1997 SOLUTIONS. Problem 1 Deepee Khosla, Lisgar Collegiate Institute, Ottawa, ON

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1 997 SOLUTIONS Problem Deepee Khosla, Lisgar Collegiate Istitute, Ottawa, ON Let p,,p deote, i icreasig order, the primes from 7 to 47 The 5! p 0 p 0 p 0 ad 50! a 3 a 5 a3 p b p b p b Note that 4, 3, 5,p,,p all divide 50!, so all its prime powers differ from those of 5! Sice x, y 50!, they are of the form x 3 p 5 y m 3 m p m 5 The max i,m i is the i th prime power i 50! ad mi i,m i is the i th prime power i 5! Sice, by the above ote, the prime powers for p ad uder differ i 5! ad 50!, there are 5 choices for x, oly half of which will be less tha y Sice for each choice of x, y is forced ad either x<yor y<x So the umber of pairs is 5 / 4

2 Problem Byug Kuy Chu, Harry Ailay Composite High School, Edmoto, AB Look at the first poit of each give uit iterval iterval This poit uiquely defies the give uit Lemma I ay iterval [x, x + there must be at least oe of these first poits 0 x 49 Proof Suppose the opposite The last first poit before x must be x ε for some ε>0 The correspodig uit iterval eds at x ε +<x+ However, the ext give uit iterval caot begi util at least x + This implies that poits x ε +,x+ are ot i set A, a cotradictio There must be a first poit i [x, x + Note that for two first poits i itervals [x, x + ad [x +,x+ 3 respectively, the correspodig uit itervals are disoit sice the itervals are i the rage [x, x+ ad [x+,x+4 respectively We ca choose a give uit iterval that begis i each of [0, [, 3 [k,k +[48, 49 Sice there are 5 of these itervals, we ca fid 5 poits which correspod to 5 disoit uit itervals Problem Coli Percival, Buraby Cetral Secodary School, Buraby, BC I prove the more geeral result, that if [0, ] i A i, A i,a i are itervals the a a, such that A ai Aa Let 0 <ε ad let b i i + ε,i The mi{b i } 0, max{b i } + ε + So all the b i are i [0, ] Let a i be such that b i A ai Sice A i [0, ], this is possible The sice b i b i + ε +ε>, ad the A i are itervals of legth, mi A ai max A a > 0, so A ai Aa Substitutig 5, we get the required result QED

3 Problem 3 Mihaela Eachescu, Dawso College, Motréal, PQ Let P The > 3 because < 3, 3 4 > 3 because 4 < 5,, > 997 because 998 < So P> Also < 3 because 3 <, 3 4 < 4 5 because 3 5 < 4 4, < because < 998 So P< }{{} P 999 Hece P < 999 < ad P< 44 The ad give 999 <P < 44 qed Problem 4 Joel Kamitzer, Earl Haig Secodary School, North York, ON O A B O D C Cosider a traslatio which maps D to A It will map 0 0 with OO DA, ad C will be mapped to B because CB DA This traslatio keeps agles ivariat, so AO B D 80 AOB AOBO is a cyclic quadrilateral ODC O AB O OB but, sice O O is parallel to BC, O OB OBC ODC OBC 3

4 Problem 4 Adria Cha, Upper Caada College, Toroto, ON A B θ 80 α α Ο 80 θ D C Let AOB θ ad B α The COD 80 θ ad AOD 80 α Sice AB CD parallelogram ad si θ si80 θ, the sie law o D ad OAB gives si CDO si80 θ si θ si ABO CD AB OA so OA Similarily, the sie law o OBC ad OAD gives si ABO si CDO so si CBO si α BC si80 α si ADO AD OA OA si ADO si CBO Equatios ad show that si ABO si CBO si ADO si CDO hece [cos ABO + CBO cos ABO CBO] [cos ADO + CDO cos ADO CDO] Sice ADC ABCparallelogram ad ADO + CDO ADC ad ABO + CBO ABC it follows that cos ABO CBO cos ADO CDO There are two cases to cosider Case i: ABO CBO ADO CDO Sice ABO + CBO ADO + CDO, subtractig gives CBO CDO so CBO CDO, ad we are doe Case ii: ABO CBO CDO ADO Sice we kow that ABO + CBO CDO + ADO, addig gives ABO CDO so ABO CDO ad CBO ADO Substitutig this ito, it follows that OA Also, ADO + ABO CBO + ABO ABC Now, ABC 80 BAD sice ABCD is a parallelogram Hece BAD + ADO + ABO 80 so DOB 80 ad D, O, B are colliear We ow have the diagram 4

5 A B D θ O 80 θ C The COD + B 80,so B θ AOB AOB is cogruet to COB SAS, OB is commo, AOB COB ad AO CO, so ABO CBO Sice also ABO CDO we coclude that CBO CDO Sice it is true i both cases, the CBO CDO QED 5

6 Problem 5 Sabi Cautis, Earl Haig Secodary School, North York, ON We first ote that k 3 +9k +6k + k + 3 k k Let S k +9k +6 The k0 S k! k! k!k + k + 3 k + 4! k + 4! k! k0 k0 k Let T S k +4 k + k0 Now, for, i i i0 0 sice Also i i i0 i i i! i! i! + 0 0! 0!! i i! i! i! i i i i i i i i i i 6

7 Substitutig i, * shows that i i i0 i 0 0 Hece T Substitutig, k +4 k + k0 k+4 +4 k + k0 k+4 +4 k k The first two terms are zero because of results * ad ** so T +3 + The S T k k k 3 +9k k0 7

Complex Numbers Solutions

Complex Numbers Solutions Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i