PROBLEMS. Day 1. Day 2

Transcription

1 PROLEMS ay 1 Problem 1. Let AC be an acute-angled triangle, and let be the foot of the altitude from C. The angle bisector of AC intersects C at E and meets the circumcircle ω of triangle AE again at F. If AF =45, show that CF is tangent to ω. Problem. A domino is a 1or1 tile. etermine in how many ways exactly n dominoes can be placed without overlapping on a n n chessboard so that every square contains at least two uncovered unit squares which lie in the same row or column. Problem 3. Let n, m be integers greater than 1, and let a 1,a,...,a m be positive integers not greater than n m. Prove that there exist positive integers b 1,b,...,b m not greater than n, such that gcd(a 1 + b 1,a + b,...,a m + b m ) <n, where gcd(x 1,x,...,x m ) denotes the greatest common divisor of x 1,x,...,x m. ay Problem 4. etermine whether there exists an infinite sequence a 1,a,a 3,... of positive integers which satisfies the equality a n+ = a n+1 + a n+1 + a n for every positive integer n. Problem 5. Let m, n be positive integers with m > 1. Anastasia partitions the integers 1,,...,m into m pairs. oris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that oris cannot make his sum equal to n. Problem 6. Let H be the orthocentre and G be the centroid of acute-angled triangle AC with A AC. The line AG intersects the circumcircle of AC at A and P. Let P be the reflection of P in the line C. Prove that CA =60 if and only if HG = GP. 4

2 ω A ω A Problem 1. Let AC be an acute-angled triangle, and let be the foot of the altitude from C. The angle bisector of AC intersects C at E and meets the circumcircle ω of triangle AE again at F. If AF =45, show that CF is tangent to ω. (Luxembourg) F G 45 C E It follows that CF is a tangent to ω, as required. Solution 1: Since CF =90 45 =45, the line F bisects CA, and so F lies on the perpendicular bisector of segment AE, which meets A at G. Let AC =β. Since AEF is cyclic, AF E =90, and hence FAE =45. Further, as F bisects AC, we have FA =90 β, and thus EA = AEG =45 β, and AE =45 + β, so GE = β. This implies that right-angled triangles EG and C are similar, and so we have G / C = E /. Thus the right-angled triangles E and GC are similar, whence GC = E = β. ut FE = AE =45 β, then GF =45 FE = β. Hence GCF is cyclic, so GF C =90, whence CF is perpendicular to the radius FG of ω. Solution : As AF =45 line F is an exterior bisector of C. Since F bisects C line CF is an exterior bisector of C. Let AC =β, so ECF =( C + C)/ =45 + β. Hence CFE = 180 ECF CE EC = 180 (45 + β +90 β + β) =45. It follows that FC = CFE, then CF is tangent to ω. Solution 3: Note that AE is diameter of circumcircle of AC since CF = 90. From AEF = AF =45 it follows that triangle AF E is right-angled and isosceles. Without loss of generality, let points A, E and F have coordinates ( 1, 0), (1, 0) and (0, 1) respectively. Points F, E, are collinear, hence have coordinates (b, 1 b) forsomeb 1. Let point C be intersection of line tangent to circumcircle of AF E at F with line E. ThusC have coordinates (c, 1) and from C E A we get c =b/(b + 1). Now vector C =(b/(b +1) b, b) =b/(b +1) (1 b, b +1), vector F =( b, b) =( 1, 1) b and vector A =( (b +1), (1 b)). Its clear that (1 b, b +1) and ( (b +1), (1 b)) are symmetric with respect to FE =( 1, 1), hence F bisects C A and C = C which completes the proof. F M 45 C E Solution 4: Again F lies on the perpendicular bisector of segment AE, so AF E is right-angled and isosceles. Let M be an intersection of C and AF. Note that AM is isosceles since F is a bisector and altitude in this triangle. Thus F is a symmetry line of AM. Then FA = FEA = MEF = 45,AF = FE = FM and AE = EMC. Let us show that EC = CM. Indeed, CEM = 180 ( AE + FEA+ MEF)=90 AE = = AE = EMC. It follows that FMCE is a kite, since EF = FM and MC = CE. Hence EFC = CFM = EF =45, so FC is tangent to ω. Solution 5: Let the tangent to ω at F intersect C at C. Let AF = FC = β. It follows that C FE =45 since C F is tangent. We have sin C sin FC sin FC sin 90 sin CF sin C = F sin C sin 45 sin(90 β) sin 45 sin β sin β = sinβ cos β sin β So by trig Ceva on triangle F, lines FC,C and C are concurrent (at C), so C = C. Hence CF is tangent to ω. =1. 5

