The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in

Transcription

1 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 9 or higher.

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3 Problem D Card Logic Four playing cards are placed in a row, from left to right. One card is a club ( ), one card is a diamond ( ), one card is a heart ( ), and one card is a spade ( ), not necessarily in that order. The number on each card is different. Using the following clues, determine the exact order of the cards, from left to right, including the suit and number. 1. The club and the spade are immediately beside each other, in that order, left to right. 2. The diamond is not the first (leftmost) card, and the heart is not the fourth (rightmost) card. 3. The 7 is somewhere to the left of the heart and the 3 is somewhere to the right of the heart. 4. The two cards whose values are still unknown add to nine. 5. From left to right, the cards are arranged from largest to smallest.

4 Problem D and Solution Card Logic Problem Four playing cards are placed in a row, from left to right. One card is a club ( ), one card is a diamond ( ), one card is a heart ( ), and one card is a spade ( ), not necessarily in that order. The number on each card is different. Using the following clues, determine the exact order of the cards, from left to right, including the suit and number. 1. The club and the spade are immediately beside each other, in that order, left to right. 2. The diamond is not the first (leftmost) card, and the heart is not the fourth (rightmost) card. 3. The 7 is somewhere to the left of the heart and the 3 is somewhere to the right of the heart. 4. The two cards whose values are still unknown add to nine. 5. From left to right, the cards are arranged from largest to smallest. Solution Let C be the club, D be the diamond, H be the heart and S be the spade. Using the first clue, we can determine that there are only three possible placements for C and S, namely {C, S,, }, {, C, S, } or {,, C, S}. We will use the second clue with each of the three possible placements. From {C, S,, } we obtain: {C, S, H, D} since H cannot go last. From {, C, S, } we obtain {H, C, S, D} since D cannot be first and H cannot be last. From {,, C, S} we obtain {H, D, C, S} since D cannot go first. The third clue says that 7 is somewhere to the left of the heart. This means that the heart cannot be furthest to the left. Therefore, we can rule out two of the possibilities that we just found leaving {C, S, H, D} as the only possible ordering of the suits. Using the third clue further, we know that a 7 is somewhere to the left of the heart and a 3 is somewhere to the right of the heart. We can conclude from this that the rightmost card is the 3 of diamonds. At this point we know the following: it is either {7C, S, H, 3D} or {C, 7S, H, 3D}. We know the 7 is in either the first or second spot, the 3 of diamonds is in the fourth spot. We still do not know the other two numbers. Using the fourth clue we know that the remaining pair of numbers add to 9. The numbers 7 and 3 are already used. Since numbers are arranged from highest to lowest and numbers are used only once, we can rule out the following pairs: 7 and 2, 6 and 3. That leaves only two possible pairs: 8 and 1, and 5 and 4. Using the fifth clue we can place the numbers to see if the solution is valid. Trying 8 and 1 in our possible solutions we obtain {7C, 8S, 1H, 3D} or {8C, 7S, 1H, 3D}. Neither of these are valid since the card values would not be arranged largest to smallest. Trying 5 and 4 in our possible solutions we obtain {7C, 5S, 4H, 3D} or {5C, 7S, 4H, 3D}. Only the first solution satisfies the condition that the numbers are arranged largest to smallest. the cards are the 7 of clubs (7 ), the five of spades (5 ), the four of hearts (4 ) and the three of diamonds (3 ).

5 Problem D An Average Bowl In his latest game, Mark Striker bowled 199 and raised his average from 177 to 178. Mark would like to raise his average to 180 after bowling his next game. What would Mark need to bowl on his next game to accomplish his goal?

6 Problem Problem D and Solutions An Average Bowl In his latest game, Mark Striker bowled 199 and raised his average from 177 to 178. Mark would like to raise his average to 180 after bowling his next game. What would Mark need to bowl on his next game to accomplish his goal? Solution 1 Let n be the number of games bowled to achieve his 177 average. His total points scored in n games is his average times n. Therefore, Mark has 177n total points in n games. To compute his average after bowling the 199 game, we take his new total points and divide by n + 1, the new number of games. Average = Total Points Games Played 178 = 177n n (n + 1) = 177n n = 177n n = 21 Prior to bowling the 199 game, Mark had bowled 21 games. So after bowling the 199 game, Mark has bowled 22 games. Mark wants to have a 180 average after bowling his 23rd game. The difference between his total points after 23 games with a 180 average and his total points after bowling 22 games with a 178 average must be his score on the 23rd game. Score on 23rd Game = = = 224 Therefore Mr. Striker must bowl 224 on his next game to raise his average from 178 to 180. Solution 2 offers a very different solution to the problem.

7 Problem In his latest game, Mark Striker bowled 199 and raised his average from 177 to 178. Mark would like to raise his average to 180 after bowling his next game. What would Mark need to bowl on his next game to accomplish his goal? Solution 2 Mark s score of 199 is = 22 points above his previous average. Mark raised his average 1 point. Therefore, his latest game with the 199 score must have been his 22nd game. To raise his average 2 points in his 23rd game he must bowl 2 23 = 46 points above his 178 average. He must bowl = 224 in his next game. Mark must bowl 224 in his next game to move his average from 178 to 180. We can verify our results: Average on 21 games is 177. Average on 22 games = = = 178 Average on 23 games = = = 180

8 Problem D What s On Your Plate? A Mathville license plate consists of exactly four digits chosen from the digits 0 to 9. What is the probability that the sum of the digits on any of these license plates is 34 or higher? MATHVILLE Yours to Uncover

9 Problem Problem D and Solution What s On Your Plate? A Mathville license plate consists of exactly four digits chosen from the digits 0 to 9. What is the probability that the sum of the digits on any of these license plates is 34 or higher? Solution First we must determine the total number of possible 4 digit number sequences. There are 10 choices for the first digit. For each of these choices there are 10 choices for the second digit. Therefore, there are = 100 choices for the first two digits. For each of these possibilities, there are 10 choices for the third digit. Therefore, there are = 1000 possibilities for the first three digits. And finally, for each of these 1000 choices for the first three digits there are 10 choices for the fourth digit. Therefore there are = possibilities for license plates with four digits in Mathville. Now we must determine the number of license plates with a digit sum of 34 or higher. We will consider cases. 1. The license contains four nines. If the digits are all nines, the sum is 36 which is acceptable. There is only one way to use all nines for digits. 2. The license contains three nines and one other digit. If three nines are used the sum is 27. In order to get 34 or higher the fourth digit must be 7 or 8. There are two choices for the fourth digit. Once the digit is chosen there are four places to put the digit. Once the digit is placed the remaining spots must be nines. Therefore there are 2 4 = 8 license plates containing three nines. (It is possible to list them: 7999, 8999, 9799, 9899, 9979, 9989, 9997, 9998.) 3. The license contains two nines. If two nines are used the sum is 18. In order to get to 34 or higher, we need a sum of = 16 or higher from the two remaining digits. The only way to do this, since we cannot use more nines, is to use two eights. The two eights can be placed in six ways and then the nines must go in the remaining spots. Therefore there are six license plates containing two nines. (It is possible to list them: 8899, 8989, 8998, 9889, 9898, 9988.) 4. The license contains one nine. If one nine is used the sum is 9. In order to get to 34 or higher, we need a sum of 34 9 = 25 or higher from the remaining three digits. But this sum would be made from three digits chosen from the digits 0 to 8. The maximum possible sum would be 24 if three eights were used. We need 25 or higher. Therefore no license plate containing only one 9 will produce a sum of 34 or higher. It should be noted that a license with no nines would not produce a sum of 34 or higher either. Therefore, the number of license plates with a digit sum of 34 or higher is = 15. To calculate the probability we divide the number of licenses with a digit sum of 34 or more by the number of possible license plates. The probability of getting a license plate in Mathville with a digit sum of 34 or 15 higher is = Another way of looking at this result is out of every cars with Mathville plates you could expect, on average, to find 3 with a digit sum of 34 or higher.

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11 Problem D Several Areas of Interest The area of ACD is twice the area of square BCDE. Square BCDE has sides of length 12 cm. AD intersects BE at F. Determine the area of quadrilateral BCDF.

12 Problem D and Solution Several Areas of Interest Problem The area of ACD is twice the area of square BCDE. Square BCDE has sides of length 12 cm. AD intersects BE at F. Determine the area of quadrilateral BCDF. Solution The area of square BCDE = = 144 cm 2. The area of ACD equals twice the area of square BCDE. Therefore area ACD = 288 cm 2. The area of a triangle is calculated using the formula base height 2. It follows that: Area ACD = (CD) (AC) = 12 AC = 6 AC 48 cm = AC But AC = AB + BC so 48 = AB + 12 and it follows that AB = 36 cm. From this point, we will present three different approaches to obtaining the required area. Method 1 Let the area of ABF be p, quadrilateral BCDF be q and DEF be r. Let F E = x. Since BE = 12, BF = 12 x. Area ACD = 288 = p + q (1) and area square BCDE = 144 = r + q (2). Subtracting (2) from (1), p r = 144 and p = r follows. Area ABF = p = r = (AB)(BF ) 2 = 36(12 x) 2 = 18(12 x) r = 18(12 x) r = x 144 r = 72 18x (3) Area DEF = r = (DE)(EF ) 2 = 12x 2 = 6x. r = 6x (4) Using (3) and (4), since r = r, 6x = 72 18x. Solving, x = 3 and r = 18 follow. The area of quadrilateral BCDF = area of square BCDE area DEF = 144 r = = 126 cm 2 Therefore the area of quadrilateral BCDF is 126 cm 2. See alternative methods on the next page.

