Download Topic 1 - Dr Frost Maths

Topic 1: Geometry
Dr J Frost ([email protected])
www.drfrostmaths.com
All Maths Challenge and Olympiad problems are
© UK Mathematics Trust (www.ukmt.org.uk)
Last modified: 27th January 2017
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
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
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?
5
4
6
?
12
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
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?
4
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 Lines
By adding extra lines to your diagram, you can often form shapes whose
properties we can exploit, or find useful angles.
r
r
R
r
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
By adding extra lines to your diagram, you can often form shapes whose
properties we can exploit, or find useful angles.
105
105
105
105
r
r
105
105
14
14
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?
14
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
#1 Adding Lines
𝑝
𝑟1
𝑟2
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:
𝒓𝟐𝟏 + 𝒑𝟐 = 𝒓𝟐𝟐
𝑨 = 𝝅𝒓𝟐𝟐 − 𝝅𝒓𝟐𝟏 = 𝝅𝒑𝟐
Source: [SMC 1999 Q18]
#1 Adding Lines
Question: What is angle x + y?
x°
A: 
270

B: 300
D: 360

E: More info
 needed
C: 
330
Adding the
appropriate extra
line makes the
problem trivial.
y°
y°
Source: [SMC 2006 Q3]
#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.
𝒙
𝒚
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!
Source: [Cayley 2010 Q5]
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
?
x
x+y
?
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: 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.
a
b
What is the expression for
the missing side?
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.
[IMC 1998 Q19] 𝐴𝐵𝐶𝐷𝐸 is a regular pentagon. 𝐹𝐴𝐵 is a
straight line. 𝐹𝐴 = 𝐴𝐵. What is the ratio 𝑥: 𝑦: 𝑧?
F
A
x
B
y
z
C
E
D
A: 1:2:3

B: 2:2:3

D: 3:4:5

E: 3:4:6

C: 2:3:4

#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?
F
A
x
B
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.
[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 
D: 10
E: 12
C: 9 
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.
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
x
3
x
x
5
4
x
x
180-x
x
2x
Angles in same
segment
Angles of a cyclic
quadrilateral
Thinking backwards
For many of the circle theorems, the CONVERSE is true…
A
B
x
180-x
x
If a circle was circumscribed
around the triangle, side AB
would be the diameter of the
circle.
If the opposite angles of a
quadrilateral add up to 180,
then the quadrilateral is a
cyclic quadrilateral.
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.
x
2x
2x
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 𝟐𝒙.
So our point isn’t necessarily at the centre.
Circle Theorems
[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
A
20
A: 8 
B: 10
D: 10√2

E: 16
If we draw the
B diameter of the
circle, we have a 90°
angle at C by our
Circle Theorems.
Then use Pythagoras.
C: 12
Circle Theorems
[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?
b°
P
50
x°
40+x
c°
R
By Alternative Segment
Theorem
By ‘Exterior Angle of
Triangle’
By Alternative Segment
Theorem
a°
Not to
scale
Q
By ‘Exterior Angle of
Triangle’
50
x
S
A: 20
B: 25
D: 35
E: 40
Angles of this triangle add up to
180, so:
2x + 110 = 180
Therefore x = 35
C: 30
Intersecting Chord Theorem
𝑎
𝑥
𝑎𝑏 = 𝑥𝑦
𝑦
𝑏
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
Ptolemy’s Theorem
A
B
𝐴𝐶 ∙ 𝐵𝐷 =
𝐴𝐵 ∙ 𝐷𝐶 + 𝐴𝐷 ∙ 𝐵𝐶
D
C
You’ll be able to practice this
in Geometry Worksheet 3.
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…
𝐵
ratio of these…
If the line 𝐴𝐷 bisects 𝐴𝐵 and 𝐴𝐶, then
𝐷
𝐵𝐷 𝐴𝐵
=
𝐷𝐶 𝐴𝐶
𝜃
𝜃
𝐶
𝐴
…is the same as the
ratio of these.
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°
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°
2?𝑥
𝑥
45°
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°
5
?
3
45°
3
5
?2
The “√2 trick”
The radius of the circle is 1. What is the side length of the square inscribed inside it?
1
2
1
or
1
2
?2
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?
Q
P
R
S
A: √15/4

