Solutions EN

Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN Solutions EN

Problems and Solutions, JBMO 2014 Problem Find all distinct prime numbers p , q and r such that 3p - 5q - 4r = 26 Solution First notice that if both primes q and r differ from , then q º r º 1(mod 3) , hence the left hand side of the given equation is congruent to zero modulo 3, which is impossible since 26 is not divisible by Thus, q = or r = We consider two cases Case q = (1) The equation reduces to 3p - 4r = 431 If p ¹ 5, by Fermat’s little - 4r º (mod 5) , or equivalently, theorem, p º (mod 5) , which yields r + º (mod 5) The last congruence is impossible in view of the fact that a residue of a square of a positive integer belongs to the set { 0, 1, } Therefore p = and r = 19 Case r = The equation becomes 3p - 5q = 62 (2) Obviously p ¹ Hence, Fermat’s little theorem gives p º (mod 5) But then 5q º (mod 5) , which is impossible Hence, the only solution of the given equation is p = , q = , r = 19 Problem Consider an acute triangle ABC with area S Let CD ^ AB ( D Ỵ AB ), DM ^ AC ( M Ỵ AC ) and DN ^ BC orthocentres of the triangles MNC ( N Ỵ BC ) Denote by H and H and MND the respectively Find the area of the quadrilateral AH 1BH in terms of S Solution Let O, P, K, R and T be the C mid-points of the segments CD, MN, R H1 K CN, CH and MH , respectively From MC and PK  MC Analogously, from DMH 1C DMNC we have that PK = we have that TR = MC and T N O P M A B D TR  MC Consequently, PK = TR and PK  TR Also OK  DN H2 (from DCDN ) and since DN ^ BC and MH ^ BC , it follows that TH  OK Since O is the circumcenter of DCMN , OP ^ MN Thus, CH ^ MN implies OP  CH We conclude DTRH @ DKPO (they have parallel sides and TR = PK ), hence RH = PO , i.e CH = 2PO and CH  PO Analogously, DH = 2PO and DH  PO From CH = 2PO = DH and CH  PO  DH the quadrilateral CH 1H 2D is a parallelogram, thus H 1H = CD and H 1H  CD AB ⋅ H 1H Therefore the AB ⋅ CD =S 2 Solution Since MH  DN area of the quadrilateral AH 1BH is = NH  CM and MH  CN and NH  DM , MDNH is a parallelogram Similarly, imply MCNH is a parallelogram Let P be the midpoint of the segment MN Then sP (D ) = H and sP (C ) = H , thus CD  H 1H and CD = H 1H From CD ^ AB we deduce AAH BH = AB ⋅ CD = S Problem Let a, b, c be positive real numbers such that abc = Prove that 2 ỉ ỉ ỉ 1ư 1ử 1ử ỗỗỗa + ữữữ + ỗỗỗb + ữữữ + çççc + ÷÷÷ ³ (a + b + c + 1) b ÷ø c ÷ø a ø÷ è è è When does equality hold? Solution By using AM-GM ( x + y + z ³ xy + yz + zx ) we have 2 ỉ ỉ ỉ ỉ ưỉ ổ ửổ ổ ửổ ỗỗa + ữữ + ỗỗb + ữữ + ỗỗc + ữữ ỗỗa + ữữ ỗỗb + ữữ + ỗỗb + ữữ ỗỗc + ữữ + ççc + ÷÷ çça + ÷÷ çè çè çè çè ÷÷ ç ÷÷ ç b ÷÷ø c ÷÷ø a ữữứ b ữữứốỗ c ữữứ ốỗ c ứố a ứữữ ốỗ a ứố b ứữữ ổ ổ ổ c a b = ỗỗỗab + + + a ữữữ + ỗỗỗbc + + + b ữữữ + ỗỗỗca + + + c ữữữ ữứ ÷ø è ÷ø è b c a è a c b + + + +a +b +c c b a b c a Notice that by AM-GM we