THE OLYMPIAD CORNER No. 184 R.E. Woodrow All ommuniation about i olumn hould be ent to Profeor R.E. Woodrow, Deartment of Maemati and Statiti, Univerity of Calgary, Calgary, Alberta, Canada. TN 1N4. We begin i number wi e roblem of e Mok Tet for e International Maematial Olymiad team of Hong Kong. My ank go to Rihard Nowakowki, Canadian Team Leader, for olleting em at e 5 IMO in Hong Kong. INTERNATIONAL MATHEMATICAL OLYMPIAD 1994 Hong Kong Committee Mok Tet, Part I Time: 4.5 hour 1. In 4ABC, we have \C = \B. Pi a oint in e interior of 4ABC atifying AP = AC and PB = PC. Show at AP triet e angle \A.. Inatable-tenni tournament of 10 ontetant, any two ontetant meet only one. We ay at ere i a winning triangle if e following ituation our: i ontetant defeated j ontetant, j ontetant defeated k ontetant, and k ontetant defeated i ontetant. Let W i and L i be reetively e number of game won and lot by e i ontetant. Suoe L i + W j 8 whenever e i ontetant beat e j ontetant. Prove at ere areexatly 40 winning triangle in i tournament.. Find all e non-negative integer x, y, and z atifying at 7 x +1= y +5 z : Mok Tet, Part II Time: 4.5 hour 1. Suoe at yz + zx + xy =1and x, y, and z 0. Prove at x(1, y )(1, z )+y(1, z )(1, x )+z(1, x )(1, y ) 4 9 :
. A funtion f (n), dened on e natural number, atie: f (n) =n,1 if n>000; and f(n) =f(f(n+ 16)) if n 000: (a) Find f (n). (b) Find all olution to f(n) =n.. Let m and n be oitive integer where m ha d digit in bae ten and d n. Find e um of all e digit (in bae ten) of e rodut (10 n, 1)m. A a eond Olymiad et we give e roblem of e Final Round of e 45 Maematial Olymiad written in Aril, 1994. My ank go to Marin E. Kuzma, Warzawa, Poland; and Rihard Nowakowki, Canadian Team leader to e 5 IMO in Hong Kong, for olleting em. 45 MATHEMATICAL OLYMPIAD IN POLAND Problem of e Final Round Aril 10{11, 1994 Firt Day Time: 5 hour 1. Determine all trile of oitive rational number (x; y; z) uh at x + y + z, x,1 + y,1 + z,1 and xyz are integer.. In e lane ere are given two arallel line k and l, and a irle dijoint from k. From a oint A on k draw e two tangent to e given irle; ey ut l at oint B and C. Let m be e line rough A and e midoint of BC. Show at all e reultant line m (orreonding to variou oint A on k) have a oint in ommon.. Let 1 be a xed integer. To eah ubet A of e et f1,, :: ::, ngwe aign a number w(a) from e et f1,, :::, g in uh a way at w(a \ B) = min(w(a);w(b)) for A; B f1; ;::: ; ng: Suoe ere area(n)uh aignment. Comute lim n!1 n a(n). Seond Day Time: 5 hour 4. We have ree bowl at our dioal, of aaitie m litre, n litre and m + n litre, reetively; m and n are mutually orime natural number. The two maller bowl are emty, e larget bowl i lled wi water. Let k be any integer wi 1 k m + n, 1. Show at by ouring water (from any one of oe bowl into any oer one, reeatedly, in an unretrited manner) we are able to meaure out exatly k litre in e ird bowl.
