Nordic (NMC) 1998 2018 EN with solutions

20 th Nordic Mathematical ContestThursday March 30, 2006 English version Time allowed: 4 hours.. The 21st Nordic Mathematical ContestMarch 29, 2007 English version Time allowed: 4 hours.

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13-th Nordic Mathematical Contest

Find all solutions of the equation f (n) = 1999.

2 A convex heptagon with all different sides is inscribed in a circle At most, how many angles equal to 120 ◦ can this heptagon have?

3 Nonnegative integers a and b are given A soldier is walking on the infinite lattice Z× Z as follows In each step, from a point (x, y) he is only allowed

to go to one of the points (x ± a, y ± b) and (x ± b, y ± a) Find all values

of a and b for which the soldier can visit every point of the lattice during his infinite walk.

4 Let a1, a2, , a n be positive numbers (n ≥ 1) Show that

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The IMO Compendium Group,

D Djukić, V Janković, I Matić, N Petrović www.imo.org.yu

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March 30, 2000

1 In how many ways can the number 2000 be written as a sum of three positive, not necesarily different integers? (The order of summands is irrelevant.)

2 The persons P1, P2, , P n sit around a table in this order, and each one has a number of coins Initially, P1has one coin more than P2, P2has one coin more than P3, etc Now P1 gives one coin to P2, who in turn gives two coins to P3, etc., up to P n who gives n coins to P1; then P1 continues

by giving n + 1 coins to P2, etc The transactions go on until someone has not enough coins to give away one coin more than he just received After this process ends, it turns out that there are two neighbors at the table one of whom has five times as many coins as the other Find the number

of persons and the number of coins circulating around the table.

3 In the triangle ABC, the bisectors of angles B and C meet the opposite sides at D and E, respectively The bisectors intersect at point O such that OD = OE Prove that either △ABC is isosceles or ∠A = 60 ◦

4 A real function defined for 0 ≤ x ≤ 1 satisfies f (0) = 0, f (1) = 1, and

1

2 ≤

f (x) − f (y)

f (y) − f (z) ≤2whenever 0 ≤ x < y < z ≤ 1 and z − y = y − x Show that 1

7 ≤ f 1

3 ≤ 4

7

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March 29, 2001

1 Let A be a finite set of unit squares in the coordinate plane, each of which has vertices at integer points Show that there exists a subset B of A consisting of at least 1/4 of the squares in A such that no two distinct squares in B have a common vertex.

2 A function f : R → R is bounded and satisfies

f

x +13

+ f

x +12

= (x) + f

x +56

for all real x Show that f is periodic.

3 Find the number of real roots of the equation

x8− x7+ 2x6− 2x5+ 3x4− 3x3+ 4x2− 4x +5

2 = 0.

4 Each of the diagonals AD, BE and CF of a convex hexagon ABCDEF divides its area into two equal parts Prove that these three diagonals pass through the same point.

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April 4, 2002

1 A trapezoid ABCD with AB k CD and AD < CD is inscribed in a circle

c Let DP be a chord parallel to AC The tangent to c at D meets the line

AB at E, and the lines P B and DC meet at Q Prove that EQ = AC.

2 Let N balls, numbered 1 through N , be distributed over two urns A ball is taken from one urn and moved to the other It turns out that the arithmetic means of the numbers on the balls in each urn increased by the same number x What is the greatest possible value of x?

3 Let a1, a2, , a n , b1, b2, , b n be real numbers with a1, , a n distinct Show that if the product (a i + b1)(a i + b2) · · · (a i + b n ) takes the same value for every i = 1, 2, , n, then the product (a1+ b j )(a2+ b j ) · · · (a n + b j ) also takes the same value for every j = 1, 2, , n.

4 Eva, Per, and Anna randomly select different nine-digit integers made of digits 1, 2, , 9 and check if they are divisible by 11 Anna claims that the probability that the number is divisible by 11 is exactly 1/11; Eva believes that this probability is less than 1/11, while Per thinks that it is greater than 1/11 Who of them is right?

