFC = BH

HC .

We have 6 EFB = 180◦ − 6 FEA = 180◦ − 6 FDA = 6 FDC and 6 FBE =6 FBA = 6 FCA = 6 FCD, so triangles BEF and CDF are similar. Thus

BF

FC=

BE

CD=

BH cos 6 EBH

CH cos 6 DCH=

BH cos(90◦ − 6 BAC)

CH cos(90◦ − 6 BAC)=

BH

CH

and the result follows.

Problem 6

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 num-

bers ai−1, ai, ai+1 is zero.

Proof. Suppose, by means of contradiction, that E is maximum and thereexists ai such that ai−1, ai, ai+1 are all nonzero (that is, ai−1aiai+1 > 0).Define A = {ai | ai > 0} and B = {ai | ai−1aiai+1 > 0}. Then B ⊂ A and

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Solutions 17

B 6= ∅. Let ak = minB and consider ak−1 and ak+1. We have the followingcases:

• ak < ak−1 and ak < ak+1. Let

a′i =

{0, if ai = 0 or i = kai +

ak|A|−1 , if ai > 0 and i 6= k

That is, we make ak be zero and distribute it among the remaining nonzeroterms. So |ai − ai+1| remains unchanged if ai, ai+1 ∈ A and k /∈ {i, i+1}, orai, ai+1 /∈ A; increases from |ai − ai+1| = max{ai, ai+1} to max{ai, ai+1} +ak

|A|−1 if ai /∈ A or ai+1 /∈ A, but not both; increases from |ak±1 − ak| =ak±1 − ak to ak±1 +

ak|A|−1 if k ∈ {i, i+ 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, weenchange (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 increasesfrom |ak−2 − ak−1| = ak−1 to |ak−2 − a′k−1| = ak−1 + ak; for i = k − 1increases from |ak−1 − ak| = ak − ak−1 to |a′k−1 − a′k| = ak−1 + ak; fori = 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 un-changed 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)| =akak+1|ak+1−ak|. Notice that the groups can be interchanged, such that wecan suppose wlog that all groups with two nonzero variables are contiguous.

17

(Solutions)

(page 18)


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 groupsare 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)| isexchanged 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 variablesare the ones with odd indices, that is, a1, a3, . . . , a2011. In this case, we obtainthe product a1a2011|a1 − a2011|a23a25 . . . a22009, and we can optimizer it locally.

Let a1 + a2011 = s and suppose wlog a1 > a2011. Let α, β be positive realnumbers to be determined. By AM–GM,

a1a2011(a1 − a2011) =1

αβ(αa1)(βa2011)(a1 − a2011)

≤ 1

αβ

(αa1 + βa2011 + (a1 − a2011)

3

)3

=1

αβ

((α+ 1)a1 + (β − 1)a2011

3

)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. Henceα =

√3− 1 and β = 1 +

√3, and

a1a2011(a1 − a2011) ≤1

αβ

((α+ 1)a1 + (β − 1)a2011

3

)3

=1

2

(√3(a1 + a2011)

3

)3

=

√3

18s3

Now we optimize the rest. If a3 + a5 + · · ·+ a2009 =20112 − s,

a23a25 . . . a

22009 ≤

(a3 + a5 + · · ·+ a2009

1004

)2008

=

( 20112 − s

1004

)2008

18

(Solutions)

(page 19)

Solutions 19

Now we finish the problem. Let γ be a positive real number to be deter-mined.

E = a1a2011|a1 − a2011|a23a25 . . . a22009

≤√3

18s3 ·

( 20112 − s

1004

)2008

=

√3

18γ3(γs)3

( 20112 − s

1004

)2008

≤√3

18γ3

(

3γs+ 2008 ·20112 −s

1004

2011

)2011

=

√3

18γ3

(2011 + (3γ − 2)s

2011

)2011

We choose γ = 2/3, so

a1a2011|a1 − a2011|a23a25 . . . a22009 ≤3√3

16

Remark: The latter part of the problem can be solved with Calculus, butwe decided to give an elementary solution. Also, the equality occurs iff

