MASSACHUSETTS MATHEMATICS LEAGUE CONTEST 3 DECEMBER 2013 ROUND 1 TRIG: RIGHT ANGLE PROBLEMS, LAWS OF SINES AND COSINES

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1 CONTEST 3 DECEMBER 03 ROUND TRIG: RIGHT ANGLE PROBLEMS, LAWS OF SINES AND COSINES ANSWERS A) B) C) A) The sides of right ΔABC are, and 7, where < < 7. A is the larger aute angle. Compute the tan( A). B) In retangle ABCD, AB = 4 and BC = 4. Point P is loated on BC suh that BP : PC = 6 : 5. Compute sinθ. A B θ P D C C) ΔABC has sides in the ratio of 4 : 5 : 6. If the area of ΔABC is 375 7, then ompute the perimeter of ABC.

2 CONTEST 3 - DECEMBER 03 ROUND ARITHMETIC/NUMBER THEORY ANSWERS A) B) 0 C) A) A and B are perfet squares. There are no perfet squares between A and B. If both A and B are 3-digit integers, what is the maimum value of A B? B) Today (-5-03) falls on a Thursday. In what year does -5 net fall on a Thursday? C) Let d be the smallest odd digit that does not appear in the deimal equivalent of 7. Consider a list of all positive odd 4-digit integers N with distint digits whih satisfies these onditions: it is a multiple of it is a multiple of d it is not divisible by 88% of the 5 primes less than 00. This list is sorted in order of inreasing digitsum. Integers with the same digitsum are sorted in inreasing order of magnitude. What is the seond integer in the list?

3 CONTEST 3 - DECEMBER 03 ROUND 3 COORDINATE GEOMETRY OF LINES AND CIRCLES ANSWERS A) (, ) B) A(, ) B(, ) C) A) The slope and the y-interept of the line with equation y + = are m and b respetively. 5 4 Compute the ordered pair (m, b). B) AB is the diameter of the irle 4 + 4y + 0y+ 8 = 0 parallel to the -ais. Compute the endpoints of A and B, given that A is to the left of B. C) Line L with a slope of passes through the point P(3, ). Line L is tangent to a irle h + y k = r. with enter C(3, ). Find the equation of this irle in the form ( ) ( )

4 CONTEST 3 - DECEMBER 03 ROUND 4 ALG : LOG & EXPONENTIAL FUNCTIONS y A) Given: For positive integers and y, = y = 6 Compute all possible values of log log y+ y. y ANSWERS A) B) = C) B) Given: b > 0 ( b ) and > 0 Solve for in terms of b. log log + log 5 = 4 b b 3 b C) Given: f( ) = A B If f( A ) = 8 and f( B ) = 4 for A > 0 and B > 0, ompute.

5 CONTEST 3 - DECEMBER 03 ROUND 5 ALG : RATIO, PROPORTION OR VARIATION ANSWERS A) lbs B) M (, ) C) r = A) Two hildren have weights of 60 lbs and 80 lbs. They sit on the right side of the seesaw at points A and B in the diagram below. A third hild sits at point C and balanes the seesaw. What is the maimum weight of the third hild?... C 0 0 A 6 B Note: The seesaw balanes if the sum of the produts of the weight and the orresponding distane from the pivot point on the right side is the equal to the produt of the weight and distane from the pivot point on the left side. B) y varies diretly as, and y varies inversely as. Speifially, 3 y = f( ) = 3+ 8 and. y = g ( ) = The graphs of f( ) and g ( ) interset in two points A and B. Compute the oordinates of the midpoint M of AB. C) On a roundtrip training run between his home (H) and the beah (B), Roky traveled over the same route both ways. Let R denote his rate when he ran from H to B. His average overall rate was only 3 R beause (due to ramps) he had to slow down returning from B to H. 4 In terms of R, ompute r, his average rate on the return trip.

