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Topic 1: Geometry

Topic 1: Geometry. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Last modified : 21 st August 2013. Slide Guidance. Key to question types:. SMC. Senior Maths Challenge. Uni. University Interview. Questions used in university interviews (possibly Oxbridge).

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Topic 1: Geometry

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  1. Topic 1: Geometry 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). The level, 1 being the easiest, 5 the hardest, will be indicated. BMO British Maths Olympiad Frost A Frosty Special 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. MAT Maths Aptitude Test STEP STEP Exam Admissions test for those applying for Maths and/or Computer Science at Oxford University. Exam used as a condition for offers to universities such as Cambridge and Bath.

  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. Part 1 – General Pointers Topic 1: Geometry a. Adding helpful sides b. Using variables for unknowns/Using known information Part 2a – Angles a. Fundamentals b. Exterior/Interior Angles of a Polygon Part 2b – Circle Theorems a. Key Theorems b. Using them backwards! c. Intersecting Chord Theorem

  5. Topic 1: Geometry Part 3 – Lengths and Area a. The “√2 trick”. b. Forming equations c. 3D Pythagoras and the “√3 trick”. d. Similar Triangles e. Area of sectors/segments f. Inscription problems Part 4 – Proofs a. Generic Tips b. Worked Examples c. Proofs involving Area

  6. ζ Topic 1 – Geometry Part 1: General Pointers General tips and tricks that will help solve more difficult geometry problems.

  7. #1 Adding Lines By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles. Simple example: What’s the area of this triangle? 12 5 ? 4 Adding the extra line in this case allows us to form a right-angled triangle, and thus we can exploit Pythagoras Theorem. 6

  8. #1 Adding Lines By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles. 2 If you were working out the length of the dotted line, what line might you add and what lengths would you identify? ? We might add the red lines so that we can use Pythagoras to work out the length of the blue. This would require us to work out the length of the orange one (we’ll see a quick trick for that later!). 4

  9. #1 Adding Lines By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles. Suppose we were trying to find the radius of the smaller circle r in terms of the radius of the larger circle R. Adding what lines/lengths might help us solve the problem? r r R r R ? By adding the radii of the smaller circle, the vertical/horizontal lines allow us to find the distance between the centres of the circle, by using the diagonal. We can form an equation comparing R with an expression just involving r.

  10. #1 Adding Lines By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles. If the radius of top large circles is 105, and the radius of the bottom circle 14. What lines might we add to find the radius of the small internal circle? 105 105 105 105 ? r Adding radii to points of contact allow us to form some triangles. And if we add the red vertical line, then we have right-angled triangles for which we can use Pythagoras! 105 105 r 14 14 14

  11. #1 Adding Lines If the indicated chord has length 2p, and we’re trying to work out the area of the shaded area in terms of p, what lines should we add to the diagram? p r1 r2 ? Again, add the radii of each circle, allowing us to form a right-angled triangle (since the chord is a tangent to the smaller circle). Then:

  12. #1 Adding Lines Question: What is angle x + y? A: 270    B: 300 C: 330 D: 360   E: More info needed x° y° Adding the appropriate extra line makes the problem trivial. SMC Level 5 Level 4 y° Level 3 Level 2 Level 1

  13. #1 Adding Lines But don’t overdo it… • Only add lines to your diagram that are likely to help. Otherwise you risk: • Making your diagram messy/unreadable, and hence make it hard to progress. • Overcomplicating the problem.

  14. #2 Introducing Variables It’s often best to introduce variables for unknown angles/sides, particularly when we can form expressions using these for other lengths. Question: A square sheet of paper ABCD is folded along FG, as shown, so that the corner B is folded onto the midpoint M of CD. Prove that the sides of triangle GCM have lengths of ratio 3 : 4 : 5. Starting point: How might I label the sides? Ensure you use information in the question! The paper is folded over, so given the square is of side 2x, and we’ve folded over at G, then clearly length GM = 2x – y. Then you’d just use Pythagoras! ? IMO Cayley Macclaurin Hamilton

  15. ζ Topic 1 – Geometry Part 2a: Angle Fundamentals Problems that involve determining or using angles.

  16. #1: Fundamentals Make sure you can rapidly apply your laws of angles. Fill in everything you know, introduce variables if necessary, and exploit equal length sides. Give an expression for each missing angle. 180°-x 2 ? x x 180°-2x ? YOU SHOULD ACTIVELY SEEK OUT OPPORTUNITIES TO USE THIS!! x x+y ? y The exterior angle of a triangle (with its extended line) is the sum of the other two interior angles.

  17. #1: Fundamentals Make sure you can rapidly apply your laws of angles. Fill in everything you know, introduce variables if necessary, and exploit equal length sides. What is the expression for the missing side? a b Angles of quadrilateral add up to 360°. ? 270 – a - b

  18. #2: Interior/Exterior Angles of Regular Polygons It’s useful to be able to quickly calculate the interior and exterior angles of a regular polygon. To work out this angle, consider that someone following this path has to turn by this angle to be in the right direction for the next edge. Once they get back to their starting point, they would have turned 360° in total. Sides = 10 The interior angle of the polygon can then be worked out using angles on a straight line. 144° ? 36° ?

