The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in Grade 11 or higher.
Problem E What s Your Angle? A E C B F A is the centre of a circle which passes through B. B is the centre of a circle which passes through A. CAE and CBF are straight line segments. If F = 78, determine the measure of C.
Problem E and Solution What s Your Angle? Problem A is the centre of a circle which passes through B. B is the centre of a circle which passes through A. CAE and CBF are straight line segments. If F = 78, determine the measure of C. C y A x y B x E 78 F Solution Construct AB and BE. A, E and F are on the circumference of the circle with centre B. Therefore, BA = BE = BF. In BEF, BE = BF. Then BEF is isosceles and BEF = F = 78. Let x = BEA and y = ABC. In BAE, BA = BE and the triangle is isosceles. Therefore, BAE = BEA = x. Since B and C are on the circle with centre A, AC = AB and ABC is isosceles. Therefore, C = ABC = y. EAB is an exterior angle to ABC. By the exterior angle theorem for triangles, EAB = C + ABC. But EAB = x and C + ABC = y + y = 2y. Therefore, x = 2y. In CEF, But x = 2y, Therefore, C = 8. C + E + F = 180 C + CEB + BEF + 78 = 180 y + x + 78 + 78 = 180 y + 2y + 156 = 180 3y = 24 y = 8
Problem E Love is in the Air In Mr. Vee s Physics class, each student gave every other student a valentine. Mr. Vee did not give or receive any valentines. The girls received a total of 540 valentines and the boys received a total of 390 valentines. How many students are in Mr. Vee s class? How many girls are in the class? How many boys are in the class?
Problem E and Solution Love is in the Air Problem In Mr. Vee s Physics class, each student gave every other student a valentine. Mr. Vee did not give or receive any valentines. The girls received a total of 540 valentines and the boys received a total of 390 valentines. How many students are in Mr. Vee s class? How many girls are in the class? How many boys are in the class? Solution Let n represent the number of students in Mr. Vee s class. Each student sends n 1 valentines and the total number of valentines sent is 540 + 390 or 930 valentines. So the number of students, n, times the number of valentines sent, n 1, equals the total number of valentines. n(n 1) = 930 n 2 n 930 = 0 (n 31)(n + 30) = 0 n = 31 or n = 30 but n > 0, n = 31 Therefore, there are 31 students in the class and each student sent 30 valentines and received 30 valentines. Since the girls received a total of 540 valentines and each student received 30 valentines, there are 540 30 or 18 girls in the class. Since the boys received a total of 390 valentines and each student received 30 valentines, there are 390 30 or 13 boys in the class. Mr. Vee s class has 31 students, 18 girls and 13 boys.
Problem E How Many Children Are There Anyway? A man died leaving some money in his estate. All of this money was to be divided among his children in the following manner: $x to the first born plus 1 16 $2x to the second born plus 1 16 $3x to the third born plus 1 16 of what remains, then of what then remains, then of what then remains, and so on. When the distribution of the money was complete, each child received the same amount and no money was left over. Determine the number of children.
Problem E and Solution How Many Children Are There Anyway? Problem A man died leaving some money in his estate. All of this money was to be divided among his children in the following manner: $x to the first born plus 1 of what remains, $2x to the 16 second born plus 1 1 of what then remains, $3x to the third born plus of what then remains, 16 16 and so on. When the distribution of the money was complete, each child received the same amount and no money was left over. Determine the number of children. Solution Let T be the total value of the estate. T > 0 since there is money in the estate. Let y be the amount of money given to each child. y > 0 since each child receives money. Then T is the number of children. y The first child receives x to begin with. There would be (T x) left at this point. The first child then receives 1 of the remaining amount (T x). Therefore, the first child receives 16 y = x + 1 (T x). 16 The second child receives 2x to begin with. There would now be (T y 2x) left in the estate. This represents the original amount minus the first child s full share minus the amount received so far by the second child. The second child then receives 1 of the remaining amount 16 (T y 2x). Therefore, the second child receives y = 2x + 1 (T y 2x). 16 But each child receives the same amount. Multiply each side of the equation by 16. x + 1 16 (T x) = 2x + 1 (T y 2x) 16 x + 1 16 T 1 16 x = 2x + 1 16 T 1 16 y 1 16 (2x) x 1 16 x = 2x 1 16 y 2 16 x Therefore each child receives $15x. 16x x = 32x y 2x 15x = 30x y y = 15x (The solution continues on the next page.)
Substituting 15x for y into the equation y = x + 1 (T x) we obtain 16 Therefore the total value of the estate is 225x. x + 1 (T x) = 15x 16 1 (T x) = 14x 16 T x = 224x T = 225x We can now determine the number of children T y = 225x 15x = 15. Therefore the father had 15 children. Notice, we did not need to know T, the total value of the estate. It turns out that once we know x, we can determine each child s share, 15x, and the total value of the estate, T = 225x.
Problem E Find the Number Five different numbers, a, b, c, d, and e, are written in order smallest to largest such that a < b < c < d < e. When the numbers are added in pairs the following sums are obtained: 726, 756, 770, 781, 795, 816, 825, 830, 860 and 885. Determine the sum of the five numbers and determine the value of the smallest number, a.
Problem E and Solution Find the Number Problem Five different numbers, a, b, c, d, and e, are written in order smallest to largest such that a < b < c < d < e. When the numbers are added in pairs the following sums are obtained: 726, 756, 770, 781, 795, 816, 825, 830, 860 and 885. Determine the sum of the five numbers and determine the value of the smallest number, a. Solution We know the sum of each pair so we can determine the sum of all of the sums. (a+b)+(a+c)+(a+d)+(a+e)+(b+c)+(b+d)+(b+e)+(c+d)+(c+e)+(d+e) = 4a + 4b + 4c + 4d + 4e = 726 + 756 + 770 + 781 + 795 + 816 + 825 + 830 + 860 + 885 = 8 044 Therefore 4a + 4b + 4c + 4d + 4e = 8 044. Dividing each side of the equation by 4, we obtain a + b + c + d + e = 2 011. That is, the sum of the five numbers is 2 011. The smallest sum is created by adding the two smallest numbers. Therefore, a + b = 726. The largest sum is created by adding the two largest numbers. Therefore, d + e = 885. If we subtract the sum of the two smallest numbers and the sum of the two largest numbers from the total sum, we will get the middle number, c. (a + b + c + d + e) (a + b) (d + e) = 2 011 726 885 c = 400 The second smallest sum is created by adding the smallest number, a, to the middle number, c. (A justification of this statement is shown on the next page.) Therefore, a + c = 756 and a = 756 c = 756 400 = 356. The sum of the five numbers is 2 011 and the smallest number is 356. The complete list is 356, 370, 400, 425 and 460. It is left as an exercise for the student to verify the correctness of this list.
Why is a + c is the second smallest sum? We know that a < b < c < d < e so the following is true: 1. Since b > a, let b = a + m, m > 0. 2. Since c > b, c = b + n = a + m + n, m, n > 0. 3. Since d > c, d = c + p = a + m + n + p, m, n, p > 0. 4. Since e > d, e = d + q = a + m + n + p + q, m, n, p, q > 0. We need to show that a + c > a + b and a + c is smaller than all other possible sums. So now let s look at the ten possible sums: 1. a + b = a + a + m = 2a + m. 2. a + c = a + a + m + n = 2a + m + n > 2a + m = a + b. So a + c > a + b. 3. a + d = a + a + m + n + p = 2a + m + n + p > 2a + m + n = a + c. So a + d > a + c. 4. a + e = a + a + m + n + p + q = 2a + m + n + p + q > 2a + m + n = a + c. So a + e > a + c. 5. b + c = a + m + a + m + n = 2a + 2m + n > 2a + m + n = a + c. So b + c > a + c. 6. b + d = a + m + a + m + n + p = 2a + 2m + n + p > 2a + m + n = a + c. So b + d > a + c. 7. b + e = a + m + a + m + n + p + q = 2a + 2m + n + p + q > 2a + m + n = a + c. So b + e > a + c. 8. c + d = a + m + n + a + m + n + p = 2a + 2m + 2n + p > 2a + m + n = a + c. So c + d > a + c. 9. c + e = a + m + n + a + m + n + p + q = 2a + 2m + 2n + p + q > 2a + m + n = a + c. So c + e > a + c. 10. d + e = a + m + n + p + a + m + n + p + q = 2a + 2m + 2n + 2p + q > 2a + m + n = a + c. So d + e > a + c. So we have shown that a + c > a + b and a + c is smaller than the eight other possible sums. Therefore, a + c is the second smallest sum. (It is also possible to use the above information to show that c + e is the second largest sum.)
