Notes on Euclidean Geometry

Paul Yiu

Department of Mathematics Florida Atlantic University

Fall 2005

Version 051007

Contents

1 Pythagoras Theorem and Its Applications 1 1.1 Pythagoras Theorem ...... 1 1.2 Construction of the regular pentagon ...... 5 1.3 The law of cosines and its applications ...... 8

2 The circumcircle and the incircle 13 2.1 The circumcircle and the law of sines ...... 13 2.2 The incircle and excircles ...... 15 2.3 Heron’s formula for the area of a triangle ...... 19 2.3.1 Exercise 2.3 ...... 20

3 The shoemaker’s knife 23 3.1 The shoemaker’s knife ...... 23 3.2 Circles in the shoemaker’s knife ...... 24 3.3 Archimedean circles in the shoemaker’s knife ...... 27

4 Introduction to Triangle Geometry 31 4.1 Preliminaries ...... 31 4.1.1 Coordinatization of points on a line ...... 31 4.1.2 Centers of similitude of two circles ...... 32 4.1.3 Tangent circles ...... 32 4.1.4 Harmonic division ...... 33 4.1.5 Homothety ...... 35 4.1.6 The power of a point with respect to a circle ...... 35 4.2 Menelaus and Ceva theorems ...... 36 4.2.1 Menelaus and Ceva Theorems ...... 36 4.2.2 Desargues Theorem ...... 36 4.2.3 The incircle and the Gergonne point ...... 38 iv CONTENTS

4.2.4 The excircles and the Nagel point ...... 41 4.3 The nine-point circle ...... 43 4.3.1 The Euler triangle as a midway triangle ...... 43 4.3.2 The orthic triangle as a pedal triangle ...... 44 4.3.3 The nine-point circle ...... 46 4.4 The OI-line ...... 49 4.4.1 The homothetic center of the intouch and excentral triangles 49 4.4.2 The centers of similitude of the circumcircle and the incircle 50 4.4.3 Reﬂection of I in O ...... 51 4.4.4 Orthocenter of intouch triangle ...... 52 4.4.5 Centroids of the excentral and intouch triangles ...... 53 4.4.6 Mixtilinear incircles ...... 57 4.5 Euler’s formula and Steiner’s porism ...... 58 4.5.1 Euler’s formula ...... 59 4.5.2 Steiner’s porism ...... 59

5 Homogeneous Barycentric Coordinates 63 5.1 Barycentric coordinates with reference to a triangle ...... 63 5.1.1 Ceva Theorem ...... 64 5.1.2 Some basic triangle centers ...... 64 5.1.3 The orthocenter ...... 66 5.2 The inferior and superior triangles ...... 68 5.2.1 More triangle centers ...... 69 5.3 Isotomic conjugates ...... 70 5.3.1 Congruent-parallelians point ...... 71 5.4 Internal center of similitudes of the circumcircle and the incircle 72 Chapter 1

Pythagoras Theorem and Its Applications

1.1 Pythagoras Theorem

Theorem 1.1 (Pythagoras). The lengths a ≤ b

Proof. (1) Dissection.

(2) Euclid’s proof. Construct the altitude at the right angle to meet AB at P and the opposite side ZZ of the square ABZZ at Q. Note that the area of the rectangle AZQP is twice of the area of triangle AZC. By rotating this triangle about A through a right angle, we obtain the congruent triangle ABX, whose area is half of the area of the square on AC. It follows that the area of rectangle AZQP is equal to the area of the square on AC. For the same reason, the area of rectangle BZQP is equal to that of the square on BC. From these, the area of the square on AB is equal to the sum of the areas of the squares on BC and CA. 2 Pythagoras Theorem and Its Applications

Y C

A B P

Z Q Z

Theorem 1.2 (Converse of Pythagoras). If the lengths of the sides of a triangles satisfy the relation a2 + b2 = c2, then the triangle contains a right angle.

Proof. Let ABC be a triangle with BC = a, CA = b, and AB = c satisfying a2 + b2 = c2. Consider another triangle XY Z with YZ = a, XZ = b, ∠XZY =90◦.

By the Pythagorean theorem, XY 2 = a2 + b2 = c2, so that XY = c. Thus the triangles ABC ≡XY Z by the SSS test. This means that ∠ACB = ∠XZY is a right angle.

Exercise 1.1 1. (a) Let ABC be a right triangle with sides a, b and hypotenuse c.Ifd is the height of on the hypotenuse, show that 1 1 1 + = . a2 b2 d2

(b) Let ABC be a right triangle with a right angle at vertex C. Let CXPY be a square with P on the hypotenuse, and X, Y on the sides. Show that 1.1 Pythagoras Theorem 3

t Y P a db t

B A C A c X

the length t of a side of this square is given by 1 1 1 = + . t a b

2. AB is a chord of length 2 in a circle O(2). C is the midpoint of the minor arc AB and M the midpoint of the chord AB. C

B A M

O √ √ √ Show that (i) CM =2− 3; (ii) BC = 6 − 2. Deduce that √ 1 √ √ 1 √ √ tan 15◦ =2− 3, sin 15◦ = ( 6− 2), cos 15◦ = ( 6+ 2). 4 4 3. ABC is a triangle with a right angle at C. If the median on the side a is the geometric mean of the sides b and c, show that c =3b. 4. ABC is a triangle with a right angle at C. If the median on the side c is the geometric mean of the sides a and b, show that one of the acute angles is 15◦. 5. In each of the following cases, ﬁnd the ratio AB : BC. 1 6. (a) The midpoint of a chord of length 2a is at a distance d from the midpoint of the minor arc it cuts out from the circle. Show that the diameter of the a2+d2 circle is d . √ √ 1 3: 3+2in the case of 4 circles. 4 Pythagoras Theorem and Its Applications

B A B A

C D C D

d b d b a a a a

(b) Two parallel chords of a circle has lengths 168 and 72, and are at a distance 64 apart. Find the radius of the circle. 2

7. (a) AB is an arc of a circle O(r), with ∠AOB = α. Find the radius of the circle tangent to the arc and the radii through A and B.

2Answer: The distance from the center to the longer chord is 13. From this, the radius of the circle is 85. More generally, if these chords has lengths 2a and 2b, and the distance between them is d, the radius r of the circle is given by [d2 +(a − b)2][d2 +(a + b)2] r2 = . 4d2 1.2 Construction of the regular pentagon 5

(b) Two congruent circles of radii a have their centers on each other. Cal- culate the radius of the circle tangent to one of them internally, the other externally, and the line joining their centers. 8. BCX and CDY are equilateral triangles inside a rectangle ABCD. The lines AX and AY are extended to intersect BC and CD respectively at P and Q. Show that (a) AP Q is an equilateral triangle; (b) AP B + ADQ = CPQ.

D Q C

Y A B

1.2 Construction of the regular pentagon

We present two ruler-and-compass constructions of the geometric means of two quantities given as lengths of segments. These are based on Euclid’s proof of the Pythagorean theorem. Construction 1.1. Given two segments of length a

Construction 1.2. Given two segments of length a, b, mark three points A, P , B on a line (P between A and B) such that PA= a, PB = b. Describe a semicircle with AB as diameter, and let the perpendicular through P intersect the semicircle at Q. Then PQ2 = PA· PB, so that the length of PQis the geometric mean of a and b. 6 Pythagoras Theorem and Its Applications

Q Q

A P P B A B

Consider a regular pentagon ADEBC. Since XB = XC by symmetry, the AC CX CAB XCB = isosceles triangles and are similar. From this, AB CB, and AC · CB = AB · CX. It follows that AX 2 = AB · XB. We say that X divides the segment AB in the golden ratio, i.e., AX : AB = XB : AX.

Y X A B P

Z Q Q P

D E A B X

Construction 1.3 (Division of a segment in the golden ratio). Given a segment AB,

1. draw a right triangle ABP with BP perpendicular to AB and half in length,

2. mark a point Q on the hypotenuse AP such that PQ= PB,

3. mark a point X on the segment AB such that AX = AQ.

Then X divides AB into the golden ratio.

Construction 1.4 (Regular pentagon). To construct a regular pentagon with a given diagonal AB, 1.2 Construction of the regular pentagon 7

1. divide a segment AB into the golden ratio at X. 2. construct the circles A(X) and X(B) to intersect at C. 3. construct a circle center C, radius AB, to meet the two circles A(X) and B(AX) at D and E respectively. Then, ACBED is a regular pentagon.

