BC Exam Solutions Texas A&M High School Math Contest October 22, 2016

Size: px

Start display at page:

Download "BC Exam Solutions Texas A&M High School Math Contest October 22, 2016"

Kenneth Morgan
5 years ago
Views:

1 BC Exam Solutions Texas A&M High School Math Contest October, 016 All answers must be simplified, if units are involved, be sure to include them. 1. Given find A + B simplifying as much as possible. 1 A = 3 B = ( 5 3)( 5 + 3), A = 1 3 = + 3 ( 3)( + 3) = + 3 ( 3) = = + 3 B = ( 5) ( 3) = 5 3. Therefore, A + B = = 9. Answer: 9. Let x y be the solutions of the system of equations 39 x 10y = 5 15 x + y = 5. Find x + y. Squaring both sides of each equation, we get 39 x 10y = 5 15 x + y = 5 39 x 10y = 5 15 x + y = 5 x + 10y = 14 x + y = 10. Adding the two equations we get 1y = 4, which implies that y =. Replacing y in the first equation gives us x + 0 = 14, which implies that x = 3. Therefore, x + y = 1. Answer: 1 3. A collection of nickels dimes has a total value of $.40 contains 35 coins. How many nickels are in the collection? Let x be the number of nickels in the collection y be the number of dimes in the collection. Then we have x + y = 35 5x + 10y = 40 5x + 5y = 175 5x + 10y = 40 x + y = 35 5y = 65 x = y = 13. Answer:

2 4. A box containing 180 cubic inches is constructed by cutting from each corner of a cardboard square a small square with side 5 inches, then turning up the sides. Find the area of the original piece of cardboard. Let x be the length (in inches) of the side of the cardboard square. The volume of the box is 5(x 10). We get that 5(x 10) = 180 (x 10) = 36 x 10 = 6 x = 16. The area of the original piece of cardboard is 16 = 56 square inches. Answer: 56 in 5. Find the largest common divisor for the numbers We have that = ( ) = ( ) = = = = ( ) = ( ) = = Therefore, the largest common divisor for the two numbers is 3 = 8. Answer: 8 6. Find the sum of all solutions of the equation x + 6x + x + 6x = 0. Let y = x + 6x. Then y + y = 0 where y 0. Solutions of y + y 0 = 0 are y = 5 or y = 4. Since y 0, we conclude that y = 4. Thus, x + 6x = 4 x + 6x = 16 x + 6x 16 = 0 x = 8 or x =. The sum of all real solutions of our initial equation is 8 + = 6. Answer: 6

3 7. Suppose that ABC is a right triangle with A = 90, AB = 5, AC = 1. On the line AB we consider the point M such that BMC is isosceles with BM = CM. Find AM. Let AM = x. Then CM = BM = AB + AM = 5 + x. By applying the Pythagorean Theorem in the right triangle CAM, we get that CM = AC + AM (5 + x) = 1 + x x + 10x + 5 = x x = 119 x = Answer: Mr. Kaye is 11 times as old as his daughter Lynn. Thirty-six years from now he will be at most twice as old as Lynn. At most, how old is Lynn? Let x be the number of years in Lynn s age now. Mr. Kaye s age now is 11x. Since Mr. Kaye s age in 36 years from now will be at most twice Lynn s age we get that 11x + 36 (x + 36) 11x + 36 x + 7 9x 36 x 4. Therefore, Lynn is at most 4 years old. Answer: 4 9. In the ABC, AB = AC A = 10. The median AD to the side BC is extended through the point D with the segment DM = 3AD. Find DMB. First, since ABC is isosceles, the median AD is also the altitude to BC the bisector of A. Let E be the midpoint of AM. Then AE = EM = AM, AD = DE = AM. In the ABE, BD then is 4 the altitude to AE the median, so ABE is isosceles. Next, BAE = BEA = 60, so ABE is equilateral, thus BE = AB = AE. So BE = EM, since E in the isosceles BEM is equal to 10, EMB = EBM = = 30. Finally, DMB = EMB = 30. Answer: 30

4 10. In the figure below AB is tangent to the circle. If AB = 8 AC exceeds AD by 1, what is AC? We notice that ABD = ACB BAD = CAB. similar. This implies AD AB = AB AC AB = AD AC. Since AB = 8 AD = AC 1 we obtain that Thus, ADB ABC are 8 = AC(AC 1) AC 1AC 64 = 0 AC = 4 or AC = 16. Since AC > 0, we conclude that AC = 16. Answer: Find the sum of all positive integers x for which x + 56 x are perfect squares. We have x + 56 = p x = q with p q positive integers. Clearly p < q. Subtracting the two equations we get = q p (q p)(q + p) = 57 (q p)(q + p) = Since 0 < q p < q + p, we have the following two possible cases q p = 1 q + p = 57 or q p = 3 p = 8, q = 9 or p = 8, q = 11 x = 78 or x = 8. q + p = 19 The sum is = 736. Answer: Consider the sum S = , + 3 n + n + 1 where n is a positive integer. If S = 10, what is the value of n? 1 3 n + 1 n S = ( 1)( + 1) + ( 3 )( 3 + ) + + ( n + 1 n)( n n) 1 3 n + 1 n = ( ) 1 + ( 3) ( ) + + ( n + 1) ( n) = n + 1 n = n Therefore, S = 10 n = 10 n + 1 = 11 n + 1 = 11 n = 10. Answer: 10

