EUROPEAN MATHEMATICAL OLYMPIAD

▲❛♥❣✉❛❣❡✿ ❉❛②✿ ❊♥❣❧✐s❤ ✶ EGMO | 2012 European Girls’ Mathematical Olympiad ❚❤✉rs❞❛②✱ ❆♣r✐❧ ✶✷✱ ✷✵✶✷ Pr♦❜❧❡♠ ✶✳ ▲❡t ABC ❜❡ ❛ tr✐❛♥❣❧❡ ✇✐t❤ ❝✐r❝✉♠❝❡♥tr❡ O✳ ❚❤❡ ♣♦✐♥ts D✱ E ❛♥❞ F ❧✐❡ ✐♥ t❤❡ ✐♥t❡r✐♦rs ♦❢ t❤❡ s✐❞❡s BC ✱ CA ❛♥❞ AB r❡s♣❡❝t✐✈❡❧②✱ s✉❝❤ t❤❛t DE ✐s ♣❡r♣❡♥❞✐❝✉❧❛r t♦ CO ❛♥❞ DF ✐s ♣❡r♣❡♥❞✐❝✉❧❛r t♦ BO✳ ✭❇② ✐♥t❡r✐♦r ✇❡ ♠❡❛♥✱ ❢♦r ❡①❛♠♣❧❡✱ t❤❛t t❤❡ ♣♦✐♥t D ❧✐❡s ♦♥ t❤❡ ❧✐♥❡ BC ❛♥❞ D ✐s ❜❡t✇❡❡♥ B ❛♥❞ C ♦♥ t❤❛t ❧✐♥❡✳✮ ▲❡t K ❜❡ t❤❡ ❝✐r❝✉♠❝❡♥tr❡ ♦❢ tr✐❛♥❣❧❡ AF E ✳ Pr♦✈❡ t❤❛t t❤❡ ❧✐♥❡s DK ❛♥❞ BC ❛r❡ ♣❡r♣❡♥❞✐❝✉❧❛r✳ ▲❡t n ❜❡ ❛ ♣♦s✐t✐✈❡ ✐♥t❡❣❡r✳ ❋✐♥❞ t❤❡ ❣r❡❛t❡st ♣♦ss✐❜❧❡ ✐♥t❡❣❡r m✱ ✐♥ t❡r♠s ♦❢ n✱ ✇✐t❤ t❤❡ ❢♦❧❧♦✇✐♥❣ ♣r♦♣❡rt②✿ ❛ t❛❜❧❡ ✇✐t❤ m r♦✇s ❛♥❞ n ❝♦❧✉♠♥s ❝❛♥ ❜❡ ✜❧❧❡❞ ✇✐t❤ r❡❛❧ ♥✉♠❜❡rs ✐♥ s✉❝❤ ❛ ♠❛♥♥❡r t❤❛t ❢♦r ❛♥② t✇♦ ❞✐✛❡r❡♥t r♦✇s [a1 , a2 , , an ] ❛♥❞ [b1 , b2 , , bn ] t❤❡ ❢♦❧❧♦✇✐♥❣ ❤♦❧❞s✿ Pr♦❜❧❡♠ ✷✳ max(|a1 − b1 |, |a2 − b2 |, , |an − bn |) = Pr♦❜❧❡♠ ✸✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t f yf (x + y) + f (x) = 4x + 2yf (x + y) ❢♦r ❛❧❧ x, y ∈ R✳ Pr♦❜❧❡♠ ✹✳ ❆ s❡t A ♦❢ ✐♥t❡❣❡rs ✐s ❝❛❧❧❡❞ s✉♠✲❢✉❧❧ ✐❢ A ⊆ A + A✱ ✐✳❡✳ ❡❛❝❤ ❡❧❡♠❡♥t a ∈ A ✐s t❤❡ s✉♠ ♦❢ s♦♠❡ ♣❛✐r ♦❢ ✭♥♦t ♥❡❝❡ss❛r✐❧② ❞✐✛❡r❡♥t✮ ❡❧❡♠❡♥ts b, c ∈ A✳ ❆ s❡t A ♦❢ ✐♥t❡❣❡rs ✐s s❛✐❞ t♦ ❜❡ ③❡r♦✲s✉♠✲❢r❡❡ ✐❢ ✐s t❤❡ ♦♥❧② ✐♥t❡❣❡r t❤❛t ❝❛♥♥♦t ❜❡ ❡①♣r❡ss❡❞ ❛s t❤❡ s✉♠ ♦❢ t❤❡ ❡❧❡♠❡♥ts ♦❢ ❛ ✜♥✐t❡ ♥♦♥❡♠♣t② s✉❜s❡t ♦❢ A✳ ❉♦❡s t❤❡r❡ ❡①✐st ❛ s✉♠✲❢✉❧❧ ③❡r♦✲s✉♠✲❢r❡❡ s❡t ♦❢ ✐♥t❡❣❡rs❄ ▲❛♥❣✉❛❣❡✿ ❊♥❣❧✐s❤ ❚✐♠❡✿ ✹ ❤♦✉rs ❛♥❞ ✸✵ ♠✐♥✉t❡s ❊❛❝❤ ♣r♦❜❧❡♠ ✐s ✇♦rt❤ ✼ ♣♦✐♥ts European Girls’ Mathematical Olympiad 2012—Day Solutions Problem Let ABC be a triangle with circumcentre O The points D, E and F lie in the interiors of the sides BC, CA and AB respectively, such that DE is perpendicular to CO and DF is perpendicular to BO (By interior we mean, for example, that the point D lies on the line BC and D is between B and C on that line.) Let K be the circumcentre of triangle AF E Prove that the lines DK and BC are perpendicular Origin Netherlands (Merlijn Staps) A K O E F B C D Solution (submitter) Let C be the tangent at C to the circumcircle of DE and C are parallel Now we find that ∠CDE = ∠(BC, C) ABC As CO ⊥ C, the lines = ∠BAC, hence the quadrilateral BDEA is cyclic Analogously, we find that the quadrilateral CDF A is cyclic As we now have ∠CDE = ∠A = ∠F DB, we conclude that the line BC is the external angle bisector of ∠EDF Furthermore, ∠EDF = 180◦ − 2∠A Since K is the circumcentre of AEF , ∠F KE = 2∠F AE = 2∠A So ∠F KE + ∠EDF = 180◦ , hence K lies on the circumcircle of DEF As |KE| = |KF |, we have that K is the midpoint of the arc EF of this circumcircle It is well known that this point lies on the internal angle bisector of ∠EDF We conclude that DK is the internal angle bisector of ∠EDF Together with the fact that BC is the external angle bisector of ∠EDF , this yields that DK ⊥ BC, as desired Solution (submitter) As in the previous solution, we show that the quadrilaterals BDEA and CDF A are both cyclic Denote by M and L respectively the circumcentres of these quadrilaterals We will show that the quadrilateral KLOM is a parallelogram The lines KL and M O are the perpendicular bisectors of the line segments AF and AB, respectively Hence both KL and M O are perpendicular to AB, which yields KL M O In the same way we can show that the lines KM and LO are both perpendicular to AC and hence parallel as well We conclude that KLOM is indeed a parallelogram Now, let K , L , O and M be the respective projections of K, L, O and M to BC We have to show that K = D As L lies on the perpendicular bisector of CD, we have that L is the midpoint of CD Similarly, M is the midpoint of BD and O is the midpoint of BC Now we are going to use directed lengths Since KLOM is a parallelogram, M K = O L As O L =O C −LC = · (BC − DC) = we find that M K = M D, hence K = D, as desired 1 · BD = M D, Solution (submitter) Denote by A , B and C the tangents at A, B and C to the circumcircle of ABC Let A be the point of intersection of B and C and define B and C analogously As in the first solution, we find that DE C and DF B Now, let Q be the point of intersection of DE and A and let R be the point of intersection of DF and A We easily find AQE ∼ AB C As |B A| = |B C|, we must have |QA| = |QE|, hence AQE is isosceles Therefore the perpendicular bisector of AE is the internal angle bisector of ∠EQA = ∠DQR Analogously, the perpendicular bisector of AF is the internal angle bisector of ∠DRQ We conclude that K is the incentre of DQR, thus DK is the angle bisector of ∠QDR Because the sides of the triangles QDR and B A C are pairwise parallel, the angle bisector DK of ∠QDR is parallel to the angle bisector of ∠B A C Finally, as the angle bisector of ∠B A C is easily seen to be perpendicular to BC (as it is the perpendicular bisector of this segment), we find that DK ⊥ BC, as desired Remark (submitter) The fact that the quadrilateral BDEA is cyclic (which is an essential part of the first two solutions) can be proven in various ways Another possibility is as follows Let P be the midpoint of BC Then, as ∠CP O = 90◦ , we have ∠P OC = 90◦ −∠OCP Let X be the point of intersection of DE and CO, then we have that ∠CDE = ∠CDX = 90◦ − ∠XCD = 90◦ − ∠OCP Hence ∠CDE = ∠P OC = 21 ∠BOC = ∠BAC From this we can conclude that BDEA is cyclic Solution (PSC) This is a simplified variant of Solution ∠COB = 2∠A (angle at centre of circle ABC) and OB = OC so ∠OBC = ∠BCO = 90◦ − ∠A Likewise ∠EKF = 2∠A and ∠KF E = ∠F EK = 90◦ − ∠A Now because DE ⊥ CO, ∠EDC = 90◦ − ∠DCO = 90◦ − ∠BCO = ∠A and similarly ∠BDF = ∠A, so ∠F DE = 180◦ − 2∠A So quadrilateral KF DE is cyclic (opposite angles), so (same segment) ∠KDE = ∠KF E = 90◦ − ∠A, so ∠KDC = 90◦ and DK is perpendicular to BC Problem Let n be a positive integer Find the greatest possible integer m, in terms of n, with the following property: a table with m rows and n columns can be filled with real numbers in such a manner that for any two different rows [a1 , a2 , , an ] and [b1 , b2 , , bn ] the following holds: max(|a1 − b1 |, |a2 − b2 |, , |an − bn |) = Origin Poland (Tomasz Kobos) Solution (submitter) The largest possible m is equal to 2n In order to see that the value 2n can be indeed achieved, consider all binary vectors of length n as rows of the table We now proceed with proving that