Winter Mathematical Competition Varna, February, 9-11, 2007

Transcription

1 Winter Mathematical Competition Varna, February, 9-11, 007 Problem 91 (Peter Boyvalenkov) Find all values of the real parameter p such that the equation x + (p + 1)x + p = has two real distinct roots x 1 and x such that x x 1 = x 1 x + 55 x x 1 x 1 x Answer p = Problem 9(Peter Boyvalenkov) In ABC with AB > BC, a point K on the side AB is such that AK = BC + BK A line l through K is perpendicular to AB Prove that l, the perpendicular bisector of AC and the external bisector of <) ABC are concurrent Solution Let the point C AB be such that BC = BC Then the external bisector of <) ABC is the perpendicular bisector of CC Since AK = BC + BK = BC +BK = KC, the line l is the perpendicular bisector of AC Hence l, the perpendicular bisector of AC and the external bisector to <) ABC are the perpendicular bisectors of the sides of AC C and thus they are concurrent Problem 93 (Ivailo Kortezov) Some of the squares of an n n table are mined In each square the number of the mined squares amongst this square and its neighbors (ie those which have common side or vertex with it) is written Is it always possible to determine which squares are mined if: а) n = 000; б) n = 007? Solution We denote the rows by i = 1,, n and the columns by j = 1,, n and let a(i; j) be the number written in the square (i; j) a) No! Consider the table A where the squares (i; j) are mined if and only if i j 1 (mod 3) and the table B where the squares (i; j) are mined if and only if i j (mod 3) Then all numbers written in the squares of A and B are equal to 1 and therefore we can not determine which squares are mined b) Yes! We first determine the mined squares in the third row It is easy to see that the number b(j) of the mined squares amongst (3; j 1), (3; j), (3; j + 1) is equal to a(; j) a(1; j) We now compare b(1) and b() and determine if the square (3;3) is mined Next we compare b(4) and b(5) and decide if (3;6) is mined The same argument shows which squares amongst (3;9), (3;1),, (3;007) are mined 1

2 We now compare b(007) and b(006) to decide if the square (3;005) is mined, then compare b(004) and b(003) to see if (3;00) is mined, etc We finally know which squares amongst (3;1999), (3;1996),, (3;1) are mined Now b(1) shows if (3;) is mined, then we compare b(3) and b(4) to determine if (3;5) is mined, etc, and find which squares amongst (3;8), (3;11),, (3;006) are mined Hence we know row 3 Analogously we can determine the mined squares in the rows 6, 9,1, 15,, 007 Using similar argument we can determine the mined squares in the rows 005, 00, 1999,, 4, 1 and the rows, 5, 8,, 006 Remark The answer is No for n (mod 3) and Yes otherwise Problem 94 (Peter Boyvalenkov) Find all positive integers x and y such that the number (x + y)(y + x) is the fifth power of a prime Solution Let (x + y)(y + x) = p 5, where p is a prime Then x + y = p s, y + x = p t, where {s, t} = {1, 4} or {, 3} In the first case we can assume without loss of generality that x < y, x + y = p and y + x = p 4 Then p = (x + y) > x + y = p 4, a contradiction Let x < y, x + y = p and y + x = p 3 Note that p > x We have p (x + y)(x y) + (y + x) = x 4 + x = x(x + 1)(x x + 1) and since p > x we see that p divides (x + 1)(x x + 1) We consider two cases Case 1 If p x+1 then p = x+1 and we easily find the solution x =, y = 5 Case If p x+1 then p x x+1 and now p x +y = (x x+1)+(x+y 1) implies that p divides x+y 1 Hence y p x+1 > p p and p 3 = y +x > p (p 1) which is impossible Finally, the solutions are (, 5) and (5, ) Remark It can be proved that (x + y)(y + x) is a prime power only for (x, y) = (1, 1), (, 5) and (5, ) Problem 101 (Ivan Landgev) The functions f(x) = x + x 4 and g(x) = x x + are given Find all real values of x such that: a) f(x) g(x) is a positive integer; b) the inequality f(x) + g(x) holds Solution a) Hint Set f(x)/g(x) = k, where k is a positive integer Then ( k)x + ( + k)x ( + k) = 0 and use the fact that the discriminant of this quadratic equation is nonnegative Answer x = , 3 33, b) Answer x (, ] [1, + )