B: 5/2

D: 2√2

E: √10

C: √6
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?
Q
2√2
2
P 2
√5
1
R
√5
S
𝟓
𝟓
𝟐
𝟐
So the height of this triangle by
Pythagoras is 3.
So that area is
1
×2 2× 3= 6
2
3D Pythagoras
Question: What’s the longest diagonal of a cube with unit length?
1
𝟑
1
𝟐
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
[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/ 3
A: 2m
2
B: 3m2
D: 5m
2
E: 6m2
C: 4m2
2
= 2 m2
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
𝒓
R
Equating lengths:
𝑅 =𝑟+𝑟 2
= 𝑟 1?+ 2
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.
4
2
2
𝒓
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.
6
𝒓
[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.
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.
𝑥+𝑦
𝑎−𝑏
𝑎−𝑐
[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
1
1
+
=
𝑎
𝑐
𝑏
𝑏−𝑐
𝑥
𝑦
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.
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
1
1
+
=
𝑎
𝑐
𝑏
𝑥+𝑦
𝑎−𝑏
𝑎−𝑐
𝑏−𝑐
𝑥
𝑥2 + 𝑎 − 𝑐
Similarly:
From from the top triangle:
𝑦
2
= 𝑎+𝑐
2
⇒
𝑥 = 2 𝑎𝑐
𝑦 = 2 𝑏𝑐
𝑥 2 + 𝑦 2 + 2𝑥𝑦 = 4𝑎𝑏
1
1
Substituting: 𝑎𝑐 + 𝑏𝑐 + 2𝑐 𝑎𝑏 = 𝑎𝑏
Dividing by 𝑎𝑏𝑐: 𝑏 + 𝑎 +
Notice that the LHS is a perfect square!
2
1
1
1
+
=
𝑐
𝑎
𝑏
2
𝑎𝑏
=
1
𝑐
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?
3
2 30°
1
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.
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.
3?
−𝑥
5
3
𝑥
𝑥
4
4−
?𝑥
𝑏
𝑑
𝑎
𝑐
𝑎 𝑐
=
𝑏 𝑑
[IMC 1998 Q25] A square is inscribed inside a
3-4-5 triangle. Determine the fraction of the
triangle occupied by the square.
𝟐𝟒
= ?
𝟒𝟗
Similar Triangles
𝐷
A particular common
occurrence is to have one
triangle embedded in
another, where the indicated
angles are the same.
𝜃
𝛽
𝐴
𝜃
𝐵
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.
𝐶
Segment of a circle
Some area related problems require us to calculate a segment.
This line is
known as a
chord.
A ‘slice’ of a
circle is
known as a
sector.
The area bound
between a chord and
the circumference is
known as a segment.
(it resembles the shape of an
orange segment!)
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.
A
r
θ
O
B
r
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.
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?
1
O
?
Area of sector = π/2
√2
1
A
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
O
C
√2
1
Therefore (by cutting the
segment area from a semicircle):
A Area of shaded area
π π
= - ( –? 1) = 1
2
2
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:
2
Using two sides and
angle between them:
3
Using three sides:
1
𝑏ℎ
2
?
1
𝑎𝑏 sin 𝐶 ?
2
𝑠 𝑠−𝑎 𝑠−𝑏 𝑠−𝑐
1
where 𝑠 = (𝑎 +? 𝑏 + 𝑐), i.e. half the
2
perimeter.
𝑎
𝑐
ℎ
𝐶
𝑏
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 𝜃)
𝜃
𝐴
𝜃
4𝜃
𝜃
𝜃
1
= 2 sin 180 − 2𝜃
2
4𝜃
?
+
×𝜋
360
𝜋𝜃
= sin 2𝜃 +
90
(Note that in general, sin 180 − 𝑥 = sin 𝑥 )
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
𝑎
𝑎
Intersection of angle bisectors.
