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1 st Annual King s College Math Competition King s College welcomes you to this year s mathematics competition and to our campus. We wish you success in this competition and in your future studies. Instructions This is a 90-minute, -problem multiple-choice exam with no calculators allowed. There are five possible responses to each question. You may mark the test booklet and on the paper provided to you. If you need more paper or an extra pencil, let one of the monitors know. When you are sure of an answer, circle the answer on the exam. Then carefully write your answer on the score sheet with a capital letter. If your answer is unreadable, then the question will be scored as incorrect. The examination will be scored on the basis of points for each correct answer, points for each omitted answer, and 0 points for each incorrect response. Note that wild guessing is likely to lower your score. Pre-selected problems will be used as tie-breakers for individual awards. These problems designated by ( ). The problems are numbered:,, 6, 8, 4 Review and check your score sheet carefully. Your name and school name should be clearly written on your score sheet. When you complete your exam, bring your pencil, scratch paper, and answer sheet to the scoring table. You may keep your copy of the exam. Your teacher will be given a copy of the solutions to the exam problems. Do not open your test until instructed to do so! Good luck!

2 King s College st Annual Math Competition Page of. ( ) Teams A and B are playing a series of games. Each team has a 0/0 chance of winning any game. If Team A must win two games or Team B must win three games to win the series, find the probability that team A wins the series. 6 B. 7 C. 8 D. E. 9. If 6 Team A can win in the following ways: We have W, W with probability 4, W, L, W or L, W, W with probability 8 = 4, or W, L, L, W, L, W, L, W, or L, L, W, W with probability 6 = 6. Therefore P (A) = 6. 4 x 9x+y = 8 and x+y y = 4, find the value of xy. B. 4 C. 6 D. E. 4 B. 4 From the first equation we have x x+y = x y = Thus x y =. From the second equation we have 9 x+y y = x+y y = x y =. So x y =. Solving this system of equations we get x = 4 and y =. Hence xy = 4.. If ab 0 and a b, find the number of distinct values of x satisfying the equation x a b + x b a = b x a + a x b

3 King s College st Annual Math Competition Page of zero B. one C. two D. three E. four D. three Forming common denominators we have a(x a) + b(x b) ab = b(x b) + a(x a). (x a)(x b) The identity can only hold if the numerators are both zero or the denominators are equal. If the numerators are zero, that implies that x = a + b, which is a real number since a + b a b. If the denominators are equal, we have ab = x (a + b)x + ab, and so (x a)(x b) = 0. Thus x = a or x = b. There are three solutions. 4. Suppose five dots are arranged in a circle. If we draw line segments joining each dot to each other dot, then a total of 0 line segments are drawn, as shown in the picture to the right. If 00 dots are arranged in a circle, and line segments are drawn joining each dot to each other dot, how many lines segments are drawn? 48 B. 490 C. 490 D. 00 E. 00 C. 490 Solution : The number of segments equals the number of distinct pairs of dots. With 00 dots, there are thus ( ) 00 = 490 segments.

4 King s College st Annual Math Competition Page 4 of Solution : Pick one dot and draw 99 segments connecting it to the other 99 dots, then move to the second dot. Draw 98 segments connecting it to the other dots (only 98, because the second dot was already connected to the first in the previous step). Repeat for the third dot, the fourth, etc. The total number of segments drawn is thus = k = k= = At his usual rate a man rows miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, find the rate of the stream s current. B. C. D. 7 E. 4 Let c denote the rate of the current and m the usual rate of the rower in still water. Then his downstream and upstream rates are m + c and m c, respectively. After he doubles his usual rate, they are m + c and m c. Since the distance is miles and the time is distance/rate, we have m + c = m c and m + c = m c Multiplying the first equation by (m + c)(m c) and the second by (m + c)(m c), yields m c = 0c and 4m c = 0c. Subtracting the second equation from the first gives m 4c = 0, so m = c. Substituting into 4m c = 0c yields 4(c) c = 0c, so c = 0. Thus c =. 6. The sum of consecutive integers is 0. What is the largest of these integers? 70 B. 7 C. 7 D. 80 E. 8

