MASSACHUSETTS MATHEMATICS LEAGUE CONTEST 2 NOVEMBER 2014 ROUND 1 COMPLEX NUMBERS (No Trig) ANSWERS

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1 ) Given: ( ) MSSHUSETTS MTHEMTIS LEGUE ONTEST NOVEMER 014 ROUN 1 OMPLEX NUMERS (No Trig) NSWERS 3+ 4i + a + bi = 3+ ki, where a and b are real. ompute k such that a= b. ) ) ) ) Given: 1 5i = x + yi for real numbers x and y. x ompute 4 y. 3 3 ( 3 3 3i ) ( i ) ) If 108 = x + yi, compute x + y. 3 3

2 ONTEST NOVEMER 014 SOLUTION KEY Round 1 ) ( ) ( ) ( ) 3+ 4i + a + bi = i + a + bi = a 7 + b + 4 i = 3+ ki Equating the real and imaginary parts, a = 10 and k = b+ 4 b= k 4. Therefore, k 4 = 10 k =14. x y = 1 ) Squaring both sides of 1 5i = x + yi, xy = 5. x + y = 13 Where did this third equation come from? We could have solved the second equation for y in terms of x and substituted in the first, but adding the first and third will be much easier. onsider 1 5i - the absolute value of the radicand. s on the real number line, the absolute value of a complex number is its distance from the origin O ( 0,0 ). Let x + yi be represented by the point P ( xy, ) in the complex plane and we have regardless of the quadrant in which P is located. Extracting, ( ) ( ) x + yi = OP = x + y, = = 169 = 13 (or recall the Pythagorean Triple ) Thus, adding the first and third equations and dividing by, x = =. 4 1 x Subtracting the same equations, y = 1 y =. Thus, = 50. Note that: 4 4 y 1 5i denotes either 5 i or 5 + i. In both cases, xy = 5 and x y = 1. ) Simply! Simplify! Simplify! Recall: ( ) ± = ± ± Thus, in the expansion, the first and third terms cancel i i = 6 i i = 3 i. Therefore, ( ) ( ) ( ) ( 3 3 3i) ( i) Since = 3 and = 1, we have 3 3 Thus, ( ) 3 = 3 x + y = 4 = 64. (( 3 i) ( ) ) 3 3+ i ( ) ( 1) 33 ( 1) i + = = 4 (, ) ( 0, 4) x yi i 4 xy =.

3 ONTEST - NOVEMER 014 ROUN LGER 1: NYTHING NSWERS ) x = ) ) (,, ) ) For some positive constant, if x = 1 is a solution of the equation x = 5, what is the other solution? ) Given: (x )(3x + ) = 0 for positive integer constants and. ompute, if + = 7 and the sum of the solutions (for x) is an integer. x y ) Given: x# y = + for integers x and y. 3 x# y = k For some minimum integer k > 10, x+ 3y = k. ompute the ordered triple ( kxy,, ).

4 ONTEST NOVEMER 014 SOLUTION KEY Round ) 1 = 5 1 =± 5 = 1± 5= 4,6 x 6 = 5 x 6 =± 5 x= 1, 11 3 ) The sum of the solutions is + =. 3 6 Testing the 6 possible ordered pairs (6, 1), (5, ), (4, 3), (3, 4), (, 5) and (1, 6), 3(4) (3) only (4, 3) produces an integer solution sum = 1 6 = 16 9 =7. ) x y + = k (1) 3x+ y = 6k 3 () x+ 3y = k 7k dding the two equations and dividing by 5, we have x+ y =. 5 Since we were given that k > 10 and x and y must be integers, k min = 15. Substituting for y in (), x+ 3( 1 x) = 15 x= 48 ( ) 15, 48, 7.

5 ONTEST - NOVEMER 014 ROUN 3 PLNE GEOMETRY: RES OF RETILINER FIGURES NSWERS ) ) ) ), a square with area 5, is subdivided into squares and rectangles by perpendiculars that intersect at point E. If E = 3, compute the area of the shaded region. E ) side of the regular octagon EFGH is. ompute the area of the square EG. H E F G ) In, the altitude is drawn to the hypotenuse of a right triangle, intersecting the hypotenuse in point. From point, altitudes are drawn to the legs, intersecting in point P and intersecting in point Q. ompute the area of rectangle PQ, as a ratio of relatively prime integers. P Q

