BHP BILLITON UNIVERSITY OF MELBOURNE SCHOOL MATHEMATICS COMPETITION, 2003: INTERMEDIATE DIVISION

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1 BHP BILLITON UNIVERSITY OF MELBOURNE SCHOOL MATHEMATICS COMPETITION, 00: INTERMEDIATE DIVISION 1. A fraction processing machine takes a fraction f and produces a new fraction 1 f. If a fraction f = p is fed into the machine, what fraction 1+f q will be produced after 00 processes? Solution: When we perform the operation for the first time, we obtain: 1 f 1 + f = 1 p q 1 + p q = q(1 p q ) q(1 + p q ) = q p q + p. When we perform the operation a second time, we obtain: 1 q p q+p 1 + q p q+p = (q + p)(1 q p q+p ) (q + p)(1 + q p q+p = q + p q + p q + p + q p = p q = p q. ) = (q + p) (q p) (q + p) + (q p) Therefore, after we perform the operation twice, we are back at our starting point. So the fraction will be p if we perform the operation q an even number of times, and q p if we perform the operation an odd q+p number of times. The number 00 is odd, so the fraction will be q p q+p if we perform the operation 00 times.. Pegs are placed in a board to form a grid as shown in the diagram below. The pegs are 5cm apart, horizontally and vertically. A rubber band is stretched over some pegs, as shown. Calculate the area enclosed by the rubber band. 1

2 Solution: We can split the figure up into triangles and a rectangle as shown. If b is the base of the triangle and h is the height, we can use the formula for the area of a triangle 1 b h. So we can work out the total area, T. T = A + B + C + D + E + F + G = = = So the total area of the rubber band is cm.. In the following multiplication problem, A, B, C and D are distinct natural numbers (1,,, 4, 5, 6, 7, 8 or 9). Find their values. ABCD 4 DCBA

3 Solution: First of all, we note that A. If A were greater than, then ABCD would be at least 000, so DCBA = 4 ABCD = So DCBA would have at least 5 digits. But we know that DCBA has four digits, so we must have A. So A is 1 or. But A can t be 1, because DCBA is a multiple of 4, and multiples of 4 never end in 1 because they are even. So A must be. BCD 4 = DCB We can see that 4 D must end in. The only digits for which this is true are and 8. But as A =, we have ABCD 000, so DCBA = 4 ABCD = So D 8, therefore D can t be. So D must be 8. BC8 4 = 8CB Now we note that B. This is because, if B, then ABCD 00, so DCBA = 4 ABCD 4 00 = 900. But we know that D = 8, so this can t occur. Therefore B. As A = and A, B, C and D are distinct (i.e. no two of them are the same), we can t have B =, so B = 1. 1C8 4 = 8C1 We now consider the tens column of this multiplication. There will be a carried from the units column (as 4 8 = ) so we have 4 C + = something ending in 1. So 4 C must end in 8, so C is either or 7. Again, as the digits are distinct and we already have A =, we must have C = 7. A quick check shows that this number does in fact work in the given equation = Two circles with centres at P and Q touch each other at C. The two circles also touch a straight line as shown in the diagram below. If the length of AC is 8cm and the length of BC is 6cm, find the length of AB. Solution: We construct a point D on AB so that CD is the common tangent of the two circles, as shown. As the two tangents from a point to a circle are equal in length, we know that CD = DA. Similarly, we have CD = DB. We now observe that, as CD = DA, DAC = DCA.

4 Similarly, DBC = DCB. Clearly ACB = DCA+ DCB. Now the angles in ABC must add up to 180, so ACB + CBA + BAC = 180 ACB + DBC + DAC = 180 ACB + DCB + DCA = 180 ACB + ACB = 180 ACB = 180 = 90 Therefore, ACB is a right angle. So we can apply Pythagoras Theorem as follows. So AB = 10 cm. AB = AC + BC AB = AB = AB = 100 AB = The pages of a book are numbered consecutively from 1 to n, where n is a natural number. When the numbers of the pages were added together one of the page numbers was mistakenly added twice, giving an incorrect total of 00. Find the number of the page that was mistakenly added twice. 4

5 Solution: We first note that, if S = n, then S = n and S = n + (n 1) + (n ) If we add these two equations, we have S = (1 + n) + ( + n 1) + ( + n ) (n + ) + (n 1 + ) + (n + 1) S = n(1 + n) S = n(n + 1) Let the number of pages in the book be n, and let the number of the page that was mistakenly added twice be x. Then the total sum of page numbers would be x (n 1) + n = n + x. We therefore have 00 = n(n + 1) + x. By trial and error, we can find the largest value of n so that n(n+1) 00. This turns out to be 6. We can then find the page number x = 00 n(n + 1) = 00 6(6 + 1) = = 50. So there were 6 pages in the book, and page 50 was repeated twice. We must now check that there are no other solutions possible. We first check n = 61 x = 00 n(n + 1) = 00 61(61 + 1) = 11 > 61. But this means that the page number that was doubled was greater than the largest page number in the book, which clearly can t occur. So no value of n that is less than or equal to 61 can work. If n = 6, then x = 00 n(n + 1) = (6 + 1) = 1.

