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1 Round #1, Problem A: (4 points/10 minutes) The perimeter of a semicircular region in centimeters is numerically equal to its area in square centimeters. What is the radius of the semicircle in centimeters? π cm Since the perimeter is equal to the area numerically, then P = A 2πr πr2 + 2r = 2 2 2πr + 4r = πr 2 2π + 4 = πr r = 2π + 4 π = π cm

2 Round #1, Problem B: (5 points/10 minutes) If the solution for x 1 + x + 2 < 5 is a < x < b, then find b a. 5 The combined distance from the center of each absolute value must be less than 5. Since the distance between 2 and 1 is 3, then you must travel 1 more unit from the edge of 2 and 1. This would give a combined distance of five for both 3 and 2. Thus, 3 < x < 2 = 2 ( 3) = distance of 3 1

3 Round #1, Problem C: (6 points/10 minutes) Amber s age is 16 more than the sum of Brandy s age and Cristina s age. The square of Amber s age is 1632 more than the square of the sum of Brandy s and Cristina s ages. What is the sum of all three girls ages? 102 The first sentence yields the equation B + C = A 16. The second sentence yields Thus, A 2 = (B + C) 2 = (A 16) 2 = A 2 32A A = 1888 A = 59 A + B + C = A + (A 16) = = 102

4 Round #2: (5 points/5 minutes) Solve for x: If b > 1, x > 0 and (2x) log b 2 (3x) log b 3 = (2x) log b 2 = (3x) log b 3 log b (2x) log b 2 = log b (3x) log b 3 log b 2 log b 2x = log b 3 log b 3x log b 2(log b 2 + log b x) = log b 3(log b 3 + log b x) (log b 2) 2 log b 2 log b x = (log b 3) 2 + log b 3 log b x (log b 2) 2 (log b 3) 2 = log b 2 log b x + log b 3 log b x (log b 2 + log b 3)(log b 2 log b 3) = (log b 2 log b 3) log b x log b 6 = log b x log b 6 1 = log b x x = 1 6

5 Round #3: (6 points/6 minutes) A vertical line divides the triangle with vertices (0,0), (1,1), and (9,1) in the xy plane into regions of equal area. What is the equation of the line? x = 3 The area of the triangle is A = 1 2 bh = 1 (8)(1) = 4. Next, find where the 2 vertical line needs to be to make C = 2. The equation of the line joining (0,0) and (9,1) is y = 1 x. Thus, the height of the triangular region C is 9 h = 1 1 x and base b = 9 x. 9 1 h = x b = 9 x It follows that 1 9 C 2 = C = 1 2 bh = 1 (1 19 ) 2 x (9 x) 36 = (9 x)(9 x) (9 x) 2 = 36 9 x = ±6 x = 9 ± 6 = 15, 3 It follows that x cannot be 15. Thus, x = 3.

6 Round #4: (7 points/7 minutes) The polynomial (x + y) 9 is expanded in decreasing powers of x. The second and third terms have equal values when evaluated at x = p and y = q, where p and q are positive numbers whose sum is one. What is the value of p? By the binomial theorem, the first three terms of (x + y) 9 are ( ) ( ) 9 9 x 9 + x 8 y 1 + x 7 y 2 = x 9 + 9x 8 y x 7 y Since the second and third terms are equal when x = p and y = q = 1 p, then it follows that 4 5 9x 8 y = 36x 7 y 2 x = 4y p = 4(1 p) = 4 4p 5p = 4 p = 4 5

7 Round #5, Problem A: (Algebra-Plane Geometry/6 points/8 minutes) The measures of the interior angles of a convex polygon are in arithmetic progression. If the smallest angle is 100 and the largest angle is 140, how many sides does the polygon have? 6 (Shorter way) The sum of the interior angles of a polygon is 180(n 2). The sum of n terms of an arithmetic progression is n (first term + last term). Thus, 2 180(n 2) = n ( ) = 180n 360 = 120n = 60n = = n = 6 (Longer way) The sum of the interior angles of a polygon is 180(n 2). Thus, (100 + d) + + (100 + (n 1)d) = 180(n 2) 100n + d + 2d + + (n 1)d = 180n n + d( n 1) = 180n 360 dn(n 1) 100n + = 180n dn(n 1) = 160n 720 Note that since the largest angle is 140, then it follows that (n 1)d = 40. Thus, dn(n 1) = 40n = 160n n = 720 n = 6

