NO CALCULATORS IN THIS ROUND

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Eric Robbins
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1 A) Evaluate 8 8. CONTEST NOVEMBER 0 ROUN COMLEX NUMBERS (No Trig) ANSWERS A) B) I II III IV C) **** NO CALCULATORS IN THIS ROUN **** B) Let z = x + yi be represented by the point (x, y). In what quadrant is, if z = 3 4i? k = 3 C) Evaluate ( 3i ). k = 0 k Refuse to be intimidated by the mathematical symbolism. Σ is just a summation operator. For example, k = 4 k = = = 30. k =

2 CONTEST - NOVEMBER 0 ROUN ALGEBRA : ANYTHING ANSWERS A) inches B) C) **** NO CALCULATORS IN THIS ROUN **** A) A rectangular piece of cardboard is 4 by 6. Squares of equal size are cut from each of the 4 corners. If the total area of the cutouts is 37.5% of the area of the original piece of cardboard, compute the length (in inches) of the side of the squares cut from each corner. B) When a two-digit number is divided by the sum of its digits, the quotient is 7 and the remainder is 0. When the same number is multiplied by the sum of its digits, the product is 567. Find this number. Joe Ford 009 When a two-digit number is divided by the sum of its digits, the quotient is 7. When the same number is multiplied by the sum of its digits, the product is 567. Find this number. Ans: 63 () 0t+ u = 7( t+ u) () (0 t+ u)( t+ u) = 567 () t = u Substituting for 0t + u in (), u = 3, t = ( t+ u) = 567 ( t+ u) = 8 t + u = 3u = 9

3 C) Suppose I drive at a certain speed S > 0 (in mph) for a certain time T > 0 (in hours). Increasing my driving speed by 0 mph and decreasing my driving time by hour, I cover the same distance. If the speed limit is 65 mph and I never break the speed limit (a governor has been installed in my vehicle), compute the number of possible ordered pairs (S, T), where S and T are both integers.

4 CONTEST - NOVEMBER 0 ROUN 3 LANE GEOMETRY: AREAS OF RECTILINEAR FIGURES ANSWERS A) in B) units C) units **** NO CALCULATORS IN THIS ROUN **** A) Rectangle ARK is 3 inches by 5 inches. T lies on KR such that ΔTK is isosceles. Compute the area of TRA. Q B) The ratio of the lengths of the sides of the two squares at the right is 3 :. The sum of their areas is 338 units. Compute the length of Q. ) If the side of a square is increased by 5%, its area equals that of a rectangle whose sides have lengths in a : 5 ratio and whose perimeter is 8 units. Compute the original length of a side of the square.

5 CONTEST - NOVEMBER 0 ROUN 4 ALG : FACTORING AN ITS ALICATIONS ANSWERS A) B) C) **** NO CALCULATORS IN THIS ROUN **** A) How many positive integer divisors does N = have? B) 4x 4 + 5x is to be completely factored over the integers, where each factor is of the form ax + b, a and b are integers and a > 0. Express the sum of the factors in terms of x. C) If x + y = 6 and x + y = 58.5, compute the numerical value of xy. 3 3

6 CONTEST - NOVEMBER 0 ROUN 5 TRIG: FUNCTIONS OF SECIAL ANGLES ANSWERS A) B) (,, ) C) **** NO CALCULATORS IN THIS ROUN **** A) ABC is a rectangle, AE B A = 5 and AE = 4 Compute cosθ. A E θ B C B) In simplest form, (sin 45 + tan 35 ) 4 = etermine the ordered triple (A, B, C). A B C, where A, B and C are positive integers. C) Compute the sum of the values of x (in degrees) that satisfy cos(70 + x) = sin( 600 ) and lie between 500 and 900 inclusive.

