***** O CALCULATORS I THIS ROU D *****

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1 CO TEST 1 OCTOBER 009 ROU D 1 VOLUME & SURFACES ***** O CALCULATORS I THIS ROU D ***** A SWERS A) A one is insribed in a sphere. The radius of the base of the one is 3 and the radius of the sphere is 5. Find the ratio of the volume of the sphere to the volume of the one. A) : B) units C) inhes 3 B) A retangular sheet of ardboard has dimensions of 9 x 11x by units. Squares x units on a side are ut from eah orner of the sheet. The sheet is then folded upward to form an open box. The volume of this box is 560 units 3. What was the perimeter of the original retangular sheet of ardboard? C) A square pyramid has a volume of 108 ubi inhes and the ratio of length of its altitude to the perimeter of its base is 3 : 8. A plane parallel to its base divides the pyramid into two solids one of whih is a smaller pyramid whose slant height is 10. Compute the volume of the smaller pyramid.

2 CO TEST 1 - OCTOBER 009 ROU D PYTHAGOREA RELATIO S I RECTILI EAR FIGURES ***** O CALCULATORS I THIS ROU D ***** A SWERS A) : B) (, ) C) A) In square ABCD, AB = 4, E and F are midpoints of BC and CD respetively. Compute the ratio of the area of AEF to the area of ABCD. B) Given: AD = 8, CD = 6 Compute the ordered pair (x, y). A D y C x B C) A right triangle has a hypotenuse of length 65. If the length of the long leg is inreased by 4 and the length of the short leg is dereased by 8, the length of the hypotenuse is unhanged. What is the perimeter of the original right triangle?

3 CO TEST 1 - OCTOBER 009 ROU D 3 ALG 1: LI EAR EQUATIO S ***** O CALCULATORS I THIS ROU D ***** A SWERS A) B) C) A) A number is 1 more than half of its opposite. Compute the reiproal of this number. B) If 3x 0.06x (10 x) =, ompute [ x]. 4 Note: The greatest integer in x, denoted [x] is the largest integer less than or equal to x. C) A, B, C and D are 4 ollinear points ordered on a line as indiated in the diagram below. AC 4 If AD = 03, BD = 9 and AB 10 =, ompute BC. CD A B C D

4 CO TEST 1 - OCTOBER 009 ROU D 4 ALG 1: FRACTIO S & MIXED UMBERS ***** O CALCULATORS I THIS ROU D ***** A SWERS A) (,, ) B) C) A) A triathlete swims miles in three hours, jogs 10 miles in four hours and biyles 40 miles in 1 1 hours. The average rate for these three events lies between two onseutive integers A and B, loser to one of these values than the other. Let C denote the loser value. Determine the ordered triple (A, C, B). B) If x is 80% of y and y is % of w, then x is k% of w. Determine the value of k to the nearest tenth. a+ b a+ b+ C) If = and = 3, ompute. b a

5 CO TEST 1 - OCTOBER 009 ROU D 5 I EQUALITIES & ABSOLUTE VALUE ***** O CALCULATORS I THIS ROU D ***** A SWERS A) B) (, ) C) A) The omplete set of x-values satisfying the inequality ( x )( x ) 4 9 > 0 B) is more than a multiple of 3, that is, = 3k + for integer values of k. It is also known that is at most 96 and at least 16. The values of k (and only those values) for whih this is true satisfy the relation k a < b, where a and b are integers. Determine the ordered pair (a, b). C) A and B are distint two-digit positive integers with digits reversed. A and B are both prime, with A < B. Let p denote the number of ordered pairs (A, B). Let C = the minimum value of A B and D = the maximum value of A B. How many integers x are there in the range p C < x < p D?

6 CO TEST 1 - OCTOBER 009 ROU D 6 ALG 1: EVALUATIO S ***** O CALCULATORS I THIS ROU D ***** A SWERS A) B) A) Compute (3 5) C) B) Given: a : b = : 3 b : = 4 : 5 a + b + = 70 Compute the value of. C) Let the binary operation ( ) 3 3 Compute 6 * * * 3 4. * be defined as follows: a + ab, when b is a proper fration a * b= b ab, when b is an improper fation