3 Problem 10 (Stoyan Boev) In acute ABC denote by M and N the midpoints of the altitudes BB 1 and CC 1, respectively, P = AM CC 1 and Q = AN BB 1 Prove that: a) the points M, N, P and Q are concyclic; b) if the points B, C, P and Q are concyclic then ABC is isosceles Solution a) Since ACC 1 ABB 1 and AN and AM are medians in these triangles we have <) ANC 1 =<) AMB 1 <) QNB =<) P MQ, ie the points M, N, P and Q are concyclic b) If the points B, C, P and Q are concyclic then <) QCP =<) QBP But <) ACC 1 =<) ABB 1 and hence <) QCA =<) P BA (1) On the other hand, since ACC 1 and ABB 1 are similar we have <) CAQ =<) CAN =<) BAM =<) BAP () Now (1) and () imply that ACQ = ABP, whence Thus AB = AC and AB = AC AC AB = AQ AP = AM AN = AB AC Problem 103 (Ivan Landgev) Find all positive integers x and y such that xy + y divides x y + xy + 8x Solution Since xy + y divides (x + y)(xy + y) y(x y + xy + 8x) = y 4xy, we conclude that xy + divides y 4x We consider two cases Case 1 Let y 4x 0 Then we have two possibilities: 11) If x then xy + > y 4x Hence y 4x = 0, ie x = a and y = a, where a is an integer Since xy + y = 4a(a + 1) divides x y + xy + 8x = 8a(a + 1), this gives a solution 1) If x = 1 then y + y divides 8, ie y = Hence in this case the solutions are (a, a), where a is a positive integer 3

4 Case Let y 4x < 0, ie 4x y > 0 If y 4, then xy + > 4x Therefore y = 1, or 3 1) If y = 1 then x + 9x = x is an integer Hence x + x + x + divides 10 and this gives the solutions x = 3, y = 1 and x = 8, y = 1 ) If y = then x + 3x x + 1 = x + is an integer which gives x = 1 x + 1 3) If y = 3 then 6x + 17x is an integer This implies that 3 x, ie x = 3k 9x + 6 for some positive integer k After simplifications we obtain that 18k + 17k = 9k + (k + 1) + 4k is an integer, which is impossible for k 1 9k + Finally, the solutions are x = a, y = a for all positive integers a and x = 3, y = 1, x = 8, y = 1 Problem 104 (Ivan Landgev) Find the minimum possible number of the edges of a graph with n vertices having the following property: a) If we draw an arbitrary new edge then a new triangle (3-clique) appears b) If we draw an arbitrary new edge then a new 4-clique appears Solution a) Let G be a graph with the required property and minimum possible number of edges If G is not connected then adding a new edge which connects two components does not give a new 3-clique, a contradiction Therefore G is connected and it has at least n 1 edges On the other hand, the graph K 1,n 1 has the required property (K m,n is the graph with m + n vertices which are divided into two sets of m and n elements, respectively, and two vertices are adjacent if and only if they belong to different sets; the edges are obviously mn) Thus the answer is n 1 b) Consider a graph with vertices u 1, u, v 1,, v n and edges all pairs u i v j, i = 1,, j = 1,, n, together with u 1 u This graph has n vertices, n 3 edges and adding any new edge increases the number of the 4-cliques Hence the required minimum number does not exceed n 3 We shall prove by induction on n that the minimum possible number of edges is n 3 and it is attained only for a graph as above The assertion is obvious for n = 4 Assume that the assertion is proved for graphs with n 1 or less vertices Let G be a graph with the required property having n 5, vertices and minimum possible number of edges 4