5 King s College st Annual Math Competition Page of D. 80 Let x be the largest of these integers. Then We have that So Dividing through by we have x + (x ) + (x ) + + (x 0) = = 0(). x 0() = 0. x = 0 = 6. So x = 80 So the largest of these integers is Solve the equation S = πrh + πr for r. r = πh ± π h 4πS π D. r = πh ± π h + 4πS π B. r = π ± π + 4πS π E. None of these C. r = πh ± πh + 4πS π D. r = πh ± π h + 4πS π Rewrite the original equation as πr + πhr S = 0 and use the quadratic formula to solve for r. 8. Let y = cos(x) + 4 cos(x). Find the minimum value of y. 9 B. 7 C. 6 D. E. C. -6

6 King s College st Annual Math Competition Page 6 of Recall by the double angle formula, cos(x) = cos (x). So, y = cos(x) + 4 cos(x) = ( cos (x) ) + 4 cos(x) = 4 cos (x) + 4 cos(x). Let a = cos(x), where a. Then completing the square we have ( y = 4a + 4a = 4 a + ) 6. The minimum will occur when a + is at it s smallest. That is when a =. When a =, y = 4( /) + 4( /) = = 6 9. Consider the parabola y = x + 4x + 7. How many units do you have to shift the parabola horizontally and vertically to obtain the parabola y = x 6x?, B., C., D., E.,, We have y = x + 4x + 7 = (x + x) + 7 = (x + x + ) + 7 = (x + ) + by completing the square. So the vertex is at (, ). The new parabola is y = (x x + 9/4) 9/ = (x /) /. The vertex is at (, ) In the horizontal direction we need to move + = to move = and in the vertical direction we need

7 King s College st Annual Math Competition Page 7 of 0. Which of the following expressions is equal to +? B. C. + D. E. 6 B. = = +. ( ) Assume that a, b, and c are real numbers. If the system { ax + 4y = c x y = has a solution, and the system { x + by = d x y = does not have a solution, which of the following is correct? a, b = 6, d 4, c arbitrary B. a =, b = 6, c, d = 4 C. a =, c =, b, d arbitrary D. a =, b = 6, c, d arbitrary E. a, b = 6, c, d arbitrary a, b = 6, d 4, c arbitrary In order for the first system to have a solution, either the lines intersect once or they are parallel. Adding the first equation and a times the second equation, we have (4 + a)y = c a. If the lines intersect once, y must have a unique value, so a. If the lines are parallel, then we must have a = and c =. If a = and c, there is no solution, so the answer is not B. Considering the second system, adding the first to twice the second we see that (b 6)y = d + 4.

8 King s College st Annual Math Competition Page 8 of If this system has no solution, then we must have b = 6 and d 4. So C, D, and E are not correct. We see that choice A satisfies all conditions, so this is the correct answer.. Consider a circle of radius centered at the origin. Draw a horizontal slice at y =. What is the area of the gray cap, as seen to the right? y = 4π B. π C. 4π + D. 4π E. π A 4π The circle s equation is x + y = 4, so the two points on the circle corresponding to y = are (, ) and (, ). Draw radii to those two points, and observe that the gray cap can be viewed as a sector of the circle minus an isosceles triangle. The isosceles triangle has base and height, hence area. The area of a circular sector is A = r θ, where angle θ is measured in radians. Our sector has θ = π/, hence area 4π/. Thus the cap s area is 4π.. Let cos θ =. Find tan θ for θ in the rd quadrant. B. C. D. E. D.

9 King s College st Annual Math Competition Page 9 of We have a triangle in the third quadrant that is a, 4, triangle as the hypotenuse, opposite, and adjacent respectively. So tan(θ) = 4. Consider a sequence of squares defined as follows: the first square S has sides of length. The second square S is obtained by doubling the sides of S, extending these sides outward in the cyclic pattern shown, and then joining the endpoints of these extended sides. The third square S is obtained from S in the same manner. If this pattern continues, find the side length of S 00. S S S 0 B. 99 C. 49 D. 49 E. 49 E. 49 The side length increases by a factor of with each iteration. For example, the sides of S are the hypotenuses of right triangles with sides of length and (see picture). And the sides of S are hypotenuses of right triangles with sides of length and (see picture again). So the side length of square S n is ( ) n. In particular, the side length of S 00 is ( ) 99 = 49.. An honest coin is tossed ten times. The probability of tossing two heads and eight tails is what?