Exercise 1.2 1. Suppose X divides AB in the golden ratio. Show that AX : XB = φ :1, where 1 √ φ = ( 5+1)≈ 1.618 ···. 2 √ AX = 1 ( 5 − 1) = φ − 1= 1 Show also that AB 2 φ . 2. If the legs and the altitude of a right triangle form the sides of another right triangle, show that the altitude divides the hypotenuse into the golden ratio. 3. ABC is an isosceles triangle with a point X on AB such that AX = CX = BC. Show that (i) ∠BAC =36◦; (ii) AX : XB = φ :1. Suppose XB =1. Let E be the midpoint of the side AC. Show that 1 √ XE = 10 + 2 5. 4 Deduce that √ 5+1 1 √ √ cos 36◦ = , sin 36◦ = 10 − 2 5, tan 36◦ = 5 − 2 5. 4 2

4. ABC is an isosceles triangle with AB = AC =4. X is a point on AB such that AX = CX = BC. Let D be the midpoint of BC. Calculate the length of AD, and deduce that √ 5 − 1 1 √ 1 √ sin 18◦ = , cos 18◦ = 10 + 2 5, tan 18◦ = 25 − 10 5. 4 4 5 8 Pythagoras Theorem and Its Applications

5. Justify the following construction of the regular pentagon. Let OA and OY be two perpendicular radii of a circle, center O. (a) Mark the midpoint M of OY and bisect angle OMA to intersect OA at P . (b) Construct the perpendicular to OA at P to intersect the circle at B and E. Y B

C M

A O P

Then A, B, E are three adjacent vertices of a regular pentagon inscribed in the circle. The remaining two vertices can be easily constructed.

1.3 The law of cosines and its applications

Given a triangle ABC, we denote by a, b, c the lengths of the sides BC, CA, AB respectively. Theorem 1.3 (The law of cosines). c2 = a2 + b2 − 2ab cos C.

Theorem 1.4 (Stuart). If X is a point on the side BC (or its extension) such that BX : XC = λ : µ, then λb2 + µc2 λµa2 AX2 = − . λ + µ (λ + µ)2

Proof. Use the cosine formula to compute the cosines of the angles AXB and AXC, and note that cos ABC = − cos AXB. 1.3 The law of cosines and its applications 9

Corollary 1.5 (Apollonius’ theorem). The length ma of the median AD is given by 1 m2 = (2b2 +2c2 − a2). a 4

Corollary 1.6. The length wa of the (internal) bisector of angle A is given by a 2 w2 = bc 1 − . a b + c

Proof. Apply Stewart’s Theorem with λ = c and µ = b. s = 1 (a + b + c) w2 = 4bcs(s−a) Remark. If we write 2 , then a (b+c)2 .

Exercise 1.3 1. Show that the (4,5,6) triangle has one angle equal to twice of another. 2. If C =2B, show that c2 =(a + b)b.

3. Find a simple relation between the sum of the areas of the three squares S1, S2, S3, and that of the squares T1, T2, T3.

S1 S2

T2 S3 T1

ABC a =12 b + c =18 cos A = 7 4. is a triangle with , , and 38 . Calculate the lengths of b and c, 3 and show that a3 = b3 + c3. √ √ 3AMM E688, P.A. Piz´a. Here, b =9− 5, and c =9+ 5. 10 Pythagoras Theorem and Its Applications

5. The lengths of the sides of a triangle are 136, 170, and 174. Calculate the lengths of its medians. 4

6. (a) mb = mc if and only if b = c. m2 + m2 + m2 = 3 (a2 + b2 + c2) (b) a b c 4 . (c) If ma : mb : mc = a : b : c, show that the triangle is equilateral.

7. Suppose mb : mc = c : b. Show that either (i) b = c,or (ii) the quadrilateral AEGF is cyclic. Show that the triangle is equilateral if both (i) and (ii) hold. [Hint: Show b2m2 − c2m2 = 1 (c − b)(c + b)(b2 + c2 − 2a2) that b c 4 ].

8. The lengths of the sides of a triangle are 84, 125, 169. Calculate the lengths of its internal bisectors. 5

Challenging Problems

1. Suppose mb : mc = c : b. Show that either (i) b = c,or (ii) the quadrilateral AEGF is cyclic. Show that the triangle is equilateral if both (i) and (ii) hold. [Hint: Show b2m2 − c2m2 = 1 (c − b)(c + b)(b2 + c2 − 2a2) that b c 4 ].

6 2. (Steiner - Lehmus Theorem) If wa = wb, then a = b.

3. Suppose wa : wb = b : a. Show that the triangle is either isosceles, or C =60◦. 7

4Answers: 158, 131, 127. 5 975 26208 12600 Answers: 7 , 253 , 209 . 6Hint: Show that a b (a − b)[(a + b + c)2 − ab] − = . (b + c)2 (c + a)2 (b + c)2(c + a)2

2 7 2 2 2 2 abc(b−a)(a+b+c) 2 2 2 a wa − b wb = (a+c)2(b+c)2 [a − ab + b − c ]. 1.3 The law of cosines and its applications 11

4. (a) Show that the length of the external angle bisector is given by a 2 4bc(s − b)(s − c) w2 = bc − 1 = . a b − c (b − c)2

(b) In triangle ABC, A =12◦, and B =36◦. Calculate the ratio of the 8 lengths of the external angle bisectors wa and wb.

5. The bisectors of angles B and C of triangle ABC intersect the median AD at E and F respectively. Suppose BE = CF. Show that triangle ABC is isosceles. 9

Appendix: Synthetic proofs of Steiner - Lehmus Theorem First proof. 10 Suppose β<γin triangle ABC. We show that the bisector BM is longer than the bisector CN. L BM ∠NCL = 1 β B N L C Choose a point on such that 2 . Then , , , are concyclic since ∠NBL = ∠NCL. Note that 1 ∠NBC = β< (β + γ)=∠LCB, 2 and both are acute angles. Since smaller chords of a circle subtend smaller acute angles, we have CN < BL. It follows that CN < BM.

Second proof. 11 Suppose the bisectors BM and CN in triangle ABC are equal. We shall show that β = γ. If not, assume β<γ. Compare the triangles CBM and BCN.

8Answer: 1:1. The counterpart of the Steiner - Lehmus theorem does not hold. See Crux Math. 2 (1976) pp. 22 – 24. D.L.MacKay (AMM E312): if the external angle bisectors of B and C of a s−a s−b s−c scalene triangle ABC are equal, then a is the geometric mean of b and c . See also Crux 1607 for examples of triangles with one internal bisector equal to one external bisector. 9Crux Math. 1897; also CMJ 629. 10Gilbert - McDonnell, American Mathematical Monthly, vol. 70 (1963) 79 – 80. 11M. Descube, 1880. 12 Pythagoras Theorem and Its Applications

A A G

N L N M M

B C B C

∠CBM = 1 β<1 γ = These have two pairs of equal sides with included angles 2 2 ∠BCN, both of which are acute. Their opposite sides therefore satisfy the relation CM < BN. Complete the parallelogram BMGN, and consider the triangle CNG. This is isosceles since CN = BM = NG. Note that 1 ∠CGN = β + ∠CGM, 2 1 ∠GCN = γ + ∠GCM. 2 Since β<γ, we conclude that ∠CGM > ∠GCM. From this, CM > GM = BN. This contradicts the relation CM < BN obtained above. Chapter 2

The circumcircle and the incircle

2.1 The circumcircle and the law of sines

The perpendicular bisectors of the three sides of a triangle are concurrent at the circumcenter of the triangle. This is the center of the circumcircle, the circle passing through the three vertices of the triangle.

A A

F O E O

B C B C D D

Theorem 2.1 (The law of sines). Let R denote the circumradius of a triangle ABC with sides a, b, c opposite to the angles A, B, C respectively. a b c = = =2R. sin A sin B sin C = 1 bc sin A Since the area of a triangle is given by 2 , the circumradius can be written as abc R = . 4 14 The circumcircle and the incircle

Exercise 2.1 1. Where is the circumcenter of a right triangle? What is its circumradius? 1

2. (The orthocenter) Given triangle ABC, construct lines through A parallel to BC, through B parallel to CA, and through C parallel to AB. These three lines bound a triangle ABC (so that A lies on BC, B on CA, and C on AB respectively). Show that the altitudes of triangle ABC are the perpendicular bisectors of the sides of ABC. Deduce that the three altitudes of a triangle always intersect at a point (called its orthocenter).

