19 η ΒΑΛΚΑΝΙΚΗ ΜΑΘΗΜΑΤΙΚΗ ΟΛΥΜΠΙΑΔΑ ΝΕΩΝ Ιουνίου, 2015, Βελιγράδι, Σερβία
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1 19 η ΒΑΛΚΑΝΙΚΗ ΜΑΘΗΜΑΤΙΚΗ ΟΛΥΜΠΙΑΔΑ ΝΕΩΝ 4-9 Ιουνίου, 015, Βελιγράδι, Σερβία Lnguge: Ελληνικά Παρασκευή, 6 Ιουνίου 015 Πρόβλημα 1. Βρείτε όλους τους πρώτους αριθμούς, b, c και όλους τους ϑετικούς ακεραίους k που ικανοποιούν την εξίσωση + b + 16c = 9k + 1 Πρόβλημα. Θεωρούμε τους ϑετικούς πραγματικούς αριθμούς, b, που είναι τέτοιοι ώστε + b + c =. Βρείτε την ελάχιστη τιμή της παράστασης A = + b b + c c Πρόβλημα. Δίνεται οξυγώνιο τρίγωνο ABC. Οι ευθείες l 1 και l είναι κάθετες στην AB στα σημεία A και B, αντίστοιχα. Οι κάθετες ευθείες από το μέσον M του AB προς τις πλευρές AC και BC του τριγώνου τέμνουν τις ευθείες l 1 και l στα σημεία E και F, αντίστοιχα. Αν D είναι το σημείο τομής των ευθειών EF και MC, να αποδείξετε ότι ADB = EMF Πρόβλημα 4. Καθένα από τα ακόλουθα τέσσερα σχήματα αποτελείται από τρία μοναδιαία τετράγωνα και καλείται L-σχήμα. Δίνονται ένας 5 5 πίνακας, αποτελούμενος από 5 μοναδιαία τετράγωνα, ένας ϑετικός ακέραιος k 5 και απεριόριστος αριθμός L-σχημάτων οποιουδήποτε τύπου. Δύο παίκτες, ο A και ο B, παίζουν το ακόλουθο παιχνίδι: Ξεκινώντας με τον A, σημειώνουν εναλλάξ σε κάθε κίνησή τους ένα τετραγώνο που δεν είναι ήδη σημειωμένο, μέχρι να σημειώσουν συνολικά k μοναδιαία τετράγωνα. Μία τοποθέτηση L-σχημάτων λέγεται καλή αν τα L-σχήματα δεν επικαλύπτονται και καθένα από αυτά καλύπτει ακριβώς τρία μοναδιαία τετράγωνα του πίνακα που δεν είναι σημειωμένα. Ο B κερδίζει αν μετά από οποιαδήποτε καλή τοποθέτηση L-σχημάτων μένουν ακάλυπτα τουλάχιστον μοναδιαία τετράγωνα που δεν είναι σημειωμένα. Προσδιορίστε την ελάχιστη τιμή του k για την οποία ο B έχει στρατηγική νίκης. Διάρκεια: 4 ώρες και 0 λεπτά Κάθε πρόβλημα βαθμολογείται με 10 μονάδες
2 June 4-9, 015, Belgrde, Serbi Problem 1. Find ll prime numbers, b, c nd positive integers k which stisfy the eqution b 16c 9k 1. 9 k 1 1 mod implies The reltion b 16c 1 mod b c 1 mod. Since 0, 1 (mod ), b 0, 1 (mod ), c 0, 1 (mod ), we hve: b c b c From the previous tble it follows tht two of three prime numbers, b, c re equl to. Cse 1. b We hve b 16c 9 k 1 9 k 16c 17 k 4c k 4c 17, k 4c 1, c, k 4c 17, k, nd (, b, (,,, ). Cse. c If (, b 0, is solution of the given eqution, then ( b 0,, is solution too. Let. We hve b 16c 9 k 1 9 k b 15 k bk b 15. Both fctors shll hve the sme prity nd we obtin only cses: k b, b 7, k b 76, k 1, nd (, b, (,7,,1); k b 4, b 17, k b 8, k 7, nd (, b, (,17,,7 ). So, the given eqution hs 5 solutions: ( 7,,,1), (17,,,7), (,7,,1), (,17,,7), (,,,). Problem. Let, b, c be positive rel numbers such tht b c. Find the minimum vlue of the expression b c A. b c
3 June 4-9, 015, Belgrde, Serbi We cn rewrite A s follows: b c A b c b c b c b bc c b bc c ( b c ) (( b c) ( b bc c)) bc bc b bc c b bc c (9 ( b bc c)) ( b bc c) 9 bc bc 1 ( b bc c) 1 9. bc Recll now the well-known inequlity ( x y z) ( xy yz zx) nd set x b, y b z c, to obtin ( b bc c) bc( b c) 9b where we hve used b c. By tking the squre roots on both sides of the lst one we obtin: b bc c bc. (1) Also by using AM-GM inequlity we get tht () bc bc Multipliction of (1) nd () gives: 1 1 ( b bc c) 1 bc 6. bc bc So A 69 nd the equlity holds if nd only if b c 1, so the minimum vlue is. Problem. Let ABC be n cute tringle. The lines l 1, l re