4 5. Let A 1 ;A ;::: ;A 8 be e vertie of a aralleleied and let O be it entre. Show at 4(OA 1 + OA + +OA 8 ) (OA 1 + OA + +OA 8 ) : 6. Suoe at n ditint real number x 1 ;x ;::: ;x n (n4) atify e ondition x 1 + x + +x n =0and x 1 + x + +x n =1.Prove at one an hooe four ditint number a, b,, d from among e x i ' in uh a way at a + b + + nab x 1 + x + +x n a+b+d+nabd: We now give ree olution to roblem given in e Marh 1996 Corner a e Teleom 199 Autralian Maematial Olymiad [1996: 58]. TELECOM 199 AUSTRALIAN MATHEMATICAL OLYMPIAD Paer 1 Tueday, 9 February, 199 (Time: 4 hour) 6. In e aute-angled triangle ABC, let D, E, F be e feet of altitude rough A, B, C, reetively, and H e oroentre. Prove at AH AD + BH BE + CH CF =: Solution by Manur Boae, tudent, St. Paul' Shool, London, England. AH AD + BH BE + CH CF =, HD AD + HE BE + HF CF [BHC] =, [ABC] + [CHA] [ABC] + [AHB] [ABC] =, [ABC] [ABC] =: 7. Let n be a oitive integer, a 1 ;a ;::: ;a n oitive real number and = a 1 + a + +a n.prove at n, n, 1 and, n(n, 1):
5 Solution by Manur Boae, tudent, St. Paul' Shool, London, England and Edward T.H. Wang, Wilfrid Laurier Univerity, Waterloo, Ontario. We give e olution by Boae., = = 1 and n X by e Cauhy{Shwarz inequality. Thu,, 1 n : a i Hene n, n = n(n, 1). a i To rove e rt inequality, rt note at 1 Hene a. i n By e Cauhy{Shwarz inequality, (, ) a i, a i =, n a i ( X 1) = n ;! = :! = : Therefore, a i P n (, ) =, n = 1 1, 1 n P n, P n a = n n, 1 : So, bo inequalitie are roved. 8. [1996: 58] Teleom 199 Autralian Maematial Olymiad. The vertie of triangle ABC in e xy{lane have integer oordinate, and it ide do not ontain any oer oint having integer oordinate. The interior of ABC ontain only one oint, G, at hnteger oordinate. Prove at G i e entroid of ABC.
6 Solution by Manur Boae, tudent, St. Paul' Shool, London, England. A C 0 r G B 0 B A 0 By Pik' Theorem 1 [ABC] = 1+ 1 [ABG] = 0+ [BCG] = 1 [CAG] = 1 : and,1 = ;,1 = 1 ; C Therefore Hene [ABG] [ABC] GA 0 = [BCG] [ABC] = [CAG] [ABC] AA 0 = GB0 BB 0 = GC0 CC 0 = 1 : = 1 : The unique oint atifying i above i well-known to be e entroid. Next we give one olution from e Jaan Maematial Olymiad 199 given in e Marh 1996 Corner.. [1996: 58] Jaan Maematial Olymiad 199. Let d(n) be e larget odd number whih divide a given number n. Suoe at D(n) and T (n) are dened by D(n) =d(1) + d() + +d(n); T(n)=1+++n: Prove at ere exit innitely many oitive number n D(n) =T(n). uh at Solution by Manur Boae, tudent, St. Paul' Shool, London, England, and Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univerity, Waterloo, Ontario. We give e olution by Boae.
