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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 Let a1 , a2 , , an be positive numbers (n ≥ 1) Show that n 1 + ···+ a1 an ≥ 1 + ··· + + a1 + an n+ 1 + ···+ a1 an When does equality hold? The IMO Compendium Group, D Djuki´c, V Jankovi´c, I Mati´c, N Petrovi´c www.imo.org.yu 14-th Nordic Mathematical Contest March 30, 2000 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.) The persons P1 , P2 , , Pn sit around a table in this order, and each one has a number of coins Initially, P1 has one coin more than P2 , P2 has one coin more than P3 , etc Now P1 gives one coin to P2 , who in turn gives two coins to P3 , etc., up to Pn who gives n coins to P1 ; then P1 continues by giving n + 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 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◦ A real function defined for ≤ x ≤ satisfies f (0) = 0, f (1) = 1, and f (x) − f (y) ≤ ≤2 f (y) − f (z) whenever ≤ x < y < z ≤ and z − y = y − x Show that ≤f ≤ 74 The IMO Compendium Group, D Djuki´c, V Jankovi´c, I Mati´c, N Petrovi´c www.imo.org.yu 15-th Nordic Mathematical Contest March 29, 2001 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 A function f : R → R is bounded and satisfies f x+ +f x+ = (x) + f x+ for all real x Show that f is periodic Find the number of real roots of the equation x8 − x7 + 2x6 − 2x5 + 3x4 − 3x3 + 4x2 − 4x + = 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 The IMO Compendium Group, D Djuki´c, V Jankovi´c, I Mati´c, N Petrovi´c www.imo.org.yu 16-th Nordic Mathematical Contest April 4, 2002 A trapezoid ABCD with AB 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 Let N balls, numbered 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? Let a1 , a2 , , an , b1 , b2 , , bn be real numbers with a1 , , an distinct Show that if the product (ai + b1 )(ai + b2 ) · · · (ai + bn) takes the same value for every i = 1, 2, , n, then the product (a1 + bj )(a2 + bj ) · · · (an + bj ) also takes the same value for every j = 1, 2, , n Eva, Per, and Anna randomly select different nine-digit integers made of digits 1, 2, , 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? The IMO Compendium Group, D Djuki´c, V Jankovi´c, I Mati´c, N Petrovi´c www.imo.org.yu 17-th Nordic Mathematical Contest April 3, 2003 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 Find all triples (x, y, z) of integers satisfying the equation x3 + y + z − 3xyz = 2003 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 Find all functions f from R \ {0} to itself satisfying f (x) + f (y) = f (xyf (x + y)) for all x, y = with x + y = The IMO Compendium Group, D Djuki´c, V Jankovi´c, I Mati´c, N Petrovi´c www.imo.org.yu 18-th Nordic Mathematical Contest April 1, 2004 Twenty-seven balls labelled from to 27 are distributed in three bowls: red, blue, and yellow What are the possible values of the number of balls in the red bowl if the average labels in the red, blue and yellow bowl are 15, 3, and 18, respectively? The Fibonacci sequence is defined by f1 = 0, f2 = 1, and fn+2 = fn+1 +fn for n ≥ Prove that there is a strictly increasing arithmetic progression whose no term is in the Fibonacci sequence Given a finite sequence x1,1 , x2,1 , , xn,1 of integers (n ≥ 2), not all equal, define the sequences x1,k , , xn,k by xi,k+1 = (xi,k + xi+1,k ), where xn+1,k = x1,k Show that if n is odd, then not all xj,k are integers Is this also true for even n? Let a, b, c be the sides and R be the circumradius of a triangle Prove that 1 1 + + ≥ ab bc ca R The IMO Compendium Group, D Djuki´c, V Jankovi´c, I Mati´c, N Petrovi´c www.imo.org.yu 19-th Nordic Mathematical Contest April 5, 2005 Find all positive integers k such that the product of the decimal digits of k equals 25 k − 211 If a, b, c are positive numbers, prove the inequality 2a2 2b2 2c2 + + ≥ a + b + c b+c c+a a+b 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 Circle C1 touches circle C2 internally at A A line through A intersects C1 again at B and C2 again at C The tangent to C1 at B intersects C2 at D and E The tangents to C1 through C touch C1 at F and G Prove that points D, E, F, G are concyclic The IMO Compendium Group, D Djuki´c, V Jankovi´c, I Mati´c, N Petrovi´c www.imo.org.yu 20th Nordic Mathematical Contest