Chapter 3

Introduction to Triangle Geometry

3.1 Preliminaries

3.1.1 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 coordi- natized in one of several ways: BX (1) the ratio of division t = XC , (2) the absolute barycentric coordinates: an expression of X as a convex com- bination 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.

P

B X C 26 Introduction to Triangle Geometry

3.1.2 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 : T I = R : r. The absolute barycentric coordinates of P with respect to OI are

R I + r O T = · · . R + r

Y 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 − 3.1.3 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. 3.1 Preliminaries 27

T ′ O T I O I

3.1.4 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 har- monically.

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 AN AM divide A, B harmonically. This is because NB = MB = 1, 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. 28 Introduction to Triangle Geometry

A

B X C X′ 2. Given a point X on the line BC, construct its harmonic associate with re- spect 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. 6 | | | | 1Make use of the notion of centers of similitude of two circles. 3.1 Preliminaries 29

3.1.5 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 :1 is 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 h r centers of similitude of two circles O(R) and I(r). Both the homotheties (Q, R ) and h(P, r ) transform the circle O(R) into I(r). − R 3.1.6 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. ·

T

O

X Y P

T ′ 30 Introduction to Triangle Geometry

3.2 Menelaus and Ceva theorems

3.2.1 Menelaus and Ceva Theorems Consider a triangle ABC with points X, Y , Z on the side lines BC, CA, AB respectively. Theorem 3.1 (Menelaus). The points X, Y , Z are collinear if and only if BX CY AZ = 1. XC · Y A · ZB −

A

Y

Z

X B C

Theorem 3.2 (Ceva). The lines AX, BY , CZ are concurrent if and only if BX CY AZ = +1. XC · Y A · ZB

3.2.2 Desargues Theorem As a simple illustration of the use of the Menelaus and Ceva theorems, we prove the following rmDesargues 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 concur- rent. 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 Y A r1 ZB r2 3.2 Menelaus and Ceva theorems 31

A

Z

Y P

B X C

X′ A

Y C P Y ′ Z X

B

Z′

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. 32 Introduction to Triangle Geometry

3.2.3 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.

A

Y

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 2 (a + b + c): AY = AZ = s a, BZ = BX = s b, CX = CY = s c. − − − ∆ Also, r = 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 XYZ is called the intouch triangle of ABC. Clearly, B + C C + A A + B X = ,Y = , Z = . 2 2 2 3.2 Menelaus and Ceva theorems 33

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.

B

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.

R r 2 s2 2 √(4 + ) + This is called the Adams circle. It is concentricwith the incircle, and has radius 4R+r r. · 34 Introduction to Triangle Geometry

A

Ge I

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 respec- tively. 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. 3.2 Menelaus and Ceva theorems 35

(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 YZ passes through the centroid G. Show that YG : GZ = c : b.

A

Y G

Z

B C

3.2.4 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. 36 Introduction to Triangle Geometry

Ib

A

Ic ′ Z Y ′ Z Na Y I

C B X X′

Ia

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

(3) The lines AMi and AT are symmetric with respect to the bisector of an- gle 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. 3.3 The nine-point circle 37

Ib

A

Ic ′ Z Y ′ Z Mi Y T I

C B X X′

Ia

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

3.3 The nine-point circle

3.3.1 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 1 homothety h(P, 2 ). The midway triangle of the orthocenter H is called the Euler

7 Crux Math. Problem 2250. The reflection of the Nagel point Na in the incenter. This is X145 of ETC. 38 Introduction to Triangle Geometry

Ib

A

Ic

Z′

Y ′ Sp

C 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.

3.3.2 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, 3.3 The nine-point circle 39

A

Y F E

Z I P

B X D C

A

A′

O

O′

P

B′ 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. 40 Introduction to Triangle Geometry

A

Y

Z P

B X C

3.3.3 The nine-point circle From 3.3.1, 3.3.2 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. 3.3 The nine-point circle 41

A

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 P A, P B, 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 P AA′, P BB′, PCC′ are concurrent at P and another point P ′. Equivalently, their cen- ters 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. 42 Introduction to Triangle Geometry

Ib

A

Ic

0 I′ I

C B

Ia A

P

P ′ C A′ B C′

B′ 3.4 The OI-line 43

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 A′B′C′ degenerates, i.e., the three points A′, B′, C′ are collinear. 11

3.4 The OI-line

3.4.1 The homothetic center of the intouch and excentral tri- angles 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 ′ The ratio of homothetic is clearly r . Their circumcenters are I = 2O I 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. 44 Introduction to Triangle Geometry

Ib

A

Ic

Y

Z T I

C B X

Ia 3.4.2 The centers of similitude of the circumcircle and the in- circle

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. 3.4 The OI-line 45 M ′

A

Y

Z I O T− T+

B X C

M 3.4.3 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 rmexcentral 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. − − 46 Introduction to Triangle Geometry

Ib

M ′ A

Ic

O I′ I

C

Xc B D Xb

M

Ia

3.4.4 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 rmintouch triangle and the rmexcentral 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. 3.4 The OI-line 47

A

Y

Z H′ I O G e Na

B X C

3.4.5 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 From the homothety h(T, r ), it is easy to see that the centroid of the − 2R 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. It is X165 of ETC. 18 The centroid of the intouch triangle is X354 of ETC. 48 Introduction to Triangle Geometry

Ib

A

Ic

Y O Z T I

C B X

Ia

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 X40 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. 3.4 The OI-line 49

ABC. Identify the homothetic center. 21

A

Y

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 XYZ, 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. 50 Introduction to Triangle Geometry

8. Let P be the point which divides OI in the ratio OP : P I = R :2r. There Rr is a circle, center P , 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.

A

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. 3.4 The OI-line 51

A

Y

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 3.4.6 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 + D D = 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). 52 Introduction to Triangle Geometry

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

A

O

I

K

B C

A′ 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 incir- cle (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.

3.5 Euler’s formula and Steiner’s porism 3.5 Euler’s formula and Steiner’s porism 53

A

O I

T−

KA

B C

A′ M 3.5.1 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 ∠ 1 ∠ and for the same reason ICB = 2 ACB. Note that A (1) IM = MB = MC =2R sin 2 , r (2) IA = A , and sin 2 (3) by the theorem of intersecting chords, OI2 R2 = the power of I with respect to the circumcircle = IA IM = 2Rr. − · − 3.5.2 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 B′C′ is always tangent to the incircle. This is the famous theorem on Steiner 54 Introduction to Triangle Geometry

A

I O

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 1. r 1 R. When does equality hold? ≤ 2 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) trian- gle(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. 3.5 Euler’s formula and Steiner’s porism 55

A

B′

Y C′

Z I O

B X C

A′ 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 A′B′C′ be a poristic triangle with the same circumcircle and incircle of triangle ABC, and let the sides of B′C′, C′A′, A′B′ touch the incircle at X, Y , Z. (i) What is the locus of the centroid of XYZ? (ii) What is the locus of the orthocenter of XYZ? (iii) What can you say about the Euler line of the triangle XYZ? 56 Introduction to Triangle Geometry