Topic 1: Algebra

1. Topic 1: Algebra Dr J Frost (jfrost@tiffin.kingston.sch.uk)

2. Slide guidance Key to question types: IMC Senior Maths Challenge Frost A Frosty Special Questions from the deep dark recesses of my head. The level, 1 being the easiest, 5 the hardest, will be indicated. Classic Classic BMO British Maths Olympiad Well known problems in maths. Those with high scores in the IMC qualify for the BMO Round 1. The top hundred students from this go through to BMO Round 2. Questions in these slides will have their round indicated.  Any box with a ? can be clicked to reveal the answer (this works particularly well with interactive whiteboards!). Make sure you’re viewing the slides in slideshow mode. ? For multiple choice questions (e.g. IMC), click your choice to check your answer (try below!) Question: The capital of Spain is:  A: London B: Paris   C: Madrid

3. Tip #1: Being adept with laws of indices You need to be well practiced with your laws of indices – many IMC questions in particular rely on this skill. Question: What is half of 220? A: 110   B: 120  C: 20 D: 210  E: 219  IMC Level 5 Half of 220 is: 220 / 2 = 220 / 21 = 219 Level 4 Level 3 Level 2 Level 1

4. Tip #1: Being adept with laws of indices You need to be well practiced with your laws of indices – many IMC questions in particular rely on this skill. You can discuss this question in groups/pairs. Raise your hand when you’ve got an answer. Question: Given that 4x+ 4x+ 4x+ 4x= 416, what is the value of x? A: 2  B: 4   C: 8  D: 12 E: 15  IMC Level 4 Level 5 Level 3 Level 2 Level 1

5. Tip #1: Being adept with laws of indices Let’s practice our laws of indices. Simplify the following. This dot means multiplication. Hint: make the base of the second power expression 2. 2 · 2x = 2x+1 ? 2x· 4y = 2x+2y ? ((x3)3)3 = x27 ? 9x· 3x = 33x = 27x ? In general, ac x bc = (ab)c 2x / 2 = 2x-1 ? 2x + 2x + 2x + 2x = 2x+2 ? √(210) = 25 ? 4x + 4x + 4x + 4x = 4x+1 ? 3√(212) = 24 ? Therefore on the previous slide 4x+1 = 416, and so x+1 = 16, and so x = 15.

6. Tip #1: Being adept with laws of indices You need to be well practiced with your laws of indices – many IMC questions in particular rely on this skill. Question: Below are three statements. Exactly which ones are true? (i) 310 is even (ii) 310 is odd (iii) 310 is square A: (i) only   B: (ii) only  C: (iii) only D: (i) + (iii)  E: (ii) + (iii)  IMC Level 5 310 is square because its square root is 35. An odd number multiplied by an odd number is odd, so 3x is always odd for (positive) x. Level 4 Level 3 Level 2 Level 1

7. Tip #1: Being adept with laws of indices You need to be well practiced with your laws of indices – many IMC questions in particular rely on this skill. Question: Given that 5p = 9, 9q = 12, 12r = 16, 16s = 20 and 20t = 25, what is the value of pqrst?  A: 1  B: 2  C: 3 D: 4   E: 5 IMC What might be going through your head: “Hmm, it seems as if the expressions are ‘chained’ together. And since I’m looking for pqrst, it indeed looks like I’ll have to combine them together.” Level 4 Level 5 Level 3 If 5p = 9 and 9q = 12, then (5p)q = 12. Continuing with this, we get ((((5p)q)r)s)t = 25, so 5pqrst = 25, and so pqrst = 2. Level 2 Level 1

8. Tip #1: Being adept with laws of indices Summary of tips: When a number is on its own, its implicit power is 1. This can help us simplify expressions. 2 · 2x = 21· 2x = 2x+1 Making numbers the same base again helps us simplify. 4x· 2y = (22)x· 2y = 22x· 2y = 22x+y Taking the square root halves the power. √(212) = √(26)2 = 26

9. Tip #2: ? Question: Simplify 1 – x2 (1+x)(1-x) ? Same algebraic questions rely on you identifying and exploiting the difference of two squares. Make sure you can spot it!