7 CO TEST 1 - OCTOBER 009 ROU D 7 TEAM QUESTIO S ***** CALCULATORS ARE PERMITTED I THIS ROU D ***** A SWERS A) : D) miles B) (,, ) E) units C) (, ) F) (, ) A) A retangular blok of ie ream has dimensions 7, 8 and 9. It is ompletely overed with hoolate (like a Klondike Bar). Suppose it is then ut up into one inh ubes by making uts parallel to the faes. Let A be the number of ubes with or 3 faes overed with hoolate. Let B be the number of ubes with 0 or 1 fae overed with hoolate. Compute the ratio of A : B. B) Several non-right triangles have all of these properties: its sides have lengths that are onseutive integers one of its altitudes has an integer length and that altitude divides the triangle into two right triangles eah with integer sides Let S be a sequene of triplets (a, b, ) representing the lengths of sides of suh triangles in inreasing order of perimeter (where a < b < ). The first term in this sequene is (13, 14, 15). The third term in this sequene is (193, 194, 195) (3-4-5) Determine the seond term in this sequene ( ) C) Find the ordered pair (x, y) that satisfies 4x 7y = 451, x and y are positive integers and x + y is the largest possible three-digit multiple of A C D B D) Alie and Barbara (starting at points A and B respetively) bike towards eah other on the trak above. Alie, biking 10 mph faster than Barbara, would meet Barbara at point D in 1 hour. If, however, Barbara inreased her speed by 10 mph and Alie dereased her speed by k mph, they would meet at point C in 40 minutes. Alie s redued speed is three-quarters of her original speed. Compute the distane (in miles) between C and D. y 15 E) Compute the area of the region ontaining all points (x, y) that satisfy x. y (x + 5) x F) In eah of the years from 1999 through 008, 5 state quarters were issued at eah of the mints in Philadelphia, Denver and San Franiso. Living on the east oast, I have found in irulation 100% of the quarters minted in Philadelphia and at least 75% of the quarters minted in Denver. San Franiso only issues mint oins and I have found at most 1.5% of these quarters. Let m and M denote the exat minimum and Maximum perentages of all the quarters minted that I have found. Compute the ordered pair (m, M).

8 Round 1 Geometry Volumes and Surfaes CO TEST 1 - OCTOBER 009 A SWERS A) 500 : 81 B) 80 units C) 4 inhes 3 Round Pythagorean Relations Round 3 Linear Equations A) 3 : 8 B) 15 9, C) 154 A) 3/ (or 1.5) B) C) 18 Round 4 Fration & Mixed numbers A) (6, 6, 7) B) 7.1 C) 7 5 (or 1.4) Round 5 Absolute value & Inequalities Round 6 Evaluations A) x < 3, < x <, x > 3 B) (18, 13) C) 143 A) 1 10 B) 30 C) or Team Round [Calulators allowed] A) 10 : 53 D) 6 miles B) (51, 5, 53) E) 15 C) (671, 319) F) 58,70 3 3

9 CO TEST 1 - OCTOBER 009 SOLUTIO KEY Round 1 4 (5) 3 π 4(15) Vsphere A) = 3 = 500 : 81 V π one (3) (4+ 5) x 11x B) The volume of the box is x x x 3 5x 7x 35x = x = = 560 = 35(16) x 3 = 64 x = 4. 4 Thus, the dimensions of the sheet of ardboard are 18 x Per = (18 + ) = 80 x x (5.5x - x) (4.5x - x) ---> C) Let the side of the base be t. Let V 1 and v 1 denote the volumes of the original and smaller pyramids respetively. Then the perimeter of the base is 8t and altitude of the pyramid is 3t. 1 ( ) V = t 3 t= 108 4t 3 = 108 t = 3 3 Computing the slant height ( l) of the original pyramid, = 90 l= 3 10 Thus, the linear dimensions of the pyramids are in 3 : 1 ratio and their volumes are is a 7 : 1 ratio. V v = 1 7 = 4 t

10 Round CO TEST 1 - OCTOBER 009 SOLUTIO KEY A) Solution #1 (arithmeti only) Square 3 right triangles! The area of ABE is 16 ( ) = 16 Thus, the required ratio of 6 : 16 = 3 : 8 Solution # (algebrai) In isoseles triangle CFE, CG = GF = GE = Sine the diagonal AC = 4, AG = 3 1 Area of AEF = ( )( 3 ) = 6 6 : 16 = 3 : 8 F C D G E A 4 B B) AC = 10. Using the Pythagorean Theorem in ABC and BCD, (1) x = ( y+ 8) () x y = 36 Expanding (1), we have x y = 16y Substituting for x y, 16y = 7 y = 9/ 81 5 Then x = 36+ = 4 4 x = 15/ and the required ordered pair is 15 9, C A x D y B Alternative method: Note ADC ~ CDB (x, y) = 15 9, AD DC AC = = 8 = 6 = 10 CD DB CB 6 y x C) Let x and y denote the lengths of the original right triangle. (1) x + y = 65 Then () ( x+ 4) + ( y 8) = 65 Expanding and subtrating () (1), 8x y + 64 = 0 x = y 10 Substituting in (1), 5y 40y = y + 8y = 13(65) 0 y 8y 85 = (y 33)(y + 5) = 0 y = 33, x = 56 Per = = 154

