Transcript Topic 1: Combinatorics & Probability
Topic 1:
Geometry
Dr J Frost ([email protected])
Last modified: 21 st August 2013
Slide Guidance
Key to question types:
SMC BMO Senior Maths Challenge The level, 1 being the easiest, 5 the hardest, will be indicated.
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.
Uni Frost Classic STEP University Interview Questions used in university interviews (possibly Oxbridge).
A Frosty Special Questions from the deep dark recesses of my head.
Classic Well known problems in maths.
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:
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.
4 6 5
Simple example: What’s the area of this triangle?
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 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
?
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 14 105 r r 14 14 105 105 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?
Adding radii to points of contact allow us to form some triangles. And if we add the red
?
right-angled triangles for which we can use Pythagoras!
#1 Adding Lines r
1
p 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?
Again, add the radii of each circle, allowing us to form a right-angled triangle (since the
?
smaller circle). Then:
𝒓 𝟐 𝟏 + 𝒑 𝟐 𝑨 = 𝝅𝒓 𝟐 = 𝒓 𝟐 𝟐 𝟐 − 𝝅𝒓 𝟐 𝟏 = 𝝅𝒑 𝟐
#1 Adding Lines x° y° y°
Question: What is angle x + y?
E:
Adding the appropriate extra line makes the problem trivial.
SMC
Level 5 Level 4
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.
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 GM = 2x – y.
?
Then you’d just use Pythagoras!
IMO
Macclaurin Hamilton
Cayley
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.
x
?
2 ° -x x 180
?
° -2x x 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.
b a
What is the expression for the missing side?
Angles of quadrilateral add up to 360°.
#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.
Sides = 10 ?
°
?
° 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.
The interior angle of the polygon can then be worked out using angles on a straight line.
#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
?
°
?
° Exterior angle = 72
?
°
?
°
#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 D
IMC
Level 5
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? 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
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.
SMC
Level 5 Level 4
Level 3
Level 2 Level 1
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
x diameter
3
x x
4
2x x
Angles in same segment
x
Alternative Segment Theorem:
The angle subtended by a chord is the same as the angle between the chord and its tangent.
x x 180-x
5 Angles of a cyclic quadrilateral
Thinking backwards
For many of the circle theorems, the CONVERSE is true…
A B If a circle was circumscribed around the triangle, side AB would be the diameter of the circle.
x x 180-x 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
2x x 2x 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
𝟐𝒙
.
So our point isn’t necessarily at the centre.
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 A 12 20 B If we draw the diameter of the circle, we have a 90° angle at C by our Circle Theorems. Then use Pythagoras.
A: 8
B: 10
E: 16
C: 12
SMC
Level 5 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?
P b
°
Q
By Alternative Segment Theorem
50
x
°
a
° Not to scale
R
A: 20
c
°
40+x x 50
S
B: 25
By ‘Exterior Angle of Triangle’ By Alternative Segment Theorem By ‘Exterior Angle of Triangle’ Angles of this triangle add up to 180, so: 2x + 110 = 180 Therefore x = 35
C: 30
E: 40
SMC Level 5
Level 4 Level 3 Level 2 Level 1
Intersecting Chord Theorem
𝑎 𝑦 𝑏 𝑥
𝑎𝑏 = 𝑥𝑦
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 D 𝐴𝐶 ∙ 𝐵𝐷 = 𝐴𝐵 ∙ 𝐷𝐶 + 𝐴𝐷 ∙ 𝐵𝐶
You’ll be able to practice this in Geometry Worksheet 3.
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…
𝐵
ratio of these… If the line 𝐴𝐷 bisects 𝐴𝐵 and 𝐴𝐶 , then
𝐷
𝐵𝐷 𝐷𝐶 = 𝐴𝐵 𝐴𝐶 𝜃 𝜃
𝐶 𝐴
…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 12 th s of the way around a full rotation, so 150 ° .
By our circle theorems, x is therefore half of this.
x
Angle = 75 °
IMC Level 5
Level 4 Level 3 Level 2 Level 1
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 ° x
?