this is the maximum value Let [aik ] be a feasible table, where i = 1, , m and k = 1, , n Let us define undirected graphs G1 , G2 , , Gn , each with vertex set {1, 2, , m}, where ij ∈ E(Gk ) if and only if |aik − ajk | = (by E(Gk ) we denote the edge set of the graph Gk ) Observe the following two properties (1) Each graph Gk is bipartite Indeed, if it contained a cycle of odd length, then the sum of ±1 along this cycle would need to be equal to 0, which contradicts the length of the cycle being odd (2) For every i = j, ij ∈ E(Gk ) for some k This follows directly from the problem statement For every graph Gk fix some bipartition (Ak , Bk ) of {1, 2, , m}, i.e., a partition of {1, 2, , m} into two disjoint sets Ak , Bk such that the edges of Gk traverse only between Ak and Bk If m > 2n , then there are two distinct indices i, j such that they belong to exactly the same parts Ak , Bk , that is, i ∈ Ak if and only if j ∈ Ak for all k = 1, 2, , n However, this means that the edge ij cannot be present in any of the graphs G1 , G2 , , Gn , which contradicts (2) Therefore, m ≤ 2n Solution (PSC) In any table with the given property, the least and greatest values in a column cannot differ by more than Thus, if each value that is neither least nor greatest in its column is changed to be equal to either the least or the greatest value in its column (arbitrarily), this does not affect any |ai − bi | = 1, nor does it increase any difference above 1, so the table still has that given property But after such a change, for any choice of what the least and greatest values in each column are, there are only two possible choices for each entry in the table (either the least or the greatest value in its column); that is, only 2n possible distinct rows, and the given property implies that all rows must be distinct As in the previous solution, we see that this number can be achieved Solution (Coordinators) We prove by induction on n that m ≤ 2n First suppose n = If real numbers x and y have |x − y| = then x and y have opposite parities and hence it is impossible to find three real numbers with all differences Thus m ≤ Suppose instead n > Let a be the smallest number appearing in the first column of the table; then every entry in the first column of the table lies in the interval [a, a + 1] Let A be the collection of rows with first entry a and B be the collection of rows with first entry in (a, a + 1] No two rows in A differ by in their first entries, so if we list the rows in A and delete their first entries we obtain a table satisfying the conditions of the problem with n replaced by n − 1; thus, by the induction hypothesis, there are at most 2n−1 rows in A Similarly, there are at most 2n−1 rows in B Hence m ≤ 2n−1 + 2n−1 = 2n As before, this number can be achieved Solution (Coordinators) Consider the rows of the table as points of Rn As the values in each column differ by at most 1, these points must lie in some n-dimensional unit cube C Consider the unit cubes centred on each of the m points The conditions of the problem imply that the interiors of these unit cubes are pairwise disjoint But now C has volume 1, and each of these cubes intersects C in volume at least 2−n : indeed, if the unit cube centred on a point of C is divided into 2n cubes of equal size then one of these cubes must lie entirely within C Hence m ≤ 2n As before, this number can be achieved Solution (Coordinators) Again consider the rows of the table as points of Rn The conditions of the problem imply that these points must all lie in some n-dimensional unit cube C, but no two of the points lie in any smaller cube Thus if C is divided into 2n equally-sized subcubes, each of these subcubes contains at most one row of the table, giving m ≤ 2n As before, this number can be achieved Problem Find all functions f : R → R such that f yf (x + y) + f (x) = 4x + 2yf (x + y) for all x, y ∈ R Origin Netherlands (Birgit van Dalen) Solution (submitter) Setting y = yields f (f (x)) = 4x, (1) from which we derive that f is a bijective function Also, we find that f (0) = f (4 · 0) = f (f (f (0))) = 4f (0), hence f (0) = Now set x = and y = in the given equation and use (1) again: = f (f (1)) = 2f (1), so f (1) = and therefore also f (2) = f (f (1)) = Finally substitute y = − x in the equation: f (2(1 − x) + f (x)) = 4x + 4(1 − x) = = f (2) for all x ∈ R As f is injective, from this it follows that f (x) = − 2(1 − x) = 2x It is easy to see that this function satisfies the original equation Hence the only solution is the function defined by f (x) = 2x for all x ∈ R Solution (Coordinators) Setting y = in the equation we see f (f (x)) = 4x so f is a bijection Let κ = f −1 (2) and set x + y = κ in the original equation to see f (2κ − 2x + f (x)) = 4κ As the right hand side is independent of x and f is injective, 2κ − 2x + f (x) is constant, i.e f (x) = 2x + α Substituting this into the original equation, we see that 2x + α is a solution to the original equation if and only if 4(y + xy + x) + (3 + 2y)α = 4(y + xy + x) + 2yα for all x, y, i.e if and only if α = Thus the unique solution to the equation is f (x) = 2x Problem A set A of integers is called sum-full if A ⊆ A + A, i.e each element a ∈ A is the sum of some pair of (not necessarily different) elements b, c ∈ A A set A of integers is said to be zero-sum-free if is the only integer that cannot be expressed as the sum of the elements of a finite nonempty subset of A Does there exist a sum-full zero-sum-free set of integers? Origin Romania (Dan Schwarz) Remark The original formulation of this problem had a weaker definition of zero-sum-free that did not require all nonzero integers to be sums of finite nonempty subsets of A Solution (submitter, adapted) The set A = {F2n : n = 1, 2, } ∪ {−F2n+1 : n = 1, 2, }, where Fk is the k th Fibonacci number (F1 = 1, F2 = 1, Fk+2 = Fk+1 + Fk for k ≥ 1) qualifies for an example We then have F2n = F2n+2 + (−F2n+1 ) and −F2n+1 = (−F2n+3 ) + F2n+2 for all n ≥ 1, so A is sum-full (and even with unique representations) On the other hand, we can never have s t F2ni − 0= i=1 F2nj +1 , j=1 owing to the fact that Zeckendorf representations are known to be unique It remains to be shown that all nonzero values can be represented as