7 CONTEST - NOVEMBER 0 ROUN 6 LANE GEOMETRY: ANGLES, TRIANGLES AN ARALLELS ANSWERS A) B) C) **** NO CALCULATORS IN THIS ROUN **** A) AB C, m = m + 5, m 3 = 73 Compute b a. A.. 3 b B.. C a B) ΔABC is isosceles with base BC. B bisects ABC and C bisects ACB. If AB > BC and m BAC = (k), compute m BA m BC in terms of k. A B C C) Two angles in an isosceles triangle measure (x + 5) and (x 30). Compute the sum of the measures of all possible vertex angles.

8 CONTEST - NOVEMBER 0 ROUN 7 TEAM QUESTIONS ANSWERS A) ) B) A B C E _ E) C) F) (, ) **** NO CALCULATORS IN THIS ROUN **** A) Given: z = a + bi and z = 7 The graph of z in the complex plane is represented by the point. Compute all possible values of a for which the distance between z+ ( i ) and z+ ( i) is 5.. (0, 0) B) Given: 0 < x A < 0, for integer values of A. Which of the following statements about the number of integer values of x satisfying the inequality are true? Circle the appropriate letters on the answer blank. A) As A increases, the number of integer values of x increases. B) As A decreases, the number of integer values of x decreases. C) Regardless of the value of A, there are 4 integer values of x. ) If A is even, there are 4 integer values of x. E) If A is odd, there are 5 integer values of x. A x C) Given: a square ABC, subdivided into two squares and two rectangles by Q and RS drawn parallel to the sides of the square. A = T = x and B = y, where x > y > 0. The sum of the areas of the two squares is k times the sum of the areas of the two rectangles. etermine a simplified expression for x in terms of k. y ) There are several positive integer values of k for which k 96 is an integer. Find all of them. E) Given: Rectangle QRS with ST = m TSR = 30, S = and Q = QT cot ST. Compute: ( ) F) Given: m A is an integer, m B= m A+ k, for some positive integer k. For each value of k, m A is as large as possible. Each angle in a convex polygon is congruent to either A or B. There are 0 of the larger angle B and 0 < n < 0 of the smaller angle A. Let m and M denote the minimum and maximum possible values of k. Compute (m, M). S R 30 T Q y. (a, b) B S C Q T R

9 CONTEST - NOVEMBER 0 ANSWERS Round Algebra : Complex Numbers (No Trig) A) B) C) 8 3 3i Round Algebra : Anything A).5 B) 63 C) Round 3 lane Geometry: Area of Rectilinear Figures A) 0.5 B) 6 34 C) Round 4 Algebra : Factoring and its Applications A) 4 B) 6x C) 35 4 ( or 8.75) Round 5 Trig: Functions of Special Angles A) 4 5 (or 0.8) B) (7,, 4) C) 490 [500, 560, 860] Round 6 lane Geometry: Angles, Triangles and arallels A) 78 B) 3k 45 or equivalent C) Team Round A) ± 4 3, ± 0 ) k = 0,, 4, 5 (in any order) B) and E only E) 3 C) k + F) (, 5) k

10 CONTEST NOVEMBER 0 SOLUTION KEY Round A) 8 8 = i 8 i 8 = i 44 = 3+ 4i 3+ 4i 3 4 B) If z = 3 4i, then z = = = + i 3 4i 3 + 4i Since x > 0 and y > 0, is located in quadrant. k k = C) ( 3i) = ( 3i) + ( 3i) + ( 3i) + ( 3i) k = 0 ( ) 3i = 3i + 3i = 3i The first three terms evaluate easily as ( ) ( ) The last term evaluates as follows: 3 ( i) ( i) ( i) ( i)( i) + 3i + 3i = 3 3i. 3 = 3 3 = 3 3 = = 8 If you recalled that the three cubes roots of 8 are, + 3 and 3, this expansion was unnecessary. Thus, the required sum is 8 3 3i.

