Upload
duongdien
View
245
Download
5
Embed Size (px)
Citation preview
(page 0)
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: Luzinalva Miranda de Amorim, Carlos Yuzo Shine
Members: Antonio Caminha Muniz Neto, Carlos Gustavo Moreira, Edmilson Luis Rodrigues Motta,
Eduardo Wagner, Eduardo Tengan, Elio Mega, Florencio Ferreira Guimaraes Filho, Luciano Guima-
raes Monteiro de Castro, Nicolau Corcao Saldanha, Onofre Campos, Pablo Rodrigo Ganassim, Paulo
Cezar Pinto Carvalho, Ronaldo Alves Garcia, Ralph Costa Teixeira, Yoshiharu Kohayakawa
Junior Members: Alex Correa Abreu, Bernardo Paulo Freitas da Costa, Bruno Holanda, Carlos Au-
gusto David Ribeiro, Carlos Stein Naves de Brito, Cıcero Thiago Magalhaes, Davi Maximo Alexan-
drino Nogueira, Einstein do Nascimento Junior, Emanuel Carneiro, Fabio Dias Moreira, Fabrıcio
Siqueira Benevides, Gabriel Tavares Bujokas, Humberto Naves, Larissa Cavalcante Lima, Marcio
Assad Cohen, Samuel Barbosa Feitosa, Telmo Correa Junior, Tertuliano Franco, Thiago Barros Ro-
drigues Costa, Rodrigo Villard, Yuri Gomes Lima
Executive Secretary: Nelly Carvajal Florez.
Assistant Secretaries: Sonia de Souza Silva de Melo
Typeset with Plain TEX.
()
(page 1)
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
(Introduction)
(page 3)
3 Problems
2.1. Grades 6–7
Problem 1
Emerald has several right triangles just like the one in the diagram.
(a) Emerald made the following diagram by coinciding part of the sidesand without superposing triangles. Find the area and the perimeter ofthis diagram.
(b) Following the same rules as above, Emerald constructed the smallestsquare with integer side. Draw a diagram showing how she can do this.
Problem 2
The cells of a 3×3 table were numbered from 1 to 9, each number appearingexactly once. For each row the cell with the greatest number is colored redand the cell with the smallest number is colored green. Let A be the smallestof the numbers in the red cells and B the greatest of the numbers in thegreen cells.
(a) Show a distribution of the numbers in the table such that A−B = 4.
(b) Show a distribution of the numbers in the table such that A−B = −3.
(c) Is it possible that A = 4 and B = 3?
3
(Problems)
(page 4)
4 XXXII Brazilian Math Olympiad 2010
Problem 3
Given a solid made of unit cubes, as in figure 1, we can write the numberof unit cubes in each direction, as shown in figure 2.
Emerald Jr made a solid out of unit cubes and drew a figure similar to figure2.
Find a, b, c, d, e, f , x and m.
Problem 4
A positive integer n is clowny if the number obtained by reversing its digitsis greater than n. For example, 2009 is clowny because 9002 is greater than2009; however, 2010 is not clowny because 0102 = 102 is less than 2010 and
4
(Problems)
(page 5)
Problems 5
3443 is not clowny because it is equal to the number obtained by reversingits digits. How many four-digit numbers are clowny?
Problem 5
(a) Show a positive integer not greater than 1000 with at least 20 positivedivisors.
(b) Does there exist a positive integer not greater than 11000 with at least200 positive divisors?
2.2. Grades 8–9
Problem 1
See problem 4, Grades 6–7.
Problem 2
Let ABCD be a paralellogram and Γ the circumcircle of the triangle ABD.Lines BC and CD meet Γ at E 6= B and F 6= D respectively. Prove thecircumcenter of the triangle CEF lies on Γ.
Problem 3
Arnold and Bernold play the following game in a m × n board: Arnoldchooses one of its cells and places a knight on it. Then Bernold and Arnoldmove the knight alternately, with the condition that the knight visits a cellat most once. The player who is unable to move the knight loses. Determine,in terms of m and n, which player has the winning strategy.
Remark: the knight always moves two cells in a row or a column and thenone cell in the perpendicular direction.
Problem 4
Let a, b and c be real numbers such that a 6= b and a2(b+ c) = b2(c+ a) =2010. Compute c2(a+ b).
Problem 5
Let O be the intersection point of the diagonals of the cyclic quadrilateralABCD. The circumcircles of triangles AOB and COD meet lines BC andAD again at M , N , P and Q. Prove that the quadrilateral MNPQ isinscribed in a circle with center O.
Problem 6
The sidelenghts and area of a triangle are all integer numbers. Find theminimum value of its area.
5(Problems)
(page 6)
6 XXXII Brazilian Math Olympiad 2010
2.3. Grades 10–12
Problem 1
Find all functions f from real numbers to real numbers such that
f(a+ b) = f(ab)
for all irrationals a, b.
Problem 2
Let P (n) be a polynomial with real coefficients. Prove that there existintegers n and k such that k has at most n digits and at least P (n) divisors.
Problem 3
What is the maximum area of the shadow cast by a unit cube? We un-derstand “area of the shadow” of something as the area of its orthogonalprojection in a given plane.
Problem 4
Let ABCD be a quadrilateral with 6 ABC 6= 90◦. Let M and N be themidpoints of AD and CD, respectively. Prove that the lines perpendicu-lar to BC passing through M and perpendicular to AB passing throughN and BD are concurrent if and only if the diagonals BD and AC areperpendicular.
Problem 5
Find all values of n such that there exists a set S of n points in plane,no three of them collinear, with the following property: one can color thepoint such that if three of such points have the same color or three differentcolors then they do not determine an obtuse angle. The number of colors isunlimited.
Problem 6
Find all positive integers a, b such that
3a = 2b2 + 1.
6
(Problems)
(page 7)
Problems 7
2.4. Undergraduates
Problem 1
Compute∫ π/4
0
x
(sinx+ cosx) cosxdx.
Problem 2
See problem 3, Grades 10–12.