6 Round 3 MSSHUSETTS MTHEMTIS LEGUE ONTEST - NOVEMER 014 SOLUTION KEY ) ( x + y) = 5 = 15 y E = 3 x = 4, y = 11 Therefore, the area of the shaded region is ( 4 11) =88. E x ) Using Pythagorean Theorem on right ΔP, = ( ) = = 4+ P H G ) The area of Δ is = 5 h h =. 5 Proceed by the Pythagorean Theorem = x + x = = = ( ) Thus, = 9 5 and = lternately, 3 P x h 5 Q E 4 F ~ ~ = 3 x = x = = or, invoking the fact 4 h that the altitude to the hypotenuse is the mean proportional between the segments on the hypotenuse, h = x( 5 x) y similar arguments for triangles and, P = and Q = =

7 ONTEST - NOVEMER 014 ROUN 4 LG 1: FTORING N ITS PPLITIONS NSWERS ) ompute the greatest common factor of 4x yz 3 and 90x 3 z 4. ) ) ) ) ompute all rational values of x satisfying ( x )( x ) = 3. ) Given: x( x ) + ( + 5) 4= 5( x+ 4) Solve for x in terms of.

8 ONTEST - NOVEMER 014 SOLUTION KEY Round 4 ) s the product of primes, 4 = (3)(7) and 90 = (3) 5. Taking the smallest exponents of the common factors, The numerical component is (3) = 6 and the literal component is x z 3. Therefore, the GF is 6x z 3. ) Since the product is not equal to zero, the factorization does not help. Multiplying out the left side and combining like terms, we have ( 3x+ 4)( 8x 5) = 3 4x + 17x+ 3 = 0. Since the coefficient of the middle term is odd, the factors of 4 cannot both be even, leaving only possibilities of 4 1 or 8 3. If these fail, we would have to use the quadratic formula and none of the solutions would be rational. Since 4x + 17x+ 3 = (8x+ 3)(3x+ 1) = 0 and we have rational solutions, namely, 3 1 x =, 8 3. ) ( ) ( ) ( ) ( ) x x = 5 x + 4 x x x 4 = 0 ( x ) ( x ) ( x )( x ) 5 4 = = 0 x = +8 or x = 3

9 ONTEST - NOVEMER 014 ROUN 5 TRIG: FUNTIONS OF SPEIL NGLES NSWERS ) ) ) (, ) tan x 1 ) etermine all values of x for which ( sin x 1 ) cos x ( tan x 3 ) interval 0 x 180. Express your answer(s) in degrees. is undefined over the ) The graph of f ( x) sin( x) sin( x) = has a period of π as is easily seen in the graph below 0 π/ π 3π/ π ompute 60π 60π π f + f f ) Given: m = 10, = = 11 EFGH is a square If the area of EFGH is M N 3, compute the ordered pair ( M, N ). E H F G

10 ONTEST NOVEMER 014 SOLUTION KEY Round 5 ) The value of the fraction is undefined only when the denominator is zero, or when one of the trig functions is undefined. We do not consider values of x for which the numerator is zero, namely 45, since for this value the denominator is not 0. 1 sin x 1 = 0 sin x= x=30,150 cos x= 0 x=90 (lso, tan x is undefined for x = 90.) tan x 3 = 0 tan x=± 3 x=60,10 ) Since sin( x) = sin(x) and sin( x) = sin(x), it follows that f ( x) = sin( x) sin( x) = sin( x) sin( x) = f ( x). Thus, since 60π = 00 π, with a period of π, we can disregard 00π f π f π f π sin π sin π = = = = = π π π f = sin sin = 1 = With these observations, the given expression evaluates to =. 4 (sk your coach/teammates about even and odd functions.) ) F + FG + G = x 3 + x= 11 3 x= = = Therefore, the area of EFGH is ( 6 3) = = ( M, N ) = ( 39,1 ) E x H x x x 3 F x G x 3

11 ONTEST - NOVEMER 014 ROUN 6 PLNE GEOMETRY: NGLES, TRINGLES N PRLLELS NSWERS ) ) ) ) E is an isosceles triangle with base E. E is a square, m E = 114 and trisects ompute m. E. F E ) In scalene triangle, m = ( 6x+ 7), m ( 8x 9) measure of ( 46) x +. ompute all possible values of x. = and the exterior angle at has a S Q ) In rectangle, m ST = 5x 11, m PQ = x + 15, ( ) ( ) where x is an integer. Given that PTS is obtuse, compute the number of possible degree-measures of PTR. P R T

