***** NO CALCULATORS ON THIS ROUND *****

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1 CONTEST 5 FEBRUARY 0 ROUND ALGEBRA : ALGEBRAIC FUNCTIONS ANSWERS A) B) : C) ***** NO CALCULATORS ON THIS ROUND ***** A) Let 4x+ 5 for x> f( x) = 0 for < x 3x 4 for 0 x f f() + f f( ) + f(0) Compute: ( ) ( ) B) f (x) is a function of minimum degree with integer coefficients and zeros of, 6 and i If the lead coefficient is positive and the greatest common factor of the coefficients is, compute the ratio f( ) : 3 f(3) C) Given f(x) = x + x Express f(x) in terms of f(x)

2 CONTEST 5 - FEBRUARY 0 ROUND ARITHMETIC / NUMBER THEORY ANSWERS A) (, ) B) C) (, ) ***** NO CALCULATORS ON THIS ROUND ***** M A) Engelbert practiced proper finger position on the P piano with his right hand Starting with his thumb he plays middle C, followed immediately T by DEFGFED, using his other fingers as follows: pointer, middle, ring, pinky (ie little), ring, middle and pointer Starting again with the thumb he continues to repeat the exercise (ad nauseum) Let F denote the finger used C D E F G T (thumb), P (pointer), M (middle), R (ring), L ( pinky, ie little) Let N denote the note being played Determine the ordered pair (F, N) when he plays his 0 th note and gets to stop for lunch R L B) Determine all primes between 300 and 500 ending in 7 whose digit sum is a multiple of? C) A spinner has 3 equally spaced positions, numbered through 3 clockwise Pointer A is initially pointing at 3 and moves clockwise 7 positions every second Pointer B is initially pointing at 3 and moves counterclockwise 5 positions every second Pointer C is initially pointing at 3 and moves clockwise positions every second Let (a, b, c) denote the numbers being referenced by the pointers A, B and C at one second intervals and S(n) = a + b + c, after n seconds have elapsed For example, at n =, (a, b, c) = (0,, 5) and S() = 6 Compute (n, m), where m = the maximum value of S(n) and n is as small as possible

3 CONTEST 5 - FEBRUARY 0 ROUND 3 TRIG: IDENTITIES AND/OR INVERSE FUNCTIONS ANSWERS A) B) C) (,, ) ***** NO CALCULATORS ON THIS ROUND ***** A) Given: Arc tan ( 075) = Arcsin ( N ) Compute N B) Compute: cos Cos 3 40 Tan 5 9 C) sin 7 cos 7 sin 37 = A B C Compute the ordered triple (A, B, C)

4 CONTEST 5 - FEBRUARY 0 ROUND 4 ALG : WORD PROBLEMS ANSWERS A) mph B) C) ***** NO CALCULATORS ON THIS ROUND ***** A) A train traveled 300 miles in t hours When the speed of the train was increased by 5 mph, it covered 0 more miles in the same amount of time Find the faster rate of the train B) There are three types of coins in the fountain, pennies, dimes and quarters worth $666 If there are 56 pennies, what is the maximum number of coins in the fountain? C) A men can complete a job in B days If C men are taken from the original crew, how many more days will be needed to complete the original job? Give your answer as a simplified expression in terms of A, B and C

5 CONTEST 5 - FEBRUARY 0 ROUND 5 PLANE GEOMETRY: CIRCLES ANSWERS A) B) C) ***** NO CALCULATORS ON THIS ROUND ***** A) Four congruent circles are inscribed in a square If the circumference of each is π, compute the area of the circle circumscribed about this square? B) Point P is in the exterior of circle O Q is the point on circle O closest to P R is the point on circle O farthest from P If PQ = 4 and PR = 0, compute the length of a tangent to circle O from point P C) Circle P with radius is tangent to circle Q with radius 3 at point T Diameter RS is perpendicular to the line of centers at point Q PR intersects the circles at points V and W W V Compute VW P T R Q S

