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Topic 1: Geometry

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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.

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

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

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

ζ

Topic 1 – Geometry

Part 1: General Pointers

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

#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

#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

#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.

#1 Adding Lines

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

#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:

#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

#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.

#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

ζ

Topic 1 – Geometry

Part 2a: Angle Fundamentals

Problems that involve determining or using angles.

#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.

#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

#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°

?

#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°

?

#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

#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.

#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

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).

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

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.

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

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

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

Intersecting Chord Theorem

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

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.

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.

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

ζ

Topic 1 – Geometry

Part 3: Lengths and Areas

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

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

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

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

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

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.

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

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

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

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

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!

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 .

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

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.

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.

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

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

?

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

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

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?

Segment of a circle

Start with sector.

Cut out these two triangles.

Which leaves this region.

Area of a Triangle

?

1

Using base and height:

Using two sides and angle between them:

2

?

Using three sides:

3

?

where , i.e. half the perimeter.

This is known as Heron’s Formula

Area of a Triangle

The circle has unit radius.

What is the area of the shaded region? (in terms of )

?

(Note that in general, )

ζ

Topic 1 – Geometry

Part 4: Proofs

Some Quick Definitions

“Inscribe”

For a shape to put inside another so that at least some of the points on the inner shape are on the perimeter of the outer shape.

“Circumscribe”

To surround a shape with a circle, such that the vertices of the shape are on the circumference of the circle.

It is possible to circumscribe any triangle and any regular polygon.

“Collinear”

Points are collinear if a single straight line can be drawn through all of them.

Centres of Triangles

Incentre

Circumcentre

Intersection of angle bisectors.

Note that the incentre is the centre of the inscribed circle (hence the name!)

Intersection of perpendicular bisectors

Similarly, this is the centre of a circumscribing circle.

Centroid

Orthocentre

Intersection of medians

Intersection of altitudes

(i.e. a line from a vertex to the opposite side such that the altitude and this side are perpendicular)

The circumcentre, centroid and orthocentre are collinear! The line that passes through these three centres is known as an Euler Line.

Golden Rules of Geometric Proofs

- Often we need to prove that some line bisects others, or that lengths/angles are the same. Here’s a few golden rules of proofs:
- Think about the significance of each piece of information given to you:
- We have a tangent?We’ll likely be able to use the Alternate Segment Theorem (which you should expect to use a lot!). If there’s a chord attached, use it immediately. If there’s isn’t a chord, consider adding an appropriate one so we can use the theorem!Also, the presence of the radius (or adding the radius) gives us a angle.
- Two circles touch?We have a tangent. The centres of the circles and the point of contact are collinear, and we can use the tips in (a).
- We’re given the diameter?The angle subtended by any point on the circumference is .

- Use variables to represent appropriate unknown angles/lengths.
- Look out for similar triangles whenever you notice angles that are the same. This allows us to compare lengths.
- As usual, look out for lengths which are the same (e.g. radii of a circle).
- Justify your assumptions. It’s incredibly easy to lose easy marks in the BMO due to lack of appropriate justification.

Example

Two circles are internally tangent at a point . A chord of the outer circle touches the inner circle at a point . Prove that bisects .

[Source: UKMT Mentoring]

Construct your diagram!

?

I’ve added the angle , so that our proof boils down to showing that .

Our usual good starting point is to label an unknown angle to help us work out other angles. But which would be best?

We have a tangent to not one but two circles! We clearly want to use the Alternate Segment Theorem. So let’s say label

?

What angle can we fill in next. Is there perhaps a line I can add to my diagram to use the Alternate Segment Theorem a second time?

By the Alternate Segment Theorem, .

But notice that the line is a chord attached to a tangent. If we added an appropriate line, we can use the theorem again: .

?

We can just use very basic angle rules (angles on a straight line, internal angles of a triangle) to find that . Now what’s the final step?

That line added is convenient a chord attached to a tangent. So we can apply the Alternate Segment Theorem a third time. .

And we’re done, because we’ve shown !

?

One more…

Two intersecting circles and have a common tangent which touches at and at . The two circles intersect at and , where is closer to than M is. Prove that the triangles and have equal areas. [Source: UKMT Mentoring]

Construct your diagram!

?

(It’s important to make your circles different sizes to keep things general)

We have a tangent, so what would be a sensible first step?

?

We also have some chords, so we should use the Alternate Segment Theorem!

We have to show the two triangles have equal area. They have the same base (i.e. ) so we need to show they have the same perpendicular height. What could we do?

?

A common strategy is to extend a line onto another. If we can show , then we’ve indirectly shown that the perpendicular distances from and to the line is the same.

Click to show this on diagram

We can see from the rectangle that if we can show is the midpoint of , then the perpendicular heights of the triangle are both half the width of the rectangle, i.e. equal.

So how could we prove that ?

?

Look out for similar triangles! Notice that triangles and both share the angle and the angle . So they’re similar. Thus . So . Similarly, . So , and thus . And we’re done!

Final Example

Two circles and touch at . They have a common tangent which meets at and and . The points and are different. Let be a diameter of . Prove that , and lie on a straight line.

[Source: BMO Round 1 - 2013]

Construct your diagram!

?

“Prove that , and lie on a straight line (i.e. are collinear).”

How could we do this?

We just need to show that

?

Now it’s a case of gradually filling in angles!

(But put in mind that we can’t assume is straight, because that’s the very thing we’re trying to prove)

2: is diameter so

1: is isosceles.

3: Either Alternate Segment Theorem, or given that

3?

5?

6?

7?

4: is isosceles since triangle formed by two tangents.

4?

2?

1?

7: is isosceles (by same reasoning)

How do we know when we’re done?

?

Other types of Geometric Proof

“A triangle has lengths of at most 2, 3 and 4 respectively. Determine, with proof, the maximum possible area of the triangle.”

[Source: BMO Round 1 – 2003]

What might be going through your head:

“Well the question wants us to maximise area, so maybe I should think about the formula for the area of a triangle?”

Formulae for area of a triangle:

Other types of Geometric Proof

“A triangle has lengths of at most 2, 3 and 4 respectively. Determine, with proof, the maximum possible area of the triangle.”

[Source: BMO Round 1 – 2003]

Method 1

Method 2

?

?

Increasing or will clearly increase , so for 2 of the sides, we can set them to the maximum length.

will be maximum when .

Consider two sides of the triangle. The height of triangle will be maximised (and hence the area) when they’re apart. And we know the making either of these two lengths larger will increase the area of the triangle. We then just have to consider right-angled triangles with sides (2, 3) or (2, 4) or (3, 4) and see if the third side is valid (we’ll do this in a second).

- The just like before, we have to consider each possible pair of sides which are fixed.
- If we have and as the base and height, then by Pythagoras, the hypotenuse is , which is less than 4, so is fine!
- If we have and , the hypotenuse is which is greater than 4, so our triangle is invalid. The same obviously happens if we use and 4.
- Thus the maximum possible area is 3.