10. Tip #2: Using the difference of two squares Question: Which of the following is equal to (1 + x + y)2 – (1 – x – y)2 for all values of x and y?  A: 4x  B: 2(x2+y2)  C: 0 D: 4xy   E: 4(x+y) IMC Note: There would be nothing stopping you expanding the original expression then simplifying, but it would waste time. Level 4 By using the difference of two squares to factorise, we get (2)(2x + 2y) = 4(x+y). Level 3 Level 5 Level 2 Level 1

11. Tip #2: Using the difference of two squares Question: What is the largest power of 2 that divides 1272 − 1?  A: 21  B: 27  C: 28 D: 263  E: 2127  IMC Using the difference of two squares, we get: 126 x 128. At this point, we look at how many factors of 2 are in each. We can divide 126 by 2 just once before we get to an odd number, but 128 is 27. So in total, we have 8 factors of 2. Level 4 Level 3 Level 5 Level 2 Level 1

12. Tip #3: More difficult simultaneous equations In the past you’ve solved simultaneous equations like 2x + 3y = 4 and x + 6y = 7. You may have solved these by scaling and then adding or subtracting the equations. But for more complicated equations, we can use substitution. x2 + y2 = 12 – 4x x – y = 2 x2 + (x-2)2 = 12 – 4x Step 3: Expand,simplify and solve as you usually would. Step 1: In the simpler equation, make one of the variables the subject of the formula. e.g. y = x - 2 Step 2: This allows us to get rid of y in the first equation by substituting y with x – 2. x2 + x2 – 4x + 4 = 12 – 4x 2x2 = 8 x2 = 4 x = 2

13. Tip #3: More difficult simultaneous equations In the past you’ve solved simultaneous equations like 2x + 3y = 4 and x + 6y = 7. For Question: Find all real values of and y that satisfy the equations: x4 – y4 = 5 x + y = 1 Use what you know! That first equation looks suspiciously like the difference of two squares...   I know how to simplify and substitute. Let’s break this up into 2 steps: What’s the simplest form we can put the first equation in? ? Answer: (x-y)(x2 + y2) = 5 IMO Macclaurin (x4 – y4) = (x2 - y2)(x2 + y2) = (x-y)(x+y)(x2 + y2) = (x-y)(x2 + y2) because x + y = 1 from the second equation. Hamilton Cayley

14. Tip #3: More difficult simultaneous equations In the past you’ve solved simultaneous equations like 2x + 3y = 4 and x + 6y = 7. For Question: Find all real values of and y that satisfy the equations: x4 – y4 = 5 x + y = 1 Use what you know! That first formula looks suspiciously like the difference of two squares...   I know how to simplify and substitute. Now let’s do the substitution... We need to get rid of y for the moment. y = 1 – x from the second equation. So substituting that into our simplified from of the first equation, we get: (2x – 1)(x2 + (1-x)2) = (2x – 1)(2x2 – 2x + 1) = 5. Expanding and simplifying gives us 2x3 – 3x2 + 2x – 3 = 0. Usually we don’t know how to factorise cubics, but it’s quite easy in this particular case: x2(2x – 3) + 1(2x – 3) = 0. So (x2 + 1)(2x – 3) = 0. Either x2 + 1 = 0 (which doesn’t have a real solution) or 2x – 3 = 0, giving x = 3/2. And since y = 1 – x, y = -1/2. IMO Macclaurin Hamilton Cayley

15. Tip #4: Introducing variables Sometimes we have to introduce variables of our choosing to help model (and subsequently solve) a problem. Question: Before the last of a series of tests, Sam calculated that a mark of 17 would enable her to average 80 over the series, but that a mark of 92 would raise her average mark over the series to 85. How many tests were in the series? Use what you know! I know the formula for the mean average.   I know how to form an expression using given information. Answer: 15 ?  Since I’ll get two equations, I know how to solve simultaneous equations. Use a variable x to represent the total from the previous papers, and n the number of papers. The using the information provided: This gives us x + 17 = 80n and x + 92 = 85n. These are simultaneous equations! Subtracting the first from the second gives 75 = 5n, so n = 25. IMO Cayley Macclaurin Hamilton

