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

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "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 11 or higher.

2

3 Problem E Mathletes in Action Four mathletes, Jen, John, JP and Judith, competed in the finals of the CEMC Mental Math Bowl. Prizes were awarded for the top three competitors as follows: first place received an anniversary Math Faculty Pink Tie, second place received a book of Brain Teasers, and third place received an I Love Math T-shirt. Three staff members of the CEMC predicted how the prizes would be awarded. Dean predicted that Judith would win the pink tie, John would win the book and JP would win the T-shirt. Ian predicted that Jen would win the pink tie, Judith would win the book and JP would win the T-shirt. Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the T-shirt. It turns out that each staff member predicted exactly one prize winner correctly. Determine precisely how the prizes were awarded.

4 Problem Problem E and Solution Mathletes in Action Four mathletes, Jen, John, JP and Judith, competed in the finals of the CEMC Mental Math Bowl. Prizes were awarded for the top three competitors as follows: first place received an anniversary Math Faculty Pink Tie, second place received a book of Brain Teasers, and third place received an I Love Math T-shirt. Three staff members of the CEMC predicted how the prizes would be awarded. Dean predicted that Judith would win the pink tie, John would win the book and JP would win the T-shirt. Ian predicted that Jen would win the pink tie, Judith would win the book and JP would win the T-shirt. Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the T-shirt. It turns out that each staff member predicted exactly one prize winner correctly. Determine precisely how the prizes were awarded. Solution Let s look at Dean s prize predictions. Assume he is correct that Judith won the pink tie. This leads to the following six possibilities. First - Pink Tie Second - Brain Book Third - T-shirt Fourth - No Prize Judith Jen John JP (1) Judith Jen JP John () Judith John Jen JP (3) Judith John JP Jen (4) Judith JP Jen John (5) Judith JP John Jen (6) Since the other two parts of Dean s prize prediction are not correct, John cannot win the book prize so we can rule out (3) and (4), and JP cannot win the T - shirt so we can also rule out (). This leaves (1), (5) and (6) as the only possibilities for Dean. First - Pink Tie Second - Brain Book Third - T-shirt Fourth - No Prize Judith Jen John JP (1) Judith JP Jen John (5) Judith JP John Jen (6) Now look at Ian s prize predictions in light of Dean s three valid possibilities. Ian makes exactly one true prize prediction. None of Dean s possibilities would ever make one of Ian s prize predictions true. Therefore, our original assumption that Dean correctly predicted Judith to win the pink tie was incorrect. So now we assume that Dean correctly predicts that John would win the book prize. This leads to the following six possibilities. First - Pink Tie Second - Brain Book Third - T-shirt Fourth - No Prize Jen John JP Judith (7) Jen John Judith JP (8) JP John Jen Judith (9) JP John Judith Jen (10) Judith John Jen JP (11) Judith John JP Jen (1)

5 Since the other two parts of Dean s prize prediction are not correct, Judith cannot win the pink tie so we can rule out (11) and (1), and JP cannot win the T - shirt so we can also rule out (7). This leaves (8), (9) and (10) as the only possibilities for Dean. First - Pink Tie Second - Brain Book Third - T-shirt Fourth - No Prize Jen John Judith JP (8) JP John Jen Judith (9) JP John Judith Jen (10) Now look at Ian s prize predictions in light of Dean s three valid possibilities. Ian makes exactly one true prize prediction. Possibility (8) is the only one of the possibilities that works for Ian. Now we must check Sandy s prize prediction to see if it is still valid with (8). Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the T-shirt. This prediction works since exactly one of Sandy s prize predictions, that Judith wins the T-shirt, is true. The other two predictions are false. We should check to see that this is the only correct solution. We do this by assuming Dean s third prediction, JP wins the T-shirt, was his only correct solution. This leads to the following six possibilities. First - Pink Tie Jen Second - Brain Book John Third - T-shirt JP Fourth - No Prize Judith (13) Jen Judith JP John (14) John Jen JP Judith (15) John Judith JP Jen (16) Judith Jen JP John (17) Judith John JP Jen (18) Since the other two parts of Dean s prize prediction are not correct, Judith cannot win the pink tie so we can rule out (17) and (18), and John cannot win the Brain Teaser Book so we can also rule out (13). This leaves (14), (15) and (16) as the only possibilities for Dean. First - Pink Tie Second - Brain Book Third - T-shirt Fourth - No Prize Jen Judith JP John (14) John Jen JP Judith (15) John Judith JP Jen (16) Now look at Ian s prize predictions in light of Dean s three valid possibilities. Ian makes exactly one true prize prediction. Possibility (15) is the only one of the possibilities that works for Ian. In both (14) and (16) Ian would make at least two correct prize predictions. Now we must check Sandy s prize prediction to see if it is still valid with (15). Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the T-shirt. Since two of Sandy s prize predictions would be true in (15) but Sandy only made one correct prize prediction, (15) does not work for Sandy. the only possibility is that Jen finishes first and wins the pink tie, John finishes second and wins the Brain Teaser book, and Judith finishes third and gets the T-shirt. JP finishes fourth and leaves with the satisfaction of making it to the finals of the Mental Math Bowl. By the way, the correct answer to the Sample Question from the CEMC Mental Math Bowl was Remember, no calculators are used to perform mental math.

6 Problem E What s Your Angle Anyway III? AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. If you need a hint, consider solving this week s Problem D, What s Your Angle Anyway II?

7 Problem E and Solution What s Your Angle Anyway III? Problem AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. Solution Join D to the centre C. Since CA, CB and CD are radii of the circle, CA = CB = CD. Since CA = CD, CAD is isosceles and CAD = CDA = x. Since CB = CD, CBD is isosceles and CBD = CDB = y. This new information is marked on the following diagram. The angles in a triangle add to 180 so in ABD ADB + DAB + DBA = 180 (x + y ) + x + y = 180 (x + y ) = 180 x + y = 90 But ADB = x + y so ADB = 90. This result is often expressed as a theorem for circles: An angle ( ADB) inscribed in a circle by the diameter (AB) of a circle is 90.

8 Problem E Partly Salted When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 salt by mass. One kilogram of water is then added to the new 3 solution resulting in a solution that is 3 10 salt by mass. What fraction of the original solution was salt? +

9 Problem E and Solution Partly Salted Problem When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 3 salt by mass. One kilogram of water is then added to the new solution resulting in a solution that is 3 10 Solution salt by mass. What fraction of the original solution was salt? Let S represent the mass of salt in the original solution. Let W represent the mass of water in the original solution. Then S + W represents the total mass of the original solution. After adding 1 kg of salt to the original solution, there is (S + 1) kg of salt and (S + W + 1) kg of solution. Since this new solution is one third salt by mass, S + 1 S + W + 1 = 1 3 3(S + 1) = 1(S + W + 1) 3S + 3 = S + W + 1 S + = W (1) After adding 1 kg of water to the new solution, there is still (S + 1) kg of salt in (S + W + ) kg of solution. Since this new solution is three tenths salt by mass, S + 1 S + W + = (S + 1) = 3(S + W + ) Substituting (1) into (), 10S + 10 = 3S + 3W + 6 7S 3W = 4 () 7S 3(S + ) = 4 7S 6S 6 = 4 S = kg Substituting for S in (1), W = 6 kg. Then the original solution was S + W = 8 kg. The fraction of the original solution that was salt was 8 = 1 4. one quarter of the original solution was salt by mass.

10 Problem E Pair Possibilities Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54.

11 Problem E and Solution Pair Possibilities Problem Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 and a + 5b 54. a + b Solution Since a and b are positive integers satisfying a + 5b 54, we could write out all ordered pairs that satisfy this inequality and then determine which ones also satisfy the first equation. There will be a large number of possibilities to check so we need to find a way to reduce the number of possibilities. Working with the first equation: Find a common denominator: b + a ab = 8 a + b Cross-multiply : (b + a)(a + b) = 8ab Expand and simplify: 4a + 4ab + b = 8ab It follows that a b = 0 and b = a. Rearrange: 4a 4ab + b = 0 Factor: (a b) = 0 Each of the ordered pairs (a, b) will look like (a, a). We substitute a for b in the inequality obtaining a + 5(a) 54 or 1a 54 or a 4.5. Since a is a positive integer, a can only take on integer values 1,, 3 and 4. The ordered pairs follow quickly: {(1, ), (, 4), (3, 6), (4, 8)}. the only ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54 are (1, ), (, 4), (3, 6), and (4, 8).

12 Problem E Get Your Analytic Skills On OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB.

13 Problem OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB. Solution Problem E and Solution Get Your Analytic Skills On In analytic geometry problems, a representative diagram is important and often provides clues for the solution of the problem. The diagram above has the given information plus a couple of pieces of information that will be justified now. Let B have coordinates (p, q). The slope of OB = q. Since AOB = p 90, OB OA and the slope of OA is the negative reciprocal of OB. Therefore the slope of OA = p. Since the q triangle is isosceles, OA = OB and it follows that the coordinates of A are ( q, p). (We can verify this by finding the length of OA and the length of OB and showing that both are equal to p + q.) Since B(p, q) is on the line x + 3y 13 = 0, it satisfies the equation of the line. p + 3q 13 = 0 (1) Since A( q, p) is on the line x + 3y 13 = 0, it satisfies the equation of the line. q + 3p 13 = 0 or 3p q 13 = 0 () Since we have two equations and two unknowns, we can use elimination to solve for p and q. (1) 4p + 6q 6 = 0 () 3 9p 6q 39 = 0 Adding, we obtain 13p 65 = 0 p = 5 Substituting in (1) q 13 = 0 3q = 3 q = 1 The point B is (5, 1) and the length of OB = = 6. Since OA = OB, OA = 6. AOB is a right triangle so we can use OB as the base and OA as the height in the formula for the area of a triangle. Then the area of AOB = OA OB = 6 6 = 13. the area of AOB is 13 units.

14 Problem E Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR.

15 Problem Problem E and Solutions Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution 1 Since P Q = P M = 7, P QM is isosceles. In P QM, draw an altitude from P to QM, intersecting at N. In an isosceles triangle, the altitude drawn to the base bisects the base. Therefore QN = NM = x. Since P M is a median in P QR, MR = QM = x. Let P N = h. P NM is a right triangle. Using Pythagoras Theorem, P N = P M NM h = 7 x h = 49 x (1) P NR is a right triangle. Using Pythagoras Theorem, P N = P R NR h = 9 (x + x) h = 81 (3x) h = 81 9x () In equations (1) and (), the left side of each equation is h. Therefore, the right side of equation (1) must equal the right side of equation (). So 49 x = 81 9x x + 9x = x = 3 x = 4 x =, x > 0 QR = QN + NM + MR = x + x + x = 4x = 8 units.

16 Problem In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution This solution is presented for students who have done some trigonometry and know the law of cosines. Since P M is a median, QM = MR = x. Then QR = x. Using the law of cosines in P QM, P M = P Q + QM (P Q)(QM)cos(Q) 7 = 7 + x (7)(x)cos(Q) 49 = 49 + x 14xcos(Q) 14xcos(Q) = x (1) Using the law of cosines in P QR, P R = P Q + QR (P Q)(QR)cos(Q) 9 = 7 + (x) (7)(x)cos(Q) 81 = x 8xcos(Q) 8xcos(Q) = 4x 3 () Using elimination to solve for x, (1) 8xcos(Q) = x () 8xcos(Q) = 4x 3 Subtracting 0 = x + 3 x = 3 x = 16 x = 4, x > 0 QR = x = 8 the length of QR is 8 units.

17 Problem E A Simple System of Equations? If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Try to see if doing this problem without a calculator affects how you think about the problem.

18 Problem E and Solutions A Simple System of Equations? Problem If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Solution 1 For this solution we will attempt to answer the question only. We will not determine the values of x and y that generate the sum. 3x = 16 y+1 3x = ( 4) y+1 3x = 4y+4 3x = 4y + 4 (1) But x = 5y 17 () (1) () x = y + 1 Rearranging x + y = 1 x + y = 1. Notice that the problem only asks for x + y, it is not necessary to find values for x and y. Solution This solution carries on from equations (1) and () in solution 1 to find the values of x and y, and then determines the sum. (1) 5 15x = 0y + 0 (3) () 4 8x = 0y 68 (4) (3) (4) 7x = 88 x = 88 ( ) 7 88 Substitue in () = 5y 17 7 Multiply by = 35y 119 As before (but with much more work) x + y = = 35y y = = 59 7 x + y = = = 1

19 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of 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. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4

20 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.

21 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.

22 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.

23 Problem E Love is Blind Valentine A heart is constructed by attaching two white semi-circles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper as illustrated below. (The dashed lines, the side measurement and the right angle symbol will not actually be on the finished card.) Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. cm

24 Problem E and Solution Love is Blind Valentine Problem A heart is constructed by attaching two white semi-circles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper. Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. Solution The semi-circles are placed along the hypotenuse of the triangle. The hypotenuse is equal in length to four radii and is therefore 4 = 8 cm long. Let x represent the length of the sides of equal length in the triangle and let h represent the height of the triangle drawn from the vertex between the two equal sides to the opposite side. Place the given information on the diagrams below. cm x h x 8 cm h x x Since the triangle is a right triangle we can use Pythagoras Theorem to find the value of x. x + x = 8 x = 64 x = 3 x = 4, since x > 0 The altitude of an isosceles, right triangle bisects the hypotenuse so we can determine the height of the triangle, h. This statement is justified at the end of the solution. h + 4 = x h + 16 = 3 h = 16 h = 4, since h > 0 We now have enough information to find the area of the heart and the entire shape.

25 The total height of the rectangle is equal to the height of the triangle plus the radius of the semi-circle. The height of the rectangle is 4 + = 6 cm. The width of the rectangle is the same as the length of the hypotenuse and is therefore equal to 8 cm. The area of the rectangle is 8 6 = 48 cm. The area of the heart is equal to the area of the triangle plus the area of two congruent semi-circles of radius cm. Since the triangle is isosceles right we can use one of the equal sides as the base and the other as the height. The area of the triangle is bh = (4 ) (4 ) = 3 = 16 cm. The area of the two semi-circles is the same as the area of one circle with radius cm. The area of the two semi-circles is πr = π() = 4π cm. The total area of the heart is (4π + 16) cm. The probability that Cupid s arrow lands on the heart is equal to the area of the heart divided by the area of the rectangle. The probability equals 4π+16 = In other words, Cupid has 48 about a 60% chance of landing his arrow on the heart. Happy Valentine s Day. Justification of the statement: The altitude of an isosceles right triangle bisects the hypotenuse. There are many ways to justify this. Since the triangle is an isosceles right triangle, the two angles opposite the sides of equal length are equal. In a triangle, the sum of the angles is 180 so the two equal angles must add to the remaining 90 and each angle is 45. Construct the altitude, h, as shown in the following diagram, splitting the opposite sides into two parts, labelled a and b. x a h b x The altitude hits the opposite side at 90 so each of the two smaller triangles contains a right angle and a 45 angle. The missing angle in each of the two smaller triangles is then 45. Each of the smaller triangles is then isosceles and it follows that h = a and h = b. a = b and the altitude bisects the hypotenuse.

26 Problem E Seeing Red A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n n n

27 Problem E and Solutions Seeing Red Problem A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n 1 n-1 A n B 1 1 n-1 n-1 C Solution 1 n This solution requires no content beyond grade ten. The second solution will use the factor theorem which is generally taught in grade twelve. In the diagram above, the sides painted red are labelled A, B, and C. We know that there are n 3 unit cubes. To determine the number of unpainted cubes we can subtract the number of cubes with some red from the total number of cubes. Side A has dimensions n by n by 1 and so contains n unit cubes with some red. Side B has dimensions n by (n 1) by 1 and so contains n (n 1) unit cubes with some red. Side C has dimensions (n 1) by (n 1) by 1 and so contains (n 1)(n 1) unit cubes with some red. The number of unpainted cubes is n 3 n n(n 1) (n 1)(n 1). We can simplify this as follows: n 3 n n(n 1) (n 1)(n 1) = n (n 1) n(n 1) (n 1)(n 1) Each term contains a common factor of (n 1) so the expression simplifies to (n 1)(n n (n 1)) = (n 1)(n n + 1). This further simplifies to (n 1) 3. If the solver pauses here to think about this, if the unit cubes on side A then side B and finally side C are removed we are left with a cube whose side lengths are (n 1) and (n 1) 3 unit cubes. But (n 1) 3 = 15, the actual number of unpainted cubes. Taking the cube root, n 1 = 5 and n = 6 follows.

28 Solution The second solution will use the factor theorem which is generally taught in grade twelve. This solution picks up from the expression giving us the number of unpainted cubes, setting it equal to 15, the number of unpainted cubes. n 3 n n(n 1) (n 1)(n 1) = 15 n 3 n n + n n + n 1 = 15 n 3 3n + 3n 16 = 0 Let f(n) = n 3 3n + 3n 16. When n = 6, f(6) = 6 3 3(6 ) + 3(6) 16 = = 4 4 = 0. Since f(6) = 0, (n 6) is a factor of f(n). After long division (or synthetic division), f(n) = (n 6)(n + 3n + 1). So (n 6)(n + 3n + 1) = 0. n + 3n + 1 = 0 has no real roots so n = 6 is the only root. Therefore the original cube has edges of length 6.

