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 youβre 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 β 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 β Angles
Part 2b β Circle Theorems
b. Using them backwards!
c. Intersecting Chord Theorem
Part 3 β Lengths and Area
a. The ββ2 trickβ.
b. Forming equations
c. 3D Pythagoras and the ββ3 trickβ.
d. Similar Triangles
Topic 1: Geometry
f. Inscription problems
e. Area of sectors/segments
Part 4 β Proofs
a. Generic Tips
b. Worked Examples
c. Proofs involving Area
ΞΆPart 1: General PointersTopic 1 β 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: Whatβs 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 (weβll 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
By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles.
#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.
?
By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles.
#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!
?
By adding extra lines to your diagram, you can often form shapes whose properties we can exploit, or find useful angles.
#1 Adding Lines
If the indicated chord has length 2p, and weβre 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?
xΒ°
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A: 270
B: 300 C: 330
D: 360 E: More info needed
yΒ°
yΒ° Adding the appropriate extra line makes the problem trivial.
#1 Adding Lines
#1 Adding Lines
But donβt overdo itβ¦Only add lines to your diagram that are likely to help. Otherwise you risk:β’ Making your diagram messy/unreadable, and hence
make it hard to progress.β’ Overcomplicating the problem.
Itβs 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 weβve folded over at G, then clearly length GM = 2x β y.Then youβd just use Pythagoras!
Macclaurin
Hamilton
Cayley
IMO
ππππβπ
?
#2 Introducing Variables
ΞΆPart 2a: Angle FundamentalsTopic 1 β 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
180Β°-x2?
180Β°-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 β a - b?
What is the expression for the missing side?
Angles of quadrilateral add up to 360Β°.
#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 360Β° in total.
Sides = 10
36Β°?144Β°? The interior angle of the polygon can
then be worked out using angles on a straight line.
Itβs useful to be able to quickly calculate the interior and exterior angles of a regular polygon.
Exterior angle = 60Β°Interior angle = 120Β°
?
?
Exterior angle = 72Β°Interior angle = 108Β°
??
#2: Interior/Exterior Angles of Regular PolygonsItβs 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
#2: Interior/Exterior Angles of Regular PolygonsItβs 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
C
D
E
Fzyx
y = 360Β° / 5 = 72Β°. So z = 180Β° β 72Β° = 108Β°.AB = AE (because itβs a regular pentagon) and weβre told FA = AB, so FA = AE. Itβs therefore an isosceles triangle, so angle AEF = x. Angles of a triangle add up to 180Β°, so x = (180Β° β 72Β°)/2 = 54Β°.The ratio is therefore 54:72:108, which when simplified is 3:4:6.
#2: Interior/Exterior Angles of Regular PolygonsItβs useful to be able to quickly calculate the interior and exterior angles of a regular polygon.
Level 2
Level 1
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Level 4
Level 3
SMC
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
#2: Interior/Exterior Angles of Regular PolygonsItβs useful to be able to quickly calculate the interior and exterior angles of a regular polygon.
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 thereβs sides.
ΞΆPart 2b: Circle TheoremsTopic 1 β Geometry
You should know most of these already. Although thereβs 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 trueβ¦
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 trueβ¦
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 βangles in the same segmentβ theorem, the angle must still be .So our point isnβt 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: 10β2 E: 16
C
BA
If we draw the diameter of the circle, we have a 90Β° 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
xΒ°
N ot toscale
aΒ°
bΒ°
cΒ° x
By βExterior Angle of Triangleβ50
50
By Alternative Segment Theorem
40+xBy βExterior Angle of Triangleβ
Angles of this triangle add up to 180, so: 2x + 110 = 180Therefore x = 35
Circle Theorems
π
π
π₯
π¦
ππ=π₯π¦
Intersecting Chord Theorem
π΅
ππ΄β ππ΅=ππΆβ ππ·
A secant is a line which passes through a circle.
