MASSACHUSETTS MATHEMATICS LEAGUE CONTEST 3 DECEMBER 2011 ROUND 1 TRIG: RIGHT ANGLE PROBLEMS, LAWS OF SINES AND COSINES ANSWERS

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1 CONTEST 3 DECEMBER 0 ROUND TRIG: RIGHT ANGLE PROBLEMS, LAWS OF SINES AND COSINES ANSWERS A) Let ΔABC be a right triangle with legs of length 5 and If A is the right angle, compute cos B+ cosc A) B) C) B) In ΔABC, sinc is a rational number with a terminating decimal representation Compute the sum of all possible positive integer values of k C A k 30 B C) In parallelogram ABCD, BD = k, where k is a positive integer Compute the positive difference between the value of cos A, when k is as large as possible, and the value of cos A, when k is as small as possible A 8 D 4 k B C

2 CONTEST 3 - DECEMBER 0 ROUND ARITHMETIC/NUMBER THEORY ANSWERS A) B) C) A) The difference between two primes is 45 Compute the sum of these two primes base ( ) B) The Gregorian calendar is now used virtually everywhere in the secular world A non-century year is a leap year (366 days) if and only if it is divisible by 4 A century year is a leap year if and only if it is divisible by 400 To the nearest integer, how many weeks (7 days) were there in the 400-year cycle from 600 through 999 inclusive, using the Gregorian calendar? C) Consider integers written in base ( ), instead of the customary base (0) In base ( ), suppose the allowable digits are only 0 and For integers, the place values are ( ) 0, ( ), ( ), etc, instead of (0) 0, (0), (0), etc Here are some examples: In base ( ), 5 (0) is expressed as 0 ( ) and 3 (0) is expressed as ( ) All base ( ) representations of base (0) integers are unique Express 00 ( ) + 00 ( ) in base ( )

3 CONTEST 3 - DECEMBER 0 ROUND 3 COORDINATE GEOMETRY OF LINES AND CIRCLES ANSWERS A) B) C) X: (, ) Y: (, ) A) A line segment in space has endpoints P(3,, 4) and Q(6, 3, 8) Compute the length of the line segment Hint: Consider triangles PSR and PRQ in the box illustrated in the diagram at the right Z P (3, -, 4) Y X S R Q(6, 3, -8) B) Compute the length of a tangent from point P(4, 3) to the circle C : (x + ) + y = 6 C) Given: L L, PQ L, P(7, ), PQ = 0 and Q is located to the right of point P If L = {( xy, ) 3x 4y+ 3 = 0}, compute the coordinates of the x- and y-intercepts of L P Q L L

4 CONTEST 3 - DECEMBER 0 ROUND 4 ALG : LOG & EXPONENTIAL FUNCTIONS ANSWERS A) (, ) B) C) 3 A) Let f( x) x + = and gx= ( ) 4 x These exponential functions intersect at P(a, b) Compute the ordered pair (a, b) B) Let P denote a point represented by the coordinates of an x-intercept of y = f( x) = log ( 8(x ) ) 5 Let Q denote a point represented by the coordinates of a y-intercept of this function Compute all possible distances PQ C) Given: A = log + log + log + + log and log 0 = k Express A as a simplified ratio in terms of k Note: log denotes common logs, ie log 0

5 CONTEST 3 - DECEMBER 0 ROUND 5 ALG : RATIO, PROPORTION OR VARIATION ANSWERS A) (, ) B) gal C) A) Determine the ordered pair (x, y) on 4x 3y = for which the x and y values are integers and in a ratio of either : or : B) A full tank holds about 40 gallons of liquid propane (LP) Suppose the percent of LP used is directly proportional to the grilling time If I start with a full tank and use 5% of the LP in 4 minutes of continuous grilling, to the nearest tenth, how many gallons of LP remain in the tank after 3 hours of grilling? x+ z C) If xy = 00 and = 0, compute yz x z