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April 3, 2003

1 The squares of a rectangular chessboard with 10 rows and 14 columns are colored alternatingly black and white in the usual manner Some stones are placed the board (possibly more than one on the same square) so that there are an odd number of stones in each row and each column Show that the total number of stones on black squares is even.

2 Find all triples (x, y, z) of integers satisfying the equation

x3+ y 3 + z 3 − 3xyz = 2003.

3 An interior point D of an equilateral triangle ABC is taken so that

∠ADC = 150 ◦ Prove that the triangle whose sides are congruent to

AD, BD and CD is right-angled.

4 Find all functions f from R \ {0} to itself satisfying

f (x) + f (y) = f (xyf (x + y)) for all x, y 6= 0 with x + y 6= 0.

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2 The Fibonacci sequence is defined by f1= 0, f2= 1, and f n +2 = f n +1 +f n

for n ≥ 1 Prove that there is a strictly increasing arithmetic progression whose no term is in the Fibonacci sequence.

3 Given a finite sequence x1,1, x2,1, , x n, 1 of integers (n ≥ 2), not all equal, define the sequences x1,k, , x n,k by

x i,k +1 =1

2(xi,k + x i +1,k ), where x n +1,k = x1,k Show that if n is odd, then not all x j,k are integers Is this also true for even n?

4 Let a, b, c be the sides and R be the circumradius of a triangle Prove that

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3 There are 2005 boys and girls sitting at a round table No more than 668

of them are boys A girl G is said to be in a strong position if, counting from G to either direction at any length (G herself included), the number

of girls is always strictly larger than the number of boys Prove that there always exists a girl in a strong position.

4 Circle C1touches circle C2internally at A A line through A intersects C1again at B and C2again at C The tangent to C1at B intersects C2at D and E The tangents to C1 through C touch C1 at F and G Prove that points D, E, F, G are concyclic.

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20 th Nordic Mathematical Contest

Thursday March 30, 2006 English version

Time allowed: 4 hours Each problem is worth 5 points

Problem 1 Let B and C be points on two fixed rays emanating from apoint A such that AB + AC is constant

Prove that there exists a point D 6= A such that the circumcircles of thetriangels ABC pass through D for every choice of B and C

Problem 2 The real numbers x, y and z are not all equal and fulfil

Problem 3 A sequence of positive integers {an} is given by

a0= m and an+1= a5n+ 487 for all n ≥ 0

Determine all values of m for which the sequence contains as many squarenumbers as possible

Problem 4 The squares of a 100 × 100 chessboard are painted with 100different colours Each square has only one colour and every colour is usedexactly 100 times

Show that there exists a row or a column on the chessboard in which at least

10 colours are used

Only writing and drawing sets are allowed

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The 21st Nordic Mathematical Contest

March 29, 2007 English version

Time allowed: 4 hours Each problem is worth 5 points Only writing and drawingsets are allowed

(ii) erase one or both of two equal numbers on the blackboard

The player who is not able to make her turn loses the game Who will win thegame if Anne begins and both players act in an optimal way?

Problem 4

A line through a point A intersects a circle in two points, B and C, in such a waythat B lies between A and C From the point A draw the two tangents to thecircle, meeting the circle at points S and T Let P be the intersection of the lines

ST and AC Show that AP/P C = 2 · AB/BC

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The 22nd Nordic Mathematical Contest

31 March 2008 English version

Time allowed is 4 hours Each problem is worth 5 points The only permittedaids are writing and drawing materials

Problem 1

Determine all real numbers A, B and C such that there exists a real function

f that satisfies

f (x + f (y)) = Ax + By + Cfor all real x and y

Problem 2

Assume that n ≥ 3 people with different names sit around a round table Wecall any unordered pair of them, say M and N , dominating, if