29 Problem E This Triangle is All Right A right-angled triangle has sides whose lengths are two-digit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. a c b

30 Problem E and Solution Problem This Triangle is All Right A right-angled triangle has sides whose lengths are two-digit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. Solution Let x represent the tens digit of the hypotenuse such that x is an integer from 1 to 9. Let y represent the units digit of the hypotenuse such that y is an integer from 1 to 9. Then the length of the hypotenuse is 10x + y. Since one of the sides has the same digits as the hypotenuse in reverse order, the length of this side is 10y + x. Let the third side be z such that z is a two digit integer. Using the Pythagorean Theorem: (10y + x) + z = (10x + y) Expanding: 100y + 0xy + x + z = 100x + 0xy + y Rearranging: z = 99x 99y Factoring: z = 99(x y)(x + y) Since z is a perfect square, 99(x + y)(x y) must also be a perfect square. But 99(x + y)(x y) = 9(11)(x + y)(x y). To be a perfect square (x + y)(x y) must be a multiple of 11 and contain a factor which is a perfect square. Since x and y are each integers from 1 to 9, x y cannot be 11, and so we must have x + y = 11. Again, since x and y are integers from 1 to 9, there are three possibilities for x y that give a perfect square; x y = 1, x y = 4 or x y = 9. These are the three possibilities: (x + y)(x y) = 11(1) or (x + y)(x y) = 11(4) or (x + y)(x y) = 11(9). If x + y = 11 and x y = 1, we solve the system of equations obtaining x = 6 and y = 5. This gives a hypotenuse of 10x + y = 65 and second side 10y + x = 56. Then solving for z, z = 99(x + y)(x y) = 99(11)(1) = 1089 and z = 33. This solution is easily confirmed but is it the only solution? If x + y = 11 and x y = 4, we solve the system of equations obtaining x = 7.5 and y = 3.5. But x and y are both integers so this solution is inadmissible. If x + y = 11 and x y = 9, we solve the system of equations obtaining x = 10 and y = 1. But x must be an integer from 1 to 9 so this solution is inadmissible. Therefore the only solution is a right triangle with hypotenuse of length 65.

31 Problem E Advanced Training A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains.

32 Problem E and Solution Advanced Training Problem A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains. Solution Let t be the time in hours until the total distance travelled by the two trains is equal to the distance between the trains. Since each train is travelling at 100 km/h, the distance travelled by each train in t hours is 100t km. The total distance travelled by the two trains is 100t = 00t km. The following diagram represents the positions of the two trains after t hours. The blue train starts at B and moves to C. The red train starts at R and moves to S. Then BC = RS = 100t and CR = t. We want the time t when CS = BC + RS = 100t + 100t = 00t. B 1000 km C 100t km R 100t km S But CRS is right angled so CS = CR + RS 0000t t = 0 (00t) = ( t) + (100t) 40000t = t t t t + 10t 50 = 0 Using the Quadratic Formula, t = 10 ± 100 4( 50) t = 10 ± 10 3 t = 5 ± 5 3 Since t > 0, is inadmissible. t = = 3.66 hours. The distance between the two trains will be equal to their total distance travelled in ( ) hours which is approximately 3 hours and 40 minutes after they leave their initial positions.

33 Problem E Median Madness A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. median In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC.

34 Problem E and Solution Median Madness Problem A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC. Solution A x N x B y M y C Since AM is a median, M is the midpoint of BC. Then BM = MC = y. Since CN is a median, N is the midpoint of AB. Then AN = NB = x. NBC is right angled since B = 90. Using the Pythagorean Theorem, NB + BC = CN x + (y) = ( 10) x + 4y = 40 (1) ABM is right angled since B = 90. Using the Pythagorean Theorem, AB + BM = AM (x) + y = 5 4x + y = 5 () Adding (1) and (), 5x + 5y = 65 Dividing by 5, x + y = 13 (3) The longest side of ABC is the hypotenuse AC. Using the Pythagorean Theorem, AC = AB + BC = (x) + (y) = 4x + 4y = 4(x + y ) Substituting from (3) above, AC = 4(13) Taking the square root, AC = 13 the length of the longest side is 13 units. Note: The solver could actually solve a system of equations to find x = and y = 3 and then proceed to solve for the longest side. The above approach was provided to expose the solver to alternate thinking about the solution of this problem.

35 Problem E A Not-So-Average Task Four numbers are selected such that when each number is added to the average of the other three the following sums are obtained: 5, 37, 43, and 51. Determine the average of the four numbers.

36 Problem Problem E and Solution A Not-So-Average Task Four numbers are selected such that when each number is added to the average of the other three the following sums are obtained: 5, 37, 43, and 51. Determine the average of the four numbers. Solution It is possible to precisely determine the four numbers but the problem only asks for their average. Let a, b, c, d represent each of the four numbers. We are looking for a+b+c+d 4. When the first number is added to the average of the other three numbers the result is 5. a + b + c + d = 5 which simplifies to 3a + b + c + d = 75 (1) 3 When the second number is added to the average of the other three numbers the result is 37. b + a + c + d = 37 which simplifies to a + 3b + c + d = 111 () 3 When the third number is added to the average of the other three numbers the result is 43. c + a + b + d = 43 which simplifies to a + b + 3c + d = 19 (3) 3 When the fourth number is added to the average of the other three numbers the result is 51. d + a + b + c = 51 which simplifies to a + b + c + 3d = 153 (4) 3 If we add (1), (), (3), and (4) we get 6a + 6b + 6c + 6d = = 468. Dividing by 6, a + b + c + d = 78. The average of the four numbers is 78 4 = Therefore the average of the four numbers is (By solving the system of equations we can actually determine that the numbers are 1.5, 16.5, 5.5 and 37.5.)

37 Problem E Five Solutions - Really? There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. x 1 = x = x 3 = x 4 = x 5 =

38 Problem E and Solution Five Solutions - Really? Problem There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. Solution Let s consider the ways that an expression of the form a b can be 1: The base, a, is 1. In this case, the exponent can be any value and we need to solve x 5x + 5 = 1. So x = 4 or x = 1. x 5x + 5 = 1 x 5x + 4 = 0 (x 4)(x 1) = 0 The exponent, b, is 0. In this case, the base can be any number other than 0 and we need to solve x + 4x 60 = 0. So x = 6 or x = 10. x + 4x 60 = 0 (x 6)(x + 10) = 0 When x = 6, the base is 6 5(6) + 5 = When x = 10, the base is ( 10) 5( 10) + 5 = The base, a, is 1 and the exponent, b, is even. We first need to solve x 5x + 5 = 1. So x = or x = 3. x 5x + 5 = 1 x 5x + 6 = 0 (x )(x 3) = 0 When x =, the exponent is + 4() 60 = 48, which is even. Therefore, when x =, (x 5x + 5) x +4x 60 = 1. When x = 3, the exponent is 3 + 4(3) 60 = 39, which is odd. Therefore, when x = 3, (x 5x + 5) x +4x 60 = 1. So x = 3 is not a solution. Therefore, the values of x that satisfy (x 5x + 5) x +4x 60 = 1 are x = 1, x =, x = 4, x = 6 and x = 10.

39 Problem E A Skinny Quadrilateral In the diagram, QP 1 R 1 is right-angled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. That is, determine the value of n so that the area of the quadrilateral with vertices P n, P n+1, R n+1, and R n is 01.

40 Problem E and Solution A Skinny Quadrilateral Problem In the diagram, QP 1 R 1 is right-angled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. Solution In order to solve the problem, looking at the calculation of a specific area may prove helpful. So let s determine the area of quadrilateral P 4 P 5 R 5 R 4. Area of quadrilateral P 4 P 5 R 5 R 4 = Area P 5 QR 5 Area P 4 QR 4 = 1 (QP 5)(QR 5 ) 1 (QP 4)(QR 4 ) = 1 ( + (5 1))(5 + (5 1)) 1 ( + (4 1))(5 + (4 1)) = 1 (6)(9) 1 (5)(8) = 7 0 = 7 units We will pattern the more general solution off the above example. Area of quad. P n P n+1 R n+1 R n = Area P n+1 QR n+1 Area P n QR n 01 = 1 (QP n+1)(qr n+1 ) 1 (QP n)(qr n ) 01 = 1 ( + ((n + 1) 1))(5 + ((n + 1) 1)) 1 ( + (n 1))(5 + (n 1)) 01 = 1 ( + n)(5 + n) 1 (1 + n)(4 + n) 404 = ( + n)(5 + n) (1 + n)(4 + n) multiplying by 404 = n + 7n + 10 (n + 5n + 4) 404 = n + 7n + 10 n 5n = n = n 009 = n Therefore the value of n is 009. Using the method of the specific example, this result is easily confirmed.

41

42 Problem E Powerful Factors at Work The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n.

43 Problem E and Solution Powerful Factors at Work Problem The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n. Solution The prime factorization of 1 is 3. We must determine how many times the factors 3 are repeated in the factorization of P. First we will count the number of factors of in P by looking at the 3 even numbers. Each of the numbers {, 4, 6,, 60, 6, 64} contains a factor of. That is a total of 3 factors of. Dividing the even numbers by, we obtain the numbers {1,, 3,, 30, 31, 3}. This list contains 16 even numbers so we gain another 16 factors of bringing the total to = 48. Dividing the even numbers by, we obtain the numbers {1,, 3,, 14, 15, 16}. This list contains 8 even numbers so we gain another 8 factors of bringing the total to = 56. Dividing the even numbers by, we obtain the numbers {1,, 3,, 6, 7, 8}. This list contains 4 even numbers so we gain another 4 factors of bringing the total to = 60. Dividing the even numbers by, we obtain the numbers {1,, 3, 4}. This list contains even numbers so we gain another factors of bringing the total to 60 + = 6. Finally, dividing the even numbers by, we obtain the numbers {1, }. This list contains 1 even number so we gain another factor of bringing the total to = 63. So when P is factored there are 63 factors of. In fact, the largest power of that P is divisible by is 63. Next we will count the number of factors of 3 in P by looking at the 1 multiples of 3, namely {3, 6, 9,, 57, 60, 63}. Each of these numbers 1 numbers contains a factor of 3. Dividing the numbers by 3, we obtain the numbers {1,, 3,, 19, 0, 1}. This list contains 7 multiples of 3 so we gain another 7 factors of 3 bringing the total to = 8. Dividing the multiples of 3 by 3, we obtain the numbers {1,, 3, 4, 5, 6, 7}. This list contains multiples of 3 so we gain another factors of 3 bringing the total to 8 + = 30. Dividing the multiples of 3 by 3, we obtain the numbers {1, }. This list contains no multiples of 3 so when P is factored there is a total of 30 factors of 3. In fact, the largest power of 3 that P is divisible by is Combining the results P is divisible by = = 3 ( ) = 3 (( ) 3) 30 = 3 (1) 30 P is divisble by 1 30 and the largest value of n is 30. (Since all of the powers of 3 have been used, none remain to combine with any of the remaining s to form an additional factor of 1.)

44 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of 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. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4

45 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.

46 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.

47 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.

48 Problem E Pick a Card - Any Card Luke has a unique deck of cards. Each card in the deck has a positive three digit number on it. There is exactly one card in the deck for every three digit positive integer. The deck is shuffled. Luke selects a card at random from the deck and calculates the sum of the digits. If the sum of the digits is 15, the card is a winner. Determine the probability that Luke selects a winning card.

49 Problem E and Solution Pick a Card - Any Card Problem Luke has a unique deck of cards. Each card in the deck has a positive three digit number on it. There is exactly one card in the deck for every three digit positive integer. The deck is shuffled. Luke selects a card at random from the deck and calculates the sum of the digits. If the sum of the digits is 15, the card is a winner. Determine the probability that Luke selects a winning card. Solution To begin, we need to determine the number of cards in the deck. Since there is a card for each three digit positive integer there are 900 cards in the deck. We must be careful calculating this number. There are 999 positive integers less than Of this set, 99 are two digit numbers. Therefore there are = 900 three digit positive integers. Next we need to determine the digit combinations on a card that have a sum of 15. We will determine the possibilities using cases. Then we will look at the specific groups of numbers that sum to 15 to count the number of cards produced from each group. 1. One of the digits on the card is a zero. The other two digits on the card must add to 15. This leads to two groups of numbers: (0,6,9) and (0,7,8).. One of the digits on the card is a one but the number does not contain a zero. The other two digits on the card must add to 14. This leads to three groups of numbers: (1,5,9), (1,6,8) and (1,7,7). 3. One of the digits on the card is a two but the number does not contain a zero or one. The other two digits on the card must add to 13. This leads to three groups of numbers: (,4,9), (,5,8) and (,6,7). 4. One of the digits on the card is a three but the number does not contain a zero, one or two. The other two digits on the card must add to 1. This leads to four groups of numbers: (3,3,9), (3,4,8), (3,5,7) and (3,6,6). 5. One of the digits on the card is a four but the number does not contain a zero, one, two or three. The other two digits on the card must add to 11. This leads to two groups of numbers: (4,4,7) and (4,5,6). 6. One of the digits on the card is a five but the number does not contain a zero, one, two, three or four. The other two digits on the card must add to 10. This leads to only one group of numbers: (5,5,5).

50 Now that we have the groups of numbers, we can determine the number of cards that can be created from each group of three numbers. We will do this again with cases: groups containing a zero, groups containing three distinct numbers but not zero, groups containing exactly two numbers the same but not zero, and groups containing three numbers the same but not zero. 1. One of the numbers in the group is zero. Earlier we found that there were two such groups: (0,6,9) and (0,7,8). This is a special case since zero cannot appear in the number as the hundreds digit for the number to be a three digit number. For each of the two groups of numbers, the zero can be placed in two ways, in the tens digit or the units digit. Once the zero is placed, the other two numbers can be placed in the remaining two spots in two ways. Each group can form = 4 three digit numbers. Since there are two groups, there are 4 = 8 cards in the deck that contain a zero and add to 15.. All three digits on the card are different but the number does not contain a zero. From the earlier cases there are eight groups in which all three numbers are different: (1,5,9), (1,6,8), (,4,9), (,5,8), (,6,7), (3,4,8), (3,5,7), and (4,5,6). For each of these groups, the hundreds digit can be filled three ways. For each of these three choices for hundreds digit, the tens digit can be filled two ways. Once the hundreds digit and tens digit are selected the units digit must get the third digit. So each group can form 3 = 6 different numbers. Since there are eight groups, there are 8 6 = 48 cards in the deck that contain three different digits other than zero and add to Two of the digits on the card are the same and the number does not contain a zero. From the earlier cases, there are four groups of numbers in which exactly two of the numbers in the group are the same: (1,7,7), (3,3,9), (3,6,6), and (4,4,7). For each of these groups, the unique number can be placed in one of three spots. Once the unique number is placed the other two numbers must go in the remaining two spots. So each group can form 3 different numbers. Since there are four groups, there are 4 3 = 1 cards in the deck that do not contain a zero but contain two digits the same and add to The three digits on the card are the same. From the earlier cases we discovered only one such group: (5,5,5). Only one card can be produced using the numbers from this group. Combining the counts from the above four cases, there are = 69 cards in the deck with a digit sum of 15. Therefore the probability that Luke selects a winning card is. This translates to approximately a 7.7% chance of winning. 69 = A game is considered fair if there is close to a 50% chance of winning. This game is definitely not fair. If you changed the game to if the card chosen has a sum that is divisible by 5, there is a 0% chance of winning. This is better but still not fair. If you changed the game to if the card chosen has a sum less than 14, there is a 46% chance of winning. This game is much fairer but not really fun. Can you create a game using this specific deck of cards that is reasonably fair and fun to play?

51 Problem E A Product for the Ages A mother has four children, each with a different age. The product of their ages is The sum of the ages of the three oldest children is 40 and the sum of the ages of the three youngest children is 3. Determine the ages of the four children.

52 Problem Problem E and Solution A Product for the Ages A mother has four children, each with a different age. The product of their ages is The sum of the ages of the three oldest children is 40 and the sum of the ages of the three youngest children is 3. Determine the ages of the four children. Solution Let the ages of the children from youngest to oldest be a, b, c, d. Since the ages of the three oldest children sum to 40, b + c + d = 40. (1) Since the ages of the three youngest children sum to 3, a + b + c = 3. () Subtracting () from (1), we obtain d a = 8. This means that the difference between the age of the oldest child and the age of the youngest child is 8. Now = = ( 3) ( 3) ( 3) 5 = Since all of the ages are different, this statement tells us that at least one child is over 1 and one child is under 10. There is a limited number of possibilities such that the difference between the oldest and youngest is 8 which also satisfy the condition that the youngest is under 10 and the oldest is over 1. The possibilities for youngest and oldest are (5,13), (6,14), (7,15), (8,16), and (9,17). No other combination would be possible since the oldest child must be over 1 and the youngest child must be under 10. The numbers 7, 13, and 17 are primes and are not factors of Therefore we can eliminate the possibilities where an age is one of 7, 13, or 17, leaving (6,14) and (8,16). But 14 contains a prime factor of 7 which is not a factor of Since there is only one possibility left, (8,16), we can conclude that the youngest is 8 and the oldest is 16. Now = Using the remaining factors 3 3 and 5, we need to create two numbers between 8 and 16. The only possibilities are 3 = 9 and 3 5 = 15. Therefore the ages of the children are 8, 9, 15, and 16. It is easy to verify that this is the correct solution.

