Topic 3: Algebra

1. Topic 3: Algebra Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 21st August 2013

2. Slide Guidance Key to question types: SMC Senior Maths Challenge Uni University Interview Questions used in university interviews (possibly Oxbridge). www.ukmt.org.uk The level, 1 being the easiest, 5 the hardest, will be indicated. Frost A Frosty Special BMO British Maths Olympiad Questions from the deep dark recesses of my head. Those with high scores in the SMC 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. Classic Classic Well known problems in maths. STEP STEP Exam MAT Maths Aptitude Test Exam used as a condition for offers to universities such as Cambridge and Bath. Admissions test for those applying for Maths and/or Computer Science at Oxford University.

3. Slide Guidance 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. SMC), click your choice to reveal the answer (try below!) Question: The capital of Spain is:  A: London  B: Paris  C: Madrid

4. ζ Topic 3 – Algebra Part 1: Recursive Expressions and Expansion Identities

5. Recursive Expressions Sometimes values are defined in terms of themselves. What is the value of the following? If we let the value of the above expression be , then Multiplying both sides by x and rearranging, we get Then . This is the golden ratio. Nice! ?

6. Recursive Expressions Question: Without explicitly calculating or , find . ? Squaring both sides, we get and . Subtracting the equations (always a sensible thing we when have two squares!), we get So , so

7. Using helpful expansion identities In solving algebraic problems, it’s often immensely useful to find expansions that involve the terms we know and that we’re trying to find. For example: would clearly be useful if we knew and , and wanted to find . We used this earlier in the Number Theory module when our equation involved , and and we wanted to factorise.

8. Using helpful expansion identities Question: We earlier found equations and and that . Now find . ? We can find by adding our two equations, giving Now look at the expansion . Then , so

9. Using helpful expansion identities Question: Four positive integers , , and are such that . What is the value of ? C: 77 A: 73  B: 75   D: 79   E: 81 SMC Level 4 The expansion we want to use is So So and must have the values 1, 2, 4 and 66 (in any order). The sum is 73. Level 5 Level 3 Level 2 Level 1

10. Factorising Strategies Factorise We might think to get these terms, we’d have something like . Either we’d need to add an , or we’d have to add an and before halving it. The expression is clearly symmetrical, so let’s try This gives Thus our factorisation is

11. Factorising Strategies Factorise Notice that there’s a symmetry here between and , but not with . We’ll have this same symmetry/asymmetry in the factorised expression. We might attempt to start factorising like so, which would give us the first four terms in the expansion… …and we can see that adding this extra term in the first bracket would give us the remaining terms. Factorising more difficult expressions is ultimately mostly about intelligent guess work, just by considering how terms combine across brackets.

12. Factorising Strategies Try factorising these: ? ? ? Pro Tip: You can check your factorisations by trying small values for your variables, e.g. or 1 or -1. This doesn’t guarantee it’s correct (you might have got lucky with the values you chose!) but at least gives you greater confidence in its validity.

13. Using helpful expansion identities Question: Find all integer solutions , and for: and You might be tempted to try and use , but this leaves the terms and which we’re unable to deal with. So looking at the first equation, what terms could we combine into some factorised expression? Then the 1st equation becomes We’ll come back to this later… ? Round 1 BMO Round 2

14. Calculating big numbers Often algebra can be used to determine large values without a calculator. Replace numbers with variables and manipulate. A quickie: What is 992? ? A quickie: What is 1013? ? What about ? You can probably get from above we get the rows of Pascal’s triangle separated by 0s, so ?

15. Calculating big numbers Question: Find the value of Solution: Replacing 2007 with , we get Expanding out the brackets (make sure you’re good on your Binomial Expansion!), we get: At this point you might make an ‘intelligent guess’ that the numerator factorises to , which means the overall value is . This is clearly much easier to calculate. ? Round 1 BMO Round 2

16. Making an Effective Substitution In many cases (particularly Olympiad problems), it’s worthwhile making a substitution that simplifies our problem. Question: Find real solutions to What substitution might aid us? (look at the relationship between the contents of the three brackets) ? and . Then Now the problem is easier to solve! ? so So But note that and are positive, and so is given that . So and must be 0, and thus and are -3 and +3.

17. ζ Topic 3 – Algebra Part 2: Simultaneous Equations & Surds

18. Surd Manipulation Question: Find the least positive integer such that: ? Solution: While we could square both sides, we’ll end up with a mixture of and . It’s better to move the to the other side first, so that after squaring, the n’s cancel. So the least positive integer is 250,000. General Tip: Before squaring both sides, if you only have two occurrences of variable(s), move one onto the other side of the equality/inequality first.