6 So the page added twice was negative. This clearly can t work either. So no value of n that is greater than or equal to 6 can work. Therefore there are 6 pages in the book, and page number 50 was added twice, and this is the only possible solution. 6. The teacher asks a member of a class to think of a three-digit number abc, where a, b and c are natural numbers. The teacher then asks the student to write down the five numbers acb, bac, bca, cab and cba, add them together and reveal the total T. When told the value of T, the teacher could then determine the original number abc. In the case when T = 194, can you find the number abc? Solution: We are given that acb + bac + bca + cab + cba = 194. We add abc to both sides of this equation (In this case, abc represents the number 100a + 10b + c, not a b c) abc + acb + bac + bca + cab + cba = abc This means that 100a + 10b + c + 100a + 10c + b + 100b + 10a + c +100b + 10c + a + 100c + 10a + b + 100c + 10b + a = abc a + b + c = abc (a + b + c) = abc Now, as abc is a -digit number, it is between 100 and 999. So abc abc (a + b + c) 419 Now, 14 = 108, which is smaller than 94, so a + b + c > 14. Also, 19 = 418, which is bigger than 419, so a + b + c < 19. So a + b + c can only be 15, 16, 17 or 18. Case 1: a + b + c = 15 (a + b + c) = abc abc = (a + b + c) 194 = = 16 6

7 But = 10, and we want a + b + c = 15, so this can t work. Case : a + b + c = 16 Similarly, we have abc = = 58 We check to see that this works in the given equation: = 194 So this is a solution. We keep checking to make sure there are no more. Case : a + b + c = 17 abc = = 580 But = 1, and we want a + b + c = 17, so this can t work. Case 4: a + b + c = 18 abc = = 80 But = 10, and we want a + b + c = 18, so this can t work. Therefore, the only solution is abc = If n is a natural number, find the values that n can take so that n 440 is a perfect square. Solution: We want to find all possible integers n so that there exists an integer x that satisfies the equation n 440 = x. We can perform some algebraic manipulation n x = 440 (n x)(n + x) = 440 So we can list all possible values of n x and n + x: (n x, n + x) = (1,440), (,0), (4,110), (5,88), (8,55), (10,44), (11,40) or (0,). Clearly n x < n + x so we list the smaller factor first. Also note that we need not consider negative factors (such as ( 5, 88)) because n is a natural number and is therefore positive, and it is clearly greater than x as n > x, so n + x and n x must both be positive. 7

8 We now note that n = (n x) + (n + x). So we can find the possible values of n from the above list of factors. The values we obtain are: , 0 + or, more simply,, , , , , 0.5, 111, 57, 46.5, 1.5, 7, , 0 +, We are only concerned with natural numbers, so this leaves us with the possible solutions 111, 57, 7 and 1. We now note that = = = = 1 So n = 111, 57, 7 or 1 are the only possible solutions for n. 8. Find the probability that a point chosen randomly inside a square of side length L cm is not more than L cm from any of the four vertices. Solution: Let us label the areas as shown in the diagram. Area A is the region in which points are at most L from each vertex. If the point is chosen randomly, the probability of landing in region A will be equal to the area of region A divided by the area of the square (which is clearly L ), so we need to find the area of A. We know the total area is L and the area of one quarter-circle is πl, so we have some equations 4 A + 4B + 4C = L A + B + C = πl 4 8

9 Unfortunately, these equations don t provide enough information to solve for the area of A. So we need another. This time we choose the region bounded by the arcs DF, EF and the segment DE, which has area A + B + C. Note that triangle DEF (with straight edges rather than arcs) must be equilateral as DE = EF = DF = L, so DEF = EDF = 60. So we can find the area of this region by adding the area of triangle DEF to the area of the two extra little segments. If one of these segments is added to the triangle, we have one-sixth of a circle (As DEF = EDF = 60 = ). So to find the total area, we add one-sixth of the area of a circle of radius L twice and take away the area of the triangle DEF because we have counted that twice. The base of DEF has length L and its height is L. So we have A + B + C = πl 6 1 L = πl L 4 L Now we have enough information to solve for A. First, we get rid of C: 4(A + B + C) (A + 4B + 4C) = 4( πl L ) L 4 A + 4B = 4πL L L 9

10 We can also see that (A + B + C) (A + 4B + 4C) = ( πl 4 ) L A + B = πl L We can now combine these two equations to find A A + 4B (A + B) = 4πL A = 4πL L L ( πl L ) L L πl + L Grouping like terms and simplifying, Now, our probability is: A = πl L + L. πl A L + L = L L = π

2008 Euclid Contest. Solutions. Canadian Mathematics Competition. Tuesday, April 15, c 2008 Centre for Education in Mathematics and Computing

2008 Euclid Contest. Solutions. Canadian Mathematics Competition. Tuesday, April 15, c 2008 Centre for Education in Mathematics and Computing Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 008 Euclid Contest Tuesday, April 5, 008 Solutions c 008