53 Problem E Show Me The Money As a cashier, Igotta Job needs to give change to people when they buy merchandise. She knows that the Canadian coin system has the following greedy property: the best way to give change is to use the fewest number of coins possible. For example, if Igotta needed to provide 30 change, she would use one quarter (5 ) and one nickel (5 ), a total of two coins. Igotta would not give three dimes (3 10 ) since this option uses three coins instead of two. She would not use two dimes ( 10 ) and two nickels ( 5 ) since this option uses four coins instead of two. And she certainly would not give thirty pennies (30 1 ). Each combination would provide the correct amount of change but only the first option uses the fewest number of coins and therefore follows the greedy property. How many different sums of money, less than one dollar, can Igotta make using the greedy property and exactly four coins chosen from pennies (1 coins), nickels (5 coins), dimes (10 coins) and quarters (5 coins)? ($1 = 100 )

54 Problem Problem E and Solution Show Me The Money As a cashier, Igotta Job needs to give change to people when they buy merchandise. She knows that the Canadian coin system has the following greedy property: the best way to give change is to use the fewest number of coins possible. For example, if Igotta needed to provide 30 change, she would use one quarter (5 ) and one nickel (5 ), a total of two coins. Igotta would not give three dimes (3 10 ) since this option uses three coins instead of two. She would not use two dimes ( 10 ) and two nickels ( 5 ) since this option uses four coins instead of two. And she certainly would not give thirty pennies (30 1 ). Each combination would provide the correct amount of change but only the first option uses the fewest number of coins and therefore follows the greedy property. How many different sums of money, less than one dollar, can Igotta make using the greedy property and exactly four coins chosen from pennies (1 coins), nickels (5 coins), dimes (10 coins) and quarters (5 coins)? ($1 = 100 ) Solution For the solution, we will consider cases to carefully count the possibilities. 4 pennies There is only one way to use four pennies for change. This gives a sum of 4. 3 pennies There are three possibilities here: three pennies and one nickel giving a sum of 8, three pennies and one dime giving a sum of 13, and three pennies and one quarter giving a sum of 8. pennies and two of one other coin. This leads to two possibilities: two pennies and two dimes for a sum of, and two pennies and two quarters for a sum of 5. Note that two pennies and two nickels gives a sum of 1. This same sum can be made using fewer coins, namely two pennies and one dime, and is therefore not valid. and one of each of two other coins. This leads to three valid possibilities: two pennies, one nickel and one dime for a sum of 17, two pennies, one nickel and one quarter for a sum of 3, and two pennies, one dime and one quarter for a sum of penny and three of one other coin. This leads to one valid possibility: one penny and three quarters for a sum of 76. The possibilities one penny and three nickels, and one penny and three dimes are both invalid since these sums can be made up with fewer coins. two of a second type of coin and one of a third type of coin. This leads to three valid possibilities: one penny, one nickel and two quarters for a sum of 56, one penny, one dime and two quarters for a sum of 61, and one penny, one quarter and two dimes for a sum of 46. The sums obtained using one penny, one nickel and two dimes, or one penny, one dime and two nickels, or one penny, one quarter and two nickels can be obtained using fewer coins and hence are invalid.

55 1 penny (continued) and one of each of the other three coins. This leads to one valid possibility: one penny, one nickel, one dime and one quarter for a sum of pennies and four of one other coin. There are no valid possibilities here since each possible sum can be made up with fewer coins. Four nickels is 0. The same sum can be made using two dimes and hence fewer coins. Four dimes is 40. The same sum can be made using one quarter, one dime and one nickel, and hence fewer coins. Four quarters would give $1=100 but we are looking for sums less than one dollar so this is invalid. three of a second type of coin and one of a third type of coin. This leads to two valid possibilities: three quarters and one nickel for a sum of 80, and three quarters and one dime for a sum of 85. Using three nickels and any other coin would be invalid since three nickels could be replaced with two coins, a dime and a nickel, to produce the same sum. Using three dimes and any other coin would be invalid since three dimes could be replaced with two coins, a quarter and a nickel, to produce the same sum. two of one coin and two of another coin. This leads to one valid possibility: two dimes and two quarters for a sum of 70. Two nickels could never be paired with any other pair since two nickels can be replaced by one dime. two of one coin and one of each of the remaining two coins. This leads to one valid possibility: two quarters, one nickel, and one dime for a sum of 65. Two dimes, one nickel and one quarter gives a sum of 50. This sum can be made with fewer coins by using two quarters. Two nickels, one dime and one quarter gives a sum of 45. This sum can be made with fewer coins by using one quarter and two dimes. The total number of different sums of money is the obtained by adding the number of possibilities from each case, namely = 18 possible sums. The 18 possible sums under one dollar, using the greedy property and four coins, are 4, 8, 13, 17,, 8, 3, 37, 41, 46, 5, 56, 61, 65, 70, 76, 80, 85.

56 Problem E A Card Game To Flip Over Too 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. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up?

57 Problem Problem E and Solution A Card Game To Flip Over Too 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. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up? Solution If a card number is a multiple of, 3, 4 and 5, it will be flipped four times. This card will go from red to yellow to red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly three of, 3, 4 and 5, it will be flipped three times. This card will go from red to yellow to red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of exactly two of, 3, 4 and 5, then it will be flipped twice. This card will go from red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly one of, 3, 4 and 5, it will be flipped once. This card will go from red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of none of, 3, 4 and 5, then this card will not be flipped and so the card will still be red once E Z has finished. To determine how many cards have the red side facing up once E Z has finished, let s determine how many cards have the yellow side facing up once E Z has finished. To do so, we need to determine how many card numbers are multiples of exactly three of, 3, 4 and 5 and how many cards are multiples of exactly one of, 3, 4 and 5. Let s consider the cases: A card number is a multiple of, 3 and 4, but not 5 If a card number is a multiple of, 3 and 4, then it must be a multiple of 1, the lowest common multiple of, 3 and 4. So we are want card numbers that are multiples of 1 but not 5. If a card number is a multiple of 1 and 5, then it is a multiple of 1 5 = 60. So we want all multiples of 1 that are not multiples of 60. There are 8 multiples of 1 from 1 to 100, but one is 60. So there are 8 1 = 7 numbers that are multiples of, 3 and 4, but not 5. A card number is a multiple of, 3 and 5, but not 4 If a card number is a multiple of, 3 and 5, then it must be a multiple of 30, the lowest common multiple of, 3 and 5. So we want all multiples of 30 that are not multiples of 4. There are 3 multiples of 30 from 1 to 100, but one is 60, which is also a multiple of 4. So there are numbers from 1 to 100 that are multiples of, 3 and 5, but not 4.

58 A card number is a multiple of, 4 and 5, but not 3 If a card number is a multiple of, 4 and 5, then it must be a multiple of 0, the lowest common multiple of, 4 and 5. So we want all multiples of 0 that are not multiples of 3. There are 5 multiples of 0 from 1 to 100, but one is 60, which is a multiple of 3. So there are 4 numbers from 1 to 100 that are multiples of, 4 and 5, but not 3. A card number is a multiple of 3, 4 and 5, but not It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of but not 3, 4, or 5 There are 50 numbers from 1 to 100 which are multiples of and 5 numbers from 1 to 100 which are multiples of 4 (and thus ). So there are 50 5 = 5 numbers from 1 to 100 multiples of by but not 4. These are {, 6, 10, 14, 18,, 6, 30, 34, 38, 4, 46, 50, 54, 58, 6, 66, 70, 74, 78, 8, 86, 90, 94, 98} We need to remove numbers that are still multiples of 3 or 5. After doing so we are left with {, 14,, 6, 34, 38, 46, 58, 6, 74, 8, 86, 94, 98} So there are 14 numbers from 1 to 100 that are multiples of but not 3, 4 or 5. A card number is a multiple of 3 but not, 4, or 5 There are 33 multiples of 3 from 1 to 100, {3, 6, 9, 1, 15,, 87, 90, 93, 96, 99}. In this group of multiples, there are 17 numbers that are odd. So there are 17 numbers from 1 to 100 that are multiples of 3 but not. These numbers are {3, 9, 15, 1, 7, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99}. We still need to remove numbers that are multiples of 5. After doing so we are left with {3, 9, 1, 7, 33, 39, 51, 57, 63, 69, 81, 87, 93, 99}. So there are 14 numbers from 1 to 100 that are multiples of 3 but not, 4 or 5. A card number is a multiple of 4 but not, 3, or 5 It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of 5 but not, 3, or 4 There are 0 multiples of 5 from 1 to 100, but half of those are multiples of. The multiples of 5 which are not multiples of are {5, 15, 5, 35, 45, 55, 65, 75, 85, 95}. We still need to remove numbers that are multiples of 3. After doing so we are left with {5, 5, 35, 55, 65, 85, 95}. So there are 7 numbers from 1 to 100 that are multiples of 5 but not, 3 or 4. Therefore, once he has finished, EZ Dealer is left with 100 ( ) = = 5 cards with the red side facing up. Extension: Suppose E Z Dealer continues flipping cards in this manner. So, after he has flipped all cards whose number is a multiple of 5, he then flips all cards whose card number is a multiple of 6, then 7, then 8, and so on until he flips all cards whose number is a multiple of 100. Once E Z has finished, how many cards will have the red side facing up?

59

60 Problem E So Many Triangles P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR.

61 Problem E and Solution So Many Triangles Problem P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR. Solution On the above diagram the lengths of the equal sides, QP = P X, P R = RZ, and RQ = QY, have been marked. Join P to Y, Q to Z, and R to X. P QR and P QY have a common altitude drawn from vertex P to the line segment RY, meeting it at A. The triangles have equal base lengths, RQ = QY. area P QR = area P QY = x. At this point we can proceed to look at various other triangles with equal areas. P QY and P XY have the same height and equal base lengths. area P XY = area P QY =x. P XR and P QR have the same height and equal base lengths. area P XR = area P QR = x. P XR and RXZ have the same height and equal base lengths. area RXZ = area P XR = x. P QR and QRZ have the same height and equal base lengths. area QRZ = area P QR = x. QRZ and QY Z have the same height and equal base lengths. area QY Z = area QRZ = x. Then the area of XY Z = area P XY +area P QY +area P QR+area P XR+area RXZ +area QRZ +area QY Z = x + x + x + x + x + x + x 7x = 1176 and x = 168 cm. the area of P QR is 168 cm.

62 Problem E What s Your Angle Anyway III? AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. If you need a hint, consider solving this week s Problem D, What s Your Angle Anyway II?

63 Problem E and Solution What s Your Angle Anyway III? Problem AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. Solution Join D to the centre C. Since CA, CB and CD are radii of the circle, CA = CB = CD. Since CA = CD, CAD is isosceles and CAD = CDA = x. Since CB = CD, CBD is isosceles and CBD = CDB = y. This new information is marked on the following diagram. The angles in a triangle add to 180 so in ABD ADB + DAB + DBA = 180 (x + y ) + x + y = 180 (x + y ) = 180 x + y = 90 But ADB = x + y so ADB = 90. This result is often expressed as a theorem for circles: An angle ( ADB) inscribed in a circle by the diameter (AB) of a circle is 90.

64 Problem E Partly Salted When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 salt by mass. One kilogram of water is then added to the new 3 solution resulting in a solution that is 3 10 salt by mass. What fraction of the original solution was salt? +

65 Problem E and Solution Partly Salted Problem When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 3 salt by mass. One kilogram of water is then added to the new solution resulting in a solution that is 3 10 Solution salt by mass. What fraction of the original solution was salt? Let S represent the mass of salt in the original solution. Let W represent the mass of water in the original solution. Then S + W represents the total mass of the original solution. After adding 1 kg of salt to the original solution, there is (S + 1) kg of salt and (S + W + 1) kg of solution. Since this new solution is one third salt by mass, S + 1 S + W + 1 = 1 3 3(S + 1) = 1(S + W + 1) 3S + 3 = S + W + 1 S + = W (1) After adding 1 kg of water to the new solution, there is still (S + 1) kg of salt in (S + W + ) kg of solution. Since this new solution is three tenths salt by mass, S + 1 S + W + = (S + 1) = 3(S + W + ) Substituting (1) into (), 10S + 10 = 3S + 3W + 6 7S 3W = 4 () 7S 3(S + ) = 4 7S 6S 6 = 4 S = kg Substituting for S in (1), W = 6 kg. Then the original solution was S + W = 8 kg. The fraction of the original solution that was salt was 8 = 1 4. one quarter of the original solution was salt by mass.

66 Problem E Pair Possibilities Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54.

67 Problem E and Solution Pair Possibilities Problem Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 and a + 5b 54. a + b Solution Since a and b are positive integers satisfying a + 5b 54, we could write out all ordered pairs that satisfy this inequality and then determine which ones also satisfy the first equation. There will be a large number of possibilities to check so we need to find a way to reduce the number of possibilities. Working with the first equation: Find a common denominator: b + a ab = 8 a + b Cross-multiply : (b + a)(a + b) = 8ab Expand and simplify: 4a + 4ab + b = 8ab It follows that a b = 0 and b = a. Rearrange: 4a 4ab + b = 0 Factor: (a b) = 0 Each of the ordered pairs (a, b) will look like (a, a). We substitute a for b in the inequality obtaining a + 5(a) 54 or 1a 54 or a 4.5. Since a is a positive integer, a can only take on integer values 1,, 3 and 4. The ordered pairs follow quickly: {(1, ), (, 4), (3, 6), (4, 8)}. the only ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54 are (1, ), (, 4), (3, 6), and (4, 8).

68 Problem E Get Your Analytic Skills On OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB.

69 Problem OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB. Solution Problem E and Solution Get Your Analytic Skills On In analytic geometry problems, a representative diagram is important and often provides clues for the solution of the problem. The diagram above has the given information plus a couple of pieces of information that will be justified now. Let B have coordinates (p, q). The slope of OB = q. Since AOB = p 90, OB OA and the slope of OA is the negative reciprocal of OB. Therefore the slope of OA = p. Since the q triangle is isosceles, OA = OB and it follows that the coordinates of A are ( q, p). (We can verify this by finding the length of OA and the length of OB and showing that both are equal to p + q.) Since B(p, q) is on the line x + 3y 13 = 0, it satisfies the equation of the line. p + 3q 13 = 0 (1) Since A( q, p) is on the line x + 3y 13 = 0, it satisfies the equation of the line. q + 3p 13 = 0 or 3p q 13 = 0 () Since we have two equations and two unknowns, we can use elimination to solve for p and q. (1) 4p + 6q 6 = 0 () 3 9p 6q 39 = 0 Adding, we obtain 13p 65 = 0 p = 5 Substituting in (1) q 13 = 0 3q = 3 q = 1 The point B is (5, 1) and the length of OB = = 6. Since OA = OB, OA = 6. AOB is a right triangle so we can use OB as the base and OA as the height in the formula for the area of a triangle. Then the area of AOB = OA OB = 6 6 = 13. the area of AOB is 13 units.

70 Problem E Sphere Pressure A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube.

71 Problem E and Solution Sphere Pressure Problem A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube. Solution Label four vertices of the cube A, B, C, D as shown in the diagram. Let x represent the side length of the cube. Then AB = BC = CD = x. The diagonals of a cube intersect in a point such that the distance from the intersection point to each vertex is equal. Since each vertex of the cube touches the sphere, the diagonal of the cube, AD, is equal in length to the diameter of the sphere. Therefore AD = (6) = 1 cm. Each face of a cube is a square so ABC = 90. Using Pythagoras Theorem, AC = AB + BC = x + x = x. In a cube the sides are perpendicular to the base. In particular, DC is perpendicular to the base and it follows that DC AC. Therefore DCA is a right angled triangle. Using Pythagoras Theorem, AD = AC + CD = x + x = 3x. But AD = 1 so AD = 144. Then 3x = 144, x = 48 and x = 4 3 since x > 0. The volume of the cube is x 3 = (4 3) 3 = 19 3 cm 3. the volume of the cube is 19 3 cm 3.

72 Problem E Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR.

73 Problem Problem E and Solutions Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution 1 Since P Q = P M = 7, P QM is isosceles. In P QM, draw an altitude from P to QM, intersecting at N. In an isosceles triangle, the altitude drawn to the base bisects the base. Therefore QN = NM = x. Since P M is a median in P QR, MR = QM = x. Let P N = h. P NM is a right triangle. Using Pythagoras Theorem, P N = P M NM h = 7 x h = 49 x (1) P NR is a right triangle. Using Pythagoras Theorem, P N = P R NR h = 9 (x + x) h = 81 (3x) h = 81 9x () In equations (1) and (), the left side of each equation is h. Therefore, the right side of equation (1) must equal the right side of equation (). So 49 x = 81 9x x + 9x = x = 3 x = 4 x =, x > 0 QR = QN + NM + MR = x + x + x = 4x = 8 units.