(i) M and N do not sit on adjacent seats, and

(ii) on one (or both) of the arcs connecting M and N along the table edge,all people have names that come alphabetically after the names of Mand N

Determine the minimal number of dominating pairs

Problem 3

Let ABC be a triangle and let D and E be points on BC and CA, respectively,such that AD and BE are angle bisectors of ABC Let F and G be points

on the circumcircle of ABC such that AF and DE are parallel and FG and

BC are parallel Show that

AG

BG =

AB + AC

AB + BC.Problem 4

The difference between the cubes of two consecutive positive integers is asquare n2, where n is a positive integer Show that n is the sum of twosquares

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The 23rd Nordic Mathematical Contest

Thursday April 2, 2009 English version

Time allowed is 4 hours Each problem is worth 5 points The only permitted aidsare writing and drawing tools

Problem 1

A point P is chosen in an arbitrary triangle Three lines are drawn through P whichare parallel to the sides of the triangle The lines divide the triangle into three smallertriangles and three parallelograms Let f be the ratio between the total area of thethree smaller triangles and the area of the given triangle Show that f ≥ 13 anddetermine those points P for which f = 13

Problem 3

The integers 1, 2, 3, 4 and 5 are written on a blackboard It is allowed to wipe outtwo integers a and b and replace them with a + b and ab Is it possible, by repeatingthis procedure, to reach a situation where three of the five integers on the blackboardare 2009?

Problem 4

There are 32 competitors in a tournament No two of them are equal in playingstrength, and in a one against one match the better one always wins Show that thegold, silver, and bronze medal winners can be found in 39 matches

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Version: English

24th Nordic Mathematical Contest

13th of April, 2010

1. A function f : Z+ → Z+, where Z+ is the set of positive integers, is non-decreasing and

satisfies f (mn) = f (m)f (n) for all relatively prime positive integers m and n Prove that

f (8)f (13) ≥ (f (10))2

2. Three circles ΓA, ΓB and ΓC share a common point of intersection O The other common

point of ΓAand ΓB is C, that of Γ Aand ΓC is B, and that of Γ C and ΓB is A The line AO

intersects the circle ΓA in the point X 6= O Similarly, the line BO intersects the circle Γ B

in the point Y 6= O, and the line CO intersects the circle Γ C in the point Z 6= O Show

a) If Richard chooses the buttons to be pressed, what is the maximum number of differentcombinations of buttons he can press until Laura can assign the buttons to the lampscorrectly?

b) Supposing that Laura will choose the combinations of buttons to be pressed, what

is the minimum number of attempts she has to do until she is able to associate thebuttons with the lamps in a correct way?

4. A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones Find the least positive integer k such that each positive integer n can be written as n = a1± a2± a3± ± a k where a1, , ak are simple

Time allowed is 4 hours.

Each problem is worth 5 points.

Only writing and drawing tools are permitted.

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The 25th Nordic Mathematical Contest

Monday 4 April 2011 English version

The time allowed is 4 hours Each problem is worth 5 points The only aids permittedare writing and drawing tools

Problem 1

When a0, a1, , a1000 denote digits, can the sum of the 1001-digit numbers

a0a1 a1000 and a1000a999 a0 have odd digits only?

Problem 2

In a triangle ABC assume AB = AC, and let D and E be points on the extension ofsegment BA beyond A and on the segment BC, respectively, such that the lines CDand AE are parallel Prove CD ≥ 4h

BC CE, where h is the height from A in triangleABC When does equality hold?

Problem 3

Find all functions f such that

f (f (x) + y) = f (x2− y) + 4yf (x)for all real numbers x and y

Problem 4

Show that for any integer n ≥ 2 the sum of the fractions 1

ab, where a and b arerelatively prime positive integers such that a < b ≤ n and a + b > n, equals 1

2.(Integers a and b are called relatively prime if the greatest common divisor of a and

b is 1.)