10 King s College st Annual Math Competition Page 0 of 4 04 B. C. D. E. None of these 4 04 Each flip has two choices, so the total number of possibilities ( ) for the ten flips is 0 = The number of ways to toss two heads and eight tails is = 4. This gives a probability 4 of ( ) Find the vertical distance y, correct to the nearest whole number. y B. C. 6 0 D. 0 E D. Solution : There are two overlapping right triangles. The larger one is a triangle, and the smaller one is a triangle. The shorter and longer leg of a triangle have a : ratio. We thus deduce that the triangle has base y 0, hence y = (y 0) y = 0. Solution : First, let x denote the hypotenuse of the triangle and use the Law of Sines to

11 King s College st Annual Math Competition Page of find x x sin(4 ) = 0 sin( ) x = 0 sin(4 ) sin( ) = 0 6, where we ve used the formula sin( ) = ( 6 )/4, which can be derived from the half-angle formulas applied to sin(0 ). Now, using the right triangle with hypotenuse x, we find y = x sin(60 ) = Dividing through by we see this agrees with solution. 7. Find sin x if tan x = ab a b, where a > b > 0 and 0 < x < 90. a b B. b a C. a b a D. a b ab E. ab a + b ab E. a + b Let x denote the acute angle that may be taken as the opposite the leg of length ab in a right triangle with other leg of length a b. Then the hypotenuse h satisfies h = (ab) + (a b ) = 4a b + a 4 a b + b 4 = a 4 + a b + b 4 Thus sin x = ab a + b. = (a + b ) = a + b. 8. ( ) Simplify ( tan 4 (θ)) cos (θ) + tan (θ). B. 0 C. D. E. None of these C. Recall: + tan (θ) = sec (θ) = cos (θ).

12 King s College st Annual Math Competition Page of So ( tan 4 (θ)) cos (θ) + tan (θ) = ( tan 4 (θ)) + tan (θ) + tan (θ) = ( tan (θ))( + tan (θ)) + tan (θ) = tan (θ) + tan (θ) = + tan (θ) 9. George R. R. Martin is writing the sixth book in the Game of Thrones series. He numbers the book beginning with page. The Onion reports that to write down all of the page numbers would require 4 digits. For example, a book with pages will have page numbers,,,...,, 4, for a total of digits. How many pages does the book contain? 880 B. 88 C. 88 D. 88 E. 884 D. 88 Pages -9 use 9 digits. Pages 0-99 use 90 = 80 digits. So pages -99 use 89 digits. That leaves 4 89 = digits for page numbers with three digits. So = 784 pages. So the total number of pages is = 88 pages. 0. What is the sum of the proper factors of 0? 6 B. 47 C. 49 D. 0 E. 06 C. 49 We see that 0 =. So + + = 49.. Let x = + and y =. Find x + y B. 8 C. 0 D. 8 E. 98

13 King s College st Annual Math Competition Page of 98 x + y = ( ) ( ) = ( + ) ( ) + ( ) ( + ) = ( + ) 4 + ( ) 4 ( ) ( + ) = ( + ) 4 + ( ) 4 = ( ) + ( ) = 98. Patrick travels from point A to point B in 8 minutes. SpongeBob travels from B to A along the same route. They start at the same time and each travels at a constant rate. If Sponge- Bob reaches point A 8 minutes after they meet, how many minutes did the entire trip take SpongeBob? 0 B.. C. 4 D.. E. 6 C. 4 Let x be the time it takes Patrick to reach the meeting point and let r be the rate he is traveling. Before they meet, Patrick travels xr and travels (8 x)r after the meet. The total distance he travels is 8r. Let r be the rate SpongeBob travels. SpongeBob travels xr before they meet and 8r after they meet. SpongeBob s total distance is also 8r. So We also have 8r = 8r + xr r = 8r 8 + x ( ) 8r xr = 8r = 8 xr = 44r 8 + x 8 + x x = x x + 8x 44 = 0 (x + 4)(x 6) = 0 We can t have a negative time, so x = 6. So SpongeBob takes = 4 minutes to complete the trip.. How many integers between and 0 are divisible by both and but not by 7?