16. Tip #5: Dealing with ratios Consider the lengths x and y of two planks of wood. The ratio of their lengths is 2:3. How could we form an equation? x y If we had 3 of the first plank, that would be the same length as 2 of the second plank. So 3x = 2y. In general, if the ratio in size of x and y is a:b, then bx = ay. Don’t get this the wrong way round! Question: The area of shapes A and B are x2 and 3x – 4. The ratio of their area is 2:1. Find the possible values of x. [Source: Frosty Special] Answer: 2 or 4. ? 1(x2) = 2(3x-4). Simplifying and rearranging, this gives us x2 – 6x + 8 = 0. Factorising, (x-4)(x-2) = 0, so x = 4 or x = 2.

17. Tip #5: Dealing with ratios Use what you know! Question: A rectangular piece of paper is cut into two shapes by a straight line passing through one corner, as shown. Given that area X : area Y = 2:7, what is the ratio of a : b? I know the area of a triangle and trapezium.   I can deal with ratios to form an equation. X a  I can introduce variables to represent unknown quantities. Y b  I can solve equations that I have formed. Answer: 4:5 ? Let’s represent the width of the rectangle by a variable w. Area of triangle: ½ wa Area of triangle: ½ ([a+b] + b)w = ½ (a + 2b)w So using the ratios, 7(½ wa) = 2(½ (a + 2b)w) Cancelling an simplifying: 7a = 2a + 4b, so 5a = 4b. Using our ‘ratio tip’ backwards, the ratio of a to b is 4:5. By checking the diagram, this answer looks sensible. IMO Cayley Macclaurin Hamilton

18. Tip #6: Cancelling/simplifying intermediate terms In some algebraic problems, either middle terms in a sum/multiplication cancel often leaving the ‘ends’, or we can easily simplify them in some way. Question: Given that the number 2006 is the correct answer to the calculation 1 − 2 + 3 − 4 + 5 − 6 + … + (n − 2) − (n − 1) + n, what is the sum of the digits of n? Advice: What happens if we group the numbers in pairs after the first term? IMC -2+3 = 1. So is -4+5. Therefore the entire sum is 1 + 1 + 1 + 1 + ... + 1. How many ones are there? ? Level 4 If we exclude the first term there’s n-1 terms. So there’s (n-1)/2 pairs. If we include the initial 1, that’s 1 + (n-1)/2 ones we have (which could simplify to (n+1)/2 ). ? Level 3 Level 5 Level 2 Level 1

19. Tip #6: Cancelling/simplifying intermediate terms In some algebraic problems, either middle terms in a sum/multiplication cancel often leaving the ‘ends’, or we can easily simplify them in some way. Question: Given that the number 2006 is the correct answer to the calculation 1 − 2 + 3 − 4 + 5 − 6 + … + (n − 2) − (n − 1) + n, what is the sum of the digits of n? We found that the above expression simplifies to (n+1)/2. What therefore is the answer? IMC  A: 3 B: 4   C: 5 Level 4 D: 6  E: 7  Level 5 Level 3 Level 2 (n+1)/2 = 2006, so solving we get n = 4011. The digits add up to 6. Level 1

20. Tip #6: Cancelling/simplifying intermediate terms In some algebraic problems, either middle terms in a sum/multiplication cancel often leaving the ‘ends’, or we can easily simplify them in some way. Frosty Special: Simplify the following. Let’s go through a potential strategy together. What will each of the brackets be as an improper fraction? 1 3 ? 2 4 ? ? 3 5 n-4 n-2 ? n-3 n-1 ? n-2 n ? ... ? ? ? ?

21. Tip #6: Cancelling/simplifying intermediate terms In some algebraic problems, either middle terms in a sum/multiplication cancel often leaving the ‘ends’, or we can easily simplify them in some way. Next job: What can we cancel in this multiplication? 1 3 2 4 3 5 4 6 n-4 n-2 n-3 n-1 n-2 n ... Then we consider what we’d have left if we cancelled all these terms. These diagonals are the same. As are these. __2__ n(n-1) ? Final answer: