Transcript 2 + - Dr Frost Maths

Topic 3: Geometry
Dr J Frost ([email protected])
Slide guidance
Key to question types:
SMC
Senior Maths Challenge
Uni
Questions used in university
interviews (possibly Oxbridge).
www.ukmt.org.uk
The level, 1 being the easiest, 5
the hardest, will be indicated.
BMO
British Maths Olympiad
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.
MAT
Maths Aptitude Test
Admissions test for those
applying for Maths and/or
Computer Science at Oxford
University.
University Interview
Frost
A Frosty Special
Questions from the deep dark
recesses of my head.
Classic
Classic
Well known problems in maths.
STEP
STEP Exam
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

Topic 3: 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 3: Geometry
Part 3 – Lengths and Area
a. The “√2 trick”.
b. Forming equations
c. 3D Pythagoras and the “√3 trick”.
d. Similar Triangles
e. Forming Equations
f. Area of sectors/segments
Part 4 – BMO Problem Guidance
[Coming soon!]
Topic 3 – 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
For problems involving circles, add radii in strategic places, e.g. towards a
point on the circumference where the circle touches another shape.
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
For problems involving circles, add radii in strategic places, e.g. towards a
point on the circumference where the circle touches another shape.
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!
#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?
By adding these lines, we can
come up with an expression for
the shaded area: π(r22 – r12)
?
And by Pythagoras:
p2 + r12 = r22, so r22 – r12 = p2.
So the shaded area is πp2.
Source: SMC
#1: Adding Lines
Question: What is angle x + y?
x°
A: 
270

B: 300
D: 360

E: More info
 needed
Adding the
appropriate extra
line makes the
problem trivial.
y°
C: 
330
SMC
Level 5
Level 4
y°
Level 3
Level 2
Level 1
#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
2x-y
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!
IMO
Macclaurin
Hamilton
Cayley
Topic 3 – 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.
Question: 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
IMC
Level 5
D
A: 1:2:3

D: 3:4:5

B: 2:2:3

E: 3:4:6

C: 2:3:4

Level 4
Level 3
Level 2
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?
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.
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 
D: 10
E: 12
C: 9 
SMC
Level 5
Let the shape have n sides. Then let k = 360 ÷ n be the
exterior angle. The interior angle is 180 - k.
So using the information:
180 – k = 4k  k = 36  n = 360 ÷ k = 10
Level 3
(Or alternatively, since the angles on a straight line are in the
ratio 1:4, we can immediately tell the exterior angle is 36°)
Level 1
Level 4
Level 2
Topic 3 – 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
Alternate Segment
Theorem:
The angle subtended by
a chord is the same as
the angle between the
chord and its tangent.
x
x
3
x
x
5
4
x
x
180-x
x
2x
Angles of a cyclic
quadrilateral
Thinking backwards
The Circle Theorems work BOTH WAYS…
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.
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
12
A
20
If we draw the
B diameter of the
circle, we have a 90°
angle at C by our
Circle Theorems.
Then use Pythagoras.
SMC
Level 5
A: 8 
D: 10√2

B: 10
E: 16
C: 12
Level 4
Level 3
Level 2
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?
b°
P
50
x°
40+x
c°
R
A: 20
D: 35
By Alternate Segment
Theorem
By ‘Exterior Angle of
Triangle’
By Alternate Segment
Theorem
a°
Not to
scale
Q
By ‘Exterior Angle of
Triangle’
50
x
S
B: 25
E: 40
Angles of this triangle add up to
180, so:
2x + 110 = 180
Therefore x = 35
C: 30
SMC
Level 5
Level 4
Level 3
Level 2
Level 1
Intersecting Chord Theorem
Version 1
Version 2
Chords intersect inside circle.
Chords intersect outside circle.
a
a
y
x
b
b
ab = xy
x
ab = xy
This is a useful tool in our arsenal particularly for Olympiad problems.
y
#3b: 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 5
Level 4
Level 3
?
Angle = 75°
Level 2
Level 1
Topic 3 – 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
1
or
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

D: 2√2

B: 5/2

E: √10

C: √6
Level 4
Level 3
Level 2
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
√5
1
R
√5
S
SMC
Level 5
√5
√5
√2
√2
So the height of this triangle by
Pythagoras is √3.
So that area is
Level 4
½ x 2√2 x √3 = √6
Level 2
Level 3
Level 1
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
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 5
A:
2m
2
D:
5m
2
B:
3m2
E:
6m2
C:
4m2
Level 4
Level 3
Level 2
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
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
Question:
What is the
radius of the
small circle?
r
2
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
IMO
Macclaurin
Hamilton
Cayley
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
a
c
a
c
b = d
Question: Find the percentage of the area of a 3:4:5 triangle occupied by a
square inscribed inside it.
3-x
?
5
3
x
4-x
?
x
?
?
3-x
_x_
=
?x
?
4-x
4
Multiplying through by the denominators give us: (4-x)(3-x)
? = x2
Expanding and solving gives us:
x=
12
7?
So the area of the square is:
24
Since the area of the big triangle is 6, the fraction the square occupies is: 49
?
144
?
49
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.
Describe a method that
would give you the area of
the shaded region.
A
r
θ
O
B
r
Method: Start with the sector AOB.
(area is (θ/360) x πr2 because θ/360
gives us the proportion of the circle
we’re using).
Then cut out the ?triangle (i.e.
subtract its area). We could work
out its area by splitting it in two
and using trigonometry.
Segment of a circle
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.
Given the arc has equation
x2 + y2 = 1, then halfway
along, 0.25 + y2 = 1, so
y = √3/2
Cut out
these two
triangles.
Which
leaves this
region.
Then just multiply this area
by 4 to finish...
Topic 3 – Geometry
Part 4: BMO Problem Guidance
I haven’t got round to making these slides yet.
Try and re-download these later in the year!