a3 = a5 = · · · = a2009 = 1, a1 = 3+√3

4 and a2011 = 3−√3

4 or it is a cyclicpermutation (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, andQ has a local minimum at (2,−16) and a local maximum at (−2, 16). SoQ(x) = −t has three (not necessarily distinct) real roots for −16 ≤ t ≤ 16and one real root for t < −16 and t > 16, which means that ∆(t) = 0 fort < −16 or t > 16. So from now on we consider only t ∈ [−16, 16]. Letu ≤ v ≤ w be the roots. Since P ′

t (x) = 0 ⇐⇒ x = −2 or x = 2 we haveu ≤ −2 ≤ v ≤ 2 ≤ w. In particular, −2 ≤ v ≤ 2, and v can assume anyvalue 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 = −vand uw = −12 − v(u + w) = v2 − 12, so (∆(t))2 = (w − u)2 = (w + u)2 −4uw = 48 − 3v2, which lies in the range [48 − 3 · 22, 48] = [36, 48]. So36 ≤ (∆(t))2 ≤ 48 ⇐⇒ 6 ≤ ∆(t) ≤ 4

√3.

So the range of ∆(t) is {0} ∪ [6, 4√3].

19

(Solutions)

(page 20)


Problem 2

First consider a triangle ABC and its circumcenter O. Then the area ofABC is R2

2 (sin 26 A + sin 26 B + sin 26 C). Notice that if 6 B > 90◦ thensin 26 B < 0.

So the sum is equal to the sum of the areas of triangles OAiAj with aplus sign or a minus sign, depending on the third vertex Ak of the triangleAiAjAk: if Ak lies on the major arc AiAj then we have a plus sign; else wehave a minus sign (it won’t matter if AiAj is a diameter, because in thatcase the area of OAiAj is zero).

Therefore, if AiAj subtend an minor arc of k · 2πn , 1 ≤ k ≤ ⌊n/2⌋, the area

of the triangle OAiAj appears with a minus sign k − 1 times and with aplus sign n− (k − 1)− 2 = n− k − 1 times. So it contributes with the sumn− k − 1− (k − 1) = n− 2k times.

Considering that there are n arcs with length k · 2πn , if θ = 2π

n the requiredsum is

S =n

2

⌊n/2⌋∑

k=1

(n− 2k) sin kθ =n2

2

⌊n/2⌋∑

k=1

sin kθ − n

⌊n/2⌋∑

k=1

k sin kθ

Consider the sums S1(θ) =∑⌊n/2⌋

k=1 sin kθ and S2(θ) =∑⌊n/2⌋

k=1 cos kθ =⇒S ′2(θ) = −∑⌊n/2⌋

k=1 k sin kθ. So we want to compute n2

2 S1(θ) + n · S ′2(θ).

ButS2(θ) + iS1(θ)

=

⌊n/2⌋∑

k=1

cos kθ + i sin kθ =

⌊n/2⌋∑

k=1

ωk = ω · ω⌊n/2⌋ − 1

ω − 1

= ω⌊n/2⌋/2+1/2ω⌊n/2⌋/2 − ω−⌊n/2⌋/2

ω1/2 − ω−1/2

=

(

cos

((⌊n/2⌋+ 1)θ

2

)

+ i sin

((⌊n/2⌋+ 1)θ

2

))

· sin(⌊n/2⌋θ/2)sin(θ/2)

,

20 (Solutions)

(page 21)

Solutions 21

where ω = cos θ + i sin θ.

So

S1 =sin(

(⌊n/2⌋+1)θ2

)

sin(

⌊n/2⌋θ2

)

sin(θ/2)=

cos(θ/2)− cos((⌊n/2⌋+ 1/2)θ)

2 sin(θ/2)

=1

2cot(θ/2)− cos((⌊n/2⌋+ 1/2)θ)

2 sin(θ/2)

S2 =cos(

(⌊n/2⌋+1)θ2

)

sin(

⌊n/2⌋θ2

)

sin(θ/2)=

sin(θ/2) + sin((⌊n/2⌋+ 1/2)θ)

2 sin(θ/2)

=1

2+

sin((⌊n/2⌋+ 1/2)θ)

2 sin(θ/2)

Finally,

S ′2(θ)