sums of distinct numbers 1, −2, 3, −5, 8, −13, 21, This may be done using a greedy algorithm: when representing n, the number largest in magnitude that is used is the element m = ±Fk of A that is closest to subject to having the same sign as n and |m| ≥ |n| That this algorithm terminates without using any member of A twice is a straightforward induction on k; the base case is k = (m = 1) and the induction hypothesis is that for all n for which the above algorithm starts with ±F with ≤ k, it terminates without having used any member of A twice and without having used any ±Fj with j > Remark (James Aaronson and Adam P Goucher) claim that the set Let n be a positive integer, and write u = 2n ; we {1, 2, 4, , 2n−1 , −u, u + 1, −(2u + 1), 3u + 2, −(5u + 3), 8u + 5, } is a sum-full zero-sum-free set The proof is similar to that used for the standard examples ▲❛♥❣✉❛❣❡✿ ❉❛②✿ ❊♥❣❧✐s❤ ✷ EGMO | 2012 European Girls’ Mathematical Olympiad ❋r✐❞❛②✱ ❆♣r✐❧ ✶✸✱ ✷✵✶✷ Pr♦❜❧❡♠ ✺✳ ❚❤❡ ♥✉♠❜❡rs p ❛♥❞ q ❛r❡ ♣r✐♠❡ ❛♥❞ s❛t✐s❢② p q+1 2n + = p+1 q n+2 ❢♦r s♦♠❡ ♣♦s✐t✐✈❡ ✐♥t❡❣❡r n✳ ❋✐♥❞ ❛❧❧ ♣♦ss✐❜❧❡ ✈❛❧✉❡s ♦❢ q − p✳ Pr♦❜❧❡♠ ✻✳ ❚❤❡r❡ ❛r❡ ✐♥✜♥✐t❡❧② ♠❛♥② ♣❡♦♣❧❡ r❡❣✐st❡r❡❞ ♦♥ t❤❡ s♦❝✐❛❧ ♥❡t✇♦r❦ ▼✉❣❜♦♦❦✳ ❙♦♠❡ ♣❛✐rs ♦❢ ✭❞✐✛❡r❡♥t✮ ✉s❡rs ❛r❡ r❡❣✐st❡r❡❞ ❛s ❢r✐❡♥❞s✱ ❜✉t ❡❛❝❤ ♣❡rs♦♥ ❤❛s ♦♥❧② ✜♥✐t❡❧② ♠❛♥② ❢r✐❡♥❞s✳ ❊✈❡r② ✉s❡r ❤❛s ❛t ❧❡❛st ♦♥❡ ❢r✐❡♥❞✳ ✭❋r✐❡♥❞s❤✐♣ ✐s s②♠♠❡tr✐❝❀ t❤❛t ✐s✱ ✐❢ A ✐s ❛ ❢r✐❡♥❞ ♦❢ B ✱ t❤❡♥ B ✐s ❛ ❢r✐❡♥❞ ♦❢ A✳✮ ❊❛❝❤ ♣❡rs♦♥ ✐s r❡q✉✐r❡❞ t♦ ❞❡s✐❣♥❛t❡ ♦♥❡ ♦❢ t❤❡✐r ❢r✐❡♥❞s ❛s t❤❡✐r ❜❡st ❢r✐❡♥❞✳ ■❢ A ❞❡s✐❣♥❛t❡s B ❛s ❤❡r ❜❡st ❢r✐❡♥❞✱ t❤❡♥ ✭✉♥❢♦rt✉♥❛t❡❧②✮ ✐t ❞♦❡s ♥♦t ❢♦❧❧♦✇ t❤❛t B ♥❡❝❡ss❛r✐❧② ❞❡s✐❣♥❛t❡s A ❛s ❤❡r ❜❡st ❢r✐❡♥❞✳ ❙♦♠❡♦♥❡ ❞❡s✐❣♥❛t❡❞ ❛s ❛ ❜❡st ❢r✐❡♥❞ ✐s ❝❛❧❧❡❞ ❛ 1✲❜❡st ❢r✐❡♥❞✳ ▼♦r❡ ❣❡♥❡r❛❧❧②✱ ✐❢ n > ✐s ❛ ♣♦s✐t✐✈❡ ✐♥t❡❣❡r✱ t❤❡♥ ❛ ✉s❡r ✐s ❛♥ n✲❜❡st ❢r✐❡♥❞ ♣r♦✈✐❞❡❞ t❤❛t t❤❡② ❤❛✈❡ ❜❡❡♥ ❞❡s✐❣♥❛t❡❞ t❤❡ ❜❡st ❢r✐❡♥❞ ♦❢ s♦♠❡♦♥❡ ✇❤♦ ✐s ❛♥ (n − 1)✲❜❡st ❢r✐❡♥❞✳ ❙♦♠❡♦♥❡ ✇❤♦ ✐s ❛ k ✲❜❡st ❢r✐❡♥❞ ❢♦r ❡✈❡r② ♣♦s✐t✐✈❡ ✐♥t❡❣❡r k ✐s ❝❛❧❧❡❞ ♣♦♣✉❧❛r✳ ✭❛✮ Pr♦✈❡ t❤❛t ❡✈❡r② ♣♦♣✉❧❛r ♣❡rs♦♥ ✐s t❤❡ ❜❡st ❢r✐❡♥❞ ♦❢ ❛ ♣♦♣✉❧❛r ♣❡rs♦♥✳ ✭❜✮ ❙❤♦✇ t❤❛t ✐❢ ♣❡♦♣❧❡ ❝❛♥ ❤❛✈❡ ✐♥✜♥✐t❡❧② ♠❛♥② ❢r✐❡♥❞s✱ t❤❡♥ ✐t ✐s ♣♦ss✐❜❧❡ t❤❛t ❛ ♣♦♣✉❧❛r ♣❡rs♦♥ ✐s ♥♦t t❤❡ ❜❡st ❢r✐❡♥❞ ♦❢ ❛ ♣♦♣✉❧❛r ♣❡rs♦♥✳ ▲❡t ABC ❜❡ ❛♥ ❛❝✉t❡✲❛♥❣❧❡❞ tr✐❛♥❣❧❡ ✇✐t❤ ❝✐r❝✉♠❝✐r❝❧❡ Γ ❛♥❞ ♦rt❤♦❝❡♥tr❡ H ✳ ▲❡t K ❜❡ ❛ ♣♦✐♥t ♦❢ Γ ♦♥ t❤❡ ♦t❤❡r s✐❞❡ ♦❢ BC ❢r♦♠ A✳ ▲❡t L ❜❡ t❤❡ r❡✢❡❝t✐♦♥ ♦❢ K ✐♥ t❤❡ ❧✐♥❡ AB ✱ ❛♥❞ ❧❡t M ❜❡ t❤❡ r❡✢❡❝t✐♦♥ ♦❢ K ✐♥ t❤❡ ❧✐♥❡ BC ✳ ▲❡t E ❜❡ t❤❡ s❡❝♦♥❞ ♣♦✐♥t ♦❢ ✐♥t❡rs❡❝t✐♦♥ ♦❢ Γ ✇✐t❤ t❤❡ ❝✐r❝✉♠❝✐r❝❧❡ ♦❢ tr✐❛♥❣❧❡ BLM ✳ ❙❤♦✇ t❤❛t t❤❡ ❧✐♥❡s KH ✱ EM ❛♥❞ BC ❛r❡ ❝♦♥❝✉rr❡♥t✳ ✭❚❤❡ ♦rt❤♦❝❡♥tr❡ ♦❢ ❛ tr✐❛♥❣❧❡ ✐s t❤❡ ♣♦✐♥t ♦♥ ❛❧❧ t❤r❡❡ Pr♦❜❧❡♠ ✼✳ ♦❢ ✐ts ❛❧t✐t✉❞❡s✳✮ Pr♦❜❧❡♠ ✽✳ ❆ ✇♦r❞ ✐s ❛ ✜♥✐t❡ s❡q✉❡♥❝❡ ♦❢ ❧❡tt❡rs ❢r♦♠ s♦♠❡ ❛❧♣❤❛❜❡t✳ ❆ ✇♦r❞ ✐s r❡♣❡t✐t✐✈❡ ✐❢ ✐t ✐s ❛ ❝♦♥✲ ❝❛t❡♥❛t✐♦♥ ♦❢ ❛t ❧❡❛st t✇♦ ✐❞❡♥t✐❝❛❧ s✉❜✇♦r❞s ✭❢♦r ❡①❛♠♣❧❡✱ ❛❜❛❜❛❜ ❛♥❞ ❛❜❝❛❜❝ ❛r❡ r❡♣❡t✐t✐✈❡✱ ❜✉t ❛❜❛❜❛ ❛♥❞ ❛❛❜❜ ❛r❡ ♥♦t✮✳ Pr♦✈❡ t❤❛t ✐❢ ❛ ✇♦r❞ ❤❛s t❤❡ ♣r♦♣❡rt② t❤❛t s✇❛♣♣✐♥❣ ❛♥② t✇♦ ❛❞❥❛❝❡♥t ❧❡tt❡rs ♠❛❦❡s t❤❡ ✇♦r❞ r❡♣❡t✐t✐✈❡✱ t❤❡♥ ❛❧❧ ✐ts ❧❡tt❡rs ❛r❡ ✐❞❡♥t✐❝❛❧✳ ✭◆♦t❡ t❤❛t ♦♥❡ ♠❛② s✇❛♣ t✇♦ ❛❞❥❛❝❡♥t ✐❞❡♥t✐❝❛❧ ❧❡tt❡rs✱ ❧❡❛✈✐♥❣ ❛ ✇♦r❞ ✉♥❝❤❛♥❣❡❞✳✮ ▲❛♥❣✉❛❣❡✿ ❊♥❣❧✐s❤ ❚✐♠❡✿ ✹ ❤♦✉rs ❛♥❞ ✸✵ ♠✐♥✉t❡s ❊❛❝❤ ♣r♦❜❧❡♠ ✐s ✇♦rt❤ ✼ ♣♦✐♥ts European Girls’ Mathematical Olympiad 2012—Day Solutions Problem The numbers p and q are prime and satisfy q+1 2n p + = p+1 q n+2 for some positive integer n Find all possible values of q − p Origin Luxembourg (Pierre Haas) Solution (submitter) Rearranging the equation, 2qn(p + 1) = (n + 2)(2pq + p + q + 1) The left hand side is even, so either n + or p + q + is even, so either p = or q = since p and q are prime, or n is even If p = 2, 6qn = (n + 2)(5q + 3), so (q − 3)(n − 10) = 36 Considering the divisors of 36 for which q is prime, we find the possible solutions (p, q, n) in this case are (2, 5, 28) and (2, 7, 19) (both of which satisfy the equation) If q = 2, 4n(p + 1) = (n + 2)(5p + 3), so n = pn + 10p + 6, a contradiction since n < pn, so there is no solution with q = Finally, suppose that n = 2k is even We may suppose also that p and q are odd primes The equation becomes 2kq(p + 1) = (k + 1)(2pq + p + q + 1) The left hand side is even and 2pq + p + q + is odd, so k + is even, so k = + is odd We now have q(p + 1)(2 + 1) = ( + 1)(2pq + p + q + 1) or equivalently q(p + 1) = ( + 1)(pq + p + 1) Note that q | pq + p + if and only if q | p + Furthermore, because (p, p + 1) = and q is prime, (p + 1, pq + p + 1) = (p + 1, pq) = (p + 1, q) > if and only if q | p + Since ( , + 1), we see that, if q p + 1, then = pq + p + and + = q(p + 1), so q = p + (and (p, p + 2, 2(2p2 + 6p + 3)) satisfies the original equation) In the contrary case, suppose p + = rq, so (p + 1) = ( + 1)(p + r), a contradiction since < + and p + ≤ p + r Thus the possible values of q − p are 2, and Solution (PSC) Subtracting and multiplying by −1, the condition is equivalent to 1 − = p+1 q n+2 Thus q > p + Rearranging, 4(p + 1)q n+2 The expression on the right is a positive integer, and q must cancel into n + else q would divide p + < q Let (n + 2)/q = u a positive integer Now 4(p + 1) q−p−1= u so uq − u(p + 1) = 4(p + 1) q−p−1= so p + divides uq However, q is prime and p + < q, therefore p + divides u Let v be the integer u/(p + 1) Now q − p = + ∈ {2, 3, 5} v All three cases can occur, where (p, q, n) is (3, 5, 78), (2, 5, 28) or (2, 7, 19) Note that all pairs of twin primes q = p + yield solutions (p, p + 2, 2(2p2 + 6p + 3)) Solution (Coordinators) Subtract from both sides to get 1 − = p+1 q n+2 From this, since n is positive, we have that q > p + Therefore q and p + are coprime, since q is prime Group the terms on the LHS to get q−p−1 = q(p + 1) n+2 Now (q, q − p − 1) = (q, p + 1) = and (p + 1, q − p − 1) = (p + 1, q) = so the fraction on the left is in lowest terms Therefore the numerator must divide the numerator on the right, which is Since q − p − is positive, it must be 1, or 4, so that q − p must be 2, or All of these can be attained, by (p, q, n) = (3, 5, 78), (2, 5, 28) and (2, 7, 19) respectively Problem There are infinitely many people registered on the social network Mugbook Some pairs of (different) users are registered as friends, but each person has only finitely many friends Every user has at least one friend (Friendship is symmetric; that is, if A is a friend of B, then B is a friend