74 Problem In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution This solution is presented for students who have done some trigonometry and know the law of cosines. Since P M is a median, QM = MR = x. Then QR = x. Using the law of cosines in P QM, P M = P Q + QM (P Q)(QM)cos(Q) 7 = 7 + x (7)(x)cos(Q) 49 = 49 + x 14xcos(Q) 14xcos(Q) = x (1) Using the law of cosines in P QR, P R = P Q + QR (P Q)(QR)cos(Q) 9 = 7 + (x) (7)(x)cos(Q) 81 = x 8xcos(Q) 8xcos(Q) = 4x 3 () Using elimination to solve for x, (1) 8xcos(Q) = x () 8xcos(Q) = 4x 3 Subtracting 0 = x + 3 x = 3 x = 16 x = 4, x > 0 QR = x = 8 the length of QR is 8 units.

75 Problem E A Simple System of Equations? If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Try to see if doing this problem without a calculator affects how you think about the problem.

76 Problem E and Solutions A Simple System of Equations? Problem If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Solution 1 For this solution we will attempt to answer the question only. We will not determine the values of x and y that generate the sum. 3x = 16 y+1 3x = ( 4) y+1 3x = 4y+4 3x = 4y + 4 (1) But x = 5y 17 () (1) () x = y + 1 Rearranging x + y = 1 x + y = 1. Notice that the problem only asks for x + y, it is not necessary to find values for x and y. Solution This solution carries on from equations (1) and () in solution 1 to find the values of x and y, and then determines the sum. (1) 5 15x = 0y + 0 (3) () 4 8x = 0y 68 (4) (3) (4) 7x = 88 x = 88 ( ) 7 88 Substitue in () = 5y 17 7 Multiply by = 35y 119 As before (but with much more work) x + y = = 35y y = = 59 7 x + y = = = 1

77 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of 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. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4

78 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.

79 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.

80 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.

81 Problem E Find The Area - No Parabolem! y B(0, 5) A(r, 0) x C(5, 0) D(p, q) In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the x-axis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC.

82 Problem Problem E and Solutions Find The Area - No Parabolem! B(0, 5) y In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the x-axis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. Solution 1 In ABC, the height is the distance from the x-axis to B(0, 5). Therefore the height is 5 units. The base is AC = 5 r. Using the formula for the area of a triangle, Area ABC = (5 r)(5) = 5. Then 5 r = and r = 3 follows. The coordinates of A are (3, 0). The axis of symmetry is a vertical line through the midpoint of AC which is (4, 0). It follows that the x-coordinate of the vertex is p = 4. Therefore, the vertex is D(4, q). Since the two x-intercepts of the parabola are 3 and 5, the equation of the parabola in factored form can be written as y = a(x 3)(x 5). Since B(0, 5) is on the parabola, we can solve for a by substituting x = 0 and y = 5 into y = a(x 3)(x 5). This leads to a = 1 3 and y = 1 3 (x 3)(x 5). To find q, the y-coordinate of D, substitute x = 4, y = q into y = 1 3 (x 3)(x 5). Then q = (4 3)(4 5) = 3. Construct a rectangle with sides parallel to the x and y-axes so that the points B, D, and C are on the rectangle. B is one vertex of the rectangle. We can determine the coordinates of the other three vertices of the rectangle. The results are shown on the diagram to the right. It is straight forward to then determine the lengths required for finding various areas: BE = = 16 3, ED = 4, DF = 1, F C = 1 3, CG = 5 and BG = 5. To find the area of DBC, subtract the areas of BED, DF C, and BCG from the area of rectangle BEF G. The area of rectangle BEF G = BE BG = = 80 3 units. The area of BED = BE ED = = 3 3 units. DF F C The area of DF C = = = 1 6 units. The area of BCG = CG BG = 5 5 = 5 units. y B(0, 5) A(r, 0) A(3, 0) D(p, q) x C(5, 0) x C(5, 0) E(0, 3) D(4, ) F(5, 3) Area DBC = Area Rectangle BEF G Area BED Area DF C Area BCG = = 10 3 units the area of DBC = 10 3 units. The second solution uses the equation of a line and is very different from the first solution. 3 G(5, 5)

83 y Problem In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the x-axis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. B(0, 5) Solution To begin this solution we will pick up a result from the first solution and do not repeat the work here. In solution 1, we found vertex D to be the point (4, 1 3 ). A(r, 0) D(p, q) x C(5, 0) Let P (t, 0) be the point where the line through B and D crosses the x-axis. We will determine the equation of the line containing B, P, and D. The slope of the line is The y-intercept of the line is 5. Therefore the equation of the line through B, P, and D is y = 4 3 x = 16 4 = 4 To find t, the x-coordinate of P we substitute x = t and y = 0 into the equation of the line. y = 4 3 x = 4 3 t + 5 4t = 15 t = 15 4 B(0, 5) y A(3, 0) P(t, 0) -1 D(4, 3) x C(5, 0) To determine the area of BDC, we find the sum of the area of BP C and the area of DP C. We will use P C as the base in both triangles. Then P C = = 5 4. In BP C, the height is the perpendicular distance from the x-axis to point B. The height is 5. The area of BP C = = 5 8 units. In DP C, the height is the perpendicular distance from the x-axis to point D. The height is 1 3. The area of DP C = = 5 4 units. the area of DBC = 10 3 units. Area DBC = Area BP C + Area DP C = = = 80 4 = 10 3 units

84 Problem E Advanced Training A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains.

85 Problem E and Solution Advanced Training Problem A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains. Solution Let t be the time in hours until the total distance travelled by the two trains is equal to the distance between the trains. Since each train is travelling at 100 km/h, the distance travelled by each train in t hours is 100t km. The total distance travelled by the two trains is 100t = 00t km. The following diagram represents the positions of the two trains after t hours. The blue train starts at B and moves to C. The red train starts at R and moves to S. Then BC = RS = 100t and CR = t. We want the time t when CS = BC + RS = 100t + 100t = 00t. B 1000 km C 100t km R 100t km S But CRS is right angled so CS = CR + RS 0000t t = 0 (00t) = ( t) + (100t) 40000t = t t t t + 10t 50 = 0 Using the Quadratic Formula, t = 10 ± 100 4( 50) t = 10 ± 10 3 t = 5 ± 5 3 Since t > 0, is inadmissible. t = = 3.66 hours. The distance between the two trains will be equal to their total distance travelled in ( ) hours which is approximately 3 hours and 40 minutes after they leave their initial positions.

86 Problem E Five Solutions - Really? There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. x 1 = x = x 3 = x 4 = x 5 =

87 Problem E and Solution Five Solutions - Really? Problem There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. Solution Let s consider the ways that an expression of the form a b can be 1: The base, a, is 1. In this case, the exponent can be any value and we need to solve x 5x + 5 = 1. So x = 4 or x = 1. x 5x + 5 = 1 x 5x + 4 = 0 (x 4)(x 1) = 0 The exponent, b, is 0. In this case, the base can be any number other than 0 and we need to solve x + 4x 60 = 0. So x = 6 or x = 10. x + 4x 60 = 0 (x 6)(x + 10) = 0 When x = 6, the base is 6 5(6) + 5 = When x = 10, the base is ( 10) 5( 10) + 5 = The base, a, is 1 and the exponent, b, is even. We first need to solve x 5x + 5 = 1. So x = or x = 3. x 5x + 5 = 1 x 5x + 6 = 0 (x )(x 3) = 0 When x =, the exponent is + 4() 60 = 48, which is even. Therefore, when x =, (x 5x + 5) x +4x 60 = 1. When x = 3, the exponent is 3 + 4(3) 60 = 39, which is odd. Therefore, when x = 3, (x 5x + 5) x +4x 60 = 1. So x = 3 is not a solution. Therefore, the values of x that satisfy (x 5x + 5) x +4x 60 = 1 are x = 1, x =, x = 4, x = 6 and x = 10.

88

89 Problem E So Many Triangles P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR.

90 Problem E and Solution So Many Triangles Problem P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR. Solution On the above diagram the lengths of the equal sides, QP = P X, P R = RZ, and RQ = QY, have been marked. Join P to Y, Q to Z, and R to X. P QR and P QY have a common altitude drawn from vertex P to the line segment RY, meeting it at A. The triangles have equal base lengths, RQ = QY. area P QR = area P QY = x. At this point we can proceed to look at various other triangles with equal areas. P QY and P XY have the same height and equal base lengths. area P XY = area P QY =x. P XR and P QR have the same height and equal base lengths. area P XR = area P QR = x. P XR and RXZ have the same height and equal base lengths. area RXZ = area P XR = x. P QR and QRZ have the same height and equal base lengths. area QRZ = area P QR = x. QRZ and QY Z have the same height and equal base lengths. area QY Z = area QRZ = x. Then the area of XY Z = area P XY +area P QY +area P QR+area P XR+area RXZ +area QRZ +area QY Z = x + x + x + x + x + x + x 7x = 1176 and x = 168 cm. the area of P QR is 168 cm.

91 Problem E Get Your Analytic Skills On OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB.

92 Problem OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB. Solution Problem E and Solution Get Your Analytic Skills On In analytic geometry problems, a representative diagram is important and often provides clues for the solution of the problem. The diagram above has the given information plus a couple of pieces of information that will be justified now. Let B have coordinates (p, q). The slope of OB = q. Since AOB = p 90, OB OA and the slope of OA is the negative reciprocal of OB. Therefore the slope of OA = p. Since the q triangle is isosceles, OA = OB and it follows that the coordinates of A are ( q, p). (We can verify this by finding the length of OA and the length of OB and showing that both are equal to p + q.) Since B(p, q) is on the line x + 3y 13 = 0, it satisfies the equation of the line. p + 3q 13 = 0 (1) Since A( q, p) is on the line x + 3y 13 = 0, it satisfies the equation of the line. q + 3p 13 = 0 or 3p q 13 = 0 () Since we have two equations and two unknowns, we can use elimination to solve for p and q. (1) 4p + 6q 6 = 0 () 3 9p 6q 39 = 0 Adding, we obtain 13p 65 = 0 p = 5 Substituting in (1) q 13 = 0 3q = 3 q = 1 The point B is (5, 1) and the length of OB = = 6. Since OA = OB, OA = 6. AOB is a right triangle so we can use OB as the base and OA as the height in the formula for the area of a triangle. Then the area of AOB = OA OB = 6 6 = 13. the area of AOB is 13 units.

93 Problem E Sphere Pressure A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube.

94 Problem E and Solution Sphere Pressure Problem A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube. Solution Label four vertices of the cube A, B, C, D as shown in the diagram. Let x represent the side length of the cube. Then AB = BC = CD = x. The diagonals of a cube intersect in a point such that the distance from the intersection point to each vertex is equal. Since each vertex of the cube touches the sphere, the diagonal of the cube, AD, is equal in length to the diameter of the sphere. Therefore AD = (6) = 1 cm. Each face of a cube is a square so ABC = 90. Using Pythagoras Theorem, AC = AB + BC = x + x = x. In a cube the sides are perpendicular to the base. In particular, DC is perpendicular to the base and it follows that DC AC. Therefore DCA is a right angled triangle. Using Pythagoras Theorem, AD = AC + CD = x + x = 3x. But AD = 1 so AD = 144. Then 3x = 144, x = 48 and x = 4 3 since x > 0. The volume of the cube is x 3 = (4 3) 3 = 19 3 cm 3. the volume of the cube is 19 3 cm 3.

95 Problem E Find The Area - No Parabolem! y B(0, 5) A(r, 0) x C(5, 0) D(p, q) In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the x-axis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC.

96 Problem Problem E and Solutions Find The Area - No Parabolem! B(0, 5) y In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the x-axis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. Solution 1 In ABC, the height is the distance from the x-axis to B(0, 5). Therefore the height is 5 units. The base is AC = 5 r. Using the formula for the area of a triangle, Area ABC = (5 r)(5) = 5. Then 5 r = and r = 3 follows. The coordinates of A are (3, 0). The axis of symmetry is a vertical line through the midpoint of AC which is (4, 0). It follows that the x-coordinate of the vertex is p = 4. Therefore, the vertex is D(4, q). Since the two x-intercepts of the parabola are 3 and 5, the equation of the parabola in factored form can be written as y = a(x 3)(x 5). Since B(0, 5) is on the parabola, we can solve for a by substituting x = 0 and y = 5 into y = a(x 3)(x 5). This leads to a = 1 3 and y = 1 3 (x 3)(x 5). To find q, the y-coordinate of D, substitute x = 4, y = q into y = 1 3 (x 3)(x 5). Then q = (4 3)(4 5) = 3. Construct a rectangle with sides parallel to the x and y-axes so that the points B, D, and C are on the rectangle. B is one vertex of the rectangle. We can determine the coordinates of the other three vertices of the rectangle. The results are shown on the diagram to the right. It is straight forward to then determine the lengths required for finding various areas: BE = = 16 3, ED = 4, DF = 1, F C = 1 3, CG = 5 and BG = 5. To find the area of DBC, subtract the areas of BED, DF C, and BCG from the area of rectangle BEF G. The area of rectangle BEF G = BE BG = = 80 3 units. The area of BED = BE ED = = 3 3 units. DF F C The area of DF C = = = 1 6 units. The area of BCG = CG BG = 5 5 = 5 units. y B(0, 5) A(r, 0) A(3, 0) D(p, q) x C(5, 0) x C(5, 0) E(0, 3) D(4, ) F(5, 3) Area DBC = Area Rectangle BEF G Area BED Area DF C Area BCG = = 10 3 units the area of DBC = 10 3 units. The second solution uses the equation of a line and is very different from the first solution. 3 G(5, 5)

97 y Problem In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the x-axis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. B(0, 5) Solution To begin this solution we will pick up a result from the first solution and do not repeat the work here. In solution 1, we found vertex D to be the point (4, 1 3 ). A(r, 0) D(p, q) x C(5, 0) Let P (t, 0) be the point where the line through B and D crosses the x-axis. We will determine the equation of the line containing B, P, and D. The slope of the line is The y-intercept of the line is 5. Therefore the equation of the line through B, P, and D is y = 4 3 x = 16 4 = 4 To find t, the x-coordinate of P we substitute x = t and y = 0 into the equation of the line. y = 4 3 x = 4 3 t + 5 4t = 15 t = 15 4 B(0, 5) y A(3, 0) P(t, 0) -1 D(4, 3) x C(5, 0) To determine the area of BDC, we find the sum of the area of BP C and the area of DP C. We will use P C as the base in both triangles. Then P C = = 5 4. In BP C, the height is the perpendicular distance from the x-axis to point B. The height is 5. The area of BP C = = 5 8 units. In DP C, the height is the perpendicular distance from the x-axis to point D. The height is 1 3. The area of DP C = = 5 4 units. the area of DBC = 10 3 units. Area DBC = Area BP C + Area DP C = = = 80 4 = 10 3 units

98 Problem E Love is Blind Valentine A heart is constructed by attaching two white semi-circles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper as illustrated below. (The dashed lines, the side measurement and the right angle symbol will not actually be on the finished card.) Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. cm

99 Problem E and Solution Love is Blind Valentine Problem A heart is constructed by attaching two white semi-circles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper. Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. Solution The semi-circles are placed along the hypotenuse of the triangle. The hypotenuse is equal in length to four radii and is therefore 4 = 8 cm long. Let x represent the length of the sides of equal length in the triangle and let h represent the height of the triangle drawn from the vertex between the two equal sides to the opposite side. Place the given information on the diagrams below. cm x h x 8 cm h x x Since the triangle is a right triangle we can use Pythagoras Theorem to find the value of x. x + x = 8 x = 64 x = 3 x = 4, since x > 0 The altitude of an isosceles, right triangle bisects the hypotenuse so we can determine the height of the triangle, h. This statement is justified at the end of the solution. h + 4 = x h + 16 = 3 h = 16 h = 4, since h > 0 We now have enough information to find the area of the heart and the entire shape.

100 The total height of the rectangle is equal to the height of the triangle plus the radius of the semi-circle. The height of the rectangle is 4 + = 6 cm. The width of the rectangle is the same as the length of the hypotenuse and is therefore equal to 8 cm. The area of the rectangle is 8 6 = 48 cm. The area of the heart is equal to the area of the triangle plus the area of two congruent semi-circles of radius cm. Since the triangle is isosceles right we can use one of the equal sides as the base and the other as the height. The area of the triangle is bh = (4 ) (4 ) = 3 = 16 cm. The area of the two semi-circles is the same as the area of one circle with radius cm. The area of the two semi-circles is πr = π() = 4π cm. The total area of the heart is (4π + 16) cm. The probability that Cupid s arrow lands on the heart is equal to the area of the heart divided by the area of the rectangle. The probability equals 4π+16 = In other words, Cupid has 48 about a 60% chance of landing his arrow on the heart. Happy Valentine s Day. Justification of the statement: The altitude of an isosceles right triangle bisects the hypotenuse. There are many ways to justify this. Since the triangle is an isosceles right triangle, the two angles opposite the sides of equal length are equal. In a triangle, the sum of the angles is 180 so the two equal angles must add to the remaining 90 and each angle is 45. Construct the altitude, h, as shown in the following diagram, splitting the opposite sides into two parts, labelled a and b. x a h b x The altitude hits the opposite side at 90 so each of the two smaller triangles contains a right angle and a 45 angle. The missing angle in each of the two smaller triangles is then 45. Each of the smaller triangles is then isosceles and it follows that h = a and h = b. a = b and the altitude bisects the hypotenuse.