Problem E A Problem For The Ages A four-digit number which is a perfect square is created by writing Jennifer s age in years followed by John s age in years. Similarly, in 31 years, their ages in the same order will again form a four-digit perfect square. How old are Jennifer and John today?
Problem Problem E and Solution A Problem For The Ages A four-digit number which is a perfect square is created by writing Jennifer s age in years followed by John s age in years. Similarly, in 31 years, their ages in the same order will again form a four-digit perfect square. How old are Jennifer and John today? Solution Both Jennifer s age and John s age must be two digit numbers. If Jennifer s age is a one-digit number, John s age would have to be a three-digit number to create the four-digit perfect square. But in 31 years, Jennifer s age would then be a two-digit number resulting in at least a five-digit number when their ages are used to form the second perfect square. A similar argument could be presented if John s age is a one-digit number. Therefore, both Jennifer and John have ages that are each two-digit numbers. Let Jennifer s present age be x and John s present age be y. Then 100x + y is the four digit number created by writing Jennifer s age followed by John s age. But 100x + y is a perfect square so let 100x + y = k 2, for some positive integer k. In 31 years, Jennifer will be (x + 31) and John will be (y + 31). The new number created by writing Jennifer s age followed by John s age is 100(x + 31) + (y + 31). This new four-digit number is also a perfect square. So 100(x + 31) + (y + 31) = m 2, for some positive integer m, m > k. This simplifies as follows: 100x + 3100 + y + 31 = m 2 100x + y + 3131 = m 2 (1) From our work above, we also have 100x + y = k 2. Substituting this into (1) we get k 2 + 3131 = m 2 or 3131 = m 2 k 2. m 2 k 2 is a difference of squares, so m 2 k 2 = (m + k)(m k) = 3131. Since m and k are positive integers, m + k is positive and m + k > m k. m k must also be positive since (m + k)(m k) = 3131. So we are looking for two positive numbers that multiply to 3131. There are two possibilities, 3131 1 or 101 31. First we will examine (m + k)(m k) = 3131 1. From this we obtain two equations in two unknowns, namely m + k = 3131 and m k = 1. Subtracting the two equations gives 2k = 3130 or k = 1565. Then k 2 = 1565 2 = 2449225. This is not a four-digit number so 3131 1 is not an admissible factorization of 3131. Next we examine (m + k)(m k) = 101 31. This leads to m + k = 101 and m k = 31. Subtracting the two equations we get 2k = 70 or k = 35. Then 100x + y = k 2 = 1225. Therefore, x = 12 and y = 25 since 1225 is the four-digit number formed by writing Jennifer s age, x, followed by John s age, y. today Jennifer is 12 and John is 25.
Problem E Trapped! A four-sided figure with one pair of parallel sides is called a trapezoid. An isosceles trapezoid is one in which the non-parallel sides are equal in length. A D x O x B C Isosceles trapezoid ABCD has parallel sides AD and BC and sides AB = DC = x. The area of the trapezoid is 510 cm 2. A circle with centre O and radius 10 cm is contained inside the trapezoid so that it is tangent to each of the four sides of the trapezoid. Determine the length of x. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent.
Problem E and Solution Trapped! Problem Isosceles trapezoid ABCD has parallel sides AD and BC and sides AB = DC = x. The area of the trapezoid is 510 cm 2. A circle with centre O and radius 10 cm is contained inside the trapezoid so that it is tangent to each of the four sides of the trapezoid. Determine the length of x. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent. Solution Draw radii OP, OQ, OR, and OS to sides AB, BC, CD, and DA, respectively. Then OP = OQ = OR = OS = 10 since they are each radii of the circle. Also, since each line drawn from the centre of the circle to a point of tangency is perpendicular to the tangent, OP A = OQB = ORD = OSA = 90. Let AP = a and DR = b. Therefore, P B = x a and RC = x b. The following diagram shows all of the given and found information. x B x-a P A a O S Q D b R x-b x C Join A to O forming two right triangles, AP O and ASO. Using the Pythagorean Theorem, AP 2 = AO 2 OP 2 and AS 2 = AO 2 OS 2. But OP = OS since they are both radii. So the two expressions are equal and AS = AP = a follows.
Using exactly the same reasoning that was used to show AP = AS = a, we can show DR = DS = b, BP = BQ = x a and CR = CQ = x b. This new information has been added to the diagram below. x x-a P A a a O S b D b R x-b x B x-a Q x-b C We can now find the area of the trapezoid. To find the area of a trapezoid, we multiply the perpendicular distance between the two parallel sides by the sum of the lengths of the two parallel sides and then divide the result by 2. Area Trapezoid ABCD = SQ (AD + BC) 2 510 = (SO + OQ) ((AS + SD) + (BQ + QC)) 2 510 = (10 + 10) ((a + b) + (x a + x b)) 2 510 = (20) (2x) 2 510 = 20x 25.5 = x Therefore, the length of x is 25.5 cm. At the end of the statement of the problem, the following fact was assumed: a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent. As an extension to this problem the solver may wish to try to prove this fact.
Problem E The Quest For Ann Sirs? Ann Sirs has already written three mathematics tests, worth 100 marks each. The average of her marks on her first two tests is calculated and then averaged with her third test mark giving an average of 86%. The average of her marks on her second and third tests is calculated and then averaged with her first test mark giving an average of 79.5%. The average of her marks on her first and third tests is calculated and then averaged with her second test mark giving an average of 81.5%. Her fourth mathematics test is tomorrow and it is also out of 100. What mark does Ann need on her test to achieve an overall average of 85% in her mathematics course?
Problem E and Solution The Quest For Ann Sirs? Problem Ann Sirs has already written three mathematics tests, worth 100 marks each. The average of her marks on her first two tests is calculated and then averaged with her third test mark giving an average of 86%. The average of her marks on her second and third tests is calculated and then averaged with her first test mark giving an average of 79.5%. The average of her marks on her first and third tests is calculated and then averaged with her second test mark giving an average of 81.5%. Her fourth mathematics test is tomorrow and it is also out of 100. What mark does Ann need on her test to achieve an overall average of 85% in her mathematics course? Solution Let a represent Ann s first mark, b represent Ann s second mark, c represent Ann s third mark and d represent Ann s fourth mark. When the average of her first and second marks is averaged with her third test mark, the new average is 86 so a+b 2 +c = 86. Multiplying by 2 gives a+b + c = 172. Multiplying by 2 again 2 2 yields a + b + 2c = 344. (1) When the average of her second and third marks is averaged with her first test mark, the new average is 79.5 so b+c 2 +a = 79.5. Multiplying by 2 gives b+c + a = 159. Multiplying by 2 again 2 2 yields b + c + 2a = 318. (2) When the average of her first and third marks is averaged with her second test mark, the new average is 81.5 so a+c 2 +b = 81.5. Multiplying by 2 gives a+c + b = 163. Multiplying by 2 again 2 2 yields a + c + 2b = 326. (3) By adding equations (1), (2) and (3) we obtain 4a + 4b + 4c = 988 which simplifies to a + b + c = 247 after dividing by 4. This means that the sum of her first three marks is 247. To obtain an average of 85% Ann needs a+b+c+d = 85. Substituting a + b + c = 247 we obtain 4 247+d = 85. Multiplying by 4 gives 247 + d = 340 or d = 93. 4 to obtain an average of 85% Ann needs 93 on tomorrow s test. It would be a straight forward process to determine Ann s first three test marks but our quest did not ask us to do this. However, for the curious, on her first test Ann got 71, on her second test Ann got 79, and on her third test Ann got 97.
Problem E How Does This Function Function? The function f(x) = x 5 2x 4 + ax 3 x 2 + bx 2 has a value of 7 when x = 1. Determine the value of the function when x = 1.