3. Let H be the orthocenter of triangle ABC. Show that (i) A is the orthocenter of triangle HBC; (ii) the triangles HAB, HBC, HCA and ABC have the same circumradius.

4. The internal bisectors of angles B and C intersect the circumcircle of ABC at B and C. (i) Show that if B = C, then BB = CC. (ii) If BB = CC, does it follow that B = C? 2

5. (Morley’s Theorem) Given three angles θ, φ, ψ such that θ + φ + ψ =60◦, and an equilateral triangle XY Z, construct outwardly triangles AY Z and BZX such that ∠AY Z =60◦ + ψ, ∠AZY =60◦ + φ ∠BZX =60◦ + θ, ∠BXZ =60◦ + ψ.

Suppose the sides of XY Z have unit length.

(a) Show that sin(60◦ + ψ) sin(60◦ + ψ) AZ = , and BZ = . sin θ sin φ

1The midpoint of the hypotenuse; half of the length of the hypotenuse. 2 π (ii) No. BB = CC if and only if B = C or A = 3 . 2.2 The incircle and excircles 15

Y Z

(b) In triangle ABZ, show that ∠ZAB = θ and ∠ZBA = φ. (c) Suppose a third triangle XY C is constructed outside XY Z such that ∠CY X =60◦ + θ and ∠CXY =60◦ + φ. Show that AY , AZ are the trisectors of angle A, BX, BZ those of angle B, and CX, CY those of angle C. (d) Show that AY · BZ · CX = AZ · BX · CY . (e) Suppose the extensions of BX and AY intersect at P . Show that the triangles PXZ and PYZ are congruent.

2.2 The incircle and excircles

The internal angle bisectors of a triangle are concurrent at the incenter of the triangle. This is the center of the incircle, the circle tangent to the three sides of the triangle. If the incircle touches the sides BC, CA and AB respectively at X, Y , and Z,

AY = AZ = s − a, BX = BZ = s − b, CX = CY = s − c. 16 The circumcircle and the incircle

Z I

B C X Denote by r the inradius of the triangle ABC. 2 r = = . a + b + c s The internal bisector of each angle and the external bisectors of the remaining two angles are concurrent at an excenter of the triangle. An excircle can be constructed with this as center, tangent to the lines containing the three sides of the triangle.

B C Y

The exradii of a triangle with sides a, b, c are given by r = ,r= ,r= . a s − a b s − b c s − c I BC I CA I AB 1 ar 1 br 1 cr The areas of the triangles a , a , and a are 2 a, 2 a, and 2 a respectively. Since

= −IaBC + IaCA + IaAB, 2.2 The incircle and excircles 17 we have 1 = r (−a + b + c)=r (s − a), 2 a a r = from which a s−a .

Exercise 2.2 1. Show that the inradius of a right triangle (of hypotenuse c)iss − c.

2. A square of side a is partitioned into 4 congruent right triangles and a small square, all with equal inradii r. Calculate r.

3. The incenter of a right triangle is equidistant from the midpoint of the hypotenuse and the vertex of the right angle. Show that the triangle contains a 30◦ angle. C

B A

4. The incircle of triangle ABC touches the sides AC and AB at Y and Z respectively. Suppose BY = CZ. Show that the triangle is isosceles.

5. A line parallel to hypotenuse AB of a right triangle ABC passes through the incenter I. The segments included between I and the sides AC and BC have lengths 3 and 4. Calculate the area of the triangle.

6. If the incenter is equidistant from the three excenters, show that the triangle is equilateral. 18 The circumcircle and the incircle

B A

7. The circle BIC intersects the sides AC, AB at E and F respectively. Show that EF is tangent to the incircle of ABC. 3

Z I F

B C X

8. (a) The triangle is isosceles and the three small circles have equal radii. Suppose the large circle has radius R. Find the radius of the small circles. 4

3Hint: Show that IF bisects angle AF E. 4Let θ be the semi-vertical angle of the isosceles triangle. The inradius of the triangle is 2R sin θ cos2 θ R 1 1+sin θ =2R sin θ(1 − sin θ). If this is equal to 2 (1 − sin θ), then sin θ = 4 . From 3 thi,s the inradius is 8 R. 2.3 Heron’s formula for the area of a triangle 19

(b) The large circle has radius R. The four small circles have equal radii. Calculate this common radius. 5 9. An equilateral triangle of side 2a is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. If the inradii of the quadrilateral and the isosceles triangle are equal, ﬁnd this inradius. 6

2.3 Heron’s formula for the area of a triangle

Consider a triangle ABC with area . Denote by r the inradius, and ra the radius of the excircle on the side BC of triangle ABC. It is convenient to introduce the s = 1 (a + b + c) semiperimeter 2 .

ra r

A Y C Y

5Let θ be the smaller acute angle of one of the right triangles. The inradius of the right triangle 2R sin θ cos θ R 5 is 1+sin θ+cos θ . If this is equal to 2 (1 − sin θ), then 5sinθ − cos θ =1. From this, sin θ = 13 , 4 R and the√ radius√ is 13 . 6( 3 − 2)a. 20 The circumcircle and the incircle

•= rs. • From the similarity of triangles AIZ and AI Z, r s − a = . ra s

• From the similarity of triangles CIY and I CY ,

r · ra =(s − b)(s − c).

• From these, (s − a)(s − b)(s − c) r = , s = s(s − a)(s − b)(s − c).

This latter is the famous Heron formula.

2.3.1 Exercise 2.3 1. The altitudes a triangle are 12, 15 and 20. What is the area of the triangle ? 7

2. Find the inradius and the exradii of the (13,14,15) triangle.

3. If one of the ex-radii of a triangle is equal to its semiperimeter, then the triangle contains a right angle. 4. 1 + 1 + 1 = 1 . ra rb rc r 2 5. rarbrc = rs . 6. Show that r + r + r = −s3+(ab+bc+ca)s (i) a b c ; (ii) (s − a)(s − b)(s − c)=−s3 +(ab + bc + ca)s − abc. Deduce that ra + rb + rc =4R + r.

7 = 150. The lengths of the sides are 25, 20 and 15. 2.3 Heron’s formula for the area of a triangle 21

7. The length of each side of the square is 6a, and the radius of each of the top and bottom circles is a. Calculate the radii of the other two circles. 8

8. (a) ABCD is a square of unit side. P is a point on BC so that the incircle of triangle ABP and the circle tangent to the lines AP , PC and CD have equal radii. Show that the length of BP satisﬁes the equation 2x3 − 2x2 +2x − 1=0.

D C D C

y P

A B A B

(b) ABCD is a square of unit side. Q is a point on BC so that the incircle of triangle ABQ and the circle tangent to AQ, QC, CD touch each other at a point on AQ. Show that the radii x and y of the circles satisfy the equations x(3 − 6x +2x2) √ √ y = , x + y =1. 1 − 2x2 Deduce that x is the root of 4x3 − 12x2 +8x − 1=0.

8 3 a and 4 a.

Chapter 3

The shoemaker’s knife

3.1 The shoemaker’s knife

Let P be a point on a segment AB. The region bounded by the three semicircles (on the same side of AB) with diameters AB, AP and PBis called a shoemaker’s knife. Suppose the smaller semicircles have radii a and b respectively. Let Q be the intersection of the largest semicircle with the perpendicular through P to AB. This perpendicular is an internal common tangent of the smaller semicircles.

H U R V K

A O1 O P O2 B A O1 O P O2 B

Exercise 1. Show that the area of the shoemaker’s knife is πab. 2. Let UV be the external common tangent of the smaller semicircles, and R the intersection of PQand UV. Show that (i) UV = PQ; (ii) UR = PR = VR = QR. Hence, with R as center, a circle can be drawn passing through P , Q, U, V . 24 The shoemaker’s knife

3. Show that the circle through U, P , Q, V has the same area as the shoemaker’s knife.

3.2 Circles in the shoemaker’s knife

Theorem 3.1 (Archimedes). The two circles each tangent to CP, the largest semicircle AB and one of the smaller semicircles have equal radii t, given by ab t = . a + b

A O1 O P O2 B A O1 O P O2 B

Proof. Consider the circle tangent to the semicircles O(a+ b), O1(a), and the line PQ. Denote by t the radius of this circle. Calculating in two ways the height of the center of this circle above the line AB,wehave

(a + b − t)2 − (a − b − t)2 =(a + t)2 − (a − t)2.

From this, ab t = . a + b The symmetry of this expression in a and b means that the circle tangent to O(a + b), O2(b), and PQhas the same radius t.