of A.) Each person is required to designate one of their friends as their best friend If A designates B as her best friend, then (unfortunately) it does not follow that B necessarily designates A as her best friend Someone designated as a best friend is called a 1-best friend More generally, if n > is a positive integer, then a user is an n-best friend provided that they have been designated the best friend of someone who is an (n − 1)-best friend Someone who is a k-best friend for every positive integer k is called popular (a) Prove that every popular person is the best friend of a popular person (b) Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person Origin Romania (Dan Schwarz) (rephrasing by Geoff Smith) Remark The original formulation of this problem was: Given a function f : X → X, let us use the notations f (X) := X, f n+1 (X) := f (f n (X)) for n ≥ 0, and also ω f (X) := f n (X) Let us now impose on f that all its fibres f −1 (y) := {x ∈ X | f (x) = y}, for y ∈ f (X), n≥0 are finite Prove that f (f ω (X)) = f ω (X) Solution (submitter, adapted) For any person A, let f (x) = x, let f (A) be A’s best friend, and define f k+1 (A) = f (f k (A)), so any person who is a k-best friend is f k (A) for some person A; clearly a k-best friend is also an -best friend for all < k Let X be a popular person For each positive integer k, let xk be a person with f k (xk ) = X Because X only has finitely many friends, infinitely many of the f k−1 (xk ) (all of whom designated X as best friend) must be the same person, who must be popular If people can have infinitely many friends, consider people Xi for positive integers i and Pi,j for i < j positive integers Xi designates Xi+1 as her best friend; Pi,i designates X1 as her best friend; Pi,j designates Pi+1,j as her best friend if i < j Then all Xi are popular, but X1 is not the best friend of a popular person Solution (submitter, adapted) For any set S of people, let f −1 (S) be the set of people who designated someone in S as their best friend Since each person has only finitely many friends, if S is finite then f −1 (S) is finite Let X be a popular person and put V0 = {X} and Vk = f −1 (Vk−1 ) All Vi are finite and (since X is popular) nonempty If any two sets Vi , Vj , with ≤ i < j are not disjoint, define f i (x) for positive integers i as in Solution It follows ∅ = f i (Vi ∩ Vj ) ⊆ f i (Vi ) ∩ f i (Vj ) ⊆ V0 ∩ Vj−i , thus X ∈ Vj−i But this means that f j−i (X) = X, therefore f n(j−i) (X) = X Furthermore, if Y = f j−i−1 (X), then f (Y ) = X and f n(j−i) (Y ) = Y , so X is the best friend of Y , who is popular If all sets Vn are disjoint, by Kă onigs infinity lemma there exists an infinite sequence of (distinct) xi , i ≥ 0, with xi ∈ Vi and xi = f (xi+1 ) for all i Now x1 is popular and her best friend is x0 = X If people can have infinitely many friends, proceed as in Solution Problem Let ABC be an acute-angled triangle with circumcircle Γ and orthocentre H Let K be a point of Γ on the other side of BC from A Let L be the reflection of K in the line AB, and let M be the reflection of K in the line BC Let E be the second point of intersection of Γ with the circumcircle of triangle BLM Show that the lines KH, EM and BC are concurrent (The orthocentre of a triangle is the point on all three of its altitudes.) Origin Luxembourg (Pierre Haas) Solution (submitter) Since the quadrilateral BM EL is cyclic, we have ∠BEM = ∠BLM By construction, |BK| = |BL| = |BM |, and so (using directed angles) ∠BLM = 90◦ − 12 ∠M BL = 90◦ − 180◦ − 12 ∠LBK − 21 ∠KBM = ∠LBK + 12 ∠KBM − 90◦ = (180◦ − ∠B) − 90◦ = 90◦ − B We see also that ∠BEM = ∠BAH, and so the point N of intersection of EM and AH lies on Γ Let X be the point of intersection of KH and BC, and let N be the point of intersection of M X and AH Since BC bisects the segment KM by construction, the triangle KXM is isosceles; as AH M K, HXN is isosceles Since AH ⊥ BC, N is the reflection of H in the line BC It is well known that this reflection lies on Γ, and so N = N Thus E, M , N and M , X, N all lie on the same line M N ; that is, EM passes through X A E Γ H M X C B L K N =N Remark (submitter) The condition that K lies on the circumcircle of ABC is not necessary; indeed, the solution above does not use it However, together with the fact that the triangle ABC is acute-angled, this condition implies that M is in the interior of Γ, which is necessary to avoid dealing with different configurations including coincident points or the point of concurrence being at infinity Solution (PSC) We work with directed angles Let HK meet BC at X Let M X meet AH at HA on Γ (where HA is the reflection of H in BC) Define E to be where HA M meets Γ (again) Our task is to show that ∠M E B = ∠M LB Observe that ∠M E B = ∠HA AB =B (angles in same segment) c Now ∠M LB = ∠HLB (Simson line, doubled) = ∠BKHC (reflecting in the line AB) = ∠BCHC (angles in the same segment) = Bc Language: English Day: Thursday, April 12, 2018 Problem A domino is a × or × tile Let n ≥ be an integer Dominoes are placed on an n × n board in such a way that each domino covers exactly two cells of the board, and dominoes not overlap The value of a row or column is the number of dominoes that cover at least one cell of this row or column The configuration is called balanced if there exists some k ≥ such that each row and each column has a value of k Prove that a balanced configuration exists for every n ≥ 3, and find the minimum number of dominoes needed in such a configuration Problem Let Γ be the circumcircle of triangle ABC A circle Ω is tangent to the line segment AB and is tangent to Γ at a point lying on the same side of the line AB as C The angle bisector of ∠BCA intersects Ω at two different points P and Q Prove that ∠ABP = ∠QBC Problem (a) Prove that for every real number t such that < t < 21 there exists a positive integer n with the following property: for every set S of n positive integers there exist two different elements x and y of S, and a non-negative integer m (i.e m ≥ 0), such that |x − my| ≤ ty (b) Determine whether for every real number t such that < t < positive integers such that |x − my| > ty there exists an infinite set S of for every pair of different elements x and y of S and every