12 ONTEST NOVEMER 014 SOLUTION KEY Round 6 ) rop a perpendicular from to. m EM = m M = 57. F E 1 m F = 114 = 38 m M = 19 3 m = m M = = 71 M ) Since the measure of the exterior angle equals the sum of the measures of the two remote interior angles, we have (6x+ 7) + (8x 9) = x + 46 x 14x+ 48 = ( x 6)( x 8) = 0. For x = 8, m = m = 55 and is isosceles. This solution is rejected. For x = 6, m = 43, m = 39, and m = 180 ( ) = 98 and is scalene. Thus, x = 6 only. ) In quadrilateral PST, 7x y = 360 y = 66 7x> 90 7x< 176 x 5 ut we also know that y < 180 P x+15 7x> 86 x 13 Thus, 13 x 5 generates all possible values of m PTR, a total of 13 different values. heck: For x = 13,...,5, ( ) y 180-y R 5x-11 T m PTR = 180 y = 7x 86 5,1 ( ),..., distinct values. S Q x+15

13 ) Let N = ( 1 i) MSSHUSETTS MTHEMTIS LEGUE ONTEST - NOVEMER 014 ROUN 7 TEM QUESTIONS NSWERS ) ) ) SUN MON TUE WE THU FRI ST E) ) F) 1 k for integer values of k. If 10 < k < 100, determine for how many values of k, N is real. ) Misao Okawa, one of the oldest living persons in the world, celebrates his birthday in March. In 014, his birthday fell on a Wednesday. On what day of the week did his birthday fall in 1898, the year he was born? Recall that there are 365 days in a year, except in leap years. The extra day (/9) is added only in non-century years divisible by 4 and in century years divisible by 400. ) Given: Trapezoid with and (,,, M) = (30, 87, 51, 4) n isosceles trapezoid PQRS has the same perimeter as, sides of integer length and an altitude equal in length to the altitude of. ompute all possible areas of trapezoid PQRS. ) In each of the squares below, consider the lattice points within the triangular regions marked I and II. The lower left vertex in each square is the origin. The upper right vertices are (n, n) and (n + 1, n + 1) respectively, where n is a positive integer. ompute all value of n for which the number of lattice points in region II is 5 more than the number of lattice points in region I. I nπ mπ E) ompute all possible values of sin +, if m and n are both positive multiples of M. N.. II. F) is scalene and acute. Its interior angles measure x, y and (3x y), where x and y are integers. If x+ y < 10, compute the number of possible values of x.

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15 ONTEST - NOVEMER 014 SOLUTION KEY Team Round k k N = = ( 1 k = = + i) ( 1 i) 1 i Since k is real for all integer values of k in the specified range, N is real whenever ( 1 i) k For k = 1 to 4, ( 1+ i) = 1 + i, i, + i, 4. ) Note that i k k For k = 5 to 8, the values obtained are obtained by multiplying the above values by 4. There is one real value of N in every block of four consecutive k-values. If k is a multiple of 4, N is real. N is real for k = 1, 16,., 96, or 4(3, 4,, 4). Thus, we have a total of different values of k. k + is real. ) 7 days before (or after) a given date will fall on the same day of the week. Since 365 = , a non-leap year consists of 5 weeks and 1 day. Therefore, from year to year, a given date advances 1 day of the week (OW), unless /9 falls between the two dates. Moving back in time, starting in 014, consider the following sequence: WE TUE MON ST FRI THU WE MON SUN ST FRI Since /9 falls between 3/011 and 3/01 and between 3/007 and 3/008, the day of the week changes by days. Over a 4 year period, OW changes by 5 days, unless the 4 year period spans a century year which is not divisible by 400 (like the year 1900). Therefore, 11 years ago (8 4-year periods), in 190, the OW has cycled through 8 5 = 140 days. Since 140 is divisible by 7, in 190, his birthday fell on the same OW, namely Wednesday. Since none of the remaining 4 intervening years were leap years, the OW changes only by 4 days. In 1898, his birthday fell on a ST. You are invited to apply Zeller s formula to his actual birthdate 3/5/1898 to confirm this result. 13m 1 y c z = d + y c 5 4 4, where d denotes the day (1..31) m denotes the month according to the following funky rule: 1 = March = pril 10 = ecember and January and February are assigned 11 and 1 respectively for the previous year c denotes the "century" in which the date falls (YYYY) y denotes the year ( YYYY), i.e Now, divide z by 7. The integer remainder determines the day of the week. (0, Sunday), (1, Monday), (, Tuesday) (6, Saturday)