7 T (n) = n(n+1) : Thu we need to rove at ere are innitely many n for whih Conider D(n) = n(n+1) o at D(n) =T(n) hold: D( n ) = d(1) + d() + +d( n, 1) + d() + d(4) + +d( n ) Now = 1+++( n,1) + d(1) + d() + +d( n,1 ) = 1+++( n,1) + D( n,1 ): 1+++( n,1) = n ( n +1), n,1 ( n,1 +1) = n,1 ( n, n,1 ) Thu D( n )=D( n,1 )+ n,. = n, : Now, D( 1 )=and we hall rove by indution at D( n )= n + for n 0. Thi hold for n =0and for n =1. Suoe it hold for n = k. Thu D( k )= k +. Then D( k+1 )=D( k )+ k = k + = 4(k )+ = k+ + + k and e reult follow by indution. Now onider D( n, ). D( n, ) = D( n ), d( n, 1), d( n ) = n + = n +,( n, 1), 1, n = n,( n )+ = (n, 1)( n, ) :
8 Thu D(x) = x(x+1) for x = n,, and ereare innitely many uh x. Next we turn to omment and olution from e reader toroblem from e Aril 1996 number of e Corner where wegave e eletion tet for e Romanian Team to e 4 IMO a well a ree ontet for e Romanian IMO Team [1996: 107{109]. SELECTION TESTS FOR THE ROMANIAN TEAM, 4 IMO. Part II Firt Contet for IMO Team 1t June, 199 1. Find e greatet real number a uh at x y + + z y z + x + i true for all oitive real number x, y, z. z x + y >a Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univerity, Waterloo, Ontario. We laim at a =. Let f (x; y; z) = x y + + z y z + x + z x + y : We how at f (x; y; z) >. Sine f(x; y; z)! a x! y and z! 0, e lower bound i har. Wiout lo of generality, aume at x y z. Sine by e arimeti{harmoni{mean inequality, we have it ue to how at or equivalently, z + x y + + y + z z z + x ; f (x; y; z) > z + x y + + y + z z z + x z z x + y > + x, x + y + z y + z, y z + x : By imle algebra, i i eaily een to be equivalent to z y + z ( z + x + x) z + z + x ( y + z + y) < 1 x + y : (1)
9 Sine y + z z = z, z + x > xand x x + y, we have z y + z ( z + x + x) < z = z(x + x) Thu to etablih (1), it remain to how at or equivalently Sine z z + x ( y + z + y) < 1 1 1 x x + y : x + y z y + z + y < z + x x + y : z + x x + y = 1, y, z x + y ; whih i an non-dereaing funtion of x, we have z + x x + y and u it ue to how at z + y y > z + y y ; z y + z + y ; or equivalently y + z + y y + z > yz: () Sine y + z yz, we have y + z + y y + z yz + y z and u () hold. Thi omlete e roof. = ( + )yz > yz. Show at if x, y, z are oitive integer uh at x + y + z = 199, en x + y + z i not a erfet quare. Solution by Manur Boae, tudent, St. Paul' Shool, London, England; and by Edward T.H. Wang, Wilfrid Laurier Univerity, Waterloo, Ontario. We give Wang' olution. We how at e reult hold for nonnegative integer x, y, and z. Wiout lo of generality, we may aume at 0 x y z. Then z x + y + z = 199
0 imlie at z 665; z 6: On e oer hand z 199 imlie at z 44 and u 6 z 44. Suoe at x + y + z = k for ome nonnegative integer k. By e Cauhy{Shwarz Inequality we have k 4 =(x+y+z) (1 +1 +1 )(x + y + z ) = 5979 and o k b 4 5979 =8. Sine k z 6, k 6. Furermore, ine x +y +z i odd, it i eaily een at x+y +z mut be odd, whih imlie at k i odd. Thu k =7and we have x + y + z =49. Let z =6+d, where 0 d 18. Then x + y =,d)y, d ) x + y y (, d) = 1058, 9d +d : (1) On e oer hand, from x + y + z = 199 we get x + y = 199, z = 199, (6 + d) = 117, 5d, d : () From (1) and (), we get or 117, 5d, d 1058, 9d +d d, 40d 59 whih i learly imoible ine d, 40d = d(d, 40) 18 14 = 5. Thi omlete e roof. Remark: It i a well-known (ough by no mean eay) reult in laial number eory at a natural number n i e um of ree quare (of nonnegative integer) if and only if n 6= 4 l (8k +7)where l and k are nonnegative integer. Sine 199 1(mod 8) it an be o exreed and u e ondition given in e roblem i not \vauouly" true. In fat, 199 an be o exreed in more an one way; for examle, 199 = 0 +1 +4 = +15 +4 = +0 + = 11 +4 +6 : Thee rereentation alo how at e onluion of e roblem i fale if we allow x, y, and z to be negative integer; e.g. if x =,, y =,0, z =en x + y + z = 199 and x + y + z =1 ; and if x =,11, y =4,z=6en x + y + z = 199 and x + y + z =49=7.