Problem 3
Let n be an integer and n1 be one of its divisors. Let A be a n×n symmetricmatrix defined by ai,i = 4, ai,i+1 = ai+1,i = −1 for all i such that 1 ≤ i ≤ n−1and i+1 is not a multiple of n1, ai,i+n1
= ai+n1,i = −1 and ai,j = 0 otherwise.
Prove that A has an inverse and that all of the entries in the inverse arepositive.
Problem 4
Define the polynomials(
xj
)
= x(x−1)(x−2)...(x−j+1)j! for j positive integer and
(
x0
)
= 1.
(a) Prove that all non-nil polynomials can be written uniquely as a linearcombination of such polynomials
(
xn
)
.
(b) Let c(n, k) be the coefficient of(
xk
)
in xn (as described in the previousitem). Compute
c(n, k) + c(n, k + 1)
c(n+ 1, k + 1).
Problem 5
For each finite subset F of the space R3, define Vr(F ) as the union of theopen spheres with center on each point of F and radius r. Prove that, for0 < r < R,
vol(VR(F )) ≤ R3
r3vol(Vr(F )).
Problem 6
Prove that if 102n+8 · 10n+1 has a prime factor of the form 60k+7, k andn both positive integers, then n and k are both even.
7
(Problems)
(page 9)
9 Solutions
3.1. Grades 6–7
Problem 1
The triangle is the 3–4–5 triangle.
(a) Since four triangles were used, the area is 4· 3·42 = 24 cm2. The perimeteris 2 · (5 + 4 + (4− 3) + 3) = 26 cm.
(b) Since the area of each triangle is 3·42 = 6 cm2, the side of the square
must be a multiple of 6. Moreover, since we are covering the sides ofthe square with the sides of the triangle, the side of the square must beequal to 3x+4y+5z, x, y, z ≥ 0. Though 6 = 2 · 3, it’s not hard to seethat it’s not possible to cover a square with side 6 cm with triangles.So the smallest square has side 12 cm and can be made of twelve 3× 4rectangles.
Problem 2
(a) For example,
7 4 18 5 29 6 3
9
(Solutions)
(page 10)
10 XXXII Brazilian Math Olympiad 2010
Notice that A = min{7, 8, 9} = 7 and B = max{1, 2, 3} = 3, so A−B =4.
(b) For example,
1 2 34 5 76 8 9
Notice that A = min{3, 7, 9} = 3 and B = max{1, 4, 6} = 6, so A−B =−3.
(c) No, it’s not possible. Since B = 3 is the greatest of the numbers ingreen cells, which are the smallest ones in their respective rows, thenumbers 1, 2 and 3 must be on different rows. Now A = 4 means that4 is the greatest number of a row, but it could only happen if two fothe numbers 1, 2, 3 are on the same row as 4, which cannot happen.
Problem 3
Consider the layer 3 × 3 × 1 to the left, with numbers 2, 2, 1 on one sideand a, 3, 2. The number of unit cubes in this layer is 2 + 2+ 1 = a+3+ 2,so a = 0. Similarly, considering the middle horizontal layer, 2 + x + 2 =3 + 3 + 1 ⇐⇒ x = 3. Analogously, 1 + 2 + 1 = m + 2 + 2 ⇐⇒ m = 0,a + 1 + d = 1 + 3 + m ⇐⇒ 0 + 1 + d = 1 + 3 + 0 ⇐⇒ d = 3. Now,2 + x + 2 = 1 + b + c ⇐⇒ b + c = 6. Since b, c ≤ 3, b = c = 3. Finally,3+b+e = 3+3+2 ⇐⇒ 3+e = 5 ⇐⇒ e = 2 and 1+c+f = 2+1+2 ⇐⇒3 + f = 4 ⇐⇒ f = 1.
Problem 4
Let (abcd) be a four-digit number a, b, c, d being its digits. So a number isclowny if and only if (dcba) > (abcd). This means that either d > a or d = aand c > b. Notice that we cannot have both d = a and b = c because itwould imply (dcba) = (abcd). There are 9·8
2 = 36 choices in the first case(recall that d > a > 0) and 9· 10·92 = 405 choices in the second case (9 choicesfor a = d and 10·9
2 for c > b ≥ 0). So there are 36 + 405 = 441 four-digitclowny numbers.
Problem 5
(a) For example, 900 = 22 · 32 · 52, which has (2 + 1) · (2 + 1) · (2 + 1) = 27positive divisors.
(b) No, there doesn’t. Let n be a number with at least 200 divisors. Ifthe i-th divisor is d, then the i-th to last divisor is n
d . Let m be the100th divisor. So m ≥ 100 and n
m > m ⇐⇒ n > m2 = 10000. Close
10
(Solutions)
(page 11)
Solutions 11
enough, but how do we fix this? First notice that if 97, 99 and 100are all divisors of n, then n ≥ lcm(97, 99, 100) > 11000. So the keyobservation is considering the 98th, 99th and 100th divisors. Let k, ℓand m be such divisors. Notice that if m ≥ 105 we are done becausethen n > m2 = 11025 > 11000. So 98 ≤ k < ℓ < m ≤ 104. Butgcd(x, y) ≤ |x − y| implies that n ≥ lcm(k, ℓ,m) ≥ k·ℓ·m
(ℓ−k)(m−ℓ)(m−k) ≥98·99·100
(104−98)·32 > 11000. Here we used the fact that if x+y ≤ 2t then xy ≤ t2
applied to x = ℓ− k, y = m− ℓ and t = 3.
3.2. Grades 8–9
Problem 1
See problem 4, Grades 6–7.
Problem 2
Consider all angles oriented and modulo 180◦. Let O be the center of thecircle. In the cyclic pentagonal ABEDF , 6 EBF = 6 EDF = 6 EDC =6 CED + 6 DCE = 6 BED + 6 DCE = 6 BAD + 6 DCE = 26 DCE. Thismenas that if M is the midpoint of the arc FE that does not contain A,6 FME = 6 EBF = 26 DCE = 26 FCE. Since ME = MF , M is thecircumcenter of triangle CEF .