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Tuesday, 27 March 2012 English VersionThe time allowed is 4 hours Each problem is worth 5 points The only

permitted aids are writing and drawing tools

Problem 1 The real numbers a, b, c are such that a2+ b2 = 2c2, and also suchthat a 6= b, c 6= −a, c 6= −b Show that

(a + b + 2c)(2a2− b2− c2)(a − b)(a + c)(b + c)

is an integer

Problem 2 Given a triangle ABC, let P lie on the circumcircle of the triangleand be the midpoint of the arc BC which does not contain A Draw a straightline l through P so that l is parallel to AB Denote by k the circle which passesthrough B, and is tangent to l at the point P Let Q be the second point ofintersection of k and the line AB (if there is no second point of intersection,choose Q = B) Prove that AQ = AC

Problem 3 Find the smallest positive integer n, such that there exist n integers

x1, x2, , xn (not necessarily different), with 1 ≤ xk ≤ n, 1 ≤ k ≤ n, and suchthat

x1+ x2+ · · · + xn = n(n + 1)

2 , and x1x2· · · xn = n!,but {x1, x2, , xn} 6= {1, 2, , n}

Problem 4 The number 1 is written on the blackboard After that a sequence

of numbers is created as follows: at each step each number a on the blackboard

is replaced by the numbers a − 1 and a + 1; if the number 0 occurs, it is erasedimmediately; if a number occurs more than once, all its occurrences are left onthe blackboard Thus the blackboard will show 1 after 0 steps; 2 after 1 step;

1, 3 after 2 steps; 2, 2, 4 after 3 steps, and so on How many numbers will there

be on the blackboard after n steps?

1

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Monday, 8 April 2013 English Version

The time allowed is 4 hours Each problem is worth 5 points.

The only permitted aids are writing and drawing tools.

Problem 1 Let (a n)n≥1 be a sequence with a1 = 1 and

an+1 =jan+√

an+ 1 2

k

for all n ≥ 1, where bxc denotes the greatest integer less than or equal to x Find all n ≤ 2013 such that a n is a perfect square

Problem 2 In a football tournament there are n teams, with n ≥ 4, and each

pair of teams meets exactly once Suppose that, at the end of the tournament,the final scores form an arithmetic sequence where each team scores 1 more pointthan the following team on the scoreboard Determine the maximum possible score

of the lowest scoring team, assuming usual scoring for football games (where thewinner of a game gets 3 points, the loser 0 points, and if there is a tie both teamsget 1 point)

Problem 3 Define a sequence (n k)k≥0 by n0 = n1 = 1, and n 2k = n k + n k−1 and

n 2k+1 = n k for k ≥ 1 Let further q k = n k /n k−1 for each k ≥ 1 Show that every positive rational number is present exactly once in the sequence (q k)k≥1

Problem 4 Let ABC be an acute angled triangle, and H a point in its interior Let the reflections of H through the sides AB and AC be called H c and H b,

respectively, and let the reflections of H through the midpoints of these same sides

be called H c0 and H b0, respectively Show that the four points H b , H b0, H c , and H c0are concyclic if and only if at least two of them coincide or H lies on the altitude from A in triangle ABC.

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Monday, 31 March 2014 English Version

The time allowed is 4 hours Each problem is worth 5 points

The only permitted aids are writing and drawing tools

Problem 1

Find all functions f : N → N (where N is the set of the natural numbers and isassumed to contain 0), such that

f (x2) − f (y2) = f (x + y)f (x − y),for all x, y ∈ N with x ≥ y

Problem 2

Given an equilateral triangle, find all points inside the triangle such that thedistance from the point to one of the sides is equal to the geometric mean of thedistances from the point to the other two sides of the triangle

[The geometric mean of two numbers x and y equals √

Problem 4

A game is played on an n × n chessboard At the beginning there are 99 stones

on each square Two players A and B take turns, where in each turn the playerchooses either a row or a column and removes one stone from each square in thechosen row or column They are only allowed to choose a row or a column, if ithas least one stone on each square The first player who cannot move, looses thegame Player A takes the first turn Determine all n for which player A has awinning strategy

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The 29th Nordic Mathematical Contest

Tuesday, March 24, 2015

Problem 1.

