No. 39. R.E. Woodrow. This issue we give another example of a team competition with the problems

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1 282 THE SKOLIAD CORNER No. 39 R.E. Woodow This issue we give anothe example of a team competition with the poblems of the 998 Floida Mathematics Olympiad, witten May 4, 998. The contest was oganized by Floida Atlantic Univesity. My thanks go to John Gant McLoughlin, Memoial Univesity of Newfoundland fo sending me the poblems. FLORIDA MATHEMATICS OLYMPIAD TEAM COMPETITION May 4, 998. Find all integes x, if any, such that 9 <x<5 and the sequence, 2, 6, 7, 9, x, 5, 8, 20 does not have thee tems in aithmetic pogession. If thee ae no such integes, wite \NONE." 2. A sequence a, a 2, a 3 ;::: is said to satisfy a linea ecuence elation of ode two if and only if thee ae numbes p and q such that, fo all positive integes n, a n+2 = pa n+ + qa n. Find the next two tems of the sequence 2, 5, 4, 4, :::, assuming that this sequence satises a linea ecuence elation of ode two. 3. Seven tests ae given and on each test no ties ae possible. Each peson who is the top scoe on at least one of the tests o who is in the top six on at least fou of these tests is given an awad, but each peson can eceive at most one awad. Find the maximum numbe of people who could be given awads, if 00 students take these tests. 4. Some pimes can be witten as a sum of two squaes. We have, fo example, that 5 = ; 3 = ; 7 = , 29 = ; 37 = ; and 4 =

2 283 The odd pimes less than 08 ae listed below; the ones that can be witten as a sum of two squaes ae boxed in. 3, 5, 7,, 3, 7, 9, 23, 29, 3, 37, 4, 43, 47, 53, 59, 6, 67, 7, 73, 79, 83, 89, 97, 0, 03, 07. The pimes that can be witten as a sum of two squaes follow a simple patten. See if you can coectly nd this patten. If you can, use this patten to detemine which of the pimes between 000 and 050 can be witten as a sum of two squaes; thee aeve of them. The pimes between 000 and 050 ae 009, 03, 09, 02, 03, 033, 039, 049. No cedit unless the coect ve pimes ae listed. 5. The sides of a tiangle ae 4, 3, and 5. Find the adius of the inscibed cicle. 6. In Athenian ciminal poceedings, odinay citizens pesented the chages, and the 500-man juies voted twice: st on guilt o innocence, and then (if the vedict was guilty) on the penalty. In 399BCE, Socates (c469{399) was chaged with dishonouing the gods and coupting the youth of Athens. He was found guilty; the penalty was death. Accoding to I.F. Stone's calculations on how the juos voted: (i) Thee wee no abstentions; (ii) Thee wee 80 moe votes fo the death penalty than thee wee fo the guilty vedict; (iii) The sum of the numbe of votes fo an innocent vedict and the numbe of votes against the death penalty equalled the numbe of votes in favou of the death penalty. a) How many of the 500 juos voted fo an innocent vedict? b) How many of the 500 juos voted in favou of the death penalty? 7. Find all x such that 0 x and tan 3 x, + cos 2 x, 3 cot 2, x You answe should be in adian measue. = 3. Last issue we gave the poblems of the Newfoundland and Labado Teaches' Association Mathematics League. Hee ae the answes.

3 284 NLTA MATH LEAGUE GAME 998{99. Find a two-digit numbe that equals twice the poduct of its digits. Solution. Denote the numbe by ab; We get 0a + b = 2ab. Tying a = ;2;::: ;9, the only integal solution is with a = 3, b = 6and 36 = The degee measues of the inteio angles of a tiangle ae A, B, C whee A B C. IfA,B, and C ae multiples of 5, how many possible values of (A; B; C) exist? Solution. Let A =5m,B=5n, and C =5p. Then 5m +5n+ 5p = 80 so m + n + p =2and m n p. Since m, n we have m + n 2 and p 0. Also 3p 2 so p 4. Fo p xed, 4 p 0 we have m + n =2,p,om=2,p,n. This leads to the solutions Thee ae twelve solutions. p n m Place an opeation (+;,; ; ) in each squae so that the expession using, 2, 3,, 9equals = 00. You may also feely place backets befoe/afte any digits in the expession. Note that the squaes must be lled in with opeational symbols only. Solution. Hee is one solution: ((( ) 6), 7) = 00. How many solutions ae thee? 4. A, B and C ae points on a line that is paallel to anothe line containing points D and E, as shown. Point F does not lie on eithe of these lines. A B C F D How many distinct tiangles can be fomed such that all thee of its vetices ae chosen fom A, B, C, D, E, and F? E

