1. Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com
Topic 1: Geometry
Dr J Frost
All Maths Challenge and Olympiad problems are © UK Mathematics Trust (
Last modified: 27th January 2017

2. Topic 1: Geometry Part 1 – General Pointers 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

3. 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

4. ζ Part 1: General Pointers Topic 1 – Geometry
General tips and tricks that will help solve more difficult geometry problems.

5. #1 Adding Lines
By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles. 4
Simple example: What’s the area of this triangle? 12 5 ?
Adding the extra line in this case allows us to form a right-angled triangle, and thus we can exploit Pythagoras Theorem. 6

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

7. #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 ?
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.

8. #1 Adding Lines
By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles.
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 14 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!
Source: UKMT

9. #1 Adding Lines
𝑟 1
𝑟 2
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 ?
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:
𝒓 𝟏 𝟐 + 𝒑 𝟐 = 𝒓 𝟐 𝟐 𝑨=𝝅 𝒓 𝟐 𝟐 −𝝅 𝒓 𝟏 𝟐 =𝝅 𝒑 𝟐
Source: [SMC 1999 Q18]

10. #1 Adding Lines A: 270   B: 300 C: 330  D: 360 
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. y°
Source: [SMC 2006 Q3]

11. But don’t overdo it… #1 Adding Lines
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.

12. #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. 𝒙
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! 𝒚
𝟐𝒙−𝒚 ?
Source: [Cayley 2010 Q5]

13. ζ Part 2a: Angle Fundamentals Topic 1 – Geometry
Problems that involve determining or using angles.

14. YOU SHOULD ACTIVELY SEEK OUT OPPORTUNITIES TO USE THIS!!
#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 ?
I’ve been through every single SMC problem on geometry since The ‘exterior angle of triangle’ law comes up very frequently.
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.

15. #1: Fundamentals a b ? What is the expression for the missing side?
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

16. #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°

17. #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° ?

18. #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.
[IMC 1998 Q19] 𝐴𝐵𝐶𝐷𝐸 is a regular pentagon. 𝐹𝐴𝐵 is a straight line. 𝐹𝐴=𝐴𝐵. What is the ratio 𝑥:𝑦:𝑧? A B C D E F z y x 
A: 1:2:3
B: 2:2:3  
C: 2:3:4 
D: 3:4:5 
E: 3:4:6

19. #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.
[IMC 1998 Q19] 𝐴𝐵𝐶𝐷𝐸 is a regular pentagon. 𝐹𝐴𝐵 is a straight line. 𝐹𝐴=𝐴𝐵. 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.

20. #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.
[SMC 1999 Q7] 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 1:4, then the exterior angle must be 180÷5=36 (since interior and exterior angle add up to 180).
Thus there’s 360÷36=10 sides.

21. ζ Part 2b: Circle Theorems Topic 1 – Geometry
You should know most of these already. Although there’s a couple you may not have used (e.g. intersecting chord theorem).

22. 1 2 3 5 4 x diameter x x x x 180-x x 2x Alternative Segment Theorem:
Chord
Tangent
Alternative Segment Theorem:
The angle subtended by a chord is the same as the angle between the chord and its tangent.
diameter 3 5 4 x x x x
180-x x 2x
Angles in same segment
Angles of a cyclic quadrilateral

23. 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.

24. 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 𝟐𝒙.
So our point isn’t necessarily at the centre. 2x 2x

25. Circle Theorems  A: 8 B: 10  C: 12  D: 10√2   E: 16
[SMC]: 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 12 20
If we draw the diameter of the circle, we have a 90° angle at C by our Circle Theorems. Then use Pythagoras. B A 
A: 8
B: 10 
C: 12 
D: 10√2  
E: 16

26. Circle Theorems x A: 20  B: 25  C: 30  D: 35  E: 40  50 50
[SMC]: 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 x
Angles of this triangle add up to 180, so: x = 180
Therefore x = 35
A: 20 
B: 25 
C: 30 
D: 35 
E: 40 

27. 𝑎𝑏=𝑥𝑦 𝑎 𝑏 𝑥 𝑦 Intersecting Chord Theorem
The proof is fairly easy if we add two lines, say connecting the leftmost point of each chord, and another connecting the rightmost points. Using the ‘angles in the same segment’ theorem, and given the opposite angles at the centre are the same, we have two similar triangles.

