MASSACHUSETTS MATHEMATICS LEAGUE CONTEST 4 - JANUARY 2016 ROUND 1 ANALYTIC GEOMETRY: ANYTHING ANSWERS

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1 ONTEST - JNURY 016 ROUND 1 NLYTI GEOMETRY: NYTHING NSWERS ) n equation of the ais of symmetry of the parabola y ( 3)( h) ompute the value of h. ) B) (, ) ) k = = + is = 8. B) The area of the region between + y = 16 and y where B > 0. Determine the ordered pair of integers (, B). + = can be epressed as ( B) π, ) The line l defined by the equation 3 + y = passes through one of the endpoints of the y major ais and one of the endpoints of minor ais of the ellipse defined by + = 1, a b where a> b. ompute the y-intercept k of the line perpendicular to l and passing through the focus of the ellipse that lies on the positive -ais.

2 ONTEST - JNURY 016 SOLUTION KEY Round 1 ) The -intercepts of the parabola y = ( + 3)( h) at ( 3,0) and (,0) other across the ais of symmetry = 8. Thus, ( ) 3 + h = 8 h = =19. 8,0 is the midpoint of B B h are images of each B) The given equations determine a circle of radius and a square with diagonals of length 8. Thus, the area of the enclosed region is 1 16π 8 8 = 16( π ) ( B, ) = (16,). - - ) The line 3 + y = passes through points (0, 6) and B(8, 0) and, since a> b, the ellipse is horizontal. Thus, the equation of the ellipse y is + = 1. The focus is at (c, 0) and since, for an ellipse, 6 36 a = b + c, we have c = 6 36 = 8 = 7. The equation of the perpendicular line must be of the form 3y = n, for some constant n. Since the focus is at ( 7,0 ), we have 3y = 8 7. Therefore, k = (0,6) F(c,0) B(8,0)

3 ONTEST - JNURY 016 ROUND LG 1: FTORING ND/OR EQUTIONS INVOLVING FTORING NSWERS ) B) ) ) Solve for over the rational numbers. 15 = 3+ + B) The difference between the squares of two positive integers and B is 7. B,, where > B. ompute all possible ordered pairs ( ) ) Factor completely over the integers using the fewest possible number of minus signs. onsider epressions like ( 1) ( + y)( y) + ( y 1) a and ( 1 a) + to each have one minus sign.

4 ONTEST - JNURY 016 SOLUTION KEY Round ) 15 = ( 3 )( + ) = = 0, provided. + 1 = = 0 ( )( ) = 3, 7. + B = 7 + = 7 = B For positive integers and B, it is always true that + B> B, so we stop at + B, B = 9,8. If + B and B were to have opposite parity, then neither nor B) B ( B)( B) ( ) ( ) B would be integers. Thus, we consider only the bolded pairs. 38 = = 19, 11, 9 B, = 18 dding the equations, we have ( ) ( 19,17 ), ( 11,7 ), ( 9,3 ) ) ( y)( y) ( y 1) + + is a sum of two terms and, since there is no common factor between these two terms (other than 1), we have no option other than first multiplying out + y y + y 1 = y + y. these terms ( )( ) ( ) Grouping the first and third, and second and fourth terms, we have 1 + y y + = + y y + = + y 1 = + y 1 +1 ( ) ( ) ( ) ( ) ( )( ) ( )( )( ) or ( + )( 1+ )( 1 ) y +. Each has eactly one minus sign. Do not accept y 1 1 ( )( )( ) +, ( + y )( 1+ )( 1 ) or ( y )( 1 )( 1 ) since each of these is considered to have three minus signs ,

5 ONTEST - JNURY 016 ROUND 3 TRIG: EQUTIONS WITH RESONBLE NUMBER OF SOLUTIONS NSWERS ) B) ) ) ompute all values of over the interval [ 0,π ] for which 3sin cos =. B) The equation cos (sin + 1) = 0 has eactly solutions over the interval 0 < k ompute the minimum value of k. ) Determine the smallest positive degree-measure of θ for which ( θ ) ( θ) sin = cos 150.