Problem 3
Suppose, without loss of generality, m ≤ n. If m = 2, Arnold has winningstrategy if ond only if n is not a multiple of 4; para m ≥ 3, Arnold haswinning strategy if and only if m and n are both odd.
11
(Solutions)
(page 12)
12 XXXII Brazilian Math Olympiad 2010
Suppose m = 2. If 4 does not divide n, Arnold can win placing the knighton the first column if n = 4k + 1 and on the second column if n = 4k + r,r = 2 or r = 3. The knight should always be moved two columns ahead oneach move, allowing exactly 2k moves.
Now consider n = 4k. Divide the table into 2 × 4 subtables, and pair thecells in such a way that it’s always possible to move the knight between cellsfrom each pair:
1 2 3 43 4 1 2
Since the whole table is divided into pairs, Bernold can always move, nomatter where Arnold places the knight: it suffices to motve the knight tothe other cell in the pair. When Arnold plays, he will move the knight to acell from another pair, and Bernold can repeat this strategy. So if n = 4kBernold has winning strategy.
This finishes the case m = 2. The case m ≥ 3 follows in a similar fashion,dividing the table into smaller subtables:
1 2 3 43 6 1 52 5 4 6
1 2 3 4 5 63 4 1 7 8 92 7 8 9 6 5
1 2 34 A 12 3 4
1 2 3 4 53 4 6 A 72 1 7 5 6
Those tables prove that Bernold has winning strategy for tables 3×n, n evenand Arnold has winning strategy for tables 3 × n, n odd: divide the tableinto one 3×3 or 3×5 subtable and several 3×4 subtables; it suffices to placethe knight on the cell marked with an A and then use the aforementionedBernold’s strategy.
The case m = 4 can be verified using several 4×2 subtables if n is even andone 4× 3 subtable and several 4× 2 subtables if n is odd. This also provesthat if either Arnold or Bernold has winning strategy for a m×n table thenhe also has winning strategy for a (m + 4) × n table, m ≥ 3. So is sufficesto solve the problem for m ∈ {3, 4, 5, 6}.The case m = 5 can be solved using the following subtables:
1 2 3 4 5 63 4 1 7 8 92 10 11 9 6 512 13 14 15 7 810 11 12 13 14 15
1 2 3 4 53 4 5 6 72 1 10 8 911 A 12 7 612 10 11 9 8
12
(Solutions)
(page 13)
Solutions 13
Notice that Bernold has winning strategy for a 5× 4 table and Arnold haswinning strategy for a 5×3 table. Thus, in the case 5×n, n even, we merge5 × 4 subtables if n is a multiple of 4 and one 5 × 6 subtable and 5 × 4subtables if n = 4k + 2; if n is odd, we merge several 5 × 4 subtables to a5× 3 or 5× 5 subtable, if n = 4k + 3 or n = 4k + 1, respectively.
The case 6× n follows directly from the case 3× n if n is even (merge two3 × n tables) and merging 6 × 4 subtables to a 6 × 3 or 6 × 5 subtable, ifn = 4k + 3 or n = 4k + 1, respectively. All the cases are covered.
Problem 4
Since a 6= b, a2(b + c) = b2(c + a) ⇐⇒ a2b + a2c − b2c − ab2 = 0 ⇐⇒ab(a−b)+c(a−b)(a+b) = 0 ⇐⇒ ab+ca+bc = 0. So (a−c)(ab+bc+ca) =0 ⇐⇒ a2b+ abc+ a2c = abc+ bc2 + ac2 ⇐⇒ c2(a+ b) = a2(b+ c) = 2010.
Problem 5
Consider angles oriented modulo 180◦. Since 6 PCO = 6 BCA = 6 BDA =6 ODN are inscribed in the circumcircle of the triangle OCD, OP = ON .Analogously, OM = OQ.
Now, 6 QOP = 6 CPO = 6 CDO = 6 CDB = 6 CAB = 6 OAB = 6 OQB =6 OQP , so OP = OQ, and analogously, OM = ON .
Since OM = ON = OP = OQ, the result follows.
Problem 6
The 3–4–5 triangle has area 3·42 = 6. We will prove that no other trian-
gle with integer sidelengths and area has smaller area. Let a, b, c be the
13 (Solutions)
(page 14)
14 XXXII Brazilian Math Olympiad 2010
sidelenghts. Then its area is S =√
s(s− a)(s− b)(s− c), where s = a+b+c2 .
Since the area is also an integer, a+ b+ c is even, and s, s− a, s− b, s− care all integers.
Now, notice that the triangle cannot be equilateral, since equilateral tri-angles with an integer side have irrational area. So, at least two of thethree integer numbers s − a, s − b, s − c are distinct and, since s =(s − a) + (s − b) + (s − c) ≥ 1 + 1 + 2 = 4, S ≥
√4 · 2 · 2 · 1 =
√8, so
S ≥ 3.
If S is odd, s ≥ 5 and s−a, s−b, s−c are all odd, so S ≥√5 · 3 · 1 · 1 =
√15,
so S ≥ 5. The only relevant case is S = 5. But this would imply two of s,s− a, s− b, s− c being equal to 5, which is impossible.
If S is even, the only relevant case is S = 4. But then all of s, s−a, s−b, s−care powers of two. So if s > 4 then s ≥ 8 and s−a, s− b, s− c would be, insome order, 1, 1, 2, which is not possible because s = (s−a)+(s−b)+(s−c).If s = 4, the only possibility would be s− a, s− b, s− c being 1, 1, 2, whichdoes not work either.
3.3. Grades 10–12
Problem 1
Let f(0) = k. Plugging a =√2 and b = −
√2 one obtain f
(√2 + (−
√2))
=
f(√
2(−√2))
⇐⇒ f(−2) = k.
Let α ∈ R\Q. Since the quadratic equation x2−αx−2 = 0 has discriminant∆ = α2+8 > 0, its roots have sum α ∈ R \Q and product −2 ∈ Q, at leastone of its roots m is irrational; the other root, n = −2/m, is also irrational.Thus we can plug m and n, obtaining f(m + n) = f(mn) ⇐⇒ f(α) =f(−2) = k.