6 CONTEST 3 - DECEMBER 0 ROUND 6 PLANE GEOMETRY: POLYGONS (no areas) ANSWERS A) B) C) A) The interior and exterior angles of a regular polygon have measures in a 44 : ratio How many diagonals may be drawn from a single vertex? B) ABCD, AEGD, EBCG and FHCG are rectangles EBHF is a square BE = 4, DF = 6 The ratio of the perimeter of EBHF to the perimeter of FHCG is 8 : 7 Compute the measure of DFE A E F B H D G C C) In ΔABC, m A= 60 L and E are located on AC and AB respectively such that CE = CL = CB and BE = EL Compute m BEL A (60 ) L E B C

7 CONTEST 3 - DECEMBER 0 ROUND 7 TEAM QUESTIONS ANSWERS A) D) B) E) C) F) A) ΔPQR is a right triangle with right angle at Q PS = a, RS = b and QS divides Q into 30 and 60 as indicated in the diagram at the right La b For positive integers L, M and N, h = Ma + Nb Compute the ordered triple (L, M, N) for which the sum L + M + N is a minimum P a Q 60 h S b R B) Consider two groups of Pythagorean triples: Group A: row : 3, 4, 5 row : 5,, 3 row 3: 7, 4, 5 row 4: 9, 40, 4 Group B: row : 8, 5, 7 row :, 35, 37 row 3: 6, 63, 65 row 4: 0, 99, 0 Compute the quotient of the third term in the 3 th row of group A to the second term in the 36 th row in group B C) Compute the coordinates of all points P(x, y), where x and y are positive integers and P is twice as far from A(5, ) as it is from B(, 8) D) Compute all possible real values of x for which 4 log x 3 logx 3 log x 5log7 x = 3x E) In the spring, the ratio of girls to boys was 7 : in the junior class at a local high school Over the summer two boys joined this class and three girls in this class moved out of the district, making the new ratio of girls to boys 5 : 8 All these students advanced to the next grade If there are no additional changes, in the fall, how many seniors will be attending this local high school? F) P and Q are regular polygons Q has five times as many sides as P The ratio of the measure of an interior angle of Q to the measure of an interior angle of P is 7 to 5 Regular polygon R has more sides than polygon Q and interior angles whose measures are an integral number of degrees Compute the minimum number of degrees in one of the interior angles of R

8 CONTEST 3 - DECEMBER 0 ANSWERS Round Trig: Right Triangles, Laws of Sine and Cosine A) 7 3 B) 08 C) 3 [3, 6, 9,, 5, 8, and 4] Round Arithmetic/Elementary Number Theory A) 49 B) 0,87 C) 0 Round 3 Coordinate Geometry of Lines and Circles A) 3 B) 7 C) X ( ) Round 4 Alg : Log and Exponential Functions 7 9, 0 Y 0, 4 A), 4 or equivalent B) 5, 7 C) 30 k + 0k Round 5 Alg : Ratio, Proportion or Variation A) ( 6, ) B) 5 gal C) 0 Round 6 Plane Geometry: Polygons (no areas) Team Round A) 87 B) 0 C) 60 A) (4, 3, ) D) + B) : 5 E) 6 C) (7, 4), (7, 6), (9, 4), (9, 6) F) 70

9 CONTEST 3 - DECEMBER 0 SOLUTION KEY Round A) Either knowing common Pythagorean Triples or using the Pythagorean Theorem, the 5 hypotenuse has length 3 Therefore, cos B+ cosc = + = sin30 sinc B) By the law of Sines, = = 4 k k k sinc = = Only fractions with denominations containing exclusively powers of, powers of 5 or products thereof have terminating decimal representations Therefore, k must be a positive multiple of 3 and less than or equal to 4 3, 6, 9,, 5, 8,, 4 08 C) Using the Law of Cosines, k = cos A= 80 64cos A 80 k Solving for cosa, cos A = 64 Applying the Triangle Inequality to ΔBAD, the minimum value of k is 5 and the maximum value is Substituting for k, cos A = =, = Thus, the positive difference is = 3