positive integer m (i.e m > 0) Language: English Time: hours and 30 minutes Each problem is worth points Problem Let ABC be a triangle with CA = CB and ∠ACB = 120◦, and let M be the midpoint of AB Let P be a variable point on the circumcircle of ABC, and let Q be the point on the segment CP such that QP = 2QC It is given that the line through P and perpendicular to AB intersects the line MQ at a unique point N Prove that there exists a fixed circle such that N lies on this circle for all possible positions of P (Velina Ivanova, Bulgaria) Solution Let O be the circumcenter of ABC From the assumption that ∠ACB = 120◦ it follows that M is the midpoint of CO Let ω denote the circle with center in C and radius CO This circle in the image of the circumcircle of ABC through the translation that sends O to C We claim that N lies on ω N C C P A Q M B A Q M B N O O P Let us consider the triangles QNP and QMC The angles in Q are equal Since NP is parallel to MC (both lines are perpendicular to AB), it turns out that ∠QNP = ∠QMC, and hence the two triangles are similar Since QP = 2QC, it follows that NP = 2MC = CO, which proves that N lies on ω Comment The possible positions of N are all the points of ω with the exception of the two points lying on the line CO Indeed, P does not lie on the line CO because otherwise the point N is not well-defined, and therefore also N does not lie on the same line Conversely, let N be any point on ω and not lying on the line CO Let P be the corresponding point on the circumcircle of ABC, namely such that NP is parallel and equal to CO Let Q be the intersection of CP and NM As before, the triangles QNP and QMC are similar, and now from the relation NP = 2MC we deduce that QP = 2QC This proves that N can be obtained from P through the construction described in the statement of the problem Alternative solution Let M ′ denote the symmetric of M with respect to O Let us consider the quadrilateral MM ′ P N The lines MM ′ and NP are parallel by construction Also the lines P M ′ and NM are parallel (homothety from C with coefficient 3) It follows that MM ′ P N is a parallelogram, and hence P N = MM ′ = OC Computational solution There are many computation approaches to this problem For example, we can set Cartesian coordinates so that A= √ − , 2 , B= √ , 2 , C = (0, 1) , M= 0, Setting P = (a, b), we obtain that Q = (a/3, (2 + b)/3) The equation of the line through P and perpendicular to AB is x = a The equation of the line MQ (if a = 0) is y− x = a +b The intersection of the two lines is therefore N = (a, + b) = P + (0, 1) This shows that the map P → N in the translation by the vector (0, 1) This result is independent of the position of P (provided that a = 0, because otherwise N is not well-defined) When P lies on the circumcircle of ABC, with the exception of the two points with a = 0, then necessarily N lies on the translated circle (which is the circle with center in C and radius 1) Problem Consider the set A= 1+ : k = 1, 2, 3, k (a) Prove that every integer x ≥ can be written as the product of one or more elements of A, which are not necessarily different (b) For every integer x ≥ 2, let f (x) denote the minimum integer such that x can be written as the product of f (x) elements of A, which are not necessarily different Prove that there exist infinitely many pairs (x, y) of integers with x ≥ 2, y ≥ 2, and f (xy) < f (x) + f (y) (Pairs (x1 , y1 ) and (x2 , y2 ) are different if x1 = x2 or y1 = y2 ) (Mihail Baluna, Romania) Solution Every integer x ≥ can be written as the telescopic product of x − elements of A as x= 1+ x−1 · 1+ x−2 · · 1+ · 1+ , which is enough to establish part (a) We now consider part (b) Notice that for any positive integer k we have f (2k + 1) ≤ k + 1, because 2k + = + 21k · 2k is a representation of 2k + as a product of k + elements of A We claim that all the pairs (x, y) of the form 24k+2 + y= x = 5, satisfy the required inequality Notice that y is an integer for any positive value of k, because 24k+2 + ≡ 16k · + ≡ ≡ (mod 5) Furthermore, f (xy) = f (24k+2 + 1) ≤ 4k + (and f (x) = f (22 + 1) ≤ 3) by the above We now need some lower bounds on the values of f Notice that no element of A exceeds 2, and therefore the product of at most k elements of A does not exceed 2k : it follows that f (n) ≥ ⌈log2 (n)⌉, (Q2.1) and in particular that f (5) = f (22 + 1) ≥ ⌈log2 (5)⌉ = We have thus proven f (x) = f (5) = We want to show f (xy) < f (x) + f (y), and since we know f (xy) ≤ 4k + and f (x) = we are reduced to showing f (y) > 4k Since y > 24k−1 , from (Q2.1) we already know that f (y) ≥ 4k, and hence we just need to exclude that f (y) = 4k Let us assume that we can represent y in the form a1 · · a4k with every in A At least one of the is not (otherwise the product would be a power of 2, while y is odd), and hence it is less than or equal to 3/2 It follows that a1 · · a4k ≤ 24k−1 · 24k−2 24k+2 = 15 · < < y, 5 which contradicts the fact that a1 · · a4k is a representation of y Note Using a similar approach one can also prove that all pairs of the form 3, 22k+1 + and 11, 210k+5 + 11 satisfy the required inequality Second solution As in the previous solution we obtain the lower bound (Q2.1) Now we claim that all the pairs of the form x = 2k + 1, y = 4k − 2k + satisfy the required inequality when k is large enough To begin with, it is easy to see that 2k + = 2k + ·2 · · 2k k and 23k + = terms 23k + · · · 2, 23k 3k terms which shows that f (2k + 1) ≤ k + and f (23k + 1) ≤ 3k + On the other hand, from (Q2.1) we deduce that the previous inequalities are actually equalities, and therefore f (x) = k + and f (xy) = 3k + Therefore, it remain to show that f (y) > 2k Since y > 22k−1 (for k ≥ 1), from (Q2.1) we already know that f (y) ≥ 2k, and hence we just need to exclude that f (y) = 2k Let us assume that we can represent y in the form a1 · · a2k At least one of the factors is not 2, and hence it is less than or equal to 3/2 Thus when k is large enough it follows that a1 · · a2k ≤ 22k−1 · 3 = · 22k < 22k − 2k < y, which contradicts the fact that a1 · · a2k is a representation of y Third solution Let’s start by showing that (x, y) = (7, 7) satisfies f (xy) < f (x) + f (y) We have f (7) ≥ since cannot be written as the product of