101 Problem E Seeing Red A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n n n

102 Problem E and Solutions Seeing Red Problem A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n 1 n-1 A n B 1 1 n-1 n-1 C Solution 1 n This solution requires no content beyond grade ten. The second solution will use the factor theorem which is generally taught in grade twelve. In the diagram above, the sides painted red are labelled A, B, and C. We know that there are n 3 unit cubes. To determine the number of unpainted cubes we can subtract the number of cubes with some red from the total number of cubes. Side A has dimensions n by n by 1 and so contains n unit cubes with some red. Side B has dimensions n by (n 1) by 1 and so contains n (n 1) unit cubes with some red. Side C has dimensions (n 1) by (n 1) by 1 and so contains (n 1)(n 1) unit cubes with some red. The number of unpainted cubes is n 3 n n(n 1) (n 1)(n 1). We can simplify this as follows: n 3 n n(n 1) (n 1)(n 1) = n (n 1) n(n 1) (n 1)(n 1) Each term contains a common factor of (n 1) so the expression simplifies to (n 1)(n n (n 1)) = (n 1)(n n + 1). This further simplifies to (n 1) 3. If the solver pauses here to think about this, if the unit cubes on side A then side B and finally side C are removed we are left with a cube whose side lengths are (n 1) and (n 1) 3 unit cubes. But (n 1) 3 = 15, the actual number of unpainted cubes. Taking the cube root, n 1 = 5 and n = 6 follows.

103 Solution The second solution will use the factor theorem which is generally taught in grade twelve. This solution picks up from the expression giving us the number of unpainted cubes, setting it equal to 15, the number of unpainted cubes. n 3 n n(n 1) (n 1)(n 1) = 15 n 3 n n + n n + n 1 = 15 n 3 3n + 3n 16 = 0 Let f(n) = n 3 3n + 3n 16. When n = 6, f(6) = 6 3 3(6 ) + 3(6) 16 = = 4 4 = 0. Since f(6) = 0, (n 6) is a factor of f(n). After long division (or synthetic division), f(n) = (n 6)(n + 3n + 1). So (n 6)(n + 3n + 1) = 0. n + 3n + 1 = 0 has no real roots so n = 6 is the only root. Therefore the original cube has edges of length 6.

104 Problem E This Triangle is All Right A right-angled triangle has sides whose lengths are two-digit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. a c b

105 Problem E and Solution Problem This Triangle is All Right A right-angled triangle has sides whose lengths are two-digit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. Solution Let x represent the tens digit of the hypotenuse such that x is an integer from 1 to 9. Let y represent the units digit of the hypotenuse such that y is an integer from 1 to 9. Then the length of the hypotenuse is 10x + y. Since one of the sides has the same digits as the hypotenuse in reverse order, the length of this side is 10y + x. Let the third side be z such that z is a two digit integer. Using the Pythagorean Theorem: (10y + x) + z = (10x + y) Expanding: 100y + 0xy + x + z = 100x + 0xy + y Rearranging: z = 99x 99y Factoring: z = 99(x y)(x + y) Since z is a perfect square, 99(x + y)(x y) must also be a perfect square. But 99(x + y)(x y) = 9(11)(x + y)(x y). To be a perfect square (x + y)(x y) must be a multiple of 11 and contain a factor which is a perfect square. Since x and y are each integers from 1 to 9, x y cannot be 11, and so we must have x + y = 11. Again, since x and y are integers from 1 to 9, there are three possibilities for x y that give a perfect square; x y = 1, x y = 4 or x y = 9. These are the three possibilities: (x + y)(x y) = 11(1) or (x + y)(x y) = 11(4) or (x + y)(x y) = 11(9). If x + y = 11 and x y = 1, we solve the system of equations obtaining x = 6 and y = 5. This gives a hypotenuse of 10x + y = 65 and second side 10y + x = 56. Then solving for z, z = 99(x + y)(x y) = 99(11)(1) = 1089 and z = 33. This solution is easily confirmed but is it the only solution? If x + y = 11 and x y = 4, we solve the system of equations obtaining x = 7.5 and y = 3.5. But x and y are both integers so this solution is inadmissible. If x + y = 11 and x y = 9, we solve the system of equations obtaining x = 10 and y = 1. But x must be an integer from 1 to 9 so this solution is inadmissible. Therefore the only solution is a right triangle with hypotenuse of length 65.

106 Problem E Median Madness A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. median In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC.

107 Problem E and Solution Median Madness Problem A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC. Solution A x N x B y M y C Since AM is a median, M is the midpoint of BC. Then BM = MC = y. Since CN is a median, N is the midpoint of AB. Then AN = NB = x. NBC is right angled since B = 90. Using the Pythagorean Theorem, NB + BC = CN x + (y) = ( 10) x + 4y = 40 (1) ABM is right angled since B = 90. Using the Pythagorean Theorem, AB + BM = AM (x) + y = 5 4x + y = 5 () Adding (1) and (), 5x + 5y = 65 Dividing by 5, x + y = 13 (3) The longest side of ABC is the hypotenuse AC. Using the Pythagorean Theorem, AC = AB + BC = (x) + (y) = 4x + 4y = 4(x + y ) Substituting from (3) above, AC = 4(13) Taking the square root, AC = 13 the length of the longest side is 13 units. Note: The solver could actually solve a system of equations to find x = and y = 3 and then proceed to solve for the longest side. The above approach was provided to expose the solver to alternate thinking about the solution of this problem.

108 Problem E A Skinny Quadrilateral In the diagram, QP 1 R 1 is right-angled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. That is, determine the value of n so that the area of the quadrilateral with vertices P n, P n+1, R n+1, and R n is 01.

109 Problem E and Solution A Skinny Quadrilateral Problem In the diagram, QP 1 R 1 is right-angled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. Solution In order to solve the problem, looking at the calculation of a specific area may prove helpful. So let s determine the area of quadrilateral P 4 P 5 R 5 R 4. Area of quadrilateral P 4 P 5 R 5 R 4 = Area P 5 QR 5 Area P 4 QR 4 = 1 (QP 5)(QR 5 ) 1 (QP 4)(QR 4 ) = 1 ( + (5 1))(5 + (5 1)) 1 ( + (4 1))(5 + (4 1)) = 1 (6)(9) 1 (5)(8) = 7 0 = 7 units We will pattern the more general solution off the above example. Area of quad. P n P n+1 R n+1 R n = Area P n+1 QR n+1 Area P n QR n 01 = 1 (QP n+1)(qr n+1 ) 1 (QP n)(qr n ) 01 = 1 ( + ((n + 1) 1))(5 + ((n + 1) 1)) 1 ( + (n 1))(5 + (n 1)) 01 = 1 ( + n)(5 + n) 1 (1 + n)(4 + n) 404 = ( + n)(5 + n) (1 + n)(4 + n) multiplying by 404 = n + 7n + 10 (n + 5n + 4) 404 = n + 7n + 10 n 5n = n = n 009 = n Therefore the value of n is 009. Using the method of the specific example, this result is easily confirmed.

110

111 Problem E Hosers Nine large hoses can fill a swimming pool in four hours and six small hoses can fill the same swimming pool in eight hours. How long will it take four large hoses and eight small hoses working together to fill the swimming pool?

112 Problem Problem E and Solutions Hosers Nine large hoses can fill a swimming pool in four hours and six small hoses can fill the same swimming pool in eight hours. How long will it take four large hoses and eight small hoses working together to fill the swimming pool? Solution 1 9 large hoses can fill 1 swimming pool in 4 hours. 9 large hoses can fill 1 of the swimming pool in 1 hour. 4 1 large hose can fill 1 1 = 1 of the swimming pool in 1 hour small hoses can fill 1 swimming pool in 8 hours. 6 small hoses can fill 1 of the swimming pool in 1 hour. 8 1 small hose can fill 1 1 = 1 of the swimming pool in 1 hour In one hour, 4 large hoses and 8 small hoses fill 4 ( ) ( ) 48 = = + 3 = 5 of the pool Since it is 5 full in 1 hour, 18 the pool will be completely full in 18 5 Solution = large hoses can fill 1 swimming pool in 4 hours. 1 large hose can fill large hoses can fill 1 9 of the swimming pool in 4 hours. of the swimming pool in 1 hour. 6 small hoses can fill 1 swimming pool in 8 hours. 1 small hose can fill small hoses can fill 1 6 of the swimming pool in 8 hours. of the swimming pool in 1 hour. Together in one hour, 4 large hoses and 8 small hoses fill = + 3 = 5 of the pool Since it is 5 full in 1 hour, 18 the pool will be completely full in 18 5 = or 3 hours and 36 minutes. or 3 hours and 36 minutes. with four large hoses and eight small hoses the pool can be filled in 3 hours and 36 minutes.

113 Problem E Partly Salted When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 salt by mass. One kilogram of water is then added to the new 3 solution resulting in a solution that is 3 10 salt by mass. What fraction of the original solution was salt? +

114 Problem E and Solution Partly Salted Problem When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 3 salt by mass. One kilogram of water is then added to the new solution resulting in a solution that is 3 10 Solution salt by mass. What fraction of the original solution was salt? Let S represent the mass of salt in the original solution. Let W represent the mass of water in the original solution. Then S + W represents the total mass of the original solution. After adding 1 kg of salt to the original solution, there is (S + 1) kg of salt and (S + W + 1) kg of solution. Since this new solution is one third salt by mass, S + 1 S + W + 1 = 1 3 3(S + 1) = 1(S + W + 1) 3S + 3 = S + W + 1 S + = W (1) After adding 1 kg of water to the new solution, there is still (S + 1) kg of salt in (S + W + ) kg of solution. Since this new solution is three tenths salt by mass, S + 1 S + W + = (S + 1) = 3(S + W + ) Substituting (1) into (), 10S + 10 = 3S + 3W + 6 7S 3W = 4 () 7S 3(S + ) = 4 7S 6S 6 = 4 S = kg Substituting for S in (1), W = 6 kg. Then the original solution was S + W = 8 kg. The fraction of the original solution that was salt was 8 = 1 4. one quarter of the original solution was salt by mass.

115 Problem E Five Prime Mates The product of five different odd prime numbers is a five-digit number of the form strst, where r = 0. Determine all possible numbers. See the next page for a summary of divisibility tests for the integers to 1.

116 Divisibility Tests Divisibility by : A number is divisible by if the last digit is even. Divisibility by 3: A number is divisible by 3 if the sum of the digits is divisible by 3. For example, 195 is not divisible by 3 since = 17 which is not divisible by 3. However, 196 is divisible by 3 since = 18 which is divisible by 3. Divisibility by 4: A number is divisible by 4 if the last two digits are divisible by 4. For example, 195 is not divisible by 4 since 95 is not divisible by 4. However, 196 is divisible by 4 since 96 is divisible by 4. Divisibility by 5: A number is divisible by 5 if the last digit is a 0 or 5. Divisibility by 6: A number is divisible by 6 if it is divisible by both and 3. The number 395 is not divisible by 6 since it is not even and hence is not divisible by. The number 86 is not divisible by 6 since it is not divisible by 3 ( = 16 which is not divisible by 3). The number 964 is divisible by 6. It is even and is therefore divisible by. It is divisible by 3 since = 1 which is divisible by 3. Since 964 is divisible by both and 3, it is divisible by 6. Divisibility by 7: We can follow an unusual algorithm to determine if an number is divisible by 7: Remove the unit s digit, double that digit and subtract it from the leftover number. If the difference is divisible by 7, the original number is divisible by seven. If unsure, repeat the algorithm with the new number. Is 1356 divisible by 7? Remove the 6, double the 6 to 1, subtract from 135 leaving 13. Is 13 divisible by 7? Remove the 3, double the 3 to 6, subtract from 1 leaving 6. 6 is not divisible by 7 and therefore 1356 is not divisible by 7. Is divisible by 7? Remove the 4, double to 8, subtract from 450 giving Repeat. Remove the 4, double to 8, subtract from 449 giving 441. Repeat. Remove the 1, double to, subtract from 44 giving 4 which is divisible by 7. Therefore, is divisible by 7. Divisibility by 8: A number is divisible by 8 if the last three digits are divisible by 8. For example, 195 is not divisible by 8 since 95 is not divisible by 8. However, 196 is divisible by 8 since 96 is divisible by 8. Divisibility by 9: A number is divisible by 9 if the sum of the digits is divisible by 9. For example, 195 is not divisible by 9 since = 17 which is not divisible by 9. However, 196 is divisible by 9 since = 18 which is divisible by 9. Divisibility by 10: A number is divisible by 10 if the last digit is a 0. Divisibility by 11: We can follow an unusual algorithm to determine if an number is divisible by 11: Add the numbers in the even positions. Add the numbers in the odd positions. Subtract the two sums. If this difference is divisible by 11, the original number is divisible by 11. Is divisible by 11? The sums are = 14 and = 13. The difference of the sums is 1, which is not divisible by 11. Therefore, the number is not divisible by 11. Is divisible by 11? The sums are = 5 and = 7. The difference of the sums is, which is divisible by 11. Therefore, the number is divisible by 11. Is divisible by 11? The sums are = 18 and = 18. The difference of the sums is 0, which is divisible by 11. Therefore, the number is divisible by 11. Divisibility by 1: A number is divisible by 1 if it is divisible by 4 and 3. The number 394 is not divisible by 1 since 94 is not divisible by 4. The number 964 is not divisible by 1 since it is not divisible by 3. (The sum of the digits is 19 which is not divisible by 3.) The number 964 is divisible by 1. The last two digits, 64, are divisible by 4 and therefore 964 is divisible by 4. The sum of the digits is 1 which is divisible by 3. Since 964 is divisible by 4 and 3, it is divisible by 1.

117 Problem E and Solutions Five Prime Mates Problem The product of five different odd prime numbers is a five-digit number of the form strst, where r = 0. Determine all possible numbers. Solution 1 Be sure to read Solution after Solution 1 A number is divisible by 11 if the difference between the sum of the even position numbers and sum of the odd position numbers is a multiple of 11. (This problem can still be done without knowing this divisibility fact but the task is made simpler with it.) The sum of the digits in the even positions is strst is t + s. The sum of the digits in the odd positions is s + r + t but r = 0 so the sum is s + t. The difference of the two sums is (t + s) (s + t) = 0 which is a multiple of 11. Therefore, st0st is divisible by 11. Only odd factors are used so the product will be odd. This means that the product looks like 10 1, 30 3, 50 5, 70 7, or So we begin to systematically look at the possibilities. First, we will examine numbers that have 3 (and 11) as a factor. To be divisible by three, the sum of the digits will be divisible by three. To be divisible by nine, the sum of the digits will be divisible by nine. But if the number is divisible by nine, it is divisible by 3 and would have a repeated prime factor which is not allowed. So we want numbers divisible by 3 but not 9. The possibilities are as follows: 101, 51051, 33033, 93093, 15015, 75075, 57057, 87087, 39039, and The sum of the digits of these numbers is divisible by three so the numbers are divisible by three. The numbers 81081, 63063, 45045, 707 and are divisible by 9 and have therefore been eliminated. Now we examine the prime factorization of each of these numbers to see which numbers satisfy the conditions. 101 = = = Since the prime factor 7 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since the prime factor 11 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number.

118 15015 = = = Since there are 5 different odd prime factors, is a valid number = = = = Since the prime factor 5 is repeated and there are six prime factors, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since the prime factor 13 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number. Second, we will examine numbers that are divisible by 5 but not 3, since divisibility by three has been examined. If a number is divisible by 5 it ends in 5 or 0. Since the number is odd, we can exclude any number ending in 0 leaving 505, 35035, 55055, 65065, and as possible numbers. (15015, 45045, were examined above and have been excluded.) Now we examine the prime factorization of each of these numbers to see which numbers satisfy the conditions. 505 = = = Since the prime factor 5 is repeated, this is not a valid number = = = Since the prime factor 7 is repeated, this is not a valid number = = = Since the prime factor 11 is repeated, this is not a valid number = = = Since the prime factor 13 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since there are 5 different odd prime factors, is a valid number. Thirdly, we will look at numbers that are divisible by 7 but not 3 or 5. If we multiply 7 by the next four odd prime numbers we get = 3333, a six digit number so we are beyond all possible solutions. Therefore there are 8 numbers of the form st0st which are the product of five different odd prime numbers, namely 51051, 93093, 15015, 57057, 87087, 69069, and Look at Solution for a much more insightful approach to the problem.