10 CONTEST 3 - DECEMBER 0 SOLUTION KEY Round A) Since primes are odd (except for ), the difference between two primes is odd if and only if one of the two primes is Therefore, the two primes must be and 47, resulting in a sum of 49 B) There are 400(365) days plus 00 leap days minus leap days in 700, 800 and 900 [400(365) ] = days or, dividing by 7, 0,87 weeks FYI: The Julian calendar (aka Julius Caesar) was instituted in 45BC, specifying a year to be exactly 3655 days long A pretty amazing result for the time! The currently accepted year length is days, ie 365 days 5 hours 49 minutes and seconds a discrepancy of 0 minutes and 48 seconds per year Not much of an error, but over the centuries it added up The Gregorian calendar we use today was adopted in most countries in 58 According to a Papal Bull issued by Pope Gregory XIII on /4/58, the day after Thursday 0/4/58 was to be Friday 0/5/58 The accumulated error had grown to 0 days Imagine the reaction from the general population!! People who accept as truth: There are three types of people: those who do math and those who don t The last country to make the transition was Greece on Thursday, March st 93 The day before had been Wednesday, February 5 th (on the old Julian calendar) The gap had grown to 3 days Great stuff! Check it out for yourself C) 00 ( ) = (6) + (4) + = 00 ( ) = (6) + ( 8) + ( ) = 6 Thus, the sum in base 0 is 7 Converting to base ( ), the place values are,, 4, 8, 6, 3, 64, Since 7 is odd and all place values are even except, we must have a in the unit s position (64) + ( 3) = 3 too large 3 +( 8) = 4 too small 4 + (4) = 8 too large 8 + ( )() = 6 too small 6 + () = 7 0 ( ) Alternate solution: (do the arithmetic in the given base!!) What is + in base ( )? (0) = 0 ( ) Why?? [ ( ) + ( ) + 0( ) 0 = (4) + ( ) + 0() = = ]

11 Round 3 CONTEST 3 - DECEMBER 0 SOLUTION KEY A) Applying the Pythagorean Theorem, in right ΔPSR (PS = 3, SR = 4 PR = 5) and in right ΔPRQ (PR = 5, RQ = PQ = 3) Alternate Solution: The Pythagorean Theorem extends to 3D as follows:,, Q x, y, z is The distance between points P( x y z ) and ( ) ( x x ) ( y y ) ( z z ) + + Thus, PQ = ( ) ( ) ( ) ( ) = = 69 =3 B) (x + ) + y = 6 Center (, 0) and r = 6 h = (4 + ) + ( 3 0) = 34 t = 34 6 = 8 t = 7 Since translations do not alter distances, translating the circle and given point one unit to the right (so that the circle is centered at the origin) gives the same result [ x + y = 6 and (5, 3) ] C) The slope of L is 3 and the slope of 4 PQ is 4 3 PR 4 Therefore, RQ = 3 5n = 00 n = Q(7 + 6, 8) = (3, 3) 0 4n The equation of L is 3 R 3n ( y 3) = ( x 3) 3x 4y = 7 Q 4 Letting x, y = 0, we have X ( 9, 0) and Y 7 0, 4 Alternate Solution (Norm Swanson) Since L L, its equation has the form 3x 4y + k = 0 Any point on L is 0 units from L Applying the point to line distance formula, the 3(7) 4() +k distance from (7, ) on L to L is given by = k = 50 k 3 = 50 k = 7,53 3x 4y+ 53 = 0 is a line above L and is rejected (Q would be to the left of P), so the equation of L is 3x 4y 7 = 0 and we proceed as above Z (-,0) P (3, -, 4) 3 P Y r h X t L 5 4 S R Q(6,3, -8) (4, -3) L

12 CONTEST 3 - DECEMBER 0 SOLUTION KEY Round 4 A) + = = + = = Thus, (a, b) =, 4 3x x x 4 3x x x B) x = 0 y = 3 5 = y = 0 log ( 8(x ) ) = 5 8(x ) = 5 = 3 x = ± x = 3, (0, ) to 3,0 5 (0, ) to,0 7 C) Using the formula log a + log b = log ab for a, b > 0, we get A = log + log + log + + log = log = log 000 = 3 + log og 0 = k = k = k log = 04 0 log( ) 0log 0k Thus, A = 3 + = 30 k+ 0k 0k