5 Since adding of a new edge increases the number of the 4-cliques, the graph G has four vertices x 1, x, x 3, x 4 which determine exactly five edges joining them Next we assume that the missing edge is x 1 x Let G be the graph which is obtained from G by cluing the vertices x 1 and x (ie we remove x 1 and x from G, add a new vertex u and save all remaining vertices; the new vertex is adjacent exactly to these vertices which were adjacent to at least one of x 1 and x, all old edges are saved) Then e(g ) e(g) n 5 = (n 1) 3 Hence the graph G has n 1 vertices, possesses the required property and has at most (n 1) 3 edges Then the induction hypothesis implies that G has exactly n 5 edges and it has the structure described above two vertices of degree n and all remaining of degree At least one of the vertices of degree n is x 3 or x 4, say x 3 Then the degree of x 3 in G is n 1 Denote by G the graph obtained from G by removing the vertex x 3 and all edges from it The graph G has at most n edges since G has at most n 3 edges Moreover, G has the property considered in a) Therefore it follows from the solution of a) that G = K 1,n It is easy to see now that G has the required structure and this completes the induction step Remark The following more general assertion is true: The minimum possible edges of a graph G with n vertices such that adding an arbitrary new edge leads to appearance of a new r-clique is ( ) r e(g) = + (n r + )(r ) Moreover, the optimal graph is G = K r +E n r+, where K r is the complete graph with r vertices and E n r+ is the graph with n r + vertices and without edges (If G 1 = (V 1, E 1 ) and G = (V, E ) are two graphs then G 1 +G is the graph with vertices V 1 V and edges E 1 E V 1 V ) Problem 111 (Alexandar Ivanov) Find all values of the real parameter a such that the equation x 3 ax + (a 1)x a + a = 0 has three distinct real roots which (in some order) form an arithmetic progression Solution Writing the equation in the form (x 1)(x + (1 a)x a + a ) = 0, 5

6 we obtain x 1 = 1 Let x and x 3 be the roots of the quadratic equation If 1 is the second term of the progression then x + x 3 =, giving a 1 =, ie a = 3 When a = 3 the roots of the quadratic equation are not real If x 1 = 1 is not the second term we may assume that 1 + x = x 3, which together with x + x 3 = a 1 implies 3x 3 = a Therefore a is a root of the 3 quadratic equation, ie ( a 3 ) + (1 a) a 3 a + a = 0 Thus, a = 0 or a = 6 7 When a = 0 we obtain x = 1, x 3 = 0 and when a = 6 7 we find x = 3 7 and x 3 = 7 Hence the desired values are a = 0 and a = 6 7 Problem 11 (Emil Kolev) In ABC with <) ACB = 60 let AA 1 and BB 1 (A 1 BC, B 1 AC) be the bisectors of <) BAC and <) ABC The line A 1 B 1 meets the circumcircle of ABC at points A and B a) If O and I are the circumcenter and the incenter of ABC prove that OI is parallel to A 1 B 1 b) If R is the midpoint of the arc AB, not containing point C, and P and Q are the midpoints of A 1 B 1 and A B, respectively, prove that RP = RQ Solution a) Since <) AOB = γ = 10 and <) AIB = 180 α + β = = 10, C the points A, O, I and B lie on a circle k Also, RI = RA (<) RIA =<) RAI = α + γ B ) B 1 Q P and analogously RI = RB Hence R is A 1 A the center of k It follows from the isosceles AOB that <) BAO = 30 and there- O I fore <) OIB 1 = 30 Since <) AIB = 10 A B we have that the quadrilateral IA 1 CB 1 is cyclic implying that <) IB 1 A 1 =<) ICA 1 = 30 and <) IA 1 B 1 =<) ICB 1 = 30 Finally, R <) OIB 1 = IB 1 A 1 gives OI A 1 B 1 b) Since OQ A B, IP A B ( A 1 IB 1 is isosceles) and OI A B, we have that OIP Q is a rectangle The perpendicular bisector of OI is also the perpendicular bisector of P Q Using that R lies on the perpendicular bisector of OI it follows that R lies on the perpendicular bisector of P Q, ie RP = RQ 6