Intersecting Secant Lengths Theorem
π΄
π
πΆπ·
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
Ptolemyβs Theorem
Youβll be able to practice this in Geometry Worksheet 3.
π΅π·π·πΆ
=π΄π΅π΄πΆ
Angle Bisector Theorem
One final theorem not to do with circlesβ¦
ππ
π΄
π΅
πΆ
π·
ratio of theseβ¦
β¦is the same as the ratio of these.
If the line bisects and , then
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IMC
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 = 75Β°?
By our circle theorems, x is therefore half of this.
This angle is much easier to work out. Itβs 5 12ths of the way around a full rotation, so 150Β°.
ΞΆPart 3: Lengths and AreasTopic 1 β Geometry
The ββ2 trickβFor 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?
45Β°
45Β°
x
x?
Therefore:If we have the non-diagonal length: multiply by β2.If we have the diagonal length: divide by β2.
The ββ2 trickβFor 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:
45Β°
45Β°
3
3? 5
?
The radius of the circle is 1. What is the side length of the square inscribed inside it?
1
1/β2
1
β2
or
β2?
The ββ2 trickβ
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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: β15/4
B: 5/2 C: β6
D: 2β2 E: β10
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
22β2
β5β5
β2 β2
β5β5So the height of this triangle by Pythagoras is β3.So that area is
Β½ x 2β2 x β3 = β6
3D Pythagoras
Question: Whatβs the longest diagonal of a cube with unit length?
By using Pythagoras twice, we get β3.The β3 trick: to get the longest diagonal of a cube, multiply the side length by β3. If getting the side length, divide by β3.
1
β2
1β3
1
?
3D Pythagoras
Question: A cube is inscribed within a sphere of diameter 1m. What is the surface area of the cube?
Level 2
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SMC
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/β3 m.Surface area = 6 x (1/β3)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 + rβ2 = r(1 + β2)
r = __r__1+ β2
?
Forming EquationsTo find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same.
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.
Forming EquationsTo find unknowns, form equations by using Pythagoras Theorem and equating length expressions where the lengths are the same.
Cayley
Hamilton
Macclaurin
IMO
Question: If and are the radii of the larger circles, and the radius of the smaller one, prove that:
π+π
π+ππ+ππβπ
πβπ
πβπ
π₯+ π¦
π₯ π¦
As always, try to find right-angled triangles. Drawing a rectangle round our triangle will create 3 of them.Fill in the lengths. We donβt 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:
π+π
π+ππ+ππβπ
πβπ
πβπ
π₯+ π¦
π₯ π¦
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
ΒΏ34
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 .
1β32 30 Β° ?
ΒΏππππ
Similar TrianglesWhen triangles are similar, we can form an equation.
5
x
4
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 ππ
=ππ
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?
?
Question: A square is inscribed inside a 3-4-5 triangle. Determine the fraction of the triangle occupied by the square.
Similar Triangles
π
π
A particular common occurrence is to have one triangle embedded in another, where the indicated angles are the same.
π΄ π΅
π·
Why are triangles and similar?
πΆ
They share a second common angle at .Weβll see an example of this later on in this module.
π½
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 βsliceβ 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 weβll see.
r
r
A
B
O
ΞΈ
#
Some area related problems require us to calculate a segment.
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...
βPerhaps I should find the radius of this other circle?β
Radius of circle centred at A: β2?
Some area related problems require us to calculate a segment.
Segment of a circle
A1
1 β2
O
B
C
Letβs put in our information first...
Whatβs the area of this sector?
Area of sector = Ο/2?
Now we need to remove this triangle from it to get the segment.
Area of triangle = 1?
Some area related problems require us to calculate a segment.
Segment of a circle
A1
1 β2
O
B
C
So area of segment = (Ο/2) - 1
Therefore (by cutting the segment area from a semicircle):
Area of shaded area
= - ( β 1) = 1
Ο 2
Ο 2 ?
Some area related problems require us to calculate a segment.
Segment of a circleSome area related problems require us to calculate a segment.