28. You may also wish to check out the Intersecting Secant Angles Theorem
Intersecting Secant Lengths Theorem
A secant is a line which passes through a circle. 𝐵 𝐴 𝑃 𝐶 𝐷
𝑃𝐴⋅𝑃𝐵=𝑃𝐶⋅𝑃𝐷
You may also wish to check out the Intersecting Secant Angles Theorem

29. Ptolemy’s Theorem A B D C 𝐴𝐶∙𝐵𝐷= 𝐴𝐵∙𝐷𝐶 + 𝐴𝐷∙𝐵𝐶
𝐴𝐶∙𝐵𝐷= 𝐴𝐵∙𝐷𝐶 + 𝐴𝐷∙𝐵𝐶
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.

30. Angle Bisector Theorem
One final theorem not to do with circles…
ratio of these… 𝐵
If the line 𝐴𝐷 bisects 𝐴𝐵 and 𝐴𝐶, then 𝐷
𝐵𝐷 𝐷𝐶 = 𝐴𝐵 𝐴𝐶 𝜃 𝜃
…is the same as the ratio of these. 𝐶 𝐴

31. Forming circles around regular polygons
By circumscribing a regular polygon, we can exploit circle theorems.
[IMC 2003 Q22] The diagram shows a regular dodecagon (a polygon with twelve equal sides and equal angles). What is the size of the marked angle?
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
Angle = 75° ?

32. ζ
Topic 1 – Geometry
Part 3: Lengths and Areas

33. 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 . ?
2 𝑥 𝑥
45° 𝑥

34. 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°
5 2 ? 3

35. 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 2 or ? 2

36. 3D Pythagoras A: √15/4  B: 5/2  C: √6   D: 2√2  E: √10
[SMC 1999 Q19] 𝑃 is a vertex of a cuboid and 𝑄, 𝑅 and 𝑆 are three points on the edges as shown. 𝑃𝑄=2𝑐𝑚, 𝑃𝑅=2𝑐𝑚 and 𝑃𝑆=1𝑐𝑚. What is the area, in cm2, of triangle 𝑄𝑅𝑆? Q P R S
A: √15/4 
B: 5/2 
C: √6  
D: 2√2 
E: √10

37. 3D Pythagoras
[SMC 1999 Q19] 𝑃 is a vertex of a cuboid and 𝑄, 𝑅 and 𝑆 are three points on the edges as shown. 𝑃𝑄=2𝑐𝑚, 𝑃𝑅=2𝑐𝑚 and 𝑃𝑆=1𝑐𝑚. What is the area, in cm2, of triangle 𝑄𝑅𝑆?
2√2 Q 2 P 2 R √5 1 √5 S
So the height of this triangle by Pythagoras is 3 .
So that area is
1 2 ×2 2 × 3 = 6 𝟓 𝟓 𝟐 𝟐

38. 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 .

39. 3D Pythagoras  A: 2m2 B: 3m2  C: 4m2   D: 5m2  E: 6m2
[SMC] 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× 1/ =2 m2 
A: 2m2
B: 3m2 
C: 4m2  
D: 5m2 
E: 6m2

40. Forming Equations Equating lengths: 𝑅=𝑟+𝑟 2 =𝑟 1+ 2 𝑟= 𝑅 1+ 2 𝒓 ? 𝑹 R
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 =𝑟 𝑟= 𝑅 1+ 2