or fewer elements of A: indeed 23 > 7, and any other product of at most three elements of A does not exceed 22 · 23 = < On the 49 other hand, f (49) ≤ since 49 = · · · · · 32 · 48 Suppose by contradiction that there exist only finitely many pairs (x, y) that satisfy f (xy) < f (x) + f (y) This implies that there exists M large enough so that whenever a > M or b > M holds we have f (ab) = f (a)+f (b) (indeed, it is clear that the reverse inequality f (ab) ≤ f (a)+f (b) is always satisfied) Now take any pair (x, y) that satisfies f (xy) < f (x) + f (y) and let n > M be any integer We obtain f (n) + f (xy) = f (nxy) = f (nx) + f (y) = f (n) + f (x) + f (y), which contradicts f (xy) < f (x) + f (y) Problem The n contestants of an EGMO are named C1 , , Cn After the competition they queue in front of the restaurant according to the following rules • The Jury chooses the initial order of the contestants in the queue • Every minute, the Jury chooses an integer i with ≤ i ≤ n – If contestant Ci has at least i other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly i positions – If contestant Ci has fewer than i other contestants in front of her, the restaurant opens and the process ends (a) Prove that the process cannot continue indefinitely, regardless of the Jury’s choices (b) Determine for every n the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves (Hungary) Solution The maximal number of euros is 2n − n − To begin with, we show that it is possible for the Jury to collect this number of euros We argue by induction Let us assume that the Jury can collect Mn euros in a configuration with n contestants Then we show that the Jury can collect at least 2Mn + n moves in a configuration with n + contestants Indeed, let us begin with all the contestants lined up in reverse order In the first Mn moves the Jury keeps Cn+1 in first position and reverses the order of the remaining contestants, then in the next n moves all contestants C1 , , Cn (in this order) jump over Cn+1 and end up in the first n positions of the line in reverse order, and finally in the last Mn moves the Jury rearranges the first n positions Since M1 = and Mn+1 ≥ 2Mn + n, an easy induction shows that Mn ≥ 2n − n − n+1 n n−1 M moves n −−− −−−→ n+1 n moves −−−−→ n−1 n n n−1 n+1 M moves n −−− −−−→ n−1 n n+1 Let us show now that at most 2n − n − moves are possible To this end, let us identify a line of contestants with a permutation σ of {1, , n} To each permutation we associate the set of reverse pairs R(σ) := {(i, j) : ≤ i < j ≤ n and σ(i) > σ(j)}, and the nonnegative integer 2i , W (σ) := (i,j)∈R(σ) which we call the total weight of the permutation We claim that the total weight decreases after any move of the contestants Indeed, let us assume that Ci moves forward in the queue, let σ be the permutation before the move, and let σ ′ denote the permutation after the move Since Ci jumps over exactly i contestants, necessarily she jumps over at least one contestant Cj with index j > i This means that the pair (i, j) is reverse with respect to σ but not with respect to σ ′ , and this yields a reduction of 2i in the total weight On the other hand, the move by Ci can create new reverse pairs of the form (k, i) with k < i, but their total contribution is at most 20 + 21 + + 2i−1 = 2i − In conclusion, when passing from σ to σ ′ , at least one term 2i disappears from the computation of the total weight, and the sum of all the new terms that might have been created is at most 2i − This shows that W (σ ′ ) ≤ W (σ) − We conclude by observing that the maximum possible value of W (σ) is realized when all pairs are reverse, in which case n W (σ) = i=1 (i − 1)2i = 2n − n − This proves that the number of moves is less than or equal to 2n − n − 1, and in particular it is finite Alternative solution As in the previous solution, the fundamental observation is again that, when a contestant Ci moves forward, necessarily she has to jump over at least one contestant Cj with j > i Let us show now that the process ends after a finite number of moves Let us assume that this is not the case Then at least one contestant moves infinitely many times Let i0 be the largest index such that Ci0 moves infinitely many times Then necessarily Ci0 jumps infinitely many times over some fixed Cj0 with j0 > i0 On the other hand, we know that Cj0 makes only a finite number of moves, and therefore she can precede Ci0 in the line only a finite number of times, which is absurd In order to estimate from above the maximal number of moves, we show that the contestant Ci can make at most 2n−i − moves Indeed, let us argue by “backward extended induction” To begin with, we observe that the estimate is trivially true for Cn because she has no legal move Let us assume now that the estimate has been proved for Ci , Ci+1 , , Cn , and let us prove it for Ci−1 When Ci−1 moves, at least one contestant Cj with j > i − must precede her in the line The initial configuration can provide at most n − i contestants with larger index in front of Ci−1 , which means at most n − i moves for Ci−1 All other moves are possible only if some contestant in the range Ci , Ci+1 , , Cn jumps over Ci−1 during her moves As a consequence, the total number of moves of Ci−1 is at most n n−i+ k=i (2n−k − 1) = 2n−i+1 − Summing over all indices we obtain that n i=1 (2n−i − 1) = 2n − n − 1, which gives an estimate for the total number of moves The same example of the first solution shows that this upper bound can actually be achieved Comment In every move of the example, the moving contestant jumps over exactly one contestant with larger index (and as a consequence over all contestants with smaller index) Problem A domino is a × or × tile Let n ≥ be an integer Dominoes are placed on an n × n board in such a way that each domino covers exactly two cells of the board, and dominoes not overlap The value of a row or column is the number of dominoes that cover at least one cell of this row or column The configuration is called balanced if there exists some k ≥ such that each row and each column has a value of k Prove that a balanced configuration exists for every n ≥ 3, and find the minimum number of dominoes needed in such a configuration (Merlijn Staps, The Netherlands) Solution