=

(⌊n2

⌋+ 1

2

)cos((⌊

n2

⌋+ 1

2

)θ)sin(θ2

)− 1

2 cos(θ2

)sin((⌊

n2

⌋+ 1

2

)θ)

2 sin2(θ/2)

=⌊n/2⌋2

cos((⌊n/2⌋+ 1/2)θ)

sin(θ/2)− 1

4

sin(⌊n/2⌋θ)sin2(θ/2)

So the required sum is

n2

2

(1

2cot(θ/2)− cos((⌊n/2⌋+ 1/2)θ)

2 sin(θ/2)

)

+

n

(⌊n/2⌋2

cos((⌊n/2⌋+ 1/2)θ)

sin(θ/2)− 1

4

sin(⌊n/2⌋θ)sin2(θ/2)

)

If n is even, ⌊n/2⌋ = n/2 and substituting θ = 2πn the sum simplifies to

n2

4cot

π

n

If n is odd, ⌊n/2⌋ = (n−1)/2 and substituting θ = 2πn the sum also simplifies

ton2

4cot

π

n

21

(Solutions)

(page 22)


So the answer isn2

4cot

π

n

Problem 3

(a) Let A = {a1, a2, . . . , a⌊n/2⌋}. Consider the sum

n−1∑

t=0

|A ∩ (A+ t)|

Now each element a ∈ A appears in a set A+ ti |A| = ⌊n/2⌋ times: chooseti = a− ai for each i = 1, 2, . . . , ⌊n/2⌋. So

n−1∑

t=0

|A ∩ (A+ t)| =(⌊n

2

⌋)2

and the average of |A ∩ (A+ t)| is

1

n

(⌊n

2

⌋)2

≤ n

4.

Since |A ∩ (A + 0)| = |A| > n4 is above average, there is a t such that

|A ∩ (A+ t)| is below average, so

f(A) ≤ |A ∩ (A+ t)| < n

4=⇒ f(A) ≤

⌈n

4

⌉

− 1.

So g(n) ≤⌈n4

⌉− 1.

(b) Let p ≡ 3 (mod 4) ne a prime, and set

A = (F×p )2 (non-zero quadratic residues modulo p)

We have to show that

|(A+ t) ∩A| ≥⌈p

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 y2 ∈ A such that

x2 = y2 + t ⇐⇒ (x− y)(x+ y) = t

⇐⇒∣∣∣∣∣

x− y = u

x+ y = u−1tfor some u ∈ F×

p

⇐⇒

∣∣∣∣∣∣∣∣

x =u+ u−1t

2

y =u−1t− u

2

for some u ∈ F×p

Therefore |A ∩ (A+ t)| equals the number of elements of Fp of the form

x2 =u2 + 2t+ u−2t2

4, u ∈ F×

p ,

with x 6= 0 and y 6= 0, that is,∣∣∣∣∣

u+ u−1t 6= 0

u−1t− u 6= 0⇐⇒ u2 6= ±t

Notice that since p ≡ 3 (mod 4)

(±t

p

)

=

(±1

p

)(t

p

)

= ±(t

p

)

and therfore there is exactly one quadratic residue in {−t, t}.On the other hand, it u2 6= v2,

u2 + 2t+ u−2t2

4=

v2 + 2t+ v−2t2

4⇐⇒ u2 − v2 =

t2

u2− t2

v2⇐⇒ u2v2 = t2

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 = v2 isexcluded since u2 · u2 = t2 ⇐⇒ u2 = ±t). Therefore

|A ∩ (A+ t)| =p−12 − 1

2=

p− 3

4=⌈p

4

⌉

− 1,

as required.

23

(Solutions)

(page 24)


Problem 4

(a) First notice that, since r is a root, then r3 + r2 − 4r + 1 = 0 ⇐⇒r2 + r − 3 = 1− 1

r . So we need to prove that

(

1− 1

r

)3

+

(

1− 1

r

)2

− 4

(

1− 1

r

)

+ 1 = 0

⇐⇒ 1− 3

r+

3

r2− 1

r3+ 1− 2

r+

1

r2− 4 +

4

r+ 1 = 0

⇐⇒ − 1− 1

r+

4

r2− 1

r3= 0

⇐⇒ r3 + r2 − 4r + 1 = 0,

which is true.