Now let q ∈ Q. The quadratic equation x2 − qx−√2 = 0 has discriminant
∆ = q2 + 4√2 > 0, sum of the roots q ∈ Q and product of the roots
−√2 ∈ R \ Q, so one of its roots r is irrational; the other root, s = q − r,
is also irrational. we can plug r and s, obtaining f(r + s) = f(rs) ⇐⇒f(q) = f
(
−√2)
= k.
So all the functions are the constant functions f(x) = k, k ∈ R.
Problem 2
Let d be the degree of P , and consider d+1 distinct primes p1, p2, . . . , pd+1.Let A = p1p2 . . . pd+1 and kN = AN . If A has r digits, then kN has at mostrN digits. On the other hand, kN has (N +1)d+1 positive divisors. Since P
14
(Solutions)
(page 15)
Solutions 15
has degree d, P (rN) < (N + 1)d+1 for all sufficiently large N , which solvesthe problem.
Problem 3
Let ABCD and EFGH be two opposite faces, AE, BF , CG and DH beingedges of the cube, and let X ′ be the orthogonal projection of point X ontothe plane. Notice that {A,G}, {B,H}, {C,E} and {D,F} are pairs ofopposite vertices. Suppose, without loss of generality, that A′ lies on theboundary of the projection of the cube. Then, considering the symmetryof the cube around the center of the cube, its symmetric point G′ lies onthe boundary as well. Two of the three neighboring vertices of A are goingto be neighbors of A′ in the projection (unless, say, face AEHD projectsonto a line; but in this case we consider a degenerate vertex inside this line).Suppose without loss that these neighbors are B′ and D′. So E ′ is insidethe projection. Again by symmetry H ′ and F ′ lie on the boundary of theprojection and C ′ lies inside the projection. Finally, since
−→AE =
−−→BF =−−→
CG =−−→DH, the projection of the cube is A′D′H ′G′F ′B′.
The faces ABCD, BCGF and CDHG project onto the parallelograms (orline segments) A′B′C ′D, B′C ′G′F and C ′D′H ′G′. Draw diagonals B′D′,B′G′ and D′G′. The area of the projection is then twice the area of thetriangle B′D′G′, which is at most the area of triangle BDG. This triangle
is equilateral with side√2, so the desired maximum is 2
(√2)
2√3
4 =√3.
Problem 4
Consider a homothety with center on D that takes M to A and N to C.So the perpendicular lines are mapped to the altitudes of the triangle ABCrelative to A and C, and the intersection P of the perpendicular lines is
15
(Solutions)
(page 16)
16 XXXII Brazilian Math Olympiad 2010
mapped to the orthocenter H of triangle ABC.
Notice that the condition that the perpendicular lines and BD are concur-rent is equivalent to B, P and D being collinear. But the homothety impliesthat D, P and H are collinear, so the three lines are concurrent if and onlyif BH and BD coincide, that is, BD ⊥ AC, since BH ⊥ AC.
Problem 5
We start with the following
Lemma. Every set with 5 or more points in the plane, no three of themcollinear, has three points that determine an obtuse angle. Besides, everyset with 4 points, no three of them determining an obtuse angle, is uniquelydefined by three of its points (that is, the position of one point can bedetermined from the other three points).
Proof. First, notice that the points should be vertices of a convex polygon.Otherwise, one of the points, say P , is inside a triangle with vertices on otherthree points A, B, C from the set, and since 6 APB+ 6 BPC+ 6 CPA = 360◦,of of the three angles 6 APB, 6 BPC, 6 CPA is greater than 90◦. Finally,since the sum of the internal angles of a convex n-gon is (n−2) ·180◦, one ofthe internal angles is greater than or equal to (n−2)·180◦
n =(
1− 2n
)
180◦ > 90◦
for n > 4. This proves the first part of the lemma.
The second part follows from the fact that if we have four points then allinternal angles of the quadrilateral with vertices on the four points must beequal to 90◦, that is, the quadrilateral is a rectangle.
Because of the lemma, there are at most four points with either the samecolor or three different colors in S. Let’s prove then that S cannot have
16
(Solutions)
(page 17)
Solutions 17
more than 12 points. If S has more than 12 points, since there are at mostfour colors and more than 3 · 4 = 12 points then there are four points withthe same color. Besides, since there are at most four points with the samecolor, there are also four points with four different colors.
Now we prove that S cannot have simultaneously four points with the samecolor and four points with four different colors. Suppose, by means of con-tradiction, the contrary. Let P = {A,B,C,D} be a set of points from Swith the same color and E, F and G points from S with three differentcolors and a different color from the points from P. So {A,E, F,G} and{B,E, F,G} are both sets of points with four different colors. But thelemma implies that E, F and G determine the position of both A and B,that is A = B, contradiction.
So S has at most 12 points. Consider the following examples with 12 points:consider the vertices of three unit squares whose centers are the vertices ofan equilateral triangle with a sufficiently large side. Color the vertices ofeach square with a single color, so we have three different colors, one foreach square. The angles determine by three points of the same color are allright angles, and the angles determined by three points with three differentcolor can be arbitrarily close to 60◦.
It is easy to obtain set S with less than 12 points deleting points from theexamples given above. So the answer is n ≤ 12.
Problem 6
The only solutions are (1, 1), (2, 2) and (5, 11).
If a is even and greater than 2, the equation can be rewritten as (3a/2 −1) · (3a/2 + 1) = 2b2. but gcd(3a/2 − 1, 3a/2 + 1) = gcd(3a/2 − 1, 2) = 2, so3a/2 + 1 = 4u2 and 3a/2 − 1 = 2v2 or 3a/2 + 1 = 2u2 and 3a/2 − 1 = 4v2.
In the former case, 3a/2 = (2v − 1)(2v + 1), and since gcd(2v − 1, 2v + 1) =gcd(2v − 1, 2) = 1, 2v − 1 = 1 ⇐⇒ v = 1 and a/2 = 1 ⇐⇒ a = 2 andthus b = 2.
In the latter case, 3a/2 = 4v2 + 1 =⇒ 0 ≡ v2 + 1 (mod 3) ⇐⇒ v2 ≡ −1(mod 3), which is impossible.
