CO TEST 6 MARCH 00 ROU D ALG : SIMULTA EOUS EQUATIO S A D DETERMI A TS A SWERS A) B) C) (, ) ***** O CALCULATORS I THIS ROU D ***** y 7 A) Given: y 3 + Determine the value of k for which this system of lines determines a y k minimum number of points of intersection B) Find all ordered pairs (, y) satisfying y + y C) Connecting the points A(3, 7), B(, ), C(3, ) and D(0, ) you have an outline of the deck in my backyard It s a nondescript conve quadrilateral and finding its area is baffling my builders If > 0 and A and C are opposite vertices, compute the coordinates of C so the area of my deck is 5 square units Note: The area of a triangle with vertices at (, ),(, ) and (, ) as y y y 3 3 y y y can be computed 3 3
4 A) Given ( 5 3 ) Note: CO TEST 6 - MARCH 00 ROU D ALG: EXPO E TS A D RADICALS A SWERS A) B) C) ***** O CALCULATORS I THIS ROU D ***** A, a reduced fraction Compute A + B B n a n a is equivalent to B) Compute 5 0 0 5 C) Given: 9 9 3 3 for some integer value of 6 5 4 Compute 4 8 3+ 4 +
CO TEST 6 MARCH 00 ROU D 3 TRIGO OMETRY: A YTHI G A SWERS A) B) C) ***** O CALCULATORS I THIS ROU D ***** A) Solve for θ, 0 θ < 3 sinθ cosθ tan(90 θ ) cot(90 + θ )sec(70 + θ )csc(70 θ ) cot(80 θ ) B) Robert Wadlow, the world s tallest man, stands feet tall Your task is to compute given the following information: At 4:00 PM, Robert s shadow is 0 feet longer than his shadow at :00 PM At 4:00 PM the sun makes a 30 with the ground, but it made a at :00 PM Recall: An eact answer is required The given information is summarized in the diagram :00 4:00 30 C) Given: k sin A 4 k+ cos A 4 5π and k > 0 Compute sec A cos( A 5π)
CO TEST 6 - MARCH 00 ROU D 4 ALG : A YTHI G A SWERS A) B) years C) ***** O CALCULATORS I THIS ROU D ***** A) Determine all possible positive integral values of for which 36 8 34 The subscripts denote the base in which the numeral is written B) Alice and her mother have the same birthday (MM/DD) Alice is 8 years younger than her mother When her mother was as old as Alice is now, her mother was twice as old as Alice was then In how many years from now will the ratio of Alice s age to her mother s age be 7 : 0? C) Compute the two rational roots of the following quadratic: 4 A(48A+ ) 5(6 0 A)
CO TEST 6 - MARCH 00 ROU D 5 PLA E GEOMETRY: A YTHI G A SWERS A) B) : C) ***** O CALCULATORS I THIS ROU D ***** A) Ten coplanar lines determine points of intersection Compute the maimum value of B) An equilateral triangle and a regular heagon have equal perimeters What is the ratio of the area of the triangle to the heagon? C) Given: PA is tangent to circle O, PA 4, CB 3PB and EP PB A Compute the area of circle O D O E P B C
CO TEST 6 - MARCH 00 ROU D 6 ALG : PROBABILITY A D THE BI OMIAL THEOREM A SWERS A) B) C) ***** O CALCULATORS I THIS ROU D ***** A) One container has 5 blue chips and 4 red chips, while another container has 4 blue chips and 5 red chips One chip is chosen at random from each container What is the probability that one chip is red and the other is blue? B) Three mathletes A, B and C are simultaneously (but independently) trying to solve a difficult math problem The probability of A correctly solving the problem is 5 and the probability of B correctly solving the same problem is 4 Let P denote the probability of the problem being solved by at least one of the mathletes Let Q denote the probability that the problem is solved by eactly one of the mathletes If P : Q 8 : 5, compute the probability that C solves the problem correctly C) The 5 th term in the epansion of + A Compute all possible ordered pairs( A, T ) Note: The st term in the epansion is 8 4 7 70 T is 8