3 Problem. A domino is a 1 or 1 tile. etermine in how many ways exactly n dominoes can be placed without overlapping on a n n chessboard so that every square contains at least two uncovered unit squares which lie in the same row or column. (Turkey) ( ) n Solution: The answer is. n ivide the schessboard into squares. There are exactly n such squares on the chessboard. Each of these squares can have at most two unit squares covered by the dominos. As the dominos cover exactly n squares, each of them must have exactly two unit squares which are covered, and these squares must lie in the same row or column. We claim that these two unit squares are covered by the same domino tile. Suppose that this is not the case for some square and one of the tiles covering one of its unit squares sticks out to the left. Then considering one of the leftmost squares in this division with this property gives a contradiction. Now consider this n n chessboard consisting of squares of the original board. efine A,, C, as the following configurations on the original chessboard, where the gray squares indicate the domino tile, and consider the covering this n n chessboard with the letters A,, C, in such a A = = C = = way that the resulting configuration on the original chessboard satisfies the condition of the question. Note that then a square below or to the right of one containing an A or must also contain an A or. Therefore the (possibly empty) region consisting of all squares containing an A or abuts the lower right corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a C or by a path which goes from the lower left corner to the upper right corner of this chessboard and which moves up or right at each step. A similar reasoning shows that the (possibly empty) region consisting of all squares containing an A or abuts the lower left corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a or C by a path which goes from the upper left corner to the lower right corner of this chessboard and which moves down or right at each step. C C C C C C C A A A A A A A A Therefore the n n chessboard is divided by these two paths into four (possibly empty) regions that consist respectively of all squares containing A or or C or. Conversely, choosing two such paths and filling the four regions separated by them with As, s, Cs ands counterclockwise starting at the bottom results in a placement of the dominos on the original board satisfying the condition of the question. placed. As each of these paths can be chosen in ( ) n ways, there are n ( ) n ways the dominos can be n 6