(b) Iterating 1− 1r , we get 1− 1

1− 1r

= 11−r . It’s not hard to see that r, 1− 1

r =r−1r and 1

1−r are all distinct. In fact, if r = 1 − 1r then r2 − r + 1 = 0,

and r3 = −1, so r2 − 4r = 0, which is not true.

So there are two possible ways of computing αβ + β

γ + γα :

• (α, β, γ) =(r, r−1

r , 11−r

):

α

β+

β

γ+

γ

α=

r2

r − 1− (r − 1)2

r− 1

r(r − 1)

=r3 − (r − 1)3 − 1

r(r − 1)=

3r(r − 1)

r(r − 1)= 3

• (α, β, γ) =(r, 1

1−r ,r−1r

):

α

β+

β

γ+

γ

α= −r(r − 1)− r

(r − 1)2+

r − 1

r2

=−r6 + 3r5 − 3r4 + r3 − 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+

β

γ+

γ

α=

−80r2 + 130r − 30

8r2 − 13r + 3= −10.

24

(Solutions)

(page 25)

Solutions 25

Another way to solve this problem is realizing that if (α, β, γ) =(r, r−1

r , 11−r

)

then the other sum is actually βα+

γβ+

αγ . By Vieta’s formula, σ1 = α+β+γ =

−1, σ2 = αβ + βγ + γα = −4 and σ3 = αβγ = −1.

α

β+

β

γ+

γ

α+

β

α+

γ

β+

α

γ=

α2β + α2γ + β2α+ β2γ + γ2α+ γ2β

αβγ

=σ1σ2 − 3σ3

σ3=

(−1) · (−4)− 3(−1)

−1= −7,

soβ

α+

γ

β+

α

γ= −7− 3 = −10.

Problem 5

(a) The probability that the cone of the four vectors is proper is 78 so the

probability that the cone is all of R3 is 18 .

Construct a vector u12 that is normal to the plane spanned by v1 and v2oriented so that v3 · u12 > 0. Then the half-space {w | w · u12 ≥ 0} containsthe cone generated by {v1, v2, v3}. It follows that if v4 · u12 > 0, the conegenerated by all four vectors will be contained in the same half-space. Soto 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 orthogonalto 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 claimthat if all three of those dot products are negative, then the cone covers allof space.

If v1, v2 and v3 are fixed, the three signs of the dot products v4 · uij arenot independent. But if we average over all choices of the vectors, then −v1occurs exactly as often as +v1 and so on. We conclude that on average thethree dot products in question are negative with probability 1

8 .

(b) Given v1, v2 and v3 then v4 lies in the interior of the cone generated bythose three if and only if v4 · uij > 0 for all three such dot products.So there is a 1

8 chance that v4 lies in C(v1, v2, v3). Similarly, there isa 1

8 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 ofprobability 1

8 , which gives a probability of 12 .

25

(Solutions)

(page 26)


Problem 6

Let (a(i)n ) be the sequence of integers obtained by shifting an by i positions,

i.e., a(i)n = an+i. Then

(a(0)n ), (a(1)n ), . . . , (a(k)n )

is a basis for the k-dimensional C-vector space of sequences (bn) satisfying

bn+k = ck−1bn+k−1 + · · ·+ c0bn (n ≥ 0) (∗)

In fact, any non-trivial C-linear relation among (a(0)n ), . . ., (a

(k−1)n ) would

imply that k is not minimal.

Now letf(x) = xk − ck−1x

k−1 − · · · − c0

andα1, . . . , αk

be the roots of f(x) (listed with multiplicity). First, observe that all co-efficients ci are rational since they are solutions to the linear system withinteger coefficients

a0 a1 . . . ak−1

a1 a2 . . . ak...

.... . .

...ak−1 ak . . . a2k−2

c0c1...

ck−1

=

akak+1...

a2k−1

(the matrix is non-singular since the (a(i)n )’s are linear independent.) There-

fore the αi’s are algebraic numbers.