Thursday March 30, 2006 English version Time allowed: hours Each problem is worth points Problem Let B and C be points on two fixed rays emanating from a point A such that AB + AC is constant Prove that there exists a point D = A such that the circumcircles of the triangels ABC pass through D for every choice of B and C Problem The real numbers x, y and z are not all equal and fulfil x+ 1 =y+ =z+ =k y z x Determine all possible values of k Problem A sequence of positive integers {an } is given by a0 = m and an+1 = a5n + 487 for all n ≥ Determine all values of m for which the sequence contains as many square numbers as possible Problem The squares of a 100 × 100 chessboard are painted with 100 different colours Each square has only one colour and every colour is used exactly 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 ❚❤❡ ✷✶st ◆♦r❞✐❝ ▼❛t❤❡♠❛t✐❝❛❧ ❈♦♥t❡st ▼❛r❝❤ ✷✾✱ ✷✵✵✼ ❊♥❣❧✐s❤ ✈❡rs✐♦♥ ❚✐♠❡ ❛❧❧♦✇❡❞✿ ✹ ❤♦✉rs✳ ❊❛❝❤ ♣r♦❜❧❡♠ ✐s ✇♦rt❤ ✺ ♣♦✐♥ts✳ ❖♥❧② ✇r✐t✐♥❣ ❛♥❞ ❞r❛✇✐♥❣ s❡ts ❛r❡ ❛❧❧♦✇❡❞✳ Pr♦❜❧❡♠ ✶ ❋✐♥❞ ♦♥❡ s♦❧✉t✐♦♥ ✐♥ ♣♦s✐t✐✈❡ ✐♥t❡❣❡rs t♦ t❤❡ ❡q✉❛t✐♦♥ x2 − 2x − 2007y = Pr♦❜❧❡♠ ✷ ❆ tr✐❛♥❣❧❡✱ ❛ ❧✐♥❡ ❛♥❞ t❤r❡❡ r❡❝t❛♥❣❧❡s✱ ✇✐t❤ ♦♥❡ s✐❞❡ ♣❛r❛❧❧❡❧ t♦ t❤❡ ❣✐✈❡♥ ❧✐♥❡✱ ❛r❡ ❣✐✈❡♥ ✐♥ s✉❝❤ ❛ ✇❛② t❤❛t t❤❡ r❡❝t❛♥❣❧❡s ❝♦♠♣❧❡t❡❧② ❝♦✈❡r t❤❡ s✐❞❡s ♦❢ t❤❡ tr✐❛♥❣❧❡✳ Pr♦✈❡ t❤❛t t❤❡ r❡❝t❛♥❣❧❡s ♠✉st ❝♦♠♣❧❡t❡❧② ❝♦✈❡r t❤❡ ✐♥t❡r✐♦r ♦❢ t❤❡ tr✐❛♥❣❧❡✳ Pr♦❜❧❡♠ ✸ ❚❤❡ ♥✉♠❜❡r 102007 ✐s ✇r✐tt❡♥ ♦♥ ❛ ❜❧❛❝❦❜♦❛r❞✳ ❆♥♥❡ ❛♥❞ ❇❡r✐t ♣❧❛② ❛ ❣❛♠❡ ✇❤❡r❡ t❤❡ ♣❧❛②❡r ✐♥ t✉r♥ ♠❛❦❡s ♦♥❡ ♦❢ t✇♦ ♦♣❡r❛t✐♦♥s✿ ✭✐✮ r❡♣❧❛❝❡ ❛ ♥✉♠❜❡r t❤❛♥ ✶ s✉❝❤ t❤❛t x ♦♥ t❤❡ ❜❧❛❝❦❜♦❛r❞ ❜② t✇♦ ✐♥t❡❣❡r ♥✉♠❜❡rs a ❛♥❞ b ❣r❡❛t❡r x = ab ; ✭✐✐✮ ❡r❛s❡ ♦♥❡ ♦r ❜♦t❤ ♦❢ t✇♦ ❡q✉❛❧ ♥✉♠❜❡rs ♦♥ t❤❡ ❜❧❛❝❦❜♦❛r❞✳ ❚❤❡ ♣❧❛②❡r ✇❤♦ ✐s ♥♦t ❛❜❧❡ t♦ ♠❛❦❡ ❤❡r t✉r♥ ❧♦s❡s t❤❡ ❣❛♠❡✳ ❲❤♦ ✇✐❧❧ ✇✐♥ t❤❡ ❣❛♠❡ ✐❢ ❆♥♥❡ ❜❡❣✐♥s ❛♥❞ ❜♦t❤ ♣❧❛②❡rs ❛❝t ✐♥ ❛♥ ♦♣t✐♠❛❧ ✇❛②❄ Pr♦❜❧❡♠ ✹ ❆ ❧✐♥❡ t❤r♦✉❣❤ ❛ ♣♦✐♥t t❤❛t B ❧✐❡s ❜❡t✇❡❡♥ A A ✐♥t❡rs❡❝ts ❛ ❝✐r❝❧❡ ✐♥ t✇♦ ♣♦✐♥ts✱ ❛♥❞ C✳ ❋r♦♠ t❤❡ ♣♦✐♥t ❝✐r❝❧❡✱ ♠❡❡t✐♥❣ t❤❡ ❝✐r❝❧❡ ❛t ♣♦✐♥ts ST ❛♥❞ AC ✳ ❙❤♦✇ t❤❛t S ❛♥❞ T ✳ ▲❡t AP/P C = · AB/BC ✳ A P B ❛♥❞ C✱ ✐♥ s✉❝❤ ❛ ✇❛② ❞r❛✇ t❤❡ t✇♦ t❛♥❣❡♥ts t♦ t❤❡ ❜❡ t❤❡ ✐♥t❡rs❡❝t✐♦♥ ♦❢ t❤❡ ❧✐♥❡s Let n > and p(x) = xn + an−1 xn−1 + · · · + a0 be a polynomial with n real roots (counted with multiplicity) Let the polynomial q be defined by 2015 p(x + j) q(x) = j=1 We know that p(2015) = 2015 Prove that q has at least 1970 different roots r1 , , r1970 such that |rj | < 2015 for all j = 1, , 1970 Solution Let hj (x) = p(x + j) Consider h2015 Like p, it has n real roots s1 , s2 , , sn , |s1 s2 · · · sn | equals 2015 Since and h2015 (0) = p(2015) = 2015 By Vi`ete, the√product √ n ≥ 2, there is at least one sj such that |sj | ≤ 2015 < 2025 = 45 Denote this sj by m Now for all j = 0, 1, , 2014, h2015−j (m + j) = p(m + j + 2015 − j) = p(m + 2015) = h2015 (m) = So m, m + 1, , m + 2014 are all roots of q Since ≤ |m| < 45, the condition |m + j| < 2015 is satisfied by at least 1970 different j, ≤ j ≤ 2014, and we are done Problem An encyclopedia consists of 2000 numbered volumes The volumes are stacked in order with number on top and 2000 in the bottom One may perform two operations with the stack: (i) For n 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 two operations repeatedly? Solution (By the proposer.) Let the positions of the books in the stack be 1, 2, 3, , 2000 from the top (and consider them modulo 2000) Notice that both operations fix the parity of the number of the book at a any given position Operation (i) subtracts an even integer from the number of the book at each position If A is an operation of type (i), and B is an operation of type (ii), then the operation A−1 BA changes the order of the books in the positions n + to n + m where n is even and m is odd This is called turning the interval Now we prove that all the volumes in odd positions can be placed in the odd positions in every way we like: If the volume we want in position is in position m1 , we turn the interval to m1 Now if the volume we want in position is in position m3 , we turn the interval to m3 , and so on In this way we can permute the volumes in odd positions exactly as we want to Now we prove that we can permute the volumes in even positions exactly as we want without changing the positions of the volumes in the odd positions: We can make a transposition of the two volumes in position 2n and 2n + 2m by turning the interval 2n + to 2n + 2m − 1, then turning the interval 2n + 2m + to 2n − 1, then turning the interval 2n + to 2n − 1, and finally adding 2m to the number of the volume in each position Since there are 1000! permutations of the volumes in the odd positions, and 1000! permutations of the volumes in the even positions, altogether we have (1000!)2 different permutations Solution We show that the volumes can be permuted so that the volumes with odd numbers are in an arbitrary order in the odd-numbered palaces and the volumes with even numbers are in an arbitrary order in the even-numbered places The main idea is to construct two combinations of the allowed operations The first one turns the volumes in a specified interval, starting and ending in an odd-numbered place, in the opposite order while keeping everything outside this interval fixed, or keeps everything fixed in an interval while turning the order of the volumes outside this interval in the opposite direction, when the counting starts below that interval and is continued from the top after reaching the bottom volume The second combined operation just exchanges two volumes in evennumbered places while keeping everything else fixed – It is clear that 2000 is not a special number, and it could be replaced by a generic even integer However, we formulate the proof according to the problem text Let E = {1, 2, , 2000} We formulate the operations described in conditions (i) and (ii), depending on an even integer n and odd integer m as functions fn : E → E and gm : E → E, defined by fn (p) = 2000 + p − n p−n for p ≥ n, for n < p and gm (p) = m−p+1 p for p ≤ m, for m < p We immediately see that fn and gm map even numbers into even numbers and odd numbers into odd numbers So the volumes can never be permuted so that an odd-numbered volume would be in an even place or an even-numbered would be in an odd place The observation f ([1, n]) = [2000 − (n + 1), 2000] easily leads to fn−1 = f2000−n Now let n be even and m odd and n + m < 2000 Consider the combined mapping fn−1 ◦ gm ◦ fn If n < n + p ≤ n + m, then fn (n + p) = p ≤ m, gm (p) = m − p + < 2000 − n and fn−1 (m − p + 1) = f2000−n (m − p + 1) = 2000 + m − p + − 2000 + n = n + m + − p Because fn ([n + 1, n + m]) = [1, m], fn maps numbers p outside the interval [n+1, n+m] into numbers outside the interval [1, m]; gm keeps these numbers fixed and fn−1 returns fn (p) into p So we have shown that for any interval [s, t] ⊂ E with s and t odd, there is a function hs,t , combined of functions of the f type and g type such that hs,t reverses the order of numbers in the interval [s, t] and is the identity function outside this interval The functions hs,t allow us to order the odd numbers in an arbitrary manner If p1 ought to be in position 1, then apply (if needed) h1,p1 ; if the number p2 which ought to be in position now is in position x, the x ≥ and we may apply (if needed) h3,x Continuing this way, we eventually arrive at the desired order of the odd numbers To construct the second one of the desired operations, we have to obtain a counterpart for hs,t for t < s To this end, consider fn−1 ◦ gm ◦ fn for m + n > 2000 By the definition of fn , fn (n + m − 2000) = 2000 + (n + m − 2000) − n = m, and so fn [n + m − 2000 + 1, n] = [m + 1, 2000] Consequently, fn−1 ◦gm ◦fn keeps numbers in the interval [n+m−2000+1, n] (with even endpoints) fixed Since gm turns the order around in [1, m] and fn−1 = f2000−n maps [1, m] onto the complement of [n+m−2000+1, n] in such a way that f2000−1 (1) = n+1, the order of numbers in the complement is reversed in the desired manner – We have shown that for odd s and t such that t < s there exists a function hs,t , combined of functions of the f type and g type such that hs, t is the identity on the interval [t + 1, s − 1], but reverses the order of the numbers outside this interval, when counting is started from s and continued through over 2000 and over to t, in other words modulo 2000 To finish the proof, we show that two numbers in the even positions can be exchanged while everything else is fixed This clearly allows us to put the even numbers in an arbitrary order without violating the order of the odd numbers To achieve this, we take two even numbers p and q, p < q, and consider the