Download XXXIII Brazilian Math Olympiad 2011

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XXXIII
Brazilian Math Olympiad
2011
Editora AOBM
Rio de Janeiro
2012
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(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.
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(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 .
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(Problems)
(page 4)
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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)
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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.
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(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)
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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)
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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