a review of geometry

A Review of Geometry

Khor Shi-Jie

May 17, 2012

Contents

1 What’s Next? 2

2 Chasing Pavements 52.1 Your Toolbox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1.1 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1.2 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.3 Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 More Powerful Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Auxiliary Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1

Chapter 1

What’s Next?

I have learnt all theorems and properties in planar geometry. What’s next for me?

Similar to the topic of inequalities in algebra, geometry is a topic in Mathematics Olympiadwhich is saturated with theorems and properties. What is even more challenging is that stu-dents must be equipped with strong visualisation skills in order to incise geometric problemswith the correct theorems and properties. This reason alone makes many student detractattention from geometry and focus on other ares instead.

Yet the mastery of geometry is not an unthinkable feat. Just like what I said in the pre-vious document on Algebra, there are two distinctive phase in the grasping of MathematicsOlympiad. The first step in studying geometry is, obviously, to study geometric theoremsand be well acquainted with basic applications of such theorems. Of course, this step isrelatively challenging in studying geometry due to the sheer number of different quantities,shapes and concepts that students have to master. On the other hand, it is no less importantfor student to learn the strategies involved in solving geometric problems. It is a cardinalsin (well, pardon my exaggeration) to neglect this step because most geometric problems incompetitions does not merely involve the application of a theorem. It is often coupled withingenious transformations and inspired combination of theorems that allows one to solve ageometric problem beautifully. In order to acquire different skills in geometry, it is impor-tant for one to do hundreds and thousands of problems to familiarise oneself with differentstrategies and pick up useful lemmas in the process. (You’ve heard it, THOUSANDS)

There are two approaches in solving geometric problems: synthetic geometry and analyt-ical geometry. Synthetic geometry involves using theorems learnt in MO to solve a geometricproblem. This is often coupled with creative construction and transformation in the processof solving the problem. The lack of a fixed approach or algorithm in solving the problem is ahuge contrast to analytical geometry. Analytical geometry, on the other hand, is also knownas the brute force method. Students can use methods such as coordinate geometry, trigonom-etry, vectors and complex numbers to quantify everything within the problem and solve theproblem algebraically. I do not have any preferred method in solving geometric problems.Synthetic geometry does help in solving a problem quickly and elegantly, but sometimes ittakes too much time in arriving at the correct approach. The analytical geometry approach

2

CHAPTER 1. WHAT’S NEXT? 3

will definitely involve heavy computation, but it is more certain that one will obtain thesolution through this method. I recommend students to master both techniques in solvinggeometric problems.

By saying basic theorems and properties in geometry, I refer to the following topics:

1. Planar Geometry

(a) Triangles

i. Similarities and congruences

ii. Area of triangles

iii. Trigonometric properties

iv. Special lines within the triangle and notable theorems (Pythagorean theorem,Mid-point theorem, Angle-bisector theorem, Stewart’s theorem, etc)

v. Centres of a triangle

(b) Circles

i. Basic circular properties

ii. Cyclic quadrilaterial

iii. Power of a point and radical axis

iv. Notable theorems (Ptolemy’s theorem, Simson’s theorem, etc)

(c) Concyclic, collinearity, concurrency

i. Menalaus’ Theorem and Ceva’s Theorem

(d) Area properties

i. Heron’s formula

ii. Area relations with circumradius and inradius

iii. Area relations with trigonometry

(e) Geometric inequalities

i. Triangle inequalities

ii. Other geometric inequalities

2. Trigonometry

(a) Trigonometric identities

(b) Sine rule and cosine rule

3. Analytical geometry

(a) Coordinate geometry

(b) Vectors

(c) Complex numbers

CHAPTER 1. WHAT’S NEXT? 4

(d) Inversive geometry

Yup, that’s a lot of things to study, considering that this is merely the first step in study-ing geometry. Note that I have omitted 3D geometry in the list above. 3D geometry iscovered in most regional Olympiads around the world. However, it is not within the syllabusof SMO and IMO. Nevertheless, it is still useful to learn the skills in 3D geometry as theseproblems do appear in AMC, AIME and Purple Comet.