Problem E and Solution How Does This Function Function? Problem The function f(x) = x 5 2x 4 + ax 3 x 2 + bx 2 has a value of 7 when x = 1. Determine the value of the function when x = 1. Solution We know that the function has a value of 7 when x = 1. Therefore, f( 1) = 7. f( 1) = 7 ( 1) 5 2( 1) 4 + a( 1) 3 ( 1) 2 + b( 1) 2 = 7 1 2 a 1 b 2 = 7 a b = 1 a + b = 1 (1) At this point we seem to have used up the given information. Maybe we can learn more by looking at precisely what we are asked to determine. In this problem, we want the value of the function when x = 1. In other words, we want f(1). f(1) = 1 5 2(1) 4 + a(1) 3 1 2 + b(1) 2 = 1 2 + a 1 + b 2 = a + b 4 But from (1) above, a + b = 1 so f(1) = (a + b) 4 = (1) 4 = 3. Therefore, the value of the function is 3 when x = 1. We are not given enough information to find the precise values of a and b but enough information is given to solve the problem.
Problem E An Uphill Battle Lisa-Mai Winn and Will Race are training for a competition by running all the way up and down a 700 m long ski slope. They each run up the slope at different constant speeds. Coming down the slope, each runs at double her or his uphill speed. Lisa-Mai reaches the top first, and immediately starts running back down, meeting Will 70 m from the top. When Lisa-Mai reaches the bottom, how far behind is Will?
Problem Problem E and Solution An Uphill Battle Lisa-Mai Winn and Will Race are training for a competition by running all the way up and down a 700 m long ski slope. They each run up the slope at different constant speeds. Coming down the slope, each runs at double her or his uphill speed. Lisa-Mai reaches the top first, and immediately starts running back down, meeting Will 70 m from the top. When Lisa-Mai reaches the bottom, how far behind is Will? Solution Let a represent Lisa-Mai s uphill speed and 2a represent her downhill speed. Let b represent Will s uphill speed and 2b represent his downhill speed. We know that distance = speed time or time = distance speed. When Lisa and Will meet 70 m from the top of the hill, their times are the same. At this time Will has run 700 70 = 630 m uphill. Lisa s Time Up 700 m + Lisa s Time Down 70 m = Will s Time Up 630 m 700 a + 70 2a = 630 b Reducing the second fraction: 700 a + 35 a = 630 b Simplifying: 735 = 630 a b By rearranging, we obtain a b = 735 630 = 7 6. (1) Solution continues on the next page.
Let x represent the distance Will is behind Lisa at the end of the run. When Lisa reaches the bottom, she has travelled 700 m up at a speed of a and 700 m down at a speed of 2a. Will has travelled 700 m up at a speed of b and (700 x) m down at a speed of 2b. When Lisa reaches the bottom and Will is x m from the bottom, their times are the same. Multiplying by 2, 700 a + 700 2a = 700 b 1400 2a + 700 2a = 1400 2b 2100 2a 2100 a It follows that a b = 2100 2100 x and a b = 7 6 = 2100 x 2b = 2100 x b + 700 x 2b from (1) earlier. + 700 x 2b Solving for x, 2100 2100 x = 7 6 12600 = 14700 7x 7x = 2100 x = 300 Therefore Will Race is 300 m behind Lisa-Mai Winn when she completes the run.
Problem E Playing With a Full Deck R S T Q P The diagram shows the plan for a deck which is to be built on the corner of a cottage. A railing is to be constructed around the four outer edges of the deck from P to Q to R to S to T. The length of the rail from P to Q is the same as the length of the rail from S to T and the length of the rail from Q to R is the same as the length of the rail from R to S. The total length of the railing is 30 m. Determine the dimensions of the deck which will have the maximum area.
Problem E and Solution Playing With a Full Deck Problem The diagram shows the plan for a deck which is to be built on the corner of a cottage. A railing is to be constructed around the four outer edges of the deck from P to Q to R to S to T. The length of the rail from P to Q is the same as the length of the rail from S to T and the length of the rail from Q to R is the same as the length of the rail from R to S. The total length of the railing is 30 m. Determine the dimensions of the deck which will have the maximum area. R y x y W V y-x S T x Solution Q x P Label the corner of the cottage V. Draw a line segment through P V to RS intersecting at W. P W RS. This makes two rectangles P QRW and W ST V. Let x represent the length of P Q and ST. Let y represent the length of QR and RS. Since P QRW is a rectangle, RW = P Q = x and W S = RS RW = y x. Let A represent the area of the deck. The total length of fencing is P Q + QR + RS + ST = x + y + y + x = 2x + 2y = 30. By dividing by 2, x + y = 15. Rearranging to solve for y we obtain y = 15 x. (1) Area of deck = Area P QRW + Area W ST V A = QR RW + W S ST = yx + (y x)x = (15 x)x + ((15 x) x)x, substituting from (1) above = (15 x)x + (15 2x)x = 15x x 2 + 15x 2x 2 = 3x 2 + 30x = 3(x 2 10x) = 3(x 2 10x + 5 2 5 2 ), by completing the square = 3(x 5) 2 + 75 This is the equation of a parabola which opens down from a vertex of (5, 75). The maximum area is 75 m 2 when x = 5 m. When x = 5, y = 15 x = 15 5 = 10 m. The side lengths of the deck are 5 m and 10 m giving a maximum area of 75 m 2.
Problem E What are the Chances of That? Use each of the digits one to nine exactly once to form a nine digit number. Determine the probability that the number formed is divisible by 4 or ends in 8.
Problem E and Solution Problem What are the Chances of That? Use each of the digits one to nine exactly once to form a nine digit number. Determine the probability that the number formed is divisible by 4 or ends in 8. Solution In creating a 9 digit number, there are 9 choices for the first digit, 8 choices for the second digit, 7 choices for the third digit, and so on. So the number of 9 digit numbers is 9 8 7 2 1 = 9! = 362 880. If the nine digit number ends in 8, there is only choice for the unit s digit. There are then 8! ways to create the remaining 8 digit number. Thus, the number of 9 digit numbers ending in 8 is 1 8! = 40 320. For a number to be divisible by 4, the last two digits of the number must be divisible by 4. There are 24 numbers less than 100 that are divisible by 4. This group includes 2 single digit numbers, 4 and 8, which must be excluded. It also includes 4 two digit numbers ending in zero, 20, 40, 60 and 80. Zero is not one of the nine possible digits so these numbers must be excluded. And finally the list includes 2 numbers, 44 and 88, which have repeated digits. Since each digit can be used only once, these numbers must be excluded. Therefore, there are 24 2 4 2 or 16 valid two digit numbers which are divisible by 4. There are then 7! ways to create the remaining 7 digit number. So the number of 9 digit numbers divisible by 4 is 16 7! = 80 640. If we added the 1 8! and the 16 7! we would be double counting the number of nine digit numbers which end in 8 and are divisible by 4. So we must subtract off the number of numbers which are divisible by 4 and end in 8 since they have been counted twice. There are 3 two digit numbers with distinct digits, 28, 48, and 68, that end in 8 and are divisible by 4. As before, there are 7! ways to create the remaining 7 digit number. So the number of 9 digit numbers with distinct digits ending in 8 and divisible by 4 is 3 7! = 15 120. We can now compute the required probability as follows: (# ending in 8) + (# divisible by 4) (# ending in 8 and divisible by 4) (# of 9 digit numbers) = 1 8! + 16 7! 3 7! 362 880 = 40 320 + 80 640 15 120 362 880 = 105 840 362 880 = 7 24. the probability that a nine digit number is divisible by 4 or ends in 8 is 7 24.
Problem E Answer the Question Asked It is given that (x + y) 2 = 16, (y + z) 2 = 36, and (x + z) 2 = 81. If x + y + z 1, determine the number of possible values for x + y + z.