119 Problem The product of five different odd prime numbers is a five-digit number of the form strst, where r = 0. Determine the number of possible numbers. Solution Since r = 0, the number is of the form st0st. Then st0st = st(1000) + st = st( ) = st(1001) This means that the number st0st is divisible by 1001 which is the product of the three odd prime factors 7, 11, and 13. So st is a two digit number which is the product of two different odd prime factors none of which can be 7, 11 or 13. It is now a straight forward matter of generating all possible two digit products, a b say, using odd prime factors other than 7, 11 and 13. Prime Factor Prime Factor st Five Different Product a b = a b Odd Primes st0st , 5, 7, 11, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, No other two digit product of two different odd prime factors other than 7, 11 and 13 exists. Therefore there are 8 numbers of the form st0st which are the product of five different odd prime numbers, namely 15015, 51051, 57057, 69069, 87087, 93093, and Look on the following page for a summary of the divisibility tests for the numbers through 1.

120 Divisibility Tests Divisibility by : A number is divisible by if the last digit is even. Divisibility by 3: A number is divisible by 3 if the sum of the digits is divisible by 3. For example, 195 is not divisible by 3 since = 17 which is not divisible by 3. However, 196 is divisible by 3 since = 18 which is divisible by 3. Divisibility by 4: A number is divisible by 4 if the last two digits are divisible by 4. For example, 195 is not divisible by 4 since 95 is not divisible by 4. However, 196 is divisible by 4 since 96 is divisible by 4. Divisibility by 5: A number is divisible by 5 if the last digit is a 0 or 5. Divisibility by 6: A number is divisible by 6 if it is divisible by both and 3. The number 395 is not divisible by 6 since it is not even and hence is not divisible by. The number 86 is not divisible by 6 since it is not divisible by 3 ( = 16 which is not divisible by 3). The number 964 is divisible by 6. It is even and is therefore divisible by. It is divisible by 3 since = 1 which is divisible by 3. Since 964 is divisible by both and 3, it is divisible by 6. Divisibility by 7: We can follow an unusual algorithm to determine if an number is divisible by 7: Remove the unit s digit, double that digit and subtract it from the leftover number. If the difference is divisible by 7, the original number is divisible by seven. If unsure, repeat the algorithm with the new number. Is 1356 divisible by 7? Remove the 6, double the 6 to 1, subtract from 135 leaving 13. Is 13 divisible by 7? Remove the 3, double the 3 to 6, subtract from 1 leaving 6. 6 is not divisible by 7 and therefore 1356 is not divisible by 7. Is divisible by 7? Remove the 4, double to 8, subtract from 450 giving Repeat. Remove the 4, double to 8, subtract from 449 giving 441. Repeat. Remove the 1, double to, subtract from 44 giving 4 which is divisible by 7. Therefore, is divisible by 7. Divisibility by 8: A number is divisible by 8 if the last three digits are divisible by 8. For example, 195 is not divisible by 8 since 95 is not divisible by 8. However, 196 is divisible by 8 since 96 is divisible by 8. Divisibility by 9: A number is divisible by 9 if the sum of the digits is divisible by 9. For example, 195 is not divisible by 9 since = 17 which is not divisible by 9. However, 196 is divisible by 9 since = 18 which is divisible by 9. Divisibility by 10: A number is divisible by 10 if the last digit is a 0. Divisibility by 11: We can follow an unusual algorithm to determine if an number is divisible by 11: Add the numbers in the even positions. Add the numbers in the odd positions. Subtract the two sums. If this difference is divisible by 11, the original number is divisible by 11. Is divisible by 11? The sums are = 14 and = 13. The difference of the sums is 1, which is not divisible by 11. Therefore, the number is not divisible by 11. Is divisible by 11? The sums are = 5 and = 7. The difference of the sums is, which is divisible by 11. Therefore, the number is divisible by 11. Is divisible by 11? The sums are = 18 and = 18. The difference of the sums is 0, which is divisible by 11. Therefore, the number is divisible by 11. Divisibility by 1: A number is divisible by 1 if it is divisible by 4 and 3. The number 394 is not divisible by 1 since 94 is not divisible by 4. The number 964 is not divisible by 1 since it is not divisible by 3. (The sum of the digits is 19 which is not divisible by 3.) The number 964 is divisible by 1. The last two digits, 64, are divisible by 4 and therefore 964 is divisible by 4. The sum of the digits is 1 which is divisible by 3. Since 964 is divisible by 4 and 3, it is divisible by 1.

121 Problem E Powerfully Perfect Squares The prime factorization of 0 is 5. The divisors of 0 are: = 1, = 5, =, = 10, 5 0 = 4, and 5 1 = 0. The number 0 has 6 divisors. Two of the divisors, 1 and 4, are perfect squares. How many divisors of are perfect squares?

122 Problem E and Solution Powerfully Perfect Squares Problem The prime factorization of 0 is 5. The divisors of 0 are = 1, = 5, =, = 10, 5 0 = 4, and 5 1 = 0. The number 0 has 6 divisors. Two of the divisors, 1 and 4, are perfect squares. How many divisors of are perfect squares? Solution First, let s look at the prime factorization of some perfect squares. 9 = 3, 16 = 4, 36 = 3, and 144 = 4 3 are the prime factorizations of four perfect squares. Note that the exponent on each of the 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. Now = ( 503) 01 = ( ) = All divisors of will be of the form k 503 n, 0 k 404, 0 n 01 where k and n are both integers. For k to be a perfect square, k must be an even integer such that 0 k 404. There are 404 = 01 even numbers from 1 to 404. The number 0 is also even so there are 013 values of k such that k is a perfect square. For 503 n to be a perfect square, n must be an even integer such that 0 n 01. There are 1006 even numbers from 1 to 01. Zero is also even so there are 1007 values of n such that 503 n is a perfect square. For each of the 013 values of k, there are 1007 values of n so there are = perfect square divisors of has perfect square divisors. (For a smaller number like 1 6 we could also determine the number of perfect square divisors using the above approach. We could verify that the approach is valid by physically listing all of the divisors and then counting the number that are perfect squares. The number 1 6 has only 91 divisors, 8 of which are perfect squares. It is not practical to list all the divisors of and then count the perfect squares. But our approach allows us to still count the perfect square divisors.)

123 Problem E Powerful Factors at Work The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n.

124 Problem E and Solution Powerful Factors at Work Problem The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n. Solution The prime factorization of 1 is 3. We must determine how many times the factors 3 are repeated in the factorization of P. First we will count the number of factors of in P by looking at the 3 even numbers. Each of the numbers {, 4, 6,, 60, 6, 64} contains a factor of. That is a total of 3 factors of. Dividing the even numbers by, we obtain the numbers {1,, 3,, 30, 31, 3}. This list contains 16 even numbers so we gain another 16 factors of bringing the total to = 48. Dividing the even numbers by, we obtain the numbers {1,, 3,, 14, 15, 16}. This list contains 8 even numbers so we gain another 8 factors of bringing the total to = 56. Dividing the even numbers by, we obtain the numbers {1,, 3,, 6, 7, 8}. This list contains 4 even numbers so we gain another 4 factors of bringing the total to = 60. Dividing the even numbers by, we obtain the numbers {1,, 3, 4}. This list contains even numbers so we gain another factors of bringing the total to 60 + = 6. Finally, dividing the even numbers by, we obtain the numbers {1, }. This list contains 1 even number so we gain another factor of bringing the total to = 63. So when P is factored there are 63 factors of. In fact, the largest power of that P is divisible by is 63. Next we will count the number of factors of 3 in P by looking at the 1 multiples of 3, namely {3, 6, 9,, 57, 60, 63}. Each of these numbers 1 numbers contains a factor of 3. Dividing the numbers by 3, we obtain the numbers {1,, 3,, 19, 0, 1}. This list contains 7 multiples of 3 so we gain another 7 factors of 3 bringing the total to = 8. Dividing the multiples of 3 by 3, we obtain the numbers {1,, 3, 4, 5, 6, 7}. This list contains multiples of 3 so we gain another factors of 3 bringing the total to 8 + = 30. Dividing the multiples of 3 by 3, we obtain the numbers {1, }. This list contains no multiples of 3 so when P is factored there is a total of 30 factors of 3. In fact, the largest power of 3 that P is divisible by is Combining the results P is divisible by = = 3 ( ) = 3 (( ) 3) 30 = 3 (1) 30 P is divisble by 1 30 and the largest value of n is 30. (Since all of the powers of 3 have been used, none remain to combine with any of the remaining s to form an additional factor of 1.)

125 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of 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. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4

126 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.

127 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.

128 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.

129 Problem E Minimum Multiplier The number 1867 is multiplied by a positive integer k. The last four digits of the product are 199. Determine the minimum value of k ??????

130 Problem E and Solution Minimum Multiplier Problem The number 1867 is multiplied by a positive integer k. The last four digits of the product are 199. Determine the minimum value of k ?????? Solution To begin with we will show that k has four digits or less. A number with five digits, pqrst for example, can be written p qrst = p qrst. The digit p in the multiplier cannot affect the final four digits in the product. Therefore the minimum k is a number with four or fewer digits. Let the multiplier be abcd such that 1867 abcd is a number whose last four digits are 199. Then multiplying 7, the units digit of 1867, by d, the units digit in abcd, produces a number ending in. The only possible value for d is 6 since 7 6 = 4. (Note the possible last digits when 7 multiplies a single digit number: 7 0 = 0, 7 1 = 7, 7 = 14, 7 3 = 1, 7 4 = 8, 7 5 = 35, 7 6 = 4, 7 7 = 49, 7 8 = 56, 7 9 = 63.) Therefore the multiplier is abc6. The second last digit in the product 199 is 9. This digit is produced by multiplying 67 from 1867 with c6 from abc c c. 9 So 0 + 7c is a number that ends in 9. The only possible value for c is 7. (Refer back to the product list given above.) Therefore the multiplier is a number of the form ab76. Is k = 76? The product, = , does not end in 199. So k is at least a three digit number. The third last digit in the product 199 is 9. This digit is produced by multiplying 867 from 1867 with b76 from ab b b So b is a number that ends in 9 and it follows that 7b is a number that ends in 1. The only possible value for b is 3. (Refer back to the product list given above.) Therefore the multiplier is a number of the form a376. Is k = 376? The product = does end in 199. So 376 multiplied by 1867 produces a number ending in 199. the smallest value of k is 376.

131 Problem E Seeing Red A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n n n

132 Problem E and Solutions Seeing Red Problem A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n 1 n-1 A n B 1 1 n-1 n-1 C Solution 1 n This solution requires no content beyond grade ten. The second solution will use the factor theorem which is generally taught in grade twelve. In the diagram above, the sides painted red are labelled A, B, and C. We know that there are n 3 unit cubes. To determine the number of unpainted cubes we can subtract the number of cubes with some red from the total number of cubes. Side A has dimensions n by n by 1 and so contains n unit cubes with some red. Side B has dimensions n by (n 1) by 1 and so contains n (n 1) unit cubes with some red. Side C has dimensions (n 1) by (n 1) by 1 and so contains (n 1)(n 1) unit cubes with some red. The number of unpainted cubes is n 3 n n(n 1) (n 1)(n 1). We can simplify this as follows: n 3 n n(n 1) (n 1)(n 1) = n (n 1) n(n 1) (n 1)(n 1) Each term contains a common factor of (n 1) so the expression simplifies to (n 1)(n n (n 1)) = (n 1)(n n + 1). This further simplifies to (n 1) 3. If the solver pauses here to think about this, if the unit cubes on side A then side B and finally side C are removed we are left with a cube whose side lengths are (n 1) and (n 1) 3 unit cubes. But (n 1) 3 = 15, the actual number of unpainted cubes. Taking the cube root, n 1 = 5 and n = 6 follows.

133 Solution The second solution will use the factor theorem which is generally taught in grade twelve. This solution picks up from the expression giving us the number of unpainted cubes, setting it equal to 15, the number of unpainted cubes. n 3 n n(n 1) (n 1)(n 1) = 15 n 3 n n + n n + n 1 = 15 n 3 3n + 3n 16 = 0 Let f(n) = n 3 3n + 3n 16. When n = 6, f(6) = 6 3 3(6 ) + 3(6) 16 = = 4 4 = 0. Since f(6) = 0, (n 6) is a factor of f(n). After long division (or synthetic division), f(n) = (n 6)(n + 3n + 1). So (n 6)(n + 3n + 1) = 0. n + 3n + 1 = 0 has no real roots so n = 6 is the only root. Therefore the original cube has edges of length 6.

134 Problem E This Triangle is All Right A right-angled triangle has sides whose lengths are two-digit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. a c b

135 Problem E and Solution Problem This Triangle is All Right A right-angled triangle has sides whose lengths are two-digit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. Solution Let x represent the tens digit of the hypotenuse such that x is an integer from 1 to 9. Let y represent the units digit of the hypotenuse such that y is an integer from 1 to 9. Then the length of the hypotenuse is 10x + y. Since one of the sides has the same digits as the hypotenuse in reverse order, the length of this side is 10y + x. Let the third side be z such that z is a two digit integer. Using the Pythagorean Theorem: (10y + x) + z = (10x + y) Expanding: 100y + 0xy + x + z = 100x + 0xy + y Rearranging: z = 99x 99y Factoring: z = 99(x y)(x + y) Since z is a perfect square, 99(x + y)(x y) must also be a perfect square. But 99(x + y)(x y) = 9(11)(x + y)(x y). To be a perfect square (x + y)(x y) must be a multiple of 11 and contain a factor which is a perfect square. Since x and y are each integers from 1 to 9, x y cannot be 11, and so we must have x + y = 11. Again, since x and y are integers from 1 to 9, there are three possibilities for x y that give a perfect square; x y = 1, x y = 4 or x y = 9. These are the three possibilities: (x + y)(x y) = 11(1) or (x + y)(x y) = 11(4) or (x + y)(x y) = 11(9). If x + y = 11 and x y = 1, we solve the system of equations obtaining x = 6 and y = 5. This gives a hypotenuse of 10x + y = 65 and second side 10y + x = 56. Then solving for z, z = 99(x + y)(x y) = 99(11)(1) = 1089 and z = 33. This solution is easily confirmed but is it the only solution? If x + y = 11 and x y = 4, we solve the system of equations obtaining x = 7.5 and y = 3.5. But x and y are both integers so this solution is inadmissible. If x + y = 11 and x y = 9, we solve the system of equations obtaining x = 10 and y = 1. But x must be an integer from 1 to 9 so this solution is inadmissible. Therefore the only solution is a right triangle with hypotenuse of length 65.

136 Problem E A Card Game To Flip Over Too 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. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up?

137 Problem Problem E and Solution A Card Game To Flip Over Too 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. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up? Solution If a card number is a multiple of, 3, 4 and 5, it will be flipped four times. This card will go from red to yellow to red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly three of, 3, 4 and 5, it will be flipped three times. This card will go from red to yellow to red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of exactly two of, 3, 4 and 5, then it will be flipped twice. This card will go from red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly one of, 3, 4 and 5, it will be flipped once. This card will go from red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of none of, 3, 4 and 5, then this card will not be flipped and so the card will still be red once E Z has finished. To determine how many cards have the red side facing up once E Z has finished, let s determine how many cards have the yellow side facing up once E Z has finished. To do so, we need to determine how many card numbers are multiples of exactly three of, 3, 4 and 5 and how many cards are multiples of exactly one of, 3, 4 and 5. Let s consider the cases: A card number is a multiple of, 3 and 4, but not 5 If a card number is a multiple of, 3 and 4, then it must be a multiple of 1, the lowest common multiple of, 3 and 4. So we are want card numbers that are multiples of 1 but not 5. If a card number is a multiple of 1 and 5, then it is a multiple of 1 5 = 60. So we want all multiples of 1 that are not multiples of 60. There are 8 multiples of 1 from 1 to 100, but one is 60. So there are 8 1 = 7 numbers that are multiples of, 3 and 4, but not 5. A card number is a multiple of, 3 and 5, but not 4 If a card number is a multiple of, 3 and 5, then it must be a multiple of 30, the lowest common multiple of, 3 and 5. So we want all multiples of 30 that are not multiples of 4. There are 3 multiples of 30 from 1 to 100, but one is 60, which is also a multiple of 4. So there are numbers from 1 to 100 that are multiples of, 3 and 5, but not 4.