4 Problem 3. Let n, m be integers greater than 1, and let a 1,a,...,a m be positive integers not greater than n m. Prove that there exist positive integers b 1,b,...,b m not greater than n, such that gcd(a 1 + b 1,a + b,...,a m + b m ) <n, where gcd(x 1,x,...,x m ) denotes the greatest common divisor of x 1,x,...,x m. (USA) Solution 1: Suppose without loss of generality that a 1 is the smallest of the a i.ifa 1 n m 1, then the problem is simple: either all the a i are equal, or a 1 = n m 1anda j = n m for some j. In the first case, we can take (say) b 1 =1,b =, and the rest of the b i can be arbitrary, and we have gcd(a 1 + b 1,a + b,...,a m + b m ) gcd(a 1 + b 1,a + b )=1. In the second case, we can take b 1 =1,b j = 1, and the rest of the b i arbitrary, and again gcd(a 1 + b 1,a + b,...,a m + b m ) gcd(a 1 + b 1,a j + b j )=1. So from now on we can suppose that a 1 n m. Now, let us suppose the desired b 1,...,b m do not exist, and seek a contradiction. Then, for any choice of b 1,...,b m {1,...,n}, wehave Also, we have gcd(a 1 + b 1,a + b,...,a m + b m ) n. gcd(a 1 + b 1,a + b,...,a m + b m ) a 1 + b 1 n m + n. Thus there are at most n m 1 possible values for the greatest common divisor. However, there are n m choices for the m-tuple (b 1,...,b m ). Then, by the pigeonhole principle, there are two m-tuples that yield the same values for the greatest common divisor, say d. ut since d n, foreachi there can be at most one choice of b i {1,,...,n} such that a i + b i is divisible by d and therefore there canbeatmostonem-tuple (b 1,b,...,b m ) yielding d as the greatest common divisor. This is the desired contradiction. Solution : Similarly to Solution 1 suppose that a 1 n m. The gcd of a 1 +1,a +1,a 3 +1,...,a m +1 is co-prime with the gcd of a 1 +1,a +,a 3 +1,...,a m +1, thusa 1 +1 n. Now change another 1 intoandsoon.afterm 1 changes we get a 1 +1 n m which gives us a contradiction. Solution 3: We will prove stronger version of this problem: For m, n > 1, leta 1,...,a m be positive integers with at least one a i n m 1. Then there are integers b 1,...,b m, each equal to 1 or, such that gcd(a 1 + b 1,...,a m + b m ) <n. Proof: Suppose otherwise. Then the m 1 integers gcd(a 1 + b 1,...,a m + b m )withb 1 =1andb i =1 or for i>1 are all pairwise coprime, since for any two of them, there is some i>1witha i +1 appearing in one and a i + in the other. Since each of these m 1 integers divides a 1 +1, andeach is n with at most one equal to n, it follows that a 1 +1 n(n +1) m 1 1 so a 1 n m 1.Thesame is true for each a i, i =1,...,n, a contradiction. Remark: Clearly the n m 1 bound can be strengthened as well. 7

5 Problem 4. etermine whether there exists an infinite sequence a 1,a,a 3,... of positive integers which satisfies the equality a n+ = a n+1 + a n+1 + a n for every positive integer n. (Japan) Solution 1: The answer is no. Suppose that there exists a sequence (a n ) of positive integers satisfying the given condition. We will show that this will lead to a contradiction. For each n define b n = a n+1 a n. Then, by assumption, for n wegetb n = a n + a n 1 so that we have b n+1 b n =(a n+1 + a n ) (a n + a n 1 )=(a n+1 a n )+(a n a n 1 )=b n + b n 1. Since each a n is a positive integer we see that b n is positive integer for n and the sequence (b n )is strictly increasing for n 3. Thus b n +b n 1 =(b n+1 b n )(b n+1 +b n ) b n+1 +b n, whence b n 1 b n+1 a contradiction to increasing of the sequence (b i ). Thus we conclude that there exists no sequence (a n ) of positive integers satisfying the given condition of the problem. Solution : Suppose that such a sequence exists. We will calculate its members one by one and get a contradiction. From the equality a 3 = a + a + a 1 it follows that a 3 >a. enote positive integers a 3 + a by b and a 3 by a, thenwehave a>b. Since a 4 = a + b and a 5 = a + b + a + b are positive integers, then a + b is positive integer. Consider a 6 = a + b + a + b + a +b + a + b. Numberc = positive integer, obviously it is greater than a + b. ut a +b + a + b must be ( a + b +1) =a + b + a + b +1=a +b + a + b +( a + b b)+1>c. So a + b<c< a + b + 1 which is impossible. Solutions 3: We will show that there is no sequence (a n ) of positive integers which consists of N>5 members and satisfies a n+ = a n+1 + a n+1 + a n (1) for all n =1,...,N. Moreover, we will describe all such sequences with five members. Since every a i is a positive integer it follows from (1) that there exists such positive integer k (obviously k depends on n) that a n+1 + a n = k. () From (1) we have (a n+ a n+1 ) = a n+1 + a n, consider this equality as a quadratic equation with respect to a n+1 : a n+1 (a n+ +1)a n+1 + a n+ a n =0. Obviously its solutions are ( ) a n+1 1, = a n+ +1±, where =4(a n + a n+ )+1. (3) Since a n+ >a n+1 we have a n+1 = a n

