topic 3: geometry

Topic 3: Geometry

Dr J Frost ([email protected])

Slide guidance

Key to question types:

SMC Senior Maths Challenge

BMO British Maths Olympiad

www.ukmt.org.uk The level, 1 being the easiest, 5 the hardest, will be indicated.

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.

Frost A Frosty SpecialQuestions from the deep dark recesses of my head.

Uni University InterviewQuestions used in university interviews (possibly Oxbridge).

Classic ClassicWell known problems in maths.

MAT Maths Aptitude TestAdmissions test for those applying for Maths and/or Computer Science at Oxford University.

STEP STEP ExamExam used as a condition for offers to universities such as Cambridge and Bath.

http://www.ukmt.org.uk/

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.

A: London B: Paris C: Madrid

For multiple choice questions (e.g. SMC), click your choice to reveal the answer (try below!)

Question: The capital of Spain is:

Topic 3: GeometryPart 1 – General Pointers

a. Adding helpful sides

b. Using variables for unknowns/Using known information

a. Fundamentals

b. Exterior/Interior Angles of a Polygon

a. Key Theorems

Part 2a – Angles

Part 2b – Circle Theorems

b. Using them backwards!

c. Intersecting Chord Theorem

Part 3 – Lengths and Area

a. The “√2 trick”.

b. Forming equations

c. 3D Pythagoras and the “√3 trick”.

d. Similar Triangles

Topic 3: Geometry

e. Forming Equations

f. Area of sectors/segments

Part 4 – BMO Problem Guidance

[Coming soon!]

ζPart 1: General PointersTopic 3 – Geometry

General tips and tricks that will help solve more difficult geometry problems.

#1: Adding linesBy 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

6

4 Adding the extra line in this case allows us to form a right-angled triangle, and thus we can exploit Pythagoras Theorem.

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!).

#1: Adding linesBy adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles.

If you were working out the length of the dotted line, what line might you add and what lengths would you identify?

?4

2

#1: Adding linesFor problems involving circles, add radii in strategic places, e.g. towards a point on the circumference where the circle touches another shape.

R

Rrr

r

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?

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.

?

#1: Adding linesFor 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?105105

1414

105 105

14

r

105

r

105

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!

?

#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?

By adding these lines, we can come up with an expression for the shaded area: π(r2

2 – r12)

And by Pythagoras:p2 + r1

2 = r22, so r2

2 – r12 = p2.

So the shaded area is πp2.

?

p

r1 r2

Source: SMC

#1: Adding Lines

Question: What is angle x + y?

x°

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A: 270

B: 300 C: 330

D: 360 E: More info needed

y°

y° Adding the appropriate extra line makes the problem trivial.

#2: Introducing variablesIt’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!

Macclaurin

Hamilton

Cayley

IMO

x

y2x-y

?

ζPart 2a: Angle FundamentalsTopic 3 – Geometry

Problems that involve determining or using angles.

#1: FundamentalsMake 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

180°-x2?

180°-2x

x

?

yx+y

The exterior angle of a triangle (with its extended line) is the sum of the other two interior angles.

?YOU SHOULD

ACTIVELY SEEK OUT OPPORTUNITIES

TO USE THIS!!

#1: FundamentalsMake sure you can rapidly apply your laws of angles. Fill in everything you know, introduce variables if necessary, and exploit equal length sides.

a

b

270 – a - b?

What is the expression for the missing side?

Angles of quadrilateral add up to 360°.

#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 = 10

36°?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.

Exterior angle = 60°Interior angle = 120°

?

?

Exterior angle = 72°Interior angle = 108°

??

#2: Interior/Exterior Angles of Regular PolygonsIt’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 = AB. What is the ratio x:y:z?

A: 1:2:3

B: 2:2:3 C: 2:3:4

D: 3:4:5 E: 3:4:6

A B

C

D

E

Fzyx


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

Fzyx

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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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


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°)

ζPart 2b: Circle TheoremsTopic 3 – Geometry

You should know most of these already. Although there’s a couple you may not have used (e.g. intersecting chord theorem).

x

x

Chord

Tangent

Alternate Segment Theorem:The angle subtended by a chord is the same as the angle between the chord and its tangent.

x x

x

2x

x

x180-x

Angles of a cyclic quadrilateral

1 2

34

5

Thinking backwardsThe Circle Theorems work BOTH WAYS…

If a circle was circumscribed around the triangle, side AB would be the diameter of the circle.

x

x180-x

If the opposite angles of a quadrilateral add up to 180, then the quadrilateral is a cyclic quadrilateral.