138 A card number is a multiple of, 4 and 5, but not 3 If a card number is a multiple of, 4 and 5, then it must be a multiple of 0, the lowest common multiple of, 4 and 5. So we want all multiples of 0 that are not multiples of 3. There are 5 multiples of 0 from 1 to 100, but one is 60, which is a multiple of 3. So there are 4 numbers from 1 to 100 that are multiples of, 4 and 5, but not 3. A card number is a multiple of 3, 4 and 5, but not It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of but not 3, 4, or 5 There are 50 numbers from 1 to 100 which are multiples of and 5 numbers from 1 to 100 which are multiples of 4 (and thus ). So there are 50 5 = 5 numbers from 1 to 100 multiples of by but not 4. These are {, 6, 10, 14, 18,, 6, 30, 34, 38, 4, 46, 50, 54, 58, 6, 66, 70, 74, 78, 8, 86, 90, 94, 98} We need to remove numbers that are still multiples of 3 or 5. After doing so we are left with {, 14,, 6, 34, 38, 46, 58, 6, 74, 8, 86, 94, 98} So there are 14 numbers from 1 to 100 that are multiples of but not 3, 4 or 5. A card number is a multiple of 3 but not, 4, or 5 There are 33 multiples of 3 from 1 to 100, {3, 6, 9, 1, 15,, 87, 90, 93, 96, 99}. In this group of multiples, there are 17 numbers that are odd. So there are 17 numbers from 1 to 100 that are multiples of 3 but not. These numbers are {3, 9, 15, 1, 7, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99}. We still need to remove numbers that are multiples of 5. After doing so we are left with {3, 9, 1, 7, 33, 39, 51, 57, 63, 69, 81, 87, 93, 99}. So there are 14 numbers from 1 to 100 that are multiples of 3 but not, 4 or 5. A card number is a multiple of 4 but not, 3, or 5 It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of 5 but not, 3, or 4 There are 0 multiples of 5 from 1 to 100, but half of those are multiples of. The multiples of 5 which are not multiples of are {5, 15, 5, 35, 45, 55, 65, 75, 85, 95}. We still need to remove numbers that are multiples of 3. After doing so we are left with {5, 5, 35, 55, 65, 85, 95}. So there are 7 numbers from 1 to 100 that are multiples of 5 but not, 3 or 4. Therefore, once he has finished, EZ Dealer is left with 100 ( ) = = 5 cards with the red side facing up. Extension: Suppose E Z Dealer continues flipping cards in this manner. So, after he has flipped all cards whose number is a multiple of 5, he then flips all cards whose card number is a multiple of 6, then 7, then 8, and so on until he flips all cards whose number is a multiple of 100. Once E Z has finished, how many cards will have the red side facing up?

139 Problem E A Not-So-Average Task Four numbers are selected such that when each number is added to the average of the other three the following sums are obtained: 5, 37, 43, and 51. Determine the average of the four numbers.

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

The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in 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. Problem D Card Logic Four playing

More information

Geometry Unit 7 (Textbook Chapter 9) Solving a right triangle: Find all missing sides and all missing angles

Geometry Unit 7 (Textbook Chapter 9) Solving a right triangle: Find all missing sides and all missing angles Geometry Unit 7 (Textbook Chapter 9) Name Objective 1: Right Triangles and Pythagorean Theorem In many geometry problems, it is necessary to find a missing side or a missing angle of a right triangle.

More information

EXPONENTS. To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS

EXPONENTS. To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires

More information

If a question asks you to find all or list all and you think there are none, write None.

If a question asks you to find all or list all and you think there are none, write None. If a question asks you to find all or list all and you think there are none, write None 1 Simplify 1/( 1 3 1 4 ) 2 The price of an item increases by 10% and then by another 10% What is the overall price

More information

Park Forest Math Team. Meet #5. Algebra. Self-study Packet

Park Forest Math Team. Meet #5. Algebra. Self-study Packet Park Forest Math Team Meet #5 Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements 3. Number

More information

4. An isosceles triangle has two sides of length 10 and one of length 12. What is its area?

4. An isosceles triangle has two sides of length 10 and one of length 12. What is its area? 1 1 2 + 1 3 + 1 5 = 2 The sum of three numbers is 17 The first is 2 times the second The third is 5 more than the second What is the value of the largest of the three numbers? 3 A chemist has 100 cc of

More information

1. By how much does 1 3 of 5 2 exceed 1 2 of 1 3? 2. What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3.

1. By how much does 1 3 of 5 2 exceed 1 2 of 1 3? 2. What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3. 1 By how much does 1 3 of 5 exceed 1 of 1 3? What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3 A ticket fee was $10, but then it was reduced The number of customers

More information

SAT Math Facts & Formulas Review Quiz

SAT Math Facts & Formulas Review Quiz Test your knowledge of SAT math facts, formulas, and vocabulary with the following quiz. Some questions are more challenging, just like a few of the questions that you ll encounter on the SAT; these questions

More information

10-4 Inscribed Angles. Find each measure. 1.

10-4 Inscribed Angles. Find each measure. 1. Find each measure. 1. 3. 2. intercepted arc. 30 Here, is a semi-circle. So, intercepted arc. So, 66 4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If, what

More information

ALGEBRA READINESS DIAGNOSTIC TEST PRACTICE

ALGEBRA READINESS DIAGNOSTIC TEST PRACTICE 1 ALGEBRA READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the examples, work the problems, then check your answers at the end of each topic. If you don t get the answer given, check your work and

More information

CSU Fresno Problem Solving Session. Geometry, 17 March 2012

CSU Fresno Problem Solving Session. Geometry, 17 March 2012 CSU Fresno Problem Solving Session Problem Solving Sessions website: http://zimmer.csufresno.edu/ mnogin/mfd-prep.html Math Field Day date: Saturday, April 21, 2012 Math Field Day website: http://www.csufresno.edu/math/news

More information

Geometry Notes PERIMETER AND AREA

Geometry Notes PERIMETER AND AREA Perimeter and Area Page 1 of 17 PERIMETER AND AREA Objectives: After completing this section, you should be able to do the following: Calculate the area of given geometric figures. Calculate the perimeter

More information

Mathematics & the PSAT

Mathematics & the PSAT Mathematics & the PSAT Mathematics 2 Sections 25 minutes each Section 2 20 multiple-choice questions Section 4 8 multiple-choice PLUS 10 grid-in questions Calculators are permitted - BUT You must think

More information

16. Let S denote the set of positive integers 18. Thus S = {1, 2,..., 18}. How many subsets of S have sum greater than 85? (You may use symbols such

16. Let S denote the set of positive integers 18. Thus S = {1, 2,..., 18}. How many subsets of S have sum greater than 85? (You may use symbols such æ. Simplify 2 + 3 + 4. 2. A quart of liquid contains 0% alcohol, and another 3-quart bottle full of liquid contains 30% alcohol. They are mixed together. What is the percentage of alcohol in the mixture?

More information

Dyffryn School Ysgol Y Dyffryn Mathematics Faculty

Dyffryn School Ysgol Y Dyffryn Mathematics Faculty Dyffryn School Ysgol Y Dyffryn Mathematics Faculty Formulae and Facts Booklet Higher Tier Number Facts Sum This means add. Difference This means take away. Product This means multiply. Share This means

More information

Solutions 1998 Gauss Contest - Grade 8. Canadian Mathematics Competition

Solutions 1998 Gauss Contest - Grade 8. Canadian Mathematics Competition s 1998 Gauss Contest - Grade 8 5th Anniversary 196 1998 Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario

More information

Grade 7/8 Math Circles March 28, 2012 Gauss Contest Preparation

Grade 7/8 Math Circles March 28, 2012 Gauss Contest Preparation 1 University of Waterloo Faculty of Mathematics Centre for Education in Mathematics and Computing Grade 7/8 Math Circles March 28, 2012 Gauss Contest Preparation Solutions: 1. Evaluate each of the following:

More information

MATHEMATICS GRADE LEVEL VOCABULARY DRAWN FROM SBAC ITEM SPECIFICATIONS VERSION 1.1 JUNE 18, 2014

MATHEMATICS GRADE LEVEL VOCABULARY DRAWN FROM SBAC ITEM SPECIFICATIONS VERSION 1.1 JUNE 18, 2014 VERSION 1.1 JUNE 18, 2014 MATHEMATICS GRADE LEVEL VOCABULARY DRAWN FROM SBAC ITEM SPECIFICATIONS PRESENTED BY: WASHINGTON STATE REGIONAL MATH COORDINATORS Smarter Balanced Vocabulary - From SBAC test/item

More information

GCSE Maths Linear Higher Tier Grade Descriptors

GCSE Maths Linear Higher Tier Grade Descriptors GSE Maths Linear Higher Tier escriptors Fractions /* Find one quantity as a fraction of another Solve problems involving fractions dd and subtract fractions dd and subtract mixed numbers Multiply and divide

More information

DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle.

DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle. DEFINITIONS Degree A degree is the 1 th part of a straight angle. 180 Right Angle A 90 angle is called a right angle. Perpendicular Two lines are called perpendicular if they form a right angle. Congruent

More information

Section 9-1. Basic Terms: Tangents, Arcs and Chords Homework Pages 330-331: 1-18

Section 9-1. Basic Terms: Tangents, Arcs and Chords Homework Pages 330-331: 1-18 Chapter 9 Circles Objectives A. Recognize and apply terms relating to circles. B. Properly use and interpret the symbols for the terms and concepts in this chapter. C. Appropriately apply the postulates,

More information

PERIMETER AND AREA OF PLANE FIGURES

PERIMETER AND AREA OF PLANE FIGURES PERIMETER AND AREA OF PLANE FIGURES Q.. Find the area of a triangle whose sides are 8 cm, 4 cm and 30 cm. Also, find the length of altitude corresponding to the largest side of the triangle. Ans. Let ABC

More information

Answer: = π cm. Solution:

Answer: = π cm. Solution: Round #1, Problem A: (4 points/10 minutes) The perimeter of a semicircular region in centimeters is numerically equal to its area in square centimeters. What is the radius of the semicircle in centimeters?

More information

1 foot (ft) = 12 inches (in) 1 yard (yd) = 3 feet (ft) 1 mile (mi) = 5280 feet (ft) Replace 1 with 1 ft/12 in. 1ft

1 foot (ft) = 12 inches (in) 1 yard (yd) = 3 feet (ft) 1 mile (mi) = 5280 feet (ft) Replace 1 with 1 ft/12 in. 1ft 2 MODULE 6. GEOMETRY AND UNIT CONVERSION 6a Applications The most common units of length in the American system are inch, foot, yard, and mile. Converting from one unit of length to another is a requisite

More information

Instructions for SA Completion

Instructions for SA Completion Instructions for SA Completion 1- Take notes on these Pythagorean Theorem Course Materials then do and check the associated practice questions for an explanation on how to do the Pythagorean Theorem Substantive

More information

UK Junior Mathematical Olympiad 2016

UK Junior Mathematical Olympiad 2016 UK Junior Mathematical Olympiad 206 Organised by The United Kingdom Mathematics Trust Tuesday 4th June 206 RULES AND GUIDELINES : READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING. Time allowed: 2 hours.

More information

4th Grade Competition Solutions

4th Grade Competition Solutions 4th Grade Competition Solutions Bergen County Academies Math Competition 19 October 008 1. Before taking the AMC, a student notices that he has two bags of Doritos and one bag of Skittles on his desk.

More information

11 8 = 11 (8) = = 100(2) = 10 2

11 8 = 11 (8) = = 100(2) = 10 2 Chapter 5 Radical Expressions and Equations Section 5.1 Working With Radicals Section 5.1 Page 78 Question 1 Mixed Radical Form Entire Radical Form 4 7 4 7 4 (7) 11 50 5() 5 50 11 8 11 8 11 (8) 968 00

More information

Expression. Variable Equation Polynomial Monomial Add. Area. Volume Surface Space Length Width. Probability. Chance Random Likely Possibility Odds

Expression. Variable Equation Polynomial Monomial Add. Area. Volume Surface Space Length Width. Probability. Chance Random Likely Possibility Odds Isosceles Triangle Congruent Leg Side Expression Equation Polynomial Monomial Radical Square Root Check Times Itself Function Relation One Domain Range Area Volume Surface Space Length Width Quantitative

More information

Identifying Triangles 5.5

Identifying Triangles 5.5 Identifying Triangles 5.5 Name Date Directions: Identify the name of each triangle below. If the triangle has more than one name, use all names. 1. 5. 2. 6. 3. 7. 4. 8. 47 Answer Key Pages 19 and 20 Name

More information

Grade 6 Math Circles March 24/25, 2015 Pythagorean Theorem Solutions

Grade 6 Math Circles March 24/25, 2015 Pythagorean Theorem Solutions Faculty of Mathematics Waterloo, Ontario NL 3G1 Centre for Education in Mathematics and Computing Grade 6 Math Circles March 4/5, 015 Pythagorean Theorem Solutions Triangles: They re Alright When They

More information

2010 Solutions. a + b. a + b 1. (a + b)2 + (b a) 2. (b2 + a 2 ) 2 (a 2 b 2 ) 2

2010 Solutions. a + b. a + b 1. (a + b)2 + (b a) 2. (b2 + a 2 ) 2 (a 2 b 2 ) 2 00 Problem If a and b are nonzero real numbers such that a b, compute the value of the expression ( ) ( b a + a a + b b b a + b a ) ( + ) a b b a + b a +. b a a b Answer: 8. Solution: Let s simplify the

More information

Parallel and Perpendicular. We show a small box in one of the angles to show that the lines are perpendicular.

Parallel and Perpendicular. We show a small box in one of the angles to show that the lines are perpendicular. CONDENSED L E S S O N. Parallel and Perpendicular In this lesson you will learn the meaning of parallel and perpendicular discover how the slopes of parallel and perpendicular lines are related use slopes

More information

Conjectures. Chapter 2. Chapter 3

Conjectures. Chapter 2. Chapter 3 Conjectures Chapter 2 C-1 Linear Pair Conjecture If two angles form a linear pair, then the measures of the angles add up to 180. (Lesson 2.5) C-2 Vertical Angles Conjecture If two angles are vertical

More information

BASIC GEOMETRY GLOSSARY

BASIC GEOMETRY GLOSSARY BASIC GEOMETRY GLOSSARY Acute angle An angle that measures between 0 and 90. Examples: Acute triangle A triangle in which each angle is an acute angle. Adjacent angles Two angles next to each other that

More information

3. If AC = 12, CD = 9 and BE = 3, find the area of trapezoid BCDE. (Mathcounts Handbooks)

3. If AC = 12, CD = 9 and BE = 3, find the area of trapezoid BCDE. (Mathcounts Handbooks) EXERCISES: Triangles 1 1. The perimeter of an equilateral triangle is units. How many units are in the length 27 of one side? (Mathcounts Competitions) 2. In the figure shown, AC = 4, CE = 5, DE = 3, and

More information

Geometry Unit 5 Practice Test Solutions

Geometry Unit 5 Practice Test Solutions Geometry Unit 5 Practice Test Solutions Problems 1 and 2 Similar Triangles Similar triangles are triangles that are the same shape, but with different sizes. When two triangles are similar, their sides

More information

NCERT. Area of the circular path formed by two concentric circles of radii. Area of the sector of a circle of radius r with central angle θ =

NCERT. Area of the circular path formed by two concentric circles of radii. Area of the sector of a circle of radius r with central angle θ = AREA RELATED TO CIRCLES (A) Main Concepts and Results CHAPTER 11 Perimeters and areas of simple closed figures. Circumference and area of a circle. Area of a circular path (i.e., ring). Sector of a circle

More information

Geometry B Solutions. Written by Ante Qu

Geometry B Solutions. Written by Ante Qu Geometry B Solutions Written by Ante Qu 1. [3] During chemistry labs, we oftentimes fold a disk-shaped filter paper twice, and then open up a flap of the quartercircle to form a cone shape, as in the diagram.

More information

SAMPLE QUESTION PAPER (Set - II) Summative Assessment II. Class-X (2015 16) Mathematics. Time: 3 hours M. M. : 90. Section A

SAMPLE QUESTION PAPER (Set - II) Summative Assessment II. Class-X (2015 16) Mathematics. Time: 3 hours M. M. : 90. Section A SAMPLE QUESTION PAPER (Set - II) Summative Assessment II Class-X (2015 16) Mathematics Time: 3 hours M. M. : 90 General Instructions: 1. All questions are compulsory. 2. The question paper consists of

More information

MATH LEVEL 1 ARITHMETIC (ACCUPLACER)

MATH LEVEL 1 ARITHMETIC (ACCUPLACER) MATH LEVEL ARITHMETIC (ACCUPLACER) 7 Questions This test measures your ability to perform basic arithmetic operations and to solve problems that involve fundamental arithmetic concepts. There are 7 questions

More information

If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?

If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question

More information

Geometry A Solutions. Written by Ante Qu

Geometry A Solutions. Written by Ante Qu Geometry A Solutions Written by Ante Qu 1. [3] Three circles, with radii of 1, 1, and, are externally tangent to each other. The minimum possible area of a quadrilateral that contains and is tangent to

More information

MATHS LEVEL DESCRIPTORS

MATHS LEVEL DESCRIPTORS MATHS LEVEL DESCRIPTORS Number Level 3 Understand the place value of numbers up to thousands. Order numbers up to 9999. Round numbers to the nearest 10 or 100. Understand the number line below zero, and

More information

(a) 5 square units. (b) 12 square units. (c) 5 3 square units. 3 square units. (d) 6. (e) 16 square units

(a) 5 square units. (b) 12 square units. (c) 5 3 square units. 3 square units. (d) 6. (e) 16 square units 1. Find the area of parallelogram ACD shown below if the measures of segments A, C, and DE are 6 units, 2 units, and 1 unit respectively and AED is a right angle. (a) 5 square units (b) 12 square units

More information

SAT Math Hard Practice Quiz. 5. How many integers between 10 and 500 begin and end in 3?