19. Surd Manipulation Question: Which of the following is equal to: A:   B:  C: D:   E: SMC Hint: Perhaps we can reorganise the contents of the outer square root such that it’s a squared expression? Level 4 Level 5 Level 3 Level 2 Level 1

20. Surd Manipulation Question: Which of the following is equal to: At this point, since we have twice the product of two things, it suggests we can perhaps get the denominator in the form , so that we can take the square root. Indeed this works very nicely: Now we can just rationalise the denominator to get .

21. Solving with Surds For what do the following equality hold: ? For negative , we’d find for example that . More generally, we have to ensure that , where is the modulus function (i.e. makes its argument positive if not already).

22. Solving with Surds What about: ? We require that . This occurs when . We could see this by sketching: | 1 1

23. Solving with Surds [Source: UKMT Mentoring] Find the values of for which: ? If we square the LHS and RHS and simplify, we end up with . This is the same as . Thus . We also require that is a real value. thus . Thus we have .

24. Simultaneous Equations What are the two main ways you would solve simultaneous equation? Substitution ? Elimination ?

25. Simultaneous Equations BMO questions essentially only use these two principles. You just have to be a bit more creative. Solve the simultaneous equations where are integers: Hint: What should we substitute, and why? ? Use substitution. Rearranging , we get . The reason we chose to make the subject is that we eliminate and from when we square and subtract it. Substituting this into : It then becomes a case of considering the factor pairs of 66 (including negative ones!) and working out the corresponding , , . Round 1 BMO Round 2

26. Simultaneous Equations Sometimes we add/subtract equations to be able to get the RHS as 0 and be able to factorise the LHS. Solve for real : Hint: They made two of the numbers a 4 for a reason! From , either or . The former contradicts and , so . We can make this substitution to obtain: and We can solve this by substituting one equation into the other. This gives , or . But we need to check these! We find because it contradicts We can use to then find for each , and any of to get . ? Round 1 BMO Round 2

27. Simultaneous Equations Sometimes we can exploit symmetry. Solve for real : Hint: Notice that Using the above tip, let’s do: So or . The first case contradicts so . If we added , then we find . (My own approach) Substitute and into . Via elimination we’d get . Substituting this into say (1), we get so or . If we worked out and for the latter, would not be satisfied, so , and , thus and . We could continue to add/subtract equations to work out more relationships between variables. ? BMO Round 2 ? Round 1

28. Simultaneous Equations Summary of Tips • You can add and subtract equations: • When you require integer solutions, the resulting expression after adding/subtracting might be factorisable in the form , where you can then reason about possible factor pairs of . • When your require real solutions, factorising in the form helps, because then we can say or . • The above might yield the relationship between just two variables. In which case, you could always then substitute into the original equations to eliminate one variable. • You can use elimination when you can one variable in terms of others. • Spot patterns in constants on the RHS. This might give clues as to what equations would be good to add/subtract. Add/subtract based on what variables you want in the resulting equation, rather than just arbitrarily.

29. ζ Topic 3 – Algebra Part 3: Inequalities

30. Starter ? Solve ? Solve Dividing or multiplying by a negative number flips the direction of the inequality. Solve ? You might think you can do and hence . You’re not allowed to multiply both sides by though, because you don’t know whether it’s positive or negative, and hence it may or may not flip the direction! You’ll learn how to solve these in FP2, but in summary, we could just consider where for the graph , the curve has a value at least 2.

31. Forming inequalities using areas and lengths By bounding a curve between two straight lines, or vice versa, we can often form some interesting inequalities by comparing lengths and areas. Question: a) Using suitable squares and circles, show that . ? a) The fact the appears between the two bounds suggests we have a circle between two squares. Then let the radius of the circle be 1 (so that its area is ). Then the outer area is 4, giving us . If we try comparing the areas of the circle and smaller square we get , which is not a tight enough bound. Comparing the lengths (of a quarter arc with one side of the square) instead gives us , i.e.

32. Forming inequalities using areas and lengths By bounding a curve between two straight lines, or vice versa, we can often form some interesting inequalities by comparing lengths and areas. Question: b) Show also (by perhaps using other shapes) that . ? b) The screams ‘TRIANGLE!’ or ‘HEXAGON!’ since in trigonometry, angles of 30 and 60 tend to lead to lengths involving this surd. A triangle will lead to lead to a worse bound, since increasing the number of sides of the surrounding polygons makes the area increasingly close to a circle, so let’s try a hexagon. Using the area of the inner hexagon, we get which is not good enough. But comparing lengths again gives . Compare areas to get the upper bound of .