The fundamental knowledge listed above will help in the following three strategies that Iwill expound in the following chapters:

1. Chasing. This technique involves using geometric theorems to ”chase” all the quantitieswithin the geometric diagram. This includes the lengths, angles and areas within theproblem. Similarity and congruences help tremendously in this technique. Often, wecannot chase all the quantities and have to set variables to simplify our work (becautious not to complicate it).

2. Construction and transformation. This technique involves constructing auxiliary lines(or circles/curves) to help us identify similarities, chase quantities or simply applygeometric theorems. We can also transform a certain part of the diagram throughreflection, rotation, translation, dilation, etc. There are often hints in the questionthat suggests us to use this technique.

3. Analytical geometry. Brute force method. Nuff said.

I have organised this set of notes based on the strategies used in solving geometric prob-lems. This set of notes is created to help students in senior section (and perhaps some injunior section), so I will try my best in choosing simple examples...

Chapter 2

Chasing Pavements

Should I give up or should I just keep chasing pavements/Even if it leads nowhere?

Chasing is one of the most important skills in solving geometric problems. From theconditions given in the problem, we ”chase” the different quantities involved in the questionuntil we arrive at what the question requires us to evaluate or prove. Let us take a look atthe following problem:

(AIME2009 P10) Four lighthouses are located at points A,B,C and D. The lighthouse at Ais 5 kilometres form the lighthouse at B, the lighthouse at B is 12 kilometres away from thelighthouse at C, and the lighthouse at A is 13 kilometres away from the lighthouse at C. Toan observer at A, the angle determined by the lights at B and D and the angle determined bythe lights at C and D are equal. To an observer at C, the angle determined by the lights at Aand B and the angle determined by the lights at D and B are equal. The number of kilometres

from A to D is given byp√r

q, where p, q and r are relatively prime positive integers, and r

is not divisible by the square of any prime. Find p + q + r.

I have copied the question exactly from AIME to preserve the integrity of the problem.To save time, I have rephrased the question as follows to cut the details.

Figure 2.0.1

As shown in the diagram above, suppose AB = 5, BC = 12, AC = 13. Point D is a point

5

CHAPTER 2. CHASING PAVEMENTS 6

such that BC is the angle bisector of ∠ACD while AD is the angle bisector of ∠BAC. Findthe length of AD.

Looks much cleaner now. Let us see what quantities we can derive from the diagram.From the fact that AD is the angle bisector of ∠BAC, we can use angle bisector theorem tofind out the length of BE and CE. So we have:

BE = 12× 5

18=

10

3CE = 12× 13

18=

26

3

Next, we can use Pythagorean theorem to find out the length of AE. That gives us:

AE =

√52 +

(10

3

)2

=5√

13

3

We can continue chasing the values of ∠ACE using cosine rule, and ∠ACD using doubleangle formula. But that is not very glamorous and takes up too much time (considering thatstudents are supposed to complete this problem in AIME). Usually when you are stuck inthe midst of chasing quantities, it is time when you should consider adding auxiliary lines tohelp solving the problem.

I have compiled a list of common auxiliary lines which are effective in solving geometricproblems in later parts of the chapter. In this scenario, it will be useful for us to extend ABand CD so that they meet at intersection point F , as shown in the diagram below:

Figure 2.0.2

Doing so is helpful because triangle ACF is an isoceles triangle, which implies thatCF = 13 and BF = 5. Alas, we can use angle bisector theorem to find out the lengthof CD and DF respectively and use the same theorem again to find the length of AD.

CD = 13× 13

23=

169

23DF = 13× 10

23=

130

23

The length of AD is given by

AD =5

3

√13×

13 + 16923

13=

60√

13

23


And that’s it, we just finished chasing all the lengths which are required with the help ofsome construction lines. The above problem is merely a simple problem that involves lengthchasing. Most of the time, it can be very challenging to chase all the quantities withoutsuitable constructions. Before we even discuss the strategies used in chasing, students mustbe very familiar with the tools which are available for them.

2.1 Your Toolbox

Here’s a list of theorems that students must be familiar with in order to do well in geometry.These theorems will greatly aid students in chasing methods. The theorems and discussionwhich follows will be tailored for students in junior section.

