A Review of Geometry

A Review of Geometry Khor Shi-Jie May 17, 2012

Contents 1 What’s Next? 2 Chasing Pavements 2.1 Your Toolbox . . . . 2.1.1 Triangles . . . 2.1.2 Circles . . . . 2.1.3 Areas . . . . 2.2 More Powerful Tools 2.3 Auxiliary Lines . . . 2.4 Problem Set . . . . .

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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 Olympiad which is saturated with theorems and properties. What is even more challenging is that students must be equipped with strong visualisation skills in order to incise geometric problems with the correct theorems and properties. This reason alone makes many student detract attention 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 previous document on Algebra, there are two distinctive phase in the grasping of Mathematics Olympiad. The first step in studying geometry is, obviously, to study geometric theorems and be well acquainted with basic applications of such theorems. Of course, this step is relatively 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 important for student to learn the strategies involved in solving geometric problems. It is a cardinal sin (well, pardon my exaggeration) to neglect this step because most geometric problems in competitions does not merely involve the application of a theorem. It is often coupled with ingenious transformations and inspired combination of theorems that allows one to solve a geometric problem beautifully. In order to acquire different skills in geometry, it is important for one to do hundreds and thousands of problems to familiarise oneself with different strategies and pick up useful lemmas in the process. (You’ve heard it, THOUSANDS) There are two approaches in solving geometric problems: synthetic geometry and analytical geometry. Synthetic geometry involves using theorems learnt in MO to solve a geometric problem. This is often coupled with creative construction and transformation in the process of solving the problem. The lack of a fixed approach or algorithm in solving the problem is a huge contrast to analytical geometry. Analytical geometry, on the other hand, is also known as the brute force method. Students can use methods such as coordinate geometry, trigonometry, vectors and complex numbers to quantify everything within the problem and solve the problem 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 it takes too much time in arriving at the correct approach. The analytical geometry approach 2

CHAPTER 1. WHAT’S NEXT?

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will definitely involve heavy computation, but it is more certain that one will obtain the solution through this method. I recommend students to master both techniques in solving geometric problems. By saying basic theorems and properties in geometry, I refer to the following topics: 1. Planar Geometry (a) Triangles i. ii. iii. iv.

Similarities and congruences Area of triangles Trigonometric properties 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. ii. iii. iv.

Basic circular properties Cyclic quadrilaterial Power of a point and radical axis 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?

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(d) Inversive geometry Yup, that’s a lot of things to study, considering that this is merely the first step in studying geometry. Note that I have omitted 3D geometry in the list above. 3D geometry is covered in most regional Olympiads around the world. However, it is not within the syllabus of SMO and IMO. Nevertheless, it is still useful to learn the skills in 3D geometry as these problems do appear in AMC, AIME and Purple Comet. The fundamental knowledge listed above will help in the following three strategies that I will expound in the following chapters: 1. Chasing. This technique involves using geometric theorems to ”chase” all the quantities within the geometric diagram. This includes the lengths, angles and areas within the problem. Similarity and congruences help tremendously in this technique. Often, we cannot chase all the quantities and have to set variables to simplify our work (be cautious 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 apply geometric theorems. We can also transform a certain part of the diagram through reflection, rotation, translation, dilation, etc. There are often hints in the question that 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 problems. This set of notes is created to help students in senior section (and perhaps some in junior 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 the conditions given in the problem, we ”chase” the different quantities involved in the question until we arrive at what the question requires us to evaluate or prove. Let us take a look at the following problem: (AIME2009 P10) Four lighthouses are located at points A, B, C and D. The lighthouse at A is 5 kilometres form the lighthouse at B, the lighthouse at B is 12 kilometres away from the lighthouse at C, and the lighthouse at A is 13 kilometres away from the lighthouse at C. To an observer at A, the angle determined by the lights at B and D and the angle determined by the lights at C and D are equal. To an observer at C, the angle determined by the lights at A and B and the angle determined by the lights at D and B are equal. The number of kilometres √ p r , where p, q and r are relatively prime positive integers, and r from A to D is given by q 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

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such that BC is the angle bisector of ∠ACD while AD is the angle bisector of ∠BAC. Find the 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 to find out the length of BE and CE. So we have: BE = 12 ×

10 5 = 18 3

CE = 12 ×

13 26 = 18 3

Next, we can use Pythagorean theorem to find out the length of AE. That gives us: s √ 2 13 10 5 AE = 52 + = 3 3 We can continue chasing the values of ∠ACE using cosine rule, and ∠ACD using double angle formula. But that is not very glamorous and takes up too much time (considering that students are supposed to complete this problem in AIME). Usually when you are stuck in the midst of chasing quantities, it is time when you should consider adding auxiliary lines to help solving the problem. I have compiled a list of common auxiliary lines which are effective in solving geometric problems in later parts of the chapter. In this scenario, it will be useful for us to extend AB and 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 that CF = 13 and BF = 5. Alas, we can use angle bisector theorem to find out the length of CD and DF respectively and use the same theorem again to find the length of AD. CD = 13 ×

13 169 = 23 23

DF = 13 ×

10 130 = 23 23

The length of AD is given by √ 13 + 169 5√ 60 13 23 AD = 13 × = 3 13 23


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And that’s it, we just finished chasing all the lengths which are required with the help of some construction lines. The above problem is merely a simple problem that involves length chasing. Most of the time, it can be very challenging to chase all the quantities without suitable constructions. Before we even discuss the strategies used in chasing, students must be 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 discussion which 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 property that students have to grasp is the similarity and congruency of triangles. The following diagram is a diagram which comes out frequently in geometry. I recommend students to spend 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 denoted as H). We are often concerned with the 5 centres of a triangle in Math Olympiad problems, namely the orthocentre, the centroid (intersection of medians), the incentre (the centre of the inscribe circle), the circumcentre (the centre of the circumscribe circle) and the excentres (the centres of the three excircles). There are many interesting properties about these centres which are often tested in Math Olympiad, but this topic will be covered in future lessons. There are many theorems which can stem from a right angle triangle and similarity of triangles. Here are a few useful theorems and corollaries: Theorem 1 (Pythagorean Theorem) Given a triangle 4ABC, we have AC 2 + AB 2 = BC 2 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.


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Corollary 2 In a triangle 4ABC and D is a point on BC. We have AB 2 − BD2 = AC 2 − CD2 if and only if AD ⊥ BC. Corollary 3 Given a quadrilateral ABCD, the diagonals AC and BD are perpendicular if and only if AB 2 + CD2 = AD2 + BC 2 . Isosceles triangles and equilateral triangles often come out in problems too and one should be familiar with its properties. For example, in an isosceles triangle, the altitude from the vertex between the two equal side both a median and an angle bisector of the triangle. One should also be aware of the likelihood of congruent triangles erected from the sides of an equilateral triangle. A cevian of a triangle is a segment connecting a vertex of the triangle to a point in the opposite side of the triangle. Examples of cevians include the altitude of the triangle, the median of the triangle and the angle bisector of the triangle. There is a very powerful theorem which allows us to calculate the length of these cevians given the lengths of the sides of the triangle.

Figure 2.1.2 Theorem 2 (Stewart’s Theorem) In triangle 4ABC, D is a point on BC. Suppose b2 m + c 2 n 2 AB = c, AC = b, BC = a, AD = d, BD = m, CD = n, we have d = − mn. a Corollary 4 (Pappus’ Theorem) If D is the midpoint of BC, then we have d2 =

2.1.2

2b2 + 2c2 − a2 . 4

Circles

There are several essential circular properties that students must know (it is covered in Sec 2 syllabus). Here’s a list of these essential properties: Property 1 Angles subtended by the same chord are equal i.e. ∠AP1 B = ∠AP2 B = ∠AP3 B. Angles subtended by chords of equal length are equal.


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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.


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

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 is defined as the difference between the square of the distance from a point P to the centre of the circle O and the square of the radius i.e. OP 2 − r2 . You can draw a line from the point such that it intersects the circle at two points Q and R. Then the power of the point OP 2 − r2 = P Q × P R. From this property, we can obtain the power of a point theorem and the intersecting chord theorem: Theorem 3 (Power of a Point) Given that P is a point outside of circle O, and that P E is tangent to circle O, we have P A × P B = P C × P D = P E 2 .


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Figure 2.1.7 Theorem 4 (Intersecting Chord Theorem) Given that P is a point in circle O, we have P A × P B = P C × P D.