13 Method 2 Let x represent the length of BF. In ABF and ACD, A is common and ABF = ACD = 90. It follows that ABF is similar to ACD and AB = BF. Therefore AC CD and x = = x 12 Since BCDE is a square, BE CD. Then quadrilateral BCDF is a trapezoid. Area Trapezoid BCDF = (BC)(BF + CD) 2 = 12(9 + 12) 2 = 6(21) = 126 cm 2 Therefore the area of quadrilateral BCDF is 126 cm 2. Method 3 Position the diagram so that C is at the origin, A and B are on the positive y-axis and D is on the positive x-axis. C has coordinates (0,0), B has coordinates (0,12),A has coordinates (0,48), and D has coordinates (12, 0). Find the equation of the line containing AD. The slope is = 4 and the line crosses the y-axis at A so the y-intercept is 48. Therefore the equation is y = 4x The line containing BE is horizontal crossing the y-axis at B. The equation of the line containing BE is y = 12. F is the intersection of y = 4x + 48 and y = 12. Since y = y at the point of intersection, 4x + 48 = 12 and x = 9 follows. Therefore F has coordinates (9,12) and BF = 9 cm. Since BCDE is a square, BE CD. Then quadrilateral BCDF is a trapezoid. Area Trapezoid BCDF = (BC)(BF + CD) 2 = 12(9 + 12) 2 = 6(21) = 126 cm 2 Therefore the area of quadrilateral BCDF is 126 cm 2.

14 Problem D What s Your Angle Anyway II? In ABD, C is on AB such that CA = CB = CD and BCD = z. Determine the measure of ADB. If you need a hint, consider solving this week s Grade 7 and 8 Problem of the Week What s Your Angle Anyway I?. If you want an extension to this problem, consider solving this week s Grade 11 and 12 What s Your Angle Anyway III?.

15 Problem D and Solution What s Your Angle Anyway II? Problem In ABD, C is on AB such that CA = CB = CD and BCD = z. Determine the measure of ADB. Solution Since ACB is a straight line, ACD + DCB = 180 but BCD = z so ACD = 180 z. In ACD, since CA = CD, ACD is isosceles and CAD = CDA = x. The angles in a triangle sum to 180 so in ACD CAD + CDA + ACD = 180 x + x z = 180 2x = z x = z 2 Similarly, in BCD, since CB = CD, CBD is isosceles and CBD = CDB = y. The angles in a triangle sum to 180 so in CBD CBD + CDB + BCD = 180 y + y + z = 180 2y = 180 z y = 180 z 2 Then ADB = CDA + CDB = x + y = z z 2 = = 90. the measure of ADB is 90. See the next page for an interesting idea.

16 It turns out that it is not necessary to find expressions for x and y in terms of z in the problem. Here is a second solution to the problem. In CAD, since CA = CD, CAD is isosceles and CAD = CDA = x. In CBD, since CB = CD, CBD is isosceles and CBD = CDB = y. The angles in a triangle sum to 180 so in ABD BAD + ADB + ABD = 180 x + (x + y ) + y = 180 2x + 2y = 180 x + y = 90 But ADB = ADC + CDB = x + y = 90. the measure of ADC is 90.

17 Problem D A Shady Region The shaded region on the diagram is bounded by the lines whose equations are 5x + 2y = 30, x + 2y = 22, x = 0, and y = 0. Determine the area of the shaded region.

18 Problem D and Solution A Shady Region Problem The shaded region on the diagram is bounded by the lines whose equations are 5x + 2y = 30, x + 2y = 22, x = 0, and y = 0. Determine the area of the shaded region. Solution On the diagram, l 1 represents the line 5x + 2y = 30 that crosses the x-axis at point R. l 2 represents the line x + 2y = 22 which crosses the y-axis at point Q. Let P represent the point of intersection of l 1 and l 2. Then h is the horizontal distance from the y-axis to P and k is the vertical distance from the x-axis to P. Let O represent the origin. To find the x-intercept of l 1 let y = 0 in 5x + 2y = 30. Therefore the x-intercept is 6 and the coordinates of R are (6, 0). To find the y-intercept of l 2 let x = 0 in x + 2y = 22. Therefore the y-intercept is 11 and the coordinates of Q are (0, 11). To find the intersection of l 1 and l 2, we can use elimination. l 1 5x + 2y = 30 l 2 x + 2y = 22 Subtracting, we obtain 4x = 8 x = 2 Substituting x = 2 in l 1, y = 30 and y = 10. The coordinates of P, the point of intersection, are (2, 10). Therefore, h = 2 and k = 10. To find the shaded area: Area P QOR = Area P QO + Area P OR = 1 2 h OQ k OR = 1 2 (2)(11) (10)(6) = = 41 Therefore the shaded area is 41 units 2.

19 Problem D Thinking Inside The Box Q R P Q is the point of intersection of the diagonals of one face of a cube whose edges have length 2 cm. Determine the length of QR.

20 Problem Problem D and Solution Thinking Inside The Box Q is the point of intersection of the diagonals of one face of a cube whose edges have length 2 cm. Determine the length of QR. T Q S Solution Label the corners S and T as shown. R P The faces of a cube are squares. The diagonals of a square right bisect each other. It follows that P Q = QT = 1 2 P T. Since the face is a square, P ST = 90 and P ST is right angled. Using the Pythagorean Theorem, P T 2 = P S 2 + ST 2 = = 8 and P T = 8. Then P Q = 1 2 P T = 8 2. Because of the 3-dimensional nature of the problem it may not be obvious to all that RP Q = 90. To help visualize this, note that RP S = 90 because the face of the cube is a square. Rotate P S counterclockwise about point P on the side face of the cube so that the image of P S lies along P Q. The corner angle will not change as a result of the rotation so RP Q = RP S = 90. We can now use the Pythagorean Theorem in RP Q to find the length RQ. ( ) 2 8 RQ 2 = RP 2 + P Q 2 = = = = 6 and RQ = 6 cm. the length of RQ is 6 cm. A couple of notes are in order at this point. First, although the mathematics required to solve this problem was fairly straight forward some students would have found it difficult because of the three dimensional nature of the problem. Second, we could have simplified P Q = 1P T = 8 to 2 as follows: = = = 2 2 = Often simplifying radicals is not a part of the curriculum at the grade 9 or 10 level. The calculation of RQ would have been simpler using P T = 2.

21 Problem D Traversing the Triangle A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. A triangle has an area of 72 cm 2. The length of one side is 12 cm and the length of the median to this side is 13 cm. Determine the length of the other two sides of the triangle.

22 Problem Problem D and Solution Traversing the Triangle A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. A triangle has an area of 72 cm 2. The length of one side is 12 cm and the length of the median to this side is 13 cm. Determine the length of the other two sides of the triangle. Solution Start with a diagram to represent the problem. Let AD be the 12 cm side and EC be the 13 cm median drawn to that side. Draw the altitude from E to side AD meeting it at B. Let EB be h. Since EC is a median, AC = CD = 1(AD) = 1 (12) = 6 cm. 2 2 Let BC, the distance along AD from the altitude to the median, be x. Then AB = 6 x. The area of the triangle is 72 cm 2 so AD EB 2 = 72. Then 12h 2 = 72 and h = 12 cm follows. EBC is right angled so, using Pythagoras Theorem, x 2 = 13 2 h 2 = = = 25 and x = 5 cm (x > 0). Then BD = BC + CD = x + 6 = 11 cm and AB = 6 x = 1 cm. EAB is right angled so, using Pythagoras Theorem, EA 2 = EB 2 + AB 2 = = 145 and EA = 145 cm. (EA > 0) EBD is right angled so, using Pythagoras Theorem, ED 2 = EB 2 + BD 2 = = = 265 and ED = 265 cm (ED > 0). Therefore the lengths of the other two sides are 145 cm and 265 cm. These lengths are approximately 12.0 cm and 16.3 cm.

23 Problem D Tangled Triangles y A(0, a) C(3, 2) O(0, 0) x D(0, -1) B(2, -1) In the diagram, A(0, a) lies on the y-axis above the origin. If ABD and COB have the same area, determine the value of a.