Theorem 3.2 (Archimedes). The circle tangent to each of the three semicircles has radius given by ab(a + b) ρ = . a2 + ab + b2 3.2 Circles in the shoemaker’s knife 25

A O1 O P O2 B

Proof. Let ∠COO2 = θ. By the cosine formula, we have

(a + ρ)2 =(a + b − ρ)2 + b2 +2b(a + b − ρ)cosθ, (b + ρ)2 =(a + b − ρ)2 + a2 − 2a(a + b − ρ)cosθ.

Eliminating θ,wehave

a(a + ρ)2 + b(b + ρ)2 =(a + b)(a + b − ρ)2 + ab2 + ba2.

The coefﬁcients of ρ2 on both sides are clearly the same. This is a linear equation in ρ: a3 + b3 +2(a2 + b2)ρ =(a + b)3 + ab(a + b) − 2(a + b)2ρ, from which

4(a2 + ab + b2)ρ =(a + b)3 + ab(a + b) − (a3 + b3)=4ab(a + b), and ρ is as above.

Theorem 3.3 (Leon Bankoff). If the incircle C(ρ) of the shoemaker’s knife touches the smaller semicircles at X and Y , then the circle through the points P , X, Y has the same radius as the Archimedean circles.

A O1 O P O2 B 26 The shoemaker’s knife

Proof. The circle through P , X, Y is clearly the incircle of the triangle CO1O2, since

CX = CY = ρ, O1X = O1P = a, O2Y = O2P = b.

The semiperimeter of the triangle CO1O2 is

ab(a + b) (a + b)3 a + b + ρ =(a + b)+ = . a2 + ab + b2 a2 + ab + b2

The inradius of the triangle is given by abρ ab · ab(a + b) ab = = . a + b + ρ (a + b)3 a + b

This is the same as t, the common radius of Archimedes’ twin circles.

Construction of incircle of shoemaker’s knife

Let Q1 and Q2 be the “highest” points of the semicircles O1(a) and O2(b) respectively. The intersection of O1Q2 and O2Q1 is a point C3 “above” P , and C P = ab = t 3 a+b . This gives a very easy construction of Bankoff’s circle in The- orem 3.3 above. From this, we obtain the points X and Y . The center of the incircle of the shoemaker’s knife is the intersection C of the lines O1X and O2Y . The incircle of the shoemaker’s knife is the circle C(X). It touches the largest semicircle of the shoemaker at Z, the intersection of OC with this semicircle.

C Q1

X Q2

C3 Y

A O1 O P O2 B

Note that C3(P ) is the Bankoff circle, which has the same radius as the Archimedean circles. 3.3 Archimedean circles in the shoemaker’s knife 27

Exercise CO O ab(a+b)2 1. Show that the area of triangle 1 2 is a2+ab+b2 . 2. Show that the center C of the incircle of the shoemaker’s knife is at a distance 2ρ from the line AB.

3. Show that the area of the shoemaker’s knife to that of the heart (bounded by semicircles O1(a), O2(b) and the lower semicircle O(a+ b))isasρ to a+ b.

Q1 C

X Y A O1 O P O2 B

A O1 O P O2 B

4. Show that the points of contact of the incircle C(ρ) with the semicircles can be located as follows: Y , Z are the intersections with Q1(A), and X, Z are the intersections with Q2(B).

3.3 Archimedean circles in the shoemaker’s knife

Let UV be the external common tangent of the semicircles O1(a) and O2(b), which extends to a chord HK of the semicircle O(a+b). Let C4 be the intersection of O1V and O2U. Since

O1U = a, O2V = b, and O1P : PO2 = a : b,

C P = ab = t C (t) P 4 a+b . This means that the circle 4 passes through and touches the common tangent HK of the semicircles at N. Let M be the midpoint of the chord HK. Since O and P are symmetric (isotomic conjugates) with respect to O1O2,

OM + PN = O1U + O2V = a + b. 28 The shoemaker’s knife

C5 H U M N V K

A O1 O P O2 B it follows that (a + b) − QM = PN =2t. From this, the circle tangent to HK and the minor arc HK of O(a + b) has radius t. This circle touches the minor arc at the point Q.

Theorem 3.4 (Thomas Schoch). The incircle of the curvilinear triangle bounded O(a + b) A(2a) B(2b) t = ab by the semicircle and the circles and has radius a+b .

A O1 O P O2 B

Proof. Denote this circle by S(x). Note that SO is a median of the triangle SO1O2. By Apollonius theorem,

(2a + x)2 +(2b + x)2 =2[(a + b)2 +(a + b − x)2].

From this, ab x = = t. a + b

Exercise 1. The circles (C1) and (C1) are each tangent to the outer semicircle of the shoemaker’s knife, and to OQ1 at Q1; similarly for the circles (C2) and (C ) t = ab 2 . Show that they have equal radii a+b . 3.3 Archimedean circles in the shoemaker’s knife 29

C2 Q1

C1 Q2 C2

A O1 O P O2 B

2. We call the semicircle with diameter O1O2 the midway semicircle of the shoemaker’s knife. Show that the circle tangent to the line PQand with center at the intersec- (O ) t = ab tion of 1 and the midway semicircle has radius a+b . Q

C C

A O1 O P O2 B 3. Show that the radius of the circle tangent to the midway semicircle, the PQ t = ab outer semicircle, and with center on the line has radius a+b . Q

A O1 O P O2 B

Chapter 4

Introduction to Triangle Geometry

4.1 Preliminaries

4.1.1 Coordinatization of points on a line Let B and C be two ﬁxed points on a line L. Every point X on L can be coordinatized in one of several ways: t = BX (1) the ratio of division XC , (2) the absolute barycentric coordinates: an expression of X as a convex combination of B and C: X =(1− t)B + tC, which expresses for an arbitrary point P outside the line L, the vector PX as a combination of the vectors PB and PC. (3) the homogeneous barycentric coordinates: the proportion XC : BX, which are masses at B and C so that the resulting system (of two particles) has balance point at X.

B X C 32 Introduction to Triangle Geometry

4.1.2 Centers of similitude of two circles

Consider two circles O(R) and I(r), whose centers O and I are at a distance d apart. Animate a point X on O(R) and construct a ray through I oppositely parallel to the ray OX to intersect the circle I(r) at a point Y . You will ﬁnd that the line XY always intersects the line OI at the same point T . This we call the internal center of similitude, or simply the insimilicenter, of the two circles. It divides the segment OI in the ratio OT : TI = R : r. The absolute barycentric coordinates of P with respect to OI are

R · I + r · O T = . R + r

T T

O I

Y X

If, on the other hand, we construct a ray through I directly parallel to the ray OX to intersect the circle I(r) at Y , the line XY always intersects OI at another point T . This is the external center of similitude, or simply the exsimilicenter,of the two circles. It divides the segment OI in the ratio OT : T I = R : −r, and has absolute barycentric coordinates

R · I − r · O T = . R − r

4.1.3 Tangent circles

If two circles are tangent to each other, the line joining their centers passes through the point of tangency, which is a center of similitude of the circles. 4.1 Preliminaries 33

T O T I O I

4.1.4 Harmonic division Two points X and Y are said to divide two other points B and C harmonically if BX BY = − . XC YC

They are harmonic conjugates of each other with respect to the segment BC.

Examples

1. For two given circles, the two centers of similitude divide the centers harmonically.

2. Given triangle ABC, let the internal bisector of angle A intersect BC at X. The harmonic conjugate of X in BC is the intersection of BC with the external bisector of angle A.

3. Let A and B be distinct points. If M is the midpoint of the segment AB, it is not possible to find a finite point N on the line AB so that M, N A B AN = − AM = −1 divide , harmonically. This is because NB MB , requiring AN = −NB = BN, and AB = BN − AN =0, a contradiction. We shall agree to say that if M and N divide A, B harmonically, then N is the infinite point of the line AB.

Exercise

1. If X, Y divide B, C harmonically, then B, C divide X, Y harmonically. 34 Introduction to Triangle Geometry

B X C X 2. Given a point X on the line BC, construct its harmonic associate with respect to the segment BC. Distinguish between two cases when X divides BC internally and externally. 1 3. The centers A and B of two circles A(a) and B(b) are at a distance d apart. The line AB intersect the circles at A and B respectively, so that A, B are between A, B.

A B A B

4. Given two ﬁxed points B and C and a positive constant k =1, the locus of the points P for which |BP| : |CP| = k is a circle.