33. Solving Inequalities using Positive Terms Prove that . ? This looks suspiciously factorisable. How therefore might I prove this inequality? The LHS is squared and therefore must be positive. Thus the above inequality clearly holds. is known as the Trivial Inequality. In general, one method of proving an inequality is to get it into the form , i.e. where the LHS is a sum of squares.

34. Using this trick in Number Theory problems Question: Find all integer solutions , and for: and Earlier we found we could simplify the first equation to: What can we determine about and ? Since and are positive (and square), then they both have to be 1 to add to 2. Then and must both be . We try each of the four possibilities to get our final solution. ? Round 1 BMO Round 2

35. Solving Inequalities using Positive Terms Practice Questions: ? Prove that ? Prove that (Harder!) Prove that ?

36. Using a substitution Sometimes making a substitution makes solving an inequality easier to solve: Prove that Make the substitution: Then: ? ?

37. Types of Mean Geometric Mean Arithmetic Mean ? Useful for average of percentage changes. e.g. If your stocks and shares account made 10% one year, and 20% the next, then the average increase is , i.e. 14.8% (not 15%!) The mean you all know and love: ? Harmonic Mean The average of rates. ?

38. AM-GM Inequality This is a hugely helpful inequality comparing Arithmetic and Geometric Means: Geometric Mean Arithmetic Mean ? ? So we can form an inequality using any sum: How could we easily prove the above using a technique we’ve seen before?

39. AM-GM Inequality Examples: ? ? ? ? ?

40. AM-GM Inequality This helps us prove certain inequalities. Prove that . By the AM-GM inequality: Similarly, and Thus ? Prove that By the AM-GM inequality, and . Thus ?

41. Cauchy-Schwarz Inequality The Cauchy-Schwarz Inequality is really quite awesome. Its form is as such: This looks rather horrible, but constructing such an inequality is simple if done is a certain way…

42. Cauchy-Schwarz Inequality Example: STEP 1: Start with a pair of brackets on the LHS, and a single squared bracket on the RHS. STEP 2: Put some sum in the RHS. STEP 3: For each term in the sum, square it, then think of two terms which multiply to give this.

43. Cauchy-Schwarz Inequality We can be slightly creative with our sum. “Prove that ” By the way, this gives us a generally handy inequality of: and similarly . In general,

44. Cauchy-Schwarz Inequality Given that are positive real numbers such that , prove that ? By the Cauchy-Schwarz Inequality: But . So Since is positive we can divide both sides by it, as well as by 2.

45. Cauchy-Schwarz Inequality Let be real numbers such that . Prove that (Hint: What is as a single fraction?) We have Multiply both sides by , we get: The hard part is working out what CS inequality to form! ? By Cauchy-Schwarz, Thus Square rooting both sides, we get the desired result. ?

46. Inequality Proofs using Geometry Sometimes there’s non-traditional and seemingly barmy ways of proving inequalities. The following problem is on your worksheet: Prove that for all real that On the worksheet I recommend using the Trivial Inequality. But the BMO model solutions mention that one student’s solution used Heron’s Formula for the area of a triangle. Let’s try it! Recall that Where

47. Inequality Proofs using Geometry Prove that for all real that Notice that from to , the decrease is . Since in the formula we have and . This suggests we should let the sides be and . Then in Heron’s formula: If we square root both sides of the inequality, the RHS matches exactly. Now we have to think about what area our triangle could be less than:

48. Inequality Proofs using Geometry Prove that for all real that We could also use the formula: for the area of the triangle. We can form the inequality: ? So far we’ve shown that using Geometry that: All that remains is to show that This gives , which is true by the Trivial Inequality.

49. Inequality Proofs using Geometry We saw that one potentially useful geometrical inequality is: Another useful one is the Triangle Inequality. Triangle Inequality i.e. Each side is less than the sum of the other two sides. (What would happen if they were equal?)

50. Summary There’s 3 main approaches to solving inequalities: 1. Trivial Inequality Put your inequality in the form This is trivially true because the sum of squares is positive. You may need a creative factorisation. 2. AM-GM Inequality The arithmetic mean is greater or equal to the geometric mean. e.g.