24 Problem Problem D and Solutions Tangled Triangles y In the diagram, A(0, a) lies on the y-axis above the origin. If ABD and COB have the same area, determine the value of a. Solution 1 Draw rectangle EDF C with sides parallel to the x and y-axes so that O(0, 0) is on ED and B(2, 1) is on DF. Since EC is parallel to the x-axis and E is on the y-axis, E has coordinates (0, 2). Since CF is parallel to the y-axis, F has the same x-coordinate as C. Since F D is parallel to the x-axis, F has the same y-coordinate as D and B. Therefore the coordinates of F are (3, 1). A(0, a) E(0, 2) O(0, 0) D(0, -1) B(2, -1) To find the area of COB, subtract the areas of CEO, ODB, and BF C from the area of rectangle EDF C. In rectangle EDF C, EC = 3 0 = 3 and ED = 2 ( 1) = 3. The area of rectangle EDF C = EC ED = 3 3 = 9 units 2. In CEO, EC = 3 and EO = 2 0 = 2. The area of ECO = EC EO = 3 2 = 3 units In ODB, OD = 0 ( 1) = 1 and DB = 2 0 = 2. The area of ODB = OD DB = 1 2 = 1 unit In BF C, BF = 3 2 = 1 and CF = 2 ( 1) = 3. The area of BF CF BF C = = 1 3 = 1.5 units Area COB = Area Rectangle EDF C CEO ODB BF C = = 3.5 units 2 But the area ABD = COB so the area of ABD = 3.5 units 2. C(3, 2) F(3, -1) x In ABD, AD = a ( 1) = a + 1 and DB = 2 0 = 2 so the value of a is 2.5. AD DB Area ABD = 2 (a + 1) = = a = a

25 Problem y In the diagram, A(0, a) lies on the y-axis above the origin. If ABD and COB have the same area, determine the value of a. Solution 2 Determine the equation of the line containing C(3, 2) and B(2, 1). The slope of the line is 2 ( 1) = 3. The equation of the line is of the 3 2 form y = 3x + b. Substitute x = 3, y = 2 to determine the value of b. 2 = 3(3) + b and b = 7 follows. Therefore the equation of the line containing C and B is y = 3x 7. A(0, a) O(0, 0) D(0, -1) Let P (p, 0) be the x-intercept of the line. Substituting into y = 3x 7 we obtain 0 = 3p 7 and p = 7 3 follows. To determine the area of COB determine the sum of the areas of COP and BOP. In COP, OP = 7 and the height is the perpendicular distance from the x-axis to C(3, 2), 3 which is 2 units. The area of COP = = units2. In BOP, OP = 7 and the height is the perpendicular distance from the x-axis to B(2, 1), 3 which is 1 unit. The area of BOP = = units2. Area COB = Area COP + Area BOP = = = 21 6 = 7 2 units2 But the area ABD = COB so the area of ABD = 7 2 units2. In ABD, AD = a ( 1) = a + 1 and DB = 2 0 = 2 so Area ABD = 7 2 AD DB 2 = (a + 1) = 2a = 2a a = 5 2 = 2.5. P(p, 0) B(2, -1) C(3, 2) x

26 Problem D Forever Mine Valentine A heart is constructed by attaching two white semi-circles to the hypotenuse of an isosceles right triangle whose equal sides measure 8 cm. The new figure is then mounted onto a rectangular piece of red construction paper as shown below. (The dashed line, the side measurement and the right angle symbol will not actually be on the finished card.) You are going to write your valentine a message in red ink on the white region of the card. Determine the total amount of area available for your special valentine greeting. 8 If the solver finds this 9-10 problem too straight forward, a similar problem called Love is Blind Valentine is presented as the for the level. This second problem may provide a greater challenge.

27 Problem D and Solution Forever Mine Valentine Problem A heart is constructed by attaching two white semi-circles to the hypotenuse of an isosceles right triangle whose equal sides measure 8 cm. The new figure is then mounted onto a rectangular piece of red construction paper as shown below. You are going to write your valentine a message in red ink on the white region of the card. Determine the total amount of area available for your special valentine greeting. Solution Let h represent the length of the hypotenuse. Let r represent the radius of the semi-circles. Since the two semi-circles lie along the hypotenuse, h = 4r or r = h 4. 8 h 8 Since the triangle is isosceles right, we can find h using Pythagoras Theorem, h 2 = ( 8) 2 + 8) 2 = = 16 and h = 4 cm follows. Then r = h 4 = 4 4 = 1 cm. Since there are two semi-circles of radius 1 cm, the total area of the two semi-circles is the same as the area of a full circle of radius 1 cm. The area of the two semi-circles is πr 2 = π(1) 2 = π cm 2. The triangle is isosceles right so we can use the lengths of the two equal sides as the base and height in the calculation of the area of the triangle. The area of the triangle is 1 2 bh = 1 2 ( 8)( 8) = 4 cm 2. The total area for writing the message is (π + 4) cm 2. This area is approximately 7.1 cm 2. Hopefully you can write that special message in a very limited space. Happy Valentine s Day.

28 Problem D Bugged A B A ladybug wishes to travel from B to A on the surface of a wooden block with dimensions as shown in the diagram. Determine the shortest distance for the ladybug to walk.

29 Problem Problem D and Solution Bugged A ladybug wishes to travel from B to A on the surface of a wooden block with dimensions as shown in the diagram. Determine the shortest distance for the ladybug to walk. Solution Many strategies could be attempted. Perhaps the bug walks along the edges and travels = 14 units. Perhaps the bug travels across the right side of the block from B to the midpoint of the top edge (marked P on the diagram below) and then across the top of the box to A. Referring to the diagram below, it can be shown that this distance is BP 2 + P A 2 = = units. But is this the shortest distance? To visualize the possible routes fold out the sides of the box so that they are laying on the same plane as the top of the box. Label the diagram as shown below. Note that as a result of folding out the sides, corner B appears twice. The second corner is labelled B. C 2 8 A 4 4 B 2 B 4 The shortest distance for the ladybug to travel is a straight line from B to A or B to A. So both cases must be considered. BA is the hypotenuse of right-angled triangle ABD. Using Pythagoras Theorem, 2 BA 2 = BD 2 + DA 2 = = 100 and BA = 10 follows. B A is the hypotenuse of right-angled triangle AB C. Using Pythagoras Theorem, (B A) 2 = (B C) 2 + CA 2 = = 116 and B A = 2 29 =10.77 follows. Since BA < B A, the shortest distance for the ladybug to travel is 10 units on the surface of the block in a straight line from B to A. This problem is quite straight forward once the three-dimensional nature of the problem is removed. 8 P 4 2 D

30 Problem D The Case of the Missing Square Rectangle DEF G has square ABCD removed leaving an area of 92 m 2. Side AE = 4 m and side CG = 8 m. Determine the original area of rectangle DEF G. E F A B D C G

31 Problem D and Solution The Case of the Missing Square Problem Rectangle DEF G has square ABCD removed leaving an area of 92 m 2. Side AE = 4 m and side CG = 8 m. Determine the original area of rectangle DEF G. Solution Let x represent the side length of square ABCD. In the diagram, extend CB to intersect EF at H. This creates rectangle AEHB and rectangle CHF G. Then F G = EA + AD = (4 + x) m and EH = DC = x m. E x H F 4 A B 4+x x D C x 8 G Area AEHB + Area CHF G = Remaining Area AE EH + CG F G = 92 4x + 8(4 + x) = 92 4x x = 92 12x + 32 = 92 12x = 60 x = 5 m Since x = 5 m, DG = 8 + x = 13 m and F G = 4 + x = 9 m. The original area of rectangle DEF G = DG F G = 13 9 = 117 m 2.

32 Problem D Walking is Good Exercise Ali, Bill and Carl are lined up such that Ali is 100 m west of Bill and Carl is 160 m east of Bill. At noon, Carl begins to walk north at a constant rate of 41 min m and Ali walks south at a constant rate of 20 min m. (Bill does not move.) At what time will the distance between Carl and Bill be the twice the distance between Ali and Bill?

33 Problem D and Solutions Walking is Good Exercise Problem Ali, Bill and Carl are lined up such that Ali is 100 m west of Bill and Carl is 160 m east of Bill. At noon, Carl begins to walk north at a constant rate of 41 min m and Ali walks south at a constant rate of 20 min m. (Bill does not move.) At what time will the distance between Carl and Bill be the twice the distance between Ali and Bill? Solution 1 Let t represent the number of minutes until Carl s distance to Bill is twice that of Ali s distance to Bill. In t minutes Ali will walk 20t m and Carl will walk 41t m. The following diagram contains the information showing Ali s position at A and Carl s position at C at time t. C 41t 20t A 100 m B In the diagram both triangles are right triangles and we can use the Pythagorean Theorem to set up an equation. BC = 2AB (BC) 2 = (2AB) 2 in to Bill. 160 m (BC) 2 = 4 (AB) 2 (41t) 2 + (160) 2 = 4 [ (20t) 2 + (100) 2] 1681t = 4 [ 400t ] 1681t = 1600t t 2 = t 2 = t = since t > 0 t = 40 3 min minutes (13 minutes 20 seconds), Carl s distance to Bill will be twice that of Ali s distance

34 Problem Ali, Bill and Carl are lined up such that Ali is 100 m west of Bill and Carl is 160 m east of Bill. At noon, Carl begins to walk north at a constant rate of 41 min m and Ali walks south at a constant rate of 20 min m. (Bill does not move.) At what time will the distance between Carl and Bill be the twice the distance between Ali and Bill? Solution 2 In this solution coordinate geometry will be used to solve the problem. Represent Ali, Bill and Carl s respective positions at noon as points on the x-axis so that Bill is positioned at the origin B(0, 0), Ali is positioned 100 units left of Bill at D( 100, 0) and Carl is positioned 160 units right of Bill at E(160, 0). Let t represent the number of minutes until Carl s distance to Bill is twice that of Ali s distance to Bill. In t minutes Ali will walk south 20t m to the point A( 100, 20t). In t minutes Carl will walk north 41t m to the point C(160, 41t). y C(160, 41t) D(-100, 0) B(0, 0) x E(160, 0) A(-100, -20t) The distance from a point P (x, y) to the origin can be found using the formula d = x 2 + y 2. Then AB = ( 100) 2 + ( 20t) 2 = t 2 and CB = (160) 2 + (41t) 2 = t 2. CB = 2AB t 2 = t 2 in to Bill. Squaring both sides, t 2 = 4( t 2 ) t 2 = t 2 81t 2 = t 2 = t = since t > 0 t = 40 3 min minutes (13 minutes 20 seconds), Carl s distance to Bill will be twice that of Ali s distance

35 Problem D Mind The Median A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. An altitude is a line segment drawn from a vertex of a triangle to the opposite side intersecting at 90. A D M C B In ABC, ACB = 90. An altitude is drawn in ABC from C to AB intersecting at D. A median is drawn in ADC from A meeting DC at M. The length of the median AM is 13 and AD is 12. Determine the length of DB.

