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1. In the diagram, two straight lines are to be drawn through O(0, 0) so that the lines divide the figure OPQRST into 3 pieces of equal area. Find the sum of the slopes of the lines. y P(0, 6) Q(6, 6) R(6, 2) O(0, 0) S(12, 2) x T(12, 0) 2. Determine all linear functions f (x) = ax + b such that if g (x) = f –1(x) for all values of x, then f(x) – g(x) = 44 for all values of x. (Note : f –1 is the inverse function of f ) 3. (a) Determine all pairs (a, b) of positive integers for which a3 + 2ab = 2013. (b) Determine all real values of x for which log2(2x–1 + 3x+1) = 2x – log2(3x). 4. Integers can be written in bases other than the usual base 10. For example, the notation (235)7 stands for the base 7 representation of the integer 2 × 72 + 3 × 7 + 5 (which equals 124 in base 10). In general, if x, y and z are integers between 0 and b – 1, inclusive, then (xyz)b = xb2 + yb + z. Find all triples (x, y, z) for which (xyz)10 = 2(xyz)7, where each of x, y and z comes from the list 1, 2, 3, 4, 5, 6. 5. Vernon starts with a first number n with 0 < n < 1. Vernon enters the first number into a machine to produce a second number. He then enters the second number back into the machine to produce a third number. Vernon continually enters the result back into the machine giving a chain of numbers. When Vernon enters a number x into the machine, 1 zz if x ≤ , the machine outputs 2x, and 2 1 zz if x > , the machine outputs 2(1 – x). 2 If the machine ever produces the number 1, Vernon stops the process. (a) A chain starts with 3 . This gives 11 3 6 10 2 → → → → .... . What are the 11 11 11 11 next four numbers in this chain? (b) Vernon enters a number x with 0 < x < 1 into the machine and the machine produces the number x. Determine the value of x. (c) The fourth number in a chain is 1. Determine all of the possible values of the first number in this chain. Solutions 1. Extend QR downwards to meet the x-axis at U(6, 0). y P(0, 6) Q(6, 6) R(6, 2) O(0, 0) U S(12, 2) x T(12, 0) The area of figure OPQRST equals the sum of the areas of square OPQU (which has side length 6, so area 36) and rectangle RSTU (which has height 2 and width 6, so area 12). Thus, the area of figure OPQRST is 48. If we are to divide the figure into three pieces of equal area, then each piece has area 16. Let V be the first point on the perimeter Mathematics TODAY | NOVEMBER ’14 73 Page 73 (measured clockwise from P) so that the line through O and V cuts off an area of 16. Note that the area of DOPQ is half of the area of square OPQU, or 18, so V is to the left of Q on PQ. Thus, V has coordinates (v, 6) for some number v. y V P(0, 6) Q(6, 6) R(6, 2) O(0, 0) U S(12, 2) x T(12, 0) Consider DOPV having base OP of length 6 and height PV of length v. Since the area of DOPV is 16, then 1 16 (6)(v) = 16 or 3v = 16 or v = . 2 3 Thus, the coordinates of W are (8, 2), and so 2 1 the slope of OW is = . 8 4 Thus, the sum of the two required slopes is 9 1 11 + = . 8 4 8 2. Since f (x) = ax + b, we can determine an expression for g(x) = f –1(x) by letting y = f (x) and to obtain y = ax + b. We then interchange x and y to obtain x = ay + b which we solve for y x b to obtain ay = x – b or y = − . a a x b Therefore, f –1 (x) = − . a a LED I A ET ER D E EF OF 6 9 OR R Therefore, the slope of OV is =M . N 16 8 FOR3 LUTIO ISSUE DAY Let W be the second desired point. ER S TO SO B M C 1 E (OT)(T S) AT I Since the area of DOTS is OV N2 M E 1 1 H theT total area) = (12) (2) = 12 (less than ofA 2 3 M 1 and the area of trapezoid ORST is (RS + OT) 2 1 1 of (ST) = (6 + 12) (2) = 18 (more than 2 3 the total area), then W lies on RS. y P(0, 6) V Q(6, 6) R O(0, 0) U W S(12, 2) x T(12, 0) Suppose that W has coordinates (w, 2) for some number w. We want the area of trapezoid WSTO to be 16. 