7 Problem 113 (Alexandar Ivanov, Emil Kolev) One cuts a paper strip of length 007 into two parts of integer lengths and writes down the two integers on the board Then cuts one of the two parts into two parts of integer lengths and writes down the two integers on the board The cutting stops when all parts are of length 1 A cut is called bad if the two parts obtained are not of equal lengths a) Find the minimum possible bad cuts b) Prove that for all cuttings with minimum possible bad cuts the number of distinct integers on the board is one and the same Solution a) Let the length of the strip be n Denote by g(n) and f(n) respectively the number of 1 s in the binary representation of n and the minimum possible number of bad cuts Let n = k 1 + k + + k l Consider the following sequence of cuts: first cut a strip of length k 1, next cut a strip of length k and so on After the last cut we obtain two strips of lengths k l 1 and k l Note that a strip whose length is a power of can be cut into strips of length 1 without bad cuts Therefore the number of bad cuts equals l 1, ie (1) f(n) g(n) 1 We prove by induction on n that f(n) g(n) 1 For n = 1 we have f(1) = 0 and g(1) = 1, ie the statement is true Suppose it is true for all n k, where k is a positive integer and let n = k Suppose the first cut is bad and it leaves two strips of lengths a and b Then a + b = k + 1 and f(k + 1) = 1 + f(a) + f(b) If the binary representations of a and b have no common digit 1 then g(k + 1) = g(a) + g(b) and therefore f(k + 1) = 1 + f(a) + f(b) 1 + g(a) 1 + g(b) 1 = g(k + 1) 1 If the binary representations of a and b have at least one common digit 1 then g(k + 1) g(a) + g(b) 1 and therefore f(k + 1) = 1 + f(a) + f(b) 1 + g(a) 1 + g(b) 1 g(k + 1) > g(k + 1) 1 Suppose the first cut is not bad, ie the strip is cut into two parts each of length a Then k + 1 = a and g(k + 1) = g(a) If g(k + 1) = 1 then f(k + 1) = 0 and the statement is true Otherwise f(k + 1) = f(a) + f(b) = f(a) g(a) = g(k + 1) > g(k + 1) 1 Thus, when g(k + 1) > 1 we have f(k + 1) > g(k + 1) 1 7

8 This proves the induction hypothesis giving () f(n) g(n) 1 It follows from (1) and () that f(n) = g(n) 1 a) Since the binary representation of 007 is , ie g(007) = 9, we obtain f(007) = 8 b) It follows from the above arguments that if the number of bad cuts is f(n) = g(n) 1 then every bad cut leaves two parts of lengths a and b such that the binary representations of a and b have no common digit 1 Moreover the good cuts are done only over strips whose lengths are powers of It is clear that rearranging the cuts we may assume that the first cuts are bad Their number equals g(n) 1 and each bad cut gives two new numbers on the table Therefore after all bad cuts we have g(n) distinct numbers The powers of that appear are all powers up to the highest power in the binary representation of n Thus, the number of distinct numbers on the table equals g(n) +k +1 = g(n) + k 1, where k is the highest power of in the binary representation of n Problem 114 (Alexandar Ivanov) For every positive integer n set a n = 0, if the number of divisors of n, greater than 007, is even and a n = 1, if this number is odd Is the number α = 0, a 1 a a 3 a k rational? Solution We prove that α is irrational Suppose α is a rational number, ie the sequence a 1, a, a 3,, a k, is periodic from some point onwards Hence there exist k 0 and T such that for any k > k 0 we have a k = a k+t Choose a positive integer m for which mt > k 0 and mt is a perfect square If T = p α 1 1 pα pαs s then m = p β 1 1 pβ pβs s, where α i + β i is even for all i = 1,,, s and β i are large enough positive integers Choose a prime number p > 007, p p i, i = 1,,, s Since pmt mt is divisible by T we have that a mt = a pmt Denote by τ(k) the number of divisors of k and by f(k) the number of divisors of k that are greater than 007 We have f(pmt ) = f(mt ) + τ(mt ) and since τ(mt ) is odd number we conclude that f(pmt ) and f(mt ) are of the same parity, a contradiction Problem 11 (Oleg Mushkarov) A plane passes through a vertex of the base of a cube of edge 1 and the centers of its two faces which do not contain that vertex Find the ratio of the volumes of the two parts of the cube cut by the plane 8

9 Solution Let P and Q be the centers of the faces BCC 1 B 1 and DCC 1 D 1 and let α = (AP Q) (Fig 1) Since P Q BD the plane α meets the plane (ABCD) at the line through A which is parallel to BD We denote by T and S the intersection points of this line with the lines CB and CD, respectively The lines T P and SQ meet the edge CC 1 at the intersection point R of α and CC 1 Let M = T R BB 1 and N = SR DD 1 Then the intersection of α and the surface of the cube is the quadrilateral AMRN (It is easy to see that it is a rhombus) It is clear that BT = BA = 1 Hence B and M are the midpoints of T C and T R, respectively Then BM RC = 1 and since BM = RC 1 = 1 RC, we find BM = 1 3 Analogously DN = 1 3 D 1 C 1 D 1 C 1 S A 1 B 1 D A N B Q M P R C A 1 B 1 D A M A N B Q P R C C T Fig 1 Fig Denote by V the volume of the polytope cut from the cube by the planes (ABCD) and (AMRN) (Fig ) Let A and C be the intersection points of AA 1 and CC 1 with the plane through MN and parallel to (ABCD) Then RC = RC CC = MB = AA = 1 3 and hence the tetrahedra NMC R and NMA A have equal volumes This shows that V = V ABCDA MC N = 1 3 Hence the required ratio is 1 : 9