36 Problem D and Solutions Mind The Median Problem A A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. An altitude is a line segment drawn from a vertex of a triangle to the opposite side intersecting at b In ABC, ACB = 90. An altitude is drawn in ABC from C to AB intersecting at D. A median is drawn in ADC from A meeting DC at M. The length of the median AM is 13 and AD is 12. Determine the length of DB. D c a 13 M a d C Solution 1 B In this solution we will use Pythagoras Theorem. Since M is a median in ADC, DM = MC = a. Let AC = b, BD = c and BC = d. The variables and the given information, AD = 12 and AM = 13, are shown on the diagram. Since ADM contains a right angle at D, DM 2 = a 2 = AM 2 AD 2 = = 25 and a = 5 follows. Then DC = 2a = 10. Since ADC contains a right angle at D, AC 2 = b 2 = AD 2 + DC 2 = = 244 and b = 244 follows. Since CDB contains a right angle at D, CB 2 = BD 2 + DC 2 = c = c d 2 = c (1) Since ABC contains a right angle at C, AB 2 = AC 2 + BC 2. (12 + c) 2 = ( 244) 2 + d 2 which simplifies to c + c 2 = d 2. This further simplifies to c c = d 2. (2) Substituting for d 2 from (1) into (2), we obtain c c = c Simplifying we get 24c = 200 and c = 25 3 follows. the length of DB is 25 3.

37 Problem A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. An altitude is a line segment drawn from a vertex of a triangle to the opposite side intersecting at 90. A 12 b In ABC, ACB = 90. An altitude is drawn in ABC from C to AB intersecting at D. A median is drawn in ADC from A meeting DC at M. The length of the median AM is 13 and AD is 12. Determine the length of DB. D c a 13 M d a C Solution 2 B In this solution we will present a solution involving similar triangles. From the first solution, we will pick up the computed values AC = 244 and DC = 10. In ADC and ACB, ADC = ACB = 90 and DAC = BAC, a common angle. So ADC ACB. From similar triangles, it follows that But c = DB, so the length of BD is AD AC = AC AB = c Cross-Multiplying, 12c = c = 100 c = 25 3

38 Problem A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. An altitude is a line segment drawn from a vertex of a triangle to the opposite side intersecting at 90. In ABC, ACB = 90. An altitude is drawn in ABC from C to AB intersecting at D. A median is drawn in ADC from A meeting DC at M. The length of the median AM is 13 and AD is 12. Determine the length of DB. Solution 3 y A(0, 12) D(0, 0) B(0, b) 13 a a x M(a, 0) C (2a, 0) In this solution, we approach the problem by superimposing the problem on the coordinate grid and solve using analytic geometry. Position ABC so that AB lies along the y-axis, C is on the positive x-axis and altitude DC lies along the positive x-axis with D at the origin. Since AM is a median in ADC, DM = MC = a. Then M has coordinates (a, 0) and C has coordinates (2a, 0). Since AD = 12, A has coordinates (0, 12). Since B is on the y-axis, let B have coordinates (0, b) with b < 0. In ADM, DM 2 = AM 2 AD 2 = = 25 and DM = 5 follows. Therefore a = 5, 2a = 10 and C has coordinates (10, 0). The slope of AB = = 6 b 0. The slope of BC = = b 10. Since ACB = 90, AC BC and their slopes should be negative reciprocals. b 10 = 5 6 6b = 50 b = 25 3 ( It then follows that the coordinates of B are 0, 25 ). The length of DB is the distance 3 from D, the origin, to the point B which is 25 3 units. the length of BD is 25 3.

39

40 Problem D Full of Beans Canada celebrates its birthday every July 1 st on Canada Day. The colours red and white, the colours on the Canadian flag, are seen everywhere. In honour of Canada Day, a company produced a jumbo bag of candy containing only red and white jelly beans in the ratio 7:11 respectively. Khan Di loves jelly beans and purchased one of the special packages of candy. He decided to make the candy last as long as possible by eating exactly two red jelly beans and three white jelly beans each day. Finally one day he ate the last two red jelly beans along with three white jelly beans leaving four white jelly beans in the bag. How many candies were in the original bag of red and white jelly beans?

41 Problem D and Solution Full of Beans Problem Canada celebrates its birthday every July 1 st on Canada Day. The colours red and white, the colours on the Canadian flag, are seen everywhere. In honour of Canada Day, a company produced a jumbo bag of candy containing only red and white jelly beans in the ratio 7:11 respectively. Khan Di loves jelly beans and purchased one of the special packages of candy. He decided to make the candy last as long as possible by eating exactly two red jelly beans and three white jelly beans each day. Finally one day he ate the last two red jelly beans along with three white jelly beans leaving four white jelly beans in the bag. How many candies were in the original bag of red and white jelly beans? Solution Let d represent the number of days Khan ate two red jelly beans and three white jelly beans. Since he ate two red jelly beans each day for d days, he ate 2d red jelly beans. Since he ate three white jelly beans each day for d days, he ate 3d white jelly beans. But there were 4 white jelly beans left so Khan had a total of 3d + 4 white jelly beans. The ratio of red jelly beans to white jelly beans is 7:11. 2d 3d + 4 = 7 11 Cross-multiplying to simplify we obtain: 22d = 21d + 28 Therefore Khan was able to follow his pattern for 28 days. d = 28 Since d = 28, 2d = 2(28) = 56 and he had 56 red jelly beans. Also 3d + 4 = 3(28) + 4 = 88 so Khan had 88 white jelly beans. The bag contained = 144 jelly beans. the special bag of candy contained 144 jelly beans. The number 144 is significant because that is how old Canada was on July 1, 2011.

45 Problem D What s Your Angle Anyway II? In ABD, C is on AB such that CA = CB = CD and BCD = z. Determine the measure of ADB. If you need a hint, consider solving this week s Grade 7 and 8 Problem of the Week What s Your Angle Anyway I?. If you want an extension to this problem, consider solving this week s Grade 11 and 12 What s Your Angle Anyway III?.

46 Problem D and Solution What s Your Angle Anyway II? Problem In ABD, C is on AB such that CA = CB = CD and BCD = z. Determine the measure of ADB. Solution Since ACB is a straight line, ACD + DCB = 180 but BCD = z so ACD = 180 z. In ACD, since CA = CD, ACD is isosceles and CAD = CDA = x. The angles in a triangle sum to 180 so in ACD CAD + CDA + ACD = 180 x + x z = 180 2x = z x = z 2 Similarly, in BCD, since CB = CD, CBD is isosceles and CBD = CDB = y. The angles in a triangle sum to 180 so in CBD CBD + CDB + BCD = 180 y + y + z = 180 2y = 180 z y = 180 z 2 Then ADB = CDA + CDB = x + y = z z 2 = = 90. the measure of ADB is 90. See the next page for an interesting idea.

47 It turns out that it is not necessary to find expressions for x and y in terms of z in the problem. Here is a second solution to the problem. In CAD, since CA = CD, CAD is isosceles and CAD = CDA = x. In CBD, since CB = CD, CBD is isosceles and CBD = CDB = y. The angles in a triangle sum to 180 so in ABD BAD + ADB + ABD = 180 x + (x + y ) + y = 180 2x + 2y = 180 x + y = 90 But ADB = ADC + CDB = x + y = 90. the measure of ADC is 90.

48 Problem D A Sour Taste A container is filled with a mixture of water and vinegar in the ratio 2:1 respectively. Another container, with twice the volume of the first container, is filled with a mixture of water and vinegar in the ratio 3:1 respectively. The contents of the two containers are emptied into a third container. Determine the ratio of water to vinegar in the third container.

49 Problem Problem D and Solutions A Sour Taste A container is filled with a mixture of water and vinegar in the ratio 2:1 respectively. Another container, with twice the volume of the first container, is filled with a mixture of water and vinegar in the ratio 3:1 respectively. The contents of the two containers are emptied into a third container. Determine the ratio of water to vinegar in the third container. Solution 1 Let V represent the volume of the first container. Then 2V represents the volume of the second container. Since the ratio of water to vinegar in the first container is 2:1 then 2 of the volume of the first 3 container is water. That is, the volume of water in the first container is 2 V and the volume of 3 vinegar in the first container is 1V. 3 Since the ratio of water to vinegar in the second container is 3:1 then 3 of the volume of the 4 second container is water. That is, the volume of water in the second container is 3(2V ) = 3V 4 2 and the volume of vinegar in the second container is 1(2V ) = 1V. 4 2 When the contents of the two containers are combined into the third container, the volume of water is 2 3 V V = 4 6 V V = 13 6 V and the volume of vinegar is 1 3 V V = 2 6 V V = 5 6 V. The ratio of water to vinegar in the third container is 13V : 5 V = 13 : Solution 2 Let 1 of the contents of the first container be a unit of volume. Since the ratio of water to 6 vinegar in the first container is 2:1, then 4 units of volume are water and 2 units of volume are vinegar, a total of 6 units of volume in the first container. Since the second container has twice the volume of the first container, the second container has 12 units of volume. The ratio of water to vinegar in the second container is 3:1 so 9 units of volume are water and 3 units of volume are vinegar. When the two containers are combined there is a total of = 18 units of volume, = 13 of which are water and = 5 of which are vinegar. the ratio of water to vinegar in the third container is 13 : 5.

50 Problem D Formidable Fractions? For positive integers a and c, ( a c + a ) ( 2 a + 2 ) = 18. c + 1 Determine the number of ordered pairs (a, c) that satisfy a + 3c 99.

