FORTY-FIFTH ANNUAL OLYMPIAD HIGH SCHOOL PRIZE COMPETITION IN MATHEMATICS. Conducted by. The Massachusetts Association of Mathematics Leagues (MAML)

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1 FORTY-FIFTH NNUL OLYMPID HIGH SCHOOL PRIZE COMPETITION IN MTHEMTICS Conducted by The Massachusetts ssociation of Mathematics Leagues (MML) Sponsored by The ctuaries Club of Boston SOLUTIONS Tuesday, October 8, 008

2 Solutions 1 ns: Let equal the number of juniors Then = (9) + (7) = 16( + 1) = 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 : Since ΔBC is isosceles, m B = m C = = 66 et angle at = = 1 and et angle at C = = 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 = = 6 B s F E t 8 10 D C

3 Therefore, FB DE = = = 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 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)( ) = (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 = R = = 0 By similar triangles, R = r r = 0

4 Thus, V = 1 0 π = 18π 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 + ( ) ( ) 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 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 = ( ) log = = 6 = ns: 1 1 p + = 1+ 1 q r + s If all variables are positive integers, then p = Eliminate p and invert both sides: q + = = + q = 1 r s 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

5 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 = 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 n, where n > P() =, P() = 6, P() = 6 7, P(6) = 6 7 8, 6 P(7) = ( n+ 1)( n+ ) In general, P(n) = Therefore, P(6) = = 7() = 188

6 B 18 ns: 11π + h = 17 and h + (1 ) = 10 Thus, h = 17 = 10 (1 ) 89 = = = 1 h = 8 and S W 17 P h Q 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 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

7 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 + = i 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 y = y = 1 = y + 7y + 1 = 0 y = 7± ±i 11 1± i 11 = = Thus, the solution set contains the ordered pairs: 1 + i i 11 1 i 11 7 i 11, 1,,, ( ) and

8 ns: 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 = 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 y = = y = lternate approach to finding y = 8

9 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 = y = + 6 is y = + 6 = + 6= + 6= 8 8 Thus, the required distance is ( 7 ) = + 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 Substituting, l = = + or l = For integer values of, the latter epression is never an integer and is rejected P y F C D

10 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 B 19 B 0 B 1 E E B D D

MT - w A.P. SET CODE MT - w - MATHEMATICS (71) GEOMETRY- SET - A (E) Time : 2 Hours Preliminary Model Answer Paper Max.

MT - w A.P. SET CODE MT - w - MATHEMATICS (71) GEOMETRY- SET - A (E) Time : 2 Hours Preliminary Model Answer Paper Max. .P. SET CODE.. Solve NY FIVE of the following : (i) ( BE) ( BD) ( BE) ( BD) BE D 6 9 MT - w 07 00 - MT - w - MTHEMTICS (7) GEOMETRY- (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40 [Triangles