Problem E and Solution Answer the Question Asked Problem It is given that (x + y) 2 = 16, (y + z) 2 = 36, and (x + z) 2 = 81. If x + y + z 1, determine the number of possible values for x + y + z. Solution Since (x + y) 2 = 16, x + y = ±4. Since (y + z) 2 = 36, y + z = ±6. And since (x + z) 2 = 81, x + z = ±9. Now (x + y) + (y + z) + (x + z) = 2x + 2y + 2z = 2(x + y + z). This quantity is two times the value of the quantity we are looking for. The following chart summarizes all possible combinations of values for x + y, y + z, and x + z and the resulting values of 2x + 2y + 2z and x + y + z. The final column of the chart states a yes or no answer to whether the value of x + y + z is 1. 1? x + y y + z x + z 2x + 2y + 2z x + y + z (yes / no) 4 6 9 19 9.5 yes 4 6 9 1 0.5 no 4 6 9 7 3.5 yes 4 6 9 11 5.5 no 4 6 9 11 5.5 yes 4 6 9 7 3.5 no 4 6 9 1 0.5 no 4 6 9 19 9.5 no Therefore there are three possible values of x + y + z such that x + y + z 1. It should be noted that for each of the three possibilities, values for x, y, and z which produce each value can be determined.
Problem E The Quest For Ann Sirs? Ann Sirs has already written three mathematics tests, worth 100 marks each. The average of her marks on her first two tests is calculated and then averaged with her third test mark giving an average of 86%. The average of her marks on her second and third tests is calculated and then averaged with her first test mark giving an average of 79.5%. The average of her marks on her first and third tests is calculated and then averaged with her second test mark giving an average of 81.5%. Her fourth mathematics test is tomorrow and it is also out of 100. What mark does Ann need on her test to achieve an overall average of 85% in her mathematics course?
Problem E and Solution The Quest For Ann Sirs? Problem Ann Sirs has already written three mathematics tests, worth 100 marks each. The average of her marks on her first two tests is calculated and then averaged with her third test mark giving an average of 86%. The average of her marks on her second and third tests is calculated and then averaged with her first test mark giving an average of 79.5%. The average of her marks on her first and third tests is calculated and then averaged with her second test mark giving an average of 81.5%. Her fourth mathematics test is tomorrow and it is also out of 100. What mark does Ann need on her test to achieve an overall average of 85% in her mathematics course? Solution Let a represent Ann s first mark, b represent Ann s second mark, c represent Ann s third mark and d represent Ann s fourth mark. When the average of her first and second marks is averaged with her third test mark, the new average is 86 so a+b 2 +c = 86. Multiplying by 2 gives a+b + c = 172. Multiplying by 2 again 2 2 yields a + b + 2c = 344. (1) When the average of her second and third marks is averaged with her first test mark, the new average is 79.5 so b+c 2 +a = 79.5. Multiplying by 2 gives b+c + a = 159. Multiplying by 2 again 2 2 yields b + c + 2a = 318. (2) When the average of her first and third marks is averaged with her second test mark, the new average is 81.5 so a+c 2 +b = 81.5. Multiplying by 2 gives a+c + b = 163. Multiplying by 2 again 2 2 yields a + c + 2b = 326. (3) By adding equations (1), (2) and (3) we obtain 4a + 4b + 4c = 988 which simplifies to a + b + c = 247 after dividing by 4. This means that the sum of her first three marks is 247. To obtain an average of 85% Ann needs a+b+c+d = 85. Substituting a + b + c = 247 we obtain 4 247+d = 85. Multiplying by 4 gives 247 + d = 340 or d = 93. 4 to obtain an average of 85% Ann needs 93 on tomorrow s test. It would be a straight forward process to determine Ann s first three test marks but our quest did not ask us to do this. However, for the curious, on her first test Ann got 71, on her second test Ann got 79, and on her third test Ann got 97.
Problem E Palindrome Possibilities A palindrome is a word or phrase which reads the same frontwards and backwards. I prefer pi is an excellent example of a palindrome phrase. Numbers which remain the same when the digits are reversed are also considered to be palindromes. For example, 1287821 and 4554 are palindromic numbers. Hannah and Habibah have a special interest in palindromic numbers. Determine, for them, the probability that a five-digit integer greater than or equal to 10 000 is a palindrome.
Problem E and Solution Palindrome Possibilities Problem A palindrome is a word or phrase which reads the same frontwards and backwards. I prefer pi is an excellent example of a palindrome phrase. Numbers which remain the same when the digits are reversed are also considered to be palindromes. For example, 1287821 and 4554 are palindromic numbers. Hannah and Habibah have a special interest in palindromic numbers. Determine, for them, the probability that a five-digit integer greater than or equal to 10 000 is a palindrome. Solution First we must determine how many five-digit numbers greater than or equal to 10 000 there are. The largest five-digit number is 99 999 so there are 99 999 numbers from 1 to the largest five-digit number. This includes the 9 999 numbers from 1 to 9 999 which must be excluded. So the number of five-digit numbers is 99 999 9 999 = 90 000. Now we can determine how many five-digit palindromes there are. These numbers look like abcba. Once the first digit is selected, the last digit is known (and there is really no choice). Once the second digit is chosen, the fourth digit is known (and again there is really no choice). There are 9 choices for the first digit. Zero cannot be used as a first digit since the number must be greater than or equal to 10 000. For each of the 9 choices for the first digit, there are 10 choices for the second digit. Therefore, there are 9 10 = 90 choices for the first two digits. For each of the 90 choices for the first two digits, there are 10 choices for the third digit. Therefore, there are 90 10 = 900 choices for the first three digits. Once the first three digits are chosen there is exactly one choice for each of the fourth and fifth digits. The fourth digit must be the same as the second digit and the fifth digit must be the same as the first digit. This means that there are 900 1 1 = 900 five-digit palindromes greater than or equal to 10 000. The probability is calculated by dividing the number of five-digit palindromes greater than or equal to 10 000 by the number of five-digit numbers greater than or equal to 10 000. the probability that a five digit number is a palindrome = 900 90 000 = 1 100.
Problem E A Shady Space A L K B M C ABC is an equilateral triangle with sides of length 2 cm. ALB, BMC, and CKA are arcs of circles having centres C, A, and B, respectively. Determine the total area of the shaded regions in the diagram.
Problem E and Solution A Shady Space Problem A L K B D M C ABC is an equilateral triangle with sides of length 2 cm. ALB, BMC, and CKA are arcs of circles having centres C, A, and B, respectively. Determine the total area of the shaded regions in the diagram. Solution First, determine the area of equilateral ABC. Construct the altitude AD. Since the triangle is equilateral, all angles are 60. Therefore, ABD is a 30, 60, 90 triangle with sides in ratio 1 : 3 : 2. Since AB = 2, BD = 1, we get AD = 3, and area ABC = (BC)(AD) 2 = (2)( 3) 2 = 3 cm 2 Each sector ABMC, BCKA and CALB has the same radius, 2 cm, and a 60 central angle. Therefore each sector has the same area, 60 360 or one-sixth the area of a circle of radius 2 cm. area ABMC = 1 6 πr2 = 1 6 π(2)2 = 2 3 π cm2 The shaded part of each sector is equal to the area of the sector minus the area of the equilateral ABC. Total Shaded Area = 3(area of any whole sector - area of the equilateral triangle) = 3(area of sector ABMC area ABC) ( 2 = 3 3 π ) 3 = (2π 3 3) cm 2
Problem E More Power When n = 6 2011 is expressed as an integer, what are its last two digits?
Problem Problem E and Solution More Power When n = 6 2011 is expressed as an integer, what are its last two digits? Solution Start by examining the last two digits of various powers of 6. 6 1 = 06 6 2 = 36 6 3 = 216 6 4 = 1 296 6 5 = 7 776 6 6 = 46 656 6 7 = 279 936 6 8 = 1 679 616 6 9 = 10 077 696 6 10 = 60 466 176 6 11 = 362 797 056 Notice that the last two digits repeat every five powers of 6 starting with the 2 nd power of 6. The pattern continues. 6 12 ends with 36, 6 13 ends with 16, 6 14 ends with 96, 6 15 ends with 76, 6 16 ends with 56, and so on. Starting with the second power of 6, every five consecutive powers of 6 will have the last two digits 36, 16, 96, 76, and 56. We need to determine the number of complete cycles in 2011 by first subtracting 1 to allow for 06 at the beginning of the list and then dividing 2011 1 or 2010 by 5. 2010 = 402 5 So there are 402 complete cycles through the digit pattern to get to 6 2011. Since 6 2011 is the last number in the pattern it ends in 56. Therefore 6 2011 ends with the digits 56.
Problem E What s Your Angle? A E C B F A is the centre of a circle which passes through B. B is the centre of a circle which passes through A. CAE and CBF are straight line segments. If F = 78, determine the measure of C.