Figure 2.1.8 The radical axis of two circles is the line whereby all the points on the line have equal circular power with respect to both circles. If the two circles overlap, the radical axis will be the line which contains the common chord of the two circles. There are many interesting properties which involves the radical axis, but this topic is usually tested only in the senior or open section of SMO.

2.1.3

Areas

There are also many formulas that concerns the area of a polygon. Denote S4ABC as the area of triangle 4ABC. We have the following theorems which are useful in evaluating the area: Theoremp5 (Heron’s Formula) Define s as the semiperimeter of the triangle. We have S4ABC = s(s − a)(s − b)(s − c). 1 Theorem 6 GIven that AC = b and AB = c, we have S4ABC = bc sin A. 2


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It is very important to relate areas with the ratio of line segments, especially in discussing triangles. Often, solution which makes use of area relationships are very elegant and short. In S4P AB PM the following 6 diagrams, the ratio of areas is given by = . More generally, if AB S4QAB QM is the common side of the two triangles 4ABP and 4ABQ and point M is the intersection S4P AB PM points of AB and P Q, then we have = . S4QAB QM

Figure 2.1.9 From theorem 6, we see that it can be useful to discuss areas if two triangles have a common angle. Based on the theorem, suppose two triangles 4ABC and 4A0 B 0 C 0 fulfils S4ABC AB · AC the condition ∠A = ∠A0 or ∠A+∠A0 = 180◦ , then we have = 0 0 0 0 . This fact S4A0 B 0 C 0 AB ·AC can be proven by superimposing the two triangles together at the equal angle and considering the 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 junior section can be greatly simplified with the use of these theorems: Theorem 7 (Angle Bisector Theorem) In 4ABC, suppose the angle bisector of ∠A inAB AC tersects side BC at D. We have = . BD CD


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Figure 2.1.10 If a question concerns angle bisector of one of the angle of the triangle, we can be quite certain that the question involves the use of angle bisector theorem. Coupled with Stewart’s theorem, 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 BX CY AZ · · = 1. respectively. Suppose AX, BY and CZ are concurrent. Then we have XC Y A ZB

Figure 2.1.11 Ceva’s theorem is often used to prove concurrency between three line segments. In junior section, this theorem can be useful in evaluating the length of segments that corresponds to the segments in the theorem. This triangle can be easily recognised and the application of Ceva’s theorem is usually obvious. Theorem 9 (Menalaus’ Theorem) Given that X, Y and Z are points on BC, AC and CX AB or their extension respectively. Suppose X, Y and Z are collinear, then we have · XB BZ AY · =1 ZA Y C

Figure 2.1.12 Menalaus’ Theorem is often used to prove collinearity of three points. I would advise people to memorise this theorem base on the direction of the line segments. This theorem can be applied often due to the common appearance of the shape that this theorem is involved in.


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b c a = = = 2R Theorem 10 (Sine Rule) Given a triangle 4ABC, we have sin A sin B sin C where R is the circumradius of the triangle. Theorem 11 (Cosine Rule) Given a triangle 4ABC, we have a2 = b2 + c2 − 2bc cos A.

2.3

Auxiliary Lines

Constructing auxiliary lines is an indispensable technique to do well in geometry. Here’s a list 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 us a bridge in chasing certain quantities in the diagram. Sometimes, the points to connect may 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 of parallel 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 to solve the problem. 4. Constructing similar shapes. We create construction lines such that the shape constructed is similar to a certain shape which is in the original diagram. 5. Constructing perpendicular lines. This allows us to use the properties of perpendicular lines as well as the Pythagorean theorem. It can be useful too in defining the ratios of lengths. 6. Geometric transformation. Transformations like rotations, translation, reflection, scaling 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 the bisector of C. Determine the area of the triangle AM C. 5. Given that P is a point in an acute angle triangle 4ABC such that ∠BP C = ∠BAC + AP × BC . Evaluate k 4 . 90◦ and AB × P C = AC × P B. Let k = AC × BP


