# 2007 Cayley Contest. Solutions

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1 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 007 Cayley Contest (Grade 10) Tuesday, February 0, 007 Solutions c 006 Waterloo Mathematics Foundation

2 007 Cayley Contest Solutions Page 1. Calculating, 8 + (3 ) = 8 + (9) = = 6.. Calculating, = 8 56 = If 3x x + x = 3 + 1, then x = or x = In 3 hours, Leona earns \$4.75, so she makes \$ = \$8.5 per hour. Therefore, in a 5 hour shift, Leona earns 5 \$8.5 = \$41.5. Answer: (B) Answer: (E) 5. The value of 1 of 100 is = We evaluate each of the five given possibilities: (A): 0% of 00 is 0.(00) = 40 (B): 10% of 50 is 0.1(50) = 5 (C): 15% of 100 is 15 (D): 5% of 50 is 0.5(50) = 1.5 (E): 5% of 300 is 0.05(300) = 15 Therefore, the correct answer is (B). (Note that we did not have to evaluate (C), (D) and (E) after seeing that (B) was correct.) Answer: (B) 6. Evaluating each of the five given possibilities with a = and b = 5, (A) a b = (B) b 5 a = 5 (C) a b = 5 = 3 (D) b a = 5 = 3 (E) 1 a = 1() = 1 The largest of these five is 3, or (D). 7. The mean of 6, 9 and 18 is = = 11. Since the mean of 1 and y is thus 11, then the sum of 1 and y is (11) =, so y = In ABC, ABC = BAC, so AC = BC. In BCD, CBD = CDB, so CD = BC. Since the perimeter of CBD is 19 and BD = 7, then 7 + BC + CD = 19 or (BC) = 1 or BC = 6. Since the perimeter of ABC is 0, BC = 6, and AC = BC, then AB +6+6 = 0 or AB = Solution 1 Since the area of rectangle ABCD is 40 and AB = 8, then BC = 5. Therefore, MBCN is a trapezoid with height 5 and parallel bases of lengths 4 and, so has area 1 (5)(4 + ) = 15. Solution Since the area of rectangle ABCD is 40 and AB = 8, then BC = 5. Draw a line from N to AB parallel to BC (and so perpendicular to AB) meeting AB at P.

3 007 Cayley Contest Solutions Page 3 A 4 M P B D N C This line divides MBCN into a rectangle P BCN of width and height 5, and a triangle MP N with base MP of length and height P N of height 5. The area of MBCN is sum of the areas of these two parts, or (5) + 1 ()(5) = = Solution 1 To get from one term to the next, we double and add 4. Since the first term is x, the second term is x + 4. Since the second term is x + 4, the third term is (x + 4) + 4 = 4x + 1. Since the third term equals 5, then 4x + 1 = 5 so 4x = 40 or x = 10. Solution To get from one term to the next, we double and add 4. Therefore, to get from one term to the previous term, we subtract 4 and divide by. Since the third term is 5, then subtracting 4 gives 48 and dividing by gives 4, which is the second term. Since the second term is 4, then subtracting 4 gives 0 and dividing by gives 10, which is the first term. Thus, x = Suppose that Ivan ran a distance of x km on Monday. Then on Tuesday, he ran x km, on Wednesday, he ran x km, on Thursday, he ran 1 x km, and on Friday he ran x km. The shortest of any of his runs was on Thursday, so 1 x = 5 or x = 10. Therefore, his runs were 10 km, 0 km, 10 km, 5 km, and 10 km, for a total of 55 km. 1. When (0, 0) is reflected in the line x = 1, the image is (, 0). y (, 4) (0, 0) (, 0) x=1 y= x When (, 0) is reflected in the line y =, the image is (, 4). Answer: (E)

4 007 Cayley Contest Solutions Page Solution 1 Since the ratio AN : AC equals the ratio AP : P B (each is 1 : ) and A is common in AP N and ABC, then AP N is similar to ABC. Since the ratio of side lengths between these two triangles is 1 :, then the ratio of areas is 1 : = 1 : 4. Thus, the area of ABC is 4 = 8 cm. Solution Since the ratio AN : AC equals the ratio AP : P B (each is 1 : ) and A is common in AP N and ABC, then AP N is similar to ABC. Therefore, ANP = ACB = 90. Similarly, P MB is similar to ACB, and P MB = ACB = 90. Therefore, AN = NC = P M = 1 AC and NP = CM = MB = 1 CB. Thus, P MB is congruent to ANP, so has area cm. Rectangle NP MC has the same width and height as ANP, so has double the area of AP N or 4 cm. Thus, the area of ACB is = 8 cm. Solution 3 Join C to P. Since CP is the diagonal of the rectangle, then CP N and P CM are congruent, so have equal areas. Since AN = NC and P NA and P NC have equal height (P N), then the areas of P NA and P NC are equal. Similarly, the areas of P MC and P MB are equal. In summary, the areas of the four small triangles are equal. Since the area of AP N is cm, then the total area of ABC is 8 cm Using a common denonimator, x x 6 = 6 x x 6 = 11 x Therefore, x 6 = 11, so x 6 =. 15. Since ABC and P QR are equilateral, then ABC = ACB = RP Q = 60. Therefore, Y BP = = 55 and Y P B = = 45. In BY P, we have BY P = 180 Y BP Y P B = = 80. Since XY C = Y BP, then XY C = 80. In CXY, we have CXY = = Since 60% of the students are in Arts, 6000 students are in Arts, and so 4000 students are in Science. Since 40% of the Science students are male, then 0.4(4000) = 1600 of the Science students are male. Since half of the total of students are male, then = 3400 of the male students are in Arts. Since there are 6000 students in Arts, then = 600 of the Arts students are female, or % 43.33% of the Arts students are female. Answer: (E)