  19. #2: Interior/Exterior Angles of Regular Polygons It’s useful to be able to quickly calculate the interior and exterior angles of a regular polygon. Exterior angle = 60° ? Interior angle = 120° ? Exterior angle = 72° ? Interior angle = 108° ?

  20. #2: Interior/Exterior Angles of Regular Polygons It’s useful to be able to quickly calculate the interior and exterior angles of a regular polygon. Question: ABCDE is a regular pentagon.FAB is a straight line. FA = AB. What is the ratio x:y:z? A B F x y z C E IMC Level 5 D  A: 1:2:3 B: 2:2:3  C: 2:3:4  Level 4 Level 3 Level 2 D: 3:4:5   E: 3:4:6 Level 1

  21. #2: Interior/Exterior Angles of Regular Polygons It’s useful to be able to quickly calculate the interior and exterior angles of a regular polygon. Question: ABCDE is a regular pentagon.FAB is a straight line. FA = FB. What is the ratio x:y:z? A B F x y z C E D y = 360° / 5 = 72°. So z = 180° – 72° = 108°. AB = AE (because it’s a regular pentagon) and we’re told FA = AB, so FA = AE. It’s therefore an isosceles triangle, so angle AEF = x. Angles of a triangle add up to 180°, so x = (180° – 72°)/2 = 54°. The ratio is therefore 54:72:108, which when simplified is 3:4:6.

  22. #2: Interior/Exterior Angles of Regular Polygons It’s useful to be able to quickly calculate the interior and exterior angles of a regular polygon. Question: The size of each exterior angle of a regular polygon is one quarter of the size of an interior angle. How many sides does the polygon have?  A: 6  B: 8  C: 9 D: 10  E: 12  SMC Level 5 If the ratio of the exterior to interior angle is , then the exterior angle must be (since interior and exterior angle add up to 180). Thus there’s sides. Level 4 Level 3 Level 2 Level 1

  23. ζ Topic 1 – Geometry Part 2b: Circle Theorems You should know most of these already. Although there’s a couple you may not have used (e.g. intersecting chord theorem).

  24. 1 2 Alternative Segment Theorem: The angle subtended by a chord is the same as the angle between the chord and its tangent. x diameter Chord 3 5 4 x x x x x 180-x x Tangent 2x Angles in same segment Angles of a cyclic quadrilateral

  25. Thinking backwards For many of the circle theorems, the CONVERSE is true… If a circle was circumscribed around the triangle, side AB would be the diameter of the circle. A B x If the opposite angles of a quadrilateral add up to 180, then the quadrilateral is a cyclic quadrilateral. 180-x x Using the theorems this way round will be particularly useful in Olympiad problems.

  26. Thinking backwards For many of the circle theorems, the CONVERSE is true… 4 We know that the angle at the centre is twice the angle at the circumference. Is the converse true, i.e. that if angle at some point inside the circle is twice that at the circumference, then it must be at the centre? x No. If we formed lines to any point on this blue circle (that goes through the centre of the outer circle), then by the ‘angles in the same segment’ theorem, the angle must still be . So our point isn’t necessarily at the centre. 2x 2x

  27. Circle Theorems Question: The smaller circle has radius 10 units; AB is a diameter. The larger circle has centre A, radius 12 units and cuts the smaller circle at C. What is the length of the chord CB? C If we draw the diameter of the circle, we have a 90° angle at C by our Circle Theorems. Then use Pythagoras. 12 B A 20 SMC Level 5 Level 4 A: 8   B: 10 C: 12  Level 3 Level 2 D: 10√2   E: 16 Level 1

  28. Circle Theorems Question: In the figure, PQ and RS are tangents to the circle. Given that a = 20, b = 30 and c = 40, what is the value of x? By Alternative Segment Theorem 50 By ‘Exterior Angle of Triangle’ By Alternative Segment Theorem By ‘Exterior Angle of Triangle’ 40+x 50 SMC x Angles of this triangle add up to 180, so: 2x + 110 = 180 Therefore x = 35 Level 4 A: 20  B: 25  C: 30  Level 5 Level 3 Level 2  D: 35 E: 40  Level 1

  29. Intersecting Chord Theorem

  30. Intersecting Secant Lengths Theorem A secant is a line which passes through a circle. You may also wish to check out the Intersecting Secant Angles Theorem

  31. Ptolemy’s Theorem A B You’ll be able to practice this in Geometry Worksheet 3. D C i.e. The product of the diagonals of a cyclic quadrilateral is the sum of the products of the pairs of opposite sides.

  32. Angle Bisector Theorem One final theorem not to do with circles… ratio of these… If the line bisects and , then …is the same as the ratio of these.