6 ONTEST - JNURY 016 SOLUTION KEY Round 3 ) 3sin + cos + = 0 ( ) 3 1 cos + cos + = 0 ( )( ) 3cos cos 5 = 3cos 5 cos + 1 = 0 5 cos =, 1 =π. 3 B) cos = 0 = n 90, 70, n sin + 1 = 0 sin = = 10, 570, or 330, 690, n rranging in order of increasing magnitude, 90, 10, 70, 330, 50, 570, 690,, we see the minimum value of k is 330. ) onverting to sine functions only, sin 3θ 160 cos 150 θ ( ) ( ) ( ) = ( ) ( ) ( ) Noting that sin sin B B n( 360 ) or 180 B n( 360 ) sin 3θ 160 = sin θ = sin θ 60 = = + = +, we eamine cases. ase 1: 3θ 160 = θ 60 + n( 360 ) θ = n( 360 ) (We ignore the co-terminal factor, since we want the smallest positive value.) ase : 3θ 160 = 180 ( θ 60 ) + n( 360 ) 5θ = ( ) + n( 360 ) = 00 + n( 360 ) θ = n. [ 80 is NOT the smallest.] n = 1 gives us the smallest positive solution θ = 8. heck: For θ = 8, sin 3θ 160 = sin 136 = sin 136 = sin 180 = sin. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) cos(150 θ ) = cos 13 = cos = cos 6 = sin 90 6 = sin. You should be able to give a reason for each of the above equalities. For eample, the sine is an odd function, or, the cosine of an angle equals the sine of its complement, etc.

7 ONTEST - JNURY 016 ROUND LG : QUDRTI EQUTIONS NSWERS ) B) ) ) It is not uncommon for an equation, like = 38 N, to have two real solutions. For eample, when N = 80, = 10 or 8. Determine the unique value of the constant N for which the quadratic equation has eactly one solution. B) ompute the product of the cubes of the rational roots of = 0. ) The equation k + k = has one root which is 3 greater than the other root. If the roots are R 1 and R, find all ordered triples ( R1, R, k ), where R1 > R.

8 ONTEST - JNURY 016 SOLUTION KEY Round ) = = 0. N To have equal roots, this trinomial must be a perfect square (like ( ) + 3 = ). Note that to have equal roots, the value in the bo must be the square of half the coefficient of the middle (i.e. linear) term. N = 361. heck: ( 19) = = 0 has eactly one root, namely 19. B) Let y = +. Then the equation becomes y 17 y + 16 = 0 y 1 y 16 y =± 1, ±. ( )( ) + =± 1 ± + = 0which has no rational solutions. + =± ± + = ( ± ) = 0 =±. ( ) ( ) 3 3 = 6. ) k+ k = ( k+ ) + k = 0 3= k+ Let R 1 = and R = 3. Then: ( 3) = k Substituting k = 5 in the second equation, 3 = ( 5) = ( 5)( ) = 0 (, ) ( 5,5 ), (, 1) Thus, ( R, R, k ) = (5,, 5), (, 1, 1). 1 k =. lternately, k k k k ( k)( ) + = ( + ) + = = 0. 1 Therefore, the roots are and k, which implies k must be ± 3=. 5

9 ONTEST - JNURY 016 ROUND 5 GEOMETRY: SIMILRITY OF POLYGONS NSWERS ) B) : ) ) Given: Right triangle B, with B = 17, B = 15 PQ B, P : P = 3 : 1 ompute the area of the trapezoid PQB. P.. Q B B) JK is a square with side JIL1L is a 3 6 rectangle, where JL < LL1 Point M is the midpoint of JK. Points, M, O, and N are collinear. ompute the ratio NO MO. J L M O K N I L 1 ) In ΔB, DE B, D = +, E = DB = + 1 and E =. ompute B., DE = 3, D.. E B

10 Round 5 ONTEST - JNURY 016 SOLUTION KEY ) B B = = 89 5 = 6 = 8 and the area of B is =. P : P = 3 : 1 P = 6 Since PQ ~ B, their areas are in the ratio of the square of their corresponding sides, namely 9 :16. Thus, the area of trapezoid PQB is ( 60) = = 105 or P.. Q B lternately, PQ ~ B with corresponding sides in a ratio of 6 : 8 or 3 : PQ = 3 5 = and the area of the trapezoid is = Therefore, ( 15) 105. LN L LN 5 JM ~ L N L N.5 JM = J 1 = =. L1L = JK + KI = 6 MI = 5. MK = 1 NO NL.5 Since MOI ~ NOL, = = = 1:. MO MI 5 B) J M O K I L N L 1 ) Since ΔDE ~ ΔB, + = + 1 ( ) D E DE = = B B (3 )( ) 0 D E = DB E = = + = =. = is etraneous. ( BD = + 1< 0) Thus, D = 6, DB = 9, DE = 10 and = 6+ 9 B B = 5. D E - B

11 B + = 10 ) Given: 3 3+ B= 3 ompute + B. ONTEST - JNURY 016 ROUND 6 LG 1: NYTHING NSWERS ) B) (,, ) ) B) ( )( ) = a+ b c, where a, b, and c are integers. ompute the ordered triple ( abc,, ). ) Two of my favorite flavors of SNPPLE iced tea mi packets were on sale at Walmart. There were T packets in the case and there were.6 times as many Peach as Mango packets. However, at ostco I was able to buy a case containing the same number of packets, but there were 50 more Mango packets in the case, resulting in there being only twice as many Peach as Mango packets. ompute T, the total number of SNPPLE drink packets in a case.