13 CONTEST 3 - DECEMBER 0 SOLUTION KEY Round 5 A) Given: 4x 3y = If x = y, then 8y 3y = and y is not an integer If y = x, then 4x 6x = x = (x, y) = ( 6, ) B) 5 x = ( 40 ) = 5 8 5(80) 5(5) x = = = 375% or used 5 8 remains in tank FYI: LP turns into a gas and it s the gas that burns when we are grilling As the gas is burned, the bottom of the tank sweats and the sweat line gives us a way of determining how full the tank is after we have been using it C) ( x+ ) ( ) x + z z y xy + yz = 0 = = 0 x z x z y xy yz 00 + yz = = 00 + yz 00 yz 00yz = 0(00) + 00 = 00(0 + ) = 00(0) yz = 0 ( yz) xy = k Generalization: You should verify that if z + x k z x = +, then yz = k +

14 CONTEST 3 - DECEMBER 0 SOLUTION KEY Round 6 A) Since the interior and exterior angles are supplementary, 44k + k = 80 k = 4 An exterior angle of 4 there must be 360/4 = 90 sides In an n-sided polygon the number of diagonals from each vertex is (n 3) 87 diagonals Recall: In the formula for the number of diagonals in a polygon with n sides, P nn ( 3) namely d =, n denoted the number of vertices from which a diagonal could start, n 3 denoted the number of vertices to which a diagonal could be drawn and division by was necessary to avoid counting each diagonal twice B) Let FG = HC = x Then 6 : (8 + x) = 8 : 7 x = 3 DF = 6 and FG = 3 m FDG = 30, m DFE = 0 A E F n = 6 B H D G C C) CL = CE = CB both ΔCLE and ΔCEB are isosceles A BE = EL CLE CEB (SSS) L Let m ECL = m ECB = x Let m CLE = m CEL = m CLB = m LCB = y Then: 4y + x = 360 x + y = 80 In ΔACE, m ACE = 80 (60 + x) = 0 x (0 x) + y = 80 or y = 60 + x Substituting, x + (60 + x) = 80 3x = 60 x = 0, y = 80, m BEL = 60 (60 ) E C B Alternative Solution (Norm Swanson) Draw BL BELC is a kite Let m CLB = m CBL = x and m ELB = m EBL = y Since ΔCLE is isosceles, m CEL = (x + y) As an exterior angle of ΔALE, m BEL = 60 + (80 (x + y)) Thus, (x + y) = 40 (x + y) x + y = 80 m BEL = 60 A (60 ) L E B C

15 CONTEST 3 - DECEMBER 0 SOLUTION KEY Team Round A) Using the Law of Sines on ΔPQS and ΔRQS, sin 30 sin P sin 60 sin R = and = a h b h Squaring each equation and adding, sin 30 sin 60 sin P sin R sin P+ cos P + = + = = a b h h h h (since P and R are complementary) 3 b + 3a + = = (L, M, N) = (4, 3, ) 4a 4b 4ab h P a Q h S b R B) In group A, the first number in each triple is k + We want the 3 rd term in the 3th row, ie we need the triple (7, x, x + ) 7 + x = ( x+ ) 79 = x + x = rd term = 365 = 5(73) In group B, the first number in each triple is 4(k + ) We want the nd term in the 36 th row, ie we need the triple (48, x, x + ) Avoid as much computation as possible 48 + x = ( x+ ) 48 = 4x + 4 4x = 48 = (48 + )(48 ) = 50(46) x = 75(73) = nd term The required ratio is 5(73) 75(73) = 5 C) ( x 5) + ( y ) = ( x ) + ( y 8) ( x 5) + ( y ) = 4 ( x ) + ( y 8) ( ) A(5, ) x 0x+ 5 + y 4y+ 4 = 4( x x+ + y 6y+ 64) = 4x 88x y 64y+ 56 3x 78x+ 3y 60y= 7 3 x 6x + 3 y 0y = = 96 ( + 69) ( + 00) ( 3) ( 0) 3 x + y = (A circle with center at (3, 0) and radius 4 ) We need to examine perfect squares which sum to 3 The list of candidates is, 4, 9, 6, 5 Only = 3! ( x 3) = 6 x 3 =± 4 x= 7,9 ( y 0) = 6 y 0 =± 4 y= 4,6 Thus, the possible points are (7, 4), (7, 6), (9, 4), (9, 6) d P d B (, 8)