11 CONTEST NOVEMBER 0 SOLUTION KEY Round A) Let x denote the side of each square cutout. 3 Since 37.5% = 3/8, we have ( 46 ) = 4 x 8 9 x = x =.5 (or 3 4 ). B) () 0t+ u = 7( t+ u) () (0 t+ u)( t+ u) = 567 () t = u Substituting for 0t + u in (), 7( t+ u) = 567 u = 3, t = 6 63 ( t+ u) = 8 t + u = 3u = 9 B)C) ST = ( S + 0)( T ) = ST S + 0T 0 S = 0(T ) S > 0 T >, S < 65 T < 4. Thus, there are 3 possible ordered pairs (S, T), namely (0, ), (40, 3) and (60, 4). However, the last ordered pair fails, since to travel 40 miles (60 mph for 4 hours), I would have to travel 80 mph for 3 hours, breaking the speed limit. Thus, there are only ordered pairs satisfying the never-break-the-speed-limit condition.

12 CONTEST NOVEMBER 0 SOLUTION KEY Round 3 K T R A) Since K is a right angle, isosceles triangle TK must be and KT = K. Thus, KR must be the longer side. K = RA = 3, KR = A = 5 TR = Use the area formula for a trapezoid directly Area = 3( + 5) = =0.5 A or determine the area of the rectangle ARK and subtract the area of the triangle KT. Area = 5 3 (5 3) = =0.5 B) ( x) ( x) 3 + = 338 x = = x = 6 Thus, the sides of the squares are 3 and 48 units. Q is the hypotenuse of a right triangle with a horizontal base of 80 and a vertical height of 48. ( 48,80, Q) is a ythagorean triple. ( Q) 48,80, = 6(3,5, x) x = 34 Q = C) Let x denote the side of the square. The side of the larger square is The perimeter of the rectangle is (k + 5k) = 8 k =. The rectangle is 4 x 0, resulting in an area of Thus, 40 4 x = x = = x = x+ x= x 4 4

13 CONTEST NOVEMBER 0 SOLUTION KEY Round A) N = 3 Any divisor (other than ) will have factors of or 3, but no other prime. Thus, the exponents of may be any integer from 0 to 5 inclusive 6 possibilities. The exponents of 3 may be any integer from 0 to 3 inclusive 4 possibilities. Choosing both exponents to be 0 gives us. Thus, there are 4(6) = 4 possible positive divisors. In general, determine the unique prime factorization of N, add to each exponent and multiply. B) 4x 4 + 5x = (4x )(x ) = (x + )(x )(x + )(x ). The sum of the factors is 6x. C) (x + y) 3 = x 3 + 3x y + 3xy + y 3 = x 3 + y 3 + 3xy(x + y) 6 3 = xy(6) = 8xy xy = 8 = 36 = 35 = The alternate solution below uses this assertion (fact): if a + b + c = 0, then a 3 + b 3 + c 3 = 3abc In other words, if the sum of 3 numbers is 0, then the sum of the cubes of the 3 numbers will always be 3 times the product of the 3 numbers. x + y = 6 x + y 6 = 0 x 3 + y 3 + ( 6) 3 = 3xy( 6) = 8xy which is the same result as above. roof of the assertion (x a)(x b)(x c) = 0 has solutions x = a, b, c Expanding, and regrouping, x 3 (a + b + c)x + (ab + ac + bc)x abc = 0 3 a ( a + b + c) a + ( ab + ac + bc) a abc = 0 3 Substituting for x: b ( a + b + c) b + ( ab + ac + bc) b abc = 0 3 c ( a + b + c) c + ( ab + ac + bc) c abc = 0 Adding these three equations, a + b + c ( a + b + c)( a + b + c ) + ( ab + ac + bc)( a + b + c) 3abc = Finally, if (a + b + c) = 0, a + b + c 3abc = 0 a + b + c = 3abc.