6 From the last equality, using that a n+1 and a n+ are positive integers, we conclude that is a square of some odd number i.e. =(m +1) for some positive integer m N, substitute this into (3): a n + a n+ = m(m +1). (4) Now adding a n to both sides of (1) and using () and (4) we get m(m +1)=k + k whence m = k. So { an + a n+1 = k, a n + a n+ = k (5) + k for some positive integer k (recall that k depends on n). Write equations (5) for n =andn = 3, then for some positive integers k and l we get Solution of this linear system is a = k l + k a + a 3 = k, a + a 4 = k + k, a 3 + a 4 = l, a 3 + a 5 = l + l., a 3 = l k, a 4 = l + k From a <a 4 we obtain k <l hence k<l. Consider a 6 : a 6 = a 5 + a 5 + a 4 = a 5 + l + l + k., a 5 = l +l + k (6). (7) Since 0 <k<lwe have l <l + l + k<(l +1). So a 6 cannot be integer i.e. there is no such sequence with six or more members. To find all required sequences with five members we must find positive integers a, a 3, a 4 and a 5 which satisfy (7) for some positive integers k<l. Its clear that k and l must be of the same parity. Vise versa, let positive integers k, l be of the same parity and satisfy k<lthen from (7) we get integers a, a 3, a 4 and a 5 then a 1 =(a 3 a ) a and it remains to verify that a 1 and a are positive i.e. k + k>l and (l k k) > k l + k. Solution 4: It is easy to see that (a n ) is increasing for large enough n. Hence and Lets define b n = a n + a n 1. Using AM-QM inequality we have a n+1 <a n + a n (1) a n <a n 1 + a n 1. () Adding (1), () and using (3): an + a n 1 an +a n 1. (3) b n+1 <b n + a n + a n 1 b n + b n. Let b n = m.since(b n ) is increasing for large enough n, we have: m <b n+1 <m +m<(m +1). So, b n+1 can t be a perfect square, so we get contradiction. 9

7 Problem 5. Let m, n be positive integers with m > 1. Anastasia partitions the integers 1,,...,m into m pairs. oris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that oris cannot make his sum equal to n. Solution 1A: efine the following ordered partitions: P 1 =({1, }, {3, 4},...,{m 1, m}), P =({1,m+1}, {,m+},...,{m, m}), P 3 =({1, m}, {,m+1},{3,m+},...,{m, m 1}). (Netherlands) For each P j we will compute the possible values for the expression s = a a m,wherea i P j,i are the chosen integers. Here, P j,i denotes the i-th coordinate of the ordered partition P j. We will denote by σ the number m i=1 i =(m + m)/. Consider the partition P 1 and a certain choice with corresponding sum s. We find that m m m = (i 1) s i = m + m. i=1 Hence, if n<m or n>m + m, this partition gives a positive answer. Consider the partition P and a certain choice with corresponding s. We find that i=1 m s i σ i=1 (mod m). Hence, if m n m + m and n σ (mod m), this partition solves the problem. Consider the partition P 3 and a certain choice with corresponding s. Weset { 0 if ai = i d i = 1, if a i i. We also put d = m i=1 d i, and note that 0 d m. Note also that if a i i, thena i i 1 (mod m). Hence, for all a i P 3,i it holds that a i i d i (mod m). Hence, m m s a i (i d i ) σ d (mod m), i=1 i=1 which can only be congruent to σ modulo m if all d i are equal, which forces s =(m + m)/ or s =(3m + m)/. Since m>1, it holds that m + m <m <m + m< 3m + m. Hence if m n m + m and n σ (mod m), then s cannot be equal to n, so partition P 3 suffices for such n. 10