16 Team Round - continued CONTEST 3 - DECEMBER 0 SOLUTION KEY log x4 log x 5log x 3 logx 3 7 D) Simplifying , the nd and 3 rd terms are clearly, x and x 5, but let s look carefully at the st term Since x occurs as a base of the logarithm, x x 3 log x = log x log 3 3 3x= log3 = log3x Thus, the st term is, in fact, x 3 log x 3 x Thus, we have x 3 + x + x 5 = 3x 4 x (x 3 3x + x + ) = 0 Since x is the argument of the log function, x > 0 and the only roots come from the cubic factor 3 By synthetic substitution, (x )(x x ) = 0 x = or x = ± 0 x = + ( < 0 and is, therefore, also extraneous) G 7 7 E) First condition: G B B = = G 3 5 Second condition : = B + 8 Cross multiplying, 8G 4 = 5B B 8 B 4 = 5B+ 0 = B+ 0 = 34 B= 374 According to the first condition, G = 7 ( 34 ) = 38 Thus, in the fall, the total number of students is (374 + ) + (38 3) = 6 F) Since Q has five times as many sides as P, the exterior angle of P 360 is five times as large as n 360 the exterior angle of Q The relationship between the interior and the exterior angles of 5n these two polygons is summarized in the following diagram: Thus, 5 j+ 5k = 80 = 7 j+ k j = k interior exterior interior Substituting, 7( k) + k = 80 k = (5j) (5k) (7j) k Since an exterior angle of Q measures, P Q Q must have 30 sides As the number of sides increases, the measure of the interior angle increases This, if R is an N-gon, N must be the smallest factor of 360 larger than 30, ie 36, producing an exterior angle of 0 and an interior angle of 70 exterior

17 Addendum #: Team A Question ΔPQR is a right triangle with right angle at Q PS = a, RS = b and QS divides Q into 30 and 60 as indicated in the diagram at the right La b For positive integers L, M and N, h = Ma + Nb Compute the ordered triple (L, M, N) P a Q 60 h S b R The original team A question specified a right triangle with PQ = 3 and QR = 4 and requested an expression for /h exclusively in terms of a and b In this case, we could have actually solved for a and b (using absinθ ) Q 60 3hsin hsin 60 = h h = = ( ) P a S b 8 6b Using the Law of Sines in triangles QPS and QRS, h= a and h = Equating, a= b, so a = ( ) and b = ( 4 3 ) But none of these values needed to come into play when solving for in terms of a and b h Depending on how you proceeded, different-looking formulas would have been possible besides the official solution Using Stewart s Theorem, we get 9b + 6a 5ab 9b + 6a 5ab 5 3 b + 4 a = h 5+ 5ab h = h = = 5 5 h 9b + 6a 5ab Using the Law of Cosines (twice) we have: 3 h = 9+ a 6acosP= 9+ a 6a 5 9a+ 6b Adding, h = 5 + a + b and we 4 5 h = 6 + b 8bcos R= 6 + b 8b 5 have a much different-looking result for h You could even find the numerical value of h, namely , and then determined coefficients to write this value as a linear combination of a = ( ) and b = ( 4 3 ) 3 3 It s quite challenging to show that all these variants are equivalent for the known values of (h, a, b) R

18 Addendum #: Team A Question The actual question used did not add the restriction the sum L + M + N is a minimum Strangely, no appeal was made and there should have been! Here is the case made by Norm Swanson Hamilton Wenham Q Since no lengths are given, we can choose any convenient lengths Q Furthermore, there are many possible locations for point Q for which 60 m PQR is a right angle, since any point on a semi-circle with center O 60 and diameter PR will suffice By choosing a specific point Q, we can h let PS (with length h) be an altitude to PR, dividing PQR into 30 and 60 angles P S S O Convenient lengths: PQ = QR = 3, PR = 4, a = PS =, b = SR = 3, h = QS = 3 La b 9L 3L M The given formula h = becomes 3 = or = or L= + 3N Ma + Nb M + 9N M + 9N 3 Since all three constants are positive integers, M must be a multiple of 3 and we get the minimum sum if we let M = 3, resulting in L= + 3N Thus, the minimum sum occurs when N = and (L, M, N) = (4, 3, ) However, (L, M, N) is not unique k+ 3 j,3 k, j, where j and k are positive integers, are solutions In fact, any triples of the form ( ) R

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