function φp,q = f2000+p−q ◦ hp+1, p−1 ◦ hq+1, p−1 ◦ hp+1,q−1 The innermost function hp+1,q−1 reverses the order on [p + 1, q − 1] and fixes everything else, the next function hq+1, p−1 fixes numbers in [p, q], hp+1, p−1 fixes p and reverses the order ( mod 2000) in E \ {p}, and f2000+p−q (p) = q The two innermost components of φp,q fix q, hp+1,p−1 takes q to a position x q − p steps ahead of p ( mod 2000) and −1 f2000+p−q = fq−p moves x q − p positions back, i.e to p If p + k is between p and q, then the innermost function maps it to q − k, the next one fixes q − k, the third function maps −1 maps 2p − q + k back to p + k q − k to p − (q − k − p) = 2p − q + k ( mod 2000), and fq−p A similar reasoning shows that φp,q also fixes numbers in E \ [p, q] Since both even and odd numbers have 1000! different permutations, the volumes can be permuted into (1000!)2 different orders by using the given operations repeatedly Solution We show by induction, that if in an ordered sequence one may exchange two consecutive elements without changing the places of any other element, then any two elements can be exchanged so that all other elements remain in place We assume that this is true for elements which are at most k steps away from each other in the sequence Assuming that a precedes b by k + steps and that c is immediately behind a, the following sequence of exchanges is allowed: , a, c, , b, → , a, b, , c, → , b, a, , c, → , b, c, , a, By assumption, all elements in the places indicated by three dots remain on their places, as does c If any two elements can be exchanged without violating the other elements, then the elements in the sequence can be arranged to any order One just gets the desired first element to its place by (at most) one exchange, and if the first k elements already are in their desired places, then the one wanted to be in place k + is not among the first k elements, and it can be moved to its place by at most one exchange, no violating the order of the first k elements We now show, that any two volumes in consecutive odd places can be exchanged The volumes on top and in place can be exchanged by operation (ii) applied to the three topmost volumes The volumes in places 2n + and 2n + can be exchanged by first applying operation (i) to the 2n topmost volumes, which moves them in the bottom but preserves their order, then applying (ii) to the three topmost volumes and finally operation (i) to the 2000 − 2n topmost volumes The last operation returns the 2n volumes to top preserving the order and returns the remaining 2000−2n volumes to the bottom, preserving the order, save the volumes in places 2n + and 2n + 3, which have changed places By the general remarks above, it is now clear that operations (i) and (ii) can be used to arrange the volumes in odd positions into any order while the volumes in even positions remain in their places We still need to show, that a similar procedure is possible for volumes in even positions First of all, the volumes in positions to can be moved to order 5, 4, 3, 2, by performing operation (ii) to the five topmost volumes Then it is possible to exchange the volumes in positions and without changing anything else So the volumes in even positions closest to the top can be exchanged For volumes on positions 2n and 2n + one can first perform operation (i) to the 2n − topmost volumes The volumes in places 2n and 2n + will be taken to places and 4, and they can be exchanged Performing operation (i) to the 2000 − (2n − 1) topmost volumes then returns everything to their previous places, except that the volumes in positions 2n and 2n + have changed places So all volumes in even positions can be put into any order by using the operations (i) and (ii), and the total number of possible orderings is (1000!)2 (We note that operation (ii) can be replaced by a weaker operation: ”It is possible to turn the order around for the and topmost volumes.”) The 30th Nordic Mathematical Contest Tuesday, April 5, 2016 Solutions Problem Determine all sequences of non-negative integers a1 , , a2016 all less than or equal to 2016 satisfying i + j | iai + jaj for all i, j ∈ {1, 2, , 2016} Solution Answer : All constant sequences of non-negative integers The condition rewrites to i + j | i(ai − aj ) Since 2k − and k are