1 4. Show at for any funtion f : P(f1; ;::: ;ng)!f1;;::: ;ng ere exit two ubet, A and B, of e et f1; ;::: ;ng, uh at A 6= B and f (A) =f(b) = maxfi j i A \ Bg. Comment by Edward T.H. Wang, Wilfrid Laurier Univerity, Waterloo, Ontario. The roblem, a tated, i learly inorret ine for maxfi : i A\Bg to make ene, we mut have A \ B 6= ;. For n =1learly ere areno ubet A and B wi A 6= B and A\B 6= ;. Aounterexamle when n = i rovided by etting f (;) = f(f1g)= f(fg)=1and f(f1; g) =. Thi ounterexamle tand if max i hanged to min. The onluion i till inorret if A \ B i hanged to A[B. Aounterexamle would be f(;) = and f (f1g) =f(fg)=ff(1; )g =1. Part III Seond Contet for IMO Team nd June, 199. Prove at for all integer number n, wi n 6, ere exit an n{oint et M in e lane uh at every oint P in M ha at leat ree oer oint in M at unit ditane to P. Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univerity, Waterloo, Ontario. The ree diagram dilayed below illutrate e exitene of uh a et. M 1 i for all n =k,m i for all n =k+1 and M i for all n =k+, where k =;;4;:::. In eah diagram, e olid line onneting two oint all have unit leng and e dotted line, alo of unit leng, indiate how to ontrut an (n +){oint et wi e deribed roerty from one wi n oint. M 1 M M (n =6;9;1;:::) (n=7;10; 1;:::) (n=8;11; 14;:::)
Part IV Third Contet for IMO Team rd June, 199 1. The equene of oitive integer fx n g n1 i dened a follow: x 1 =1, e next two term are e even number and 4, e next ree term are e ree odd number 5, 7, 9, e next four term are e even number 10, 1, 14, 16 and o on. Find a formula for x n. Solution by Manur Boae, tudent, St. Paul' Shool, London, England; and by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univerity, Waterloo, Ontario. We give e olution of Shan and Wang. We arrange e term of e equene fx n g in a triangular array aording to e given rule o at e 1 level onit of ngle 1 and for t all k, e k level onit ofekoneutive even (odd) integer at follow e lat odd (even) integer in e (k, 1) level: t 1 4 5 7 9 10 1 14 16 17 19 1 5 at i, For any given n N, let l n denote e unique integer uh at ln ln +1 <n ; (l n, 1)l n < n l n(l n +1) : (1) Note at l n i imly e number of e level whih x n i on. We laim at x n =n,l n : () When n =1, learly l 1 =1and u n, l n =1=x 1. Suoe () hold for ome n 1. There are two oible ae: Cae (i): If n +1, ln+1 ; at i, if xn+1 and x n are on e ame level, en l n+1 = l n and hene x n+1 = x n +=n,l n +=(n+1),l n+1 :
, ln+1 Cae (ii): Ifn+1> en xn i e lat number on e l level n and x n+1 i e rt number on e l n+1 level. Thu l n+1 = l n +1 and x n+1 = x n +1=n,l n +1=(n+1),l n+1 : Hene, by indution, () i etablihed. From (1) we get l n, l n < n (l n +1),(l n +1): Solving e inequalitie, we eaily obtain Hene l n < 1+ 1+8n l n = & 1+ 1+8n l n +1: ',1 () where dxe denote e leat integer greater an or equal to x (at i, e eiling funtion). From () and () we onlude at & ' 1+ 1+8n x n =n+1, : That omlete i number of e Corner. Send me your nie olution a well a Olymiad ontet.