1Make use of the notion of centers of similitude of two circles. 4.1 Preliminaries 35

4.1.5 Homothety Given a point T and a nonzero constant k, the similarity transformation h(T,k) which carries a point X to the point X on the line TX satisfying TX : TX = k :1is called the homothety with center T and ratio k. Explicitly, h(T,k)(P )=(1− k)T + kP.

Any two circles are homothetic. Let P and Q be the internal and external O(R) I(r) h(Q, r ) centers of similitude of two circles and . Both the homotheties R h(P, − r ) O(R) I(r) and R transform the circle into .

4.1.6 The power of a point with respect to a circle The power of a point P with respect to a circle C = O(R) is the quantity C(P ):=OP 2 − R2.

This is positive, zero, or negative according as P is outside, on, or inside the circle C. If it is positive, it is the square of the length of a tangent from P to the circle.

Theorem (Intersecting chords) If a line L through P intersects a circle C at two points X and Y , the product PX· PY (of signed lengths) is equal to the power of P with respect to the circle.

X Y P

T 36 Introduction to Triangle Geometry

4.2 Menelaus and Ceva theorems

4.2.1 Menelaus and Ceva Theorems Consider a triangle ABC with points X, Y , Z on the side lines BC, CA, AB respectively. Theorem 4.1 (Menelaus). The points X, Y , Z are collinear if and only if BX CY AZ · · = −1. XC YA ZB

X B C

Theorem 4.2 (Ceva). The lines AX, BY , CZ are concurrent if and only if BX CY AZ · · =+1. XC YA ZB

4.2.2 Desargues Theorem As a simple illustration of the use of the Menelaus and Ceva theorems, we prove the following Desargues Theorem: Given three circles, the exsimilicenters of the three pairs of circles are collinear. Likewise, the three lines each joining the insimilicenter of a pair of circles to the center of the remaining circle are concurrent. We prove the second statement only. Given three circles A(r1), B(r2) and C(r3), the insimilicenters X of (B) and (C), Y of (C), (A), and Z of (A), (B) are the points which divide BC, CA, AB in the ratios BX r CY r AZ r = 2 , = 3 , = 1 . XC r3 YA r1 ZB r2 4.2 Menelaus and Ceva theorems 37

Y P

B X C

X A

Y C P Y

Z X

It is clear that the product of these three ratio is +1, and it follows from the Ceva theorem that AX, BY , CZ are concurrent. 38 Introduction to Triangle Geometry

4.2.3 The incircle and the Gergonne point The incircle is tangent to each of the three sides BC, CA, AB (without extension). Its center, the incenter I, is the intersection of the bisectors of the three angles. The inradius r is related to the area ∆ by S =(a + b + c)r.

Z G e I

B X C

If the incircle is tangent to the sides BC at X, CA at Y , and AB at Z, then b + c − a c + a − b a + b − c AY = AZ = ,BZ= BX = ,CX= CY = . 2 2 2 These expressions are usually simpliﬁed by introducing the semiperimeter s = 1 (a + b + c) 2 : AY = AZ = s − a, BZ = BX = s − b, CX = CY = s − c.

r = ∆ Also, s . It follows easily from the Ceva theorem that AX, BY , CZ are concurrent. The point of concurrency Ge is called the Gergonne point of triangle ABC. Triangle XY Z is called the intouch triangle of ABC. Clearly, B + C C + A A + B X = ,Y= ,Z= . 2 2 2 4.2 Menelaus and Ceva theorems 39

It is always acute angled, and A B C YZ =2r cos ,ZX=2r cos ,XY=2r cos . 2 2 2

Exercise

1. Given three points A, B, C not on the same line, construct three circles, with centers at A, B, C, mutually tangent to each other externally.

Y X

C Z A

2. Construct the three circles each passing through the Gergonne point and tangent to two sides of triangle ABC. The 6 points of tangency lie on a circle. 2

3. Two circles are orthogonal to each other if their tangents at an intersection are perpendicular to each other. Given three points A, B, C not on a line, construct three circles with these as centers and orthogonal to each other. (1) Construct the tangents from A to the circle B(b), and the circle tangent to these two lines and to A(a) internally. (2) Construct the tangents from B to the circle A(a), and the circle tangent to these two lines and to B(b) internally. (3) The two circles in (1) and (2) are congruent. √ (4R+r)2+s2 2 This is called the Adams circle. It is concentric with the incircle, and has radius 4R+r · r. 40 Introduction to Triangle Geometry

I Ge

B C

4. Given a point Z on a line segment AB, construct a right-angled triangle ABC whose incircle touches the hypotenuse AB at Z. 3

5. Let ABC be a triangle with incenter I. (1a) Construct a tangent to the incircle at the point diametrically opposite to its point of contact with the side BC. Let this tangent intersect CA at Y1 and AB at Z1. (1b) Same in part (a), for the side CA, and let the tangent intersect AB at Z2 and BC at X2. (1c) Same in part (a), for the side AB, and let the tangent intersect BC at X3 and CA at Y3.

(2) Note that AY3 = AZ2. Construct the circle tangent to AC and AB at Y3 and Z2. How does this circle intersect the circumcircle of triangle ABC?

6. The incircle of ABC touches the sides BC, CA, AB at D, E, F respectively. X is a point inside ABC such that the incircle of XBC touches BC at D also, and touches CX and XB at Y and Z respectively.

3P. Yiu, G. Leversha, and T. Seimiya, Problem 2415 and solution, Crux Math. 25 (1999) 110; 26 (2000) 62 – 64. 4.2 Menelaus and Ceva theorems 41

(1) The four points E, F , Z, Y are concyclic. 4 (2) What is the locus of the center of the circle EFZY ? 5

7. Given triangle ABC, construct a circle tangent to AC at Y and AB at Z such that the line YZpasses through the centroid G. Show that YG: GZ = c : b.

Y G

B C

4.2.4 The excircles and the Nagel point Let X , Y , Z be the points of tangency of the excircles (Ia), (Ib), (Ic) with the corresponding sides of triangle ABC. The lines AX , BY , CZ are concurrent. The common point Na is called the Nagel point of triangle ABC.

Exercise 1. Construct the tritangent circles of a triangle ABC. (1) Join each excenter to the midpoint of the corresponding side of ABC. These three lines intersect at a point Mi. (This is called the Mittenpunkt of the triangle).

4International Mathematical Olympiad 1996. 5IMO 1996. 42 Introduction to Triangle Geometry

Z Y

Z Na Y I

B X X

(2) Join each excenter to the point of tangency of the incircle with the corresponding side. These three lines are concurrent at another point T .

(3) The lines AMi and AT are symmetric with respect to the bisector of angle A; so are the lines BMi, BT and CMi, CT (with respect to the bisectors of angles B and C).

2. Construct the excircles of a triangle ABC. (1) Let D, E, F be the midpoints of the sides BC, CA, AB. Construct the 6 incenter Sp of triangle DEF, and the tangents from S to each of the three excircles. (2) The 6 points of tangency are on a circle, which is orthogonal to each of the excircles. 6This is called the Spieker point of triangle ABC. 4.3 The nine-point circle 43

Z Y

Z Mi Y I T

B X X

3. Let D, E, F be the midpoints of the sides BC, CA, AB, and let the incircle touch these sides at X, Y , Z respectively. The lines through X parallel to ID, through Y to IE and through Z to IF are concurrent. 7

4.3 The nine-point circle

4.3.1 The Euler triangle as a midway triangle Let P be a point in the plane of triangle ABC. The midpoints of the segments AP , BP, CP form the midway triangle of P . It is the image of ABC under the (P, 1 ) H homothety h 2 . The midway triangle of the orthocenter is called the Euler

7 Crux Math. Problem 2250. The reﬂection of the Nagel point N a in the incenter. This is X145 of ETC. 44 Introduction to Triangle Geometry

Y Sp

B X

Ia triangle. The circumcenter of the midway triangle of P is the midpoint of OP.In particular, the circumcenter of the Euler triangle is the midpoint of OH, which is the same as N, the circumcenter of the medial triangle. (See Exercise ??.). The medial triangle and the Euler triangle have the same circumcircle.