The minimal number of dominoes required in a balanced configuration is 2n/3 if n is a multiple of 3, and 2n otherwise In order to show that this number is necessary, we count in two different ways the number of elements of the set S of all pairs (ℓ, d), where ℓ is a row or a column of the board, and d is a domino that covers at least one cell of that row or column On the one hand, since each row or column intersects the same number k of dominoes, the set S has 2nk elements On the other hand, since each domino intersects rows/columns, the set S has 3D elements, where D is the total number of dominoes on the board This leads to the equality 2nk = 3D If n is a multiple of 3, from the trivial inequality k ≥ we obtain that D ≥ 2n/3 If n is not a multiple of 3, then k is a multiple of 3, which means that k ≥ and hence D ≥ 2n Now we need to exhibit a balanced configuration with this number of dominoes The following diagram shows a balanced configuration with n = and k = If n is any multiple of 3, we can obtain a balanced configuration with k = by using n/3 of these × blocks along the principal diagonal of the board The following diagrams show balanced configurations with k = and n ∈ {4, 5, 6, 7} Any n ≥ can be written in the form 4A + r where A is a positive integer and r ∈ {4, 5, 6, 7} Therefore, we can obtain a balanced configuration with n ≥ and k = by using one block with size r × r, and A blocks with size × along the principal diagonal of the board In particular, this construction covers all the cases where n is not a multiple of Problem Let Γ be the circumcircle of triangle ABC A circle Ω is tangent to the line segment AB and is tangent to Γ at a point lying on the same side of the line AB as C The angle bisector of ∠BCA intersects Ω at two different points P and Q Prove that ∠ABP = ∠QBC (Dominika Regiec, Poland) Solution Let M be the midpoint of the arc AB that does not contain C, let V be the intersection of Ω and Γ, and let U be the intersection of Ω and AB C Γ Ω Q V U P A B M The proof can be divided in two steps: Proving that MP · MQ = MB It is well-known that V , U and M are collinear (indeed the homothety with center in V that sends Ω to Γ sends U to the point of Γ where the tangent to Γ is parallel to AB, and this point is M), and MV · MU = MA2 = MB This follows from the similitude between the triangles △MA V and △MUA Alternatively, it is a consequence of the following well-known lemma: Given a circle Γ with a chord AB, let M be the middle point of one of the two arcs AB Take a line through M which intersects Γ again at X and AB at Y Then MX · MY is independent of the choice of the line Computing the power of M with respect to Ω we obtain that MP · MQ = MU · MV = MB Conclude the proof given that MP · MQ = MB The relation MP · MQ = MB in turn implies that triangle △MBP is similar to triangle △MQB, and in particular ∠MBP = ∠MQB Keeping into account that ∠MCB = ∠MBA, we finally conclude that as required ∠QBC = ∠MQB − ∠MCB = ∠MBP − ∠MBA = ∠P BA, Solution The second solution is in fact a different proof of the first part of Solution Let us consider the inversion with respect to circle with center M and radius MA = MB This inversion switches AB and Γ, and fixes the line passing through M, U, V As a consequence, it keeps Ω fixed, and therefore it switches P and Q This is because they are the intersections between the fixed line MC and Ω, and the only fixed point on the segment MC is its intersection with the inversion circle (thus P and Q are switched) This implies that MP · MQ = MB Solution This solution is instead a different proof of the second step of Solution Let I and J be the incenter and the C-excenter of △ABC respectively It is well-known that MA = MI = MJ, therefore the relation MP · MQ = MA2 implies that (P, Q, I, J) = −1 Now observe that ∠IBJ = 90◦ , thus BI is the angle bisector of ∠P BQ as it is well-known from the theory of harmonic pencils, and this leads easily to the conclusion Solution Let D denote the intersection of AB and CM Let us consider an inversion with respect to B, and let us use primes to denote corresponding points in the transformed diagram, with the gentlemen agreement that B ′ = B ′ Q′ C ω′ ′ Ω P′ D′ A′ B′ M′ Since inversion preserves angles, it turns out that ∠A′ B ′ M ′ = ∠A′ M ′ B ′ = ∠ACB, and in particular triangle A′ B ′ M ′ is isosceles with basis B ′ M ′ The image of CM is the circumcircle of B ′ C ′ M ′ , which we denote by ω ′ It follows that the centers of both ω ′ and the image Ω′ of Ω lie on the perpendicular bisector of B ′ M ′ Therefore, the whole transformed diagram is symmetric with respect to the perpendicular bisector of B ′ M ′ , and in particular the arcs D ′ P ′ and Q′ C ′ of ω ′ are equal This is enough to conclude that ∠D ′ B ′ P ′ = ∠Q′ B ′ C ′ , which implies the conclusion Problem (a) Prove that for every real number t such that < t < 21 there exists a positive integer n with the following property: for every set S of n positive integers there exist two different elements x and y of S, and a non-negative integer m (i.e m ≥ 0), such that |x − my| ≤ ty (b) Determine whether for every real number t such that < t < S of positive integers such that |x − my| > ty there exists an infinite set for every pair of different elements x and y of S and every positive integer m (i.e m > 0) (Merlijn Staps, The Netherlands) Solution Part (a) Let n be any positive integer such that (1 + t)n−1 ≥ t (this inequality is actually true for every large enough n due to Bernoulli’s inequality) Let S be any set of n distinct positive integers, which we denote by s1 < s2 < < sn We distinguish two cases • If si+1 ≤ (1 + t)si for some i ∈ {1, , n − 1}, then |si+1 − si | = si+1 − si ≤ tsi , and therefore the required inequality is satisfied with x = si+1 , y = si , and m = • If si+1 > (1 + t)si for every i ∈ {1, , n − 1}, then by induction we obtain that sn > (1 + t)n−1 s1 As a consequence, from (Q6.1) it follows that |s1 | = s1 < · sn ≤ tsn , (1 + t)n−1 and therefore the required inequality is satisfied with x = s1 , y = sn , and m = 10 (Q6.1) Part (b) (Explicit formula) We claim that an infinite set with the required property exists To this end, we rewrite the required condition in the form x − m > t