Problem E and Solution What s Your Angle? Problem A is the centre of a circle which passes through B. B is the centre of a circle which passes through A. CAE and CBF are straight line segments. If F = 78, determine the measure of C. C y A x y B x E 78 F Solution Construct AB and BE. A, E and F are on the circumference of the circle with centre B. Therefore, BA = BE = BF. In BEF, BE = BF. Then BEF is isosceles and BEF = F = 78. Let x = BEA and y = ABC. In BAE, BA = BE and the triangle is isosceles. Therefore, BAE = BEA = x. Since B and C are on the circle with centre A, AC = AB and ABC is isosceles. Therefore, C = ABC = y. EAB is an exterior angle to ABC. By the exterior angle theorem for triangles, EAB = C + ABC. But EAB = x and C + ABC = y + y = 2y. Therefore, x = 2y. In CEF, But x = 2y, Therefore, C = 8. C + E + F = 180 C + CEB + BEF + 78 = 180 y + x + 78 + 78 = 180 y + 2y + 156 = 180 3y = 24 y = 8
Problem E Answer the Question Asked It is given that (x + y) 2 = 16, (y + z) 2 = 36, and (x + z) 2 = 81. If x + y + z 1, determine the number of possible values for x + y + z.
Problem E and Solution Answer the Question Asked Problem It is given that (x + y) 2 = 16, (y + z) 2 = 36, and (x + z) 2 = 81. If x + y + z 1, determine the number of possible values for x + y + z. Solution Since (x + y) 2 = 16, x + y = ±4. Since (y + z) 2 = 36, y + z = ±6. And since (x + z) 2 = 81, x + z = ±9. Now (x + y) + (y + z) + (x + z) = 2x + 2y + 2z = 2(x + y + z). This quantity is two times the value of the quantity we are looking for. The following chart summarizes all possible combinations of values for x + y, y + z, and x + z and the resulting values of 2x + 2y + 2z and x + y + z. The final column of the chart states a yes or no answer to whether the value of x + y + z is 1. 1? x + y y + z x + z 2x + 2y + 2z x + y + z (yes / no) 4 6 9 19 9.5 yes 4 6 9 1 0.5 no 4 6 9 7 3.5 yes 4 6 9 11 5.5 no 4 6 9 11 5.5 yes 4 6 9 7 3.5 no 4 6 9 1 0.5 no 4 6 9 19 9.5 no Therefore there are three possible values of x + y + z such that x + y + z 1. It should be noted that for each of the three possibilities, values for x, y, and z which produce each value can be determined.
Problem E Love is in the Air In Mr. Vee s Physics class, each student gave every other student a valentine. Mr. Vee did not give or receive any valentines. The girls received a total of 540 valentines and the boys received a total of 390 valentines. How many students are in Mr. Vee s class? How many girls are in the class? How many boys are in the class?
Problem E and Solution Love is in the Air Problem In Mr. Vee s Physics class, each student gave every other student a valentine. Mr. Vee did not give or receive any valentines. The girls received a total of 540 valentines and the boys received a total of 390 valentines. How many students are in Mr. Vee s class? How many girls are in the class? How many boys are in the class? Solution Let n represent the number of students in Mr. Vee s class. Each student sends n 1 valentines and the total number of valentines sent is 540 + 390 or 930 valentines. So the number of students, n, times the number of valentines sent, n 1, equals the total number of valentines. n(n 1) = 930 n 2 n 930 = 0 (n 31)(n + 30) = 0 n = 31 or n = 30 but n > 0, n = 31 Therefore, there are 31 students in the class and each student sent 30 valentines and received 30 valentines. Since the girls received a total of 540 valentines and each student received 30 valentines, there are 540 30 or 18 girls in the class. Since the boys received a total of 390 valentines and each student received 30 valentines, there are 390 30 or 13 boys in the class. Mr. Vee s class has 31 students, 18 girls and 13 boys.
Problem E How Many Children Are There Anyway? A man died leaving some money in his estate. All of this money was to be divided among his children in the following manner: $x to the first born plus 1 16 $2x to the second born plus 1 16 $3x to the third born plus 1 16 of what remains, then of what then remains, then of what then remains, and so on. When the distribution of the money was complete, each child received the same amount and no money was left over. Determine the number of children.
Problem E and Solution How Many Children Are There Anyway? Problem A man died leaving some money in his estate. All of this money was to be divided among his children in the following manner: $x to the first born plus 1 of what remains, $2x to the 16 second born plus 1 1 of what then remains, $3x to the third born plus of what then remains, 16 16 and so on. When the distribution of the money was complete, each child received the same amount and no money was left over. Determine the number of children. Solution Let T be the total value of the estate. T > 0 since there is money in the estate. Let y be the amount of money given to each child. y > 0 since each child receives money. Then T is the number of children. y The first child receives x to begin with. There would be (T x) left at this point. The first child then receives 1 of the remaining amount (T x). Therefore, the first child receives 16 y = x + 1 (T x). 16 The second child receives 2x to begin with. There would now be (T y 2x) left in the estate. This represents the original amount minus the first child s full share minus the amount received so far by the second child. The second child then receives 1 of the remaining amount 16 (T y 2x). Therefore, the second child receives y = 2x + 1 (T y 2x). 16 But each child receives the same amount. Multiply each side of the equation by 16. x + 1 16 (T x) = 2x + 1 (T y 2x) 16 x + 1 16 T 1 16 x = 2x + 1 16 T 1 16 y 1 16 (2x) x 1 16 x = 2x 1 16 y 2 16 x Therefore each child receives $15x. 16x x = 32x y 2x 15x = 30x y y = 15x (The solution continues on the next page.)
Substituting 15x for y into the equation y = x + 1 (T x) we obtain 16 Therefore the total value of the estate is 225x. x + 1 (T x) = 15x 16 1 (T x) = 14x 16 T x = 224x T = 225x We can now determine the number of children T y = 225x 15x = 15. Therefore the father had 15 children. Notice, we did not need to know T, the total value of the estate. It turns out that once we know x, we can determine each child s share, 15x, and the total value of the estate, T = 225x.
Problem E A Problem For The Ages A four-digit number which is a perfect square is created by writing Jennifer s age in years followed by John s age in years. Similarly, in 31 years, their ages in the same order will again form a four-digit perfect square. How old are Jennifer and John today?
Problem Problem E and Solution A Problem For The Ages A four-digit number which is a perfect square is created by writing Jennifer s age in years followed by John s age in years. Similarly, in 31 years, their ages in the same order will again form a four-digit perfect square. How old are Jennifer and John today? Solution Both Jennifer s age and John s age must be two digit numbers. If Jennifer s age is a one-digit number, John s age would have to be a three-digit number to create the four-digit perfect square. But in 31 years, Jennifer s age would then be a two-digit number resulting in at least a five-digit number when their ages are used to form the second perfect square. A similar argument could be presented if John s age is a one-digit number. Therefore, both Jennifer and John have ages that are each two-digit numbers. Let Jennifer s present age be x and John s present age be y. Then 100x + y is the four digit number created by writing Jennifer s age followed by John s age. But 100x + y is a perfect square so let 100x + y = k 2, for some positive integer k. In 31 years, Jennifer will be (x + 31) and John will be (y + 31). The new number created by writing Jennifer s age followed by John s age is 100(x + 31) + (y + 31). This new four-digit number is also a perfect square. So 100(x + 31) + (y + 31) = m 2, for some positive integer m, m > k. This simplifies as follows: 100x + 3100 + y + 31 = m 2 100x + y + 3131 = m 2 (1) From our work above, we also have 100x + y = k 2. Substituting this into (1) we get k 2 + 3131 = m 2 or 3131 = m 2 k 2. m 2 k 2 is a difference of squares, so m 2 k 2 = (m + k)(m k) = 3131. Since m and k are positive integers, m + k is positive and m + k > m k. m k must also be positive since (m + k)(m k) = 3131. So we are looking for two positive numbers that multiply to 3131. There are two possibilities, 3131 1 or 101 31. First we will examine (m + k)(m k) = 3131 1. From this we obtain two equations in two unknowns, namely m + k = 3131 and m k = 1. Subtracting the two equations gives 2k = 3130 or k = 1565. Then k 2 = 1565 2 = 2449225. This is not a four-digit number so 3131 1 is not an admissible factorization of 3131. Next we examine (m + k)(m k) = 101 31. This leads to m + k = 101 and m k = 31. Subtracting the two equations we get 2k = 70 or k = 35. Then 100x + y = k 2 = 1225. Therefore, x = 12 and y = 25 since 1225 is the four-digit number formed by writing Jennifer s age, x, followed by John s age, y. today Jennifer is 12 and John is 25.
