FORTY-FIFTH NNUL OLYMPID HIGH SCHOOL PRIZE COMPETITION IN MTHEMTICS 008 009 Conducted by The Massachusetts ssociation of Mathematics Leagues (MML) Sponsored by The ctuaries Club of Boston SOLUTIONS Tuesday, October 8, 008
Solutions 1 ns: Let equal the number of juniors Then 1 9 + 7 = 8 1 + 0(9) + (7) = 16( + 1) 190 + 10 = 11( + 1) = ns: 1 Let (B, G) denote the original number of boys and girls, respectively Then: B B 6 B (1) = and () = From (1) we have G = G G B 6 8B Substituting in () we have = B 18= 9B = 8B B = B / G = 18 () + = 1 1 ns: 180 Let n be a typical number in set S ccording to the stated rule, n becomes 6(n 6) + 6 = 6n 0 and the sum S becomes 6S 0p (a, b) = (6, 0) ab = 180 ns: 1 1 = 1 1 = ( 1) ( ) ( 1) + = 1 1 = 1 ns: 19 : 180 8 Since ΔBC is isosceles, m B = m C = = 66 et angle at = 180 8 = 1 and et angle at C = 180 66 = 11 Thus, the required ratio is 11 : 1 = 7 : 66 = 19 : 6 ns: 1 0 Since D is a midpoint, DC = t 6 1 ΔCDE ~ ΔCB = t = 8 The area of ΔCB may be computed as 1 68 or 1 10 s, implying s = 68 10 = 6 B s F E t 8 10 D C
1 96 7 Therefore, FB DE = = = 1 0 0 7 ns: 777 Let 10 + y denote the original two-digit number Then: (10 + y)(y) = yyy = 100y + 10y + y = 111y (10 + y)() = 111 = (7) Thus, = and 10 + y = 7 y = 7 and the three digit number is 777 8 ns: 7, 1 st pass: (N) + N = 10N nd pass: 10N(N) + N = 0N + N = 0 10N 7 + N 1 = (N 7)(N + ) = 0 N =, 9 ns: ( ) + ( 1)( ) + 1 9 = (9 1 ) = ( ) + ( 1)( ) + ( ) ( ) = ( )( ) + + = 10 ns: 0 Let (G, D) denote the lengths of the giant s step and Jack s step (in feet), respectively Therefore, G = 11D, and without any loss of generality, we may let G = 11 and D = In 1 unit of time, the giant takes steps ( feet) and the dwarf takes 8 steps (16 feet) Suppose it takes T units of time for the giant to catch up to Jack who had an 8 - step head start Then 8() + 16T = T T = 10 In 10 units of time, the giant takes 0 steps 11 ns: 18π Since the ratio of the sides of the two cones is 1 :, then the ratio of the volumes of the cones is 1: 8 Since the volume of the larger cone is 1π cm, the volume of the smaller cone is 1 8 π cm = 18π cm lternate solution: 1 1 π Rh= π Rh= h = 10 16 6 R = = 0 By similar triangles, R = r r = 0
Thus, V = 1 0 π = 18π 1 1 1 or since V' = πr h= πr h' and V = π ( r ) ( h' ) 1 ns: y =± Completing the square we see that C 1 is a circle 0+ + y = 6 + ( ) ( ) 100 100 10 + y = 6 This is a circle with center at (10, 0) and radius 6 Knowing that tangents to a circle are perpendicular to a radius drawn to the point of contact, we notice the 6 8 10 triangle and, therefore, tangents through the origin will have slopes of ± Thus, the equations are: y =± V 1 8 = 1 8π = 18π π rh' 8 6 θ θ C(10, 0) 1 ns: 6 log 6 + log + log 6 = 8 1+ log + + log = 8 6 6 6 ( ) log = = 6 = 6 6 6 6 1 ns: 1 1 p + = 1+ 1 q + 68 1 r + s If all variables are positive integers, then p = 1 1 68 Eliminate p and invert both sides: q + = = + q = 1 r + 1 1 s 1 1 1 Eliminate and invert again: r + s = = + r = and s = Thus, pq + rs = 1() + () = 1 ns: 1 ΔPQT ~ ΔRST and the ratio of corresponding sides is 9 = PT = 8 11 PR
But PB = PR BT = 11 PR = 1 PR PT /11 9 BT = 1/ = 1 S : Q = RB : BP = 1 : ΔPQT ~ ΔBT PQ = PT B BT 9 9 = B = 1 B 1 S P 9 T Q B R 16 ns: a+ b= a + c = 7 (a, b, c) = (,, ) b + c = 6 Using Hero s formula to find the area of ΔBC, = 9()()() = 6 6 Using 1 sin ab θ, = 1 6 6sin B = 6 6 sin B = 1 cos B = or using the law of cosines directly on ΔBC, 9 = + 6 6 cosb = 61 60cosB cosb = 1 B b F a b D a E c c C Now in ΔBD: = + cosb = + = = 17 ns: 188 Let P(n) denote the product 1 1 1 1 1 1 + + + + + + 1+ 6 7 6 n, where n > P() =, P() = 6, P() = 6 7, P(6) = 6 7 8, 6 P(7) = 7 8 9 ( n+ 1)( n+ ) In general, P(n) = 7 7 8 Therefore, P(6) = = 7() = 188