y This is equivalent to saying that the distance between the ratio x/y and the set of positive integers is greater than t Now we construct an increasing sequence sn of odd coprime positive integers satisfying 1 − >t 2sn ∀n ≥ 1, (Q6.2) and such that for every j > i it turns out that si < sj and t< sj si < , (Q6.3) where {α} denotes the fractional part of α This is enough to show that the set S := {sn : n ≥ 1} has the required property To this end, we consider the sequence defined recursively by sn+1 (s1 · · sn )2 + = , with s1 large enough An easy induction shows that this is an increasing sequence of odd positive integers For every i ∈ {1, , n} it turns out that si sn+1 ≤ 2 ≤ < si s1 because s1 is large enough, which proves the first relation in (Q6.3) Moreover, it turns out that sn+1 (s1 · · sn )2 = + si 2si 2si The first term is a positive integer plus 1/2, from which it follows that the distance of sn+1 /si from the positive integers is greater than or equal to 1 1 − ≥ − , 2si 2s1 which is greater than t if s1 is large enough This proves the second relation in (Q6.3) Part (b) (Arithmetic approach) We produce an increasing sequence sn of odd and coprime positive integers that satisfies (Q6.3) every j > i As in the previous solution, this is enough to conclude We argue by induction To begin with, we choose s1 to be any odd integer satisfying the inequality in (Q6.2) Let us assume now that s1 , , sn have already been chosen, and let us choose sn+1 in such a way that sn+1 ≡ si − (mod si ) 11 ∀i ∈ {1, , n} We can solve this system because the previously chosen integers are odd and coprime Moreover, any solution of this system is coprime with s1 , , sn Indeed, for every ≤ i ≤ n it turns out that si − + k i si sn+1 = for some positive integer ki Therefore, any prime p that divides both sn+1 and si divides also (2ki + 1)si − 2sn+1 = 1, which is absurd Finally, we observe that we can assume that sn+1 is odd and large enough In this way we can guarantee that si < ∀i ∈ {1, , n}, sn+1 which is the first requirement in (Q6.3), and sn+1 1 = < ki + ∀i ∈ {1, , n}, ki + t < ki + − 2si si which implies the second requirement in (Q6.3) Part (b) (Algebraic approach) Again we produce an increasing sequence sn of positive integers that satisfies (Q6.3) every j > i To this end, for every positive integer x, we define its security region (n + t)x, (n + 21 )x S(x) := n≥1 The security region S(x) is a periodic countable union of intervals of length ( 21 − t)x, whose left-hand or right-hand endpoints form an arithmetic sequence It has the property that y < ∀y ∈ S(x) t< x Now we prove by induction that we can choose a sequence sn of positive integers satisfying (Q6.3) and in addition the fact that every interval of the security region S(sn ) contains at least one interval of S(sn−1) To begin with, we choose s1 large enough so that the length of the intervals of S(s1 ) is larger than This guarantees that any interval of S(s1 ) contains at least a positive integer Now let us choose a positive integer s2 ∈ S(s1 ) that is large enough This guarantees that s1 /s2 is small enough, that the fractional part of s2 /s1 is in (t, 1/2), and that every interval of the security region S(s2 ) contains at least one interval of S(s1 ), and hence at least one positive integer Let us now assume that s1 , , sn have been already chosen with the required properties We know that every interval of S(sn ) contains at least one interval of S(sn−1 ), which in turn contains an interval in S(sn−2 ), and so on up to S(s1 ) As a consequence, we can choose a large enough positive integer sn+1 that lies in S(sk ) for every k ∈ {1, , n} Since sn+1 is large enough, we are sure that sk ∠ABC, and let I be its incentre Let D be the point on segment BC such that ∠CAD = ∠ABC Let ω be the circle tangent to AC at A and passing through I Let X be the second point of intersection of ω and the circumcircle of ABC Prove that the angle bisectors of ∠DAB and ∠CXB intersect at a point on line BC Language: English Time: hours and 30 minutes Each problem is worth points Language: English European Girls’ Mathematical Olympiad Day: Kyiv, Ukraine 2019 Wednesday, April 10, 2019 Problem Let ABC be a triangle with incentre I The circle through B tangent to AI at I meets side AB again at P The circle through C tangent to AI at I meets side AC again at Q Prove that P Q is tangent to the incircle of ABC Problem Let n ≥ be an integer, and let a1 , a2 , , an be positive integers Show that there exist positive integers b1 , b2 , , bn satisfying the following three conditions: (A) ≤ bi for i = 1, 2, , n; (B) the remainders of b1 , b2 , , bn on division by n are pairwise different; and (C) b1 + · · · + bn ≤ n n−1 a1 + · · · + an + n (Here, x denotes the integer part of real number x, that is, the largest integer that does not exceed x.) Problem On a circle, Alina draws 2019 chords, the endpoints of which are all different A point is considered marked if it is either (i) one of the 4038 endpoints of a chord; or (ii) an intersection point of at least two chords Alina labels each marked point Of the 4038 points meeting criterion (i), Alina labels 2019 points with a and the other 2019 points with a She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive) Along each chord, Alina considers the segments connecting two consecutive marked points (A chord with k marked points has k − such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference Alina finds that the N + yellow labels take each value 0, 1, , N exactly once Show that at least one blue label is a multiple of (A chord is a line segment joining two different points on a circle.) Language: English Time: hours and 30 minutes Each problem is worth points ...▲❛♥❣✉❛❣❡✿ ❉❛②✿ ❊♥❣❧✐s❤ ✶ EGMO | 2012 European Girls’ Mathematical Olympiad ❚❤✉rs❞❛②✱ ❆♣r✐❧ ✶✷✱ ✷✵✶✷ Pr♦❜❧❡♠ ✶✳ ▲❡t ABC ❜❡ ❛... zero-sum-free set The proof is similar to that used for the standard examples ▲❛♥❣✉❛❣❡✿ ❉❛②✿ ❊♥❣❧✐s❤ ✷ EGMO | 2012 European Girls’ Mathematical Olympiad ❋r✐❞❛②✱ ❆♣r✐❧ ✶✸✱ ✷✵✶✷ Pr♦❜❧❡♠ ✺✳ ❚❤❡ ♥✉♠❜❡rs p ❛♥❞... 2012 Problem Selection and Coordination Problem Selection Committee Dr Geoff Smith MBE (Chair) Dr Joseph Myers The following countries submitted a total of 33 problems for consideration for EGMO