Problem E Trapped! A four-sided figure with one pair of parallel sides is called a trapezoid. An isosceles trapezoid is one in which the non-parallel sides are equal in length. A D x O x B C Isosceles trapezoid ABCD has parallel sides AD and BC and sides AB = DC = x. The area of the trapezoid is 510 cm 2. A circle with centre O and radius 10 cm is contained inside the trapezoid so that it is tangent to each of the four sides of the trapezoid. Determine the length of x. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent.
Problem E and Solution Trapped! Problem Isosceles trapezoid ABCD has parallel sides AD and BC and sides AB = DC = x. The area of the trapezoid is 510 cm 2. A circle with centre O and radius 10 cm is contained inside the trapezoid so that it is tangent to each of the four sides of the trapezoid. Determine the length of x. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent. Solution Draw radii OP, OQ, OR, and OS to sides AB, BC, CD, and DA, respectively. Then OP = OQ = OR = OS = 10 since they are each radii of the circle. Also, since each line drawn from the centre of the circle to a point of tangency is perpendicular to the tangent, OP A = OQB = ORD = OSA = 90. Let AP = a and DR = b. Therefore, P B = x a and RC = x b. The following diagram shows all of the given and found information. x B x-a P A a O S Q D b R x-b x C Join A to O forming two right triangles, AP O and ASO. Using the Pythagorean Theorem, AP 2 = AO 2 OP 2 and AS 2 = AO 2 OS 2. But OP = OS since they are both radii. So the two expressions are equal and AS = AP = a follows.
Using exactly the same reasoning that was used to show AP = AS = a, we can show DR = DS = b, BP = BQ = x a and CR = CQ = x b. This new information has been added to the diagram below. x x-a P A a a O S b D b R x-b x B x-a Q x-b C We can now find the area of the trapezoid. To find the area of a trapezoid, we multiply the perpendicular distance between the two parallel sides by the sum of the lengths of the two parallel sides and then divide the result by 2. Area Trapezoid ABCD = SQ (AD + BC) 2 510 = (SO + OQ) ((AS + SD) + (BQ + QC)) 2 510 = (10 + 10) ((a + b) + (x a + x b)) 2 510 = (20) (2x) 2 510 = 20x 25.5 = x Therefore, the length of x is 25.5 cm. At the end of the statement of the problem, the following fact was assumed: a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent. As an extension to this problem the solver may wish to try to prove this fact.
Problem E How Does This Function Function? The function f(x) = x 5 2x 4 + ax 3 x 2 + bx 2 has a value of 7 when x = 1. Determine the value of the function when x = 1.
Problem E and Solution How Does This Function Function? Problem The function f(x) = x 5 2x 4 + ax 3 x 2 + bx 2 has a value of 7 when x = 1. Determine the value of the function when x = 1. Solution We know that the function has a value of 7 when x = 1. Therefore, f( 1) = 7. f( 1) = 7 ( 1) 5 2( 1) 4 + a( 1) 3 ( 1) 2 + b( 1) 2 = 7 1 2 a 1 b 2 = 7 a b = 1 a + b = 1 (1) At this point we seem to have used up the given information. Maybe we can learn more by looking at precisely what we are asked to determine. In this problem, we want the value of the function when x = 1. In other words, we want f(1). f(1) = 1 5 2(1) 4 + a(1) 3 1 2 + b(1) 2 = 1 2 + a 1 + b 2 = a + b 4 But from (1) above, a + b = 1 so f(1) = (a + b) 4 = (1) 4 = 3. Therefore, the value of the function is 3 when x = 1. We are not given enough information to find the precise values of a and b but enough information is given to solve the problem.
Problem E An Uphill Battle Lisa-Mai Winn and Will Race are training for a competition by running all the way up and down a 700 m long ski slope. They each run up the slope at different constant speeds. Coming down the slope, each runs at double her or his uphill speed. Lisa-Mai reaches the top first, and immediately starts running back down, meeting Will 70 m from the top. When Lisa-Mai reaches the bottom, how far behind is Will?
Problem Problem E and Solution An Uphill Battle Lisa-Mai Winn and Will Race are training for a competition by running all the way up and down a 700 m long ski slope. They each run up the slope at different constant speeds. Coming down the slope, each runs at double her or his uphill speed. Lisa-Mai reaches the top first, and immediately starts running back down, meeting Will 70 m from the top. When Lisa-Mai reaches the bottom, how far behind is Will? Solution Let a represent Lisa-Mai s uphill speed and 2a represent her downhill speed. Let b represent Will s uphill speed and 2b represent his downhill speed. We know that distance = speed time or time = distance speed. When Lisa and Will meet 70 m from the top of the hill, their times are the same. At this time Will has run 700 70 = 630 m uphill. Lisa s Time Up 700 m + Lisa s Time Down 70 m = Will s Time Up 630 m 700 a + 70 2a = 630 b Reducing the second fraction: 700 a + 35 a = 630 b Simplifying: 735 = 630 a b By rearranging, we obtain a b = 735 630 = 7 6. (1) Solution continues on the next page.
Let x represent the distance Will is behind Lisa at the end of the run. When Lisa reaches the bottom, she has travelled 700 m up at a speed of a and 700 m down at a speed of 2a. Will has travelled 700 m up at a speed of b and (700 x) m down at a speed of 2b. When Lisa reaches the bottom and Will is x m from the bottom, their times are the same. Multiplying by 2, 700 a + 700 2a = 700 b 1400 2a + 700 2a = 1400 2b 2100 2a 2100 a It follows that a b = 2100 2100 x and a b = 7 6 = 2100 x 2b = 2100 x b + 700 x 2b from (1) earlier. + 700 x 2b Solving for x, 2100 2100 x = 7 6 12600 = 14700 7x 7x = 2100 x = 300 Therefore Will Race is 300 m behind Lisa-Mai Winn when she completes the run.
Problem E Cutting Corners A cube with side lengths 4 cm is to be cut in the following manner: the corner of the cube is removed by making a cut through the midpoints of three adjacent sides. The diagram shows the resulting figure after the first cut. Similar cuts are made at each of the remaining corners of the cube. Determine the increase or decrease in total surface area as a result of slicing the eight corners off the original cube.
Problem Problem E and Solution Cutting Corners A cube with side lengths 4 cm is to be cut in the following manner: the corner of the cube is removed by making a cut through the midpoints of three adjacent sides. The diagram shows the resulting figure after the first cut. Similar cuts are made at each of the remaining corners of the cube. Determine the increase or decrease in total surface area as a result of slicing the eight corners off the original cube. Solution To determine the increase or decrease in surface area, we need only look at one corner, find the surface area increase or decrease there and multiply the result by eight to account for the eight corners. At each corner, the surface area of three right angled triangles is removed and replaced by the surface area of a single triangle. Since the cut is made from the midpoints of three adjacent sides, each right triangle has base and height with length 2 cm. The resulting area of one of the right triangles is 1(2)(2) = 2 2 cm2. The hypotenuse of the right triangle is found using Pythagoras Theorem, 2 2 + 2 2 = 8 = 2 2. Since the length of the hypotenuse is the same in each of the three right triangles, the remaining triangle in the corner is equilateral with side lengths 2 2. The triangle looks like the bottom triangle shown to the right. The altitude of an equilateral triangle right bisects the base. Let h be the height of the triangle. Using Pythagoras Theorem, h 2 = (2 2) 2 ( 2) 2 = 8 2 = 6. Therefore h = 6. The area of the remaining triangle is 1 2 (2 2)( 6) = 12 = 2 3. 2 2 h 2 2 2 2 In each corner, the new surface area is the area of the equilateral triangle minus the area of the three right triangles. the remaining corner area is 2 3 3 (2) = 2 3 6. Since 2 3 < 6, this result is negative and the surface area is decreased in each corner. Therefore the surface area is reduced by 6 2 3 in each corner. Since there are eight corners, the total decrease in surface area is 8 (6 2 3) = 48 16 3. = 20.3 cm 2. Therefore, as a result of cutting off each of the corners, the total surface area decreases by (48 16 3) cm 2.