Definetn = αn

1 + αn2 + · · ·+ αn

k (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

tn = r0a(0)n + r1a

(1)n + · · ·+ rk−1a

(k−1)n

Clearly ri ∈ Q since tn, a(i)n ∈ 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 polynomialsin αn as polynomials with rational coefficients of

tn = αn1 + · · ·+ αn

k

t2n = α2n1 + · · ·+ α2n

k

...

tkn = αkn1 + · · ·+ αkn

k

Therefore the minimal polynomials of αni 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 ∆ · αn

i are algebraic integersfor all n. But that implies that the ring Z[αi] is contained in the finitelygenerated Z-module 1

∆OQ(αi), where OQ(αi) is the ring of algebraic integersin 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, togetherwith the structure theorem of finitely generated modules over a PID). Henceαi is an algebraic integer, as required.

Alternatively, suppose that αi is not an algebraic integer, so that it hasprime ideal factorization

(αi) = pei1 . . . pess

with ei < 0 for some i, say e1.

Then(∆αn

i ) = (∆) · pne11 . . . pness

would have a negative p1-exponent for n sufficiently large, contradicting thefact that ∆ · αn

i is an algebraic integer for all n ≥ 0.

27

(Solutions)

(page 28)

(Solutions)

(page 29)

29 Winners in 2011

4.1. Grades 6–7

Gold medals

Pedro Henrique Sacramento de Oliveira

Rogerio Aristida Guimaraes Junior

Mateus Siqueira Thimoteo

William Hideki Kondo

Bruna Malvar Castello Branco

Nathan Bonetti Teodoro

Silver medals

Mariana Miwa Okuma Miyashiro

Lucas dos Anjos Dantas Teixeira

Maria Julia Costa Medeiros

Mateus Pereira

Carolina Carvalho Silva

Laura Mello D’Urso Vianna

Henrique Gontijo Chiari

Nicolas Wolaniuk do Amaral Carvalho

Lucas Diniz Goncalves Villas Boas

Bronze medals

Leonardo de Matos Fellippetti Mariano

Lucia Veronica Copque Aguiar de Souza

Adriana Mayumi Shiguihara

Daniel Akira Hasimoto

Rodrigo Goncalves Correa

Cesar Ricardo Silva Filippi

Marina Maciel Ansanelli

Henrique Bittencourt Netto Monteiro

Julia Perdigao Saltiel

Jonathan Pereira Maria

Lucas Tokio Kawahara

Sandra Ayumi Nihama

Joao Guilherme Madeira Araujo

Andrey Jhen Shan Chen

29

(Winners in 2011)

(page 30)


Bruno Brasil Meinhart

Daniel Quintao de Moraes

Diene Xie

Honorable mention

Felipe Reyel Feitosa

Henrique Corato Zanarella

Alıcia Fortes Machado

Andre 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 Ramme

Iara Rohn Kombrink Davies

Victor Alves Benevides

Samuel Sena Galvao

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

Joao Alberto Moreira Serodio

Loc Dominguez

Vinıcius Soares de Abreu Silva

Breno Maia Baptista

Luısa Holler Lee

Brendon Diniz Borck

Eduardo Emilio Costa Trunci

Bernardo Gabriele Collaco

Lucas Hideki Takeuchi Okamura

Plinio Melo Guimaraes Valerio

Rodrigo Vieira Casanova Monteiro

Victor M. K. Tsuda

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Arthur Henrique Craveiro Costa

Pedro Orii Antonacio

Gabriel Moura Brauna

Victoria Santos Duarte Ramos

Italo Rennan Lima

Amanda Barbosa Schirmbeck

Thiago Ferreira Teixeira

Gabriel Dante Cawamura Seppelfelt

Lucas Siqueira Aragao

Milena Delarete Drummond Marques

Rodrigo Moutinho Faria

Daniel Lopes de Castro

Joao 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

Luıze Mello DUrso Vianna

Silver medals

Daniel Lima Braga

Fabio da Silva Soares

Joao 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 Emılia Hernandes Dib

Bronze medals

Pedro Henrique Alencar Costa

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Pedro Augusto Brasileiro Lins Barbosa