Let ABC be a triangle and Γ the circle with diameter AB The bisectors of ∠BAC and

∠ABC intersect Γ (also) at D and E, respectively The incircle of ABC meets BC and

AC at F and G, respectively Prove that D, E, F and G are collinear.

Problem 2.

Find the primesp, q, r, given that one of the numbers pqr and p + q + r is 101 times the

other

Problem 3.

Letn > 1 and p(x) = x n+a n−1 x n−1+· · · + a0 be a polynomial withn real roots (counted

with multiplicity) Let the polynomialq be deﬁned by

q(x) =

2015

j=1

p(x + j).

We know thatp(2015) = 2015 Prove that q has at least 1970 diﬀerent roots r1, , r1970

such that |r j | < 2015 for all j = 1, , 1970.

Problem 4.

An encyclopedia consists of 2000 numbered volumes The volumes are stacked in orderwith number 1 on top and 2000 in the bottom One may perform two operations with thestack:

(i) Forn even, one may take the top n volumes and put them in the bottom of the stack

without changing the order

(ii) For n odd, one may take the top n volumes, turn the order around and put them on

top of the stack again

How many diﬀerent permutations of the volumes can be obtained by using these twooperations repeatedly?

Time allowed: 4 hours.

Each problem is worth 7 points.

Only writing and drawing tools are allowed.

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Tuesday, April 5, 2016 English version

Only writing and drawing tools are allowed

Find all a ∈ R for which there exists a function f : R → R, such that

(i) f (f (x)) = f (x) + x, for all x ∈ R,

(ii) f (f (x) − x) = f (x) + ax, for all x ∈ R

Problem 4

King George has decided to connect the 1680 islands in his kingdom by bridges nately the rebel movement will destroy two bridges after all the bridges have been built,but not two bridges from the same island

Unfortu-What is the minimal number of bridges the King has to build in order to make sure that

it is still possible to travel by bridges between any two of the 1680 islands after the rebelmovement has destroyed two bridges?

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The 31st Nordic Mathematical Contest

Monday, 3 April 2017 English version

Time allowed: 4 hours Each problem is worth 7 points.

Only writing and drawing tools are allowed.

Problem 1 Let n be a positive integer Show that there exist positive integers a and b

ab cos(α − β) ≤√(1− a2)(1− b2),

then

a cos α + b sin β ≤ 1 + ab sin(β − α).

Problem 3 Let M and N be the midpoints of the sides AC and AB, respectively, of

an acute triangle ABC, AB ̸= AC Let ωB be the circle centered at M passing through

B, and let ω C be the circle centered at N passing through C Let the point D be such that ABCD is an isosceles trapezoid with AD parallel to BC Assume that ω B and ω C intersect in two distinct points P and Q Show that D lies on the line P Q.

Problem 4 Find all integers n and m, n > m > 2, and such that a regular n-sided polygon can be inscribed in a regular m-sided polygon so that all the vertices of the n-gon lie on the sides of the m-gon.