perpendiculr to AB t the points A, B respectively. The perpendiculr lines from the midpoint M of AB to the sides of the tringle AC, BC intersect the lines l 1, l t the points E, F, respectively. If D is the intersection point of EF nd MC, prove tht ADB EMF. Let H, G be the points of intersection of ME, MF with AC, BC respectively. From the MH MA similrity of tringles MHA nd MAE we get, thus MA ME MA MH ME. (1) MB MG Similrly, from the similrity of tringles MBG nd MFB we get, thus MF MB MB MF MG. () Since MA MB, from (1), (), we conclude tht the points E, H, G, F re concyclic.
4 June 4-9, 015, Belgrde, Serbi l 1 C F D l E H G A M B Therefore, we get tht FEH FEM HGM. Also, the qudrilterl CHMG is cycli so CMH HGC. We hve FEH CMH HGM HGC 90, thuscm EF. Now, from the cyclic qudrilterls FDMB nd EAMD, we get tht DFM DBM nd DEM DAM. Therefore, the tringles EMF nd ADB re similr, so ADB EMF. Problem 4. An -figure is one of the following four pieces, ech consisting of three unit squres: A 5 5 bord, consisting of 5 unit squres, positive integer k 5 nd n unlimited supply L-figures re given. Two plyers, A nd B, ply the following gme: strting with A they lterntively mrk previously unmrked unit squre until they mrked totl of k unit squres. We sy tht plcement of L-figures on unmrked unit squres is clled good if the L-figure do not overlp nd ech of them covers exctly three unmrked unit squres of the bord. B wins if every good plcement of L-figures leves uncovered t lest three unmrked unit squres. Determine the minimum vlue of k for which B hs winning strtegy. We will show tht plyer A wins if k = 1,,, but plyer B wins if k 4. Thus the smllest k for which B hs winning strtegy exists nd is equl to 4. If k 1, plyer A mrks the upper left corner of the squre nd then fills it s follows.
5 June 4-9, 015, Belgrde, Serbi If k, plyer A mrks the upper left corner of the squre. Whtever squre plyer B mrks, then plyer A cn fill in the squre in exctly the sme pttern s bove except tht he doesn't put the L-figure which covers the mrked squre of B. Plyer A wins becuse he hs left only two unmrked squres uncovered. For k, plyer A wins by following the sme strtegy. When he hs to mrk squre for the second time, he mrks ny yet unmrked squre of the L-figure tht covers the mrked squre of B. Let us now show tht for k 4 plyer B hs winning strtegy. Since there will be 1 unmrked squres, plyer A will need to cover ll of them with seven L-figures. We cn ssume tht in his first move, plyer A does not mrk ny squre in the bottom two rows of the chessbord (otherwise just rotte the chessbord). In his first move plyer B mrks the squre lbeled 1 in the following figure. If plyer A in his next move does not mrk ny of the squres lbeled, nd 4 then plyer B mrks the squre lbeled. Plyer B wins s the squre lbeled is left unmrked but cnnot be covered with n L-figure. If plyer A in his next move mrks the squre lbeled, then plyer B mrks the squre lbeled 5. Plyer B wins s the squre lbeled is left unmrked but cnnot be covered with n L-figure. Finlly, if plyer A in his next move mrks one of the squres lbeled or 4, plyer B mrks the other of these two squres. Plyer B wins s the squre lbeled is left unmrked but cnnot be covered with n L-figure. Since we hve covered ll possible cses, plyer B wins when k 4.
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