14 CONTEST NOVEMBER 0 SOLUTION KEY Round 5 A θ B AB A) In ΔBA, cosθ = B A = 5 and AE = 4 E = 3 ΔABE ΔBA (by AA) AB = AE = 4 B A 5 [Using the dimensions of right ΔABE is a distraction, since computing and 0 AB = is unnecessary.] 3 E 6 BE = 3 C B) C) = = = = (A, B, C) = (7,, 4) n cos(70 + x) = sin( 600 ) sin x= sin(0 ) x= n n 900 n n= 4(only) n 900 n n= 4,5 500, Adding, 490. Alternate Solution: cos(70 + x) = sin( 600 ) cos(70 + x) = sin( ) = sin(0 ) = sin 60 = cos( ± 30 ) , 480,840,00,560, x=± n x = + n n 60, 40,780,40,500,

15 CONTEST NOVEMBER 0 SOLUTION KEY Round 6 A) Let m = x and m = (x + 5) m 3 = 73 (x + 5) + x + 07 = 80 x = 9 m = 44, m = 9 b = 5 Since a is vertical to the alternate interior angle of 3, a = 73. Thus, b a = 5 73 = b.. C A a B B) In isosceles ΔABC, a vertex angle of (k) leaves base angles of (90 k) k k Thus, m AB = 45 m BA = 35 The required difference is 35 ( 90 k ) and = ( 90 + ) k 3k + = 45 m BC k or equivalent. A B C C) Case - Given angles are the base angles: x + 5 = x 30 x = 35 base angles 40, vertex angle 00 Case - Both base angles measure (x + 5) : (x + 5) + x 30 = 80 4x = 00 x = 50 vertex angle = (50) 30 = 70 Case 3 Both base angles measure (x 30) : (x 30) + (x + 5) = 80 5x = 35 x = 47 vertex angle = = 5 Thus, the required sum is.

16 CONTEST - NOVEMBER 0 SOLUTION KEY Team Round A) Let ( x, y ) denote z+ ( i ) = ( a ) + ( b+ ) i. Then: Let Q( x, y ) denote z+ ( i) = ( a+ ) + ( b ) i. Then: Q = 5 (( a ) ( a+ ) ) + (( b+ ) ( b ) ) = 5 ( ) ( b ) 4( b + ) = 6 b = ± =, 3 z = a + b = 49 a = 49 or 49 9 a = ±4 3, ± 0 x = a, y = b+ x = a+, y = b = 5 B) If 0 < x A < 0, then the solutions in x lie strictly between A/ and A/ + 5. If A is even, then so are the coordinates of the endpoints and Q. Since this a strict inequality, the endpoints are excluded and there are always 4 integer solutions Q. A/ A/ + 5 If A is odd (say k + for some integer k), then the coordinates of the endpoints and Q are k + and k + 5. There are 5 integer solutions, namely (k + ) (k + 5)..... Q.. k+ k+ k+3 k+4 k+5 Thus, only and E are true statements. C) x y k xy + = ( ) Moving all terms to the left side and dividing both sides by y, we get a quadratic equation in x y : A x y B x x k + = 0 y y Using the quadratic formula, x k± ( k) 4 = = k± k. R y Since x > y, the required ratio is greater than and x y = k+ k The same result is obtained if x ( ky) x + y = 0 is treated as a quadratic in x and solved for x in terms of k and y. Verify that you get the same result follows. x ky y k T Q = ± and S C

17 Team Round - continued ) CONTEST - NOVEMBER 0 SOLUTION KEY k 96 must be positive, so k > 0 for starters. k 96 must also be a perfect square, i.e. there is an integer N such that k 96 = N or k N = 96 Let s examine a list of consecutive perfect squares and look for gaps of The gap between consecutive entries grows uniformly by, so eventually the gap can be any odd integer. Consecutive perfect squares always differ by an odd amount, so we are looking for nonconsecutive perfect squares. We are looking for perfect squares which sandwich an odd number of consecutive perfect squares. (erfect squares sandwiching an even number of perfect squares would differ by an odd number.) The minimum difference if there are 9 intermediate numbers is - = 0, so 7 is the maximum number of intermediate perfect squares. There are 4 possible answers = 96 k = 0 5 intermediate numbers 6 gaps a, a +, a +0 6a + 30 = 96 a = 5 ( ) k = k = 3 intermediate numbers 4 gaps a, a +, a + 6 4a + = 96 a = 00 ( 44 69) 96 k = 96 k = 4 intermediate number a + (a + ) = 96 a = 47 which occurs between.. hmm? Continuing the list of perfect squares, 400, 44, 484, 59, (576), 65 k = 65 k = 5 Alternate solution: (Aaargh! - Why didn t I go down this road first!!!) k 96 = N k N = 96 (k + N)(k N) = 96 So just look at factor pairs of 96, namely (, 9), (, 48), (3, 3), (4, 4), (6, 6), (8, ) Equating k N with the smaller factor and k + N with the larger, we have: k N = k + N = Adding and ignoring odd sums, we have k = 50, 8,, 0 k = 5, 4, and 0 (in any order).