6 007 Cayley Contest Solutions Page 6 and DO = 4. The area of ABCD equals the area of P RQO minus the areas of triangles AP B, BRC, CQD, and DOA. P RQO is a rectangle, so has area 3 5 = 15. Triangles AP B and CQD have bases P B and QD of length 1 and heights AP and CQ of length, so each has area 1 (1)() = 1. Triangles BRC and DOA have bases BR and DO of length 4 and heights CR and AO of length 1, so each has area 1 (4)(1) =. Thus, the area of ABCD is = 9. (Alternatively, we could notice that ABCD is a parallelogram. Therefore, if we draw the diagonal AC, the area is split into two equal pieces. Dropping a perpendicular from C to Q on the x-axis produces a trapezoid ACQO from which only two triangles need to be removed to determine half of the area of ABCD.) 0. Since 3(n 007 ) < , then n 007 < = But = (3 ) 007 = so we have n 007 < Therefore, n < 9 and so the largest integer n that works is n = Since T has played 5 matches so far, then T has played P, Q, R, S, and W (ie. each of the other teams). Since P has played only 1 match and has played T, then P has played no more matches. Since S has played 4 matches and has not played P, then S must have played each of the remaining 4 teams (namely, Q, R, T, and W ). Since Q has played only matches and has played T and S, then Q has played no more matches. Since R has played 3 matches, has played T and S but has not played P or Q, then R must have played W as well. Therefore, W has played T, S and R, or 3 matches in total. (Since we have considered all possible opponents for W, then W has played no more matches.). Suppose the first integer in the list is n. Then the remaining four integers are n + 3, n + 6, n + 9, and n + 1. Since the fifth number is a multiple of the first, then n + 1 n Since is an integer, then = n n + 1 n = n is an integer, or n is a positive divisor of 1. is an integer. n n The positive divisors of 1 are 1,, 3, 4, 6, and 1, so there are 6 possible values of n and so 6 different lists. (We can check that each of 6 values of n produces a different list, each of which has the required property.) 3. Since the same region (AEHD) is unshaded inside each rectangle, then the two shaded regions have equal area, since the rectangles have equal area. Thus, the total shaded area is twice the area of AEHCB. Draw a horizontal line through E, meeting AB at X and HC at Y.

7 007 Cayley Contest Solutions Page 7 G A D H F X B E Y C We know BAE = 30 and AE = 1, so EX = 1AE = 6 and AX = 3EX = 6 3, by the ratios in a triangle. Thus, the area of EXA is 1(6)(6 3) = Since AB = 1 and AX = 6 3, then XB = Therefore, rectangle BXY C has height and width 18, so has area 18(1 6 3) or Since XY = BC = 18 and EX = 6, then EY = 1. Since AEB = 60 and AEH = 90, then HEY = 30. Since EY = 1 and HEY = 30, then HY = 1 = Thus, EY H has area 1 (1)(4 3) = 4 3. So the total area of AEHCB is ( ) = Therefore, the total shaded area is ( ) = To form such a collection of integers, our strategy is to include some integers larger than 1 whose product is 007 and then add enough 1s to the collection to make the sum 007. In order to make n as small as possible, we would like to include as few 1s as possible, and so make the sum of the initial integers (whose product is 007) as large as possible. Since we would like to consider integers whose product is 007, we should find the divisors of 007. We see that 007 = = 3 3 3, and 3 is a prime number. So the collections of positive integers larger than 1 whose product is 007 are {3, 669}, {3, 3, 3}, and {9, 3}. The collection with the largest sum is {3, 669} whose sum is 67. To get a sum of 007, we must add = 1335 copies of 1, which means that we are using = 1337 integers. Therefore, the smallest value of n is Answer: (B) 5. Since W AX = 90 regardless of the position of square ABCD, then A always lies on the semi-circle with diameter W X. The centre of this semi-circle is the midpoint, M, of W X. To get from P to M, we must go up 4 units and to the left 3 units (since W X = ), so P M = = 5 or P M = 5. Since the semi-circle with diameter W X has diameter, it has radius 1, so AM = 1. So we have AM = 1 and MP = 5.

8 007 Cayley Contest Solutions Page 8 W M A X P Therefore, the maximum possible length of AP is = 6, when A, M, P lie on a straight line. Answer: (E)

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