THE 08 09 KENNESW STTE UNIVERSITY HIGH SHOOL MTHEMTIS OMPETITION PRT I MULTIPLE HOIE For ech of the following questions, crefully blcken the pproprite box on the nswer sheet with # pencil. o not fold, bend, or write stry mrks on either side of the nswer sheet. Ech correct nswer is worth 6 points. Two points re given if no box is mrked. Zero points re given for n incorrect nswer or if multiple boxes re mrked. Note tht wild guessing is likely to lower your score. When the exm is over, give your nswer sheet to your proctor. You my keep your copy of the questions. NO LULTORS 90 MINUTES. Most of Hrry Potter s clssmtes t Hogwrts School of Witchcrft nd Wizrdry pln to ttend the school s nnul Hlloween Prty. Wht is the minimum number of students who would hve to ttend in order to gurntee tht there re t lest students born on the sme dy of the week nd in the sme month of the yer. () 8 () () () 68 (E) 69. two-digit positive integer N hs n interesting property. The sum of the digits of N, N itself, nd the number obtined by reversing the digits of N form n rithmetic sequence in tht order. ompute the sum of ll such integers N. () () 96 () 08 () 0 (E). The digrm t the right consists of eight line segments with djoining endpoints. ompute the sum of the degree mesures of ngles,,,, E, F, G, nd H. () 0 () 60 () 0 () 80 (E) 900 F E G H. The number of miles per hour n utomobile is trveling is less thn the number of feet per second it is trveling. ompute the number of miles per hour in the rte of the utomobile. (Note: mile =,80 feet) () () 8 () 0 () (E). If one of the girls t Hlloween prty leves, then 0% of the people remining t the prty re girls. If, insted, nother girl rrives t the prty, then % of the people t the prty re girls. How mny boys re t the prty? () 6 () 8 () 0 () (E)
6. The symbols rr nd ss represent two-digit numbers in bses r nd s respectively. If rr = ss wht is the smllest possible positive vlue for r + s? () 8 () 9 () 0 () (E). How mny integers x stisfy the eqution (xx xx ) xx+ =? () () () () (E) 8. cr trvels mile t miles per hour. Wht must its verge speed be, in miles per hour, for the next miles in order to verge 60 miles per hour for the miles? () 68 () 0 () () (E) None of these 9. One of the roots of the eqution xx xx + 6xx + cc = 0 is the verge of the other two roots. ompute the vlue of c. () 6 () 0 () 86 () (E) 0 6 6 6 6 0. Let p nd q be two prime numbers for which + =, with p < q. ompute p q p q the lrgest possible vlue of the product pq. () () 0 () 99 () (E). The sides of tringle hve lengths n, n, nd n +, nd its re is nn nn. Wht is the vlue of n? () (). () 6 () 6. (E). The vlue of x exceeds y by the squre root of y. The vlue of y exceeds z by the squre root of z. If the vlue of x is one greter thn the vlue of z, compute the sum x + y + z. () () 6 () () (E) 8. Which of the following vlues of n stisfies the eqution ( + i) n = n, where i is the imginry unit? () () 8 () () 6 (E) 0. Given the pir of equtions x y = nd x + y = b, where b is n integer nd b 00. For how mny vlues of b will x + y be the squre of n integer? () () () () (E)
. The line y = mx + (m > 0) intersects the ellipse xx + yy = in exctly one point. If θ is the cute ngle this line mkes with the x-xis, compute sin θθ. () () () () (E) 6. ebbie nd on were compring their stcks of pennies. ebbie sid If you gve me certin number of pennies from your stck, then I d hve six times s mny s you, but if I gve you tht number, you d hve one-third s mny s me. Wht is the smllest number of pennies tht ebbie could hve hd? () 8 () () 0 () (E). The prllel sides of trpezoid re nd inches long, nd the nonprllel sides re nd inches long. ompute the re of the trpezoid. (). ().8 (). () 8. (E) None of these 8. plindrome is number tht reds the sme forwrds nd bckwrds. For exmple, is plindrome. The digits,,,,,,,, nd re ll to be used in writing nine digit number. If the digits re rndomly plced in row, wht is the probbility tht the number will be plindrome? () () () () 9 (E) None of these 9. In the figure, M nd N re midpoints of two djcent sides of squre. ompute tn( MN). () () () () (E) M N 0. The points P ( x, y) nd Q ( x, y ) lie on the grph y = log xx, nd x > x. The horizontl line through the midpoint of segment PQ intersects the grph of y = log xx t the point N x, ). Which of the following is n equivlent expression for x? ( y () (xx + xx ) () log ( xx + xx ) () (log xx + log xx ) () (xx )(xx ) (E) x x