5 13. Find the product of all solutions of the equation (3x 4x + 1) 3 + (x + 4x 5) 3 = 64(x 1) 3. If we denote 3x 4x + 1 = a x + 4x 5 = b then a + b = 4x 4 so our equation becomes Then we have a 3 + b 3 = (a + b) 3 a 3 + b 3 = a 3 + 3a b + 3ab + b 3 3a b + 3ab = 0 3ab(a + b) = 0 a = 0 or b = 0 or a + b = 0. a = 0 3x 4x + 1 = 0 (3x 1)(x 1) = 0 x = 1 3 or x = 1 b = 0 x + 4x 5 = 0 (x 1)(x + 5) = 0 x = 5 or x = 1 a + b = 0 x 1 = 0 x = 1 or x = 1. The solutions of our equation are 5, 1, 1 3, 1. The product of all solutions is 5 3. Answer: A circle whose center is on the x-axis passes through the points (3, 5) (6, 4). Find the radius of the circle. Let C(h, k) be the center of the circle r be the radius of the circle. Since the center is on the x-axis we get that k = 0. Then an equation of the circle has the form (x h) + y = r. The points (3, 5) (6, 4) being on the circle gives us the system of equations (3 h) + 5 = r If we subtract the two equations we get (6 h) + 16 = r. (3 h) (6 h) + 9 = 0 9 6h + h h h + 9 = 0 6h 18 = 0 h = 3. So (3 3) + 5 = r r = 5 r = 5. Answer: Find the sum of all integers N with the property that N 71 is divisible by 7N If (N 71)/(7N + 55) = M where M is an integer, then N 7MN (55M + 71) = 0. Solving for N, we find N = 7M ± 49M + 0M The number under the radical must be a perfect square to produce an integer N. Notice that (7M + 15) < 49M + 0M + 84 < (7M + 17) thus 49M +0M +84 = (7M +16). Then, solving 49M +0M +84 = (7M +16), we get M = 7, so N = 57 or N = 8. The sum is 57 + ( 8) = 49. Answer: 49

6 16. In the ABC, BD is the median to the side AC, DG is parallel to the base BC (G is the point of intersection of the parallel with AB). In the ABD, AE is the median to the side BD F is the intersection point of DG AE. Find BC F G. Since DG is parallel to CB AD = DC, we conclude that BG = GA. Thus DG is the midsegment of the ABC, which means that GD = 1 BC. In the ABD, DG AE are two medians, F is their intersection point, thus F G = 1 3 DG. Therefore, F G = 1 BC BC, which implies that 6 F G = 6. Answer: The function f(x) = x + (x + ) + (x + 98) [ (x + 1) + (x + 3) + (x + 99) ] is a linear function, f(x) = ax + b. Find a b. We notice that f(x) = [x (x + 1) ] + [(x + ) (x + 3) ] + + [(x + 98) (x + 99) ] = (x x 1)(x + x + 1) + (x + x 3)(x + + x + 3) + + (x + 98 x 99)(x x + 99) = (x + 1) (x + 5) (x + 197) = 50 x ( ) = 100x = 100x Therefore, a = 100, b = 4950, a b = Answer: In the isosceles ABC with AB = AC, let AM be the median to the side BC let BD be the altitude to the side AC. If AMD = 4 BDM find ACB. Let BDM = α. Then AMD = 4α. Since BDC is a right triangle M is the midpoint of the hypotenuse BC, we get that DM = BM = MC. This implies that MDB is isosceles thus MBD = BDM = α.

7 Because ABC is isosceles M is the midpoint of BC we obtain that AM is perpendicular onto BC, which implies that BMD = α. In MBD we have In AMD we have MBD + BDM + BMD = 180 α + α α = 180 α = 15. MAD = 180 ( AMD + ADM) = 180 (4α α) = 90 5α = 15. Therefore, in AMC we see that which implies that ACB = 75. Answer: Let f g be two linear functions such that for all numbers x. Find g(5). ACM = 90 MAC = 90 MAD = = 75, For x = we get f(1) = 3 + g(1) f(1) g(1) = g(1) f(1) f(x 1) = x 3 + g(1) f(1) g(x 1) = 4x + 5 g(1) f(1), f(1) g(1) = 1 f(1) + g(1) = 13 f(1) = 3 g(1) = 5. So, g(x 1) = 4x = 4x 3 = 4(x 1) + 1, which implies that g(x) = 4x + 1. Thus, g(5) = 1. Answer: 1 0. Consider ABC with B = C = 70. On the sides AB AC we take the points F E, respectively, so that ABE = 15 ACF = 30. Find AEF. We see that In BCE we have EBC = ABC ABE = = 55 BCF = ACB ACF = = 40. BEC = 180 ( EBC + ECB) = = 55.

8 So, BEC = EBC = 55, which implies that BCE is isosceles. Thus BC = EC (1). In BCF we have BF C = 180 ( F BC + BCF ) = = 70. Therefore, BF C = F BC = 70, which implies that BCF is isosceles. We get that BC = F C (). From (1) () we deduce that EC = F C which makes the CEF isosceles. Since ECF = 30, we get that F EC = EF C = = 75, AEF = 180 F EC = = 105. Answer: 105

1) Exercise 1 In the diagram, ABC = AED, AD = 3, DB = 2 and AE = 2. Determine the length of EC. Solution:

1) Exercise 1 In the diagram, ABC = AED, AD = 3, DB = 2 and AE = 2. Determine the length of EC. Solution: First, we show that AED and ABC are similar. Since DAE = BAC and ABC = AED, we have that AED is