41. Forming Equations
To find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same.
[Maclaurin 2006 Q3] Two circles are drawn in a rectangle of 6 by 4, such that the larger circle touches three sides of the rectangle, whereas the smaller one only touches 2. Determine the radius of the smaller circle.
This is a less obvious line to add, but allows us to use Pythagoras to form an equation.
𝟒−𝒓 2 𝒓 4
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

42. Forming Equations
To find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same.
As always, draw lines between the centres of touching circles.
[Maclaurin 2011 Q5] If 𝑎 and 𝑏 are the radii of the larger circles, and 𝑐 the radius of the smaller one, prove that:
1 𝑎 𝑏 = 1 𝑐
𝑎−𝑐
𝑏−𝑐
𝑎−𝑏
𝑥+𝑦 𝑥 𝑦
𝑏+𝑐
𝑎+𝑐
𝑎+𝑏
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 𝑥 and 𝑦. This would mean the width of the top triangle is 𝑥+𝑦.

43. Forming Equations
[Maclaurin 2011 Q5] If 𝑎 and 𝑏 are the radii of the larger circles, and 𝑐 the radius of the smaller one, prove that:
1 𝑎 𝑏 = 1 𝑐
𝑎−𝑐
𝑏−𝑐
𝑎−𝑏
𝑥+𝑦 𝑥 𝑦
𝑏+𝑐
𝑎+𝑐
𝑎+𝑏
𝑥 2 + 𝑎−𝑐 2 = 𝑎+𝑐 ⇒ 𝑥=2 𝑎𝑐
Similarly: 𝑦=2 𝑏𝑐
From from the top triangle: 𝑥 2 + 𝑦 2 +2𝑥𝑦=4𝑎𝑏
Substituting: 𝑎𝑐+𝑏𝑐+2𝑐 𝑎𝑏 =𝑎𝑏 Dividing by 𝑎𝑏𝑐: 1 𝑏 + 1 𝑎 𝑎𝑏 = 1 𝑐
Notice that the LHS is a perfect square!
1 𝑎 𝑏 2 = 1 𝑐

44. Inscription Problems = 3 4 ?
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? 1
3 2
30°
= 3 4 ?
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 3 /2.
The proportion taken up by the smaller circle is therefore 3/4.

45. Similar Triangles 𝑎 𝑏 = 𝑐 𝑑 ? = 𝟐𝟒 𝟒𝟗 ? ?
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
[IMC 1998 Q25] A square is inscribed inside a triangle. Determine the fraction of the triangle occupied by the square. 3 𝑥
= 𝟐𝟒 𝟒𝟗 ? 𝑥
4−𝑥 ? 4

46. 𝐷 𝐴 𝐵 𝐶 Similar Triangles 𝜃 𝛽 𝜃 Why are triangles 𝑨𝑩𝑫 and 𝑨𝑫𝑪 similar?
A particular common occurrence is to have one triangle embedded in another, where the indicated angles are the same. 𝜃 𝛽 𝜃 𝐴 𝐵 𝐶
Why are triangles 𝑨𝑩𝑫 and 𝑨𝑫𝑪 similar?
They share a second common angle at 𝐴.
We’ll see an example of this later on in this module.

47. 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.

48. # Segment of a circle r r θ
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

49. 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 ?

50. 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

51. 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

52. 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?

53. Segment of a circle Start with sector. Cut out these two triangles.
Which leaves this region.

54. Area of a Triangle 𝑎 𝑐 ℎ 𝐶 𝑏 1 2 𝑏ℎ ? 1 2 𝑎𝑏 sin 𝐶 ? 𝑠 𝑠−𝑎 𝑠−𝑏 𝑠−𝑐 ?
Using base and height: 2
Using two sides and angle between them:
1 2 𝑎𝑏 sin 𝐶 ? 3
Using three sides:
𝑠 𝑠−𝑎 𝑠−𝑏 𝑠−𝑐 ?
where 𝑠= 1 2 (𝑎+𝑏+𝑐), i.e. half the perimeter.
This is known as Heron’s Formula 𝑎 𝑐 ℎ 𝐶 𝑏