8 Round #5, Problem B: (Trigonometry-Analytical Geometry/6 points/8 minutes) If sin x + cos x = 1, where 0 x < π, what is tan x? If sin x + cos x = 1 5, then cos x = 1 5 sin x. Because cos2 x = 1 sin 2 x, then 1 sin 2 x = cos 2 x = ( ) sin x 1 sin 2 x = sin x + sin2 x 0 = sin x + 2 sin2 x 0 = 25 sin 2 x 5 sin x 12 0 = (5 sin x 4)(5 sin x + 3) Since sin x 0, then sin x = 4 5 and cos x = = 3 5. Thus, tan x = sin x cos x = 4 3 (Alternative) Since 0 x < π, then sin x 0. Next, squaring both sides of the equation gives (sin x + cos x) 2 = = sin2 x + cos 2 x + 2 sin x cos x = 1 25 = 2 sin x cos x = Similarly, squaring the equation sin x cos x = y yields (sin x cos x) 2 = y 2 = 2 sin x cos x = 1 y 2

9 Thus, 1 y 2 = = y2 = = y = ±7 5 So adding sin x + cos x = 1 5 and sin x cos x = ±7 5 gives Since sin x > 0, then sin x = 4 5 Thus, 2 sin x = 1 5 ± 7 5 = 6 5 or 8 5 Therefore, cos x = 1 5 = cos x = 3 5 tan x = sin x cos x = 4/5 3/5 = 4 3

10 Round #5, Problem C: (Probability-Statistics/6 points/8 minutes) A point (x, y), x and y both integers with absolute value less than or equal to four, is chosen at random, with all such points having an equal chance of being chosen. What is the probability that the distance from the point to the origin is at most two units? Proof in pictures: Thus, counting the number of points on the circle or inside it results in 13 81

11 Round #6: (9 points/9 minutes) If r is the remainder when each of 1059, 1417, and 2312 is divided by d, where d is an integer greater than one, what is d r? 15 Since 1059, 1417, and 2312 all have a remainder of r, then they can all be written as 1059 = q 1 d + r 1417 = q 2 d + r 2312 = q 3 d + r Further, subtracting q 2 d q 1 d and q 3 d q 1 d yields 358 = (q 2 q 1 )d 895 = (q 3 q 1 )d It follows that d must be a factor of both 358 and 895. The prime factorization of each is 358 = = It follows that d = 179. Finally, to find r, divide any of the three given numbers by 179 and find the remainder. Thus, = It follows that r = 164 and d = = 15.

12 Round #7, Problem A: (4 points/15 minutes) In a triangle with sides a, b, and c, (a + b + c)(a + b c) = 3ab. What is the measure (in radians) of the angle opposite side c? π 3 By the law of cosines, c 2 = a 2 + b 2 2ab cos C Rearranging the given equation to look like the law of cosines gives (a + b + c)(a + b c) = 3ab a 2 + 2ab + b 2 c 2 = 3ab c 2 = a 2 + b 2 ab Thus, 2 cos C = 1 = cos C = 1 2 = C = π 3

13 Round #7, Problem B: (6 points/15 minutes) A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn without replacement with each coin having an equal chance of being chosen. What is the probability that the value of the coins drawn is at least 50 cents? The possible ways to get at least 50 cents are: 6 dimes 60 cents 5 dimes & 1 penny 55 cents 5 dimes & 1 nickel 51 cents 4 dimes, 2 nickles 50 cents The total number of ways to pick 6 coins from 12 is ( 12) 6. The number of ways to get 50 cents is ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) = }{{} 6 dimes }{{} 5 dimes, 1 nickel = 1 + 6(4) + 6(2) + 15(6) = }{{} 5 dimes, 1 penny }{{} 4 dimes, 2 nickels Thus, P (at least 50 cents) =