1 Therefore, (WS + OT) (ST) = 16 2 1 ⇒ (12 – w + 12) (2)= 16 2 ⇒ 24 – w = 16 ⇒ w = 8 74 Note that a ≠ 0. (This makes sense since the function f(x) = b has a graph which is a horizontal line, and so cannot be invertible.) Therefore, the equation f(x) – g(x) = 44 x b becomes (ax + b) – − = 44 or a a 1 b = 0x + 44, and this a − a x + b + a = 44 equation is true for all x. Since the equation 1 b a − a x + b + a = 0x + 44 is true for all x, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side. 1 b Therefore, a − = 0 and b + = 44. a a From the first of these equations, we obtain 1 or a2 = 1, which gives a = 1 or a = – 1. a b If a = 1, the equation b + = 44 becomes a b + b = 44, which gives b = 22. a= b = 44 becomes a b – b = 44, which is not possible. Therefore, we must have a = 1 and b = 22, and so f(x) = x + 22. If a = – 1, the equation b + Mathematics TODAY | NOVEMBER ’14 Page 74 3. (a) First, we factor the left side of the given equation to obtain a(a2 + 2b) = 2013. Next, we factor the integer 2013 as 2013 = 3 × 671 = 3 × 11 × 61. Note that each of 3, 11 and 61 is prime, so we can factor 2013 no further (We can find the factors of 3 and 11 using tests for divisibility by 3 and 11, or by systematic trial and error.) Since 2013 = 3 × 11 × 61, then the positive divisors of 2013 are 1, 3, 11, 33, 61, 183, 671, 2013 Since a and b are positive integers, then a and a2 + 2b are both positive integers. Since a and b are positive integers, then a2 ≥ a and 2b > 0, so a2 + 2b > a. Since a(a 2 + 2b) = 2013, then a and a2 + 2b must be a divisor pair of 2013 (that is, a pair of positive integers whose product is 2013) with a < a2 + 2b. We make a table of the possibilities: a a2 + 2b 2b b 1 2013 2012 1006 3 671 662 331 11 183 62 31 33 61 –1028 N/A Note that the last case is not possible, since b must be positive. Therefore, the three pairs of positive integers that satisfy the equation are (1, 1006), (3, 331), (11, 31). (We can verify by substitution that each is a solution of the original equation.) (b) We successively manipulate the given equation to produce equivalent equations : log2(2x–1 + 3x+1) = 2x – log2(3x) log2(2x–1 + 3x+1) + log2(3x) = 2x log2((2x–1 + 3x+1)3x) = 2x (using log2 A + log2 B = log2 AB) (2x–1 + 3x+1)3x = 22x (exponentiating both sides) 2–12x3x + 313x3x = 22x 1 x x ⋅ 2 3 + 3 ⋅ 32x = 22x 2 2x3x + 6 ⋅ 32x = 2 ⋅ 22x (multiplying by 2) 2x3x2–2x + 6 ⋅ 32x2–2x = 2 (dividing both sides by 22x ≠ 0) 2–x3x + 6 ⋅ 32x ⋅ 2–2x = 2 x 3 3 2 + 6 2 2x =2 x 3 Next, we make the substitution t = , 2 2 2x x 3 3 noting that = = t 2. Thus, 2 2 we obtain the equivalent equations t + 6t2 = 2 6t2 + t – 2 = 0 (3t + 2) (2t – 1) = 0 1 2 Therefore, t = − or t = . 2 3 x 3 Since t = > 0, then we must have x 2 1 3 t= = . 2 2 Thus, log(1 / 2) log 1 − log 2 = log(3 / 2) log 3 − log 2 log 2 − log 2 = = log 3 − log 2 log 2 − log 3 x = log 3/2(1 / 2) = 4. Since we are told that (xyz)b = xb2 + yb + z, then (xyz)10 = 102x + 10y + z = 100x + 10y + z and (xyz)7 = 72x+ 7y + z = 49x + 7y + z. From the given information (xyz)10 = 2(xyz)7 100x + 10y + z = 2(49x + 7y + z) 100x + 10y + z = 98x + 14y + 2z 2x = 4y + z Mathematics TODAY | NOVEMBER ’14 75 Page 75 Since the left side of this equation is an even integer (2x), then the right side must also be an even integer. Since 4y is an even integer, then for 4y + z to be an even integer, it must be the case that z is an even integer. This gives us three possibilities: z = 2, z = 4 and z = 6. Case 1 : z = 2 Here, 2x = 4y + 2 or x = 2y + 1. We try the possible values for y : zz y = 1 gives