4.3.2 The orthic triangle as a pedal triangle The pedals of a point are the intersections of the sidelines with the corresponding perpendiculars through P . They form the pedal triangle of P . The pedal triangle of the orthocenter H is called the orthic triangle of ABC. The pedal X of the orthocenter H on the side BC is also the pedal of A on the same line, and can be regarded as the reﬂection of A in the line EF. It follows that ∠EXF = ∠EAF = ∠EDF, 4.3 The nine-point circle 45

F E

Z I

B X D C

B C

since AEDF is a parallelogram. From this, the point X lies on the circumcircle of the medial triangle DEF; similarly for the pedals Y and Z of H on the other two sides CA and AB. 46 Introduction to Triangle Geometry

Z P

B X C

4.3.3 The nine-point circle From §§4.3.1, 4.3.2 above, the medial triangle, the Euler triangle, and the orthic triangle all have the same circumcircle. This is called the nine-point circle of triangle ABC. Its center N, the midpoint of OH, is called the nine-point center of triangle ABC.

Exercise 1. Show that (i) the incenter is the orthocenter of the excentral triangle (ii) the circumcircle is the nine-point circle of the excentral triangle, (iii) the circumcenter of the excentral triangle is the reﬂection of I in O.

2. Let P be a point on the circumcircle. What is the locus of the midpoint of HP? Why?

3. If the midpoints of AP , BP, CP are all on the nine-point circle, must P be the orthocenter of triangle ABC? 8

8P. Yiu and J. Young, Problem 2437 and solution, Crux Math. 25 (1999) 173; 26 (2000) 192. 4.3 The nine-point circle 47

A Y

F E

O N

Z H

C B

B X D C

4. Let ABC be a triangle and P a point. The perpendiculars at P to PA, PB, PC intersect BC, CA, AB respectively at A, B, C. (1) A, B, C are collinear. 9 (2) The nine-point circles of the (right-angled) triangles PAA, PBB, PCC are concurrent at P and another point P . Equivalently, their centers are collinear. 10

5. (Triangles with nine-point center on the circumcircle) Begin with a circle, center O and a point N on it, and construct a family of triangles with (O) as circumcircle and N as nine-point center. (1) Construct the nine-point circle, which has center N, and passes through

9B. Gibert, Hyacinthos 1158, 8/5/00. 10A.P. Hatzipolakis, Hyacinthos 3166, 6/27/01. The three midpoints of AA , BB, CC are collinear. The three nine-point circles intersect at P and its reﬂection in this line. 48 Introduction to Triangle Geometry

0 I

C B

C A B

B 4.4 The OI-line 49

the midpoint M of ON. (2) Animate a point D on the minor arc of the nine-point circle inside the circumcircle. (3) Construct the chord BC of the circumcircle with D as midpoint. (This is simply the perpendicular to OD at D). (4) Let X be the point on the nine-point circle antipodal to D. Complete the parallelogram ODXA (by translating the vector DO to X). The point A lies on the circumcircle and the triangle ABC has nine-point center N on the circumcircle. Here is a curious property of triangles constructed in this way: let A, B, C be the reﬂections of A, B, C in their own opposite sides. The reﬂection triangle ABC degenerates, i.e., the three points A, B, C are collinear. 11

4.4 The OI-line

4.4.1 The homothetic center of the intouch and excentral triangles The OI-line of a triangle is the line joining the circumcenter and the incenter. We consider several interesting triangle centers on this line, which arise from the homothety of the intouch and excentral triangles. These triangles are homothetic since their corresponding sides are parallel, being perpendicular to the same angle bisector of the reference triangle. 2R I =2O − I The ratio of homothetic is clearly r . Their circumcenters are and I. The homothetic center is the point T such that 2R 2R 2R 2O − I = h T, (I)= 1 − T + · I. r r r

This gives (2R + r)I − 2r · O T = . 2R − r It is the point which divides OI externally in the ratio 2R + r : −2r. 12

11O. Bottema, Hoofdstukken uit de Elementaire Meetkunde, Chapter 16. 12 The point T is X57 in ETC. 50 Introduction to Triangle Geometry

Z T I

C B X

Ia 4.4.2 The centers of similitude of the circumcircle and the incircle

These are the points T+ and T− which divide the segment OI harmonically in the ratio of the circumradius and the inradius. 13

1 T = (r · O + R · I), + R + r 1 T = (−r · O + R · I). − R − r

13 T+ and T− are respectively X55 and X56 in ETC. 4.4 The OI-line 51

Z O I T − T+

B X C

M 4.4.3 Reﬂection of I in O The reﬂection I of I in O is the circumcenter of the excentral triangle. It is the intersections of the perpendiculars from the excenters to the sidelines. 14 The midpoint M of the arc BC is also the midpoint of IIa. Since IM and I Ia are parallel, I Ia =2R. Similarly, I Ib = Pc =2R. This shows that the excentral triangle has circumcenter I =2O − I and circumradius 2R. Since I is the orthocenter of IaIbIc, its follows that O, the midpoint of I and I , is the nine-point center of the excentral triangle. In other words, the circumcircle of triangle ABC is the nine-point circle of the excentral triangle. Apart from A, the second intersection of IbIc with the circumcircle of ABC is the midpoint M of 15 IbIc. From this we deduce the following interesting formula:

ra + rb + rc =4R + r.

14 I appears as X40 in ETC. 15 Proof. ra +rb +rc =2MD+(I Ia −I X )=2R+2OD+I Ia −I X =4R+IX =4R+r. 52 Introduction to Triangle Geometry

O I

C B D Xc Xb

4.4.4 Orthocenter of intouch triangle

The OI-line is the Euler line of the excentral triangle, since O and I are the nine-point center and orthocenter respectively. The corresponding sides of the intouch triangle and the excentral triangle are parallel, being perpendicular to the respective angle bisectors. Their Euler lines are parallel. Since the intouch triangle has circumcenter I, its Euler line is actually the line OI, which therefore contains its orthocenter. This is the point which divides OI in the ratio R + r : −r. 16 It also lies on the line joining the Gergonne and Nagel points.

16 This is the point X65 in ETC. 4.4 The OI-line 53

Z H I O Ge Na

B X C

4.4.5 Centroids of the excentral and intouch triangles

The centroid of the excentral triangle is the point which divides OI in the ratio −1:4 17 h(T, r ) . From the homothety 2R , it is easy to see that the centroid of the intouch triangle is the point which divides OI in the ratio 3R + r : −r. 18

Exercises

1. Can any of the centers of similitude of (O) and (I) lie outside triangle ABC?

17 This is the point which divides the segment I I in to the ratio 1:2.ItisX165 of ETC. 18 The centroid of the intouch triangle is X354 of ETC. 54 Introduction to Triangle Geometry

O Z T I

C B X

2. Show that the distance between the circumcenter and the Nagel point is R − 2r. 19

3. Identify the midpoint of the Nagel point and the deLongchamps point as a point on the OI-line. 20

4. Construct the orthic triangle of the intouch triangle. This is homothetic to

19Feuerbach’s theorem. 20 Reﬂection of I in O. This is because the pedal triangle of X 40 is the cevian triangle of the Nagel point, and the reﬂections of the pedals of the Nagel point in the respective traces form the pedals of the de Longchamps point. 4.4 The OI-line 55

ABC. Identify the homothetic center. 21

Z O H I

B X C

5. Construct the external common tangent of each pair of excircles. These three external common tangents bound a triangle. (i) This triangle is perpsective with ABC. Identify the perspector. 22 (ii) Identify the incenter of this triangle. 23

6. Extend AB and AC by length a and join the two points by a line. Similarly deﬁne the two other lines. The three lines bound a triangle with perspector 24 X65

7. Let H be the orthocenter of the intouch triangle XY Z, and X , Y , Z its pedals on the sides YZ, ZX, XY respectively. Identify the common point of the three lines AX , BY , CZ as a point on the OI-line. Homothetic center of intouch and excentral triangles.

21T . 22Orthocenter of intouch triangle. 23Reﬂection of I in O. 242/18/03. 56 Introduction to Triangle Geometry

8. Let P be the point which divides OI in the ratio OP : PI = R :2r. There P Rr is a circle, center , radius R+2r , which is tangent to three congruent circles of the same radius, each tangent to two sides of the triangle. Construct these circles. 25

9. There are three circles each tangent internally to the circumcircle at a vertex, and externally to the incircle. It is known that the three lines joining the points of tangency of each circle with (O) and (I) pass through the internal 26 center T+ of similitude of (O) and (I). Construct these three circles.

