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Topic 2: Geometry Dr J Frost

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Slide guidance Key to question types: IMCSenior Maths Challenge BMOBritish Maths Olympiad The level, 1 being the easiest, 5 the hardest, will be indicated. 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. FrostA Frosty Special Questions from the deep dark recesses of my head. Classic Well known problems in maths. ? 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. A: London For multiple choice questions (e.g. IMC), click your choice to check your answer (try below!) Question: The capital of Spain is: B: ParisC: Madrid

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Topic 2: Geometry Part 1 – Problems involving angles a. Fundamentals b. Interior/Exterior Angles of Regular Polygons c. Circle Theorems d. Forming circles around regular polygons

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Topic 2: Geometry Part 2 – Problems involving lengths and areas a. Quickly getting diagonals b. Cutting and reforming c. Similar Triangles f. Segments of a circle e. Adding circle radii g. Equating areas d. Forming an equation

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ζ Part 1: Problems Involving Angles Topic 2 – Geometry Here, we recap the basic laws of angles and how they can be used in more involved settings.

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#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. x x x 180 ° - x ? 2 ? 180 ° -2x x ? y x+y The exterior angle of a triangle (with its extended line) is the sum of the other two interior angles. Remember this! ?

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Level 2 Level 1 Level 5 Level 4 Level 3 IMC Question: ABC and DEF are equilateral triangles. Find x? A: 70 ° B: 60 ° C: 50 ° D: 40 ° E: 30 ° Use what you know! Be sure to pick out key information (i.e. ‘equilateral’) #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. 75 ° 65 ° x°x° A B C D E F a) Fill in extra key information. 60 ° b) Derive other angles. 45 ° 55 ° 80 °

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Level 2 Level 1 Level 5 Level 4 Level 3 IMC Question: What is angle BAC? A: 60 ° B: 75 ° C: 90 ° D: 20 ° E: More info needed. Use what you know! I can introduce a variable for an unknown angle. I can exploit the fact these triangles are isosceles. A BCD Let’s introduce an x. x x 2x ? Angle ADB is 2x : (from previous law) So angle BAD is (180-2x)/2 = 90 – x. So BAC = (90-x) + x = 90 I know the exterior angle of a triangle. #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.

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Level 2 Level 1 Level 5 Level 4 Level 3 IMC Which of these triangles are right-angled? 132 ° 42 ° 6x 4x5x 4x x3x YesNo This angle is 48, so remaining angle in triangle is 90. YesNo If this angle is half of the total, it must be half of 180 which is 90. But 6x is not half of 15x. YesNo Using the same reasoning as before, 4x IS half of 8x. YesNo It is was right-angled, then by Pythagoras Theorem, = 6 2. But this is not the case. #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.

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#2: Interior/Exterior Angles of Regular Polygons 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 = ° ? 144 ° ? The interior angle of the polygon can then be worked out using angles on a straight line. It’s useful to be able to quickly calculate the interior and exterior angles of a regular polygon.

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

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Level 2 Level 1 Level 5 Level 3 Level 4 IMC Question: ABCDE is a regular pentagon.FAB is a straight line. FA = FB. What is the ratio x:y:z? A: 1:2:3 B: 2:2:3C: 2:3:4 D: 3:4:5E: 3:4:6 AB C D E F zyx Use what you know! I can now find interior/exterior angles. The angles of a triangle add up to 180 °. The fact FA = FB must be significant. #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.

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Question: ABCDE is a regular pentagon.FAB is a straight line. FA = FB. What is the ratio x:y:z? AB C D E F zyx 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. #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.

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#3a: Circle Theorems Look for opportunities to use these theorems. Sometimes it might not always be obvious when you can do so. 2x x The angle between two points on the circumference from the centre is twice the area from any point on the circumference. Use this when: You have some angle formed by 3 points on the circumference, and know the angle in the centre formed by 2 of them.

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#3a: Circle Theorems Look for opportunities to use these theorems. Sometimes it might not always be obvious when you can do so. The angle from a point on the circumference to the end of the diameter is 90 °. Use this when: You have three equal length sides emanating from a single point. E.g. We could have used it earlier: A BCD Since BC, AD and DC are of equal length, we can draw a circle around ABC with centre D. By the theorem, angle BAC must be 90 °.

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#3a: Circle Theorems Look for opportunities to use these theorems. Sometimes it might not always be obvious when you can do so. x x Alternative Segment Theorem: The angle subtended by a chord is the same as the angle between the chord and its tangent. Chord Tangent

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Level 2 Level 1 Level 4 Level 3 Level 5 IMC Question: What is the angle within this regular dodecagon? (12 sides) #3b: Forming circles around regular polygons By drawing a circle around a regular polygon, we can exploit circle theorems. x Angle = 75 ° ? Use what you know! I can form a circle around a regular polygon. I have an angle between three points on the circumference. By our circle theorems, x is therefore half of this. This angle is much easier to work out. It’s 5 12 th s of the way around a full rotation, so 150 °.

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ζ Part 2: Problems Involving Lengths/Areas Topic 2 – Geometry There’s a whole lot of tips and tricks we can use to simplify problems involving lengths and areas of different kinds of shapes.

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#1: Quickly getting a diagonal For an isoceles 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 ° x x ? Therefore: If we have the non-diagonal length: multiply by √ 2. If we have the diagonal length: divide by √ 2.

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#1: Quickly getting a diagonal For an isoceles 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 ° 3 3 ? ? 5

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Level 2 Level 1 Level 4 Level 3 Level 5 IMC Question: The sides of this regular octagon are 2cm. What is the difference between the area of the shaded region and the unshaded region? #1: Quickly getting a diagonal Using this makes our life much easier for some questions! I’ve handily broken this shape up for you on the next slide...