coprime, we see that 2k −1 | ak −ak−1 Thus if 2k −1 > 2016, then ak = ak−1 since ak and ak−1 are non-negative and at most 2016 All together a1009 = a1010 = · · · = a2016 If i < 1009 we know that i is coprime to one of the number 2016, 2015, , 2017 − i say j Then i + j | − aj and since i + j > 2016 we conclude as before that = aj = a2016 So any such sequence is constant Problem Let ABCD be a cyclic quadrilateral satisfying AB = AD and AB + BC = CD Determine ∠CDA Solution Answer : ∠CDA = 60◦ Choose the point E on the segment CD such that DE = AD Then CE = CD − AD = CD − AB = BC, and hence the triangle CEB is isosceles A D B E C Now, since AB = AD then ∠BCA = ∠ACD This shows that CA is the bisector of ∠BCD = ∠BCE In an isosceles triangle, the bisector of the apex angle is also the perpendicular bisector of the base Hence A is on the perpendicular bisector of BE, and AE = AB = AD = DE This shows that triangle AED is equilateral, and thus ∠CDA = 60◦ Problem 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 √ 1± Solution Answer : a = From (i) we get f (f (f (x)) − f (x)) = f (x) On the other hand (ii) gives f (f (f (x)) − f (x)) = f (f (x)) + af (x) Thus we have (1 − a)f (x) = f (f (x)) Now it follows by (i) that (1 − a)f (x) = f (x) + x, and hence f (x) = − x, since a = obviously does not give a solution a We now need to check whether (i) and (ii) hold for this function for some values of a and all real x We have 1 a−1 x f (f (x)) = − f (x) = x, and f (x) + x = − x + x = a a a a √ a−1 1± Thus (i) will hold for all real x iff = , i.e iff a = For these values of a a a we have 1 f (f (x) − x) = − (f (x) − x) = − a a − x−x a and = 1 + a a x= a+1 x = x, a2 a2 − 1 x = x, f (x) + ax = − x + ax = a a so that for these two values of a both (i) and (ii) hold for all real x Thus the √ values of a 1± such that there exists a function f with the desired properties are a = 2 Problem King George has decided to connect the 1680 islands in his kingdom by bridges Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island 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 rebel movement has destroyed two bridges? Solution Answer: 2016 An island cannot be connected with just one bridge, since this bridge could be destroyed Consider the case of two islands, each with only two bridges, connected by a bridge (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands no matter what.) If they are also connected to two separate islands, then they would be isolated if the rebel movement destroys the two bridges from these islands not connecting the two So the two bridges not connecting them must go to the same island That third island must have at least two other bridges, otherwise the rebel movement could cut off these three islands Suppose there is a pair of islands with exactly two bridges that are connected to each other From the above it is easy to see that removing the pair (and the three bridges connected to them) must leave a set of islands with the same properties Continue removing such pairs, until there are none left (Note that the reduced set of islands could have a new such pair and that also needs to be removed.) Suppose we are left with n islands and since two islands are removed at a time, n must be an even number And from the argument above it is clear that n ≥ Consider the remaining set of islands and let x be the number of islands with exactly two bridges (which now are not connected to each other) Then n − x islands have at least three bridges each Let B be the number of bridges in the reduced set Now B ≥ 2x and , and thus B ≥ 6n 2B ≥ 2x + 3(n − x) = 3n − x Hence 2B ≥ max(4x, 3n − x) ≥ · 3n 5 Now let B be the number of bridges in the original set Then B =B +3· 1680 − n 6n 6(1680 − n) · 1680 ≥ + ≥ = 2016 5 It is possible to construct an example with exactly 2016 bridges: Take 672 of the islands and number them 0, 1, 2, 671 Connect island number i with the islands numbered i−1, i + and i + 336 (modulo 672) This gives 1008 bridges We now have a circular path of 672 bridges: − − − · · · − 671 − If one of these 672 bridges are destroyed, the 672 islands are still connected If two of these bridges are destroyed, the path is broken into two parts Let i be an island on the shortest path (if they have the same length, just pick a random