SAT Math Hard Practice Quiz. 5. How many integers between 10 and 500 begin and end in 3? SAT Math Hard Practice Quiz Numbers and Operations 5. How many integers between 10 and 500 begin and end in 3? 1. A bag contains tomatoes that are either green or red. The ratio of green tomatoes to red

More information

2004 Solutions Ga lois Contest (Grade 10)

2004 Solutions Ga lois Contest (Grade 10) Canadian Mathematics Competition An activity of The Centre for Education in Ma thematics and Computing, University of W aterloo, Wa terloo, Ontario 2004 Solutions Ga lois Contest (Grade 10) 2004 Waterloo

More information

Placement Test Review Materials for

Placement Test Review Materials for Placement Test Review Materials for 1 To The Student This workbook will provide a review of some of the skills tested on the COMPASS placement test. Skills covered in this workbook will be used on the

More information

Winter 2016 Math 213 Final Exam. Points Possible. Subtotal 100. Total 100

Winter 2016 Math 213 Final Exam. Points Possible. Subtotal 100. Total 100 Winter 2016 Math 213 Final Exam Name Instructions: Show ALL work. Simplify wherever possible. Clearly indicate your final answer. Problem Number Points Possible Score 1 25 2 25 3 25 4 25 Subtotal 100 Extra

More information

How do you compare numbers? On a number line, larger numbers are to the right and smaller numbers are to the left.

How do you compare numbers? On a number line, larger numbers are to the right and smaller numbers are to the left. The verbal answers to all of the following questions should be memorized before completion of pre-algebra. Answers that are not memorized will hinder your ability to succeed in algebra 1. Number Basics

More information

Each pair of opposite sides of a parallelogram is congruent to each other.

Each pair of opposite sides of a parallelogram is congruent to each other. Find the perimeter and area of each parallelogram or triangle. Round to the nearest tenth if necessary. 1. Use the Pythagorean Theorem to find the height h, of the parallelogram. 2. Each pair of opposite

More information

PERIMETER AND AREA. In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures.

PERIMETER AND AREA. In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. PERIMETER AND AREA In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. Perimeter Perimeter The perimeter of a polygon, denoted by P, is the

More information

3 5th Anniversary

3 5th Anniversary th Anniversar 96 998 Canadian Mathematics Competition An activit of The Centre for Education in Mathematics and Computing, Universit of Waterloo, Waterloo, Ontario 998 s Fermat Contest (Grade ) for the

More information

Grade 7/8 Math Circles Greek Constructions - Solutions October 6/7, 2015

Grade 7/8 Math Circles Greek Constructions - Solutions October 6/7, 2015 Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 7/8 Math Circles Greek Constructions - Solutions October 6/7, 2015 Mathematics Without Numbers The

More information

ROUND TOSSUP: What is the cost of five cans of soup at 98 per can? 6. TOSSUP: What is the formula for the volume of a right circular cylinder?

ROUND TOSSUP: What is the cost of five cans of soup at 98 per can? 6. TOSSUP: What is the formula for the volume of a right circular cylinder? ROUND 1 1. TOSSUP: The fraction 3 is equivalent to what percent? 4 BONUS: (a) How much money is 10,000% of one penny? (b) What is your weekly salary if you get a 5% increase, and your previous wages were

More information

Remember that the information below is always provided on the formula sheet at the start of your exam paper

Remember that the information below is always provided on the formula sheet at the start of your exam paper Maths GCSE Linear HIGHER Things to Remember Remember that the information below is always provided on the formula sheet at the start of your exam paper In addition to these formulae, you also need to learn

More information

Higher Geometry Problems

Higher Geometry Problems Higher Geometry Problems ( Look up Eucidean Geometry on Wikipedia, and write down the English translation given of each of the first four postulates of Euclid. Rewrite each postulate as a clear statement

More information

Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers.

Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers. Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used

More information

San Jose Math Circle April 25 - May 2, 2009 ANGLE BISECTORS

San Jose Math Circle April 25 - May 2, 2009 ANGLE BISECTORS San Jose Math Circle April 25 - May 2, 2009 ANGLE BISECTORS Recall that the bisector of an angle is the ray that divides the angle into two congruent angles. The most important results about angle bisectors

More information

11-1 Areas of Parallelograms and Triangles. Find the perimeter and area of each parallelogram or triangle. Round to the nearest tenth if necessary.

11-1 Areas of Parallelograms and Triangles. Find the perimeter and area of each parallelogram or triangle. Round to the nearest tenth if necessary. Find the perimeter and area of each parallelogram or triangle. Round to the nearest tenth if necessary. 2. 1. Use the Pythagorean Theorem to find the height h, of the parallelogram. Each pair of opposite

More information

National 4: Expressions and Formulae

National 4: Expressions and Formulae Biggar High School Mathematics Department National 4 Pupil Booklet National 4: Expressions and Formulae Learning Intention Success Criteria I can simplify algebraic expressions. I can simplify and carry

More information

MATH Fundamental Mathematics II.

MATH Fundamental Mathematics II. MATH 10032 Fundamental Mathematics II http://www.math.kent.edu/ebooks/10032/fun-math-2.pdf Department of Mathematical Sciences Kent State University December 29, 2008 2 Contents 1 Fundamental Mathematics

More information

CHAPTER 8: ACUTE TRIANGLE TRIGONOMETRY

CHAPTER 8: ACUTE TRIANGLE TRIGONOMETRY CHAPTER 8: ACUTE TRIANGLE TRIGONOMETRY Specific Expectations Addressed in the Chapter Explore the development of the sine law within acute triangles (e.g., use dynamic geometry software to determine that

More information

SAT Math Medium Practice Quiz (A) 3 (B) 4 (C) 5 (D) 6 (E) 7

SAT Math Medium Practice Quiz (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 SAT Math Medium Practice Quiz Numbers and Operations 1. How many positive integers less than 100 are divisible by 3, 5, and 7? (A) None (B) One (C) Two (D) Three (E) Four 2. Twice an integer is added to

More information

BC Exam Solutions Texas A&M High School Math Contest November 8, 2014

BC Exam Solutions Texas A&M High School Math Contest November 8, 2014 Exam Solutions Texas &M High School Math ontest November 8, 014 If units are involved, include them in your answer, which needs to be simplified. 1. If an arc of 60 on circle I has the same length as an

More information

Coach Monks s MathCounts Playbook! Secret facts that every Mathlete should know in order to win!

Coach Monks s MathCounts Playbook! Secret facts that every Mathlete should know in order to win! Coach Monks s MathCounts Playbook! Secret facts that every Mathlete should know in order to win! Learn the items marked with a first. Then once you have mastered them try to learn the other topics.. Squares

More information

CAMI Education linked to CAPS: Mathematics

CAMI Education linked to CAPS: Mathematics - 1 - TOPIC 1.1 Whole numbers _CAPS curriculum TERM 1 CONTENT Mental calculations Revise: Multiplication of whole numbers to at least 12 12 Ordering and comparing whole numbers Revise prime numbers to

More information

11-4 Areas of Regular Polygons and Composite Figures

11-4 Areas of Regular Polygons and Composite Figures 1. In the figure, square ABDC is inscribed in F. Identify the center, a radius, an apothem, and a central angle of the polygon. Then find the measure of a central angle. Center: point F, radius:, apothem:,

More information

13-3 Geometric Probability. 1. P(X is on ) SOLUTION: 2. P(X is on ) SOLUTION:

13-3 Geometric Probability. 1. P(X is on ) SOLUTION: 2. P(X is on ) SOLUTION: Point X is chosen at random on. Find the probability of each event. 1. P(X is on ) 2. P(X is on ) 3. CARDS In a game of cards, 43 cards are used, including one joker. Four players are each dealt 10 cards

More information

Applications of Trigonometry

Applications of Trigonometry chapter 6 Tides on a Florida beach follow a periodic pattern modeled by trigonometric functions. Applications of Trigonometry This chapter focuses on applications of the trigonometry that was introduced

More information

MATH DAY 2013 at FAU Competition A Individual

MATH DAY 2013 at FAU Competition A Individual MATH DAY 01 at FAU Competition A Individual SOLUTIONS NOTE: Two of the problems had problems. A typo occurred in the choice of possible answers for Problem # 6 and Problem # 16 had two possible correct

More information

7 th Grade Pre Algebra A Vocabulary Chronological List

7 th Grade Pre Algebra A Vocabulary Chronological List SUM sum the answer to an addition problem Ex. 4 + 5 = 9 The sum is 9. DIFFERENCE difference the answer to a subtraction problem Ex. 8 2 = 6 The difference is 6. PRODUCT product the answer to a multiplication

More information

ALGEBRA. sequence, term, nth term, consecutive, rule, relationship, generate, predict, continue increase, decrease finite, infinite

ALGEBRA. sequence, term, nth term, consecutive, rule, relationship, generate, predict, continue increase, decrease finite, infinite ALGEBRA Pupils should be taught to: Generate and describe sequences As outcomes, Year 7 pupils should, for example: Use, read and write, spelling correctly: sequence, term, nth term, consecutive, rule,

More information

MDPT - Geometry Practice Problems. 1. ABC is an isosceles triangle with base BC. L1 and L2 are parallel. 1=80. Find 4.

MDPT - Geometry Practice Problems. 1. ABC is an isosceles triangle with base BC. L1 and L2 are parallel. 1=80. Find 4. MDPT - Geometry Practice Problems 1. C is an isosceles triangle with base C. L1 and L are parallel. 1=80. Find 4. L1 1 4 a. 80 b. 50 c. 45 d. 60. In the figure, the measure of arc C is 7 π / 4 and O is

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Tuesday, August 13, 2013 8:30 to 11:30 a.m., only.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Tuesday, August 13, 2013 8:30 to 11:30 a.m., only. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Tuesday, August 13, 2013 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications

More information

Pilot Flyskole AS Hangarveien 13 N-3241 Sandefjord Tlf Epost Preparatory Course.

Pilot Flyskole AS Hangarveien 13 N-3241 Sandefjord Tlf Epost  Preparatory Course. Pilot Flyskole AS Hangarveien 13 N-3241 Sandefjord Tlf +47 9705 6840 Epost post@pilot.no www.pilot.no Preparatory Course Mathematics Pilot Flight School 2014 Order of operations Operations means things

More information

Mathematics Pre-Test Sample Questions A. { 11, 7} B. { 7,0,7} C. { 7, 7} D. { 11, 11}

Mathematics Pre-Test Sample Questions A. { 11, 7} B. { 7,0,7} C. { 7, 7} D. { 11, 11} Mathematics Pre-Test Sample Questions 1. Which of the following sets is closed under division? I. {½, 1,, 4} II. {-1, 1} III. {-1, 0, 1} A. I only B. II only C. III only D. I and II. Which of the following

More information

SOLVED PROBLEMS REVIEW COORDINATE GEOMETRY. 2.1 Use the slopes, distances, line equations to verify your guesses

SOLVED PROBLEMS REVIEW COORDINATE GEOMETRY. 2.1 Use the slopes, distances, line equations to verify your guesses CHAPTER SOLVED PROBLEMS REVIEW COORDINATE GEOMETRY For the review sessions, I will try to post some of the solved homework since I find that at this age both taking notes and proofs are still a burgeoning

More information

Geometry in a Nutshell

Geometry in a Nutshell Geometry in a Nutshell Henry Liu, 26 November 2007 This short handout is a list of some of the very basic ideas and results in pure geometry. Draw your own diagrams with a pencil, ruler and compass where

More information

High School Math Contest

High School Math Contest High School Math Contest University of South Carolina January 31st, 015 Problem 1. The figure below depicts a rectangle divided into two perfect squares and a smaller rectangle. If the dimensions of this

More information

Right Triangles and Quadrilaterals

Right Triangles and Quadrilaterals CHATER. RIGHT TRIANGLE AND UADRILATERAL 18 1 5 11 Choose always the way that seems the best, however rough it may be; custom will soon render it easy and agreeable. ythagoras CHATER Right Triangles and

More information

0810ge. Geometry Regents Exam 0810

0810ge. Geometry Regents Exam 0810 0810ge 1 In the diagram below, ABC XYZ. 3 In the diagram below, the vertices of DEF are the midpoints of the sides of equilateral triangle ABC, and the perimeter of ABC is 36 cm. Which two statements identify

More information

Mary s top book shelf holds five books with the 1. following widths, in centimeters: 6,

Mary s top book shelf holds five books with the 1. following widths, in centimeters: 6, Mary s top book shelf holds five books with the 1 following widths, in centimeters: 6, 2, 1, 2.5, and 5. What is the average book width, in centimeters? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 2010 AMC 10 A, Problem

More information

The integer is the base number and the exponent (or power). The exponent tells how many times the base number is multiplied by itself.

The integer is the base number and the exponent (or power). The exponent tells how many times the base number is multiplied by itself. Exponents An integer is multiplied by itself one or more times. The integer is the base number and the exponent (or power). The exponent tells how many times the base number is multiplied by itself. Example:

More information

10.1: Areas of Parallelograms and Triangles

10.1: Areas of Parallelograms and Triangles 10.1: Areas of Parallelograms and Triangles Important Vocabulary: By the end of this lesson, you should be able to define these terms: Base of a Parallelogram, Altitude of a Parallelogram, Height of a

More information

Chapter 8. Right Triangles

Chapter 8. Right Triangles Chapter 8 Right Triangles Objectives A. Use the terms defined in the chapter correctly. B. Properly use and interpret the symbols for the terms and concepts in this chapter. C. Appropriately apply the

More information

Junior Math Circles November 18, D Geometry II

Junior Math Circles November 18, D Geometry II 1 University of Waterloo Faculty of Mathematics Junior Math Circles November 18, 009 D Geometry II Centre for Education in Mathematics and Computing Two-dimensional shapes have a perimeter and an area.

More information

4. How many integers between 2004 and 4002 are perfect squares?

4. How many integers between 2004 and 4002 are perfect squares? 5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2009 8:30 to 11:30 a.m., only.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2009 8:30 to 11:30 a.m., only. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 13, 2009 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of your

More information

Session 6 The Pythagorean Theorem

Session 6 The Pythagorean Theorem Session 6 The Pythagorean Theorem Key Terms for This Session Previously Introduced altitude perpendicular bisector right triangle side-angle-side (SAS) congruence New in This Session converse coordinates

More information

Unit 3: Triangle Bisectors and Quadrilaterals

Unit 3: Triangle Bisectors and Quadrilaterals Unit 3: Triangle Bisectors and Quadrilaterals Unit Objectives Identify triangle bisectors Compare measurements of a triangle Utilize the triangle inequality theorem Classify Polygons Apply the properties

More information

(Mathematics Syllabus Form 3 Track 3 for Secondary Schools June 2013) Page 1 of 9

(Mathematics Syllabus Form 3 Track 3 for Secondary Schools June 2013) Page 1 of 9 (Mathematics Syllabus Form 3 Track 3 for Secondary Schools June 2013) Page 1 of 9 Contents Pages Number and Applications 3-4 Algebra 5-6 Shape Space and Measurement 7-8 Data Handling 9 (Mathematics Syllabus

More information

22 Proof Proof questions. This chapter will show you how to:

22 Proof Proof questions. This chapter will show you how to: 22 Ch22 537-544.qxd 23/9/05 12:21 Page 537 4 4 3 9 22 537 2 2 a 2 b This chapter will show you how to: tell the difference between 'verify' and 'proof' prove results using simple, step-by-step chains of

More information

Geometry Regents Review

Geometry Regents Review Name: Class: Date: Geometry Regents Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. If MNP VWX and PM is the shortest side of MNP, what is the shortest

More information

as a fraction and as a decimal to the nearest hundredth.

as a fraction and as a decimal to the nearest hundredth. Express each ratio as a fraction and as a decimal to the nearest hundredth. 1. sin A The sine of an angle is defined as the ratio of the opposite side to the hypotenuse. So, 2. tan C The tangent of an

More information

2000 Solutions Fermat Contest (Grade 11)

2000 Solutions Fermat Contest (Grade 11) anadian Mathematics ompetition n activity of The entre for Education in Mathematics and omputing, University of Waterloo, Waterloo, Ontario 000 s Fermat ontest (Grade ) for The ENTRE for EDUTION in MTHEMTIS

More information

Factoring Polynomials

Factoring Polynomials UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can

More information

Biggar High School Mathematics Department. National 4 Learning Intentions & Success Criteria: Assessing My Progress

Biggar High School Mathematics Department. National 4 Learning Intentions & Success Criteria: Assessing My Progress Biggar High School Mathematics Department National 4 Learning Intentions & Success Criteria: Assessing My Progress Expressions and Formulae Topic Learning Intention Success Criteria I understand this Algebra

More information

12-6 Surface Area and Volumes of Spheres. Find the surface area of each sphere or hemisphere. Round to the nearest tenth. SOLUTION: ANSWER: 1017.

12-6 Surface Area and Volumes of Spheres. Find the surface area of each sphere or hemisphere. Round to the nearest tenth. SOLUTION: ANSWER: 1017. Find the surface area of each sphere or hemisphere. Round to the nearest tenth. 3. sphere: area of great circle = 36π yd 2 We know that the area of a great circle is r.. Find 1. Now find the surface area.

More information

State the assumption you would make to start an indirect proof of each statement.

State the assumption you would make to start an indirect proof of each statement. 1. State the assumption you would make to start an indirect proof of each statement. Identify the conclusion you wish to prove. The assumption is that this conclusion is false. 2. is a scalene triangle.

More information