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**Dr J Frost (jamie@drfrostmaths.com)**

Topic 2: Geometry Dr J Frost

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** Slide guidance ? Question: The capital of Spain is:**

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

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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 d. Forming an equation e. Adding circle radii f. Segments of a circle g. Equating areas

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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. 180°-x 2 ? x ? 180° - x x x x ? x+y 180°-2x ? 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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**#1: Fundamentals A: 70° B: 60° C: 50° D: 40° E: 30°**

Make sure you can rapidly apply your laws of angles. Fill in everything you know, introduce variables if necessary, and exploit equal length sides. Use what you know! Question: ABC and DEF are equilateral triangles. Find x? Be sure to pick out key information (i.e. ‘equilateral’) D a) Fill in extra key information. A F b) Derive other angles. x° 60° C 80° 80° 60° 60° 75° 45° 55° 65° IMC B E Level 5 Level 4 A: 70° B: 60° C: 50° Level 3 Level 2 D: 40° E: 30° Level 1

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**#1: Fundamentals A: 60° B: 75° C: 90° D: 20° **

Make sure you can rapidly apply your laws of angles. Fill in everything you know, introduce variables if necessary, and exploit equal length sides. Use what you know! Question: What is angle BAC? Let’s introduce an x. I can introduce a variable for an unknown angle. Angle ADB is 2x : (from previous law) A So angle BAD is (180-2x)/2 = 90 – x. x 2x ? I can exploit the fact these triangles are isosceles. So BAC = (90-x) + x = 90 I know the exterior angle of a triangle. B D C IMC Level 5 Level 4 A: 60° B: 75° C: 90° Level 3 Level 2 D: 20° E: More info needed. Level 1

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**#1: Fundamentals Yes No Yes No Yes No Yes No **

Make sure you can rapidly apply your laws of angles. Fill in everything you know, introduce variables if necessary, and exploit equal length sides. Which of these triangles are right-angled? Yes No If this angle is half of the total, it must be half of 180 which is 90. But 6x is not half of 15x. 42° 6x 132° 4x 5x This angle is 48, so remaining angle in triangle is 90. It is was right-angled, then by Pythagoras Theorem, = 62. But this is not the case. Yes No IMC 6 Level 5 5 Level 4 x 3x Level 3 4x 4 Level 2 Using the same reasoning as before, 4x IS half of 8x. Yes No Yes No Level 1

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**#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°

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**#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° ?

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

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**#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 C D E F z y x 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.

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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 between two points on the circumference from the centre is twice the area from any point on the circumference. x 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. 2x

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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 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°. B D C

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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. Alternative Segment Theorem: The angle subtended by a chord is the same as the angle between the chord and its tangent. x Chord x Tangent

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**#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) Use what you know! I can form a circle around a regular polygon. This angle is much easier to work out. It’s 5 12ths of the way around a full rotation, so 150°. I have an angle between three points on the circumference. By our circle theorems, x is therefore half of this. x IMC Level 5 Level 4 Level 3 Level 2 Angle = 75° ? Level 1

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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° Therefore: If we have the non-diagonal length: multiply by √2. If we have the diagonal length: divide by √2. ? x 45° x

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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° 45° ? 5 3 45° 45° 3 ?

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**#1: Quickly getting a diagonal**

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

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**Tip #1: Quickly getting a diagonal**

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

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**#2: Cutting and reforming**

Frequently, we can chop a shape into bits and put them back together in some other way, clearly with the same area. 2cm There’s a faster way we could have solved this problem without needing to calculate the shaded/unshaded area! 2cm 2cm 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!

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**#2: Cutting and reforming**

Frequently, we can chop a shape into bits and put them back together in some other way, clearly with the same area. Question: The diagram shows three semicircles, each of radius 1. What is the size of the total shaded 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. IMC Level 5 Level 4 A: π + 2 B: 5 C: (3/2)π + 1 Level 3 Level 2 D: 4 E: 2π - 1 Level 1

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**#3: Similar Triangles A: 12/25 B: 24/49 C: 1/2 D: 25/49 **

When triangles are similar, we can form an equation. Question: A square is inscribed in a right-angled triangle as shown. What fraction of the triangle does it occupy? We know these triangles are similar because their angles are the same. 5 3 4 IMC 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. Level 5 Level 4 A: 12/25 B: 24/49 C: 1/2 Level 3 Level 2 D: 25/49 E: 13/25 Level 1

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**#3: Similar Triangles = ? 3-x x ? _x_ 4-x ? = ? ? ? ? ? ? ? a b c d**

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 ? 3-x 5 3 3-x x x ? _x_ 4-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:

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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. Use what you know! 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. I’m told in the question the shape is a square. I can introduce variables for unknown quantities. IMO Macclaurin Hamilton Cayley

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**#4: Forming an equation! x y 2x-y x2 + y2 = (2x-y)2**

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. Since we’re trying to find the ratio of sides around triangle GCM, introduce unknowns around here. x Since the paper is square and it was folded, GM = GB. And GB = 2x – y y 2x-y Using Pythagoras Theorem, we can form an equation to give the relationship of x and y. x2 + y2 = (2x-y)2 So ratio of x:y is 4:3, and using Pythag, GM would be 5. Simplifying: x = y 4 3

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**#5: Adding circle radii x r r**

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? x r Add and label lines for the radii of the big circle that you think might help us solve the problem. >> Click here to reveal r Do the same for the smaller circle, labelling these lengths x. >> Click here to reveal

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**#5: Adding circle radii ___r___ x = 1 + √ 2 x r ? r ?**

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? x r ? Answer: x + x√2 = (1+ √2)x r Therefore: ___r___ 1 + √ 2 ? x =

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**#6: Segment of a circle This line is known as a chord.**

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.

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**# #6: Segment of a circle r ? r θ**

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.

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**#6: Segment of a circle A Radius of circle centred at A: √2 ?**

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 ?

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**#6: Segment of a circle B 1 √2 A O 1 C Area of sector = π/2 ?**

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

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**#6: Segment of a circle B 1 √2 A O 1 C So area of segment = (π/2) - 1**

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

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**r h #7: Equating areas x° Answer: π/4 ?**

When you’re told areas are equal, form an equation and simplify. Use what you know! Question: The area of the semi-circle and isosceles triangle are the same. What is tan x? I can form an expression for tan x for a right-angled triangle. Answer: π/4 ? I can find the area of an isosceles triangle by splitting it in two. x° As usual, introduce some variables for unknowns, and split our isosceles triangle so we can use trigonometry.. I can introduce variables to represent unknown quantities. r IMC So tan x = h/r Level 5 h Area of triangle = 2rh Angle of semi-circle = ½ π r2 Level 4 Level 3 Equating and simplifying: h = ¼ π r Level 2 So tan x = (¼ π r) / r = ¼ π Level 1

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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 R2 ¾ √3 R2 ? (4 – π)R2 ? R2 ? ? (π – 2)R2

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