6 CONTEST 3 - DECEMBER 03 ROUND 6 PLANE GEOMETRY: POLYGONS (no areas) ANSWERS A) A regular polygon has 88 more diagonals than sides. Compute the degree-measure of an eterior angle of this polygon. A) B) C) B) Regular pentagon MINDY and regular deagon BLACKSMITH lie on opposite sides of line MI. List the three degree-measures of the angles in TIN from smallest to largest. C) Square ABCD has a side of length. Equilateral triangle CEF has side of length. Points B, C and F are ollinear, as are eah of these sets of three points: P, B and A A, D and R R, F and Q Q, E and P The diagonal in retangle PQRA has length d. d Compute the numerial value of the ratio as a single fration with a rationalized denominator. P E Q B C F A D R

7 CONTEST 3 - DECEMBER 03 ROUND 7 TEAM QUESTIONS ANSWERS A) D) B) E) C) F) A) The sides of a unique right triangle ABC are, in order of inreasing magnitude,, 8 and. The sides are integer lengths with no ommon fator (other than ). Compute the perimeter of this triangle. Note: [ N ] denotes the greatest integer less than or equal to N. For positive real numbers, the frational part is trunated (dropped). B) Definitions: y = + y (arithmeti average) and y = y + y Under whih of the following ondition(s) does the operation distribute over (harmoni average). the operation, i.e. a (b ) = (a b) (a ), where a, b and are non-negative integers for whih both sides of the equality are defined. Cirle your hoie(s) above. ) a = 0 ) b = 0 3) = 0 4) a = b 5) a = 6) b = 7) none of the above C) A line L through the point Q(, 3) divides the irle P with equation + y 8+ 0y 3 = 0 into two ongruent regions. A line L passes through the enter of the irle perpendiular to L. Let R be the -interept of L and S be the y-interept of L respetively. Compute the area of quadrilateral PROS, where O denotes the origin. 4 log 0 D) For 0 < < 0000, define the funtion f( ) =. Let M be the minimum value and N be the maimum value that f() may take on. Compute M N. E) The area of a ertain quadrilateral varies jointly as the distane D between two parallel sides and the sum S of the lengths of these parallel sides. Two suh quadrilaterals are initially ongruent and, therefore, have the same area. In the first quadrilateral, D is multiplied by a fator of 9n and S is unhanged. In the seond quadrilateral, S is multiplied by a fator of 7n 0 and D is unhanged. Compute all possible values of n for whih the areas of these quadrilaterals remain equal. F) A regular polygon has verties VVV,, 3,..., V. n 7 diagonals an be drawn from eah verte. Let P be the point of intersetion between diagonal VV i i + 3 and Vi + Vi + 5, where i 0. Compute m Vi+ PVi+ 3.

8 CONTEST 3 - DECEMBER 03 ANSWERS Round Trig: Right Triangles, Laws of Sine and Cosine A) 4 3 B) C) 50 Round Arithmeti/Elementary Number Theory A) 6 B) 09 C) 03 Round 3 Coordinate Geometry of Lines and Cirles A) 8,4 5 B) A,, B,, C) ( ) ( y ) = 0 Round 4 Alg : Log and Eponential Funtions A) ±.5 B) = Round 5 Alg : Ratio, Proportion or Variation 60 3 b C) A) 94 4 B),4 3 C) 3 R 3 (or R or 0.6R ) 5 5 (The zero is not required.) Round 6 Plane Geometry: Polygons (no areas) A).5 B) C) (or equivalent) (Aept A) and B) with or without the degree symbol.) Team Round A) 306 D) 0,000 B) and 6 only E) 4 5, 3 3 C) 33 8 (or 6.65) F) 53

9 CONTEST 3 - DECEMBER 03 SOLUTION KEY Round A) Sine 7 must be the length of the hypotenuse, = 4 3. Sine A is the larger aute angle, it must be opposite the longer side. SOHCAHTOA tan( A) = 4 3. B) BP : PC = 6 : 5 and BC = 4 BC = 3, CP = 0. Using speial Pythagorean triples of and 5--3, we have in ΔPCD, (0, 4,? ) = (5,,?) PD = 6 in ΔABP, (4, 3,? ) = 8(3, 4,?) AP = 40 In ΔPCD, os( ) = 4 = 6 3 Angles and are omplementary, so sin( ) = os( ) = 3. Using the Law of Sines in ΔAPD, sinθ sin 4 = sinθ = = C) Let the three sides have lengths 4k, 5k and 6k. The smallest angle θ will be opposite the side of length 4k. Using the law of Cosines, 6k = 5k + 36k 60k osθ. 3 7 k 0 60os θ = ( ) = 45 osθ = sinθ = + (sine θ must be aute) 4 4 The area of any triangle an be omputed as absinc, where C is the inluded angle. 7 Thus, the area of ΔABC is ( 5k)( 6k) = k 4 = 4(375). k = 0 and the perimeter is 50. A 4 D 4 4 θ C B P

