No. 51. R.E. Woodrow. In the November 2000 number of the Corner we gave the problems of. Junior High School Mathematics Contest

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1 7 THE SKOLIAD CORNER No. 51 R.E. Woodrow In the November 000 number of the Corner we gave the problem of the Final Round of the Britih Columbia College Junior High School Mathematic Contet, May 5, 000. In thi iue we give the olution. BRITISH COLUMBIA COLLEGES Junior High School Mathematic Contet Part A Final Round May 5, The lat (one) digit of a perfect quare cannot be: Solution. The anwer i E. If we quare the digit from 0 to 9 and conider the nal digit of the quare we get only the digit 0, 1, 4, 9, 6, and 5. Since thereare no other, we ee that 8 i NOT the nal digit of any quare.. Suppoe a tring art deign i contructed by connecting nail on a vertical axi and on a horizontal axi by line egment a follow: The nail furthet from the origin on the vertical axi i connected to the nail nearet the origin on the horizontal axi. Then proceed toward the origin on the vertical axi and away on the horizontal axi a hown in the diagram. If thi were done on a project with 10 nail on each axi, the number of point of interection of line egment would be: Solution. The anwer i A. Each of the 10 traight line interect each of the other exactly once. Thi make for 90 interection however, each of the interection i counted twice in thi approach, depending upon which of the two line we conider rt. To get the correct number of interection we imply divide 90 by to get Aume there i an unlimited upply of pennie, nickel, dime, and quarter. An amount (in cent) which cannot be made uing exactly 6 of thee coin i:

2 8 Solution. The anwer i E. Let u try ucceively to make up each of the given amount uing 6 coin: 91 = , 87 = , 78 = , 51 = Thu each of the rt 4 choice can be made up with 6 coin. In order to make up 49 we would need to ue 4 pennie. Thi would require u to make up the total of 45 cent with only coin, which i clearly impoible. 4. Given x + y =8and xy =14, the value of x ; y equal: Solution. The anwer i B. Firt oberve that (x ; y) = x + y ; xy = 8 ; (14) = 0. Thi mean that x ; y = 0 that i, x = y. In that event we clearly have x ; y =0. 5. Given that 0 <x<y<0, the number of integer olution (x y) to the equation 3x +3y =50i: Solution. The anwer i E. Atute reader will notice that thi problem and olution were preented in previou iue of the Corner a Quetion #10 of the Britih Columbia College Senior High School Preliminary exam given late latyear [000 : 343, 476]. 6. The number 1, 3, 6, 10, 15 ::: are known a triangular number. Each triangular number can be expreed a n(n+1) where n i a natural number. The larget triangular number le than 500 i: that Solution. The anwer i C. We are eeking the larget integer n uch n(n +1) 500 or n(n +1) 1000 Since 3 = 104 we ee that n<3. Checking n =31we nd 313 = 99. Thu the integer n we eek i 31. The triangular number aociated with thi value of n i 1 (99) = An 80 mrope i upended at it two end from the top of two 50 m agpole. If the lowet point to which the mid-point of the rope can be pulled i 36 mfrom the ground, then the ditance, in metre, between the agpole i:

3 9 Solution. The anwer i C. Inorder for the rope to be at the lowet poible point, that point mut be the middle of the rope. Thu we are faced with olving a right-angled triangle with hypotenue 40 m and one ide of 50 ; 36 = 14 m. By the Theorem of Pythagora the third ide (x in the diagram at the right) i p 40 ; 14 = p 1404 = 6 p 39. The ditance between the two agpole i x =1 p 39. x - 50 m 50 m 36 m 8. Atacertain party, the rt time the door bell rang 1 guet arrived. On each ucceeding ring two more guet arrived than on the previou ring. After 0 ring the number of guet at the party wa: Solution. The anwer i E. Let a n be the number of people who arrive at the n th ring of the door bell. Then a n =n ; 1. Letb n be the number of people who have arrived after the n th ring of the door bell. Then we have Thi can be rewritten a b 1 = 1, b n+1 = b n + a n+1 for n 1, = b n +n +1. b 1 = 1, b n+1 ; b n = n +1 for n 1. If we write out the rt 0 of thee we get b 1 = 1, b ; b 1 = 3, b 3 ; b = 5,. b 0 ; b 19 = 39. When we add all 0 of the above equation together we get b 0 = = 1 (0) ( 1+19 ) = 400, where we have ued the well-known formula for the um of an arithmetic progreion with n term, having rt term a and common dierence d: n(a +(n ; 1)d). 1

4 30 9. An operation i dened uch that The value of (;1) i: A B = A B ; B A Solution. The anwer i C. According to the denition of the operation, we have (;1) = ;1 ; (;1) = 1 ; 1=; Three circle with a common centre P are drawn a hown with PQ = QR = RS. The ratio of the area of the region between the inner and middle circle (haded with quare) to the area of the region between the middle and outer circle (haded with line) i: P Q r r R r rs Solution. The anwer i D. Let a be the length of PQ, QR, andrs. Then the radii of the 3 circle are a, a, and 3a. The area between the inner and middle circle i then (a) ; a =3a, and the area between the middle and outer circle i (3a) ; (a) =5a. Thu the ratio we want i 3a :5a = 3 5. Part B Final Round May 5, (a) How many 3-digit number can be formed uing only the digit 1,, and 3 if both of the following condition hold: (i) repetition i allowed (ii) no digit can have alarger digit to it left. (b) Repeat for a 4-digit number uing the digit 1,, 3, and 4. Solution. (a) For thi part of the quetion, the implet method i imply to lit all the poible number. In increaing order they are: 111, 11, 113, 1, 13, 133,, 3, 33, and 333, for a total of 10 number. (b) Again, mot junior tudent will imply try to lit all the poible integer. In increaing order they are:

5 , 111, 1113, 1114, 11, 113, 114, 1133, 1134, 1144, 1, 13, 14, 133, 134, 144, 1333, 1334, 1344, 1444,, 3, 4, 33, 34, 44, 333, 334, 344, 444, for a total of 35 number. 3333, 3334, 3344, 3444, and 4444, A more ophiticated approach (which can be generalized) follow: We rt dene n(k d) to be the number of k-digit integer ending with the digit d and atifying the two condition (i) and (ii) in the problem tatement. Since a k-digit number ending with the digit d conit of appending the digit d to all (k ; 1)-digit number ending with a digit le than or equal to d, we have n(k d) = n(k ; 1 1) + n(k ; 1 ) + + n(k ; 1 d) () Furthermore, we alo have n(1 d)=1for all digit d and n(k 1) = 1 for all integer k. The relationhip () allow u to create the following table of value for n(k d): d k Each entry in the table i the um of the entrie in the previou row up to and including the column containing the given entry (note the preence of Pacal' Triangle in the table). From that table, the anwer to part (a) and (b) are: (a) : n(3 1) + n(3 ) + n(3 3) = = 10, (b) : n(4 1) + n(4 ) + n(4 3) + n(4 4) = = 35. Clearly thi table could have beenextended to deal with any number k and with any digit d 9.. The quare ABCD i incribed in a circle of radiu one unit. ABP i a traight line, PC i tangent to the circle. Find the length of PD. Make ure you explain thoroughly how you got all the thing you ued to nd your olution! P B C q O D A

6 3 Solution. Since ABCD i a quare, the line AC and BD are perpendicular. Since the circle had radiu 1 unit, the Theorem of Pythagora tell u that AB = BC = CD = DA = p. The tangent PC at C i perpendicular to the diameter AC thu \PCB = 45. Since PA? BC we alo have \CPB = 45. Thi make 4PBC iocele, which mean that PB = BC = p. Applying the Theorem of Pythagora to 4AP D we have PD = AP + AD = ( p ) + p = 8 + = 10, from which we ee that PD = p If a diagonal i drawn in a 3 6 rectangle, it pae through four vertice of maller quare. How many vertice doe the diagonal of a rectangle pa through?.r r r r Solution. Since the rectangle ha it ide in the proportion 3:,wewillconider rt looking at a 3 rectangle, in which there are vertice which lie on the diagonal. In the original rectangle we need only conider the fteen 3 rectangle which traddle the diagonal in quetion. The lower left of thee ha it lower leftmot vertex on the diagonal, and each of thee 3 rectangle add a further vertex to the count for it upper rightmot corner. Thi give u a total of = 16 vertice on the diagonal. 4. Leta and b be any real number. Then (a;b) i alo a real number, and conequently (a ; b) 0. Expanding give a ; ab + b 0. Ifwe add ab to both ide of the inequality, we get a + b ab. Thu, for any real number a and b, wehave a + b ab. Prove that for any real number a, b, c, d: (a) abcd b c + a d. (b) 6abcd a b + a c + a d + b c + b d + c d. Solution. (a) We will ue the proof in the problem tatement a a model. Conider bc ; ad. Clearly (bc ; ad) 0. Expanding give b c ; abcd + a d 0. Thi i eaily rearranged to yield b c + a d abcd. (b) We will ue part (a) to prove part (b). Since a, b, c, and d are arbitrary real number, the inequality in part (a) remain true for any rearrangement of the letter in particular we have: abcd b c + a d, abdc b d + a c, adcb d c + a b.

7 33 Recognizing that multiplication i commutative for real number, we can reorganize the product in each of the above inequalitie and um the three inequalitie to get the deired reult. 5. A circular coin i placed on a table.then identical coin are placed around it o that each coin touche the rt coin and it other two neighbour. (a) If the outer coin have the ame radiu a the inner coin, how that there will be exactly 6 coin around the outide. (b) If the radiu of all 7 coin i 1, nd the total area of the pace between the inner coin and the 6 outer coin. Solution. (a) Let u place (outer) coin next to the original coin o that they touch each other. Then the centre of the 3 coin form an equilateral triangle with ide length equal to twice the radiu of a ingle coin. Therefore the angle between the centre of the (outer) coin meaured at the centre of the rt coin i 60. Since 6 uch angle make upafullrevolution around the inner coin, we can have exactly 6 outer coin each touching the original (inner) coin and alo touching it other two neighbour. (b) There are 6 non-overlapping pace whoe area we mut add each i found between 3 coin which imultaneouly touch other, and whoe centre form the equilateral triangle mentioned in part (a) above. Thi equilateral triangle ha ide length, incewearegiven the radii of the coin a 1. Our trategy to compute the area of one uch pace ito nd the area of the equilateral triangle and ubtract the area of the 3 circular ector found within the triangle. The altitude of the equilateral triangle with ide length can be eaily found (Theorem of Pythagora) a p 3. Thu the area of the triangle itelf i 1 p 3= p 3. The area of a ingle coin i 1 =. The circular ector within the equilateral triangle are each one-ixth of the area of the coin thereare 3 uch ector which give u a total area of one-half the area of a ingle coin to be ubtracted from the area of the equilateral triangle. Thu the area of a ingle pace i p 3 ; (=). Since there are 6 uch pace, we have atotal area of 6 p 3 ; 3 quare unit. That complete the Skoliad Corner for thi iue. Send me uitable contet material for future ue in CRUX with MAYHEM.