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Topic 3: Geometry

Topic 3: Geometry. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Slide guidance. Key to question types:. SMC. Senior Maths Challenge. Uni. University Interview. Questions used in university interviews (possibly Oxbridge). www.ukmt.org.uk

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Topic 3: Geometry

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  1. Topic 3: Geometry Dr J Frost (jfrost@tiffin.kingston.sch.uk)

  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. Part 1 – General Pointers Topic 3: 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 3: Geometry Part 3 – Lengths and Area a. The “√2 trick”. b. Forming equations c. 3D Pythagoras and the “√3 trick”. d. Similar Triangles e. Forming Equations f. Area of sectors/segments Part 4 – BMO Problem Guidance [Coming soon!]

  6. ζ Topic 3 – 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 For problems involving circles, add radii in strategic places, e.g. towards a point on the circumference where the circle touches another shape. 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 For problems involving circles, add radii in strategic places, e.g. towards a point on the circumference where the circle touches another shape. 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 ? By adding these lines, we can come up with an expression for the shaded area: π(r22 – r12) And by Pythagoras: p2 + r12 = r22, so r22 – r12 = p2. So the shaded area is πp2. Source: SMC

  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. #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! x y 2x-y ? IMO Cayley Macclaurin Hamilton

  14. ζ Topic 3 – Geometry Part 2a: Angle Fundamentals Problems that involve determining or using angles.

  15. #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.

  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. What is the expression for the missing side? a b Angles of quadrilateral add up to 360°. ? 270 – a - b

  17. #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° ?

  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. Exterior angle = 60° ? Interior angle = 120° ? Exterior angle = 72° ? Interior angle = 108° ?

  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. 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

  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 = 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.

  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: 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 Let the shape have n sides. Then let k = 360 ÷ n be the exterior angle. The interior angle is 180 - k. So using the information: 180 – k = 4k  k = 36  n = 360 ÷ k = 10 (Or alternatively, since the angles on a straight line are in the ratio 1:4, we can immediately tell the exterior angle is 36°) Level 4 Level 3 Level 2 Level 1

  22. ζ Topic 3 – 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).

  23. 1 2 Alternate Segment Theorem: The angle subtended by a chord is the same as the angle between the chord and its tangent. x Chord 3 5 4 x x x x x 180-x x Tangent 2x Angles of a cyclic quadrilateral

  24. Thinking backwards The Circle Theorems work BOTH WAYS… 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.

  25. 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

  26. 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 Alternate Segment Theorem 50 By ‘Exterior Angle of Triangle’ By Alternate 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

  27. Intersecting Chord Theorem Version 1 Version 2 Chords intersect inside circle. Chords intersect outside circle. a a x b b y y x ab = xy ab = xy This is a useful tool in our arsenal particularly for Olympiad problems.

  28. #3b: 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

  29. ζ Topic 3 – Geometry Part 3: Lengths and Areas

  30. 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

  31. 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

  32. 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

  33. 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

  34. 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

  35. 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.

  36. 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

  37. 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

  38. 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

  39. 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. a b c d b d = c a Question: Find the percentage of the area of a 3:4:5 triangle occupied by a square inscribed inside it. ? 3-x 5 3 3-x x _x_ 4-x ? ? x = ? ? x ? 4-x 4 Multiplying through by the denominators give us: (4-x)(3-x) = x2 ? 12 7 144 49 ? ? Expanding and solving gives us: x = So the area of the square is: 24 49 ? Since the area of the big triangle is 6, the fraction the square occupies is:

  40. 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.

  41. Segment of a circle Some area related problems require us to calculate a segment. # Describe a method that would give you the area of the shaded region. A r ? Method: Start with the sector AOB. (area is (θ/360) x πr2 because θ/360 gives us the proportion of the circle we’re using). θ B r O Then cut out the triangle (i.e. subtract its area). We could work out its area by splitting it in two and using trigonometry.

  42. Segment of a circle Some area related problems require us to calculate a segment. The radius of the circle is 1. The arc is formed by a circle whose centre is the point A. What is the area shaded? A What might be going through your head at this stage... “Perhaps I should find the radius of this other circle?” Radius of circle centred at A: √2 ?

  43. Segment of a circle Some area related problems require us to calculate a segment. B Let’s put in our information first... What’s the area of this sector? Area of sector = π/2 1 √2 ? A O 1 Now we need to remove this triangle from it to get the segment. Area of triangle = 1 ? C

  44. Segment of a circle Some area related problems require us to calculate a segment. B So area of segment = (π/2)- 1 1 √2 Therefore (by cutting the segment area from a semicircle): Area of shaded area = - ( – 1) = 1 A O 1 π 2 π 2 ? C

  45. Segment of a circle Some area related problems require us to calculate a segment. Question: Here are 4 overlapping quarter circles of unit radius. What’s the area of the shaded region?

  46. Segment of a circle Start with sector. Cut out these two triangles. Given the arc has equation x2 + y2 = 1, then halfway along, 0.25 + y2 = 1, so y = √3/2 Which leaves this region. Then just multiply this area by 4 to finish...

  47. ζ Topic 3 – Geometry Part 4: BMO Problem Guidance I haven’t got round to making these slides yet. Try and re-download these later in the year!

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