Question: Here are 4 overlapping quarter circles of unit radius. Whatβs 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:
πΆ
π
π
π
Using two sides and angle between them:
Using three sides:
12hπ
12ππ sinπΆ
β π (π βπ ) (π βπ ) (π βπ )where , i.e. half the perimeter.
?
?
?
1
2
3
This is known as Heronβs 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 )
π
π4 π
π
π
(Note that in general, )
?
ΞΆPart 4: ProofsTopic 1 β Geometry
Some Quick Definitions
βInscribeβFor 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.
βCircumscribeβTo 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.
βCollinearβPoints 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!)
ππ
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. Hereβs a few golden rules of proofs:
1. Think about the significance of each piece of information given to you:a. We have a tangent?
Weβll likely be able to use the Alternate Segment Theorem (which you should expect to use a lot!). If thereβs a chord attached, use it immediately. If thereβs isnβt 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. Weβre 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. Itβs 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]
π
π΅
π΄
π
π
Iβve 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 letβs say label
π
π΅
π΄
π
π
Our usual good starting point is to label an unknown angle to help us work out other angles. But which would be best?
π
π
?
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: .
π
π΅
π΄
π
π
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?
π
π
?
π
π+ππ
That line added is convenient a chord attached to a tangent. So we can apply the Alternate Segment Theorem a third time. .And weβre done, because weβve shown !
π
π΅
π΄
π
π
We can just use very basic angle rules (angles on a straight line, internal angles of a triangle) to find that . Now whatβs the final step?
π
π
?
π
π+πππ
π
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]
ππ
π
π
(Itβs important to make your circles different sizes to keep things general)
Construct your diagram!
?
One moreβ¦
We have a tangent, so what would be a sensible first step?
ππ
π
π
We also have some chords, so we should use the Alternate Segment Theorem!?
π π
ππ
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?
ππ
π
π
A common strategy is to extend a line onto another. If we can show , then weβve indirectly shown that the perpendicular distances from and to the line is the same.
π π
ππ
π
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 ?
ππ
π
π
Look out for similar triangles! Notice that triangles and both share the angle and the angle . So theyβre similar. Thus . So . Similarly, . So , and thus . And weβre done!
π π
ππ
π
?
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]
π΄π΅
π
π
π
Construct your diagram!
?
Final Example
βProve that , and lie on a straight line (i.e. are collinear).βHow could we do this?
We just need to show that
π΄π΅
π
π
π
?
Now itβs a case of gradually filling in angles!(But put in mind that we canβt assume is straight, because thatβs the very thing weβre trying to prove)
π΄π΅
π
π
π
π90βπ
π
π
180β2π2π
90βπ
1?
2?
3?
4?
5? 6?
7?
1: is isosceles.
π π
2: is diameter so
3: Either Alternate Segment Theorem, or given that
4: is isosceles since triangle formed by two tangents.
π
7: is isosceles (by same reasoning)
How do we know when weβre done?
?
βA triangle has lengths of at most 2, 3 and 4 respectively. Determine, with proof, the maximum possible area of the triangle.β[Source: BMO Round 1 β 2003]
What might be going through your head:βWell the question wants us to maximise area, so maybe I should think about the formula for the area of a triangle?β
Formulae for area of a triangle:
π΄=πππ πΓh hπππ π‘ π΄=12ππsinπΆ
Other types of Geometric Proof
βA triangle has lengths of at most 2, 3 and 4 respectively. Determine, with proof, the maximum possible area of the triangle.β[Source: BMO Round 1 β 2003]
π΄=πππ πΓh hπππ π‘
Consider two sides of the triangle. The height of triangle will be maximised (and hence the area) when theyβre 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 (weβll do this in a second).
π΄=12ππsinπΆ
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. β’ If we have and as the base and height,
then by Pythagoras, the hypotenuse is , which is less than 4, so is fine!
β’ 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.
β’ Thus the maximum possible area is 3.
Method 1 Method 2
??
Other types of Geometric Proof