12 ONTEST - JNURY 016 SOLUTION KEY Round 6 ) B + = B= B= 15 + B=. 3+ B= 3 3+ B= 3 It was not necessary to first solve for and B separately!! B) Grouping the first terms in each trinomial, we view each trinomial as a binomial and we have the product of a sum and a difference (which is equivalent to the difference of perfect squares!) = ( )( ) ( ) ( ) ( 3 + 5) ( 15) = = ( ) 7,,15. ) Let P and M denote the number of Peach and Mango packets purchased. Let T = P+ M. P 13 P 50 P=.6M = and = P= M M 5 M Substituting, M = 13 10M = 13M M = 50, P= 650 T =900. M 5

13 ONTEST - JNURY 016 ROUND 7 TEM QUESTIONS NSWERS ) D) B) { } E) ) F) ) Given a hyperbola defined by ( y ) ( ) = 1. The major ais of an ellipse is the transverse (or major) ais of this hyperbola. (The vertices on the major ais of the ellipse coincide with the vertices of the hyperbola.) The major ais of the ellipse is twice as long as its minor ais. ompute the length of the line segment on y = whose endpoints are on the ellipse. { } = whose B) (, y) y ( y ) = 6, where y 1 defines a real-valued function y f ( ) domain is a subset of the real numbers. Specify the domain of f. For disjoint intervals, the use of the proper connector ( and or or ) is required. lternatively, the symbols " "and " " may be used. ) ompute all possible solutions, where [ 0, π ], to the equation D) Given: BD, a square with side 6. Beetle Bailey received these marching orders: Start at. Proceed along. Stop at point P which is twice as far from B as it is from. tan cot 1 D + =. P. These instructions are way above Beetle s pay grade, but with your help he can prove how smart he is to his commanding officer. In simplified form, P ( N 1) =, where N is an integer. ompute N. B E) Given: m D = m D, B ~ BD, B = 1, =. ompute D. F) My little brother JJ and I are very competitive. JJ s best mile time to date is :30. JJ s goal is to be able to beat my time. ssuming he improves his time k% each year for the net two years, he will beat my current best time of :03. Determine the smallest integer value of k which allows him to reach his goal. [ helpful fact: ] B D

14 Team Round ) ( y ) ( ) ONTEST - JNURY 016 SOLUTION KEY ( y+ ) ( ) = 1 = 1 1, 3, This is a vertical hyperbola with center at ( ) a =, b= 3 and c =. Thus, the vertices of the hyperbola are at, 3 V, 1, 3 = V, 5 ( + ) = 1 ( ) and ( ) ( ) The center of the ellipse is (, 3), a, b 1 = = (since the major ais is twice as long as the minor ais) and, therefore, y = 1. Substituting, when y =, we have its equation is ( ) ( ) ( ) ( ) = 1 = = ± and the 3 3 length of the segment PQ is + = 3. FYI: R and S are the foci of the ellipse, while T and W are the foci of the hyperbola. V 3 and V are the endpoints of the minor ais of the ellipse. You should be able to specify the coordinates of these points, as well as the equations of the lines containing the diagonals of the rectangle which are referred to as the asymptotes of the hyperbola. lthough neither branch of the hyperbola ever intersects these lines, as ±, the distance between the hyperbola and these lines becomes arbitrarily small V 3 P T V 1 R S V W 5 Q V y = - B) y ( y ) y y ( y ) ( + )( ) = 6 = 1 = + 1= =. Thus, ( + 3)( ) ( )( ) + 3 y = 1± and since it was given that y 1, we have y = f ( ) = 1+. The critical values are 3, 0, and, as each critical value is passed from left to right, there is one less negative factor in the radicand. On the left (as ) all three factors are negative; whereas, on the right (as + ) all three factors are positive. The following chart summarizes this discussion. Radicand: Thus, the domain is { 3 <0or } 3 neg neg 1 neg all pos The symbol " " may be used in place of or.