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The 32nd Nordic Mathematical Contest

Monday, 9 April 2018 English version

Only writing and drawing tools are allowed

Problem 1 Let k be a positive integer and P a point in the plane We wish

to draw lines, none passing through P , in such a way that any ray startingfrom P intersects at least k of these lines Determine the smallest number oflines needed

Problem 2 A sequence of primes p1, p2, is given by two initial primes p1

and p2, and pn+2 being the greatest prime divisor of pn+ pn+1+ 2018 for all

n ≥ 1 Prove that the sequence only contains nitely many primes for allpossible values of p1 and p2

Problem 3 Let ABC be a triangle with AB < AC Let D and E be on thelines CA and BA, respectively, such that CD = AB, BE = AC, and A, Dand E lie on the same side of BC Let I be the incentre of triangle ABC,and let H be the orthocentre of triangle BCI Show that D, E, and H arecollinear

Problem 4 Let f = f(x, y, z) be a polynomial in three variables x, y, z suchthat

f (w, w, w) = 0for all w ∈ R Show that there exist three polynomials A, B, C in these samethree variables such that A + B + C = 0 and

f (x, y, z) = A(x, y, z) · (x − y) + B(x, y, z) · (y − z) + C(x, y, z) · (z − x)

Is there any polynomial f for which these A, B, C are uniquely determined?

1

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14th Nordic Mathematical Contest

Solutions

Problem 1 Set x = the number of sums with 3 different integers, y = the numbers ofsums with 2 different integers Consider a sequence of 3999 numbered boxes where everyodd-numbered box contains a red ball Every placement of blue balls in any two of the even-numbered boxes produces a division of 2000 in three parts There are 1999

2

= 999 · 1999ways of placing the two balls Now every division of 2000 in three parts of different size

is produced by 3! = 6 different placements, and every division having two equal parts

is produced by 3 different placements So 6x + 3y = 1999 · 999 But y = 999, sincethe two equal parts can assume any size from 1 to 999 Solving, we get x = 998 · 333,

x + y = 1001 · 333 = 333333

Problem 2 Assume Pn originally has m coins, Pn−1 m + 1 coins, , P1 m + n − 1coins In every move, a person receives k coins and gives k + 1 coins, so in total his fortunediminishes by one coin After the first round, when Pn has left n coins to P1, Pn has m − 1coins, Pn−1 has m coins, etc., after two rounds Pn has m − 2 coins, Pn−1 has m − 1 coins,etc We can continue like this for m rounds, and after that Pn has no money, Pn−1 hasone coin etc Now in round m + 1 every person who has money can receive money andgive away money as before, except Pn who was bankrupt He receives n(m + 1) − 1 coinsfrom Pn−1, but cannot give n(m + 1) coins away In this situation Pn−1 has one coin,and P1 has n − 2 coins The only pair of neighbors where one player can have 5 times

as many coins as the other is (P1, Pn) Because n − 2 < n(m + 1) − 1, we must have5(n − 2) = n(m + 1) − 1 or n(4 − m) = 9 Since n > 1, either n = 3, m = 1 or n = 9,

m = 3 Both alternatives work: in the first case, the number of coins is 3 + 2 + 1 = 6, inthe second, 11 + 10 + · · · + 3 = 63

Problem 3, solution 1 Consider triangles AOE and AOD They have two equalsides, and equal angles opposite to one pair of equal sides Then either AOE and AODare congruent or 6 AEO = 180◦ −6 ADO In the first case, 6 BEO = 6 CDO, and thetriangles EBO and DCO are congruent Altogether, then, AB = AC In the secondcase, denote the angles at A, B, and C by 2α, 2β, and 2γ, respectively, and 6 AEO by δ.Using the theorem of the adjacent angle in a triangle,we get 6 BOE = 6 DOC = β + γ,

δ = 2β + γ, 180◦ − δ = β + 2γ Adding these,we have 3(β + γ) = 180◦, β + γ = 60◦.Combining this with 2(α + β + γ) = 180◦, we get 2α = 60◦

Problem 3, solution 2 Let β, γ be as above Using the sine theorem in 4BEO and4CDO, we obtain

OEsin β =

OBsin(180◦ − 2β − γ),

ODsin γ =

OCsin(180◦ − β − 2γ).These combine to

OB

OC =

sin(2β + γ) sin γsin(β + 2γ) sin β.

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Using the theorem of sines to 4BOC, we obtain

OB

OC =

sin γsin β.