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6. 4ABC is a right-angled triangle with ∠C = 90◦ . M is the midpoint of AB while D and E are points on BC and AC respectively such that ∠DM E = 90◦ . Given that DB = 20 and EC = 21, evaluate M D2 + M E 2 . 7. Given a convex quadrilateral ABCD, AB = CD, E, F are the midpoints of AD and BC respectively. Extend EF to meet the extension of BA at S and the extension of CD at K. Prove that ∠ASF = ∠DKF . 8. (AIME 1984 P3) A point P is chosen in the interior of 4ABC such that when lines are drawn 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 smaller triangles by lines drawn from the vertices through a common interior point. The areas of 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 bisectors of ∠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 is perpendicular to BC. If BD = 3 cm and DC = 2 cm, and the area of the 4ABC is x cm2 , find the value of x. 12. (SMO(J) 2007 P1) In the trapezium ABCD, AB k CD, O is the intersection of AC and BD, AB = b, CD = a and a < b. Let S be the area of the trapezium ABCD. Suppose the area of 4BOC is 2S . Find the value of ab . 9


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13. (SMO(J) 2007 P2) Equilateral triangles 4ABE and 4BCF are erected externally on the 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 that AD = 1. Find the value of BD · DC. 15. (SMO(J) 2008 P3) In the quadrilateral P QRS, A, B, C and D are midpoints of the sides P Q, QR, RS and SP respectively, and M is the midpoint of CD. Suppose H is the 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 centres A, B and C in a row. If the line W Z is tangent to the third circle, find the length of XY.

17. (SMO(J) 2009 P15) 4ABC is a right-angled triangle with ∠BAC = 90◦ . A square is constructed on side AB and BC as shown. The area of square ABDE is 8 cm2 and the area of the square BCF G 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 sides BC, 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 are on 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 quadrilateral DBEF , find the area of triangle 4AEC in cm2 .


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20. (SMO(S) 2009 P16) 4ABC is a triangle and D is a point on side BC. Point E is on side AB such that DE is the angle bisector of ∠ADB, and point F is on side AC such AE BD CF · · . that DF is the angle bisector of ∠ADC. Find the value of EB DC F A 21. (SMO(S) 2009 P29) ABCD is a rectangle, E is the midpoint of AD and F is the midpoint of CE. If the area of triangle 4BDF is 12 cm2 , find the area of rectangle ABCD in cm2 . 22. (SMO(O) 2009 P2) Let A1 , A2 , · · · A6 be 6 points [ [ [ [ [ [ A 1 A2 = A2 A3 , A3 A4 = A4 A5 , A5 A6 = A6 A1 , where arc A1 A2 etc. It is also known that ∠A1 A3 A5 = degrees.

on a circle in this order such that [ A 1 A2 denotes the arc length of the 72◦ . Find the size of ∠A4 A6 A2 in

23. (SMO(O) 2009 P4) Let P1 , P2 , · · · P4 1 be 41 distinct on segment BC of a triangle Ppoints 4 1i=1 (APi2 + Pi B · Pi C). 4ABC, where AB = AC = 7. Evaluate the sum 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 and CD respectively, such that BE = CF = 13 AB. G is the intersection of BF and DE. Area of ABGD m If = is in its lowest term, find the value of m + n. Area of ABCD n 26. (SMO(J) 2010 P1) Let the diagonals of the square ABCD intersect at S and let P be the midpoint of AB. Let M be the intersection of AC and P D and N the intersection of BD and P C. A circle is inscribed in the quadrilateral P M SN . Prove that the radius of the circle is M P − M S. 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 cm and BC + CD + DA = 17 2 2 cm, find the area of trapezium ABCD in cm . 28. (SMO(S) 2010 P22) Given a circle with diameter AB, C and D are points on the circle on the same side of AB such thath BD bisects ∠CBA. The chords AC and BD intersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, find the value of x.


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29. (SMO(S) 2010 P1) In the triangle 4ABC with AC > AB, D is the foot of perpendicular from A onto BC and E is the foot of perpendicular from D onto AC. Let F be the point on the line DE such that EF · DC = BD · DE. Prove that AF is perpendicular to BF . 30. (SMO(J) 2011 P33) In the following diagram, ABCD is a square, BD k CE and BE = BD. Let ∠E = x◦ . Find x.

31. (SMO(J) 2011 P2) Two circles Γ1 , Γ2 with radii r1 , r2 respectively, touch internally at point P . A tangent parallel to the diameter through P touches Γ1 at R and intersects Γ2 at M and N . Prove that P R bisects ∠M P N . 32. (SMO(S) 2011 P21) ABCD is a convex quadrilateral such that AC intersects BD at the 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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