T+ O I

B C

10. Let T+ be the insimilicenter of (O) and (I), with pedals Y and Z on CA and AB respectively. If Y and Z are the pedals of Y and Z on BC, calculate the length of Y Z. 27

11. Let P be the centroid of the excentral triangle, with pedals X, Y , Z on the

25A. P. Hatzipolakis, Hyacinthos message 793, April 18, 2000. 26A.P. Hatzipolakis and P. Yiu, Triads of circles, preprint. 27A.P. Hatzipolakis and P. Yiu, Pedal triangles and their shadows, Forum Geom., 1 (2001) 81 – 90. 4.4 The OI-line 57

Z T+ O I

B Z X Y C

sides BC, CA, AB respectively. Show that 28 1 AY + AZ = BZ + BX = CX + CY = (a + b + c). 3

4.4.6 Mixtilinear incircles A mixtilinear incircle of triangle ABC is one that is tangent to two sides of the triangle and to the circumcircle internally. Denote by A the point of tangency of the mixtilinear incircle K(ρ) in angle A with the circumcircle. The center K clearly lies on the bisector of angle A, and AK : KI = ρ : −(ρ − r). In terms of barycentric coordinates, 1 K = (−(ρ − r)A + ρI) . r 28The projections of O and I on the side BC are the midpoint D of BC, and the point of a 1 tangency D of the incircle with this side. Clearly, BD = 2 and BD = 2 (c + a − b). It follows that 4 4 1 1 BX = BD + DD = BD − BD = (3a + b − c). 3 3 3 6 1 1 Similarly, BZ = 6 (3c + b − a), and BX + BZ = 3 (a + b + c). A similar calculation shows that 1 AY + AZ = CX + CY = 3 (a + b + c). 58 Introduction to Triangle Geometry

Also, since the circumcircle O(A) and the mixtilinear incircle K(A) touch each other at A,wehaveOK : KA = R − ρ : ρ, where R is the circumradius. From this, 1 K = (ρO +(R − ρ)A) . R

B C

Comparing these two equations, we obtain, by rearranging terms, RI − rO R(ρ − r)A + r(R − ρ)A = . R − r ρ(R − r)

We note some interesting consequences of this formula. First of all, it gives the intersection of the lines joining AA and OI. Note that the point on the line OI represented by the left hand side is T−, the external center of similitude of the circumcircle and the incircle. This leads to a simple construction of the mixtilinear incircle. Given a triangle ABC, let P be the external center of similitude of the circumcircle (O) and incircle (I). Extend AP to intersect the circumcircle at A. The intersection of AI and A O is the center KA of the mixtilinear incircle in angle A. The other two mixtilinear incircles can be constructed similarly.

4.5 Euler’s formula and Steiner’s porism 4.5 Euler’s formula and Steiner’s porism 59

O I

T−

B C

A M 4.5.1 Euler’s formula The distance between the circumcenter and the incenter of a triangle is given by OI2 = R2 − 2Rr.

Let the bisector of angle A intersect the circumcircle at M. Construct the circle M(B) to intersect this bisector at a point I. This is the incenter since 1 1 1 ∠IBC = ∠IMC = ∠AMC = ∠ABC, 2 2 2 ∠ICB = 1 ∠ACB and for the same reason 2 . Note that IM = MB = MC =2R sin A (1) 2 , IA = r (2) sin A , and 2 (3) by the theorem of intersecting chords, OI2 −R2 = the power of I with respect to the circumcircle = IA · IM = −2Rr.

4.5.2 Steiner’s porism Construct the circumcircle (O) and the incircle (I) of triangle ABC. Animate a point A on the circumcircle, and construct the tangents from A to the incircle (I). Extend these tangents to intersect the circumcircle again at B and C. The lines BC is always tangent to the incircle. This is the famous theorem on Steiner 60 Introduction to Triangle Geometry

O I

B C

M porism: if two given circles are the circumcircle and incircle of one triangle, then they are the circumcircle and incircle of a continuous family of poristic triangles.

Exercises r ≤ 1 R 1. 2 . When does equality hold? 2. Suppose OI = d. Show that there is a right-angled triangle whose sides are d, r and R − r. Which one of these is the hypotenuse?

3. Given a point I inside a circle O(R), construct a circle I(r) so that O(R) and I(r) are the circumcircle and incircle of a (family of poristic) triangle(s).

4. Given the circumcenter, incenter, and one vertex of a triangle, construct the triangle.

5. Construct an animation picture of a triangle whose circumcenter lies on the incircle. 29 29Hint: OI = r. 4.5 Euler’s formula and Steiner’s porism 61

Y C

Z I O

B X C

6. What is the locus of the centroids of the poristic triangles with the same circumcircle and incircle of triangle ABC? How about the orthocenter?

7. Let ABC be a poristic triangle with the same circumcircle and incircle of triangle ABC, and let the sides of BC, CA, AB touch the incircle at X, Y , Z. (i) What is the locus of the centroid of XY Z? (ii) What is the locus of the orthocenter of XY Z? (iii) What can you say about the Euler line of the triangle XY Z?

Chapter 5

Homogeneous Barycentric Coordinates

5.1 Barycentric coordinates with reference to a triangle

The notion of barycentric coordinates dates back to M¨obius. In a given triangle ABC, every point P is coordinatized by a triple of numbers (x : y : z) in such a way that the system of masses x at A, y at B, and z at C will have its balance P y B z C X = yB+zC point at . A mass at and a mass at will balance at the point y+z on the line BC. A mass x at A and a mass y + z at X will balance at the point

xA +(y + z)X xA + yB + zC P = = . x +(y + z) x + y + z

These are the absolute homogeneous barycentric coordinates of P with reference to triangle ABC. Because of the fundamental importance of the Ceva theorem in triangle geometry, we shall follow traditions and call the three lines joining a point P to the vertices of the reference triangle ABC the cevians of P . The intersections X, Y , Z of these cevians with the side lines are called the traces of P . The coordinates of the traces can be very easily written down:

X =(0:y : z),Y=(x :0:z),Z=(x : y :0). 64 Homogeneous Barycentric Coordinates

Z X P

C Y A Figure 5.1: Traces of a point

5.1.1 Ceva Theorem Three points X, Y , Z on BC, CA, AB respectively are the traces of a point if and only if they have coordinates of the form X =0:y : z, Y = x :0:z, Z = x : y :0, for some x, y, z.

5.1.2 Some basic triangle centers The centroid The midpoint points of the sides have coordinates

X =(0:1:1), Y =(1:0:1), Z =(1:1:0).

The centroid G has coordinates (1:1:1). 1

The incenter The traces of the incenter have coordinates

X =(0:b : c),

1The centroid appears in Kimberling’s Encyclopedia of Triangle centers (ETC, available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html) as the point X2. 5.1 Barycentric coordinates with reference to a triangle 65

B B

Z X Z X G I

C Y A C Y A Figure 5.2: (a) Centroid (b) Incenter

Y =(a :0:c), Z =(a : b :0).

The incenter I has coordinates I =(a : b : c). 2

The Gergonne point

The points of tangency of the incircle with the side lines are

X =0:s − c : s − b, Y = s − c :0:s − a, Z = s − b : s − a :0.

These can be reorganized as

X =0:1 : 1 , s−b s−c Y = 1 :0:1 , s−a s−c Z = 1 : 1 :0. s−a s−b

It follows that AX, BY , CZ intersect at a point with coordinates 1 1 1 : : . s − a s − b s − c

3 This is called the Gergonne point Ge of triangle ABC.

2 The incenter appears in ETC as the point X1. 3 The Gergonne point appears in ETC as the point X 7. 66 Homogeneous Barycentric Coordinates

X I Ge

C Y A Figure 5.3: The Gergonne point

The Nagel point

The points of tangency of the excircles with the corresponding sides have coordinates

X =(0:s − b : s − c), Y =(s − a :0:s − c), Z =(s − a : s − b :0).

These are the traces of the point with coordinates

(s − a : s − b : s − c).

4 This is the Nagel point Na of triangle ABC.

5.1.3 The orthocenter The trace X of the orthocenter H divides BC in the ratio c2 + a2 − b2 a2 + b2 − c2 BX : XC = c cos B : b cos C = : = c2+a2−b2 : a2+b2−c2. 2a 2a

It has homogeneous barycentric coordinates 1 1 0: : . c2 + a2 − b2 a2 + b2 − c2

4 The Nagel point appears in ETC as the point X8. 5.1 Barycentric coordinates with reference to a triangle 67

Z Na

C Y A

Figure 5.4: The Nagel point

Similarly, the other two traces are 1 1 Y = :0: , b2 + c2 − a2 a2 + b2 − c2 1 1 Z = : :0 . b2 + c2 − a2 c2 + a2 − b2

The orthocenter is the point 5 1 1 1 H = : : . b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2

For convenience of computation, we often write b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 S = ,S= ,S= . A 2 B 2 C 2 and shorten products like SASB simply as SAB. In terms of these, 1 1 1 H = : : =(SBC : SCA : SAB). SA SB SC

5 The orthocenter appears in ETC as the point X 4. 68 Homogeneous Barycentric Coordinates

H X

C Y A Figure 5.5: The orthocenter point

5.2 The inferior and superior triangles

The triangle whose vertices are the midpoints of the sides of triangle ABC is called the inferior triangle of ABC. Its vertices are the points

A =(0:1:1),B =(1:0:1),C =(1:1:0).