If a is odd, the equation is equivalent to 3 · (3(a−1)/2)2 − 2b2 = 1. Letc = 3(a−1)/2. Let’s find the solutions to 3c2 − 2b2 = 1 (∗). Since
(√3 +
√2)(
√3−
√2) = 1 =⇒ (
√3 +
√2)2k+1(
√3−
√2)2k+1 = 1
and
(√3 +
√2)2k+1 = ck
√3 + bk
√2 y (
√3−
√2)2k+1 = ck
√3− bk
√2 (∗∗)
17
(Solutions)
(page 18)
18 XXXII Brazilian Math Olympiad 2010
(ck, bk) are solutions to (∗), for integers k ≥ 0. Now suppose that (α, β) isa different solution. Then there exists an integer k such that
(√3 +
√2)2k−1 < α
√3 + β
√2 < (
√3 +
√2)2k+1
⇐⇒√3 +
√2 <
α√3 + β
√2
(5 + 2√6)k−1
< 9√3 + 11
√2
⇐⇒√3 +
√2 < (α
√3 + β
√2) · (5− 2
√6)k−1 < 9
√3 + 11
√2
It’s not hard to prove by induction that (α√3 + β
√2) · (5 − 2
√6)k−1 =
θ√3+φ
√2, θ and φ both integers. Moreover, θ
√3+φ
√2 > 1 > 1
θ√3+φ
√2=
θ√3− φ
√2 > 0, so θ
√3 > φ
√2 > 0. Hence (θ, φ) is a solution to (∗), with
1 < θ < 9. But it can be verified that there are no solutions with 1 < θ < 9,contradiction.
Now suppose that c > 9, that is, k > 1. By (∗∗) and the binomial theorem,
c =
k∑
m=0
(
2k + 1
2m+ 1
)
·3m ·2k−m = (2k+1) ·2k +(
2k + 1
3
)
·3 ·2k−1+ · · · (∗∗∗)
Thus 3 divides 2k + 1. Let 3t be the largest power of 3 that divides 2k + 1.Since
(
2k+13
)
· 3 · 2k−1 = (2k + 1) · k · (2k − 1) · 22k−1, the largest power of3 that divides the second term of the right hand side of (∗ ∗ ∗) is also 3t
and, letting 3s be the largest power of 3 that divides 2m+1, the m-th term(
2k+12m+1
)
· 3m · 2k−m = 2k+12m+1
(
2k2m
)
· 3m · 2k−m has at least t+m− s ≥ t+ 3s+12 − s
factors 3 (t from 2k + 1, m from 3m, minus s from 2m + 1). Since 3s =(1 + 2)s ≥ 1 + 2s ⇐⇒ 3s+1
2 − s ≥ 1 for s ≥ 0, with equality if and only ifs = 0 or s = 1, the m-th term has more than t + 1 factors 3 except whenm = 3s+1
2 . If s = 0 we have m = 1 ⇐⇒ 2m + 1 = 3, a known solution; ifs = 1 we have m = 2 which has no factors 3. Thus, since k > 1, all termsstarting from the third one have at least t+ 2 factors 3, and
c = (2k + 1) · 2k + (2k + 1) · k · (2k − 1) · 2k−1 + 3t+2 ·N⇐⇒ 3(a−1)/2 = (2k + 1) · 2k−1 · [(2k + 1)(k − 1) + 3] + 3t+2 ·N
Notice that since 3 divides (2k+ 1) then it also divides (2k+ 1)(k− 1) + 3;moreover, 9 divides (2k + 1)(k − 1), and then the largest power of 3 thatdivides (2k + 1)(k − 1) + 3 is 3. Hence the largest power of 3 that divides(2k+1) · 2k−1 · [(2k+1)(k− 1)+3] is t+1 (t factors from 2k+1 and 1 from(2k + 1)(k − 1) + 3). Finally,
3(a−1)/2 = c = 3t+1(1 + 3N),
which is impossible because N > 0.
So there are no more solutions.
18 (Solutions)
(page 19)
Solutions 19
3.4. Undergraduates
Problem 1
First notice the identity sinx + cosx =√2(
cosx cos π4 + sinx sin π
4
)
=√2 cos
(
π4 − x
)
. So
I =
∫ π/4
0
x
(sinx+ cosx) cosxdx =
∫ π/4
0
x√2 cos
(
π4 − x
)
cosxdx
Now consider the substitution y = π4 − x. So dx = −dy and
∫ π/4
0
x√2 cos
(
π4 − x
)
cosxdx =
∫ 0
π/4
−π4 − y√
2 cos y cos(
π4 − y
)dy
=
∫ π/4
0
π4 − y√
2 cos y cos(
π4 − y
)dy =
∫ π/4
0
π4 − x√
2 cos(
π4 − x
)
cosxdx
Summing the two integrals, we find
2I =
∫ π/4
0
x√2 cos
(
π4 − x
)
cosxdx+
∫ π/4
0
π4 − x√
2 cos(
π4 − x
)
cosxdx
=
∫ π/4
0
π4 − x+ x√
2 cos(
π4 − x
)
cosxdx =
∫ π/4
0
π4√
2 cos(
π4 − x
)
cosxdx
So I = π8
∫ π/4
01
(sinx+cosx) cosxdx.
Let f(x) = ln(sinx + cosx) and g(x) = ln(cosx). Notice that f ′(x) =cosx−sinxsinx+cosx = f ′(x) = cos2 x−sinx cosx
(sinx+cosx) cosx and g′(x) = − sinxcosx = − sin2 x+sinx cosx
(sinx+cosx) cosx . So
f ′(x)− g′(x) = cos2 x−sinx cosx+sin2 x+sinx cosx(sinx+cosx) cosx = 1
(sinx+cosx) cosx . Hence
I =π
8
∫ π/4
0
1
(sinx+ cosx) cosxdx =
π
8
(
f(π
4
)
− g(π
4
)
− (f(0)− g(0)))
=π
8
(
ln(√2)− ln
(√2
2
)
− ln 1 + ln 1
)
=π ln 2
8
Problem 2
See problem 3, grades 10–12.