55. Area of a Triangle 𝜃 𝜃 4𝜃 𝜃 ? The circle has unit radius.
What is the area of the shaded region? (in terms of 𝜃) 4𝜃 𝜃 ? 𝜃
𝐴= sin 180−2𝜃 𝜃 360 ×𝜋
= sin 2𝜃 + 𝜋𝜃 90 𝜃
(Note that in general, sin 180−𝑥 = sin 𝑥 )

56. ζ
Topic 1 – Geometry
Part 4: Proofs

57. 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.

58. Incentre Circumcentre Centroid Orthocentre Centres of Triangles 𝑎 𝑎
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.

59. 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 90° 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 90°.
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.

60. Construct your diagram!
Example
Two circles are internally tangent at a point 𝑇. A chord 𝐴𝐵 of the outer circle touches the inner circle at a point 𝑃. Prove that 𝑇𝑃 bisects ∠𝐴𝑇𝐵.
[Source: UKMT Mentoring]
Construct your diagram! ? 𝐵
I’ve added the angle 𝑎, so that our proof boils down to showing that ∠𝑃𝑇𝐵=𝑎. 𝑃 𝐴 𝑎 𝑇

61. 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 ∠𝐴𝑇𝑋=𝑏 𝐵 𝑃 𝐴 𝑎 𝑋 𝑏 𝑇

62. What angle can we fill in next
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, ∠𝑋𝑇𝐴=∠𝐴𝐵𝑇.
But notice that the line 𝑃𝑇 is a chord attached to a tangent. If we added an appropriate line, we can use the theorem again: ∠𝑋𝑇𝑃=∠𝑃𝑈𝑇. 𝐵 𝑃 𝒃
𝒂+𝒃 𝑈 𝐴 𝑎 𝑋 𝑏 𝑇

63. We can just use very basic angle rules (angles on a straight line, internal angles of a triangle) to find that ∠𝑈𝑃𝐵=𝑎. Now what’s the final step?
That line 𝑃𝑈 added is convenient a chord attached to a tangent. So we can apply the Alternate Segment Theorem a third time. ∠𝐵𝑃𝑈=∠𝑃𝑇𝑈.
And we’re done, because we’ve shown ∠𝐴𝑇𝑃=∠𝑃𝑇𝑈! ? 𝐵 𝑃 𝒃
𝒂+𝒃 𝑈 𝒂 𝐴 𝒂 𝑎 𝑋 𝑏 𝑇

64. Construct your diagram!
One more…
Two intersecting circles 𝐶 1 and 𝐶 2 have a common tangent which touches 𝐶 1 at 𝑃 and 𝐶 2 at 𝑄. The two circles intersect at 𝑀 and 𝑁, where 𝑁 is closer to 𝑃𝑄 than M is. Prove that the triangles 𝑀𝑁𝑃 and 𝑀𝑁𝑄 have equal areas. [Source: UKMT Mentoring]
Construct your diagram! ? 𝑃 𝑄 𝑁
(It’s important to make your circles different sizes to keep things general) 𝑀

65. 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! 𝑃 𝑄 𝑎 𝑏 𝑁 𝑀

66. Click to show this on diagram
We have to show the two triangles have equal area. They have the same base (i.e. 𝑀𝑁) 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 𝑃𝑋=𝑋𝑄, then we’ve indirectly shown that the perpendicular distances from 𝑃 and 𝑄 to the line 𝑀𝑁 is the same.
Click to show this on diagram 𝑃 𝑋
We can see from the rectangle that if we can show 𝑋 is the midpoint of 𝑃𝑄, then the perpendicular heights of the triangle are both half the width of the rectangle, i.e. equal. 𝑄 𝑎 𝑏 𝑁 𝑀