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Question: The sides of this regular octagon are 2cm. What is the difference between the area of the shaded region and the unshaded region? Tip #1: Quickly getting a diagonal 2cm cm 2cm cm 2cm ? ? Area of each shaded triangle: Area of each rectangle: Total shaded: Total unshaded: (So the difference is 0!) ? ? ? ?

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There’s a faster way we could have solved this problem without needing to calculate the shaded/unshaded area! #2: Cutting and reforming 2cm Frequently, we can chop a shape into bits and put them back together in some other way, clearly with the same area. 2cm If we use all four of these shaded triangles, we can form a square with sides 2cm. Does this shape occur elsewhere in the diagram? The triangles combined have the same area as the centre square. And the shaded rectangles have the same area as the unshaded ones. So the overall difference in area is 0! 2cm

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Level 2 Level 1 Level 4 Level 5 Level 3 IMC Question: The diagram shows three semicircles, each of radius 1. What is the size of the total shaded area? #2: Cutting and reforming A: π + 2 B: 5C: (3/2) π + 1 D: 4E: 2 π - 1 Frequently, we can chop a shape into bits and put them back together in some other way, clearly with the same area. The key is cutting this into shapes we know the area of. We have a 2x1 rectangle, and the remaining bits can combine to form a circle of area π.

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Level 2 Level 1 Level 4 Level 3 Level 5 IMC Question: A square is inscribed in a right-angled triangle as shown. What fraction of the triangle does it occupy? #3: Similar Triangles Hint: Let the width of the square be x (see Tip #4 in Algebra), and then form an equation by using the fact the triangles are similar. When triangles are similar, we can form an equation A: 12/25 B: 24/49C: 1/2 D: 25/49E: 13/25 We know these triangles are similar because their angles are the same.

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#3: Similar Triangles When triangles are similar, we can form an equation. 5 x x 3-x x x _x_ 4-x ? ? ? ? = Key Theory: If two triangles are similar, then their ratio of width to height is the same. a b c d abab = cdcd ? ? Multiplying through by the denominators give us: (4-x)(3-x) = x 2 Expanding and solving gives us: x = 12 7 So the area of the square is: Since the area of the big triangle is 6, the fraction the square occupies is: ? ?? ?

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#4: Form an equation! There are many other circumstances in which you can form an equation, if you’re equating lengths using different expressions, or using Pythagoras. 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. Macclaurin Hamilton Cayley IMO Use what you know! I’m told in the question the shape is a square. I can introduce variables for unknown quantities.

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#4: Forming an equation! There are many other circumstances where you should form an equation, if you’re equating lengths using different expressions, or using Pythagoras. 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. x Since we’re trying to find the ratio of sides around triangle GCM, introduce unknowns around here. y Since the paper is square and it was folded, GM = GB. And GB = 2x – y 2x-y Using Pythagoras Theorem, we can form an equation to give the relationship of x and y. x 2 + y 2 = (2x-y) 2 Simplifying: x = y 4343 So ratio of x:y is 4:3, and using Pythag, GM would be 5.

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r #5: Adding circle radii Sometimes it helps to add the radii of different circles to help us compare lengths. Question: If the radius of the big circle is r, what is the radius of the small circle? Add and label lines for the radii of the big circle that you think might help us solve the problem. >> Click here to reveal Do the same for the smaller circle, labelling these lengths x. >> Click here to reveal r x x x

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r #5: Adding circle radii Sometimes it helps to add the radii of different circles to help us compare lengths. At this point, we might form an equation by giving a key length in terms of r, and in terms of x. The bold black line is clearly of length r. But what is it in terms of x? r x x x Answer: x + x √ 2 = (1+ √ 2)x Therefore: x = ___r___ 1 + √ 2 ? ?

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

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#6: Segment of a circle Describe a method that would give you the area of the shaded region. Method: Start with the sector AOB. (area is ( θ /360) x π r 2 because θ /360 gives us the proportion of the circle we’re using). r r A B O θ Then cut out the triangle (i.e. subtract its area). We could work out its area by splitting it in two and using trigonometry. ? Some area related problems require us to calculate a segment.

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#6: Segment of a circle A 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? 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 ? Some area related problems require us to calculate a segment.

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

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#6: Segment of a circle A 1 1 √2√2 O B C So area of segment = ( π /2) - 1 Therefore (by cutting the segment area from a semicircle): Area of shaded area = - ( – 1) = 1 π2π2 π2π2 ? Some area related problems require us to calculate a segment.

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#7: Equating areas When you’re told areas are equal, form an equation and simplify. x°x° Question: The area of the semi-circle and isosceles triangle are the same. What is tan x? Level 2 Level 1 Level 4 Level 3 Level 5 IMC Use what you know! I can form an expression for tan x for a right-angled triangle. I can find the area of an isosceles triangle by splitting it in two. I can introduce variables to represent unknown quantities. Answer: π /4 As usual, introduce some variables for unknowns, and split our isosceles triangle so we can use trigonometry.. r h So tan x = h/r Area of triangle = 2rh Angle of semi-circle = ½ π r 2 Equating and simplifying: h = ¼ π r So tan x = (¼ π r) / r = ¼ π ?

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A little bit of everything! Get into teams of 4 or 5, and see which team can calculate all the areas of the black regions first. The radius of the main circle in each case is R. ½ π R 2 ¾ √ 3 R 2 (4 – π )R 2 R2R2 ( π – 2)R 2 ?? ? ? ?

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