Problem E A Touchy Triangle D C A O P B Two circles with centres O and P, each with a radius of 2, are tangent to each other. A straight line is drawn through O and P meeting the circles at A and B. Two other sides of ABC are drawn such that side AC is tangent to the circle with centre P at D and side CB is tangent to the circle with centre P at B. Determine the length of BC. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent.
Problem Problem E and Solution A Touchy Triangle Two circles with centres O and P, each with a radius of 2, are tangent to each other. A straight line is drawn through O and P meeting the circles at A and B. Two other sides of ABC are drawn such that side AC is tangent to the circle with centre P at D and side CB is tangent to the circle with centre P at B. Determine the length of BC. Solution A D 2 O 2 T 2 Let T be the point of tangency of the two circles. Then AO = OT = T P = P B = 2. Since AC is tangent to the circle with centre P at D, CD P D. Since CB is tangent to the circle with centre P at B, CB P B. Since P D and P B are radii of the circle with centre P, P D = P B = 2. This information has been added to the diagram. ADP is right angled at D since AD P D. So AD 2 = AP 2 P D 2 = 6 2 2 2 = 36 4 = 32. Since AD > 0, AD = 32 = 4 2. At this point there are many ways to find the length of CB. An interesting (and possibly different) way to find CB is to use basic trigonometry in the right triangles. In AP D, tan(a) = P D CB AD and in ACB, tan(a) = AB. P D AD = CB AB 2 4 2 = CB 8 CB = 16 4 2 = 4 = 4 2 = 2 2 2 2 Therefore the length of CB is 2 2. There are many other solutions to this problem. The solver can use Pythagoras Theorem in ABC. Another solution uses similar triangles AP D and ACB. 2 P 2 C B
Problem E Playing With a Full Deck R S T Q P The diagram shows the plan for a deck which is to be built on the corner of a cottage. A railing is to be constructed around the four outer edges of the deck from P to Q to R to S to T. The length of the rail from P to Q is the same as the length of the rail from S to T and the length of the rail from Q to R is the same as the length of the rail from R to S. The total length of the railing is 30 m. Determine the dimensions of the deck which will have the maximum area.
Problem E and Solution Playing With a Full Deck Problem The diagram shows the plan for a deck which is to be built on the corner of a cottage. A railing is to be constructed around the four outer edges of the deck from P to Q to R to S to T. The length of the rail from P to Q is the same as the length of the rail from S to T and the length of the rail from Q to R is the same as the length of the rail from R to S. The total length of the railing is 30 m. Determine the dimensions of the deck which will have the maximum area. R y x y W V y-x S T x Solution Q x P Label the corner of the cottage V. Draw a line segment through P V to RS intersecting at W. P W RS. This makes two rectangles P QRW and W ST V. Let x represent the length of P Q and ST. Let y represent the length of QR and RS. Since P QRW is a rectangle, RW = P Q = x and W S = RS RW = y x. Let A represent the area of the deck. The total length of fencing is P Q + QR + RS + ST = x + y + y + x = 2x + 2y = 30. By dividing by 2, x + y = 15. Rearranging to solve for y we obtain y = 15 x. (1) Area of deck = Area P QRW + Area W ST V A = QR RW + W S ST = yx + (y x)x = (15 x)x + ((15 x) x)x, substituting from (1) above = (15 x)x + (15 2x)x = 15x x 2 + 15x 2x 2 = 3x 2 + 30x = 3(x 2 10x) = 3(x 2 10x + 5 2 5 2 ), by completing the square = 3(x 5) 2 + 75 This is the equation of a parabola which opens down from a vertex of (5, 75). The maximum area is 75 m 2 when x = 5 m. When x = 5, y = 15 x = 15 5 = 10 m. The side lengths of the deck are 5 m and 10 m giving a maximum area of 75 m 2.
Problem E A Shady Space A L K B M C ABC is an equilateral triangle with sides of length 2 cm. ALB, BMC, and CKA are arcs of circles having centres C, A, and B, respectively. Determine the total area of the shaded regions in the diagram.
Problem E and Solution A Shady Space Problem A L K B D M C ABC is an equilateral triangle with sides of length 2 cm. ALB, BMC, and CKA are arcs of circles having centres C, A, and B, respectively. Determine the total area of the shaded regions in the diagram. Solution First, determine the area of equilateral ABC. Construct the altitude AD. Since the triangle is equilateral, all angles are 60. Therefore, ABD is a 30, 60, 90 triangle with sides in ratio 1 : 3 : 2. Since AB = 2, BD = 1, we get AD = 3, and area ABC = (BC)(AD) 2 = (2)( 3) 2 = 3 cm 2 Each sector ABMC, BCKA and CALB has the same radius, 2 cm, and a 60 central angle. Therefore each sector has the same area, 60 360 or one-sixth the area of a circle of radius 2 cm. area ABMC = 1 6 πr2 = 1 6 π(2)2 = 2 3 π cm2 The shaded part of each sector is equal to the area of the sector minus the area of the equilateral ABC. Total Shaded Area = 3(area of any whole sector - area of the equilateral triangle) = 3(area of sector ABMC area ABC) ( 2 = 3 3 π ) 3 = (2π 3 3) cm 2
Problem E Trapped! A four-sided figure with one pair of parallel sides is called a trapezoid. An isosceles trapezoid is one in which the non-parallel sides are equal in length. A D x O x B C Isosceles trapezoid ABCD has parallel sides AD and BC and sides AB = DC = x. The area of the trapezoid is 510 cm 2. A circle with centre O and radius 10 cm is contained inside the trapezoid so that it is tangent to each of the four sides of the trapezoid. Determine the length of x. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent.
Problem E and Solution Trapped! Problem Isosceles trapezoid ABCD has parallel sides AD and BC and sides AB = DC = x. The area of the trapezoid is 510 cm 2. A circle with centre O and radius 10 cm is contained inside the trapezoid so that it is tangent to each of the four sides of the trapezoid. Determine the length of x. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent. Solution Draw radii OP, OQ, OR, and OS to sides AB, BC, CD, and DA, respectively. Then OP = OQ = OR = OS = 10 since they are each radii of the circle. Also, since each line drawn from the centre of the circle to a point of tangency is perpendicular to the tangent, OP A = OQB = ORD = OSA = 90. Let AP = a and DR = b. Therefore, P B = x a and RC = x b. The following diagram shows all of the given and found information. x B x-a P A a O S Q D b R x-b x C Join A to O forming two right triangles, AP O and ASO. Using the Pythagorean Theorem, AP 2 = AO 2 OP 2 and AS 2 = AO 2 OS 2. But OP = OS since they are both radii. So the two expressions are equal and AS = AP = a follows.
Using exactly the same reasoning that was used to show AP = AS = a, we can show DR = DS = b, BP = BQ = x a and CR = CQ = x b. This new information has been added to the diagram below. x x-a P A a a O S b D b R x-b x B x-a Q x-b C We can now find the area of the trapezoid. To find the area of a trapezoid, we multiply the perpendicular distance between the two parallel sides by the sum of the lengths of the two parallel sides and then divide the result by 2. Area Trapezoid ABCD = SQ (AD + BC) 2 510 = (SO + OQ) ((AS + SD) + (BQ + QC)) 2 510 = (10 + 10) ((a + b) + (x a + x b)) 2 510 = (20) (2x) 2 510 = 20x 25.5 = x Therefore, the length of x is 25.5 cm. At the end of the statement of the problem, the following fact was assumed: a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent. As an extension to this problem the solver may wish to try to prove this fact.
Problem E An Uphill Battle Lisa-Mai Winn and Will Race are training for a competition by running all the way up and down a 700 m long ski slope. They each run up the slope at different constant speeds. Coming down the slope, each runs at double her or his uphill speed. Lisa-Mai reaches the top first, and immediately starts running back down, meeting Will 70 m from the top. When Lisa-Mai reaches the bottom, how far behind is Will?