Note that the incentre is the centre of the inscribed
circle (hence the name!)
Centroid
Intersection of perpendicular bisectors
Similarly, this is the centre of a circumscribing circle.
Orthocentre
Intersection of medians
Intersection of altitudes
The circumcentre, centroid and orthocentre
are collinear! The line that passes through
these three centres is known as an Euler Line.
(i.e. a line from a vertex to the opposite
side such that the altitude and this side
are perpendicular)
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:
1.
Think about the significance of each piece of information given to you:
a.
b.
c.
2.
3.
4.
5.
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.
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]
𝐵
𝐴
𝑃
Construct your diagram!
?
𝑎
𝑇
I’ve added the
angle 𝑎, so
that our proof
boils down to
showing that
∠𝑃𝑇𝐵 = 𝑎.
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
𝑏
𝑋
𝑇
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
𝐴
𝑎
𝑏
𝑋
𝑇
𝑈
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
𝒂
𝒂+𝒃
𝐴
𝑎 𝒂
𝑏
𝑋
𝒃
𝑇
𝑈
One more…
Two intersecting circles 𝐶1 and 𝐶2 have a common tangent which touches 𝐶1 at 𝑃
and 𝐶2 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]
𝑃
𝑄
𝑁
Construct your diagram!
?
𝑀
(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!
𝑃
𝑎
𝑏
𝑁
𝑎
𝑏
𝑀
𝑄
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
? distances from 𝑃 and 𝑄 to the line 𝑀𝑁 is
we’ve indirectly shown that the perpendicular
Click to show this on diagram
the same.
𝑃
𝑋
𝑎
𝑏
𝑄
𝑁
𝑎
𝑏
𝑀
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.
So how could we prove that 𝑃𝑋 = 𝑋𝑄?
Look out for similar triangles! Notice that triangles 𝑃𝑋𝑁 and 𝑃𝑋𝑀 both share the angle 𝑎
𝑋𝑃
𝑋𝑀
and the angle ∠𝑃𝑋𝑀. So they’re similar. Thus
= 𝑋𝑄 . So 𝑋𝑃2 = 𝑋𝑀 ∙ 𝑋𝑁. Similarly,
? 𝑋𝑁
𝑋𝑄2 = 𝑋𝑀 ∙ 𝑋𝑁. So 𝑋𝑃2 = 𝑋𝑄2 , and thus 𝑋𝑃 = 𝑋𝑄. And we’re done!
𝑃
𝑋
𝑎
𝑏
𝑁
𝑎
𝑏
𝑀
𝑄
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 1 - 2013]
𝐴
𝐵
Construct your diagram!
?
𝑋
𝑎
𝑃
“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
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
1: 𝑂𝑃𝑋 is isosceles.
𝐴
𝑎3?
𝑌
1805?− 2𝑎 2𝑎
6?
902?− 𝑎
𝑂
1?𝑎
𝑎
𝑃
∠𝐴𝑋𝑃 = 90°
𝐵
90 7?
−𝑎
4?
𝑎
𝑋
𝑄
3: Either Alternate
Segment Theorem,
or given that
∠𝐵𝐴𝑂 = 90°
4: 𝐴𝑌𝑋 is isosceles
since triangle
formed by two
tangents.
7: 𝑌𝑋𝐵 is isosceles
(by same reasoning)
How do we know when we’re done?
∠𝑃𝑋𝐵 = ∠𝐴𝑋𝑃 + ∠𝐴𝑋𝑌 + ∠𝑌𝑋𝐵
?
= 90° + 𝑎 + 90° − 𝑎 = 180°
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
1
𝐴 = 𝑎𝑏 sin 𝐶
2
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
𝐴 = 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
Method 2
𝐴=
1
𝑎𝑏 sin 𝐶
2
Increasing 𝑎 or 𝑏 will
clearly increase 𝐴, so for 2
of the sides, we can set
them to the maximum
length.
Consider two sides of the triangle. The height
? (and hence the
of triangle will be maximised
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).
sin 𝐶 will be
maximum when
𝐶 = 90°.
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.