19
(Solutions)
(page 20)
20 XXXII Brazilian Math Olympiad 2010
Problem 3
Let’s find the inverse of A/4 = I −X, where all entries in X are either 0 or1/4. We will use the series
(I −X)−1 = I +X +X2 +X3 + · · ·
First let’s prove that this series converges. It suffices to show that themaximum M such that ‖Xw‖ ≤ M‖w‖ for all column vectors w of X is lessthan 1, so the sum of the entries always decrease by a factor smaller than 1if you multiply a vector by X2; then we sum the series as (I +X)(I +X2 +X4 + · · ·). Notice that every row of X has at most four nonzero entries, allof which are equal to 1/4. So if w = (a1, a2, . . . , an) then every entry of Xw
is of the formar1+ar2+···+ars
4 , s ≤ 4. By the Cauchy-Schwarz inequality, its
square is at most s16 (a
2r1+a2r2 + · · ·+a2rs) ≤
a2r1+a2r2+···+a2rs4 , with equality if and
only if ar1 = ar2 = · · · = ars and s = 4. Summing over all the rows, the sumof squares of the coordinates of Xw is at most a21 + a22 + · · · + a2n = ‖w‖2,because all columns of X have at most four nonzero entries. But equalitywould only happen if all entries ai are equal and s = 4 always, which doesnot happen for, say, the first row. So M < 1 and the series converges.
Consider the graph whose vertices are the numbers v1, v2, . . . , vn and weconnect vi and vj if and only if the entry xij in X is 1/4. By the defini-tion of the matrix A, this graph has a lattice-like configuration: it can besplit in several paths v1v2 . . . vn1−1, vkn1
vkn1+1 . . . v(k+1)n1−1, 1 ≤ k ≤ nn1
− 1,vrvr+n1
vr+2n1. . . vr+n−n1
, 1 ≤ r ≤ n1. It is clear that this graph is connected.Since the entry mij in Xk is nonzero if and only if there exists a circuitfrom i to j with k edges, for all i, j there is k such that the correspondingentry mij in Xk is nonzero. This proves that all entries in the inverse of Ais positive.
Problem 4
(a) Induct on the degree n of the polynomial P (x) =∑n
k=0 akxk. It is
immediate if n = 0. Since the degree of(
xk
)
, k < n, is k, the coefficient
on(
xn
)
on P (x) is an ·n!. Now consider the polynomial P (x)−an ·n!(
xn
)
.
This is a polynomial with degree less than n, so P (x) − an · n!(
xn
)
can
be uniquely represented as a linear combination of the polynomials(
xk
)
,and the result follows.
20
(Solutions)
(page 21)
Solutions 21
(b) We have xn =∑n
k=0 c(n, k)(
xk
)
, so
xn+1 =
n∑
k=0
c(n, k)
(
x
k
)
x
=
n∑
k=0
c(n, k)
(
x
k
)
(x+ 1)−n∑
k=0
c(n, k)
(
x
k
)
=
n∑
k=0
c(n, k)(k + 1)
(
x+ 1
k + 1
)
−n∑
k=0
c(n, k)
(
x
k
)
=
n∑
k=0
c(n, k)(k + 1)
((
x
k + 1
)
+
(
x
k
))
−n∑
k=0
c(n, k)
(
x
k
)
=
n∑
k=0
(c(n, k − 1)k + (k + 1− 1)c(n, k))
(
x
k
)
=
n∑
k=0
k(c(n, k − 1) + c(n, k))
(
x
k
)
in which we consider c(n,−1) = 0.
Thus c(n + 1, k) = k(c(n, k − 1) + c(n, k)) ⇐⇒ c(n,k)+c(n,k−1)c(n+1,k) = 1
k , andc(n,k+1)+c(n,k)
c(n+1,k+1) = 1k+1 .
Problem 5
Let F = {P1, P2, . . . , Pn} and let αij be the perpendicular plane bisector ofPi and Pj for i 6= j. Those planes define n convex regions R1, R2, . . . , Rn,where Ri is the intersection of the half-spaces determined by αij that containPi. Finally, let Ar(i) be the intersection of Ri with the sphere with centerPi and radius r. Thus, since the (disjoint) union of the regions Ri is thewhole space, Vr(F ) is the disjoint union of Ar(1), Ar(2), . . ., Ar(n).
Now, for each point Pi, apply a homothety with center Pi and ratio r/R < 1.It is clear that the image of AR(i) is contained in Ar(i), since a sphere withradius R is taken to a sphere with radius r and the planes αij are taken to
planes α′ij closer to Pi. Thus vol(Ar(i)) ≥
(
rR
)3vol(Ar(i)) and the result
follows by summing up these inequalities for i = 1, 2, . . . , n.
Problem 6
Let p = 60k+7 be such a prime. Then 102n+8 ·10n+1 ≡ 0 (mod p) ⇐⇒(10n − 1)2 ≡ −10n+1 (mod p). Now suppose n is odd. Then (10n − 1)2 ≡
21
(Solutions)
(page 22)
22 XXXII Brazilian Math Olympiad 2010
−1 · (10(n+1)/2)2 (mod p), and −1 is a quadratic residue. But by the Eulercriterion,
(−1p
)
= (−1)(p−1)/2 = −1, a contradiction. So n is even and,
moreover,(−10
p
)
= 1 ⇐⇒(
2p
)(
5p
)
= −1 ⇐⇒(
2p
)
= −(
5p
)
.
By the quadratic reciprocity lemma,(
5p
)(
p5
)
= (−1)5−12 · p−1
2 = 1 ⇐⇒(
5p
)
=(
p5
)
=(
25
)
= −1. So(
2p
)
= 1 ⇐⇒ (−1)p2−18 = 1, so p2−1
8 = (30k+3)(15k+2)is even. Since 30k + 3 is odd, 15k + 2 is even, that is, k is even.
Notice that we did not need the factor 3, that is, the problem holds forp = 20k + 7.