Therefore: If we have the non-diagonal length: multiply by √ 2.
If we have the diagonal length: divide by √ 2.
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
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 cm 2 , of triangle QRS?
Q P S R
C: √6
SMC
Level 5
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 cm 2 , of triangle QRS?
Q √5 2 1 P 2√2 2 √5 S R √5 √2 √2 √5
So the height of this triangle by Pythagoras is √3.
So that area is
½ x 2√2 x √3 = √6 SMC
Level 5
Level 4
Level 3 Level 2 Level 1
3D Pythagoras
Question: What’s the longest diagonal of a cube with unit length?
1 1 √2 √3 1
By using Pythagoras twice, we get √3.
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 = 2m 2 2 2 2 2 2
SMC Level 5
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.
r r R r 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
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 2 4-r (2-r) 2 + (4-r) 2 = (2+r) 2
6 make the smaller circle larger than the big one.
r r IMO Macclaurin
Hamilton Cayley
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.
𝑎 − 𝑐 𝑥 + 𝑦 𝑎 − 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 .
Question: If 𝑎 and 𝑏 are the radii of the larger circles, and 𝑐 the radius of the smaller one, prove that: 1 + 1 𝑎 𝑏 = 1 𝑐
IMO Macclaurin
Hamilton Cayley
Forming Equations
𝑎 − 𝑐 𝑥 𝑥 + 𝑦 𝑦 𝑎 − 𝑏 𝑏 − 𝑐 Question: If 𝑎 and 𝑏 are the radii of the larger circles, and 𝑐 the radius of the smaller one, prove that: 1 + 1 𝑎 𝑏 = 1 𝑐 𝑥 2 + 𝑎 − 𝑐 Similarly: From from the top triangle: 2 = 𝑎 + 𝑐 2 ⇒ 𝑥 = 2 𝑎𝑐 𝑦 = 2 𝑏𝑐 𝑥 2 + 𝑦 2 + 2𝑥𝑦 = 4𝑎𝑏 Substituting: 𝑎𝑐 + 𝑏𝑐 + 2𝑐 𝑎𝑏 = 𝑎𝑏 Notice that the LHS is a perfect square!
1 𝑎 + 1 𝑏 Dividing by 2 = 1 𝑐 𝑎𝑏𝑐 : 1 𝑏 + 1 𝑎 + 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?
2 3 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.
b a d c 3 x x 4 5
𝑎 𝑏 𝑐 = 𝑑
Question: 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 𝜃 A particular common occurrence is to have one triangle embedded in another, where the indicated angles are the same.
𝛽 𝜃 𝐴 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.
A ‘slice’ of a circle is known as a sector.
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!)
Segment of a circle
Some area related problems require us to calculate a segment.
A
r
O θ
r
B 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.
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 1 O
√
1 2
Let’s put in our information first...
What’s the area of this sector?
Area of sector = π /2
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
√
2 O 1
Therefore (by cutting the segment area from a semicircle):
A
Area of shaded area 2 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:
2
Using two sides and angle between them:
3
Using three sides:
𝑎 ℎ 𝑐
1 2 𝑏ℎ ?
1 2 𝑎𝑏 sin 𝐶 ?
𝑠 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐
where 𝑠 = 1
?
, i.e. half the 2 perimeter.
This is known as Heron’s Formula 𝐶 𝑏
Area of a Triangle
𝜃 𝜃 𝜃 4𝜃 𝜃 The circle has unit radius.
What is the area of the shaded region? (in terms of 𝜃 ) 𝐴 = 2 + 1 2 sin 180 − 2𝜃 4𝜃 × 𝜋
?
360 = sin 2𝜃 + 𝜋𝜃 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 The circumcentre, centroid and orthocentre are collinear! The line that passes through these three centres is known as an Euler Line. Intersection of altitudes (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.
2.
3.
4.
5.
Think about the significance of each piece of information given to you:
a.
We have a tangent?
b.
c.
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 touches the inner circle at a point [Source: UKMT Mentoring] 𝑃 . Prove that 𝑇𝑃 of the outer circle bisects ∠𝐴𝑇𝐵 .