16 ONTEST NOVEMER 014 SOLUTION KEY ) rop perpendiculars from and to. M = 18 (18, 4, 30) = 6(3, 4, 5) N = 45 (4, 45, 51) = 3(8, 15, 17) 4 4 = 87 MN = = ( ) = 4 Per() = 19 M N Per(PQRS) = x + y + c = 19 x + y + c = 96 and (y, 4, c) is a Pythagorean triple The possible triples (with a leg of 4) are: x P S 1) 6(3, 4, 5) (18, 4, 30) ) 8(3, 4, 5) (4, 3, 40) c c 3) (5, 1, 13) (10, 4, 6) 4 4) (7, 4, 5) 5) 3(8, 15, 17) (4, 45, 51) Q y x y R 6) (4, 143, 145) side: a= m n The system b = mn may be used to generate any Pythagorean triple. The triple will be c = m + n primitive whenever the greatest common factor of m and n is 1. The primitive triples 1) through 6) above were generated by (m, n) = (, 1), (3, ), (4, 3), (4,1) and (1, 1). From the list above, the possible values of y + c are: 48 for (y, c) = (18, 30), 7 for (y, c) = (3, 40), 36 for (y, c) = (10, 6), 3 for (y, c) = (7, 5), 96 for (y, c) = (45, 51) rejected ( x = 0) 88 for (y, c) = (143, 145) also rejected x = 96 (y + c) the allowable values of x are: 48, 4, 60, 64 Thus, the allowable corresponding values of (x, y) are: (48, 18), (4, 3), (60, 10) and (64, 7) Since the area(pqrs) = 4(x + y), we have , 1344, 1680,

17 ONTEST NOVEMER 014 SOLUTION KEY ) If a is odd, the diagonals do not intersect at a lattice point; for a even, they do. ontinuing to draw pics would quickly become tedious. Let s generalize. ase a - odd 16 pts (3,3) ase a - even 5 pts (4,4) ase even (a = 4): There are (a + 1) lattice points, but we must exclude: points on the boundary: 4(a + 1) 4 = 4a (0,0) (4 points on each edge 4 corners which have been counted twice) points on the diagonals (not already counted): ( ( a+ 1) ) 1= a 3 (minus center point which is on both diagonals and has been counted twice) Simplifying, ( ) a+ 1 4 a (a 3) = a 4a+ 4 = ( a ) ( a ) Since each of the 4 regions has the same number of lattice points, we have (a even) 4 ase odd (a = 3): There are (a + 1) lattice points, but we must exclude: points on the boundary: 4a points on the diagonals: (a + 1) 4 = (a 1) Simplifying, ( ) ( a 1)( a 3) a+ 1 4a ( a 1) = a 4a+ 3 = ( a 1)( a 3) (a odd) 4 n odd use (n + 1) for a in the even formula: (( n + 1) ) ( n 1)( n 3) = 5 ( n n+ 1) ( n 4n+ 3) = 0 n= n= n even use (n + 1) for a in the odd formula: (( n+ 1) 1 )(( n+ 1) 3 ) ( n ) = 5 ( n n) ( n 4n+ 4) = 0 n= 4 n=1 4 4 sk you teammates/coach about Pic s Theorem. (0,0)

18 ONTEST NOVEMER 014 SOLUTION KEY E) Let m= 3 j and n= 3k for positive integers j and k. nπ mπ π π π + = ( n+ m) = ( 6j+ 3k) = ( j+ k) Since j and k are positive integers, so is j+ k. an we get all non-coterminal multiples of π? Yes! j+ k produces all and only quadrantal values 3π 3π ( jk, ) = ( 1,1) and sin = 1 4π ( jk, ) = ( 1, ) and sin( π ) = 0 5π 5π π ( jk, ) = (,1) and sin = sin = + 1 6π ( jk, ) = (,) and sin( 3π) = sin( π) = 0 F) Let (,, ) denote the interior angles with measures ( xy,,3x y) Triangle Sum 4x y = 180 y = 4x 180 Substituting in x+ y< 10, 5x < 300 x < 60. Thus, our starting point is 59 x = ( 59,56, = 65). ecreasing x by 1, decreases y by 4, and consequently the third interior angle will increase by 5. We must stop when the largest angle becomes a right angle. Possible measures of are 65, 70, 75, 80 and 85, implying we have 5 possible x-values.

19 ONTEST - NOVEMER 014 NSWERS Round 1 lgebra : omplex Numbers (No Trig) ) 14 ) 50 ) 64 Round lgebra 1: nything ) 11 ) 7 ) (15, 48, 7) Round 3 Plane Geometry: rea of Rectilinear Figures ) 88 ) 4+ ) Round 4 lgebra: Factoring and its pplications ) 6x z 3 ) 3 1, ) x = + 8, Round 5 Trig: Functions of Special ngles ) 30, 60, 90, 10, 150 ) ) (39, 1) (answers with degree symbols are acceptable) Round 6 Plane Geometry: ngles, Triangles and Parallels Team Round ) 71 ) 6 only ) 13 [8 is extraneous.] ) ) 11 or 1 ) ST E) 0, ± 1 ) 1344, 1584, 1680, 1704 F) 5 (in any order)