51 Problem D and Solutions Formidable Fractions? Problem For positive integers a and c, that satisfy a + 3c 99. Solution 1 ( a c + a ) ( 2 a + 2 ) = 18. Determine the number of ordered pairs (a, c) c + 1 ( a c + a ) First we should simplify the fractional equation, ( 2 a + 2 ) = 18 c + 1 ( 2a 2c + ac 2c + 2c ) 2c Finding common denominators, ( 2c ac + 2a ac + ac ) = 18 ac ( ) 2a + ac + 2c Multiplying by the reciprocal, Simplifying, (2a + ac + 2c) 2c 2c ( ) = 18 2c + 2a + ac ac ac (2c + 2a + ac) = 18 Since the bracketed numerator and bracketed denominator are the same and cannot equal zero, we can simplify to ac = 18. Since c 0, the expression 2c further simplifies to a = 18 or a = 18(2) = 36. Substituting a = 36 into 2 a + 3c 99 we obtain c 99 which simplifies to 3c 63 and c 21 follows. But c 1 and c is an integer so 1 c 21. The value of a is 36 for each of the 21 possible values of c. there are 21 ordered pairs (a, c) that satisfy the problem. Solution 2 is on the next page.

52 Solution 2 First we should simplify the fractional equation, Multiply the numerator and the denominator by 2ac: ( a c + a ) ( 2 a + 2 ) = 18 c + 1 ( a c + a ) ( 2 a + 2 ) 2ac c + 1 2ac = 18 The equation simplifies to: Factoring, we obtain: 2a 2 + a 2 c + 2ac 4c + 4a + 2ac = 18 a(2a + ac + 2c) 2(2c + 2a + ac) = 18 Since the bracketed numerator and bracketed denominator are the same and cannot equal zero, we can simplify to a = 18 and a = 36 follows. 2 Substituting a = 36 into a + 3c 99 we obtain c 99 which simplifies to 3c 63 and c 21. But c 1 and c is an integer so 1 c 21. The value of a is 36 for each of the 21 possible values of c. there are 21 ordered pairs (a, c) that satisfy the problem.

55 Problem D Thinking Inside The Box Q R P Q is the point of intersection of the diagonals of one face of a cube whose edges have length 2 cm. Determine the length of QR.

56 Problem Problem D and Solution Thinking Inside The Box Q is the point of intersection of the diagonals of one face of a cube whose edges have length 2 cm. Determine the length of QR. T Q S Solution Label the corners S and T as shown. R P The faces of a cube are squares. The diagonals of a square right bisect each other. It follows that P Q = QT = 1 2 P T. Since the face is a square, P ST = 90 and P ST is right angled. Using the Pythagorean Theorem, P T 2 = P S 2 + ST 2 = = 8 and P T = 8. Then P Q = 1 2 P T = 8 2. Because of the 3-dimensional nature of the problem it may not be obvious to all that RP Q = 90. To help visualize this, note that RP S = 90 because the face of the cube is a square. Rotate P S counterclockwise about point P on the side face of the cube so that the image of P S lies along P Q. The corner angle will not change as a result of the rotation so RP Q = RP S = 90. We can now use the Pythagorean Theorem in RP Q to find the length RQ. ( ) 2 8 RQ 2 = RP 2 + P Q 2 = = = = 6 and RQ = 6 cm. the length of RQ is 6 cm. A couple of notes are in order at this point. First, although the mathematics required to solve this problem was fairly straight forward some students would have found it difficult because of the three dimensional nature of the problem. Second, we could have simplified P Q = 1P T = 8 to 2 as follows: = = = 2 2 = Often simplifying radicals is not a part of the curriculum at the grade 9 or 10 level. The calculation of RQ would have been simpler using P T = 2.

57 Problem D The Key Factor A specific type of six-digit number is formed by repeating a three-digit number, for example, or or Determine the largest integer which will divide all such numbers.

58 Problem D and Solution The Key Factor Problem A specific type of six-digit number is formed by repeating a three-digit number, for example, or or Determine the largest integer which will divide all such numbers. Solution To get started look at the prime factorization of each of the given numbers = = = = = = = = = = = = = All of the numbers are divisible by = These are the only three factors common to all three numbers. Pick another six-digit number formed by repeating a three-digit number and test to see if it is also divisible by The number , for example, is It would appear that our conjecture (guess) is correct but it has not been proven. Let abc abc be any six digit number formed by repeating the three-digit number abc. Then abc abc = abc000 + abc = 1000 abc + abc = 1000 abc + 1 abc = abc Since abc abc = abc, it is divisible by A specific number abc abc may also have other divisors but is the largest divisor common to all such numbers. In the first example = and in the third example = Both numbers have other factors but no other common factors. In some cases there will be other common factors but not in general. This problem is not hard if you initially get it. The solution presented shows an approach that can be taken when you may not be certain where to begin. Try some specific examples and then attempt to generalize based on what you observe from the specific examples. Also note that discovering that worked for the three given examples and the test example is not sufficient to make a general conclusion that divides all such numbers.

59 Problem D An Average Bowl In his latest game, Mark Striker bowled 199 and raised his average from 177 to 178. Mark would like to raise his average to 180 after bowling his next game. What would Mark need to bowl on his next game to accomplish his goal?

60 Problem Problem D and Solutions An Average Bowl In his latest game, Mark Striker bowled 199 and raised his average from 177 to 178. Mark would like to raise his average to 180 after bowling his next game. What would Mark need to bowl on his next game to accomplish his goal? Solution 1 Let n be the number of games bowled to achieve his 177 average. His total points scored in n games is his average times n. Therefore, Mark has 177n total points in n games. To compute his average after bowling the 199 game, we take his new total points and divide by n + 1, the new number of games. Average = Total Points Games Played 178 = 177n n (n + 1) = 177n n = 177n n = 21 Prior to bowling the 199 game, Mark had bowled 21 games. So after bowling the 199 game, Mark has bowled 22 games. Mark wants to have a 180 average after bowling his 23rd game. The difference between his total points after 23 games with a 180 average and his total points after bowling 22 games with a 178 average must be his score on the 23rd game. Score on 23rd Game = = = 224 Therefore Mr. Striker must bowl 224 on his next game to raise his average from 178 to 180. Solution 2 offers a very different solution to the problem.

61 Problem In his latest game, Mark Striker bowled 199 and raised his average from 177 to 178. Mark would like to raise his average to 180 after bowling his next game. What would Mark need to bowl on his next game to accomplish his goal? Solution 2 Mark s score of 199 is = 22 points above his previous average. Mark raised his average 1 point. Therefore, his latest game with the 199 score must have been his 22nd game. To raise his average 2 points in his 23rd game he must bowl 2 23 = 46 points above his 178 average. He must bowl = 224 in his next game. Mark must bowl 224 in his next game to move his average from 178 to 180. We can verify our results: Average on 21 games is 177. Average on 22 games = = = 178 Average on 23 games = = = 180

62 Problem D Just Perfect A perfect square is an integer that is the square of another integer. For example, 36 is a perfect square since 36 = 6 2. Determine the smallest perfect square greater than that is divisible by 392.

63 Problem Problem D and Solution Just Perfect A perfect square is an integer that is the square of another integer. For example, 36 is a perfect square since 36 = 6 2. Determine the smallest perfect square greater than that is divisible by 392. Solution In order to understand the nature of perfect squares we should examine the prime factorization of a few perfect squares. From the example, 36 = 6 2 = (2 3) 2 = The number 144 = 12 2 = (3 4) 2 = 3 2 (2 2 ) 2 = From the above examples we note that, for each perfect square, the exponent on each of its prime factors is even. For some integer a, if m is an even integer greater than or equal to zero then a m is a perfect square. For example, = = = The number 392 = 8 49 = This clearly is not a perfect square since the exponent of 2 is odd. We require another factor of 2 to obtain a multiple of 392 that is a perfect square, namely = 784. We now must multiply 784 by a perfect square so that the product is greater than Since = 5.1 we must multiply 784 by the smallest perfect square greater than 5.1 to create a perfect square greater than In this case that perfect square is 9. So the smallest perfect square greater than that is divisible by 392 is = = We can also show that = 18 to verify that 7056 is divisible by 392. the smallest perfect square greater than that is divisible by 392 is

64 Problem D Looking for Leftovers 3 5 means and equals 243 when expressed as an integer. When the number is divided by 5, what is the remainder?

65 Problem Problem D and Solution Looking for Leftovers 3 5 means and equals 243 when expressed as an integer. When the number is divided by 5, what is the remainder? Solution We looked at a similar problem during the second half of the last school year that asked for the pattern of the final digits. It turns out that we can use the same idea to find the last digit of and then apply the divisibility rule for division by 5. When a number ends in 0 or 5, the remainder is 0 when divided by 5. If a number ends in 1 or 6, the remainder is 1 when divided by 5. For example, 56 5 = 11 R1. If a number ends in 2 or 7, the remainder is 2 when divided by 5. If a number ends in 3 or 8, the remainder is 3 when divided by 5. If a number ends in 4 or 9, the remainder is 4 when divided by 5. So let s examine the pattern of the last digits on the first eight powers of 3. Exponent Power Value Final Digit It would appear that there is a pattern in the final digits that repeats every four numbers. The pattern would continue: the final digits of 3 9, 3 10, 3 11, 3 12 would be 3, 9, 7, 1, respectively. So we need to determine how many complete groups of four are in = 502 R3. There are 502 complete repetitions of the pattern of final digits and 3 numbers into the next pattern. This means that the final digit of is 7, the third number in the pattern of the final digits. Since the final digit is 7, the remainder when is divided by 5 is 2. when is divided by 5, the remainder is 2.