CO TEST 6 - MARCH 00 ROU D 7 TEAM QUESTIO S A SWERS A) D) (, ) B) (,, ) E) C) F) ***** O CALCULATORS ARE PERMITTED I THIS ROU D ***** A) Let functions f and g be defined by f ( ) k + A and g( ) 4 k 5B k k If k >, A, B and the functions f and g have the same zeros, compute the k k area of the region bounded by f and g B) Suppose 4 a 3 6 3 3 b r p q 3, each radical is in simplest form and all variable denote positive integers Compute the number of distinct values of the product pqr, the maimum product and the minimum product Epress your answer as an ordered triple (count, ma, min) Q A C) Points P and Q lie on the hypotenuse of right triangle ABC 3 7 If sinθ and sinδ, compute the value of sinβ 5 5 D) Two bottles of equal volume contain rubbing alcohol and water The B θ β ratios of alcohol to water in the two bottles are 3 : and A : B, where A and B are relatively prime integers The contents of the two bottles are mied and the new ratio of alcohol to water is 7:3 Determine the ordered pair (A, B) P δ C E) Given square ABCD below with sides as indicated Compute, the length of C Diagram is not necessarily drawn to scale A B N 9 F) Let be the positive difference between the two 3a+ b largest coefficients in the epansion of ( ) Determine the prime factorization of When writing the product, list the primes from smallest to largest D C M
CO TEST 6 MARCH 00 A SWERS Round Alg : Simultaneous Equations and Determinants A) B) (,0),, C) (9, 3) 3 3 Round Alg : Eponents and Radicals A) 3 B) C) 64 Round 3 Trigonometry: Anything Round 4 Alg : Anything A) 35, 35 B) 5 3 feet C) 5 (or 5) 4 A) B) 6 C) 6 5, 3 64 Round 5 Plane Geometry: Anything A) 45 B) : 3 C) 5π Round 6 Alg : Probability and the Binomial Theorem A) 4 8 B) 4 5 ( 8 or 08) C) ±, 3 Team Round A) 500 3 (or equivalent) D) (3, ) B) (9, 40, ) E) 35 C) 3 5 ( 6 or 06) F) 5 7 3 (the last eponent is not required)
Round CO TEST 6 MARCH 00 SOLUTIO KEY A) If the three lines were parallel, then no points of intersection would be determined This is not possible, since the first two lines have different slopes and the third line is horizontal The (unique) minimum will occur when the horizontal line passes through the point of intersection of the first two lines 7 3 + 3 y k B) The st + condition describes a line y ( slope of / and an -intercept of (, 0) ) The nd condition describes square connecting (in order): (, 0), (0, ), (, 0) and (0, ) Clearly, (, 0) is a solution The equation of the side of the square in quadrant 3 lies on the line y B 3, 3 3 C Area( ABCD) Area ABD + Area BCD C) ( ) ( ) Using the determinant method, 3 7 Area( ABCD) + 3 0 0 (( 6 + 70) ( 7+ 0+ 3) ) + (( + 3+ 0) ( 6 0 ) ) 5 ( 59 + ( 8 + )) 5 8 + 80 04 3 (9, 3) Alternative #: 3 7 The area is given by 3 This determinant is evaluated by the 0 3 7 lattice method Take the absolute value of the difference between -7 6-0 3-4 - 4 the sum of the diagonal down products and the sum of the diagonal up products Thus, the area is ( 4 + 76 ( 4 4 ) ) 5 3 (9, 3) This technique works for finding the area of any conve polygonal region, given given a clockwise or counterclockwise listing of the coordinates of its vertices, repeating the starting coordinates See if it works for some concave regions! Does it work for all polygonal regions conve or concave? The simplicity and generality of this method is astounding E D 3 7-3 - 0 3 7 A 6 3 70 4 + 76