4 285 Solution. Any choice of thee of the six vetices detemines a tiangle except when they lie on a line; that is, except fo the one choice fa; B; Cg. The total is thus, 6 3, =20,=9. 5. Michael, Jane and Bet enjoyed a picnic lunch. The thee of them wee to contibute an equal amount of money towad the cost of the food. Michael spent twice as much money as Jane did buying food fo lunch. Bet did not spend any money on food. Instead, Bet bought $6 which exactly coveed his shae. How much (in dollas) of Bet's contibution should be given to Michael? Solution. Bet bought $6, which exactly coveed his shae, so the total cost of food is 3 $6 = $8. Now Michael spent twice as much as Jane so he spent $2 and she spent $6. The total amount of $6 bought by Bet should go to Michael. 6. Two semicicles of adius 3 ae inscibed in a semicicle of adius 6. A cicle of adius R is tangent to all thee semicicles, as shown. Find R. R 3 R R h Solution. Join the centes of the two smalle semicicles and the cente of the cicle. This foms an isosceles tiangle with equal sides 3+Rand base 6 units. Call the altitude of this tiangle h. The altitude extends to a adius of the lage semicicle, so h + R =6. By Pythagoas, h =(R+3) 2,so The adius of the small cicle is 2. (6, R) = (R+3) 2, 36, 2R + R 2 +9 = R 2 +6R+9, 36 = 8R, 2 = R. 7. If5 A =3and 9 B = 25, nd the value of AB. Solution. Now 5 A =3,so5 2A =3 2 =9and so 2AB =3and AB = AB = (5 2A ) B = 9 B = 25 = 5 3,

5 The legs of a ight angled tiangle ae 0 and 24 cm espectively. Let A = the length (cm) of the hypotenuse, B = the peimete (cm) of the tiangle, C = the aea (cm 2 ) of the tiangle. Detemine the lowest common multiple of A, B, and C. Solution. Then A = p = 26 B = = 60 C = = 20 lcm(26; 60; 20) = = A lattice point is a point (x; y) such that both x and y ae integes. Fo example, (2;,) is a lattice point, wheeas,, 3; 2 and,, 3 ; 2 3 ae not. How many lattice points lie inside the cicle dened by x 2 + y 2 =20? (Do NOT count lattice points that lie on the cicumfeence of the cicle.) Solution. Since 4 2 < 20 < 5 2 we have that,4 x 4. Fo a xed x in the ange we must have, p 20, x 2 <y< p 20, x 2 fo intege x, y solutions coesponding to inteio points of the cicle. p x 20, x p 2 2 p 6 p 9 0 p 20 y,, 0,,3, :::,3,3, :::, 3,4, :::,4,4, :::,4 Lattice pts. 2 3=6 27=4 27=4 29=8 9 The total numbe is then =6. 0. The quadatic equation x 2 + bx + c =0has oots and 2 that have a sum which equals 3 times thei poduct. Suppose that ( +5)and ( 2 +5)ae the oots of anothe quadatic equation x 2 + ex + f =0. Given that the atio of e : f =:23, detemine the values of b and c in the oiginal quadatic equation. Solution. Now + 2 =,b and 2 = c, soas + 2 =3 2 we get 3c =,b. Similaly we have Thus =,e, ( + 5)( 2 +5) = f. 0, b =,e and b, 0 = e, 2 +5( + 2 )+25 = f, c,5b+25 = f.

6 287 Fom e f =, 23(b, 0) = c, 5b +25. Using b =,3c 23 23(,3c, 0) = c +5c+25, so that, 69c, 230 = 6c +25,,255 = 85c,,3 = c, and 9 = b. The quadatic is x 2 +9x,3=0. RELAY R. Opeations and ae dened as follows: A B = AB + B A A + B and A B = AB, B A A, B. Simplify N =(32) (3 2). Wite the value of N in Box #oftheelay answe sheet. Solution. 3 2 = = 32, 2 3 N = (3 2) (3 2) = 3, 2 = = 9, = = 7 5, =,, () 7= =. R2. A squae has a peimete of P cm and an aea of Q sq.cm. Given that 3NP =2Q, detemine the value of P. Wite the value of P in Box #2 of the elay answe sheet. Solution. Fom R, N =,so3p =2Q. Also the side length is s = P, 4 so Q = ( P 4 )2 = P2 P2. So 3P =2 o P = P 2. This gives P =0o P =24. We use P =24, assuming the squae is not degeneate. R3. List all two-digit numbes that have digits whose poduct is P. Call the sum of these two-digit numbes S. Wite the value of S in Box #3 of the elay answe sheet. Solution. P =24 = 2 2=46=83. The two-digit numbes ae 46, 64, 38 and 83; so S = 23.

7 288 R4. How many integes between 6 and 24 shae no common factos with S that ae geate than? Solution. Now 23 = 2 = 7 3. The numbes between 6 and 24 that shae nocommon facto with 23 ae Thee ae 8of them. Find the maximum value of 8, 0, 3, 6, 7, 9, 20, 23. TIE-BREAKER f (x) = 4, p x 2, 6x +25. Solution. f (x) = 4, p x 2, 6x +25 = 4p x 2, 6x +9,9+25 = 4, p (x, 3) Fo f (x) to be maximum we want (x, 3) 2 +6to be minimum. This occus when x =3, and f (3) = 4, p 6 = 4, 4 = 0. That completes the Skoliad Cone fo this numbe. Send me you comments, suggestions, and especially suitable mateial fo use in the Cone. Announcement The second volume in the ATOM seies has just been published. Algeba Intemediate Methods by Buce Shawye. Contents include: Mathematical Induction, Seies, Binomial Coecients, Solutions of Polynomial Equations, and Vectos and Matices. Fo moe infomation, contact the CMS Oce addess on outside back cove.

No. 32. R.E. Woodrow. As a contest this issue we give the Junior High School Mathematics

No. 32. R.E. Woodrow. As a contest this issue we give the Junior High School Mathematics 334 THE SKOLIAD CORNER No. 32 R.E. Woodow As a contest this issue we give the Junio High School Mathematics Contest, Peliminay Round 1998 of the Bitish Columbia Colleges which was witten Mach 11, 1998.