one) Then island i + 336 (modulo 672) must be on the other part of the path, and the bridge connecting these two islands will connect the two paths Hence no matter which two bridges the rebel movement destroys, it is possible to travel between any of the 672 islands Now for every of the 1008 bridges above, replace it with two bridges with a new island between the two This increases the number of bridges to 2016 and the number of islands to 672 + 1008 = 1680 completing the construction Since the rebel movement does not destroy two bridges from the same island, the same argument as above shows that with this construction it is possible to travel between any of the 1680 islands after the destruction of the two bridges The 31st Nordic Mathematical Contest Monday, April 2017 Solutions Problem Let n be a positive integer Show that there exist positive integers a and b such that: a2 + a + = n2 + n + b2 + b + Solution Let P (x) = x2 + x + We have P (n)P (n + 1) = (n2 + n + 1)(n2 + 3n + 3) = n4 + 4n3 + 7n2 + 6n + Also, P ((n + 1)2 ) = n4 + 4n3 + 7n2 + 6n + By choosing a = (n + 1)2 and b = n + we get P (a)/P (b) = P (n) as desired Problem Let a, b, α, β be real numbers such that ≤ a, b ≤ 1, and ≤ α, β ≤ π2 Show that if ab cos(α − β) ≤ (1 − a2 )(1 − b2 ), then a cos α + b sin β ≤ + ab sin(β − α) Solution The condition can be rewritten as ab cos (α − β) = ab cos α cos β + ab sin α sin β ≤ (1 − a2 )(1 − b2 ) Set x = a cos α, y = b sin β, z = b cos β, t = a sin α We can now rewrite the condition as xz + yt ≤ (1 − x2 − t2 )(1 − y − z ), whereas the inequality we need to prove now looks like x + y ≤ + xy − zt Since x, y, z, t ≥ 0, and + xy − zt = + ab sin (β − α) ≥ 0, we can square both sides of both inequalities, and get equivalent ones After a couple of cancelations the condition yields 2xyzt ≤ − x2 − y − z − t2 + x2 y + z t2 , so that x2 + y + z + t2 ≤ (xy − zt)2 + 1, which is equivalent to x2 + y + z + t2 + 2xy − 2zt ≤ (1 + xy − zt)2 , or (x + y)2 + (z − t)2 ≤ (1 + xy − zt)2 Since (x + y)2 ≤ (x + y)2 + (z − t)2 , the desired inequality follows Problem 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 ωC 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 ωB and ωC intersect in two distinct points P and Q Show that D lies on the line P Q Solution Let E be such that ABEC is a parallelogram with AB and let ω be the circumscribed circle of ABC with centre O CE and AC BE, It is known that the radical axis of two circles is perpendicular to the line connecting the two centres Since BE ⊥ M O and CE ⊥ N O, this means that BE and CE are the radical axes of ω and ωB , and of ω and ωC , respectively, so E is the radical centre of ω, ωB , and ωC Now as BE = AC = BD and CE = AB = CD we find that BC is the perpendicular bisector of DE Most importantly we have DE ⊥ BC Denote by t the radical axis of ωB and ωC , i.e t = P Q Then since t ⊥ M N we find that t and DE are parallel Therefore since E lies on t we get that D also lies on t Alternative solution Reflect B across M to a point B forming a parallelogram ABCB Then B lies on ωB diagonally opposite B, and since AB BC it lies on AD Similarly reflect C across N to a point C , which satisfies analogous properties Note that CB = AB = CD, so we find that triangle CDB and similarly triangle BDC are isosceles Let B and C be the orthogonal projections of B and C onto AD Since BB is a diameter of ωB we get that B lies on ωB , and similarly C lies on ωC Moreover BB is an altitude of the isosceles triangle BDC with BD = BC , hence it coincides with the median from B, so B is in fact the midpoint of DC Similarly C is the midpoint of DB From this we get DC DB 2= = DB DC which rearranges as DC · DC = DB · DB This means that D has same the power with respect to ωB and ωC , hence it lies on their radical axis P Q Problem 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 Solution It works only for n = 2m, and for m = and n = To begin with let’s see why it works for n = 2m For a 2m-gon we can choose two points on each side, symmetrically, so that the distance between the two of them is equal to the distance between two close points on adjacent sides For n = and m = we need to inscribe a square in an equilateral triangle, by choosing two vertices of the square symmetrically on one of the sides of the