8 Note that all n are treated in one of the cases above, so we are done. Common notes for solutions 1 and 1C: Given the analysis of P 1 and P as in the solution 1A, we may conclude (noting that σ m(m +1)/ (modm)) that if m is odd then m and m + m are the only candidates for counterexamples n, while if m is even then m + m is the only candidate. There are now various ways to proceed as alternatives to the partition P 3. Solution 1: Consider the partition ({1,m+}, {,m+3},...,{m 1, m}, {m, m+1}). We consider possible sums mod m+1. For the first m 1 pairs, the elements of each pair are congruent mod m+1, so the sum of one element of each pair is (mod m+1) congruent to 1 m(m+1) m, which is congruent to1ifm +1isoddand1+ m +1 if m + 1 is even. Now the elements of the last pair are congruent to 1 and 0, so any achievable value of n is congruent to 0 or 1 if m + 1 is odd, and to 0 or 1 plus m +1 if m + 1 is even. If m is even then m + m 1+m, which is not congruent to 0 or 1. If m is odd then m 1andm + m 0, neither of which can equal 0 or 1 plus m +1. Solution 1C: Similarly, consider the partition ({1,m},{,m+1},...,{m 1, m }, {m 1, m}), this time considering sums of elements of pairs mod m 1. If m 1 is odd, the sum is congruent to 1or;ifm 1 is even, to 1 or plus m 1.Ifmis even then m + m 1+m, and this can only be congruent to 1 or when m =. Ifm is odd, m and m + m are congruent to 1 and, and these can only be congruent to 1 or plus m 1 when m = 3. Now the cases of m =andm = 3 need considering separately (by finding explicit partitions excluding each n). Solution : This solution does not use modulo arguments. Use only P 1 from the solution 1A to conclude that m n m + m. Now consider the partition ({1, m}, {, 3}, {4, 5},...,{m, m 1}). If 1 is chosen from the first pair, the sum is at most m ; if m is chosen, the sum is at least m + m. So either n = m or n = m + m. Now consider the partition ({1, m 1}, {, m}, {3, 4}, {5, 6},...,{m 3, m }). Sums of one element from each of the last m pairs are in the range from (m )m = m m to (m )(m +1)=m m inclusive. Sums of one element from each of the first two pairs are 3, m +1and4m 1. Inthefirstcasewehave n m m+1 <m, in the second m +1 n m +m 1 and in the third n m +m 1 >m +m. So these three partitions together have eliminated all n. 11

9 Problem 6. Let H be the orthocenter and G be the centroid of acute-angled triangle AC with A AC. The line AG intersects the circumcircle of AC at A and P. Let P be the reflection of P in the line C. Prove that CA =60 if and only if HG = GP. (Ukraine) C ω ω P P M O O G A H Solution 1: Let ω be the circumcircle of AC. Reflecting ω in line C, we obtain circle ω which, obviously, contains points H and P. Let M be the midpoint of C. As triangle AC is acute-angled, then H and O lie inside this triangle. Let us assume that CA =60. Since CO = CA = 10 = = 180 CA = CH, hence O lies on ω. Reflecting O in line C, we obtain point O which lies on ω and this point is the center of ω. Then OO =OM =R cos CA = AH, so AH = OO = HO = AO = R, where R is the radius of ω and, naturally, of ω. Then quadrilateral AHO O is a rhombus, so A and O are symmetric to each other with respect to HO. As H, G and O are collinear (Euler line), then GAH = HO G. iagonals of quadrilateral GOP O intersects at M. Since OM =60, so OM = MO =ctg60 M = M 3. As 3MO MO = M = M MC = MP MA =3MG MP, then GOP O is a cyclic. Since C is a perpendicular bisector of OO, so the circumcircle of quadrilateral GOP O is symmetrical with respect to C. Thus P also belongs to the circumcircle of GOP O, hence GO P = GP P. Note that GP P = GAH since AH PP. And as it was proved GAH = HO G, then HO G = GO P. Thus triangles HO G and GO P are equal and hence HG = GP. NowwewillprovethatifHG = GP then CA =60. Reflecting A with respect to M, we get A. Then, as it was said in the first part of solution, points,c,h and P belong to ω. Also it is clear that A belongs to ω. Note that HC CA since A CA and hence HA is a diameter of ω. Obviously, the center O of circle ω is midpoint of HA. From HG = GP it follows that HGO is equal to P GO. Therefore H and P are symmetric with respect to GO. Hence GO HP and GO A P. Let HG intersect A P at K and K O since A AC. We conclude that HG = GK, because line GO is midline of the triangle HKA. Note that GO = HG. since HO is Euler line of triangle AC. So O is midpoint of segment GK. ecause of CMP = CMP, then GMO = OMP. Line OM, that passes through O, is an external angle bisector of P MA. Also we know that P O = O A, then O is the midpoint of arc P MA of the circumcircle of triangle P MA. It 1