10 Problem 1 (Nikolai Nikolov) Let ABC be a right triangle with <) ACB = 90, AC = 1 and BC = Given a point A 1 BC such that A 1 C 1 3 we construct a sequence of points A n BC, n in the following way Let B 1 be the intersection point of AC and the line through A 1 and parallel to AB, and C 1 be the foot of the perpendicular from B 1 to AB Then A is the intersection point of BC and the line through C 1 and parallel to AC Using A we construct A 3 in the same way, etc Find: a) 3A C 1 3A 1 C 1 ; b) lim S A n nb nc n Solution a) Set A n C = x n Then it follows from A n B n C BAC that B n C = x n Since AB nc n ABC, we find that AC n = x n 5 Hence A n+1 C = x n+1 = x n and therefore 3x n x n 1 = b) Since A n B n = x n, B n C n = x n and A n B n B n C n we get 5 S An B n C n = x n( x n ) But a) implies that {x n } n=1 is a geometric progression with ratio 1 Hence lim 5 x n = 1 n 3 and therefore lim n S A n B n C n = 5 36 Problem 13 (Nikolai Nikolov) Alexander and Denitza play the following game Alexander cuts (if possible) a band of positive integer length to three bands of positive integer lengths such that the largest band is unique Then Denitza cuts (if possible) the largest band in the same way and so on The winner is the one who makes the last move Consider the bands whose lengths are integers of the form a b, a 1, b 1 N For which of them Denitza has a winning strategy Solution Consider a band of length n It is clear that no move is possible for n = 1,, 3 For 4 n 7 = Alexander has a winning move For n = 8 and 9 after the first move of Alexander the largest length is between 4 and 7 and then Denitza has a winning move Similarly, for 10 n = 5, Alexander has a winning strategy since he can cut the band in such a way that the largest length is either 8 or 9 and so on It follows by induction that Denitza has a winning strategy if and only if n = 3 k or n = 3 k 1 for some integer k > 1 The numbers 3 k and 3 1 = 3 obviously have the desired form We shall prove that there are no other solutions Assume that a b = 3 k 1 Since a 0, 1 (mod 3), the number b is even Then (a + 1)(a b a + 1) = 3 k and we 10

11 have a + 1 = 3 i, and 3 k i = a b a + 1 = A(a + 1) + b, for some integers i and k, 0 < i < k Hence 3 divides b and setting c = a b/3 we have 3 k = (c + 1)((c + 1) 3c) Then c + 1 = 3 j and (c + 1) 3c = 3 k j, 0 < j < k In particular, 9 divides (c + 1), but does not divide 3c We consecutively find k j = 1, c =, a = k = and b = 3 Remark The fact proved in the second part of the solution is a particular case of the famous Catalan equation a b c d = 1 It has been recently proved that the only nontrivial solutions in positive integers of this equation are a = d = 3, b = c = Problem 14 (Oleg Mushkarov) Find all positive integers n such that if a, b, c 0 and a + b + c = 3, then abc(a n + b n + c n ) 3 Solution For a =, b = c = 1 and n 3 the inequality is not satisfied On the other hand, for n = 1 it is equivalent to the AM-GM inequality It remains to consider the case n = We shall prove that the inequality is true for n = First solution Set x = bc Then abc(a + b + c ) = ax(a + (b + c) x) The function x(p x) is increasing for x p 4 Since bc (b + c) 4 a + (b + c), 4 it follows that if b + c = b + c and bc b c, then (1) abc(a + b + c ) ab c (a + b + c ) Without loss of generality we can assume that b 1 c Set b = 1, c = b + c 1 Since b + c = b + c and bc b c = (b 1)(c 1) 0, (1) implies that () abc(a + b + c ) a( a)(a ( a) ) Set d = (a 1) Then a( a)(a ( a) ) = (1 d)(3 + d) = 3 d d 3 and the given inequality (for n = ) follows from () Second solution Let n = and set ab + bc + ac = x Then we have a + b + c = 9 x and 9abc x (this follows from the inequality (ab + bc + ac) 3abc(a+b+c)) Hence we have to prove that (9 x)x 7, which is equivalent to the obvious (x + 3)(x 3) 0 11