Gabriel Mayrink Verdun

Leonardo Santos Matiello

Matheus Carius Castro

Lucca Morais de Arruda Siaudzionis

Luiz Claudio Sampaio Ramos

Matheus Carioca Sampaio

Jose Wanderclesson Nobre Damasceno Filho

Suzane Eberhart Ribeiro da Silva

Estevao Waldow

Erika Rizzo Aquino

Pedro Jorge Luz Alves Cronemberger

Alexandre Mendonca Cardoso

Ricardo Ken Wang Tsuzuki

Honorable mention

Leonardo Alves Ramalho

Ana Paula Lopes Schuch

Flavia 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 Picanco Costa

Guilherme Anitele Silva

Mateus Arraes Feitosa Borges

Rodrigo Zanette de Magalhaes

Luis Eduardo de Sousa Lima

Gabriel Vidigal de Paula Santos

Bruno Almeida Costa

Joao Baptista de Paula e Silva

Gabriel Ribeiro Barbosa

Kevin Eiji Inashita

Dimas Macedo de Albuquerque

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Mauricio Najjar da Silveira

Juliano Petry Pesarico

Bruna Caroline Pimentel Goncalves

Gustavo Torres da Silva

Artur Corassa Martins

Italo Lesione de Paiva Rocha

Nathan Antonio de Azevedo Milagres

Juliana Amoedo Amoedo Placido

Victoria Moreira Reis Cogo

Leandro Alves Cordeiro

Romulo Gabriel Lima da Costa

Bruno Vasconcelos Silva

Alexandro Vıtor Serafim de Carvalho

Cristhian Mafalda

Douglas Matos Gomes

Gabriel Diniz Vieira e Sousa

Enrico Pascucci Loffel

Ricardo Vidal Mota Peixoto

4.3. Grades 10–12

Gold medals

Joao Lucas Camelo Sa

Henrique Gasparini Fiuza do Nascimento

Rafael Kazuhiro Miyazaki

Andre 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 Correa Rodrigues

Franco Matheus de Alencar Severo

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Gabriel Ilharco Magalhaes

Bronze medals

Lucas Lourenco Hernandes

Ivan Tadeu Ferreira Antunes Filho

Kayo de Franca Gurgel

Michel Rozenberg Zelazny

Alexandre Perozim de Faveri

Davi Coelho Amorim

Marcos Massayuki Kawakami

Daniel dos Santos Bossle

Gabriel Militao Vinhas Lopes

Mateus Henrique Ramos de Souza

Victor Venturi

Ramon Silva de Lima

Gabriel Jose Moreira da Costa Silva

Otavio Augusto de Oliveira Mendes

Marcelo Luiz Goncalves

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

Jose Ney Alves Feitosa Neto

Andre Bandeira Pinheiro

Fernando Lima Saraiva Filho

Rafael Tedeschi Eugenio Pontes Barone

Vinıcius Canto Costa

Lincoln de Queiroz Vieira

Thiago Poeiras Silva

Andre Amaral de Souza

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Carlos Alexandre Silva dos Santos

Felipe Viana Sousa

Liara Guinsberg

Otavio Araujo de Aguiar

Rodolfo Rodrigues da Costa

Caıque Porto Lira

Kelvin Azevedo Santos

Eric Tada de Souza

Marcelo Cargnelutti Rossato

Marina Pessoa Mota

4.4. Undergraduates

Gold medals

Renan Henrique Finder

Rafael Tupynamba Dutra

Regis Prado Barbosa

Guilherme Rodrigues Nogueira de Souza

Davi Lopes Alves de Medeiros

Silver medals

Gabriel Luis Mello Dalalio

Felipe Goncalves Assis

Darcy Gabriel Augusto de Camargo Cunha

Hugo Fonseca Araujo

Matheus Secco Torres da Silva

Erik Fernando de Amorim

Lucas Colucci Cavalcante de Souza

Bronze medals

Jose 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 Sergio de Castro Moreira

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Alexandre Azevedo Cesar

Ricardo Turolla Bortolotti

Roberio 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

Joao 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

Jose Armando Barbosa Filho

Alysson Espındola de Sa Silveira

Fernando Nascimento Coelho

Gabriel Caser Brito

Fernando Fonseca Andrade Oliveira

Breno Vieira de Aguiar

Thales Graca Athanasio

Tiago Fonseca

Gabriel Queiroz de Brito Melo

Rafael Pereira de Paula de Lucas Simon

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Documents

XXXIII Brazilian Math Olympiad 2011 - OBM