2.1.1 Triangles

Triangles form a core component in geometry for junior section. The most important propertythat students have to grasp is the similarity and congruency of triangles. The followingdiagram is a diagram which comes out frequently in geometry. I recommend students tospend some time in locating all similar triangles found in the diagram below:

Figure 2.1.1

The intersection of three altitudes is known as the orthocentre of the triangle (often de-noted as H). We are often concerned with the 5 centres of a triangle in Math Olympiadproblems, namely the orthocentre, the centroid (intersection of medians), the incentre (thecentre of the inscribe circle), the circumcentre (the centre of the circumscribe circle) and theexcentres (the centres of the three excircles). There are many interesting properties aboutthese centres which are often tested in Math Olympiad, but this topic will be covered infuture lessons.

There are many theorems which can stem from a right angle triangle and similarity oftriangles. Here are a few useful theorems and corollaries:

Theorem 1 (Pythagorean Theorem) Given a triangle 4ABC, we have AC2 + AB2 =BC2 if and only if ∠A is a right angle.

Corollary 1 Given a triangle 4ABC where ∠A is a right angle and D is a point on BC.We have AD2 = CD ×BD if and only if AD ⊥ BC.


Corollary 2 In a triangle 4ABC and D is a point on BC. We have AB2 − BD2 =AC2 − CD2 if and only if AD ⊥ BC.

Corollary 3 Given a quadrilateral ABCD, the diagonals AC and BD are perpendicular ifand only if AB2 + CD2 = AD2 + BC2.

Isosceles triangles and equilateral triangles often come out in problems too and one shouldbe familiar with its properties. For example, in an isosceles triangle, the altitude from thevertex between the two equal side both a median and an angle bisector of the triangle. Oneshould also be aware of the likelihood of congruent triangles erected from the sides of anequilateral triangle.

A cevian of a triangle is a segment connecting a vertex of the triangle to a point in theopposite side of the triangle. Examples of cevians include the altitude of the triangle, themedian of the triangle and the angle bisector of the triangle. There is a very powerful theoremwhich allows us to calculate the length of these cevians given the lengths of the sides of thetriangle.

Figure 2.1.2

Theorem 2 (Stewart’s Theorem) In triangle 4ABC, D is a point on BC. Suppose

AB = c, AC = b, BC = a,AD = d,BD = m,CD = n, we have d2 =b2m + c2n

a−mn.

Corollary 4 (Pappus’ Theorem) If D is the midpoint of BC, then we have d2 =2b2 + 2c2 − a2

4.

2.1.2 Circles

There are several essential circular properties that students must know (it is covered in Sec2 syllabus). Here’s a list of these essential properties:

Property 1 Angles subtended by the same chord are equal i.e. ∠AP1B = ∠AP2B =∠AP3B. Angles subtended by chords of equal length are equal.


Figure 2.1.3

Property 2 Angle subtended by a diameter is 90◦.

Property 3 Angles at the centre of the circle is twice the angle at the opposite arc i.e.∠AOB = 2∠ACB, reflex ∠AOB = 2∠ADB.

Figure 2.1.4

Property 4 Angles in opposite segments are supplementary i.e. ∠ABC + ∠ADC = 360◦.

Figure 2.1.5

Property 5 A tangent to the circle is perpendicular to the radius of the circle.

Property 6 The two tangents drawn from an external point to a circle is equal in length.


Property 7 The radius which is perpendicular to a chord bisects the chord into two equalsegments.

Property 8 (Alternate Segment Theorem) Angles in alternate segments are equal i.e.∠BAD = ∠BCA,∠ABC = ∠CAE.

Figure 2.1.6

In discussing circles, we are also concerned with the circular power of a point, which isdefined as the difference between the square of the distance from a point P to the centreof the circle O and the square of the radius i.e. OP 2 − r2. You can draw a line from thepoint such that it intersects the circle at two points Q and R. Then the power of the pointOP 2− r2 = PQ×PR. From this property, we can obtain the power of a point theorem andthe intersecting chord theorem:

Theorem 3 (Power of a Point) Given that P is a point outside of circle �O, and thatPE is tangent to circle �O, we have PA× PB = PC × PD = PE2.