66 Problem D Sum Time Soon When fifty consecutive even integers are added together, their sum is Determine the value of the largest number. The formula for the sum of the first n positive integers is (n)(n+1) 2. This formula may be helpful in the solution of the above problem. The symbol shown below is the uppercase Greek letter sigma and is often used in mathematics in problems involving sums.

67 Problem D and Solutions Sum Time Soon Problem When fifty consecutive even integers are added together, their sum is Determine the value of the largest number. Solution 1 Let a represent the smallest number. Since the numbers are even, they increase by 2. So the second number is (a + 2), the third is (a + 4), the fourth is (a + 6), and so on. What does the fiftieth number look like? A closer look at the numbers reveals that the second number is (a + 1(2)), the third is (a + 2(2)), the fourth is (a + 3(2)), and so on. Following the pattern, the fiftieth number is (a + 49(2)) = a Then a + (a + 2) + (a + 4) + (a + 6) + + (a + 98) = 3250 the largest number is a = a + 2( ) = 3250 ( ) a + 2 = 3250, using the helpful formula. 2 50a = a = 800 a = 16 a + 98 = 114 In the above solution, we looked at patterns. We also used the formula for the sum of the first n positive integers (n)(n+1) 2 to find the sum of the positive integers In Solution 2 we will use averages. In Solution 3 we will use arithmetic sequences. This solution is presented last since many students in grade 9 or 10 have not encountered arithmetic sequences yet.

68 Problem When fifty consecutive even integers are added together, their sum is Determine the value of the largest number. Solution 2 In this solution we will use averages to solve the problem Let A represent the average of the fifty numbers. The average times the number of integers equals the sum of the integers. Since the sum of the fifty integers is 3 250, then 50A = and A = 65. Now the numbers in the sequence are consecutive even integers. The average is odd. It follows that 25 numbers are below the average and 25 numbers are above. We are looking for the 25 th even number after the average. In fact we want the 25 th even number after the even number 64, the first even number below the average. This number is easily found, (2) = = 114. the largest number is 114. In Solution 3, on the next page, arithmetic sequences are used.

69 Problem When fifty consecutive even integers are added together, their sum is Determine the value of the largest number. Solution 3 In this solution we will use arithmetic sequence formulas to solve the problem. An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. The general term, t n, of an arithmetic sequence is t n = a + (n 1)d where a is the first term, d is the difference between consecutive terms and n is the number of terms. The sum of n terms of an arithmetic sequence, S n, can be found using the formula S n = n 2 (2a + (n 1)d). a, d, and n are the same variables used in the general term formula. Let a represent the first term in the sequence. Since the numbers in the sequence are consecutive even integers, the numbers go up by two. Therefore d = 2. Since there are 50 terms in the sequence, n = 50. The sum of the fifty numbers in the sequence is 3250 so S 50 = S n = n (2a + (n 1)d) = 50 (2a + (50 1)(2)) = 25(2a + 49(2)) Dividing by 25, 130 = 2a = 2a 16 = a Since we want the largest number in the sequence, we are looking for the fiftieth term. Using t n = a + (n 1)d, with a = 16, d = 2, n = 50 t 50 = (2) t 50 = 114 the largest number is 114.

73 Problem D Forever Mine Valentine A heart is constructed by attaching two white semi-circles to the hypotenuse of an isosceles right triangle whose equal sides measure 8 cm. The new figure is then mounted onto a rectangular piece of red construction paper as shown below. (The dashed line, the side measurement and the right angle symbol will not actually be on the finished card.) You are going to write your valentine a message in red ink on the white region of the card. Determine the total amount of area available for your special valentine greeting. 8 If the solver finds this 9-10 problem too straight forward, a similar problem called Love is Blind Valentine is presented as the for the level. This second problem may provide a greater challenge.

74 Problem D and Solution Forever Mine Valentine Problem A heart is constructed by attaching two white semi-circles to the hypotenuse of an isosceles right triangle whose equal sides measure 8 cm. The new figure is then mounted onto a rectangular piece of red construction paper as shown below. You are going to write your valentine a message in red ink on the white region of the card. Determine the total amount of area available for your special valentine greeting. Solution Let h represent the length of the hypotenuse. Let r represent the radius of the semi-circles. Since the two semi-circles lie along the hypotenuse, h = 4r or r = h 4. 8 h 8 Since the triangle is isosceles right, we can find h using Pythagoras Theorem, h 2 = ( 8) 2 + 8) 2 = = 16 and h = 4 cm follows. Then r = h 4 = 4 4 = 1 cm. Since there are two semi-circles of radius 1 cm, the total area of the two semi-circles is the same as the area of a full circle of radius 1 cm. The area of the two semi-circles is πr 2 = π(1) 2 = π cm 2. The triangle is isosceles right so we can use the lengths of the two equal sides as the base and height in the calculation of the area of the triangle. The area of the triangle is 1 2 bh = 1 2 ( 8)( 8) = 4 cm 2. The total area for writing the message is (π + 4) cm 2. This area is approximately 7.1 cm 2. Hopefully you can write that special message in a very limited space. Happy Valentine s Day.

75 Problem D Turkey Time Among grandfather s papers a faded old piece of paper was found with the following written on it: 72 turkeys $ The first and the last digits of the number that represented the total price of the turkeys have been replaced here with dashes as the digits had faded and were no longer legible. Each dash represents one digit. Determine the missing digits.

76 Problem Problem D and Solution Turkey Time Among grandfather s papers a faded old piece of paper was found with the following written on it: 72 turkeys $ The first and the last digits of the number that represented the total price of the turkeys have been replaced here with dashes as the digits had faded and were no longer legible. Each dash represents one digit. Determine the missing digits. Solution Let the total price of 72 turkeys be A679B cents. We know the total value of 72 turkeys. We could find the value of 1 turkey by dividing the total value by 72. Since the total value of the turkeys is divisible by 72, it is also divisible by the divisors of 72, namely 1,2,3,4,6,8,9,12,18,24,36 and 72. If a number is divisible by 4, the last two digits of the number are divisible by 4. Therefore 9B is divisible by 4. The only two digit numbers beginning with 9 that are divisible by 4 are 92 and 96. So the only possible values for B are 2 and 6. The value of the 72 turkeys is either A6792 or A6796. But the number must also be divisible by 8. To be divisible by 8, the last three digits of the number must be divisible by 8. Of the two possible numbers, 792 and 796, only 792 is divisible by 8. Therefore the last digit of the price is 2 and we now know that 72 turkeys cost A6792 cents. If a number is divisible by 9, the sum of the digits of the number is divisible by 9. So A = A + 24 must be divisible by 9. A is a single digit from 0 to 9. The sum A + 24 is therefore an integer from 24 to 33. The only number in this range divisible by 9 is 27. It follows that A + 24 = 27 and A = turkeys cost $ and the missing digits are 3 and 2. Each turkey costs $ or $5.11. Note: This approach is very efficient but the solver must be careful. The number 4 divides 72 and any number that is divisible by 72. It does not follow that a number divisible by 4 is also divisible by 72. In this solution we found a number divisible by both 8 and 9. Since 8 and 9 have no common factors, a number divisible by 8 and 9 is also divisible by 72. It is possible to approach this question systematically using multiplication. In essence you can check possible values for B by performing the multiplication. In this case, since B = 2, the solution can be found reasonably quickly. However, in general, this would not be an effective strategy.

77 Problem D The Case of the Missing Square Rectangle DEF G has square ABCD removed leaving an area of 92 m 2. Side AE = 4 m and side CG = 8 m. Determine the original area of rectangle DEF G. E F A B D C G

78 Problem D and Solution The Case of the Missing Square Problem Rectangle DEF G has square ABCD removed leaving an area of 92 m 2. Side AE = 4 m and side CG = 8 m. Determine the original area of rectangle DEF G. Solution Let x represent the side length of square ABCD. In the diagram, extend CB to intersect EF at H. This creates rectangle AEHB and rectangle CHF G. Then F G = EA + AD = (4 + x) m and EH = DC = x m. E x H F 4 A B 4+x x D C x 8 G Area AEHB + Area CHF G = Remaining Area AE EH + CG F G = 92 4x + 8(4 + x) = 92 4x x = 92 12x + 32 = 92 12x = 60 x = 5 m Since x = 5 m, DG = 8 + x = 13 m and F G = 4 + x = 9 m. The original area of rectangle DEF G = DG F G = 13 9 = 117 m 2.

79 Problem D Walking is Good Exercise Ali, Bill and Carl are lined up such that Ali is 100 m west of Bill and Carl is 160 m east of Bill. At noon, Carl begins to walk north at a constant rate of 41 min m and Ali walks south at a constant rate of 20 min m. (Bill does not move.) At what time will the distance between Carl and Bill be the twice the distance between Ali and Bill?