B 18 ns: 11π + h = 17 and h + (1 ) = 10 Thus, h = 17 = 10 (1 ) 89 = 100 1 + 60 = = 1 h = 8 and S W 17 P h Q 10 1- The rotation produces a cylinder with a circular base whose diameter is BR and whose height is B Therefore, volume = π (8) () = 1600π However, we must eliminate two cones that have equal base diameters ( S and BR ), but different heights PW and QT From the diagram we see that PW = 1 and QT = 6 and the required volumes that must be deducted are: 1 1 π(8) 6 + π(8) 1 = 1 (6)(1) π = 8π Therefore, the net volume is 11π 19 ns: 6 t t = 1, the runners are at (0, ) and (, 10) d = t t =, the runners are at (0, 8) and (6, 10) d = 0 t t =, the runners are at (0, 10) and (7, 10) d = 7 d = 6 Clearly a minimum has occurred, but where? ssume the minimum occurs at time t, when the runners are at (0, t) and (t, 10) Let s eamine d d = ( t) + (10 t) = t 80t+ 100 8 Completing the square, t + 6 and a minimum value occurs at t = 8/ Substituting, d = 18 + = 76 + = 900 = 6 = 6 0 ns: 10, B, C, D and E agree to meet every,,, and 6 days respectively There are C = 10 different threesomes {,, 6} meet every 6 days (BE) {,, }, {,, 6}, {,, 6} meet every 1 days (BC, CE, BCE) {,, } meet every 0 days (CD) {,, }, {,, 6}, {,, 6} meet every 0 days (BD, DE, BDE) {,, }, {,, 6}meet every 60 days (BCD, CDE) Thus, we look at: multiples of 6 that are not multiples of 1, 0, 0 or 60-6, 18,,, 66, 78 T R
multiples of 0 that are not multiples of 6, 1, 0 or 60 0, 0, 80, 100 There are no multiples of 1 that are not multiples of 6, 0, 0 and 60 and likewise for multiples of 0 and multiples of 60 Summing, K = 6 + = 10 1+ i 11 7+ i 11 1 i 11 7 i 11 1 ns: (, 1 ),, and, y+ y y = 1 y+ y y = ( + ) 6 y+ 6y y = 18 y+ y y = 6 ( y) = 6 y= = y+ ( )y+ y = 10 y ( + y) = 0 Substituting, (y + )y(y + + y) = 0 ( y+ ) y( y+ ) = 1 y + 6y + 8y 1 = 0 By synthetic division we see that y = 1 is a root and we have a quadratic factor 1 6 8 1 1 7 1 y = 1 1 7 1 0 y = 1 = y + 7y + 1 = 0 y = 7± 9 60 7±i 11 1± i 11 = = Thus, the solution set contains the ordered pairs: 1 + i 11 7 + i 11 1 i 11 7 i 11, 1,,, ( ) and
ns: 167 88 Train Let BCD represent all of the times that Mufasa and the train can arrive at the station Mufasa can catch the train at all points satisfying m t > that lie in the square Let the lower right corner represent the origin In region #1, Mufasa arrives before the train (and has to wait ) long C, the train and Mufasa arrive at the same time In region #, Mufasa arrives within minutes of the train (and gets to board ) In region #, Mufasa arrives more than minutes after the train P(, ) (and misses his ride ) Hence, the probability is 1 1 60 = 167 88 B (m, t) #1 Q(10, ) # # C D Mufasa ns: 7 The slope of C is slope of BN is The equation of BN is y = and the equation of D is = 1 Therefore, the coordinates of H are, The area of ΔBC can be computed using Hero s formula rea = 1(1 1)(1 1)(1 1) = 1(8)(7)(6) = 8 (, 1) 1 N 1 H P(7,y) B(0, 0) C(1, 0) D(, 0) 9 1 abc The radius R of the circumscribed circle can be computed as = 1(1)(1) = 6 (8) 8 Since the center of the circumscribed circle lies on the point of concurrency of the perpendicular bisectors of the sides of the triangle, this center lies on the line = 7, say at 6 the point P(7, y), where y > 0 Therefore, R = = PB = 7 + y 8 6 6 y = = y = 8 8 8 lternate approach to finding y = 8
By finding the perpendicular bisectors of side BC ( = 7) and the perpendicular bisector of C Midpoint of C is 19,6, slope of 1 0 C = = Therefore the slope 9 perpendicular bisector of C is The equation of the perpendicular bisector is 19 y 6 = or 19 y = + 6 The point of intersection of and = 7 19 7 19 1 y = + 6 is y = + 6 = + 6= + 6= 8 8 Thus, the required distance is ( 7 ) 1 + 8 = + 8 = 6 8 6 + 8 = 7 ns: + Let F = y Then: 1 w ( + y ) 1 l ( + y) l = = wl ( y+ l ) ( l y) 17 + y = l l ly l + + y 0 = l ly l B E y+ l 0 0 = l y l = From the fact that both l and w are integers, we conclude that ΔEPF has sides of lengths 8, 1, and 17 Thus, PF = 8 or 1 and y = + 8 or + 1 8+ 1 + Substituting, l = = + or l = For integer values of, the latter epression is never an integer and is rejected P y F C D
ns: 6 In theory, the hour and the minute hands could coincide only on the positive or negative y-ais ctual coincidences occur on the positive y-ais only There are two such occurrences in hours (at M and PM) In theory, the hour hand and the second hand could coincide only on the positive or negative z-ais ctual coincidences never occur since the second hand passes the z-ais at hh:mm:1 and hh:mm: In theory, the minute hand and the second hand could coincide only on the positive or negative -ais ctual coincidences occur Hour hand only on the positive -ais, which corresponds to 1 minutes past the hour There are such occurrences in H hours Thus, the answer is + = 6 z Y M Minute hand S Second hand NSWERS 1 D E D C 6 7 D 8 9 E 10 E 11 1 D 1 B 1 D 1 D 16 17 18 B 19 B 0 B 1 E E B D D