10 C A T K O P G M P O ω A H follows that quadrilateral P MO A is cyclic, then O MA = O P A = O A P. Let OM and P A intersect at T. Triangles TO A and A O M are similar, hence O A /O M = O T/O A. In the other words, O M O T = O A. Using Menelaus theorem for triangle HKA and line TO, we obtain that A O O H HO OK KT TA =3 KT TA =1. It follows that KT/TA =1/3 andka =KT. Using Menelaus theorem for triangle TO A and line HK we get that 1= O H HA A K KT TO OO = 1 TO OO = TO OO. It means that TO = OO, so O A = O M O T = OO. Hence O A = OO and, consequently, O ω. Finally we conclude that CA = OC = 180 CA, so CA =60. C ω H F G O P P ε G F O O δ G A ε H H O Solution : Let O and G denote the reflection of O and G, respectively, with respect to the line C. We then need to show CA =60 iff G H = G P. Note that H OP is isosceles and hence 13

11 G H = G P is equivalent to G lying on the bisector H OP. Let H AP = ε. y the assumption A AC, we have ε 0. Then H OP = H AP =ε, hence G H = G P iff G OH = ε. ut GO H = G OH. Let be the midpoint of OO. It is known that GO = GAH = ε. Let F be the midpoint of HG. Then HG = FO (Euler line). Let GO H = δ. We then have to show δ = ε iff CA =60. ut by similarity ( GO FO O)wehave FO O = ε. Consider the circumcircles of the triangles FO O and GO H. y the sine law and since the segments HG and FO are of equal length we deduce that the circumcircles of the triangles FO O and GO H are symmetric with respect to the perpendicular bisector of the segment FG iff δ = ε. Obviously, O is the common point of these two circles. Hence O must be fixed after the symmetry about the perpendicular bisector of the segment FG iff δ = ε so we have ε = δ iff HOO is isosceles. ut HO = H O = R, and so ε = δ OO = R O = R cos CA = 1 CA =60. Solution 3: Let H and G denote the reflection of points H and G with respect to the line C. It is known that H belongs to the circumcircle of AC. The equality HG = GP is equivalent to H G = G P. As in the Solution, it is equivalent to the statement that point G belongs to the perpendicular bisector of H P, which is equivalent to OG H P,whereO is the circumcenter of AC. Let points A(a), (b), and C(c = b) belong to the unit circle in the complex plane. Point G have coordinate g =(a + b b)/3. Since C is parallel to the real axis point H have coordinate h = a =1/a. Point P (p) belongs to the unit circle, so p =1/p. Since a, p, g are collinear we have p a g a = ( ) p a. After computation we get p = g a g a 1 ga.sinceg (g) is the reflection of G with respect to the chord C, wehaveg = b +( b) b( b)g = b b + g. Letb b = d. Wehaved = d. So g = a + d 3, g = a d 3, g = d + g = a +d 3 It is easy to see that OG H P is equivalent to, g = a d 3 and p = g a 1 ga = d a +ad. (1) g ( g h p = ) h = g p 1 h 1 p = g h p h p since h and p belong to the unit circle (note that H P because A AC). This is equivalent to g = g h p and from (1), after easy computations, this is equivalent to a g + a + d +1 = (a +1)(d +1)=0. We cannot have a + 1 = 0, because then a = ±i, but A AC. Hence d = b b = ±i, andthe pair {b, c = b} is either { 3/+i/, 3/+i/} or { 3/ i/, 3/ i/}. oth cases are equivalent to AC =60 which completes the proof. 14