14 King s College st Annual Math Competition Page 4 of B. 4 C. 47 D. 49 E. A number is divisible by and if it is divisible by 4. We have 0 = 4(4) +. So there are 4 numbers between and 0 divisible by 4, namely 4, 4, 4 etc. Since 4 isn t divisible by 7, the only numbers of this list divisible by 7 is 7 4 and 4 4. Thus there are numbers. 4. If the sum of two numbers is and their product is, then what is the sum of their cubes? B. i 4 C. 0 D. i 4 E. E. Let x and y be numbers such that x + y = xy =. Then Thus x + y =. (x + y) = x + y + xy(x + y) = x + y + ()() =.. In the figure below, a rectangle of dimensions x has been partitioned into a square (on the left) and a second rectangle (shaded, on the right). Suppose the second rectangle is similar to the original rectangle. What is the length x? x + B. C. D. E. +

15 King s College st Annual Math Competition Page of + If the inner rectangle is similar to the original, outer rectangle, then the rectangles length-to-width ratios are proportional: x = x. Clear denominators and bring everything to the left, and we get x x = 0, hence Since x > 0, we take the positive solution, so x = ±. which is the Golden Ratio. x = +, 6. Suppose sin(θ) + cos(θ) =. Find sin(θ) cos(θ). 9 B. 9 C. 4 9 D. 9 E. 9 C. 4 9 Squaring both sides of sin(θ) + cos(θ) =, we have ( ) (sin(θ) + cos(θ)) = sin (θ) + sin(θ) cos(θ) + cos (θ) = 9 sin(θ) cos(θ) + = 9 sin(θ) cos(θ) = 8 8 = Find the smallest positive integer that is divisible by every integer from up to. 4,60 B.,44 C. 9,40 D.,860 E. 7,70

16 King s College st Annual Math Competition Page 6 of E. 7,70 The solution is the least common multiple of,,,...,. Find the prime factorization of each of these integers, and choose the necessary exponents to find our solution S = 7 = 7, A lot is in the shape of a right triangle. The shorter leg measures 0 meters. The hypotenuse is 40 meters longer that the length of the longer leg. How long (in meters) is the longer leg? 60 B. 40 C. 0 D. 00 E Use the Pythagorean Thm with a = 0, b = x, and c = x+40. Then x +0 = (x+40), which gives x = Consider a sequence of integers a, a, a,... where each term in the sequence is determined from the previous term according to the formula { a n / if a n is even a n+ = a n + if a n is odd For example, if the first term were a = 7, then the next five terms would be a =, a =, a 4 = 4, a = 7, a 6 =. Supposing the first term is a =, find a 0. B. C. 4 D. 8 E. Unable to determine B. We have a = 4, a = 7, a 4 =, a = 6, a 6 =, a 7 = 40, a 8 = 0, a 9 = 0, a 0 =, a = 6, a = 8, a = 4, a 4 =, a = Eventually we reach a power of, after which point the subsequent terms eventually enter a cycle, 4,,, 4,... With the first term (a.k.a. the seed) set to a =, it takes terms to enter that cycle, i.e. a =. Thus for all n = + k (k N), we ll have a n =. In other words, for all n, a n = if n is divisible by. Note 0 is divisible by, so a 0 =, and thus a 04 = 4, and finally, a 0 =.