So we must have sin(β + 2γ) = sin(2β + γ) So either β + 2γ = 2β + γ or β + 2γ =

180◦− 2β − γ The first equation implies β = γ, or the isosceles case, while the second onegives β + γ = 60◦, which easily leads to 6 BAC = 60◦

Problem 4 Assuming 0 ≤ x < y < z ≤ 1 and y − x = z − y, we have

f (z) − f (y) ≤ 2f (y) − 2f (x)

f (y) − f (x) ≤ 2f (z) − 2f (z) − f (y),or

1

3b ≤ a ≤

2

3b2

13

2

3a +

13

,

fron which one solves for a to obtain 1

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NMC 2002Solutions

1 The arcs AD and BC are equal Since AD < CD the line PB will intersectthe line DC between D and C Also, since AB||DC and DP ||AC, we have

6 CAB =6 P DC and the arcs P C and CB are equal Since DE is tangent to

c, and AD, P C are equal, 6 EDA =6 ACD = 6 P BC = 6 QBC As ABCD

is inscribed in c, 6 QCB = 180◦ − 6 DAB = 6 EAD Seeing that ABCD

is an isosceles trapezoid, AD = CB So the triangles ADE and CBQ arecongruent But then QC = EA Now EACQ is a quadrilateral with a pair

of opposite sides equal and parallel Thus, EACQ is a parallelogram and

EQ = AC

2 Let n be the original number of balls in the urn from which the ball is moved,and let a denote the sum of the numbers of these balls Further, let m bethe original number of balls in the other urn, and let b denote the sum of thenumbers of the corresponding balls, i.e n + m = N and

b + q

m + 1 =

b

m + x,from which we get

2 − x(n − m − 1) = N + 1

2 − x(N − 2m − 1),i.e

Trang 35

Furthermore, as b≥ m(m + 1)/2, we have mn(1

2 − x) ≥ 0, or x ≤ 1

2.The maximum value, x = 12, is taken when b = m(m + 1)/2, i.e when theballs with the numbers 1, 2, , m are in the “receiving” urn, the balls withthe numbers m + 1, m + 2, , N are in the “transmitting” urn and, from (3),when q = m + 1

3 Since the polynomial

= a0+ (11− 1)a1+ (99 + 1)a2+ (1001− 1)a3+ (9999 + 1)a4

+ (100001− 1)a5 + (999999 + 1)a6+ (10000001− 1)a7+ (99999999 + 1)a8

= (a0− a1+ a2− a3+ a4− a5+ a6 − a7+ a8)

+ 11(a1+ 9a2+ 91a3+ 909a4+ 9091a5+ 90909a6+ 909091a7+ 9090909a8),i.e every number n can be written as

(a0− a1+ a2− a3+ a4− a5+ a6− a7+ a8) + 11k,where k is an integer We find that

1 + 2 + 3 + 4≤ s ≤ 6 + 7 + 8 + 9, i.e 10 ≤ s ≤ 30,

2

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If t ≥ 7, then s ≥ 39, which is impossible.

Now it remains to examine the different cases giving s=17 and s=28 Fors=17 we have the following possible cases (except for permutations):

(a2, a4, a6, a8) = (1, 2, 5, 9), (1, 2, 6, 8), (1, 3, 4, 9), (1, 3, 5, 8), (1, 3, 6, 7),(1, 4, 5, 7), (2, 3, 4, 8), (2, 3, 5, 7), (2, 4, 5, 6) For s=28 we have the cases(4, 7, 8, 9) and (5, 6, 8, 9) In total we have 11 different ordered cases But thetotal number of ordered 9-digit numbers is 9!/(4!·5!) = 126, so the probability

of choosing a number, which is a multiple of 11, is 11/126 < 11/121 = 1/11.Thus, the probability is less than 1/11, which means that Eva is correct

