The Viking Battle - Part Version: Finnish

Transkriptio

1 The Viking Battle - Part Version: Finnish Tehtävä 1 Olkoon kokonaisluku, ja olkoon A n joukko A n = { n k k Z, 0 k < n}. Selvitä suurin kokonaisluku M n, jota ei voi kirjoittaa yhden tai useamman joukon A n alkion summana. (Alkioiden ei välttämättä tarvitse olla erisuuria.) Tehtävä Määritellään funktio f : (0, 1) (0, 1) asettamalla { x + 1, x < 1 f(x) = x, x 1 Olkoot a 0 ja b 0 kaksi reaalilukua, joille 0 < a 0 < b 0 < 1. Määritellään jonot a n ja b n asettamalla a n = f(a n 1 ) ja b n = f(b n 1 ) kaikille n = 1,, 3,.... Osoita, että on olemassa positiivinen kokonaisluku n, jolle (a n a n 1 ) (b n b n 1 ) < 0. Tehtävä 3 Oletetaan, että teräväkärkisessä kolmiossa ABC pätee AB > BC. Olkoon Ω kolmion ABC ympäripiirretty ympyrä, ja olkoon O ympyrän Ω keskipiste. Kulman ABC kulmanpuolittaja leikkaa ympyrän Ω pisteessä M B. Olkoon Γ se ympyrä, jonka eräs halkaisija on BM. Kulman AOB puolittaja leikkaa ympyrää Γ pisteessä P, ja kulman BOC puolittaja leikkaa ympyrää Γ pisteessä Q. Suoralta P Q on valittu piste R siten, että BR = MR. Osoita, että BR AC. (Tässä oletamme aina, että kulmanpuolittaja on puolisuora.) 1

2 Solution to problem 1 Answer: M n = (n ) n + 1. Part 1. First we prove that every integer greater than (n ) n +1 can be represented as such a sum. This is achieved by induction on n. For n =, the set A n = {, 3}. Every positive integer m except 1 can be represented as a sum of elements of A n : as m = if m is even, and as m = if m is odd. Now consider some n > and assume the induction hypothesis holds for n 1. Take an integer m > (n ) n + 1. If m is even, then Hence by the induction hypothesis m > (n )n 1 > ((n 1) ) n m = (n 1 k 1 ) + ( n 1 k ) + + ( n 1 kr ) for some k i, with 0 k i < n 1. It follows that m = ( n k 1+1 ) + ( n k +1 ) + + ( n kr+1 ), giving us the desired representation as a sum of elements of A n. If m is odd, we consider m ( n 1) > (n )n + 1 ( n 1) = (n 3) n By the induction hypothesis there is a representation of the form m ( n 1) = ( n 1 k 1 ) + ( n 1 k ) + + ( n 1 kr ) for some k i, with 0 k i < n 1. It follows that m = ( n k 1+1 ) + ( n k +1 ) + + ( n kr+1 ) + ( n 1), giving us the desired representation of m once again. Part. It remains to prove that there is no representation of M n = (n ) n + 1. Let N be the smallest positive integer that satisfies N 1 (mod n ), and which can be represented as a sum of elements of A n. Consider the representation of N, i.e. N = ( n k 1 ) + ( n k ) + + ( n kr ), where 0 k 1, k,..., k r < n. If k i = k j = n 1, then we can simply remove these two terms from the sum to get a representation for N ( n n 1 ) = N n as a sum

3 of elements of A n, which contradicts our choice of N. If k i = k j = k < n 1, replace the two terms by n k+1, which is also an element of A n, to get a representation for N ( n k )+ n k+1 = N n. This is a contradiction once again. Therefor, all k i have to be distinct, which means that k 1 + k + + kr n 1 = n 1. On the other hand ( ) k 1 + k + + kr ( n k 1 )+( n k )+ +( n kr ) = N 1 (mod n ) Thus we must have k 1 + k + + kr = n 1, which is only possible if each element of {0, 1,,..., n 1} occurs as one of the k i. This gives us N = n n ( n 1 ) = (n 1) n + 1. In particular this means that (n ) n +1 cannot be represented as a sum of elements of A n. 3

4 Solution to problem Note that f(x) x = 1 > 0 if x < 1 f(x) x = x x < 0 if x 1. We consider the interval (0, 1) divided into the two subintervals I 1 I = [ 1, 1). The inequality = (0, 1 ) and 0 > (a n a n 1 ) (b n b n 1 ) = (f(a n 1 ) a n 1 )(f(b n 1 b n 1 ) holds if and only if a n 1 and b n 1 lie in distinct subintervals. Let us now assume, to the contrary, that a k and b k always lie in the same subinterval. Consider the distance d k = a k b k. If both a k and b k lie in I 1, then d k+1 = a k+1 b k+1 = a k + 1 ( b k + 1 ) = dk. If, on the other hand, a k and b k both lie in I, then a k + b k d k = 1 + d k, which implies d k+1 = a k+1 b k+1 = a k b k = (a k b k )(a k + b k ) d k (1 + d k ). This means that the difference d k is non-decreasing, and particular d k d 0 > 0 for all k. If a k and b k lie in I, then d k+ d k+1 d k (1 + d k ) d k (1 + d 0 ). If a k and b k lie in I 1, then a k+1 and b k+1 both lie in I, and so we have d k+ d k+1 (1 + d k+1 ) d k+1 (1 + d 0 ) d k (1 + d 0 ). In either case, d k+ d k (1 + d 0 ), and inductively we get d m d 0 (1 + d 0 ) m. For sufficiently large m, the right-hand side is greater than 1, a contradiction. Thus there must be a positive integer n such that a n 1 and b n 1 do not lie in the same subinterval, which proves the desired statement. 4

5 Solution to problem 3 Let K be the midpoint of BC, i.e. the centre of Γ. Notice that AB BC implies that K O. Clearly the lines OM and OK are perpendicular bisectors of AC and BM, respectively. Therefore, R is the intersection point of P Q and OK. Let N be the second point of intersection of Γ with the line OM. Hence BN AC, and it suffices to prove that BN passes trough R. Our plan for doing this is to interpret the lines BN, OK and P Q as the radical axes of three appropriate circles. Let ω be the circle with diameter BO. Since BNO = BKO = 90, the points N and K lie on ω. Next we show that the points O, K, P, and Q are concyclic. To this end, let D and E be the midpoints of BC and AB, respectively. By our assumption of triangle ABC, the points B, E, O, K, and D lie on ω is this order. It follows that EOR = EBK = KBD = KOD, so the line KO externally bisects the angle P OQ. Since the point K is the centre of Γ, it also lies on the perpendicular bisector of P Q. So K coincides with the midpoint of the arc P OQ of the circumcircle γ of triangle P OQ. Thus the lines OK, BN, and P Q are pairwise radical axis of the circles ω, γ and Γ. Hence they are concurrent at R, as required. R P N B γ E D Q Ω A O K ω C Γ M 5