ABC G − 1 This triangle is homothetic to at the centroid , with ratio 2 . Equivalently, ABC h(G, − 1 ) we say that it is the image of under the homothety 2 . This means h(G, − 1 ) that 2 is a one-to-one correspondence of points on the plane such that P and P have the same homogeneous barycentric coordinates with reference to ABC and ABC whenever X, X and G are collinear and

GP 1 = − . GP 2

We call P the inferior of P . More explicitly, P divides PG in the ratio PP : P G =3:−1 P = 1 (3G−P ) P , so that 2 . Suppose has homogeneous barycentric (x : y : z) ABC P = xA+yBzC 6 coordinates with reference to . It is the point x+y+z .

6The conversion from homogeneous barycentric coordinates to absolute barycentric coordinates is called normalization. 5.2 The inferior and superior triangles 69

Thus, 1 P = (3G − P ) 2 1 xA + yBzC = A + B + C − 2 x + y + z 1 (x + y + z)(A + B + C) − (xA + yB + zC) = 2 x + y + z (y + z)A +(z + x)B +(x + y)C = . 2(x + y + z)

It follows that the homogenous coordinates of P are (y + z : z + x : x + y). The superior triangle of ABC is its image under the homothety h(G, −2). Its vertices are the points

A =(−1:1:1),B =(1:−1:1),C =(1:1:−1).

5.2.1 More triangle centers

The circumcenter

The circumcenter is the orthocenter of the inferior triangle. It is therefore the inferior of the orthocenter, and has coordinates 7

(SCA + SAB : SAB + SBC : SCA + SAB)

=(SA(SB + SC ):SB(SC + SA):SC (SA + SB)) 2 2 2 =(a SA : b SB : c SC ).

The nine-point center

The nine-point circle is the circumcircle of the inferior triangle. Its center N is the N = 3G−O O = 1 (3G − H) inferior of the circumcenter. From 2 and 2 , we obtain N = 1 (O + H) N OH 8 2 . The nine-point center is the midpoint of .

7 The circumcenter appears in ETC as the point X3. 8 The nine-point center appears in ETC as the point X 5. 70 Homogeneous Barycentric Coordinates

Exercise

1. The Nagel point Na lies on the line joining the incenter to the centroid; it divides IG in the ratio INa : NaG =3:−2.

2. Find the coordinates of the circumcenter O by using the fact that OG : GH =1:2.

P P GP GP = −2 3. The superior of is the point on the line for which GP . Show that if P =(x : y : z), then P =(−x+y +z : x−y +z : x+y −z).

4. Find the coordinates of the superior of the incenter.

5. Find the coordinates of the inferior of the incenter. Show that it is the centroid of the perimeter of triangle ABC.

6. Calculate the coordinates of the nine-point center in terms of SA, SB, SC .

7. The midway triangle of a point P is the triangle whose vertices are the midpoints of AP , BP, CP.IfP =(x : y : z), ﬁnd the coordinates of the centroid and the incenter of the midway triangle with reference to ABC.

5.3 Isotomic conjugates

The Gergonne and Nagel points are examples of isotomic conjugates. Two points P and Q (not on any of the side lines of the reference triangle) are said to be isotomic conjugates if their respective traces are symmetric with respect to the midpoints of the corresponding sides. Thus,

BX = XC, CY = Y A, AZ = ZB.

We shall denote the isotomic conjugate of P by P •.IfP =(x : y : z), then

1 1 1 P • =( : : ) x y z

. 5.3 Isotomic conjugates 71

X Z

X Z • P P

C Y Y A Figure 5.6: Isotomic conjugates

5.3.1 Congruent-parallelians point Given triangle ABC, we want to construct a point P the three lines through which parallel to the sides cut out equal intercepts. Let P = xA + yB + zC in absolute barycentric coordinates. The parallel to BC cuts out an intercept of length (1 − x)a. It follows that the three intercepts parallel to the sides are equal if and only if 1 1 1 1 − x :1− y :1− z = : : . a b c

The right hand side clearly gives the homogeneous barycentric coordinates of I •, the isotomic conjugate of the incenter I. 9 This is a point we can easily construct. Now, translating into absolute barycentric coordinates: 1 1 I• = [(1 − x)A +(1− y)B +(1− z)C]= (3G − P ). 2 2 we obtain P =3G − 2I •, and can be easily constructed as the point dividing the segment I •G externally in the ratio I •P : PG =3:−2. The point P is called the congruent-parallelians point of triangle ABC. 10

Exercises 1. Calculate the homogeneous barycentric coordinates of the congruent-parallelian point and the length of the equal parallelians. 11

9 The isotomic conjugate of the incenter appears in ETC as the point X 75. 10 This point appears in ETC as the point X192. 11(ca + ab − bc : ab + bc − ca : bc + ca − ab). The common length of the equal parallelians is 2abc ab+bc+ca . 72 Homogeneous Barycentric Coordinates

P G

I•

C A Figure 5.7: Congruent parallelians point

2. Let ABC be the midway triangle of a point P . The line BC intersects CA at Ba = B C ∩ CA, Ca = B C ∩ AB, Cb = C A ∩ AB, Ab = C A ∩ BC, Ac = A B ∩ BC, Bc = A B ∩ CA.

Determine P for which the three segments BaCa, CbAb and AcBc have equal lengths.

5.4 Internal center of similitudes of the circumcircle and the incircle

These points, T and T , divide the segment OI harmonically in the ratio of the R = abc S R : r = abc : S = sabc : S2 circumradius 2S and the inradius 2s . Note that 2S 2s . Since O =(a2(b2 + c2 − a2):···: ···) with coordinates sum 4S2 and I =(a : b : c) with coordinates sum 2s,we equalize their sums and work with

O =(sa2(b2 + c2 − a2):···: ···), I =(2S2a :2S2b :2S2c).

The internal center of similitude T divides OI in the ratio OT : TI = R : r, the a-component of its homogeneous barycentric coordinates can be taken as

S2 · sa2(b2 + c2 − a2)+sabc · 2S2a. 5.4 Internal center of similitudes of the circumcircle and the incircle 73

The simpliﬁcation turns out to be easier than we would normally expect: S2 · sa2(b2 + c2 − a2)+sabc · 2S2a = sS2a2(b2 + c2 − a2 +2bc) = sS2a2((b + c)2 − a2) = sS2a2(b + c + a)(b + c − a) =2s2S2 · a2(b + c − a).

The other two components have similar expressions obtained by cyclically per- muting a, b, c. It is clear that 2s2S2 is a factor common to the three components. Thus, the homogeneous barycentric coordinates of the internal center of similitude are 12 (a2(b + c − a):b2(c + a − b):c2(a + b − c)).

Exercises 1. The external center of similitude of (O) and (I) has homogeneous barycentric coordinates 13 (a2(a+b−c)(c+a−b):b2(b+c−a)(a+b−c):c2(c+a−b)(b+c−a)),

which can be taken as a2 b2 c2 : : . b + c − a c + a − b a + b − c

2. The orthocenter H lies on the Euler line and divides the segment OG externally in the ratio OH : HG =3:−2. 14 Show that its homogeneous barycentric coordinates can be written as H = (tan A :tanB :tanC),

or equivalently, 1 1 1 H = : : . b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2

12In ETC, the internal center of similitude of the circumcircle and the incircle appears as the point X55. 13In ETC, the external center of similitude of the circumcircle and the incircle appears as the point X56. 14 In ETC, the orthocenter appears as the point X 4. 74 Homogeneous Barycentric Coordinates

3. Make use of the fact that the nine-point center N divides the segment OG in the ratio ON : GN =3:−1 to show that its barycentric coordinates can be written as 15 N =(a cos(B − C):b cos(C − A):c cos(A − B)).

15 In ETC, the nine-point center appears as the point X 5.