Figure 2.1.7

Theorem 4 (Intersecting Chord Theorem) Given that P is a point in circle �O, wehave PA× PB = PC × PD.

Figure 2.1.8

The radical axis of two circles is the line whereby all the points on the line have equalcircular power with respect to both circles. If the two circles overlap, the radical axis willbe the line which contains the common chord of the two circles. There are many interestingproperties which involves the radical axis, but this topic is usually tested only in the senioror open section of SMO.

2.1.3 Areas

There are also many formulas that concerns the area of a polygon. Denote S4ABC as thearea of triangle 4ABC. We have the following theorems which are useful in evaluating thearea:

Theorem 5 (Heron’s Formula) Define s as the semiperimeter of the triangle. We haveS4ABC =

√s(s− a)(s− b)(s− c).

Theorem 6 GIven that AC = b and AB = c, we have S4ABC =1

2bc sinA.


It is very important to relate areas with the ratio of line segments, especially in discussingtriangles. Often, solution which makes use of area relationships are very elegant and short. In

the following 6 diagrams, the ratio of areas is given byS4PAB

S4QAB

=PM

QM. More generally, if AB

is the common side of the two triangles 4ABP and 4ABQ and point M is the intersection

points of AB and PQ, then we haveS4PAB

S4QAB

=PM

QM.

Figure 2.1.9

From theorem 6, we see that it can be useful to discuss areas if two triangles have acommon angle. Based on the theorem, suppose two triangles 4ABC and 4A′B′C ′ fulfils

the condition ∠A = ∠A′ or ∠A+∠A′ = 180◦, then we haveS4ABC

S4A′B′C′=

AB · ACA′B′ · A′C ′

. This fact

can be proven by superimposing the two triangles together at the equal angle and consideringthe areas of the triangle based on the ratio of lengths.

2.2 More Powerful Tools

The following theorems are not required in junior section, but many problems in juniorsection can be greatly simplified with the use of these theorems:

Theorem 7 (Angle Bisector Theorem) In 4ABC, suppose the angle bisector of ∠A in-

tersects side BC at D. We haveAB

BD=

AC

CD.


Figure 2.1.10

If a question concerns angle bisector of one of the angle of the triangle, we can be quitecertain that the question involves the use of angle bisector theorem. Coupled with Stewart’stheorem, we can find the length of AD easily given the length of sides AB,BC and AC.

Theorem 8 (Ceva’s Theorem) Given that X, Y and Z are points on BC, AC and AB

respectively. Suppose AX, BY and CZ are concurrent. Then we haveBX

XC· CY

Y A· AZZB

= 1.

Figure 2.1.11

Ceva’s theorem is often used to prove concurrency between three line segments. In juniorsection, this theorem can be useful in evaluating the length of segments that corresponds tothe segments in the theorem. This triangle can be easily recognised and the application ofCeva’s theorem is usually obvious.

Theorem 9 (Menalaus’ Theorem) Given that X, Y and Z are points on BC, AC and

AB or their extension respectively. Suppose X, Y and Z are collinear, then we haveCX

XB·

BZ

ZA· AYY C

= 1

Figure 2.1.12

Menalaus’ Theorem is often used to prove collinearity of three points. I would advise peopleto memorise this theorem base on the direction of the line segments. This theorem can beapplied often due to the common appearance of the shape that this theorem is involved in.


Theorem 10 (Sine Rule) Given a triangle 4ABC, we havea

sinA=

b

sinB=

c

sinC= 2R

where R is the circumradius of the triangle.

Theorem 11 (Cosine Rule) Given a triangle 4ABC, we have a2 = b2 + c2 − 2bc cosA.

2.3 Auxiliary Lines

Constructing auxiliary lines is an indispensable technique to do well in geometry. Here’s alist of possible auxiliary lines that one can construct to solve a geometric problem:

1. Connecting points. Often, connecting important points in the diagram may provide usa bridge in chasing certain quantities in the diagram. Sometimes, the points to connectmay not be very obvious as it may not be mentioned in the problem itself.

