topic 1: geometry

Topic 1: GeometryDr J Frost ([email protected])

Last modified: 21st August 2013

Key to question types:

SMC Senior Maths Challenge

BMO British Maths Olympiad

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.

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:

Slide Guidance

Topic 1: 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 1: Geometry

f. Inscription problems

e. Area of sectors/segments

Part 4 – Proofs

a. Generic Tips

b. Worked Examples

c. Proofs involving Area

ζPart 1: General PointersTopic 1 – Geometry

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

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

6

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

#1 Adding Lines

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!).

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 Lines

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 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?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?

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:

?

p

r1 r2

#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.

#1 Adding Lines

#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.

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!

Macclaurin

Hamilton

Cayley

IMO

𝒙𝒚𝟐 𝒙−𝒚

?

#2 Introducing Variables

ζPart 2a: Angle FundamentalsTopic 1 – 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


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.

ζPart 2b: Circle TheoremsTopic 1 – 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

Alternative 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

Angles in same segment

diameter

Thinking backwardsFor 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.

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.

Thinking backwards

4

x

2x2x

For many of the circle theorems, the CONVERSE is true…

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?

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.

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

C

BA

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

Circle Theorems

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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

By Alternative Segment TheoremP

Q

R S

x°

Not toscale

a°

b°

c° x

By ‘Exterior Angle of Triangle’50

50

By Alternative Segment Theorem

40+xBy ‘Exterior Angle of Triangle’

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

Circle Theorems

𝑎

𝑏

𝑥

𝑦

𝑎𝑏=𝑥𝑦

Intersecting Chord Theorem

𝐵

𝑃𝐴⋅𝑃𝐵=𝑃𝐶⋅𝑃𝐷

A secant is a line which passes through a circle.

Intersecting Secant Lengths Theorem

𝐴𝑃

𝐶𝐷

You may also wish to check out the Intersecting Secant Angles Theorem

A

i.e. The product of the diagonals of a cyclic quadrilateral is the sum of the products of the pairs of opposite sides.

B

CD

Ptolemy’s Theorem

You’ll be able to practice this in Geometry Worksheet 3.

𝐵𝐷𝐷𝐶=

𝐴𝐵𝐴𝐶

Angle Bisector Theorem

One final theorem not to do with circles…

𝜃𝜃

𝐴

𝐵

𝐶

𝐷

ratio of these…

…is the same as the ratio of these.

If the line bisects and , then

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

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

1

1/√2

1

√2or

√2?

The “√2 trick”

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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

P

S

RQ

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?

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

3D Pythagoras

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

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

?

3D Pythagoras

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

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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

3D Pythagoras

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

?


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.


Cayley

Hamilton

Macclaurin

IMO

Question: If and are the radii of the larger circles, and the radius of the smaller one, prove that:

𝑏+𝑐

𝑎+𝑐𝑎+𝑏𝑎−𝑐

𝑏−𝑐

𝑎−𝑏𝑥+𝑦

𝑥 𝑦As 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 .

As always, draw lines between the centres of touching circles.

Question: If and are the radii of the larger circles, and the radius of the smaller one, prove that:

𝑏+𝑐

𝑎+𝑐𝑎+𝑏𝑎−𝑐

𝑏−𝑐

𝑎−𝑏𝑥+𝑦

𝑥 𝑦

Similarly: From from the top triangle:

Substituting: Dividing by : Notice that the LHS is a perfect square!

Forming Equations

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?

Inscription Problems

¿34

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 .

1√32 30 ° ?

¿𝟐𝟒𝟒𝟗

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

5

x4

3

4-x

3-x

x

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

bc

d 𝑎𝑏=

𝑐𝑑

?

?

?

Question: A square is inscribed inside a 3-4-5 triangle. Determine the fraction of the triangle occupied by the square.

Similar Triangles

𝜃

𝜃

A particular common occurrence is to have one triangle embedded in another, where the indicated angles are the same.

𝐴 𝐵

𝐷

Why are triangles and similar?