have ab + ³ 2b, bc + ³ 2c, and ca + ³ 2a a b c Thus , = ab + bc + ca + 2 æ ö æ ö æ ö æ ö æ ö æ ỗỗa + ữữ + ỗỗb + ữữ + ỗỗc + ữữ ỗỗab + b ữữ + ỗỗbc + c ữữ + ỗỗca + a ữữ + + a + b + c ³ 3(a + b + c + 1) ỗố b ữứữ c ứữữ a ứữữ a ứữữ ốỗ b ứữữ ốỗ c ứữữ ốỗ ốỗ ốỗ The equality holds if and only if a = b = c = Solution From QM-AM we obtain (a + ) b + (b + c1 ) + (c + a1 ) 2 ổ ổ ỗỗa + ữữ + ỗỗb + ữữ ỗố b ứữữ c ứữữ ốỗ From AM-GM we have a + b1 + b + c1 + c + a1  2 ỉ ư÷ a + b1 + b + c1 + c + a1 ) ( + ỗỗỗc + ữữ (1) a ứữ ố 1 1 + + ³ 33 = , and substituting in (1) we get a b c abc ỉ ỉ ỉ a + b1 + b + c1 + c + a1 ) (a + b + c + 3) ỗỗa + ữữ + ỗỗb + ữữ + ỗỗc + ữữ ( = ỗố ỗố ỗố b ữữứ c ÷÷ø a ÷÷ø 3 (a + b + c )(a + b + c ) + (a + b + c ) + (a + b + c ) 3 abc + (a + b + c ) + = ³ = 3 (a + b + c ) + = = (a + b + c + 1) The equality holds if and only if a = b = c = 2 Solution By using x + y + z ³ xy + yz + zx 2 2 ổ ổ ổ ỗỗa + ữữ + ỗỗb + ữữ + ỗỗc + ữữ = a + b + c + + + + 2a + 2b + 2c ỗố ỗố ỗố b ữữứ c ữữứ a ÷÷ø b c a b2 c2 a 1 2a 2b 2c ³ ab + ac + bc + + + + + + bc ca ab b c a Clearly 1 abc abc abc + + = + + = a +b +c, bc ca ab bc ca ab a b c ab + + bc + + ca + ³ 2a + 2b + 2c, b c a a b c a b c + + ³ 3 ⋅ ⋅ = b c a b c a Hence 2 æ ö æ ö æ ö æ ö æ ö ổ ỗỗa + ữữ + ỗỗb + ữữ + ỗỗc + ữữ ỗỗab + a ữữ + ỗỗac + c ữữ + ỗỗbc + b ÷÷ + a + b + c + a + b + c ỗố ỗố b ữứữ c ứữữ a ứữữ b ứữữ ốỗ a ứữữ ốỗ c ứữữ b c a ốỗ ốỗ 2a + 2b + 2c + a + b + c + = (a + b + c + 1) The equality holds if and only if a = b = c = Solution a = x y z ,b= ,c= y z x 2 ỉx z æy x ö æz y ö æx y z ö ççç + ÷÷÷ + ççç + ÷÷÷ + ççç + ữữữ ỗỗỗ + + + 1ữữữ ố y y ÷ø è z z ø÷ è x x ø÷ èy z x ø÷ (x + z )2 x 2z + (y + x )2 y 2x + (z + y )2 z 2y ³ 3xyz (x 2z + y 2x + z 2y + xyz ) x 4z + 2x 3z + x 2z + x 2y + 2x 3y + x 4y + y 2z + 2y 3z + y 4z ³ 3x 3yz + 3x 2y 3z + 3xy 2z + 3x 2y 2z 1)x 3y + y 3z + z 3x ³ 3x 2y 2z 2)x 4z + z 4x + x 3y ³ 3x 3z 2y üïï ï 3)x 4y + y 4x + y 3z ³ 3y 3x 2z ïý ï 4)z 4y + y 4z + x 3z 3z 3y 2x ùùù ỵ Equality holds when x = y = z , i.e., a = b = c = Solution å (a + b ) cyc ³ 3å a + cyc ỉ a  2å + ỗỗa + - 3a - 1ữữữ ữứ ỗ a cyc b cyc ố 2ồ cyc a abc ³ 63 =6 b bca (1) - 3a ³ - a a  a - 3a + 4a - 3a + ³ "a > 0, a + ( )  (a - 1) a - a + ổ ỗốỗỗa cyc + ư÷ 1 a ÷÷÷ ³ 3å a - 15 ³ abc - 15 = -6 a ø cyc Using (1) and (2) we obtain 2å cyc ỉ a + å ççça + - 3a - 1÷÷÷ ³ - = b a è ø÷ Equality holds when a = b = c = (2) Problem For a positive integer n , two players A and B play the following game: Given a pile of s stones, the players take turn alternatively with A going first On each turn the player is allowed to take either one stone, or a prime number of stones, or a multiple of n stones The winner is the one who takes the last stone Assuming both A and B play perfectly, for how many values of s the player A cannot win? Solution Denote by k the sought number and let {s1, s2 , , sk } be the corresponding values for s We call each si a losing number and every other nonnegative integer a winning numbers Clearly every multiple of n is a winning number Suppose there are two different losing numbers si > s j , which are congruent modulo n Then, on his first turn of play, player A may remove si - s j stones (since n si - s j ), leaving a pile with s j stones for B This is in contradiction with both si and s j being losing numbers Hence, there are at most n - losing numbers, i.e k £ n - Suppose there exists an integer r Ỵ {1,2, , n - 1} , such that mn + r is a winning number for every m Ỵ  Let us denote by u the greatest losing number (if k > ) or (if k = ), and let s = LCM (2, 3, , u + n + 1) Note that all the numbers s + , s + , …, s +u +n +1 are composite Let m ' Ỵ 0, be such that s + u + £ m ' n + r £ s + u + n + In order for m ' n + r to be a winning number, there must exist an integer p , which is either one, or prime, or a positive multiple of n , such that m ' n + r - p is a losing number or 0, and hence lesser than or equal to u Since s + £ m ' n + r - u £ p £ m ' n + r £ s + u + n + , p must be a composite, hence p is a multiple of n (say p = qn ) But then m ' n + r - p = (m '- q ) n + r must be a winning number, according to our assumption This contradicts our assumption that all numbers mn + r , m Ỵ  are winning Hence, each nonzero residue class modulo n contains a loosing number There are exactly n - losing numbers Lemma: No pair (u, n ) of positive integers satisfies the following property: (*) In  exists an arithmetic progression (at )t¥=1 with difference n such that each segment éa - u, a + u ù contains a prime êë i úû i Proof of the lemma: ¥ t t =1 progression (a ) Suppose such a pair (u, n ) and a corresponding arithmetic exist In  exist arbitrarily long patches of consecutive composites Take such a patch P of length 3un Then, at least one segment éêëai - u, + u ùúû is fully contained in P , a contradiction Suppose such a nonzero residue class modulo n exists (hence n > ) Let u Ỵ  be greater than every loosing number Consider the members of the supposed residue class which are greater than u They form an arithmetic progression with the property (*) , a contradiction (by the lemma)  ... such a patch P of length 3un Then, at least one segment éêëai - u, + u ùúû is fully contained in P , a contradiction Suppose such a nonzero residue class modulo n exists (hence n > ) Let u Ỵ... number or 0, and hence lesser than or equal to u Since s + £ m ' n + r - u £ p £ m ' n + r £ s + u + n + , p must be a composite, hence p is a multiple of n (say p = qn ) But then m ' n + r - p... every multiple of n is a winning number Suppose there are two different losing numbers si > s j , which are congruent modulo n Then, on his first turn of play, player A may remove si - s j stones