15 ONTEST - JNURY 016 SOLUTION KEY Team Round - continued sin cos ) tan + cot = 1 1 cos + sin = sin + cos = 1sin cos sin sin cos cos 16sin cos + + = ( ) sin + cos = 16sin cos 16sin cos = 1. Dividing through by and applying the double angle identity, sin θ = sinθcosθ, ( ) ( ) 1 1 sin cos = sin cos = sin = sin = ±. Thinking the 30 -family of related angles in all quadrants, i.e. 6 π radians, we have π π nπ + nπ ( quadrants 1,3) π 1+ 6n = = =. 5π 5π nπ n + nπ ( quadrants, ) π 5π 7π 11π For n = 0,1, we have =,,, lternately, tan + cot = 1 tan + cot + = 16 ( tan + 1) 1 sec 1 cos 1 tan + = = = = = 16 ( ) tan tan tan cos sin sin cos 1 = = 16. Thus, sin = and the same result follows. ( sin cos ) sin D D) Using the dimensions in the diagram at the right, apply Stewart s Theorem. ( ) ( ) ( ) = = = = 0. pplying the quadratic formula, ( ) = P = + + = + = 1+ 7 N = 7. lternately, without resorting to Stewart s Theorem, drop a perpendicular from P to B. Let E = and mark the other sides as indicated. In BEP, + ( 6 ) = ( ) + = + = = P ( ) = 1+ 7 N =7. D 6 P. P. E B B

16 ONTEST - JNURY 016 SOLUTION KEY Team Round - continued E) Let D =, BD = y and m D = m D = θ. s an eterior angle of D, m BD = θ 1 θ θ m D = m D D = D. B ~ BD m B = m BD = θ. Thus, θ D is an B y D y B + y angle bisector and = = y. B ~ BD = =. 1 B D ross multiplying, + y = y + y = y = y = D = = F) JJ s current mile time is 70 seconds; big brother s is 3. fter one year, JJ s time will be 100 k 70( 1 k% ) = 70. fter years, his time will be k 100 k 100 k 70 = We require that k < 3 ( 100 k ) < 10 = ( 100 k ) < 9 10 = k > Therefore, k must be at least 6. FYI: 6% improvement over years would drop little brother s time to seconds and he would break the :00 mile (0 seconds), whereas a 5% improvement would result in a mile-time of seconds and he would fall just short of beating big brother. Some real world perspective: The first merican to run a sub -minute mile was Don Bowdon in 1957 (3:58.7). Jim Ryan was the first merican high school runner to break the -minute mile (3:55.3 in Witchita, KS). 5 others have joined him since - Tim Danielson (1966), Marty Liquori (1967), lan Webb (001), Lukas Verzbicas (011), and Matthew Maton (015). Only Webb has had a faster time (3:53.3). The current world record time for the mile of 3:3.13 (set in 1999 in Rome) belongs to Morocco s Hicham El Guerrouj who bested Kenyan Noah Ngeny s time of 3:3.0 in the same race with quarter-mile splits of 55.6, 56.0, 56.3 and 55.. Two runners bettered the then world record of lgerian Noureddine Morceli (3:.39) by almost a full second! s of this date (016), these three times are still the fastest mile times ever recorded! You can watch the 1999 record setting race on Youtube. The link is θ

17 Team Question : enter (, 3) Since a=, b= 3 and c = for the hyperbola, and a =, b= 1and c = 3 for the ellipse. R (, ), S(, 3 3) T(, 1 ), W(, 7) V ( 3, 3 ), V ( 5, 3) 3 Since the slope of asymptotes are ± 3 3 =± 3, 3 + =±. 3 In the diagram below, graphically, it certainly appears as if the hyperbola approaches arbitrarily close to these lines. nalytically, why is this the case? the point-slope equations of the asymptotes are ( y 3) ( ) ( y+ ) ( ) 3 Staring with 3( y+ 3) ( ) = 1 = 1, solve for y. 1 ( ) ( y + 3) = + 3 s ±, that is, as becomes an arbitrarily large positive number (moving to the right on the graph) or an arbitrarily small negative number (moving to the left on the graph), the becomes negligible and we have ( ) 1 ( y+ 3) y+ 3 ± ( ) 3 3 Thus, as ±, the y-coordinate of a point on the straight lines (the asymptotes) becomes an increasingly better approimation of the y-coordinate of the corresponding point on corresponding branch of the hyperbola V 3 P T V 1 R S V W 5 Q V y = - Q.E.D ( Quod erat demonstratum, Latin for enough said, literally that which was to be demonstrated ) V

18 ONTEST - JNURY 016 NSWERS Round 1 nalytic Geometry: nything ) 19 B) (16, ) ) Round lg: Factoring ) 3, 7 B) ( 19,17 ), ( 11,7 ), ( 9,3 ) ) ( + y )( 1)( + 1) in any order or ( + y )( 1+ )( 1+ ) in any order Round 3 Trig: Equations ) π B) 330 ) 8 Round lg : Quadratic Equations ) 361 B) 6 ) (5,, 5) and (, 1, 1) Round 5 Geometry: Similarity ) 6.5 or Round 6 lg 1: nything 105 B) 1 : ) 5 ) B) ( 7,,15) ) 900 Team Round ) 3 D) 7 B) 3 < 0 or E) 3 3 ) π 5π 7π 11π,,, F) 6