17 King s College st Annual Math Competition Page 7 of 0. A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. At least chips are white or blue. Find the minimum number of red chips. 4 B. 7 C. D. 4 E. 4 B. 7 Let r, w, and b denote the number of red, white, and blue chips, respectively. Then w b r and w + b. Since w b, b + b = b. Thus b. Since b is an integer, b 6 = 9. Since r b, r 7.. Which of the following is a factor of p(x) = 4x + 4x 4 4x 4x + x +? x B. x C. x D. x E. x D. x Note: (x r) is a (linear) factor of p if and only if r is a root of p. Assuming that this strange fifth-degree polynomial must have at least one nice root, we use the rational root theorem. If r = a is a rational root, then a must divide the constant term and b must b divide the leading coefficient. So possible roots are ±, ±, ± 4. Now, x = is a root, meaning q(x) = x+ is a factor of p. This lets us start the factoring, and we then find So x is a factor. p(x) = (x + )(4x 4 4x + ) = (x + )(x ) = 4(x + )(x /) (x + /)

18 King s College st Annual Math Competition Page 8 of. Compute the probability that a randomly chosen basic trigonometric function, when divided by a different randomly chosen basic trigonometric function, will result in a quotient that is itself one of the six basic trigonometric functions. B. C. 7 D. 6 E. None of these B. There are six choices for the numerator and five choices for the denominator (since we need a different trigonometric function). So there are a total of 0 possible quotients. Using sine, cosine, and tangent as the numerator we see sin cos = tan, sin tan = cos, cos sin = tan, cos cot = sin, tan sin = csc, tan csc = sin are the combinations that give a basic trigonometric function. Each of these can be replaced with it s reciprocal quotient and will give the reciprocal trigonometric function. So there are a total of, and the probability is 0 =.. Three men, Andrew, Brian, and Connor, working together, do a job in 6 hours less time than Andrew alone, in one hour less time than Brian alone, and in one-half the time needed by Connor when working alone. Find the number of hours needed by Andrew and Brian, working together, to do the job. B. C. 4 D. 4 E. 4 C. 4 Let a, b, and c denote the number of hours it takes for Andrew, Brian, and Connor, respectively, to complete the job alone. Then each man completes a, b, and of the job c in one hour. So together they complete a + b + c

19 King s College st Annual Math Competition Page 9 of of the job in one hour. Thus, Setting () equal to (), we have Likewise setting () equal to (), we have a + b + c = a 6 () a + b + c = b () a + b + c = c/ = c. () a 6 = c a = c +. b = c b = c +. Let h denote the number of hours needed by Andrew and Brian together to complete the job. Then a + b = h. Since and so h = c. Then a + b + c = c a + b = c, a + b = c + + c + = c. This yields c + 4c 4 = 0, which implies that (c 4)(c + 6) = 0, for which the positive solution is c = 4. Hence h = ( ) Pictured to the right is the unit circle of radius centered at the origin. The tangent lines to two points on the circle have been drawn. These lines are not parallel, so if extended into the first quadrant they will eventually meet. Find the x-coordinate of the point where these two tangent lines meet B. + C. + D. + + E. +

20 King s College st Annual Math Competition Page 0 of B. + Let l denote the top tangent line, and l the bottom tangent line. Using the special angles, the points of tangency are A( /, /) and B( /, /), respectively. Next we ll need the lines slopes. The radius extending up to point A has slope, hence l has slope m = / = /. The radius extending down to point B has slope, hence l has slope. Now, knowing points and slopes, we can get the lines equations. We find l : y = x + and l : y = x. Equate lines to get the x-coordinate of their point of intersection. We find the lines meet when x = +.. The figure to the right represents a 4 block of city streets. Assuming you can only move north and east, how many different paths are there from the starting location to the ending location? End Start 0 B. 4 C. D. 6 E. 40 C. Solution : Any valid path consists of exactly 7 steps, and all such paths are completely determined by choosing which three steps to go north, so ( 7 ) = You could also choose which four steps to go east, yielding the answer ( ( 7 4). Lucky for us, 7 ( ) = 7 ) 4 Solution :

21 King s College st Annual Math Competition Page of Any solution begins by first choosing to go east or north, and once that initial choice has been made, the problem has been reduced to a smaller version of itself. Go north first, and now you re counting northeast paths in a 4 grid. Go east first, and now you re counting northeast paths in a grid. Letting N a b equal the number of solutions to the generalized counting problem on an a b grid, we discover the recurrence N 4 = N 4 + N. Going further, we discover recurrences for N 4 and N, etc.

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