6 CONTEST 5 - FEBRUARY 0 ROUND 6 ALG : SEQUENCES AND SERIES ANSWERS A) B) : C) ***** NO CALCULATORS ON THIS ROUND ***** A) 4x+, 7 x, 8x+ 3 form an arithmetic progression Find the sum of the first 30 terms of this progression B) The following three terms (x + ), (4x + ) and (x + 6) are the second, third and fourth terms of a geometric sequence When is added to the middle term, this sequence of three terms becomes the first three terms of an arithmetic sequence Compute the ratio of the fifth term of the arithmetic sequence to the sixth term of the geometric sequence C) Given: t n =,,,,, A= ( tn) and n = A Compute as a reduced quotient B 3 B n n k = ( i), where i = k =

7 CONTEST 5 - FEBRUARY 0 ROUND 7 TEAM QUESTIONS ANSWERS A) D) B) E) C) (,, ) F) ***** NO CALCULATORS ON THIS ROUND ***** A) Given: f( x) x x 4 3 = +, f hx x x x x 4 3 ( ( )) = If h( x) = Ax + B, where A and B are integer constants, compute h (3) B) Let M(b) be the base 0 representation of the minimum natural number in base b that has a digit sum greater than 0 For example, M(0) = 9 and M(3) = (3) = 845 (0) Compute b = 9 b = 4 Mb () Recall: is the summation symbol (Ex: x= 4 x = = 30 ) x= C) Compute the ordered triple (A, B, C) for which the following equation is an identity, for all values of x for which both sides of the equation are defined tan x( tan x) = A + Bsin ( Cx) ( + tan x) D) In a special summer session of Hogwarts School of Witchcraft and Wizardry, three courses were offered to new students: Charms (C), Potions (P) and Flying (F) Every student chose to take at least one course and some chose to take multiple courses Let XY denote taking both course X and course Y Let X + Y denote taking either course X or course Y (or both) Let n(x) denote the number of students signed up for course X Given: n(c) = 30, n(c + P + F) = 6, n(cpf) = 6 and n(cp) : n(cf) : n(pf) = n(c) : n(p) : n(f) = : 4 : 5 Compute the largest possible number of students who could have signed up just for flying A D P E) AB and CD are chords in circle O that intersect at point E A secant line through points A and D and a line tangent to the circle at point B intersect at point P If m DBA = m ADC + 0, m P = 5 and m AED : m BED = 4 : 5, compute m EBO C E O B F) The sum of an infinite geometric progression with first term a and common multiplier r, is one more than the sum of its first two terms If < a < 6, compute all possible values of r

8 Round Alg : Algebraic Functions MASSACHUSETTS MATHEMATICS LEAGUE CONTEST 5 FEBRUARY 0 ANSWERS A) B) : 8 C) f(x) = 3 f ( x ) + f( x) + 3 Round Arithmetic/ Number Theory A) (R, F) B) 37 C) (5, 9) Round 3 Trig Identities and/or Inverse Functions A) 06 (or 3 5 ) B) 87 C) (, 3, 4) 05 Round 4 Alg : Word Problems A) 80 mph B) 04 C) BC A C Round 5 Geometry: Circles A) 88π B) 4 5 C) 04 (or 5 ) Round 6 Alg : Sequences and Series A) 445 B) 7 8 C) 3 5 i 65 Team Round A) D) 40 B) 476 E) 0 C) 0,,4 F) < r or r 3

9 CONTEST 5 FEBRUARY 0 SOLUTION KEY Round f f() + f f( ) + f(0) = f(0) + f( 0) + f(0) = 45 + (-34) + 0 = A) ( ) ( ) B) f ( x) = ( x+ ) ( x 6)( x + ) ( ) ( ) f ( ) 5 0 = = = 3 f (3) x + C) Replace f(x) with y, ie let y = and solve for x in terms of y x 8 xy y = x + Solving for x y + = xy x = x(y ) x = y + Thus, f(x) = x + + x = y ( y+ ) + ( y ) = = 3 y + y + ( y+ ) ( y ) y + 3 y Round y + y = 3 f ( x )+ f( x)+3 A) The notes are being played in repeating blocks of 8 in the sequence CDEFGFED 0 = 5, r = 4 8 Thus, Engelbert completed 5 blocks and stops on the 4 th note in the 5 nd block (F, N) = (R, F) B) The 3-digit number must be of the form 3x7 or 4x7 digit sum is 0 + x or + x For integers in the 300s, we must test only 37 For integers in the 400s, we must test only 407 Since 407 = (37), we need only rigorously test 37 We must test for divisibility only by primes smaller than 37 Since 9 = 36 > 37, the divisors to be tested are:, 3, 5, 7,, 3 and 7 The first three divisors are easily eliminated 7 r = r = 9 3 r = 5 7 r = Thus, 37 is prime C) Each pointer cycles through the 3 numbered positions, before returned to #3 Pointer A: Pointer B: Pointer C: Sums: Thus, the maximum sum is 9 and it occurs first for n = 5 (5, 9)

10 CONTEST 5 FEBRUARY 0 SOLUTION KEY Round 3 Y A) Think right triangle! Arc tan ( 075) is the measure of the smaller acute angle, y OPP 3 namely k, since tan(k ) = = = 075 = x ADJ 4 3 sin(k ) = (or 06) Recall: SOH-CAH-TOA 5 k X B) cos(a B) = cosacosb + sinasinb = = = C) sin 7 cos 7 sin 37 = sin5 cos75 = sin 5 = ( sin(45 30) ) = ( sin 45cos30 sin 30cos 45) = 8 6 = = 3 4 = (A, B, C) = (, 3, 4) (-3, 4) A B 4 (9, - 40) Round A) + 5 = t = 30 t = 4 t t Thus, the faster rate is 30 = 80 mph 4 B) Let D and Q denote the number of dimes and quarters respectively Then: 0D + 5Q = = 50 D + 5Q = 0 (D, Q) = (, 0), (6, 8),, (46, ), (5, 0) Since there are 3 types of coins in the fountain (5, 0) is rejected and the maximum number of coins in the fountain is = 04 C) Let X denote the number of days it would take for (A C) men to complete the job The time it takes to complete a job is inversely proportional to the size of the works force The larger A X AB the workforce, the less time it takes to complete the job Thus, = X = A C B A C AB AB B( A C) BC Additional days = x B = B = = A C A C A C

11 CONTEST 5 FEBRUARY 0 SOLUTION KEY Round 5 A) C = π radius r = 6 diameter d = side of square s = 4 diagonal of the square = 4 radius of cc circle = area = 88π T B) Draw PO This line intersects the circle twice The point between P and O is the closet point, Q The other point of intersection is point R PR = PQ + QR 0 = 4 + QR Since QR is a diameter, the radius of circle O is 8 Using the Pythagorean Theorem on TPO, PT + 8 = PT = 80 =4 5 P Q r O r R C) RQP is a right triangle (with area 6) Thus, ( QX )( PR) = 6 QX 5= 6 QX = 5 Note since RXQ ~ RQP, RXQ is a scaled version of a triangle 3,,3 = (,,5) = (4,,5) Rather than grinding out the Pythagorean Theorem for RXQ, we see that XR = 3 (3) = 9 = Therefore, WR = 36 and VW = 5 36 = 04 P X W V T 3 R S 3 Q

12 CONTEST 5 FEBRUARY 0 SOLUTION KEY Round 6 A) 8x+ 3 7x= 7 x (4x+ ) x = 4 x = AP: 9, 4, 9, 30 a = 9, d = 5 and n = 30 S 30 = ( (9) + (30 )5 ) = 5( ) = 5(63) = 445 B) The first three terms of the AP are (x + ), (4x + 4) and (x + 6) The common difference d = (4x + 4) (x + ) = (x + 6) (4x + 4) 3x + = 8x 8 x = 4 and d = 4 The AP is 6, 30, 54, and the GP is 6, 8, 54, The required ratio is ( ) : (54 9) = 0 : 486 = 7 : 8 C) t n generates a geometric sequence, where r = and a = 3 a 6 Since r <, the infinite geometric series converges to = = r B3 = ( i) + ( i) + ( i) = i+ ( i+ ) + ( i ) = 5i A 6 + 5i 6 30i 3 5i = = = B 5 5i + 5i 5(6) 65 3