80 Problem D and Solutions Walking is Good Exercise Problem Ali, Bill and Carl are lined up such that Ali is 100 m west of Bill and Carl is 160 m east of Bill. At noon, Carl begins to walk north at a constant rate of 41 min m and Ali walks south at a constant rate of 20 min m. (Bill does not move.) At what time will the distance between Carl and Bill be the twice the distance between Ali and Bill? Solution 1 Let t represent the number of minutes until Carl s distance to Bill is twice that of Ali s distance to Bill. In t minutes Ali will walk 20t m and Carl will walk 41t m. The following diagram contains the information showing Ali s position at A and Carl s position at C at time t. C 41t 20t A 100 m B In the diagram both triangles are right triangles and we can use the Pythagorean Theorem to set up an equation. BC = 2AB (BC) 2 = (2AB) 2 in to Bill. 160 m (BC) 2 = 4 (AB) 2 (41t) 2 + (160) 2 = 4 [ (20t) 2 + (100) 2] 1681t = 4 [ 400t ] 1681t = 1600t t 2 = t 2 = t = since t > 0 t = 40 3 min minutes (13 minutes 20 seconds), Carl s distance to Bill will be twice that of Ali s distance

81 Problem Ali, Bill and Carl are lined up such that Ali is 100 m west of Bill and Carl is 160 m east of Bill. At noon, Carl begins to walk north at a constant rate of 41 min m and Ali walks south at a constant rate of 20 min m. (Bill does not move.) At what time will the distance between Carl and Bill be the twice the distance between Ali and Bill? Solution 2 In this solution coordinate geometry will be used to solve the problem. Represent Ali, Bill and Carl s respective positions at noon as points on the x-axis so that Bill is positioned at the origin B(0, 0), Ali is positioned 100 units left of Bill at D( 100, 0) and Carl is positioned 160 units right of Bill at E(160, 0). Let t represent the number of minutes until Carl s distance to Bill is twice that of Ali s distance to Bill. In t minutes Ali will walk south 20t m to the point A( 100, 20t). In t minutes Carl will walk north 41t m to the point C(160, 41t). y C(160, 41t) D(-100, 0) B(0, 0) x E(160, 0) A(-100, -20t) The distance from a point P (x, y) to the origin can be found using the formula d = x 2 + y 2. Then AB = ( 100) 2 + ( 20t) 2 = t 2 and CB = (160) 2 + (41t) 2 = t 2. CB = 2AB t 2 = t 2 in to Bill. Squaring both sides, t 2 = 4( t 2 ) t 2 = t 2 81t 2 = t 2 = t = since t > 0 t = 40 3 min minutes (13 minutes 20 seconds), Carl s distance to Bill will be twice that of Ali s distance

82 Problem D A Problem for the Ages Grandmother has four grandchildren, two boys and two girls. When the ages of the grandchildren are multiplied together, the product of their ages is Only one of the grandchildren is a teenager and the oldest grandchild is under 40. The difference in ages between the oldest and youngest grandchild is thirty years. Determine the ages of Grandmother s grandchildren.

83 Problem Problem D and Solution A Problem for the Ages Grandmother has four grandchildren, two boys and two girls. When the ages of the grandchildren are multiplied together, the product of their ages is Only one of the grandchildren is a teenager and the oldest grandchild is under 40. The difference in ages between the oldest and youngest grandchild is thirty years. Determine the ages of Grandmother s grandchildren. Solution The first task is to factor Since the number is even, we can divide out powers of 2 such that = After some trial we discover that = Notice the inclusion of the number 1 as one of the factors. (One of the grandchildren could be 1.) We can rule out 1 as a possible age since no matter how we combine the other factors to create the three remaining ages, at least one of the ages will be greater than 40. It should also be noted that multiplying any of the factors 13, 17, 19 together, in any combination, will produce a number greater than 40. Since all of the grandchildren are under 40, we can eliminate these possibilities. We are left with combining various powers of 2 with the other factors to form the four ages. If 2 4 is one of the ages, the only possibility for the ages is {13,16,17,19}. This is not a valid combination because we know only one grandchild is a teenager. If 2 3 is one of the ages, we can multiply the fourth factor of 2 by one of the other ages to produce three possible sets of ages: {8,26,17,19}, {8,13,34,19}, and {8,13,17,38}. In each of these cases there are two teenagers so none of these sets produces valid ages. If 2 2 is one of the ages, we could multiply one of the ages by 4 or multiply two out of three of the other ages by 2. Multiplying either 13, 17 or 19 by 4 produces a product greater than 40 so this is not a possibility. Multiplying two out of three ages 13, 17, 19 by 2 produces the following results: {4,26,34,19}, {4,26,17,38}, and {4,13,34,38}. In each of these cases exactly one of the grandchildren is a teenager. But only the first possibility also satisfies the condition that the difference in ages between the oldest and the youngest is 30. Therefore, {4,26,34,19} is a possibility for the ages. There is one final possibility to consider. If one of the grandchildren is 2, we could double each of the remaining ages to create the set {2,26,34,38}. This possibility is quickly dismissed since none of the ages is in the range 13 to 19. That is, there are no teenagers in this list. Therefore the only possible ages for Grandmother s grandchildren are 4, 19, 26 and 34.

88 Problem D Change Adds Up In a jar I have a collection of 30 coins consisting of five pennies (Canadian one cent coins), five nickels (Canadian five cent coins), five dimes (Canadian ten cent coins), five quarters (Canadian twenty-five cent coins), five loonies (Canadian one dollar coins worth one hundred cents each) and five toonies (Canadian two dollar coins worth two hundred cents each). I need exactly five dollars. Without looking, I reach into my coin jar and take out a handful of coins. I take out anywhere from one coin (a very small handful) to all thirty coins (quite a fistful of coins). What is the largest sum of money I can remove from the jar so that I am not able to create exactly five dollars? For example, if I remove all thirty coins I can easily create five dollars using two of the toonies and one of the loonies. But if I remove only three toonies and three quarters, although I have pulled out more than five dollars, no matter how I combine these coins I cannot create exactly five dollars from them.

89 Problem Problem D and Solution Change Adds Up In a jar I have a collection of 30 coins consisting of five pennies (Canadian one cent coins), five nickels (Canadian five cent coins), five dimes (Canadian ten cent coins), five quarters (Canadian twenty-five cent coins), five loonies (Canadian one dollar coins worth one hundred cents each) and five toonies (Canadian two dollar coins worth two hundred cents each). I need exactly five dollars. Without looking, I reach into my coin jar and take out a handful of coins. I take out anywhere from one coin (a very small handful) to all thirty coins (quite a fistful of coins). What is the largest sum of money I can remove from the jar so that I am not able to create exactly five dollars? For example, if I remove all thirty coins I can easily create five dollars using two of the toonies and one of the loonies. But if I remove only three toonies and three quarters, although I have pulled out more than five dollars, no matter how I combine these coins I cannot create exactly five dollars from them. Solution If approached correctly, this is not a difficult problem. I want to create a handful of coins with the greatest value such that no matter how I combine the coins I cannot create exactly five dollars from them. In order to maximize the sum, start with the coin of greatest value. If I start with a coin of lesser value, I may not be able to use the higher value coins. For example, my handful may consist of four loonies. I cannot select five loonies since that is exactly five dollars. But by choosing four loonies I cannot select even one toonie because one toonie would combine with three of the loonies to create exactly five dollars. Therefore start by making sure all five toonies are in the handful. I would have $10 or and not be able to create exactly five dollars. But now I cannot use any loonies. If I pick one loonie, I could combine it with two of the toonies to create exactly five dollars (1 $1 + 2 $2 = $5). The next coin with the greatest value is the quarter. If I select four quarters, I have one dollar. This could be combined with two toonies to make exactly five dollars. Therefore the maximum number of quarters is three. I now have $10.75 or and cannot create exactly five dollars. Next, determine the number of dimes that can be selected. If five dimes are chosen, they could be combined with two of the quarters and two toonies to make exactly five dollars. Therefore only four dimes can be selected. The total value of the coins is $ $0.40 = $11.15 or = No nickels can be selected. If one nickel is chosen, it could be combined with two dimes, three quarters and two toonies to make exactly five dollars. Four pennies can be selected. If all five pennies are chosen, they could be combined with two dimes, three quarters and two toonies to make exactly five dollars. Therefore the greatest sum of money that can be made using the coins given is $ $0.04 = $11.19 or = This handful of money consists of five toonies, three quarters, four dimes and four pennies, a total of 16 of the coins from the jar.

90 Problem D A Card Game To Flip Over E Z Dealer has a deck consisting of 100 cards numbered from 1 to 100. Each card has the same number printed on both sides. One side of the card is yellow and the other side of the card is red. E Z places all the cards, red side up, on the table. He first turns over every card that has a number which is a multiple of 2. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. After E Z has finished, how many cards have the red side facing up?

91 Problem Problem D and Solution A Card Game To Flip Over E Z Dealer has a deck consisting of 100 cards numbered from 1 to 100. Each card has the same number printed on both sides. One side of the card is yellow and the other side of the card is red. E Z places all the cards, red side up, on the table. He first turns over every card that has a number which is a multiple of 2. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. After E Z has finished, how many cards have the red side facing up? Solution After flipping over all of the cards with numbers that are multiples of 2, 50 red cards are facing up and 50 yellow cards are facing up. All of the red cards facing up are odd. All of the yellow cards facing up are even. Next, in the second round of flips, EZ flips over every card that has a number which is a multiple of 3. Let us look at how many red cards will be flipped over to yellow and how many yellow cards will be flipped over to red. There are 33 multiples of three from 1 to 100, {3, 6, 9, 12, 15,, 87, 90, 93, 96, 99}. In this group of multiples, there are 17 numbers that are odd and 16 numbers that are even. But the 17 odd multiples of 3 are currently red and therefore are flipped over to become yellow. The 16 even multiples of 3 are currently yellow and are therefore flipped over to become red (again). So after the first flip there were 50 red and 50 yellow. Of the 50 red, 17 were flipped to yellow. Of the 50 yellow, 16 of them were flipped to red. Therefore after the second round, there were = 49 red cards.