10 CONTEST 3 - DECEMBER 03 SOLUTION KEY Round A) A and B must be squares of onseutive integers. Sine the gap between onseutives squares grows as the squares get larger, we take the two largest 3-digit perfet squares. 30 = 900, 3 = 96, but 3 = 04, so the maimum differene is = 6. B) Today is /5/03. Let DOW denote day of the week. The DOW sequene is MonTueWedThuFriSatSunMon. There are 365 days in a year (unless it s a leap year, in whih ase February 9 th makes days). In a 365 day year, there are = 5 7 weeks, plus etra day. Thus, from one year to the net, a speifi date advanes one day of the week, unless there is an intervening leapday! 0 was a leap year and 06, 00, 04 will be as well. For eample, /5/06 falls DOWs after /5/05, beause of the etra day /9/06. The sequene of DOWs for /5, starting in 03 is Thu, Fri(04), Sat(05), Mon(06), Tue (07), Wed(08), Thu(09). C) = d = 3 Therefore, N is an odd multiple of 33. There are 5 primes less than 00:,3,5,7,,3,7,9, 3,9, 3,37, 4,43,47, 53,59, 6,67, 7,73,79, 83,89, 97 88% of 5 is N is divisible by eatly 3 distint primes, inluding 3 and. The smallest digit sum is 6 if N an be formed using the digits 0,, and 3. All other sets of possible digits with a digit sum of 6 will have at least one repeated digit. A 4-digit integer onsisting of digits 0,, and 3 will always be divisible by 3. There are 4 arrangements of these N-values divisible, only 8 are 4-digit numbers and only of these are odd. Divisibility by narrows the field to : 03 and 03 The smallest value 03 = 3 3 and the net smallest is 03 = Thus, N = 03. (Without the restrition that the digits were distint, we would have had to onsider N-values formed from {,,, }, {,,, 3}, {,, 0, 4} and {,,, 0}. Only the first set of digits produes a multiple of. In this ase, the seond integer in the list would have been.) We ould also have proeeded by brute fore, eamining produts of the form 3, where 000 is a prime suh that > = : 03 (smallest) 37: (rejeted, repeated digits) 4: 353 (rejeted, digit sum = ) 43: 49 (rejeted, digit sum = 5) 47: 55 (rejeted, digit sum = ) 53: 749 (rejeted digit sum = ) 59: 947 (rejeted, digit sum = ) 6: 03 Bingo!

11 CONTEST 3 - DECEMBER 03 SOLUTION KEY Round 3 A) Multiplying by 60, y + = 8+ 5y = 60 y = Sine the equation is now in y = m + b form, the required order pair is 8,4. 5 B) y + 5y = y+ = Center:, and radius 3 5 ±, A,, B,, C) The equation of line L is ( y+ ) = ( 3) + y = 9 A radius through the enter drawn to the point of ontat will have slope +. Its equation is ( y+ ) = ( 3) y = 8 The point of tangeny T is the intersetion of these two lines. + y = 9 + y = 9 5= 5 ( y, ) = (5,) y = 8 4 y = 6 The radius of this irle is the distane from C to T, namely ( ) + + (5 3) = 0. Therefore, the required equation is ( 3 ) + ( y + ) = 0. Alternately, the point-to-line distane formula ould have been used to find the radius. 3 + ( ) + ( 9) 0 The distane from ( 3, ) to + y 9= 0 is = = 5 whih is the + 5 radius of the given irle, resulting in the same equation as above.

12 CONTEST 3 - DECEMBER 03 SOLUTION KEY Round 4 4 A) Sine = 6 and 4 = 6, we have (, y) = (,4) or (4, ). log y+ log y log4 + log4 + Therefore, = = = ±.5. y ± ± How do you argue that there are no other ordered pairs of positive integers??? /3 /5 B) logb logb( ) logb( ) b + = 4 4 3/5 log b = Raising eah side to the 5/3 th power, = /5 /3 8/5 = 4 log b 5/ b 3/5 = log b ( ) = 4 C) If f( ) = = k, then k k k ( ) k = = = 0 = 0 If N =, then we have N k± k + 4 kn = 0 or N = A 8± For = A and k = 8, we have N = = = B 4± For = B and k = 4, we have N = = = + 5 Thus, 4 7 ( 5) A B = + + =