CO TEST 6 MARCH 00 SOLUTIO KEY Round - continued Alternative # (Utilizing only basic area formulas) Study the diagram at the right Draw a pair of verticals and a pair of horizontals through opposite pairs of vertices 5(3+ ) 0+ + + (0 3 ) + 5 0() B(-, ) 5(3 + ) + (0 3 ) 0 83 7 5+ 5+ 40 54 3 9 3 (9, 3) Ạ(3, 7) 4 7 0 5 6 5 5(3+)/ 5 3 + D(0, ) 4 (0-3) 0-3 C(3, -) Alternate #3 Here is a (more involved) solution which utilizes the triangle area formula for any quadrilateral, conve or concave, namely half the product of the diagonals times the sine of the included Ạ(3, 7) c angle We start with an area formula for the triangle ab sin θ, B(-, ) where θ denotes the included angle 80 θ a Since sin(80 θ ) sinθ, we have Area(ABCD) sin θ ( ac+ ad+ bd+ bc) sin θ ( a ( c+ d ) + b ( c+ d ) ) sin θ ( ( a+ b )( c+ d ) ) sinθ AC BD Let s show that (9, 3) produces an area of 5 θ b d Using the distance formula, AC ( 6) + 0 36 34 and BD + ( ) 5 The slope of AC and the slope of BD 3 m m We need the angle between the two lines If θ denotes the acute angle, then tanθ + m m 5 + 3 55 5 6 In this case, tan 3 + 6 θ sinθ and the area is 5 + 33+ 5 38 9 037 3 6 34 But does this equal 5????? 037 6 34 6 ( 7) ( 6) 3(4) 5 037 6 7 Amazing (albeit tedious) solution The basic area formulas above are worth remembering D(0, ) C(3, -)
CO TEST 6 MARCH 00 SOLUTIO KEY Round 4 4 4 3 5 4 4 48 5 3 5 3 5 3 5 3 5 3 65 A) ( ) B) A + B 3 5 0 0 5 ( ) ( ) 0 4 50+ 0 30 4 50 30 0 0 3 0 50+ 5 5 50 5 0 5 3 C) First and foremost - 9 9 3 3 6 5 4 9 9 3 3!!! Squaring both sides, 6 5 4 5 9 9 3 3 9 3 3 9 9 9 9 + + 6 5 4 6 4 6 9(5 6) 9 9 9 9 + + 6(5) 6 6(5) 6 9 5 + 6(5) 6(5) 5 Multiplying through by 50, 5 50 5 + 50 ( 5)( 0) 0 0 4 8 3+ 4 + 6+ 8 6+ 6 + Since 0, we have 6 64
CO TEST 6 MARCH 00 SOLUTIO KEY Round 3 A) Rather than evaluating by direct substitution, let s simplify the epression and show that, if the epression is defined, the product P is invariant, ie has the same value, namely Using reduction formulas, θ is a reference value (angle) in quadrant and when we see (90 θ) think Q, (90 + θ) think Q, (70 θ) think Q3, (70 + θ) think Q4 The mneumonic ASTC identifies the sign of the trig functions in quadrants 4 Simplifying the lefthand side, we have sinθ cos θ (cot θ )( tan θ )(csc θ )( sec θ ) Regrouping, ( sinθ cscθ)( cosθ secθ)( cotθ tanθ) Thus, cot(80 θ) cot(θ ) θ belongs to 45 family in quadrants and 4 35, 35 B) Let s denote the length of Shorty s shadow at :00 Then: tan s s 3 At 4:00 tan 30 3 s+ 0 s+ 0 3 Multiplying by 3, 3 s 3+ 0 3 3 + 0 3 5 3 Note: Wadlow was a real person and his actual maimum adult height was 8 feet inch He died at the age of and is acknowledged to be the modern world s tallest man Goto http://wwwmaniacworldcom/worlds_tallest_manhtm to learn more or Google Robert Wadlow As a3 year old Boy Scout 7' 4" k sin A 4 k ( k+ ) C) Since, + k+ (5, 4) 4 4 cos A 4 4 4 k + k 40 k k 0 ( k+ 5)( k 4) 0 k > 0 k 4 5π Simplifying, sec cos 5 A ( A 5π) π π cos A sec A cos( A π) sec A cos( π A) csc A cos A cot A 5 sin A 4