Problem Problem E and Solution An Uphill Battle Lisa-Mai Winn and Will Race are training for a competition by running all the way up and down a 700 m long ski slope. They each run up the slope at different constant speeds. Coming down the slope, each runs at double her or his uphill speed. Lisa-Mai reaches the top first, and immediately starts running back down, meeting Will 70 m from the top. When Lisa-Mai reaches the bottom, how far behind is Will? Solution Let a represent Lisa-Mai s uphill speed and 2a represent her downhill speed. Let b represent Will s uphill speed and 2b represent his downhill speed. We know that distance = speed time or time = distance speed. When Lisa and Will meet 70 m from the top of the hill, their times are the same. At this time Will has run 700 70 = 630 m uphill. Lisa s Time Up 700 m + Lisa s Time Down 70 m = Will s Time Up 630 m 700 a + 70 2a = 630 b Reducing the second fraction: 700 a + 35 a = 630 b Simplifying: 735 = 630 a b By rearranging, we obtain a b = 735 630 = 7 6. (1) Solution continues on the next page.
Let x represent the distance Will is behind Lisa at the end of the run. When Lisa reaches the bottom, she has travelled 700 m up at a speed of a and 700 m down at a speed of 2a. Will has travelled 700 m up at a speed of b and (700 x) m down at a speed of 2b. When Lisa reaches the bottom and Will is x m from the bottom, their times are the same. Multiplying by 2, 700 a + 700 2a = 700 b 1400 2a + 700 2a = 1400 2b 2100 2a 2100 a It follows that a b = 2100 2100 x and a b = 7 6 = 2100 x 2b = 2100 x b + 700 x 2b from (1) earlier. + 700 x 2b Solving for x, 2100 2100 x = 7 6 12600 = 14700 7x 7x = 2100 x = 300 Therefore Will Race is 300 m behind Lisa-Mai Winn when she completes the run.
Problem E Cutting Corners A cube with side lengths 4 cm is to be cut in the following manner: the corner of the cube is removed by making a cut through the midpoints of three adjacent sides. The diagram shows the resulting figure after the first cut. Similar cuts are made at each of the remaining corners of the cube. Determine the increase or decrease in total surface area as a result of slicing the eight corners off the original cube.
Problem Problem E and Solution Cutting Corners A cube with side lengths 4 cm is to be cut in the following manner: the corner of the cube is removed by making a cut through the midpoints of three adjacent sides. The diagram shows the resulting figure after the first cut. Similar cuts are made at each of the remaining corners of the cube. Determine the increase or decrease in total surface area as a result of slicing the eight corners off the original cube. Solution To determine the increase or decrease in surface area, we need only look at one corner, find the surface area increase or decrease there and multiply the result by eight to account for the eight corners. At each corner, the surface area of three right angled triangles is removed and replaced by the surface area of a single triangle. Since the cut is made from the midpoints of three adjacent sides, each right triangle has base and height with length 2 cm. The resulting area of one of the right triangles is 1(2)(2) = 2 2 cm2. The hypotenuse of the right triangle is found using Pythagoras Theorem, 2 2 + 2 2 = 8 = 2 2. Since the length of the hypotenuse is the same in each of the three right triangles, the remaining triangle in the corner is equilateral with side lengths 2 2. The triangle looks like the bottom triangle shown to the right. The altitude of an equilateral triangle right bisects the base. Let h be the height of the triangle. Using Pythagoras Theorem, h 2 = (2 2) 2 ( 2) 2 = 8 2 = 6. Therefore h = 6. The area of the remaining triangle is 1 2 (2 2)( 6) = 12 = 2 3. 2 2 h 2 2 2 2 In each corner, the new surface area is the area of the equilateral triangle minus the area of the three right triangles. the remaining corner area is 2 3 3 (2) = 2 3 6. Since 2 3 < 6, this result is negative and the surface area is decreased in each corner. Therefore the surface area is reduced by 6 2 3 in each corner. Since there are eight corners, the total decrease in surface area is 8 (6 2 3) = 48 16 3. = 20.3 cm 2. Therefore, as a result of cutting off each of the corners, the total surface area decreases by (48 16 3) cm 2.
Problem E A Touchy Triangle D C A O P B Two circles with centres O and P, each with a radius of 2, are tangent to each other. A straight line is drawn through O and P meeting the circles at A and B. Two other sides of ABC are drawn such that side AC is tangent to the circle with centre P at D and side CB is tangent to the circle with centre P at B. Determine the length of BC. For purposes of this problem accept the fact that a line drawn from the centre of a circle to a point of tangency is perpendicular to the tangent.
Problem Problem E and Solution A Touchy Triangle Two circles with centres O and P, each with a radius of 2, are tangent to each other. A straight line is drawn through O and P meeting the circles at A and B. Two other sides of ABC are drawn such that side AC is tangent to the circle with centre P at D and side CB is tangent to the circle with centre P at B. Determine the length of BC. Solution A D 2 O 2 T 2 Let T be the point of tangency of the two circles. Then AO = OT = T P = P B = 2. Since AC is tangent to the circle with centre P at D, CD P D. Since CB is tangent to the circle with centre P at B, CB P B. Since P D and P B are radii of the circle with centre P, P D = P B = 2. This information has been added to the diagram. ADP is right angled at D since AD P D. So AD 2 = AP 2 P D 2 = 6 2 2 2 = 36 4 = 32. Since AD > 0, AD = 32 = 4 2. At this point there are many ways to find the length of CB. An interesting (and possibly different) way to find CB is to use basic trigonometry in the right triangles. In AP D, tan(a) = P D CB AD and in ACB, tan(a) = AB. P D AD = CB AB 2 4 2 = CB 8 CB = 16 4 2 = 4 = 4 2 = 2 2 2 2 Therefore the length of CB is 2 2. There are many other solutions to this problem. The solver can use Pythagoras Theorem in ABC. Another solution uses similar triangles AP D and ACB. 2 P 2 C B
Problem E Playing With a Full Deck R S T Q P The diagram shows the plan for a deck which is to be built on the corner of a cottage. A railing is to be constructed around the four outer edges of the deck from P to Q to R to S to T. The length of the rail from P to Q is the same as the length of the rail from S to T and the length of the rail from Q to R is the same as the length of the rail from R to S. The total length of the railing is 30 m. Determine the dimensions of the deck which will have the maximum area.
Problem E and Solution Playing With a Full Deck Problem The diagram shows the plan for a deck which is to be built on the corner of a cottage. A railing is to be constructed around the four outer edges of the deck from P to Q to R to S to T. The length of the rail from P to Q is the same as the length of the rail from S to T and the length of the rail from Q to R is the same as the length of the rail from R to S. The total length of the railing is 30 m. Determine the dimensions of the deck which will have the maximum area. R y x y W V y-x S T x Solution Q x P Label the corner of the cottage V. Draw a line segment through P V to RS intersecting at W. P W RS. This makes two rectangles P QRW and W ST V. Let x represent the length of P Q and ST. Let y represent the length of QR and RS. Since P QRW is a rectangle, RW = P Q = x and W S = RS RW = y x. Let A represent the area of the deck. The total length of fencing is P Q + QR + RS + ST = x + y + y + x = 2x + 2y = 30. By dividing by 2, x + y = 15. Rearranging to solve for y we obtain y = 15 x. (1) Area of deck = Area P QRW + Area W ST V A = QR RW + W S ST = yx + (y x)x = (15 x)x + ((15 x) x)x, substituting from (1) above = (15 x)x + (15 2x)x = 15x x 2 + 15x 2x 2 = 3x 2 + 30x = 3(x 2 10x) = 3(x 2 10x + 5 2 5 2 ), by completing the square = 3(x 5) 2 + 75 This is the equation of a parabola which opens down from a vertex of (5, 75). The maximum area is 75 m 2 when x = 5 m. When x = 5, y = 15 x = 15 5 = 10 m. The side lengths of the deck are 5 m and 10 m giving a maximum area of 75 m 2.
Problem E Endless Possibilities? The digits 1, 2, 3, 4, and 5 are each used once to create a five digit number abcde which satisfies the following conditions: (i) the three digit number abc is odd; (ii) the three digit number bcd is divisible by 5; and (iii) the three digit number cde is divisible by 3. Determine every five digit number formed using each of the digits 1 to 5 exactly once that satisfies all three conditions.