3

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r + b + y = 2715r + 3b + 18y = 27 · 14From this we get

4r + 5y = 99

b = 27 − r − y

b ≤ 5The only positive solutions are (r, b, y) = (11, 5, 11), (16, 4, 7), (21, 3, 3)

The 3 values of r are all possible

r = 11 : Blue : {1, 2, 3, 4, 5} Red : {10, 11, · · · , 18, 19, 20}

r = 16 : Blue : {1, 2, 4, 5} Red : {7, 8, · · · , 14, 16, 17, · · · , 23}

r = 21 : Blue : {2, 3, 4} Red : {5, 6, · · · , 25}

Problem 2

A sequence {ak}is arithmetic if ak+1− ak= d for all k, where d is some constant,

so ak = dk + a0 Notice that the arithmetic sequence is constant modulo its xedincrease, ak ≡ a0 (mod d) for all k So to nd an increasing arithmetic sequencewith no term in common with the Fibonacci sequence, it suces to nd integers

d > 0and a0 such that fn is never equivalent to a0 modulo d

Calculate the Fibonacci sequence modulo 8:

0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1,

We have again reached: 0, 1, and because of the relation fn+2 = fn+1 + fnthe sequence now repeats itself modulo 8 Notice that 4 does not appear, sothe arithmetic sequence ak = 8k + 4 has no term in common with the Fibonaccisequence

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Problem 3

Let Mk= maxjxjk and mk = minjxjk It is clear that Mk is a non-increasing and

mk a non-decreasing sequence Also, Mk+1 = Mk only if xjk = xj+1,k = Mk forsome j If exactly p "adjacent" xik's equal Mk, then only p − 1 adjacent xi,k+1'sequal Mk Eventually we reach a step, where Mk+1 < Mk Similarly, mk+1 > mk

at some stage Now if all the numbers in all the sequences are integers, so must

be the maxima and minima After a nite number of steps the maximum andminimum are equal, and so are all the numbers We then have for some k

x1k+ x2k = x2k+ x3k = · · · = xnk + x1k

If n is odd, this gives x1k = x3k = · · · = xnk = x2k = · · · = xn−1,k Workingbackwards, the numbers in the starting sequence have to be equal

But if n is even, then the sequence 0, 2, 0, 2, , 0, 2 is a counterexample because

in next step all the numbers will be equal to 1 and we never get a number that isnot an integer

Remark: It can be shown that the only counterexamples are sequences of thetype a, b, a, b, , a, b with a ≡ b (mod 2):

In the argument above, if n is even, we get

x1k = x3k = · · · = xn−1,k = a and x2k = x4k = · · · = xnk = b

But if k > 1, then x1k = 1

2(x1,k−1+ x2,k−1), x2k = 1

2(x2,k−1+ x3,k−1) etc., andn

2a = x1k+ x3k+ · · · + xn−1,k =

1

2(x1,k−1+ x2,k−1+ · · · + xn,k−1),n

xi,k−2+ xi+1,k−2 ≡ xi+1,k−2+ xi+2,k−2 (mod 4) for all i

Since n is odd, it follows that xi,k−2 ≡ xi+1,k−2 (mod 4) for all i By induction itfollows that xi,k−j ≡ xi+1,k−j (mod 2j) for all i and all j < k Hence if xi,k is aninteger for all i and k, then xi,1 = xi+1,1 (mod 2j) for all i and j, thus they mustall be equal

14th Nordic Mathematical Contest

Solutions

Problem Set x = the number of sums with different integers, y = the numbers ofsums with different integers Consider a sequence of 3999... 2000 in three parts of different size

is produced by 3! = different placements, and every division having two equal parts

is produced by different placements So 6x + 3y = 1999 · 999... value, x = 12, is taken when b = m(m + 1)/2, i.e when theballs with the numbers 1, 2, , m are in the “receiving” urn, the balls withthe numbers m + 1, m + 2, , N are

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