67. ? 𝑃 𝑋 𝑄 𝑁 𝑀 So how could we prove that 𝑃𝑋=𝑋𝑄? 𝑎 𝑏
Look out for similar triangles! Notice that triangles 𝑃𝑋𝑁 and 𝑃𝑋𝑀 both share the angle 𝑎 and the angle ∠𝑃𝑋𝑀. So they’re similar. Thus 𝑋𝑃 𝑋𝑁 = 𝑋𝑀 𝑋𝑄 . So 𝑋 𝑃 2 =𝑋𝑀∙𝑋𝑁. Similarly, 𝑋 𝑄 2 =𝑋𝑀∙𝑋𝑁. So 𝑋 𝑃 2 =𝑋 𝑄 2 , and thus 𝑋𝑃=𝑋𝑄. And we’re done! 𝑃 𝑋 𝑄 𝑎 𝑏 𝑁 𝑀

68. Construct your diagram!
Final Example
Two circles 𝑆 and 𝑇 touch at 𝑋. They have a common tangent which meets 𝑆 at 𝐴 and 𝑇 and 𝐵. The points 𝐴 and 𝐵 are different. Let 𝐴𝑃 be a diameter of 𝑆. Prove that 𝐵, 𝑋 and 𝑃 lie on a straight line.
[Source: BMO Round ] 𝐴
Construct your diagram! ? 𝐵 𝑋 𝑎 𝑃

69. “Prove that 𝐵, 𝑋 and 𝑃 lie on a straight line (i.e. are collinear).”
How could we do this?
We just need to show that ∠𝑷𝑿𝑩=𝟏𝟖𝟎° ? 𝐴 𝐵 𝑋 𝑎 𝑃

70. 𝐴 𝑌 𝐵 𝑂 𝑄 𝑋 𝑃 2? 1? ? Now it’s a case of gradually filling in angles!
(But put in mind that we can’t assume 𝑃𝑋𝐵 is straight, because that’s the very thing we’re trying to prove)
2: 𝐴𝑃 is diameter so ∠𝐴𝑋𝑃=90°
1: 𝑂𝑃𝑋 is isosceles. 𝐴
3: Either Alternate Segment Theorem, or given that ∠𝐵𝐴𝑂=90° 𝑌 𝑎 3? 𝐵
180−2𝑎 5? 6? 2𝑎
90−𝑎 7?
4: 𝐴𝑌𝑋 is isosceles since triangle formed by two tangents. 𝑎 4?
90−𝑎 2? 𝑂 𝑋 𝑄 1? 𝑎
7: 𝑌𝑋𝐵 is isosceles (by same reasoning) 𝑎
How do we know when we’re done?
∠𝑃𝑋𝐵=∠𝐴𝑋𝑃+∠𝐴𝑋𝑌+∠𝑌𝑋𝐵=90°+𝑎+ 90°−𝑎 =180° 𝑃 ?

71. 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:
𝐴=𝑏𝑎𝑠𝑒×ℎ𝑒𝑖𝑔ℎ𝑡
𝐴= 1 2 𝑎𝑏 sin 𝐶

72. 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
𝐴=𝑏𝑎𝑠𝑒×ℎ𝑒𝑖𝑔ℎ𝑡
𝐴= 1 2 𝑎𝑏 sin 𝐶 ? ?
Increasing 𝑎 or 𝑏 will clearly increase 𝐴, so for 2 of the sides, we can set them to the maximum length.
sin 𝐶 will be maximum when 𝐶=90°.
Consider two sides of the triangle. The height of triangle will be maximised (and hence the area) when they’re 90° 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 2 and 3 as the base and height, then by Pythagoras, the hypotenuse is 13 , which is less than 4, so is fine!
If we have 2 and 4, the hypotenuse is √20 which is greater than 4, so our triangle is invalid. The same obviously happens if we use 3 and 4.
Thus the maximum possible area is 3.