A B

Using the theorems this way round will be particularly useful in Olympiad problems.

12

20

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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?

A: 8

B: 10 C: 12

D: 10√2 E: 16

Circle Theorems

C

BA

If we draw the diameter of the circle, we have a 90° angle at C by our Circle Theorems. Then use Pythagoras.

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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?

A: 20

B: 25 C: 30

D: 35 E: 40

Circle Theorems

By Alternate Segment TheoremP

Q

R S

x°

Not toscale

a°

b°

c° x

By ‘Exterior Angle of Triangle’50

50

By Alternate Segment Theorem

40+xBy ‘Exterior Angle of Triangle’

Angles of this triangle add up to 180, so: 2x + 110 = 180Therefore x = 35

a

Version 1

Intersecting Chord Theorem

Version 2

b

x

y

ab = xy

a

b

xy

ab = xy

Chords intersect inside circle. Chords intersect outside circle.

This is a useful tool in our arsenal particularly for Olympiad problems.

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IMC

Question: What is the angle within this regular dodecagon? (12 sides)

#3b: Forming circles around regular polygonsBy drawing a circle around a regular polygon, we can exploit circle theorems.

x

Angle = 75°?

By our circle theorems, x is therefore half of this.

This angle is much easier to work out. It’s 5 12ths of the way around a full rotation, so 150°.

ζPart 3: Lengths and AreasTopic 3 – Geometry

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°

45°

x

x?

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

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°

45°

3

3? 5

?

The “√2 trick”

The radius of the circle is 1. What is the side length of the square inscribed inside it?

1

1/√2

1

√2or

√2?

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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?

A: √15/4

B: 5/2 C: √6

D: 2√2 E: √10

3D Pythagoras

P

S

RQ

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?

3D Pythagoras

P

S

RQ

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21

22√2

√5√5

√2 √2

√5√5So the height of this triangle by Pythagoras is √3.So that area is ½ x 2√2 x √3 = √6

Question: What’s the longest diagonal of a cube with unit length?

3D Pythagoras

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.

1

√2

1√3

1

?

Question: A cube is inscribed within a sphere of diameter 1m. What is the surface area of the cube?

3D Pythagoras

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A: 2m2

B: 3m2 C: 4m2

D: 5m2 E: 6m2

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

Forming EquationsTo find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same.

R

Rrr

r

Returning to this previous problem, what is r in terms of R?

Equating lengths:R = r + r√2 = r(1 + √2)

r = __r__1+ √2

?

Forming EquationsTo find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same.

Cayley

Hamilton

Macclaurin

IMO

Question: What is the radius of the small circle?

6

4

2r

r2

This is a less obvious line to add, but allows us to use Pythagoras to form an equation.

4-r

(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.

Similar TrianglesWhen triangles are similar, we can form an equation.

5

x4

3

4-x

3-x

x 3-xx

_x_4-x

??

??=

Key Theory: If two triangles are similar, then their ratio of width to height is the same. a

bc

d ab =

cd

?

?

Multiplying through by the denominators give us: (4-x)(3-x) = x2

Expanding and solving gives us: x = 127 So the area of the square is: 144

49

Since the area of the big triangle is 6, the fraction the square occupies is:2449

?? ?

?

Question: Find the percentage of the area of a 3:4:5 triangle occupied by a square inscribed inside it.

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.

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 πr2 because θ/360 gives us the proportion of the circle we’re using).

r

r

A

B

O

θThen cut out the triangle (i.e. subtract its area). We could work out its area by splitting it in two and using trigonometry.

#

?


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?


Segment of a circle

A1

1 √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?


Segment of a circle

A1

1 √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

π 2

π 2 ?


Segment of a circleSome 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?

Start with sector.

Cut out these two triangles.

Which leaves this region.

Segment of a circle

Given the arc has equation x2 + y2 = 1, then halfway along, 0.25 + y2 = 1, so y = √3/2

Then just multiply this area by 4 to finish...

ζPart 4: BMO Problem GuidanceTopic 3 – Geometry

I haven’t got round to making these slides yet.Try and re-download these later in the year!

topic 3: geometry

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