13 CONTEST 3 - DECEMBER 03 SOLUTION KEY Round 5 A) The heavier hild should sit farther from the pivot point. 60(0) + 80(0 + 6) = W(0) = 880 = 0W W = 94. Note: If the heavier hild sits loser to the pivot point on the right side, we would have 60(0 + 6) + 80(0) = W(0) W = = 88, a smaller value.... W C 0 0 A 6 B 3 B) y = 3+ 8 and y = A B ( 3) 9 ( ) Applying the midpoint formula, M 3 + 4, =,4. 3 ( 3 + 8) = = ( 3 )( + 3) = 0 =, 3,9, ( 3, ) rr C) His overall average rate is the harmoni average of his rates - NOT the arithmeti r+ r r+ r average. A good topi of disussion with your teammates/oah! Sine r denotes his average return rate, we have rr 3 3R R r R = 4 = Cross multiplying, 8rR = 3Rr + 3R. Dividing through by R, + 4 8r = 3r+ 3R 5r = 3R or r = 3. 5R (Aeptable alternative forms are listed on the answer key.)

14 CONTEST 3 - DECEMBER 03 SOLUTION KEY Round 6 nn ( 3) A) A regular polygon with n sides (sometimes referred to as an n-gon) has and the eterior angle ontains 360. Thus, we require that n nn ( 3) = n+ 88 n 5n 76 = 0 Fatoring, we have ( n+ )( n 6) = 0 n= 6 and the eterior angle measure is = 45 ( or.5 ). diagonals B) The angles of a regular pentagon are eah 80(5 ) = 08 ; the angles of a regular deagon are 5 80(0 ) = 44. Therefore, 0 K m TIN = 360 ( m MIN + m MIT ) = 360 ( ) = 08. S C Sine TIN is isoseles, its base angles eah measure Y M D A = = 36. I Thus, the required sequene is 36, 36, 08. N L T B H C) Sine EM = 3, AR =, we have d ( ) = d = = = + 3 = P E Q d C B F M A D R

15 CONTEST 3 - DECEMBER 03 SOLUTION KEY Team Round A) Your first line of attak might be to try the Pythagorean Theorem. + ( ) = 8 = ( ) = 8 8 But then what?? A quiker solution would be to list the Pythagorean triples where the length of the hypotenuse is more than the long leg and note a pattern The gap between the long legs is growing by 4 as the gap between the short legs remains onstant at. Sine , 5 8, 7 8, and 9 8, we must ontinue the pattern = 7 = = 0 = = 4 = = 8 = 7 8 Bingo! Therefore, the perimeter of ABC is =306.

16 CONTEST 3 - DECEMBER 03 SOLUTION KEY Team Round + y B) Definitions: y = (arithmeti average) and y = y (harmoni average) + y Method #: Let (a, b, ) = (, 0, ). (Testing onditions and 5) (0 ) = 0 = / and ( 0) ( ) = (/) = /3 Sine the results are unequal, in general, onditions ) and 5) fail. Let (a, b, ) = (,, 0). (Testing onditions 3 and 4) ( 0) = 0 = / and ( ) ( 0) = (/) = /3 Sine the results are unequal, in general, onditions 3) and 4) fail. All my attempts to eliminate and/or 6 have failed. I an assume and 6 always results in equality. But what if I missed ordered triples whih would have eliminated one or both onditions? Method #: (brute fore substitution) Compute formulas for a (b ) and (a b) (a ) and then substitute for eah of the onditions and find formulas for eah epression. b a (b ) = a b a + = b+ ab + a + b = b+ ( b+ ) (#) a+ b a+ (a b) (a ) = = ( a+ b )( a+ ) a+ b a+ a+ b+ + (#) b+ Both sides are defined provided, provided b and a. Verdit b = b b (0 b) (0 ) = ( b+ ) b+ b+ ok = a aa ( + ) (a 0) (a ) = a+ fails b = a (b 0) (a b) (a ( a+ ba ) 0) = a+ b fails b + 3b (b b) (b ) = ( bb+ ) ( b+ ) 3b+ fails + 3b ( b) ( ) = ( + b ) ( b+ ) b+ 3 fails a + (a ) (a ) = ( a+ )( a+ ) 4 a+ + = ( ) a+ a+ ( a+ ) a+ = ( 0) = = (a + 0) 4 ( a+ ) ok ) a = 0 0 (b ) = ) b = 0 a (0 ) = a 3) = 0 a (b 0) = ab 4) a = b b (b ) = 5) a = (b ) = 6) b = a ( ) =