2. Constructing parallel lines. This can be useful as it allows us to use the properties ofparallel lines. Especially useful if there are several equal angles involved in the problem.

3. Constructing auxiliary circles. This allows us to use the many circular properties tosolve the problem.

4. Constructing similar shapes. We create construction lines such that the shape con-structed is similar to a certain shape which is in the original diagram.

5. Constructing perpendicular lines. This allows us to use the properties of perpendicularlines as well as the Pythagorean theorem. It can be useful too in defining the ratios oflengths.

6. Geometric transformation. Transformations like rotations, translation, reflection, scal-ing can be very useful.

7. Others. Lots of practice is required to identify the right construction.

2.4 Problem Set

1. Prove the angle bisector theorem.

2. Prove Ceva’s theorem.

3. Prove Menalaus’ theorem.

4. Given a triangle 4ABC with area 2012. Let BM be the perpendicular from B to thebisector of C. Determine the area of the triangle AMC.

5. Given that P is a point in an acute angle triangle4ABC such that ∠BPC = ∠BAC+

90◦ and AB × PC = AC × PB. Let k =AP ×BC

AC ×BP. Evaluate k4.


6. 4ABC is a right-angled triangle with ∠C = 90◦. M is the midpoint of AB while Dand E are points on BC and AC respectively such that ∠DME = 90◦. Given thatDB = 20 and EC = 21, evaluate MD2 + ME2.

7. Given a convex quadrilateral ABCD, AB = CD, E,F are the midpoints of AD andBC respectively. Extend EF to meet the extension of BA at S and the extension ofCD at K. Prove that ∠ASF = ∠DKF .

8. (AIME 1984 P3) A point P is chosen in the interior of 4ABC such that when lines aredrawn through P parallel to the sides of 4ABC, the resulting smaller triangles t1, t2,and t3 in the figure, have areas 4, 9, and 49, respectively. Find the area of 4ABC.

9. (AIME 1985 P6) As shown in the figure, triangle 4ABC is divided into six smallertriangles by lines drawn from the vertices through a common interior point. The areasof four of these triangles are as indicated. Find the area of triangle 4ABC.

10. (SMO(J) 2006 P4) In 4ABC, the bisector of ∠B meets AC at D and the bisectorsof ∠C meets AB at E. These bisectors intersect at O and OD = OE. If AD 6= AE,prove that ∠A = 60◦.

11. (SMO(J) 2007 P30) In 4ABC, ∠BAC = 45◦. D is a point on BC such that AD isperpendicular to BC. If BD = 3 cm and DC = 2 cm, and the area of the 4ABC is xcm2, find the value of x.

12. (SMO(J) 2007 P1) In the trapezium ABCD, AB ‖ CD, O is the intersection of ACand BD, AB = b, CD = a and a < b. Let S be the area of the trapezium ABCD.Suppose the area of 4BOC is 2S

9. Find the value of a

b.


13. (SMO(J) 2007 P2) Equilateral triangles 4ABE and 4BCF are erected externally onthe sides AB and BC of a parallelogram ABCD. Prove that 4DEF is equilateral.

14. (SMO(J) 2007 P31) In 4ABC, AB = AC =√

3 and D is a point on BC such thatAD = 1. Find the value of BD ·DC.

15. (SMO(J) 2008 P3) In the quadrilateral PQRS, A,B,C and D are midpoints of thesides PQ,QR,RS and SP respectively, and M is the midpoint of CD. Suppose H isthe point on the line AM such that HC = BC. Prove that ∠BHM = 90◦.

16. (SMO(J) 2009 P4) Three circles of radius 20 are arranged with their respective centresA,B and C in a row. If the line WZ is tangent to the third circle, find the length ofXY.

17. (SMO(J) 2009 P15) 4ABC is a right-angled triangle with ∠BAC = 90◦. A square isconstructed on side AB and BC as shown. The area of square ABDE is 8 cm2 andthe area of the square BCFG is 26 cm2, Find the area of triangle 4DBG in cm2.