13 CONTEST 5 - FEBRUARY 0 SOLUTION KEY Team Round A) f x x x x x ( ) = + = ( + ) 3 3 Thus, ( ) ( ) ( ) f( hx ( )) = hx ( ) hx ( ) + = Ax+ B ( Ax+ (B+ )) Expanding ( Ax + B) 3 ( Ax + (B + )), the lead coefficient would be 4 A and the constant term would be 3 4 B (B+ ) Therefore, A = 3 A=± and B (B+ ) = B + B = 0 B= Consequently, hx ( ) = x or x+ However, checking the other coefficients of f(h(x)), only A = produces the correct coefficients for x 3, x and x and h(x) = x only x + 4 Thus, h ( x) = and h (3) = = B) Base 4: 333 (4) = (4 3 ) + (4 3 ) = 3(64) = 9 Base 5: 344 (5) = 3(5 ) + (5 ) = 99 Base 6: 55 (6) = = 7 Base 7: 56 (7) = = 4 Base 8: 47 (8) = = 39 Base 9: 38 (9) = = 35 Total : 476 C) sin x cos x sin x tan x tan x cos x cos x cos x sin x = = cos ( ) sin x 4 3 ( + tan x) sec x cos x ( ) ( ) = sin xcos x cos x sin x = sin xcosx= sin xcosx = sin 4x Thus, (A, B, C) = 0,,4 4 x

14 CONTEST 5 - FEBRUARY 0 SOLUTION KEY Team Round - continued C P D) (A classic Venn Diagram problem) a 6 a b 6 b a 6 a b d e a+ 6 5a+ 8 = 5a+ 30 = c+ c= c To insure that c is an integer, a must be even 6 Since the total number of students involved is 6, to maximize f, we must minimize a Let n(c) = N, n(p) = 4N and n(f) = 5N b c If a =, (b, c) = (0, 4) f d + 8 = N e + = 4N and 30 + f 5N F ( ) + d + e+ f + 6 = 6 d + e+ f = 84 Consequently, N 8 + 4N + 5N 30 = 84 N = 54 N = 4 f = =40 ( ) ( ) ( ) E) Since angles AED and BED are supplementary, a 4 : 5 ratio implies m AED = 80 and m BED = 00 Let m ADC = x and m DBA = (x + 0) and m BD A = y ( ) As arcs subtended by inscribed angles, m AC = x and m AD = x + 0 ( ) As a leftover arc, ( ) ( ) ( ) ( ) m BC = x y As an angle formed by intersecting chords, m BED = ( x + y) = 00 x + y = 00 As an angle formed by a tangent and a secant line m P= (( 340 x y) y) = x y = 0 x+ y = 65 Thus, x = 35, y = 30 m( AD) = 90 m( ADB) = 0 If BO intersects the circle in point X, then m AX = = As an inscribed angle, m EBO = 0 C E O ( ) D B P

15 CONTEST 5 - FEBRUARY 0 SOLUTION KEY Team Round - continued a F) An infinite geometric series converges to a sum if and only if r < r a r Thus, = a + ar + = a( + r) + a = a( r ) + ( r) ar + r = 0 a = r r r r + r 0 This means 6 r r and r 6r or r 6r + r 0 We must take the intersection of these two conditions r + r 0 (r )( r+ ) 0 6r + r 0 (r+ )(3r ) 0 The first condition requires that x, but convergence requires r ; hence, < x The second condition requires r or r, but convergence requires < r < ; hence, 3 < r or r < 3 - / - -/ /3 Carefully taking the intersection, we have < r or r 3