92 Problem D Actress or Comedian - Dare to Compare Holly Woods is a popular young actress and Joe King is an up and coming young comedian. Joe has an income which is five-eighths of Holly s income. Joe s expenses are one-half those of Holly, and Joe saves 40% of his income. Determine the percentage of her income that Holly Woods saves. OR

93 Problem Problem D and Solutions Actress or Comedian - Dare to Compare Holly Woods is a popular young actress and Joe King is an up and coming young comedian. Joe has an income which is five-eighths of Holly s income. Joe s expenses are one-half those of Holly, and Joe saves 40% of his income. Determine the percentage of her income that Holly Woods saves. Solution 1 Using only one variable Let h represent Holly s income. Then Joe s income is 5 8 h. Since Joe saves 40% of his income, his expenses are 100% 40% = 60% of his income. Therefore, his expenses are 60% 5 8 h = h = 3 8 h. Joe s expenses are one-half of Holly s expenses so Holly s expenses are twice Joe s expenses. Therefore, Holly s expenses are 2 3h = 3 h = 0.75h = 75% of h. Since Holly s expenses are % of her income, she saves 100% 75% = 25% of her income. Holly Woods saves 25% of her income. Solution 2 Using two variables Let x represent Holly s income and y represent her expenses. Then Joe s income is 5 8 x and his expenses are 1 2 y. Since Joe saves 40% of his income, his expenses are 60% of his income. 1 2 y = x 1 2 y = x 1 2 y = 3 8 x y = 3 4 x Holly saves whatever is left of her income after expenses. Therefore Holly saves Holly Woods saves 25% of her income. x y = x 3 4 x = 1 x = 0.25x = 25% of x. 4 See the next page for a variation on solution 2.

94 Solution 3 Using two variables a bit differently Let 8x represent Holly s income and 2y represent her expenses. Then Joe s income is 5(8x) = 5x and his expenses are 1 (2y) = y. 8 2 Since Joe saves 40% of his income, his expenses are 60% of his income. y = x y = x y = 3x Holly earns 8x and her expenses are 2y so her savings are 8x 2y. We want the ratio of her 8x 2y 8x 2(3x) savings to her income, = = 2x 8x 8x 8x = 1 or 25%. 4 Holly Woods saves 25% of her income.

95 Problem D Higher Powers If a, b and c are distinct positive integers such that abc = 16, then determine the largest possible value of a b b c + c a.

96 Problem Problem D and Solution Higher Powers If a, b and c are distinct positive integers such that abc = 16, then determine the largest possible value of a b b c + c a. Solution Let s first look at the different ways to factor 16 as the product of three positive integers: 16 = = = = We are given that a, b and c are distinct, so the required factorization is 16 = We could, at this point, try all possible assignments of the values 1, 2, and 8 to a, b, and c, respectively. There are six possible ways to assign the values. However, by applying the following logic we can reduce the number of combinations that must be considered. To make the expression a b b c + c a as large as possible, we need to make b c as small as possible. We can do so by setting b = 1. So we either have a = 2 and c = 8 or a = 8 and c = 2. If a = 2, b = 1 and c = 8, then a b b c + c a = = 65. If a = 8, b = 1 and c = 2, then a b b c + c a = = 263. Therefore, a b b c + c a is maximized when a = 8, b = 1 and c = 2 producing a maximum value of 263. Something to Think About Would the solution change if the word distinct was removed from the original problem statement?

97 Problem D A Pane - Full Experience A contractor is building new houses. Each house has a large window with width 2 feet and height 8 feet to be made up of eight identical 1 foot by 2 feet glass panes. The contractor wants each house to look slightly different. The window size is fixed, but he plans on arranging the glass panes in the windows so that no two windows have the same configuration. Two possible arrangements of the glass panes are given below. Assuming that the glass panes cannot be cut, how many different arrangements can be made?

98 Problem Problem D and Solution A Pane - Full Experience A contractor is building new houses. Each house has a large window with width 2 feet and height 8 feet to be made up of eight identical 1 foot by 2 feet glass panes. The contractor wants each house to look slightly different. The window size is fixed, but he plans on arranging the glass panes in the windows so that no two windows have the same configuration. Assuming that the glass panes cannot be cut, how many different arrangements can be made? Solution Let s consider the ways that the glass panes can be arranged. First, notice that there must always be an even number of window panes that have a vertical orientation (standing on end). All glass panes are horizontal, none are vertical This can only be done one way. Six glass panes are horizontal, two are vertical There could be 0, 1, 2, 3, 4, 5, or 6 horizontal panes below the two vertical panes. So there are 7 ways that six glass panes are horizontal and two are vertical. Four glass panes are horizontal, four are vertical We need to consider sub cases: Case 1: There are no horizontal panes between the vertical panes. There could be 0, 1, 2, 3, or 4 horizontal panes below the four vertical panes. So there are 5 ways that four glass panes are horizontal, four are vertical and there are no horizontal panes between the vertical panes. Case 2: There is one horizontal pane between the vertical panes. There could be 0, 1, 2, or 3 horizontal panes below the bottom two vertical panes. So there are 4 ways that four glass panes are horizontal, four are vertical and there is one horizontal pane between the vertical panes. Case 3: There are two horizontal panes between the vertical panes. There could be 0, 1 or 2 horizontal panes below the bottom two vertical panes. So there are 3 ways that four glass panes are horizontal, four are vertical and there are two horizontal panes between the vertical panes.

99 Case 4: There are three horizontal panes between the vertical panes. There could be 0 or 1 horizontal pane below the bottom two vertical panes. So there are 2 ways that four glass panes are horizontal, four are vertical and there are three horizontal panes between the vertical panes. Case 5: There are four horizontal panes between the vertical panes. Since there are no more glass panes left, there can only can be 0 horizontal panes below the bottom two vertical panes. So there is 1 way that four glass panes are horizontal, four are vertical and there are four horizontal panes between the vertical panes. Two glass panes are horizontal, six are vertical Case 1: There are no vertical panes between the horizontal panes. There could be 0, 2, 4 or 6 vertical panes below the bottom horizontal pane. So there are 4 ways that two glass panes are horizontal, six are vertical and there are no vertical panes between the horizontal panes. Case 2: There are two vertical panes between the horizontal panes. There could be 0, 2 or 4 vertical panes below the bottom horizontal pane. So there are 3 ways that two glass panes are horizontal, six are vertical and there are two vertical panes between the horizontal panes. Case 3: There are four vertical panes between the horizontal panes. There could be 0 or 2 vertical panes below the bottom horizontal pane. So there are 2 ways that two glass panes are horizontal, six are vertical and there are four vertical panes between the horizontal panes. Case 4: There are six vertical panes between the horizontal panes. Since there are no more glass panes left, there can only can be 0 vertical panes below the bottom horizontal pane. So there is 1 way that two glass panes are horizontal, six are vertical and there are six vertical panes between the horizontal panes. All glass panes are vertical, none are horizontal This can only be done one way. Therefore, the total number of different configurations of the glass panes is ( ) + ( ) + 1 = 34. So the contractor can build 34 houses before he has to start duplicating window patterns.

100

104 Problem D Formidable Fractions? For positive integers a and c, ( a c + a ) ( 2 a + 2 ) = 18. c + 1 Determine the number of ordered pairs (a, c) that satisfy a + 3c 99.

105 Problem D and Solutions Formidable Fractions? Problem For positive integers a and c, that satisfy a + 3c 99. Solution 1 ( a c + a ) ( 2 a + 2 ) = 18. Determine the number of ordered pairs (a, c) c + 1 ( a c + a ) First we should simplify the fractional equation, ( 2 a + 2 ) = 18 c + 1 ( 2a 2c + ac 2c + 2c ) 2c Finding common denominators, ( 2c ac + 2a ac + ac ) = 18 ac ( ) 2a + ac + 2c Multiplying by the reciprocal, Simplifying, (2a + ac + 2c) 2c 2c ( ) = 18 2c + 2a + ac ac ac (2c + 2a + ac) = 18 Since the bracketed numerator and bracketed denominator are the same and cannot equal zero, we can simplify to ac = 18. Since c 0, the expression 2c further simplifies to a = 18 or a = 18(2) = 36. Substituting a = 36 into 2 a + 3c 99 we obtain c 99 which simplifies to 3c 63 and c 21 follows. But c 1 and c is an integer so 1 c 21. The value of a is 36 for each of the 21 possible values of c. there are 21 ordered pairs (a, c) that satisfy the problem. Solution 2 is on the next page.

106 Solution 2 First we should simplify the fractional equation, Multiply the numerator and the denominator by 2ac: ( a c + a ) ( 2 a + 2 ) = 18 c + 1 ( a c + a ) ( 2 a + 2 ) 2ac c + 1 2ac = 18 The equation simplifies to: Factoring, we obtain: 2a 2 + a 2 c + 2ac 4c + 4a + 2ac = 18 a(2a + ac + 2c) 2(2c + 2a + ac) = 18 Since the bracketed numerator and bracketed denominator are the same and cannot equal zero, we can simplify to a = 18 and a = 36 follows. 2 Substituting a = 36 into a + 3c 99 we obtain c 99 which simplifies to 3c 63 and c 21. But c 1 and c is an integer so 1 c 21. The value of a is 36 for each of the 21 possible values of c. there are 21 ordered pairs (a, c) that satisfy the problem.

116 Problem D Actress or Comedian - Dare to Compare Holly Woods is a popular young actress and Joe King is an up and coming young comedian. Joe has an income which is five-eighths of Holly s income. Joe s expenses are one-half those of Holly, and Joe saves 40% of his income. Determine the percentage of her income that Holly Woods saves. OR