18. (SMO(J) 2009 P1) In 4ABC, ∠A = 2∠B. Let a, b, c be the lengths of its sidesBC,CA,AB, respectively. Prove that a2 = b(b + c)

19. (SMO(S) 2009 P6) The area of a triangle 4ABC is 40 cm2. Points D,E and F areon sides AB,BC and CA respectively, as shown in the figure below. If AD = 3 cm,DB = 5 cm, and the area of triangle 4ABE is equal to the area of quadrilateralDBEF , find the area of triangle 4AEC in cm2.


20. (SMO(S) 2009 P16) 4ABC is a triangle and D is a point on side BC. Point E is onside AB such that DE is the angle bisector of ∠ADB, and point F is on side AC such

that DF is the angle bisector of ∠ADC. Find the value ofAE

EB· BD

DC· CF

FA.

21. (SMO(S) 2009 P29) ABCD is a rectangle, E is the midpoint of AD and F is themidpoint of CE. If the area of triangle 4BDF is 12 cm2, find the area of rectangleABCD in cm2.

22. (SMO(O) 2009 P2) Let A1, A2, · · ·A6 be 6 points on a circle in this order such that

A1A2 = A2A3, A3A4 = A4A5, A5A6 = A6A1, where A1A2 denotes the arc length of thearc A1A2 etc. It is also known that ∠A1A3A5 = 72◦. Find the size of ∠A4A6A2 indegrees.

23. (SMO(O) 2009 P4) Let P1, P2, · · ·P41 be 41 distinct points on segment BC of a triangle4ABC, where AB = AC = 7. Evaluate the sum

∑4 1i=1(AP2i + PiB · PiC).

24. (SMO(J) 2010 P14) In triangle 4ABC, AB = 32 cm, AC = 36 cm and BC = 44 cm.If M is the midpoint of BC, find the length of AM in cm.

25. (SMO(J) 2010 P32) Given ABCD is a square. Points E and F lie on the side BC andCD respectively, such that BE = CF = 1

3AB. G is the intersection of BF and DE.

IfArea of ABGD

Area of ABCD=

m

nis in its lowest term, find the value of m + n.

26. (SMO(J) 2010 P1) Let the diagonals of the square ABCD intersect at S and let P bethe midpoint of AB. Let M be the intersection of AC and PD and N the intersectionof BD and PC. A circle is inscribed in the quadrilateral PMSN . Prove that the radiusof the circle is MP −MS.

27. (SMO(S) 2010 P10) Let ABCD be a trapezium with AD parallel to BC and ∠ADC =90◦. Given that M is the midpoint of AB with CM = 13

2cm and BC +CD+DA = 17

cm, find the area of trapezium ABCD in cm2.

28. (SMO(S) 2010 P22) Given a circle with diameter AB, C and D are points on thecircle on the same side of AB such thath BD bisects ∠CBA. The chords AC and BDintersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, findthe value of x.


29. (SMO(S) 2010 P1) In the triangle 4ABC with AC > AB, D is the foot of perpendicu-lar from A onto BC and E is the foot of perpendicular from D onto AC. Let F be thepoint on the line DE such that EF ·DC = BD ·DE. Prove that AF is perpendicularto BF .

30. (SMO(J) 2011 P33) In the following diagram, ABCD is a square, BD ‖ CE andBE = BD. Let ∠E = x◦. Find x.

31. (SMO(J) 2011 P2) Two circles Γ1,Γ2 with radii r1, r2 respectively, touch internally atpoint P . A tangent parallel to the diameter through P touches Γ1 at R and intersectsΓ2 at M and N . Prove that PR bisects ∠MPN .

32. (SMO(S) 2011 P21) ABCD is a convex quadrilateral such that AC intersects BD atthe midpoint E of BD. The point H is the foot of perpendicular from A onto DE,and H lies in the interior of the segment DE. Suppose ∠BCA = 90◦, CE = 12 cm,EH = 15 cm, AH = 40 cm and CD = x cm. Find the value of x.

33. (SMO(O) 2011 P4) Given an isosceles triangle 4ABC with AB = AC and ∠A = 20◦.The point D lies on AC such that AD = BC. Determine ∠ABD in degrees.

a review of geometry

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