14 Round #7, Problem C: (10 points/15 minutes) Triangle ABC has AB = 4 and AC = 8. If M is the midpoint of BC and AM = 3, what is the length of BC? 2 31 (Easy Method) A Create a parallelogram from the triangle having adjacent sides AB and AC. Because the sum of the squares of the sides of a parallelogram equals the sum of the squares of its diagonals, then B a b M a + b X C 2( ) = 36 + (BC) 2 = (BC) 2 = 124 Thus, BC = 2 31 (Pythagoean Theorem) Drop a perpendicular from A to X. Let a = BX and b = XM. Then MC = a + b. By the Pythagorean Theorem, Thus, Therefore, 64 (a + 2b) 2 = 16 a 2 and 16 a 2 = 9 b 2 64 a 2 4ab 4b 2 = 16 a = a 2 b 2 7 = 48 = 4ab + 4b 2 7 = a 2 b 2 12 = a(a + b) 7 = (a b)(a + b) 12 a = a + b 7 a b = a + b 12 b = 7 a b = a = b ( ) b b 2 = b2 = b 2 = = b = It follows that ( 19 BC = 2(a + b) = ) ( ) 31 = 2 =

15 Round #8: (10 points/10 minutes) The lengths of the sides of a triangle are consecutive integers and the largest angle is twice the smallest angle. What is the cosine of the smallest angle? 3 4 The side that is opposite the smallest angle is the smallest side and the side opposite the largest angle 2α is the largest side. By the law of sines, x + 2 sin(2α) = By the law of cosines, x sin α = x sin α cos α = x sin α x 2α x + 2 = cos α = x + 2 2x x + 1 x 2 = (x + 1) 2 + (x + 2) 2 2(x + 1)(x + 2) cos α 2(x + 1)(x + 2) cos α = (x + 1) 2 + (x + 2) 2 x 2 = x 2 + 2x x 2 + 4x + 4 x 2 = x 2 + 6x + 5 2(x + 1)(x + 2) cos α = (x + 1)(x + 5) ( ) x + 2 2(x + 2) = x + 5 2x x 2 + 4x + 4 = x 2 + 5x x = 4 α Thus, cos α = (4) = 6 8 = 3 4

16 Tie-Breaker #1 What is the sum of all integers between 50 and 350 which end in 1? 5,880 Adding the sum to itself (and backwards) gives Thus, 2S is 392 times 30, and S = S = S = }{{} 30 of them S = 392(30) 2 = 5880

17 Tie-Breaker #2 What is the smallest integral value of k, so that 2x(kx 4) x = 0 has no real roots? 2 We need to find when the discriminant of the polynomial is less than zero. The polynomial simplifies to Thus, 2x(kx 4) x = 0 = (2k 1)x 2 8x + 6 = 0 0 > b 2 4ac = ( 8) 2 4(2k 1)6 = 88 48k 48k > 88 k > = The next largest integer is 2.

18 Tie-Breaker #3 Let f(x) = ax 7 + bx 3 + cx 5. If f( 7) = 7, what is f(7)? 17 Note that f( 7) = a( 7) 7 + b( 7) 3 + c( 7) 5 = 7 = a(7) 7 b(7) 3 c(7) 5 = 7 It follows that a(7) 7 + b(7) 3 + c(7) = 7 5 = 12 Thus, f(7) = a(7) 7 + b(7) 3 + c(7) 5 = 12 5 = 17

19 Tie-Breaker #4 The Kahuku Superette prices an item in dollars and cents so that when a 4% sales tax is added, no rounding is necessary because the result is exactly n dollars, where n is a positive integer. What is the smallest possible value of n? 13 Suppose x is the cost of the item andthat n is the price paid. It follows that Converting to fractions yields: C +.04C = 1.04C = n C = n = C = n = n Note that to make C a number that needs no rounding, the 13 in the bottom must be eliminated. Thus, n = 13.

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