triangle It is easy to calculate the side length of the square so that its remaining two vertices lie on the remaining two sides of the triangle We need to show that it cannot be done in any other way One side of the m-gon can contain at most two of the vertices of the n-gon, so that n ≤ 2m For n ≥ m + at least two of the sides of the m-gon must contain two of the vertices of the n-gon each By symmetry the midpoints, and thus the perpendicular bisectors, of such a side of the m-gon and of the side of the n-gon it contains must coincide If two such sides of the m-gon are not opposite to each other the corresponding perpendicular bisectors will intersect, and we can deduce that the centres of the circumscribed circles of the m-gon and of the n-gon must coincide If two such sides are opposite, then the centres of the circumscribed circles will coincide with the midpoint of the segment between the midpoints of the sides, and thus the two circumscribed circles will once again have the same centre Denote the radii of the two circumscribed circles by R and r, where R > r The smaller circle intersects the sides of the m-gon at 2m points, among which are the possible vertices for the n-gon Denote these points by P1 , P2 , , P2m clockwise, where P1 and P2 are vertices of the n-gon If the side length of the n-gon is s, we now have |P1 P2 | = s, and thus |P3 P4 | = s If only one of the points P3 and P4 is a vertex in the n-gon, it would have to be P4 , but the distance between P2 and P4 is greater than s, which means P2 P4 cannot be a side We can now deduce that both P3 and P4 are vertices of the n-gon, and by symmetry all the 2m points of intersection will be vertices of the n-gon, so that n = 2m We now need to handle the case n = m + > Denote the vertices of the m-gon by Q1 , Q2 , , Qm clockwise Now only one of the sides of the m-gon contains two of the vertices of the n-gon, let this side be Qm Q1 Denote the vertices of the (m + 1)-gon by P1 , P2 , , Pm+1 clockwise, where P1 and Pm+1 lie on the side Qm Q1 of the m-gon Let α = π − 2π/(m + 1), and β = π − 2π/m be the angles of the n-gon and the mgon, respectively We now get a number of triangles P1 Q1 P2 , P2 Q2 P3 , Pm Qm Pm+1 We begin by establishing a connection between the sizes of the angles To begin with we have ∠Q1 P1 P2 = π − α, and ∠P1 Q1 P2 = β, so that ∠P1 P2 Q1 = α − β We now proceed to get ∠Q2 P2 P3 = π − α − (α − β), and ∠P2 Q2 P3 = β, which gives ∠P2 P3 Q2 = 2(α − β), and so on If now γ = α − β = 2π/m(m + 1), we have ∠Pk Pk+1 Qk = kγ, and ∠Qk Pk Pk+1 = (m + − k)γ for k = 1, 2, , m Since s is the side length of the n-gon, i.e s = |P1 P2 | = |P1 P2 | = · · · = |Pm+1 P1 |, according to the law of sines we get |Pk Qk | |Qk Pk+1 | s = = , sin β sin(Pk Pk+1 Qk ) sin(Qk Pk Pk+1 ) i.e |Pk Qk | |Qk Pk+1 | s = = sin β sin(kγ) sin((m + − k)γ) Since |Qk Qk+1 | = |Qk Pk+1 | + |Pk+1 Qk+1 |, we get sin((k + 1)γ) + sin((m + − k)γ) = σ sin β s where σ = |Q1 Q2 | = |Q2 Q3 | = · · · = |Qm Q1 | is the side length of the m-gon Using the above for k = and k = 2, we get sin 2γ + sin mγ = sin 3γ + sin(m − 1)γ For m ≥ the angles 3γ and (m − 1)γ are both in the interval between 2γ and mγ, which means we can’t have equality as the sine function is concave (convex from above) in the interval [0, π/2] We therefore deduce that it is impossible to inscribe an (m + 1)-gon in an m-gon for m ≥ 4 ❙♦❧✉t✐♦♥s ◆▼❈ ✷✵✶✽ ❙♦❧✉t✐♦♥ t♦ ♣r♦❜❧❡♠ ✶ ■t ✐s ❝❧❡❛r t❤❛t 2k ❧✐♥❡s ❞♦ ♥♦t s✉✣❝❡✱ ❛s t❤❡ t✇♦ r❛②s ♣❛r❛❧❧❡❧ t♦ ❛♥② ✜①❡❞ ❧✐♥❡ 2k − ❧✐♥❡s✱ s♦ ♦♥❡ ✐♥t❡rs❡❝ts ❛t ♠♦st k − 1✳ ❲❡ s❤❛❧❧ s❤♦✇ 2k + ❧✐♥❡s ❛r❡ s✉✣❝✐❡♥t✳ ▲❡t P ❜❡ ❛t t❤❡ ♦r✐❣✐♥✳ ●✐✈❡♥ ❛♥ ❛♥❣❧❡ v ✱ ❧❡t s(v) ❞❡♥♦t❡ t❤❡ r❛② ❢r♦♠ P t❤❛t ✐♥t❡rs❡❝ts t❤❡ ✉♥✐t ❝✐r❝❧❡ ❛t t❤❡ ♣♦✐♥t (cos v, sin v)✳ ❋♦r ❡❛❝❤ ❛♥❣❧❡ vi ❛♥❞ ❡❛❝❤ > 0✱ ❜② ❞r❛✇✐♥❣ t✇♦ ✑❛❧♠♦st ♣❛r❛❧❧❡❧✑ ❧✐♥❡s ♦♥ ❡❛❝❤ s✐❞❡ ♦❢ P ❣✉❛r❛♥t❡❡ t❤❛t ❛❧❧ r❛②s ❢r♦♠ P t❤❛t ❛r❡ ♥♦t s(v) ❢♦r s♦♠❡ vi ≤ v ≤ vi + ✐♥t❡rs❡❝t ❛t ❧❡❛st ♦♥❡ ♦❢ t❤❡s❡ t✇♦ ❧✐♥❡s✳ ■❢ ✇❡ ❢♦r ❡①❛♠♣❧❡ ❝❤♦♦s❡ vi = 2i ❛♥❞ s✉✣❝✐❡♥t❧② s♠❛❧❧✱ ✇❡ ♠❛② s❡❡ t❤❛t [vi , vi + ]✱ i ∈ {1, , k} ❛r❡ ♣❛✐r✇✐s❡ ❞✐s❥♦✐♥t ❛♥❞ < vi < π − ✳ ❲❡ t❤❡♥ ❤❛✈❡ 2k ❧✐♥❡s ✇✐t❤ t❤❡ ♣r♦♣❡rt② t❤❛t ❡❛❝❤ s(v)✱ < v < π ✱ ✐♥t❡rs❡❝ts ❛t ❧❡❛st k − ❧✐♥❡s ❛♥❞ ❡❛❝❤ s(v)✱ π ≤ v ≤ 2π ✱ ✐♥t❡rs❡❝ts k ❧✐♥❡s✳ ◆♦✇ ❛❞❞ t❤❡ ❧✐♥❡ t❤❛t ✐s t❛♥❣❡♥t t♦ t❤❡ ✉♥✐t ❝✐r❝❧❡ ❛t (0, 1) ❛♥❞ ✇❡ ❤❛✈❡ 2k + ❧✐♥❡s ✇✐t❤ t❤❡ ✐♥t❡rs❡❝t ❛ t♦t❛❧ ♦❢ t❤❛t r❡q✉✐r❡❞ ♣r♦♣❡rt✐❡s✳ ❆♥ ❛❧t❡r♥❛t✐✈❡ ❞❡s❝r✐♣t✐♦♥ ✐s ❛s ❢♦❧❧♦✇s✳ P✱ P✐❝❦ ❛♥② ❧✐♥❡ t❤❛t ❞♦❡s ♥♦t ❣♦ Q1 , Q2 , , Q2k ♦♥ s♦ t❤❛t t❤❡② ❛♣♣❡❛r ✐♥ t❤✐s ♦r❞❡r✳ ❋♦r ❡❛❝❤ ≤ j ≤ k ✱ ❝❤♦♦s❡ ❛ ♣♦✐♥t Sj s♦ t❤❛t t❤❡ tr✐❛♥❣❧❡ Q2j−1 Q2j Sj ❝♦♥t❛✐♥s P ✐♥ ✐ts ✐♥t❡r✐♦r✳ ❖✉r r❡♠❛✐♥✐♥❣ k ❧✐♥❡s ❛r❡ t❤❡♥ ❣♦✐♥❣ t♦ ❜❡ Q2j−1 Sj ❛♥❞ Q2j Sj ✇✐t❤ ≤ j ≤ k ✳ ◆♦t❡ t❤❛t ❢♦r ❛♥② r❛② s ❢r♦♠ P ❛♥❞ ✐♥❞❡① j ✱ t❤❡ r❛② s ✐♥t❡rs❡❝ts ❛t ❧❡❛st ♦♥❡ ♦❢ t❤❡ ❧✐♥❡s Q2j−1 Sj ❛♥❞ Q2j Sj ✉♥❧❡ss ✐t ✐♥t❡rs❡❝ts t❤❡ s❡❣♠❡♥t Q2j−1 Q2j ✳ ❙✐♥❝❡ t❤❡s❡ s❡❣♠❡♥ts ❛r❡ ♣❛✐r✇✐s❡ ❞✐s❥♦✐♥t✱ t❤✐s ♠❛② ❤❛♣♣❡♥ ❛t ♠♦st ♦♥❝❡✱ ✐♥ ✇❤✐❝❤ ❝❛s❡ ♣r♦✈✐❞❡s t❤❡ k ✲t❤ ✐♥t❡rs❡❝t✐♦♥✳ t❤r♦✉❣❤ ❛♥❞ ❝❛❧❧ ✐t ✳ ❚❤❡♥ ❝❤♦♦s❡ ♣♦✐♥ts ❙♦❧✉t✐♦♥ t♦ ♣r♦❜❧❡♠ ✷ pn+2 ≤ max(pn , pn + 1) + 2020 ❢♦r ❛❧❧ ♣♦s✐t✐✈❡ ✐♥t❡❣❡rs n✳ pn , pn+1 > 2✳ ❚❤❡♥ pn + pn+1 + 2018 ✐s ❡✈❡♥✱ ♦❣ ❛♥❞ ❤❡♥❝❡ ❋✐rst ✇❡ ♣r♦✈❡ t❤❛t ❆ss✉♠❡ t❤❛t pn+2 ≤ ■❢ pn ♦r pn+1 pn + pn+1 + 1009 ≤ max(pn , pn+1 ) + 2020 ✐s ✷✱ t❤❡♥ pn+2 ≤ pn + pn+1 + 2018 = max(pn , pn+1 ) + 2020 m s✉❝❤ t❤❛t M = m · 2021! M + 2, M + 3, , M + 2021 ❛r❡ ✷✵✷✵ ❈❤♦♦s❡ ✐s ❣r❡❛t❡r t❤❛♥ ❜♦t❤ p1 ❛♥❞ p2 ✳ ◆♦✇ ❝♦♥s❡❝✉t✐✈❡ ✐♥t❡❣❡rs ❛♥❞ ♥♦♥❡ ♦❢ ✇❤✐❝❤ ❛r❡ ♣r✐♠❡s✳ ❇② ✐♥❞✉❝t✐♦♥ ✐t ❢♦❧❧♦✇s t❤❛t ❛❧❧ ♣r✐♠❡s ♦❢ t❤❡ s❡q✉♥❝❡ ❛r❡ ❧❡ss t❤❛♥ ♦r ❡q✉❛❧ t♦ M + 1✿ ❇② ❝♦♥str✉❝t✐♦♥ p1 ❛♥❞ p2 ❛r❡ ❧❡ss t❤❛♥ M + 1✳ ❆ss✉♠❡ t❤❛t p1 , p2 , , pn+1 M + ❢♦r s♦♠❡ n ≥ 1✳ ❚❤❡♥ ❛❧❧ ❛r❡ ❧❡ss t❤❛♥ ♦r ❡q✉❛❧ t♦ pn+2 ≤ max(pn , pn + 1) + 2020 ≤ M + 2021 ❙✐♥❝❡ pn+2 ✐s ❛ ♣r✐♠❡✱ ❛♥❞ ♥♦♥❡ ♦❢ t❤❡ ✐♥t❡❣❡rs ♣r✐♠❡s✱ ✇❡ ❝♦♥❝❧✉❞❡ M + 2, M + 3, , M + 2021 pn+2 ≤ M + 1✳ ❍❡♥❝❡ t❤❡ s❡q✉❡♥❝❡ ♦♥❧② ❝♦♥t❛✐♥s ✜♥✐t❡❧② ♠❛♥② ♣r✐♠❡s✳ ✶ ❛r❡ ❙♦❧✉t✐♦♥ t♦ ♣r♦❜❧❡♠ ✸ ▲❡t t❤❡ ♣♦✐♥t AC A B✳ A ABA C ✐s ❛ α = ∠BAC = ∠CA B ✳ ❜❡ s✉❝❤ t❤❛t ❉❡♥♦t❡ ♣❛r❛❧❧❡❧♦❣r❛♠ ✇✐t❤ AB AC ❛♥❞ ❙✐♥❝❡ CD = AB = CA ✱ ✇❡ ✜♥❞ t❤❛t CDA ✐s ❛♥ ✐s♦s❝❡❧❡s tr✐❛♥❣❧❡✳ ❆s ∠DCA = 180◦ − α✱ ✇❡ ❞❡❞✉❝❡ t❤❛t ∠A DC = ∠CA D = α2 ✱ s♦ t❤❛t D ❧✐❡s ♦♥ t❤❡ ❛♥❣❧❡ ❜✐s❡❝t♦r ♦❢ ∠CA B ✳ ❙✐♠✐❧❛r❧② E ❧✐❡s ♦♥ ✳ ∠CBH = 90◦ − ∠BCI = 90◦ − 21 ∠BCA = 90◦ − 12 ∠A BC ✱ s♦ BH ✐s t❤❡ ❡①t❡r✐♦r ❛♥❣❧❡ ❜✐s❡❝t♦r ♦❢ ∠A BC ✳ ❙✐♠✐❧❛r❧② CH ✐s t❤❡ ❡①t❡r✐♦r ❛♥❣❧❡ ❜✐s❡❝t♦r ♦❢ ∠A CB ✱ s♦ H ✐s ✐♥ ❢❛❝t t❤❡ ❡①❝❡♥t❡r ♦❢ tr✐❛♥❣❧❡ A BC ♦♣♣♦s✐t❡ A ✳ ❚❤❡r❡❢♦r❡ ✇❡ ❝♦♥❝❧✉❞❡ t❤❛t D✱ E ✱ ❛♥❞ H ❛❧❧ ❧✐❡ ♦♥ ✳ ◆❡①t ♥♦t✐❝❡ t❤❛t ❙♦❧✉t✐♦♥ t♦ ♣r♦❜❧❡♠ ✹ ■♥tr♦❞✉❝✐♥❣ ♥❡✇ ✈❛r✐❛❜❧❡s B1 (X, Y, z)Y + c(z)✳ X = x−y, Y = y−z ✱ ✇❡ ♠❛② ✇r✐t❡ f ❛s A1 (X, Y, z)X+ ❚♦ s❡❡ t❤✐s✱ ♥♦t❡ t❤❛t t❤✐s tr❛♥s❢♦r♠❛t✐♦♥ ✐s ✐♥✈❡rt✐❜❧❡✱ ✐✳❡✳✱ ✇❡ ❝❛♥ ❡①♣r❡ss t❤❡ ♦r✐❣✐♥❛❧ ✈❛r✐❛❜❧❡s ❛s ♣♦❧②♥♦♠✐❛❧s ♦❢ t❤❡ ♥❡✇ ♦♥❡s✿ x = X + Y + z ❛♥❞ y = Y + z ✱ ❛♥❞ ♣❧✉❣❣✐♥❣ ✐♥ ②✐❡❧❞s ❛ ♣♦❧②♥♦♠✐❛❧ ✐♥ t❤❡ X, Y, z ✳ ❚❤❡♥ ✇❡ ❣r♦✉♣ t❤❡ t❡r♠s ✐♥ t❤❡ ♥❡✇ ❡①♣r❡ss✐♦♥ ✇✐t❤ t❤♦s❡ ❤❛✈✐♥❣ ❛ ❢❛❝t♦r X ✐♥ t❤❡ ✜rst ❣r♦✉♣✱ ❛♠♦♥❣ t❤❡ r❡♠❛✐♥✐♥❣ ♦♥❡s✱ t❤♦s❡ ❝♦♥t❛✐♥✐♥❣ ❛ ❢❛❝t♦r Y ✐♥ t❤❡ s❡❝♦♥❞✱ ❛♥❞ t❤❡ r❡st ✐♥ t❤❡ t❤✐r❞ ❣r♦✉♣✳ ❚❤✐s ②✐❡❧❞s t❤r❡❡ ✈❛r✐❛❜❧❡s t❤❡ ❞❡s✐r❡❞ ❡①♣r❡ss✐♦♥ ✭♥♦t❡ t❤❛t ✇❡ ❤❛✈❡ s♦♠❡ ❢r❡❡❞♦♠ ❛s t♦ ✇❤❡r❡ ✇❡ ✇❛♥t t♦ ♣✉t t❡r♠s ✇✐t❤ ❢❛❝t♦r ❳❨✮✳ ✷ f ❛s ❛ ♣♦❧②♥♦♠✐❛❧ ✐♥ ♦♥❡ ✈❛r✐❛❜❧❡✱ x✱ ❛♥❞ y, z ❛s ♣❛r❛♠❡t❡rs✳ ❚❤❡♥ ♣❡r❢♦r♠✐♥❣ ❛ ♣♦❧②♥♦♠✐❛❧ ❞✐✈✐s✐♦♥ ✇✐t❤ x − y ✱ ✇❡ ♦❜t❛✐♥ f (x, y, z) = g(x, y, z) · (x − y) + r(x, y, z)✳ ◆♦t❡ t❤❛t r ❤❛s t♦ ❤❛✈❡ ❞❡❣r❡❡ ③❡r♦ ❛s ❛ ♣♦❧②♥♦♠✐❛❧ ✐♥ x✱ s♦ ✐♥ ❢❛❝t r(x, y, z) = r(y, z)✳ ◆♦✇ r❡♣❡❛t t❤✐s ✇✐t❤ r ❛s ❛ ♣♦❧②♥♦♠✐❛❧ ✐♥ y ✇✐t❤ ♣❛r❛♠❡t❡r z ✱ ❞✐✈✐❞❡❞ ❜② y − z ✳ ❚❤✐s ②✐❡❧❞s t❤❡ s❛♠❡ ❆❧t❡r♥❛t✐✈❡❧②✱ ✇❡ ♠❛② ❝♦♥s✐❞❡r ✇✐t❤ r❡s✉❧t ❛s t❤❡ ❛❜♦✈❡ ❛♣♣r♦❛❝❤✳ f = A2 (x, y, z)(x − y) + B2 (x, y, z)(y − z) + c(z)✳ ❙✐♥❝❡ f (w, w, w) = 0✱ ,A = c(w) = ❢♦r ❛❧❧ w✱ t❤❛t ✐s c(z) = 0✳ ▲❡t C(x, y, z) = − A2 +B A2 +C, B = B2 +C ✳ ❚❤❡♥ A+B +C = A2 +B2 +3C = A2 +B2 −(A2 +B2 ) = 0✱ ❚❤✉s ✇❡ ♦❜t❛✐♥ ❛♥❞ f = (A − C)(x − y) + (B − C)(y − z) = A(x − y) + B(y − z) + C(z − x) ❚❤❡ r❡♣r❡s❡♥t❛t✐♦♥ ✐s ♥❡✈❡r ✉♥✐q✉❡✳ ❚♦ s❤♦✇ t❤✐s✱ ✐t ✐s ❜② ❛❞❞✐t✐✈✐t② ❡♥♦✉❣❤ t♦ s❤♦✇ t❤❛t t❤❡ r❡♣r❡s❡♥t❛t✐♦♥ ✐s ♥♦t ✉♥✐q✉❡ ❢♦r t❤❡ ③❡r♦ ♣♦❧②♥♦♠✐❛❧✳ ▲❡t A = 2z − x − y = (z − x) − (y − z); B = 2x − y − z = (x − y) − (z − x); C = 2y − x − z = (y − z) − (x − y), ✇❤❡r❡ A + B + C = A · (x − y) + B · (y − z) + C · (z − x) = 0✱ s❤♦✇✐♥❣ t❤❛t ❤❛s t✇♦ ❞✐✛❡r❡♥t r❡♣r❡s❡♥t❛t✐♦♥s ✭t❤❡ ♦t❤❡r ❜❡✐♥❣ t❤❡ tr✐✈✐❛❧ ♦♥❡✮✳ ✸ ... ❦✏❫❧❦❷❳➏❪✺❳❩❨✰❬❲❣✥♥➙❱✸➑④⑤ 14th Nordic Mathematical Contest Solutions Problem Set x = the number of sums with different integers, y = the numbers of sums with different integers Consider a sequence of 3999 numbered... 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... division of 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 But y