𝐴 𝐵 𝑃 Construct your diagram!
I’ve added the angle 𝑎 , so that our proof boils down to showing that ∠𝑃𝑇𝐵 = 𝑎 .
𝑎 ?
𝑇
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
?
∠𝐴𝑇𝑋 = 𝑏
𝐵 𝑃 𝐴 𝑋
𝑏 𝑎
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 appropriate line, we can use the theorem again: ∠𝑋𝑇𝑃 = ∠𝑃𝑈𝑇 .
𝐴 𝑃
𝒃 𝒂 + 𝒃
𝑈 𝐵 𝑋
𝑏 𝑎
𝑇
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
?
∠𝐵𝑃𝑈 = ∠𝑃𝑇𝑈 .
And we’re done, because we’ve shown ∠𝐴𝑇𝑃 = ∠𝑃𝑇𝑈 !
𝐴 𝑃
𝒂 𝒃 𝒂 + 𝒃
𝑈 𝐵 𝑋
𝑏 𝑎 𝒂
𝑇
One more…
Two intersecting circles 𝐶 1 and 𝐶 2 at 𝑄 and 𝐶 2 have a common tangent which touches . The two circles intersect at 𝑀 and 𝑁 , where 𝑁 is closer to 𝑃𝑄 𝐶 1 at 𝑃 than M is. Prove that the triangles 𝑀𝑁𝑃 and 𝑀𝑁𝑄 have equal areas. [Source: UKMT Mentoring]
𝑃 𝑄
?
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?
𝑃
𝑎
𝑁
𝑏
𝑄
𝑎 𝑏
𝑀
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
?
𝑃 and 𝑄 to the line 𝑀𝑁 the same.
Click to show this on diagram
is
𝑃 𝑋
𝑎
𝑄
𝑏
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 and the angle 𝑋𝑄 2 ∠𝑃𝑋𝑀 = 𝑋𝑀 ∙ 𝑋𝑁 . So 𝑋𝑃 2
?
𝑋𝑃 𝑋𝑁 = 𝑋𝑀 . So 𝑋𝑃 2 𝑋𝑄 = 𝑋𝑀 ∙ 𝑋𝑁 . Similarly, = 𝑋𝑄 2 , and thus 𝑋𝑃 = 𝑋𝑄 . And we’re done!
𝑃
𝑎
𝑋 𝑁
𝑏
𝑄
𝑎 𝑏
𝑀
Final Example
Two circles 𝑆 and 𝑇 and 𝐵 and 𝑇 touch at . The points 𝐴 𝑋 and . They have a common tangent which meets 𝐵 are different. Let 𝐴𝑃 be a diameter of 𝑆 𝑆 at . Prove 𝐴 that 𝐵 , 𝑋 and 𝑃 lie on a straight line.
[Source: BMO Round 1 - 2013]
𝐴 𝐵 Construct your diagram!
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𝑋
𝑎
𝑃
“Prove that 𝐵 , 𝑋 and 𝑃 How could we do this?
lie on a straight line (i.e. are collinear).”
We just need to show that
∠𝑷𝑿𝑩 = 𝟏𝟖𝟎°
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𝐴 𝐵 𝑋 𝑃
𝑎
Now it’s a case of gradually filling in angles!
(But put in mind that we can’t assume 𝑃𝑋𝐵 thing we’re trying to prove) 1: 𝑂𝑃𝑋 is isosceles. is straight, because that’s the very 2: 𝐴𝑃 is diameter so ∠𝐴𝑋𝑃 = 90°
𝐴
𝑎 5?
𝑌
6?
𝐵
3: Either Alternate Segment Theorem, or given that ∠𝐵𝐴𝑂 = 90°
𝑂
𝑎 1?
𝑎
𝑋 𝑄
4: 𝐴𝑌𝑋 is isosceles since triangle formed by two tangents. 7: 𝑌𝑋𝐵 is isosceles (by same reasoning)
𝑃
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: 𝐴 = 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 𝐴 = 1 2 𝑎𝑏 sin 𝐶
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 ?
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 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.