CO TEST 6 - MARCH 00 SOLUTIO KEY Round 4 A) (8 ) + 3(8) + 6 + 3+ 4 58 + 3 + 4 + 3 54 ( )( + 4) 0, 4 B) Mom and Alice are now + 8 and years old respectively Clearly, 8 years ago, Mom was as old as Alice is now and Alice was only 8 years old ( 8) 36 Thus, in n years the required ratio is satisfied, namely 36 + n 7 54+ n 0 3 + 0n 378 + 7n n 6 C) 4 A(48A+ ) 5(6 0 A) 9A + 4A 80 005A 9A + 009A 80 0 No doubt the coefficients are intimidating, BUT Since the roots are rational, this quadratic must factor How do we avoid tedious trial and error? The key observations are: 6 4 9 3, 80 5 (each with eactly one odd factor) and the coefficient of the middle term of the trinomial is odd Since the OI in FOIL produces the middle term, we must have an odd outer product and an even inner product (or vice versa) Thus, the factorization must be (3A + 6)(64A 5) 0 A 6, 5 3 64 Round 5 A) Two coplanar lines determine at most one point of intersection A third line will determine at most two more, if it crosses both of the eisting lines Likewise, each subsequent line adds a maimum number of new intersection points if it crosses each of the eisting lines Thus, for 0 lines ma + + 3 + 4 + + 9 9(0)/ 45 B) Let p perimeter Area of triangle ½(base)(height) ( )( ) p p 3 3 3 p 3 36 Ratio of triangle to heagon is : 3 Area of heagon 6(/)(base)(height) 6( )( )( ) p p 3 6 6 p 3 4
Round 5 - continued CO TEST 6 - MARCH 00 SOLUTIO KEY A C) Let PB and BC 3 According to the tangent-secant relation, PA PB(PC), ie outer(outer + inner) (4) 4 7 PE 7 Let OE OD r Then 4 7( 7 r) + D O E P B 96 8+ 4 7r 68 4 r 6 7 4 7 7 Thus, the area of circle O is π ( 6 7) 5π C Round 6 A) The desired outcome could be either R B or B R OR ADD 4 4 5 5 6+ 5 Thus, the probability is + 4 9 9 9 9 8 8 4 3 5 4 B) P P(none of them solve the problem) P(not A) P(not B) P(not C) ( ) + 3 8 + 5 0 Q P(A notb notc or nota B notc or nota notb C) 3 4 4 3 ( ) + ( ) + 7 7 + 5 ( ) + 5 4 5 4 5 4 0 0 0 Thus, P : Q 8 : 5 8 + 8 40 + 56 + 40 0 6 7+ 5 5 3 4 7 4 7 6 5 4 6 4 35A 70 4 A A 3 8 8 8 4 6 T and A 8 (A, T) ±, 3 C) The 5th term is ( ) T 4 5
Team Round CO TEST 6 - MARCH 00 SOLUTIO KEY A) The graph of the bounded region is represented in the diagram at the right: A Q(, 0) is,0 k from f s point of view and 5 D,0 from g s point of 4k view Evaluating the determinants, equating and canceling out the common factors of k in the denominator, B) ( k ) + k 5 4 + 4k 5k 5 k 9 k 3 k 4k 0 0 P,0, Q, 0, R(0,0) and S(0, 40) 3 3 PRQS is a quadrilateral w/perpendicular diagonals; hence its area is given by d d PQ RS 0 ( ) 0 500 0 ( 40) 50 3 3 3 4 a 3 6 3 3 b b 5 3 3 a 4 6 ( a ) ( b 3 3 ) 3 ( 3) 3 3 3a b ( a ) ( b ) ( 3) 3 3 3 3 4 8 4 3 5 3 b a ( 3 ) r p Since this epression must equal 3 q, we know that r, p b 5 <, q 3a < Since each of the original radicals was also in simplest form, a,, 3 and b,, 3, 4, 5 Thus, p, 3 or 5 and q, 4, 7 Since pqr pq, the minimum value occurs for (p, q) (,) and the maimum occurs for (p, q) (5, 7) (count, Ma, min) (9, 40, ) C) Think θ: 3 4 5 and δ: 7 4 5 Let α θ + δ Then 4 4 3 7 75 3 4 cosα cos( θ+ δ) cosθ cosδ sinθ sinδ sinα 5 5 5 5 5 5 5 sin β+ ( θ+ δ ) sin β+ α sinβ cosα+ cosβ sinα sin(90 ) ( ) ( ) If sinβ k, then cosβ k 3 4 k + k 5 5 k k 6( k ) 5 30k 9k 4 5 3 5k 30k+ 9 (5k 3) 0 k sinβ 3 5 + Does this imply that CP actually bisects QCB? R(0, A) P(-, 0) Q(, 0) B P S(0, -5B) Q θ β δ A C