𝐶

They share a second common angle at .We’ll see an example of this later on in this module.

𝛽

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

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.

r

r

A

B

O

θ

#


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

Using base and height:

𝐶

𝑎

𝑏

𝑐

Using two sides and angle between them:

Using three sides:

12 h𝑏

12 𝑎𝑏 sin𝐶

√ 𝑠 (𝑠−𝑎 ) (𝑠−𝑏 ) (𝑠−𝑐 )where , i.e. half the perimeter.

?

?

?

1

2

3

This is known as Heron’s Formula

h

Area of a Triangle

Area of a Triangle

The circle has unit radius.What is the area of the shaded region? (in terms of )

𝜃𝜃

4 𝜃

𝜃

𝜃

(Note that in general, )

?

ζPart 4: ProofsTopic 1 – Geometry

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

Intersection of angle bisectors.Note that the incentre is the centre of the inscribed circle (hence the name!)

𝑎𝑎

Intersection of perpendicular bisectorsSimilarly, this is the centre of a circumscribing circle.

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)

Circumcentre

Centroid Orthocentre

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

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:

1. Think about the significance of each piece of information given to you:a. 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.

b. 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).

c. We’re given the diameter?The angle subtended by any point on the circumference is .

2. Use variables to represent appropriate unknown angles/lengths.3. Look out for similar triangles whenever you notice angles that are the

same. This allows us to compare lengths.4. As usual, look out for lengths which are the same (e.g. radii of a circle).5. Justify your assumptions. It’s incredibly easy to lose easy marks in the

BMO due to lack of appropriate justification.

Golden Rules of Geometric Proofs

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]

𝑇

𝐵

𝐴𝑃

𝑎

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

Construct your diagram!

?

Example

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

𝑇

𝐵

𝐴𝑃

𝑎

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

𝑋𝑏

?

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: .

𝑇

𝐵

𝐴𝑃

𝑎

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?

𝑋𝑏

?

𝒃𝒂+𝒃𝑈

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 !

𝑇

𝐵

𝐴𝑃

𝑎

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?

𝑋𝑏

?

𝒃𝒂+𝒃𝑈𝒂

𝒂

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]

𝑃𝑄

𝑁

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


?

One more…

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

𝑃𝑄

𝑁

𝑀

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

𝑎 𝑏

𝑎𝑏

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?

𝑃𝑄

𝑁

𝑀

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.

𝑎 𝑏

𝑎𝑏

𝑋

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.

Click to show this on diagram?

So how could we prove that ?

𝑃𝑄

𝑁

𝑀

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!

𝑎 𝑏

𝑎𝑏

𝑋

?

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]

𝐴𝐵

𝑋

𝑃

𝑎


?

Final Example

“Prove that , and lie on a straight line (i.e. are collinear).”How could we do this?

We just need to show that

𝐴𝐵

𝑋

𝑃

𝑎

?

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)

𝐴𝐵

𝑋

𝑃

𝑎

𝑎90−𝑎

𝑎

𝑎

180−2𝑎2𝑎

90−𝑎

1?

2?

3?

4?

5? 6?

7?

1: is isosceles.

𝑂 𝑄

2: is diameter so

3: Either Alternate Segment Theorem, or given that

4: is isosceles since triangle formed by two tangents.

𝑌

7: is isosceles (by same reasoning)

How do we know when we’re done?

?

“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:

𝐴=𝑏𝑎𝑠𝑒×h h𝑒𝑖𝑔 𝑡 𝐴=12 𝑎𝑏sin𝐶

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]

𝐴=𝑏𝑎𝑠𝑒×h h𝑒𝑖𝑔 𝑡

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).

𝐴=12 𝑎𝑏sin𝐶

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

will be maximum when .

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.

Method 1 Method 2

??

Other types of Geometric Proof

topic 1: geometry

Documents

extra lines

useful angles

larger circle

smaller circle r

verticalhorizontal lines

red lines

bmo round

pythagoras theorem