22
(Solutions)
(page 23)
23 Winners in 2010
4.1. Grades 6–7
Gold medals
Ana Emılia Hernandes Dib
Pedro Henrique Alencar Costa
Ryunosuke Watanabe Tagami
Helena Veronique Rios
Italo Lesione de Paiva Rocha
Jose Henrique Carvalho
Silver medals
Juliana Bacelar de Freitas
Daniel Lima Braga
Hermes Lins e Nascimento
Laıs Monteiro Pinto
Lucca Morais de Arruda Siaudzionis
Leandro Alves Cordeiro
Henrique Gontijo Chiari
Andre Akinaga Benites
Gabriel Diniz Vieira e Sousa
Rafael Seiji Uezu Higa
Adriana de Sousa Figueiredo
Gustavo Figueiredo Serra
Bronze medals
Matheus Uchoa Constante
Kristian Holanda Nogueira
Fabio Itikama
Loic Dominguez
Jiang Zhi
Ricardo Ken Wang Tsuzuki
Ana Caroline Obana da Cruz
Ana Paula Lopes Schuch
Jose Marcio Machado de Brito
Lucas Bastos Germano
Victoria Moreira Reis Cogo
23
(Winners in 2010)
(page 24)
24 XXXII Brazilian Math Olympiad 2010
Thiago Araujo Oliveira
Gabriel Toneatti Vercelli
Nathan Bonetti Teodoro
Jefferson Daxian Hong
Cristobal Sciutto Rodriguez
Aruana Almeida Correa
Cynthia Lacroix Herkenhoff
Honorable mention
Kaıque Maestrini Sacchi
Igor de Lacerda
Rafael Reple Geromee
Leonardo de Matos Felippetti Mariano
Gabriel Passamani Correa
Daniel de Almeida Souza
Diego Teixeira Nogueira Fidalgo
Natan Novelli Tu
Ricardo Borsari Brinati
Rafael Neves Vieira
Juliano Pecica Negri
Gustavo Rodrigues Machado
Zoltan Flamarion Glueck Carvalho
Gabriel Ribeiro Barbosa
Pedro Henrique Rocha de Freitas
Pedro Henrique Sacramento de Oliveira
Guilherme Goulart Kowalczuk
Pedro de Vasconcellos Oporto
Aryssa Victoria Shitara
Ives Vaz Caldeira Lopes
Marcos Vinıcius de Oliveira Soares
Jessica Carolina Zilio
Joao Pedro Graca Melo Vieira
Henrique Medici Pontieri
Gabriel Caino Castilho Rodrigues
Tamara P. de A. Moraes
Karine Quaresma Lima
Natalia Brasileiro Lins Barbosa
Lucki Li
Heloısa Antunes de Medeiros
24 (Winners in 2010)
(page 25)
Winners in 2010 25
Iuri Grangeiro Carvalho
Lara Sampaio Pinheiro de Freitas
Maria Julia Costa Medeiros
Kevin Korpasch
Sofıa Leite Correia Lima
Joao Baptista de Paula e Silva
Bernardo Puetter Schaeffer
Julia Bertelli
Rafael Purim de Azevedo
Pedro Henrique da Silva Dias
Marcelo Bandeira de Melo Boavista
Gabriel Branco Frizzo
Maria Eduarda Muller Eyng
Henrique Martınez Rocamora
Felipe Roz Barscevicius
Joao Vitor Vaz Oliveira
Mateus Siqueira Thimoteo
Ebenezer Pinto Bandeira Neto
Maria Clara Vasconcelos Andrade
Rafael Beck
Arthur Monteiro Dos Santos
Julia Wotzasek Pereira
Gabriel Oliveira Rigo
Leonardo Galante Barco
Bruno Scatolini
Lucas Pereira Galvao de Barros
Vıtor Ossamu Rodrigues Okamura
4.2. Grades 8–9
Gold medals
Rafael Rodrigues Rocha de Melo
Vinıcius Canto Costa
Henrique Vieira G. Vaz
Fellipe Sebastiam da Silva P. Pereira
Roberto Tadeu Abrantes de Araujo
Pedro Victor Falci de Rezende
25
(Winners in 2010)
(page 26)
26 XXXII Brazilian Math Olympiad 2010
Silver medals
Alessandro A. de Oliveira Pacanowski
Lincoln de Queiroz Vieira
Tadeu Pires de Matos Belford Neto
Vitor Ramos de Paula
Francisco Markan Nobre de Souza Filho
Jair Gomes Soares Junior
Breno Soares da Costa Vieira
Gabriel Jose Moreira da Costa Silva
Pedro Morais de Arruda Siaudzionis
Gabriel Sena Galvao
Fabio da Silva Soares
Michel Rozenberg Zelazny
Bruno Eidi Nishimoto
Franco Matheus de Alencar Severo
Aime Parente de Sousa
Bronze medals
Marcos Paulo Nunes de Lima Silva
Gabriel Nogueira Coelho de Togni de Souza
Rafael Tedeschi Eugenio Pontes Barone
Murilo Corato Zanarella
Rodrigo Sanches Angelo
Alexandre Perozim de Faveri
Luıze Mello D’urso Vianna
Maria Clara Cardoso
Liara Guinsberg
Lucas Cawai Juliao Pereira
Luis Guilherme Gomes Aguiar
Carlos Adriano Vieira
Daniel Santana Rocha
Raphael Mendes de Oliveira
Samuel Brasil de Albuquerque
Gustavo Souto Henriques Campelo
Honorable mention
Lucas de Moura Herlin
Vitor Dias Gomes Barrios Marin
Joao Pedro Sedeu Godoi
26
(Winners in 2010)
(page 27)
Winners in 2010 27
Suzane Eberhart Ribeiro da Silva
Icaro Sampaio Viana
Pedro Henrique Bortolozo Maria
Fabio Kenji Arai
Guilherme de Oliveira Rodrigues
Alexandre Mendonca Cardoso
Leyberson Pereira Assuncao
Rubens Martins Bezerra Farias
Joao Vıtor Fernandes Paiva
Bruno Almeida Costa
Daniel Lima Santanelli
Marılia Nascimento Monteiro