. In r. Grner s mth clss, the number of students receiving s is t lest one-fifth of the number of students receiving s nd t most one-sixth the number of students receiving s. The number of students receiving s or s is t lest. If,, nd re the only grdes given in the clss, wht is the minimum number of students receiving s? () 8 () 0 () () (E). In the digrm, semicircle is inscribed in qurter circle. second semicircle, tngent to the first, is inscribed in the qurter circle, s shown. ompute the rtio of the dimeter of the smller semicircle to the dimeter of the lrger semicircle. () () () 8 () 0 (E) None of these. Let x nd y be two positive integers such tht xx + yy is exctly 08 more thn xx + yy. ompute the sum of ll possible vlues of x. () () 6 () 60 () 68 (E) 696. Let x represent the gretest integer less thn or equl to x. For exmple, π = nd 8 = 8. How mny positive integers n, with n 08, yield solution for x (where x is rel) in the eqution x + x + x = n? () 6 () 98 () 60 () 08 (E) None of these P. The digrm shows cube whose edge is inches long. line segment is drwn from vertex of the cube to point P on the min digonl,, so tht the length of segment P is lso inches. ompute the rtio P P. () () () () (E) 8
Solutions. E The gretest number of people group cn hve with no two born in the sme month nd on the sme dy of the week is ()() = 8. The gretest number of people group cn hve with no three born in the sme month nd on the sme dy of the week is (8) = 68. dd one more person, nd there must be t lest born in the sme month nd on the sme dy. Therefore, the nswer is 69.. Let N = 0t + u. Then the rithmetic sequence is t + u, 0t +u, 0u + t. Then 0t +u (t + u) = 0u + t (0t +u) 9t = 9u 9t u = t. Therefore, the only possible vlues of N re,, 6, nd 8, with sum of 0.. onstruct segment E. The sum of the mesures of the ngles of pentgon E is 80( ) = 0. The sum of the mesures of ngles F, G, nd H is 60 x. The sum of the mesures of HE nd FE is 80 x. The desired sum is 0 (80 x) + (60 x) = 0 F x x E G H. mile per hour = 88 feet per minute = feet per second. Therefore, x mph = x feet per x second. We must solve x = x = nd x =.. E Let w be the number of girls t the prty nd m be the number of boys. w w + Then = w m = nd = w m =. w + m w + m + Solving the two equtions, w = nd m =. Therefore, there were boys t the prty. 6. E First note tht r > nd s >. Since r = r + nd s = s +, we get r + = s + nd r = s +. The first vlue of s > for which this is true is s = 9 nd r =. Therefore, the desired sum is. b. There re three wys in which cn equl when nd b re integers: (i) =, (ii) = nd b is even, or (iii) b = 0 nd 0. (i) x x = x = or. (ii) x x = nd x + is even. x x = x = 0 or. If x = 0, x + is even, but if x =, x + is odd. (iii) x + = 0 nd x x 0. When x =, x x = 0. Therefore, there re totl of solutions (nmely,,, 0, nd ). 8. Let x be the speed for the next two miles. verge speed = = 60 + x nd solving, x =. totl distnce. Therefore, totl time
9. In ny eqution of the form x + x + bx + c = 0 with roots p, q, nd r, (i) p + q + r =, (ii) pq + pr + qr = b, nd (iii) pqr = c The three roots of the given eqution xx xx + 6xx + cc = 0 my be represented by p d, p, nd p + d. Using (i) (p d) + p + (p + d) = nd p =. The vlue of c cn be found by substituting in the eqution, or using (ii) ( d) + ( + d) + ( d)( + d) = 6 nd d = ±. Therefore, the three roots re,, nd 8. Using (iii) c = ( )()(8) = 0. 6 6 6 6 0. + = p q p q 6q + 6p = 6 p + q = 6. The combintions of two primes (p, q) tht stisfy this eqution with p < q re (, ), (, 9), (, ), nd (, 9). Of these the lrgest product pq is ()(9) =.. The semiperimeter of the tringle is n n n n n n = + n. Using Heron s Formul, Simplifying, n 6n = 0 giving (n +)(n 6) = 0, so tht n = 6.. x = y + y, y = z + z, nd x = z + z + = y + y = z + z + y = z + z + z z = z + z. Squring both sides of the lst eqution, z + z = z + z or z = nd z =. 9 0 Therefore, x = z + =, nd y = z + z =. Hence, x + y + z = 0 + + =. 9 9 9 9 9. Look for pttern: ( + i) = i ( + i) = (i) = ( + i) 8 = ( ) = 6 ( + i) 6 = 6 Therefore, n = 6 b. Eliminte y from the given equtions to obtin x = 6 + b nd x = +. b Eliminting x from the given equtions nd solving for y, we obtin y = +. b Therefore, x + y =. In order for x + y to be n integer, b must be multiple of. The only multiples of less thn or equl to 00 for which x + y is squre re: b =, x + y = = ; b = 6, x + y = 9 = ; b = 0, x + y = 9 = ; nd b = 69, x + y = 6 = 8. Thus, there re.