x = 2(1) + 1 = 3; this gives the triple (x, y, z) = (3, 1, 2) zz y = 2 gives x = 2(2) + 1 = 5; this gives the triple (x, y, z) = (5, 2, 2) zz If y is at least 3, then x = 2y + 1 is at least 7, which is impossible. Therefore, there are two triples that work when z = 2. Case 2: z = 4 Here, 2x = 4y + 4 or x = 2y + 2. We try the possible values for y : zz y = 1 gives x = 2(1) + 2 = 4; this gives the triple (x, y, z) = (4, 1, 4) zz y = 2 gives x = 2(2) + 2 = 6; this gives the triple (x, y, z) = (6, 2, 4) zz If y is at least 3, then x = 2y + 2 is at least 8, which is impossible. Therefore, there are two triples that work when z = 4. Case 3: z = 6 Here, 2x = 4y + 6 or x = 2y + 3. We try the possible values for y : zz y = 1 gives x = 2(1) + 3 = 5; this gives the triple (x, y, z) = (5, 1, 6) zz If y is at least 2, then x = 2y + 3 is at least 7, which is impossible. Therefore, there is one triple that works when z = 6. Finally, the triples that satisfy the equation are (x, y, z) = (3, 1, 2), (5, 2, 2), (4, 1, 4), (6, 2, 4), (5, 1, 6). 5. (a) Since the fourth number in the chain is 2 1 2 and is less than , then the fifth 11 2 11 76 2 4 number is 2 = . 11 11 4 Since the fifth number in the chain is 11 4 1 is less than , then the sixth and 11 2 4 8 number is 2 = . 11 11 8 Since the sixth number in the chain is 11 8 1 is larger than , then the seventh and 11 2 8 3 6 number is 2 1 − = 2 = . 11 11 11 6 Since the seventh number in the chain is 11 1 6 is larger than , then the eighth and 2 11 6 5 10 number is 2 1 − = 2 = . 11 11 11 Therefore, the next four numbers in the 4 8 6 10 chain are , , , . 11 11 11 11 (b) Therefore are two possibilities : 1 1 x≤ and x > . 2 2 1 If x ≤ and x is entered into the machine, 2 then the machine outputs 2x. Since the input is to be the same as the output, then x = 2x or x = 0. This is impossible since x > 0. 1 If x > and x is entered into the machine, 2 then the machine outputs 2(1 – x). Since the input is to be the same as the output, then x = 2(1 – x) or x = 2 – 2x. 2 Thus, 3x = 2 and so x = , which satisfies 3 the restrictions. Therefore, if x is entered into the machine 2 and x is produced, then x = . 3 (c) Suppose that the chain is a, b, c, 1. In this part, we have to produce the chain Mathematics TODAY | NOVEMBER ’14 Page 76 Mathematics TODAY | NOVEMBER ’14 77 Page 77 backwards (that is, we have to determine c, b and a). We make the following general observation: Suppose that x is entered into the machine and y is produced. 1 If x ≤ , then y = 2x. Solving for x in 2 1 terms of y, we obtain x = y. (This gives 2 us the input in terms of the output) 1 If x > , then y = 2(1 – x). Solving for 2 x in terms of y, we obtain y = 2 – 2x or 1 2x = 2 – y or x = (2 − y). 2 Since the third number in the chain is c and the fourth is 1, then from above, 1 1 1 1 c = (1) = or c = 2 (2 − 1) = 2 . 2 2 1 In either case, the chain is a, b, , 1. 2 Since the second number in the chain is 1 b and the third is , then from above, 2 1 1 1 1 1 1 3 3 b = = or b = 2 − = = . 2 2 4 2 2 2 2 4 78 Therefore, the chain is either a, 1 1 , ,1 4 2 1 1 3 1 , , 1. If the chain is a, , , 1, 4 2 4 2 then the first number is a and the second 1 number is . 4 1 1 1 Thus, either a = = or 2 4 8 1 1 1 7 7 a = 2 − = = . 2 4 2 4 8 or a, 3 1 , , 1, then the first 4 2 number is a and the second number is 3 . 4 1 3 3 Thus, either a = = or 2 4 8 1 3 1 5 5 a = 2 − = = . 2 4 2 4 8 If the chain is a, Therefore, the possible first numbers in 1 3 5 7 the chain are , , , . 8 8 8 8 We can double check that each of these gives a fourth number of 1: 1 1 1 3 3 1 → → →1 → → →1 8 4 2 8 4 2 5 3 1 7 1 1 → → →1 → → →1 8 4 2 8 4 2 nn Mathematics TODAY | NOVEMBER ’14 Page 78