CO TEST 6 - MARCH 00 SOLUTIO KEY Team Round - continued D) Bottle # contains 4 parts and bottle # contains (A + B) parts To insure equal volumes multiply the bottle # ratio by (A + B) 3(A + B) : (A + B) and the bottle # ratio by 4 4A : 4B The alcohol : water ratio, after miing, is (7A + 3B) : (A + 5B) 7 : 3 9A + 39B 7A + 35B 64A 96B A 3B The only relatively prime pair of values satisfying this equality is (A, B) (3, ) E) Since BA ~ CM, we require C B CM BA N Solution #: (Banking on a hunch) A triangle with sides 5 3 is a right triangle and since 9 7(3), we might guess that C 7(5) 35, D C 5 CM 7() 84, CM C B 5 5 B 35 5 and, so our hunch was correct, 35 BA In the absence of a hunch, here are two analytical solutions Solution #: (Thanks to Andrew Geng Westford Academy/MIT) Let y By applying the Pythagorean Theorem and similarity relations: + y y ( + y ) 9 y (***) 9 Rewriting + y as ( ) y + 0y and noting that y, we have + y + 0y This substitution will make it possible to reduce equation (***) to a quadratic + 0y 9 y 4 + 0 y 9 y 0 ( ) The quadratic formula can be used to solve for (in terms of y) or alternately, the left side can be factored by noticing that 9 is the product of 7 and 3, which differ by 0 ( ) 0 0 4 9 y± y + y y± y + 9 y y y ± 88 09 y± y 49y, 69y Since both and y must be positive, 69y can be discarded Applying the substitution y and using the quadratic formula (or factoring again) finishes the problem: 49( ) + 49 49 0 ( 49 49 ) 4 49 ± + ( 49 7 49 40) ± + 7 ( 7 89) ± 7 ( 7 7 ), 84 ± 35 A B 9 M
CO TEST 6 - MARCH 00 SOLUTIO KEY Team Round - continued E) Solution #3 (Norm Swanson Hamilton-Wenham/MIT) Since AB ~ MC, let B k, then A + k k + and CM k By the Pythagorean Theorem, we have + 9 k k+ Since 9 7(3), let r and rewrite the 7 r r equation as + 3 Recalling that 5--3 is a pythagorean triple, k k+ r we try k and r k+ 5 (the larger the denominator, the smaller the fraction) Since k +, solving for k + will give us r 5 / 7 5 k+ Thus, k+ k+ 7 / 7 35 D B C N 9 M
Team Round - continued CO TEST 6 - MARCH 00 SOLUTIO KEY F) Which coefficients are the largest coefficients is not so obvious How do we avoid calculating all the coefficients in order to decide? The combinatorial coefficients are: 55 65 330 46 46 330 65 55 55 The last si coefficients are identical to the first si since i j whenever i, j > 0 and i + j Think of successive combinatorial coefficients in terms of multiplicative factors of, 5, 3,, 7/5,, etc Successive coefficients in the epansion of ( ) 3a+ b introduce one more factor of and one less factor of 3, ie a multiplicative factor of /3 Successive coefficients will continue to increase as long as the multiplicative factor is greater than Combining these multipliers we can determine which coefficient is the largest, without a great deal of tedious arithmetic 0 The first coefficient is 3 0 The composite multipliers are: (/3) /3, 5(/3) 0/3, 3(/3), (/3) 4/3, (7/5)(/3) 4/5 Specifically, the 5 th term is 4/3 times the 4 th, but the 6 th term is only 4/5 times the 5 th Therefore, the 5 th and 6 th terms are the two largest coefficients 7 4 c 5 (3) () 7 4 6 5 330 (3) () and c 6 (3) () 6 5 46 (3) () 4 5 4 6 4 6 4 6 5 7 3 330 3 46 3 (990 94) 3 66 3 c 5 c 6 ( ) Notice we did not bother to find the numerical value of either c 5 or c 6 or their difference