11 Round Pythagorean Theorem - CAVEATS Here are the two nie relations derived from applying the Pythagorean Theorem mentioned last year. If you tried justifying them to yourself, ompare the following with your results. These two results may be used in any future ontests. Ignore them at your own peril! #1 P d a R S #1: If PQ RS, then a + = b + d In any quadrilateral with perpendiular diagonals, the sums of the squares of the opposite sides are equal. (Of ourse, this is true in any square, in any rhombus and in any kite, but it s not very useful in these ases. It is, however, very useful in those ases when neither diagonal bisets the other diagonal! Suh is the ase in the diagram below. It ould be a trapezoid, but in general, it s just got 4 sides.) The proof: Rt s 1, : a = m + p, = q + n Rt s 3, 4: b = p + n, d = m + q Adding, we have the required result: a + = m + p + q + n = b + d Q.E.D. (or That s all folks!) b Q # S P a d P d a m #1 #4 S R p q #3 # b n b R Q Q # IF PQRS is a retangle, a + = b + d From any interior point in a retangle, the sums of the squares of the distanes to opposite verties are equal. The proof: a = r + q, b = q + s, = p + s and d = r + s Adding a + = r + q + p + s = b + d Q.E.D. (or That s all folks!)*** Hmmmm? I wonder if it s true for exterior points, or points on the retangle? How about other quadrilaterals? Something to ponder in your spare time. S P r a d p q L b s R W Q *** This abbreviation is for the Latin quod erat demonstratum whih translates literally to whih was to be proven. It was ommonly used at the end of proofs in surviving Latin translations of Eulid s 13 volume geometri masterpiee The Elements. (ISBN: )

12 Round 3 CO TEST 1 - OCTOBER 009 SOLUTIO KEY A) B) 1 x= 1 + ( x) x = x x = /3 1/x = 3/ 3x x+.15(10 x) = 4 6x x = 75x 84x = 150 x = 1. atual digits unimportant Thus, [ x] = [ 1. ] = - -x..x A B C D AC 4 C) Sine =, let AC = 4x, BD = 9x and the overlap BC = k. Then: BD 9 AB = 4x k, CD = 9x k 13x k = 03 AB 4x k 10 = = 18x = 17k CD 9x k 7 17 Substituting for x in the first equation, k k = ( )k = k = k = 18.

13 CO TEST 1 - OCTOBER 009 SOLUTIO KEY Round 4 A) Sine R T = D, i.e. (Rate)(Time) = Distane, the average rate is the total distane traveled divided by the total time it took to travel that distane. r ave = = = < = 6 < 7 (A, C, B) = (6, 6, 7) x = y B) x= w x 1 15 = w 5 Sine k% = k k 16, we have = y = w k = k = = 7 = C) (1) a+ b= b = 3b 4b = 3 () a+ = 3b It s reasonable to assume that the value of b + is unique, i.e. is the same for all values of a, a b and that satisfy the initial onditions. Thus, take b = 3, = 4 a = = 5 5 Of ourse two equations in three unknowns do not nail down a unique solution. We need to also show that the value of the required fration is invariant, i.e. the same for all hoies of a, b and. Substituting b = 3 4 in (1), we have 3 a+ = = b Now, 4 + = = = a

14 CO TEST 1 - OCTOBER 009 SOLUTIO KEY Round 5 x 4 x 9 = ( x+ )( x )( x+ 3)( x 3) A) ( )( ) Test eah of the 5 setions on the number line determined by the ritial values, ±, ±3 The number of negative fators determines the sign of the produt B) 16 < 3k + < < 3k < < 3k < 93 5 < k < 31 This set of values is shown in the diagram at the right. A, 5 M, 18 B, 31 Sine distane between two points on the number line is the absolute value of the differene of the oordinates of the points involved, we note the midpoint of the interval has oordinate 18 and the distane to eah endpoint is 13. Therefore, the equivalent absolute value representation is k 18 < 13 (a, b) = (18, 13) C) If the units digit of A were 5 or any even digit, then A would not be prime. The only digits that an be used to form A and B are 1, 3, 7 and 9. Thus, there are 4 possible ordered pairs (A, B): (13, 31), (17, 71), (37, 73) and (79, 97) The values of A B are: 18, 54, 36 and 18 C = 18 and D = 54 The interval 7 < x < 16 ontains = 143 integers Round (64) A) + (3 5) = + = + = + = = (5) 15 9(5) B) produt -----> neg 3 neg neg 1 neg all pos a = b a= = b = 5 Substituting and multiplying through by 15, = 35 = 15(70) = 30 a+ ab, when b is a proper fration C) Reall: a * b= b ab, when b is an improper fation Sine 6 * = (6 + 4) = 10 and * = = =, we have * * * 3 4 = 3 10* 8 = = 10+ = 13 or

15 Team Round CO TEST 1 - OCTOBER 009 SOLUTIO KEY A) 3 faes: orner ubes (at the 8 verties) 8 faes: edge ubes (1 edges) 7 x 8:, 7 x 9: 4, 8 x 9: fae: enter ubes (6 faes) [(5)(6) + (5)(7) + (6)(7)] 14 0 faes: interior ubes only 5(6)(7) 10 This totals 504 unit ubes in total and there should be 7(8)(9) = 504 ubes. Thus, (8 + 7) : ( ) 10 : 53 (1) a + h = x Q B) In right triangles PQS and RQS, we have () b + h = y y = x+ (3) z= a+ b x h Subtrating () (1), b a = (b + a)(b a) = y x = (x + ) x = 4(x + 1) P But a + b = x + 1!!! a S z = x + 1 Caneling, b a = 4 b = a + 4 b Substituting, a + 4 = x + 1 x = a + 3 It remains to find h in terms of a. In PQS, a + h = x a + h = (a + 3) h = 3(a + 4a + 3) h = 3( a+ 1)( a+ 3) R Clearly, we want values of a for whih 3( a+ 1)( a+ 3) is a perfet square, i.e (a + 1) is a perfet square and (a + 3) is 3 times a perfet square or vie versa. a + 3 The first and third term are generated by a = 5 and 95 respetively, so we restrit our searh to integer values of a between 6 and 94 inlusive. a = = 45 BINGO! Thus, the seond term is (51, 5, 53). P Q h a S a+4 a + 4 a + 5 R Chek: PQS: (4, 45, 51) 3(8, 15, 17) and RQS: (8, 45, 53) = 809 = 53

16 CO TEST 1 - OCTOBER 009 SOLUTIO KEY Team Round y 3(1 + y) C) Solving for x, x= = 11+ y Thus, y must be piked so that 3(1 + y ) is an integer. 4 This ours when y = 3 and orrespondingly x = = 118. Sine the given equation is a straight line with slope 4/7, an inrease of 7 in the value of x orresponds to an inrease of 4 in the value of y. Thus, a general solution is ( t, 3 + 4t) and x + y = t = 11(11 + t) The expression is a multiple of 3 for t = 1, 4, 7, 10,, i.e for t of the form 3k (1 + 3k) < 1000 k < 6.3 k = 6 t = 79 (x, y) = (671, 319) Alternate solution: The largest possible value of x + y is 999. If we try 999, we notie that 4(999 y) 7y = 451 or 4(999) 451 = 11y requires divisibility by 11. 4(999) 451 = = 3545 whih fails [sine (5 + 5) (3 + 4) = 3] 4(996) 451 = 3533 whih fails [sine (5 + 3) (3 + 3) = ] 4(993) 451 = 351 whih fails [sine (5 + 1) (3 + ) = 1] 4(990) 451 = 3509 BINGO! [sine (5 + 9) (3 + 0) = 11, a multiple of 11] 4x 7 y= 451 Thus, (x, y) = (671, 319) x + y = 990 A.... C D) Let x denote Barbara s original rate (in mph). Then AB = AD + DB = (x + 10)(1) + x(1) = x + 10 Also AB = AC + CB = (x + 10 k)(/3) + (x + 10)(/3) (x+ 0 k) Equating these expressions for AB, x+ 10= x = 10 k x = 5 k 3 x+ 10 k 3 Alie s redued rate = 4x k = 3k + 30 x = 4k 10 x k = 4k 10 k = 3, x =, AB = 14, AD = 1, DB =, AC = ( )(/3) = 6 CD = 14 6 = 6 D B

17 CO TEST 1 - OCTOBER 009 SOLUTIO KEY Team Round E) y = x + 5 is a line with slope and y-interept at (0, 5) and x-interept at (.5, 0) the dotted line x =± 1 depending on whether x is positive or negative. x The following diagram shows the region in question. For positive x, y was positive and it stays positive. For negative x, y hanged sign, i.e. the graph was refleted aross the x-axis. The area of region I is 1 (10)(0) = 100 The area of region II is 1 (5)(10) = 5 Thus, the total area is 15 F) (10)(5)(3) = 150 oins available. I have 50 P quarters. I have at least 3 50= 37.5 at least 38 D quarters. 4 E(-10,15) D(5, 15) II I y = x+5 y = -x- 5 B(-.5,0) A(0,5) C(0,-5) I have at most 1 50 = 6.5 at most 6 S quarters 8 Thus, minimum and maximum number of oins I have are = 88 and = and (m, M) =,, %, % = = = 58 %,70 %