CONTEST 3 - DECEMBER 03 SOLUTION KEY Team Round B) ontinued ab + a + b Method #3: In general, a (b ) = (a b) (a ) = ( a+ b )( a+ ) ( b+ ) a+ b+ Instead of substituting for 6 speial ases, algebraially

17 CONTEST 3 - DECEMBER 03 SOLUTION KEY Team Round B) ontinued ab + a + b Method #3: In general, a (b ) = (a b) (a ) = ( a+ b )( a+ ) ( b+ ) a+ b+ Instead of substituting for 6 speial ases, algebraially manipulate the equality. Cross multiplying, we have equality if and only if (ab + a + b)(a + b + ) = (b + )(a + b)(a + ) or a b + a + 4ab + ab + ab + b + ab + a + b = (a + a + ab + b)(b + ) = a b + ab + ab + b + a + a + ab + b 6ab + ab + a = 4ab + ab + a 0 = ab + ab + a 0 = a(b b + ) = 0 0 = a(b ) a = 0 or b = Thus, the distributive property is satisfied under onditions and 6. C) + y 8+ 0y 3 = 0 ( 4) + ( y+ 5) = 64 Sine the line L must divide the irle into semi-irles, it must pass through the enter of the irle. Thus, L passes through Q(, 3) and P(4, 5). Its equation is ( y 3) = ( + ) y 3 = ( + ) 4+ 3y = 4 ( ) (slope ). L has slope +, passes through (4, 5) and 3 4 has equation 3 4y = 3. The y-interepts are and 8. 3 The area of quadrilateral PROS equals the area of PTS minus the area of ORT, namely, = = = = The following proedure works for any onve polygon whose verties are known. Start at any verte and list the verties in order (lokwise or ounterlokwise your hoie). Repeat the oordinates of the starting verte. The area is given by half the absolute value of the sum of the downward diagonal produts minus the sum of the upward diagonal produts. 0 0 / = 3= =

18 CONTEST 3 - DECEMBER 03 SOLUTION KEY Team Round 4 log 0 D) f( ) = By inspetion, we are tempted to let = 0, getting M = f O 4 (0) θ θ θ θ θ P = =, 4 4 let = 0000, getting N = f (0000) = 4 = and assume these are the maimum and minimum values respetively. However, this is not the ase. log 0 f( ) = ( 4 log0 ) log0 Taking the logarithm (base 0) of both sides, we have = log + 4log ( ) 0 0 This is a quadrati epression in log0. For larity, let A= log0. Completing the square, we have ( ) ( ) Think parabola opening down with A + 4A= A 4A = A + 4 maimum value at the verte (, 4). Thus, the funtion attains a maimum when A= log0 = or = f (00) = 00 = 0000 and M N = E) Joint variation implies as the produt. Let A and A denote the areas of the transformed quadrilaterals. The new area A is given by k( 9nD) S, while A = kd(7n 0) S A= A' k9n D S = k D (7n 0) S 9n 7n+ 0 = 0 (3n 4)(3n 5) = 0 n= 4, F) If there are 7 diagonals from eah verte, then n 3 = 7 and the polygon has 0 sides. Here s a sketh of the pertinent portion of the 0-gon. Eah side of the 0-gon subtends (uts off) 360 = 8 of ar, 0 i.e. 0 of the irumsribed irle. Therefore, eah V 4 V 5 V 6 V entral angle θ measures 8. We require m V 3 PV4. V Consider quadrilateral PVV 3V 4. At V, the insribed angle VVV 6 3 measures ( 3 8 ) = 7. At V 3, m VVV 3 4 = ( 8 8 ) = 6. At V 4, m VVV 4 3= ( 8 ) = 8. Therefore, m P= ( ) = = 53. Alternate solution: (Angle formed by two hords in a irle) Let denote an ar ut off by two suessive verties of the 0-gon. = 360 / 0 = 8 The pair of vertial angles at P in whih we are interested ut off (subtend) ars of and 5. m P= ( ) = 7 8 = 7 9 =53. V

To derive the other Pythagorean Identities, divide the entire equation by + = θ = sin. sinθ cosθ tanθ = 1

To derive the other Pythagorean Identities, divide the entire equation by + = θ = sin. sinθ cosθ tanθ = 1 Syllabus Objetives: 3.3 The student will simplify trigonometri expressions and prove trigonometri identities (fundamental identities). 3.4 The student will solve trigonometri equations with and without