  33. Forming circles around regular polygons By drawing a circle around a regular polygon, we can exploit circle theorems. Question: What is the angle within this regular dodecagon? (12 sides) This angle is much easier to work out. It’s 5 12ths of the way around a full rotation, so 150°. By our circle theorems, x is therefore half of this. x IMC Level 4 Level 3 Level 5 Level 2 Angle = 75° ? Level 1

  34. ζ Topic 1 – Geometry Part 3: Lengths and Areas

  35. The “√2 trick” For an isosceles right-angled triangle (i.e. with angles 90, 45, 45), you can very quickly get the non-diagonal length from the diagonal, or vice versa. Question: What factor bigger is the diagonal relative to the other sides? 45° Therefore: If we have the non-diagonal length: multiply by √2. If we have the diagonal length: divide by √2. ? x 45° x

  36. The “√2 trick” For an isosceles right-angled triangle (i.e. with angles 90, 45, 45), you can very quickly get the non-diagonal length from the diagonal, or vice versa. Find the length of the middle side without computation: 45° 5 ? 3 45° ? 3

  37. The “√2 trick” The radius of the circle is 1. What is the side length of the square inscribed inside it? 1/√2 or 1 1 √2 ? √2

  38. 3D Pythagoras Question: P is a vertex of a cuboid and Q, R and S are three points on the edges as shown. PQ = 2 cm, PR = 2 cm and PS = 1 cm. What is the area, in cm2, of triangle QRS? Q P R S SMC Level 5  A: √15/4 B: 5/2  C: √6  Level 4 Level 3 Level 2 D: 2√2   E: √10 Level 1

  39. 3D Pythagoras Question: P is a vertex of a cuboid and Q, R and S are three points on the edges as shown. PQ = 2 cm, PR = 2 cm and PS = 1 cm. What is the area, in cm2, of triangle QRS? 2√2 2 Q P 2 R √5 1 √5 S SMC Level 5 So the height of this triangle by Pythagoras is √3. So that area is ½ x 2√2 x √3 = √6 √5 √5 Level 4 Level 3 Level 2 Level 1 √2 √2

  40. 3D Pythagoras Question: What’s the longest diagonal of a cube with unit length? 1 √3 √2 1 1 ? By using Pythagoras twice, we get √3. The √3 trick: to get the longest diagonal of a cube, multiply the side length by √3. If getting the side length, divide by √3.

  41. 3D Pythagoras Question: A cube is inscribed within a sphere of diameter 1m. What is the surface area of the cube? Longest diagonal of the cube is the diameter of the sphere (1m). So side length of cube is 1/√3 m. Surface area = 6 x (1/√3)2 = 2m2 SMC Level 4 A: 2m2   B: 3m2 C: 4m2  Level 5 Level 3 Level 2  D: 5m2  E: 6m2 Level 1

  42. Forming Equations To find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same. Returning to this previous problem, what is r in terms of R? r r R ? Equating lengths: R = r + r√2 = r(1 + √2) r = r R __r__ 1+√2

  43. Forming Equations To find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same. This is a less obvious line to add, but allows us to use Pythagoras to form an equation. Question: What is the radius of the small circle? 2 4 r 2 4-r r IMO Macclaurin (2-r)2 + (4-r)2 = (2+r)2 This gives us two solutions: reject the one that would make the smaller circle larger than the big one. 6 Hamilton Cayley

  44. Forming Equations To find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same. Question: If and are the radii of the larger circles, and the radius of the smaller one, prove that: As always, draw lines between the centres of touching circles. IMO Macclaurin As always, try to find right-angled triangles. Drawing a rectangle round our triangle will create 3 of them. Fill in the lengths. We don’t know the bases of the two bottom triangles, so just call them and . This would mean the width of the top triangle is . Hamilton Cayley

  45. Forming Equations Question: If and are the radii of the larger circles, and the radius of the smaller one, prove that: Similarly: From from the top triangle: Substituting: Dividing by : Notice that the LHS is a perfect square!

  46. Inscription Problems Question: A circle is inscribed inside a regular hexagon, which is in turn inscribed in another circle. What fraction of the outer circle is taken up by the inner circle? ? You might as well make the radius of the outer circle 1. Using the triangle and simple trigonometry, the radius of the smaller circle is therefore . The proportion taken up by the smaller circle is therefore .

  47. Similar Triangles When triangles are similar, we can form an equation. Key Theory: If two triangles are similar, then their ratio of width to height is the same. b d c a 3-x ? 5 Question: A square is inscribed inside a 3-4-5 triangle. Determine the fraction of the triangle occupied by the square. 3 x ? x ? 4-x 4

  48. Similar Triangles A particular common occurrence is to have one triangle embedded in another, where the indicated angles are the same. Why are triangles and similar? They share a second common angle at . We’ll see an example of this later on in this module.

  49. Segment of a circle Some area related problems require us to calculate a segment. This line is known as a chord. The area bound between a chord and the circumference is known as a segment. (it resembles the shape of an orange segment!) A ‘slice’ of a circle is known as a sector.

  50. Segment of a circle Some area related problems require us to calculate a segment. # Remember that we can find the area of a segment by starting with the sector and cutting out the triangle. But this technique of cutting out a straight edged polygon from a sector can be used to find areas of more complex shapes also, as we’ll see. A r θ B r O

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