18 Team Round - continued CONTEST - NOVEMBER 0 SOLUTION KEY E) ST = and m TSR = 30 TR = and SR = 3. Isosceles triangles S and QT m ST = 90 Let Q = QT = x S = = x + Q = x + T QT x cot ( ST) = S = = x + 3 Q = SR x + = 3 x = 3+ S (x + ) = 3 ( 3 ) 3 Substituting, cot ( ST) = = = = Alternate Solution: Let = x and Q = y. The remaining lengths are as indicated in the diagram. Since ST = and m TSR = 30, we have the following system of x y = equations: x x + y = ( xy, ) =, and S T y y cot ( ST) = = = = = = 3 S x x x 30 x x + y y y Q T R Q y T x - y R Note: m ST = 80 ( ) = 75, but is was not necessary to invoke the expansions cos( ) for sin(a + B) and cos(a + B) to evaluate cot(75 ) as to get an exact answer. sin( ) This makes the problem solvable by anyone with minimal knowledge of right triangle trig, i.e. SOHCAHTOA, reciprocal relationships and special angles (30, 60, 45 ).

19 Team Round - continued CONTEST - NOVEMBER 0 SOLUTION KEY F) Let x = m A. has (0 + n) sides, i.e. at least sides and at most 9 sides. Thus, 80((0 + n) ) = 0(x + k) + nx = x(0 + n) + 0k k (80 x)(0 + n) = k x = n As n increases from to 9, the denominator of the fraction increases from to 9. If n =, then the numerator must be the smallest possible multiple of. This occurs for k = 8 and we have x = A = / = 40. k = 9 also insures divisibility by, but x = / = 30 and it s not as large as possible k 5k n = : = 30 + k = 6, x= 80 35, (A, B) = (45,5) 6 n = 3: k 35 0 k+ = + 9 k = 3, x= 80 30, (A, B) = (50,53) k 5( 36 + k ) n = 4: = k = 6, x= 80 30, (A, B) = (50,56) k k n = 5: = 4 + k = 3, x= 80 6, (A, B) = (54,57) 5 3 n = 6: k 5(36 + k = ) k = 4, x= 80 5, (A, B) = (55,59) 6 8 n = 7: k k+ = + 3 k = 5, x= 80 30, (A, B) = (50,65) k 5k n = 8: = 0 + k = 9, x= 80 5, (A, B) = (55,64) 8 9 n = 9: k k =, x= 80 0, (A, B) = (60,6) 9 Thus, (m, M) = (, 5). Checks that 0(m B) + n(m A) = 80(n ) (not necessary, but included just to be on the safe side!) n = : 50 + (45) = ( )80 = 800 n = 3: (50) = (3 )80 = 980 n = 4: (50) = (4 )80 = 60 n = 5: (54) = (5 )80 = 340 n = 6: (55) = (6 )80 = 50 n = 7: (50) = (7 )80 = 700 n = 8: (55) = (8 )80 = 880 n = 9: (60) = (9 )80 = 3060

Glossary. Glossary 981. Hawkes Learning Systems. All rights reserved.

Glossary. Glossary 981. Hawkes Learning Systems. All rights reserved. A Glossary Absolute value The distance a number is from 0 on a number line Acute angle An angle whose measure is between 0 and 90 Addends The numbers being added in an addition problem Addition principle