Igor Albuquerque Araujo
Josue Knorst
Ricardo Vieira Marques
Julio Cesar de Barros
Thomas Akio Ikeda Valvassori
Gabriel Fazoli Domingos
Henrique Luan Gomes Pereira Braga
Beatriz Yumi Ota
Kiane Sassaki Menezes
Eric Gripa Marques
Samuel Kuo Chen Shao
Pedro Henrique Jagosenit Vilaca
Caio de Souza Camara
Lucas David Noveline
Lucas Rebelo Vieira da Silva
Elias Brito Oliveira
Guilherme Ryu Odaguiri Kobori
Mariana Souza de Araujo
Francisco Claudio Coelho
Murilo Leao Pereira
Jadi Diniz Guimaraes de Queiroz
Caio Lima Albuquerque
Carolina Lima Guimaraes
27
(Winners in 2010)
(page 28)
28 XXXII Brazilian Math Olympiad 2010
4.3. Grades 10–12
Gold medals
Gustavo Lisboa Empinotti
Marcelo Tadeu de Sa Oliveira Sales
Joao Lucas Camelo Sa
Hanon Guy Lima Rossi
Maria Clara Mendes Silva
Silver medals
Matheus Secco Torres da Silva
Lucas Lourenco Hernandes
Deborah Barbosa Alves
Henrique Gasparini Fiuza do Nascimento
Luiz Filipe Martins Ramos
Andre Macieira Braga Costa
Thiago Saksanian Hallak
Victor Juca Martins
Caıque Porto Lira
Gustavo Haddad Francisco e Sampaio Braga
Alvaro Lopes Pedroso
Andre Amaral de Sousa
Bronze medals
Marcos Massayuki Kawakami
Carlos Henrique de Andrade Silva
Rafael Kazuhiro Miyazaki
Andre Saraiva Nobre dos Santos
Daniel Eiti Nishida Kawai
Lucas de Freitas Smaira
Cassio dos Santos Sousa
Alessandro Macedo de Araujo
Breno Vieira da Silva Passos
Iago Dalmaso Brasil Dias
Isabella Amorim Goncalez
Daniel dos Santos Bossle
Davi Coelho Amorim
Lucas Mestres Mendes
Vinıcius Gomes Pereira
28(Winners in 2010)
(page 29)
Winners in 2010 29
Renan Pablo da Cruz
Jonas Rocha Lima Amaro
Iuri Rezende Souza
Honorable mention
Matheus Araujo Marins
Felipe Vieira de Paula
Rafael Farias Marinheiro
Elvis Falcao de Araujo
Pablo Almeida Gomes
Paulo Gabriel Ramos Monteiro
Victor de Oliveira Bitaraes
Daniel Caueh Dunaiski Figueira Leal
Raphael Julio Barcelos
Fernando Fonseca Andrade Oliveira
Felipe Mendes dos Santos
Felipe Abella Cavalcante Mendonca de Souza
Francisco Raul Lobo Rodrigues
Gabriel Leite de Carvalho
Andre Austregesilo Scussel
Victorio Takahashi Chu
Victor Jose Tiburtius Franco
Matheus Cavalcante Lima
Cleberton de Santana Oliveira
Mauro Brito Junior
Gabriel Jose Guimaraes Barbosa
Lucas Colucci Cavalcante de Souza
Sarah Villanova Borges
Ivan Tadeu Ferreira Antunes Filho
Dalton Felipe de Menezes
Thiago de Paula Vasconcelos
Jardiel Freitas Cunha
Ana Beatriz Prudencio de A. Reboucas
Rafael Sussumu Yamaguti Miada
Davi Sampaio de Alencar
Bruno Ferri de Moraes
29
(Winners in 2010)
(page 30)
30 XXXII Brazilian Math Olympiad 2010
4.4. Undergraduates
Gold medals
Rafael Tupynamba Dutra
Renan Henrique Finder
Regis Prado Barbosa
Ramon Moreira Nunes
Thomas Yoiti Sasaki Hoshina
Silver medals
Guilherme Rodrigues Nogueira de Souza
Jorge Henrique Craveiro de Andrade
Rafael Assato Ando
Gabriel Luıs Mello Dalalio
Charles Barbosa de Macedo Brito
Leonardo Ribeiro de Castro Carvalho
Marcelo Matheus Gauy
Leandro Farias Maia
Bronze medals
Adenilson Arcajo de Moura Junior
Paulo Andre Carvalho de Melo
Joas Elias dos Santos Rocha
Guilherme Lourenco Mejia
Reinan Ribeiro Souza Santos
Rafael Alves da Ponte
Davi Lopes Alves de Medeiros
Luca Mattos Moller
Renato Reboucas de Medeiros
Danilo Furlan Kaio
Rafael Endlich Pimentel
Paulo Sergio de Castro Moreira
Honorable mention
Carlos Coelho Lechner
Thiago Ribeiro Ramos
Hugo Fonseca Araujo
Alysson Espındola de Sa Silveira
Jordan Freitas Piva
30
(Winners in 2010)
(page 31)
Winners in 2010 31
Erik Fernando de Amorim
Daniel Ungaretti Borges
Antonio Deromir Neves Silva Junior
Rafael Parpinel Cavina
Isaque Santa Brigida Pimentel
Mateus Oliveira de Figueiredo
Davi Dos Santos Lima
Bruno da Silva Santos
Francisco Osman Pontes Neto
Breno Vieira de Aguiar
Ricardo Turolla Bortolotti
Guilherme Philippe Figueiredo
Daniel de Barros Soares
Hudson do Nascimento Lima
Eduardo Fischer
Luty Rodrigues Ribeiro
Jose Leandro Pinheiro
Caio Ishizaka Costa
Gabriel Caser Brito
Leonardo Donisete da Silva
Alan Anderson da Silva Pereira
Diego Andres de Barros Lima Barbosa
Renato Dias Costa
Ivan Guilhon Mitoso Rocha
Willy George do Amaral Petrenko
Leonardo Borges Avelino
Jose Armando Barbosa Filho
31
(Winners in 2010)