. E Substituting y = mx + into x + y = nd rerrnging terms, we obtin (mm + )xx + 8mmmm + = 0 In order for the two grphs to intersect exctly once, the discriminnt must be zero. ( 8m) ( m + )( ) = 6m = 0 nd m =. Since the slope of line is the tngent of the ngle the line mkes with the positive x-xis, tn θ = from which sin θ =. θ 6. E Let nd b be, respectively, the number of pennies tht ebbie nd on hd in their stcks, nd let x be the certin number of pennies. From the given informtion, we obtin the following two equtions: + x = 6( b x) nd x = (b + x). From the first eqution, = 6b x, nd, from the second eqution, =b +x. Therefore, 6b x =b + x, from which b = x. Since, b, nd x re required to be positive integers, the smllest possible vlue for x is. Then b = nd = 6() () =. Therefore, is the smllest number of pennies tht ebbie could hve hd.. Method : In the digrm, construct line through prllel to leg. Qudrilterl E is prllelogrm, mking E -- right tringle. The re E of E is ()() = 6. Thus, the ltitude from to E hs length =.. Therefore, the re of trpezoid is. (.)( + ) =.8 Method : In the digrm, drw ltitudes from nd nd represent the lengths of the segments s shown. Then, h = 9 xx = 6 ( xx) from which xx = 9 =.8. Thus, h = 9 (.8) =., nd the re of trpezoid is (.)( + ) =. 8 h h x -x
9! 8. The number of distinct rrngements of the nine digits is = 80. (!)(!)(!) plindrome will hve in the fifth position nd two s nd one ech of the s nd s in positions. Positions 6 9 will mirror. The number of distinct wys of! rrnging these numbers is =. The probbility tht the number is plindrome is! =. 80 9. Method : Since M N, M N. Represent the mesure of ech s α. Without loss of generlity, let the length of the sides of the squre be. Then M = N = nd N = M =, nd cos α =. Sin ( MN) = sin(90 α) = cosα = cos α = Thus cos( MN) = nd tn( MN) =. =. α α M N Method : M Let m( M) = β. onstruct MP. Then tn α = nd tn β =. Since m( MN) = β α, tn( MN) = tnβ tnα +tnβtnα = +() = α P N 0. The coordintes of the midpoint of segment PQ re the verge of the coordintes of P x + x log x + log x nd Q,,. Since the line through the midpoint of segment PQ nd point N is horizontl, the y-coordinte of point N is the sme s the y-coordinte of log x + log x the midpoint. Therefore, y = log x =. Using the lws of logrithms, log x = log [( x )( x )], from which x = x )( ). ( x. E Let,, nd represent the number of students receiving ech of the three grdes. Then 6 nd +. Since,, nd 6 +, then 6 nd. 6 ut must be positive integer, so. Since 6, we hve nd. The vlues =, =, nd = 0 re consistent with 6 the given informtion. Therefore, the minimum number of students receiving s is.
. Represent the dimeter of the lrger semicircle s, nd the dimeter of the smller semicircle s b. Note tht the line joining the centers of the two semicircles psses through their common point. Using the Pythgoren Theorem, ( b) + = ( + b) b + b + = + b + b. b Simplifying, = 6b =. -b b b b. We re looking for the solutions to the eqution ( + y) ( x + y ) = 08 Rewriting the left side s ( x y ) ( x y) x. nd fctoring, the eqution becomes ( x y)( x + y ) = 08. Since x nd y, then x + y is positive, nd thus x y is positive. Therefore, x y nd x + y re positive integrl fctors of 08, with the first smller thn the second. There re two possibilities. Either x y = nd x + y = 08, or x y = nd x + y = 009 (note tht 009 is prime). In the first cse, x = 00 nd y = 009. In the second cse, x = 06 nd y = 0. The desired sum is 6.. Let k be non-negtive integer. Let f(x) = x + x + x. If k x < k +, then f(x) = 6k. If k + x < k +, then f(x) = 6k +. If k + x < k +, then f(x) = 6k +. If k + x < k +, then f(x) = 6k +. Therefore, there is only solution if n is congruent to 0,,, or mod 6. Thus, there re (06) + = 6 such 6 positive integers n.. Method : rw segment. Using the Pythgoren Theorem on isosceles right tringle, =. Since segments nd re perpendiculr, right tringle, nd =. Let P = α. Then cos α = = =. onstruct the ltitude M to side P of isosceles P. Then in right tringle M, cos α = = M = M, nd M = = (P). Thus P = P =, P =, nd the desired rtio =. P P M α
Method : rw segment. s in method, =, =, nd cos α =. Since P is isosceles, P = α. Then, P cos θ = cos(80 α) = cos α = (cos α ) = = Using the Lw of osines in P, P = + P ()(P)c os θ = + ()() = nd P =. Thus, P =, nd the desired rtio P =. P. θ α