Chapter 4 Introduction to Triangle Geometrymath.fau.edu/yiu/PSRM2015/yiu/New Folder (4)/Downloaded...

Chapter 4

Introduction to Triangle Geometry

4.1 Preliminaries

4.1.1 Coordinatization of points on a line

LetB andC be two fixed points on a lineL. Every pointX onL can be coordi-natized in one of several ways:

(1) the ratio of divisiont = BXXC

,(2) theabsolute barycentric coordinates: an expression ofX as aconvexcom-

bination ofB andC:X = (1 − t)B + tC,

which expresses for an arbitrary pointP outside the lineL, the vectorPX as acombination of the vectorsPB andPC.

(3) the homogeneous barycentric coordinates: the proportionXC : BX,which are masses atB andC so that the resulting system (of two particles) hasbalance pointatX.

P

B CX

32 Introduction to Triangle Geometry

4.1.2 Centers of similitude of two circles

Consider two circlesO(R) and I(r), whose centersO and I are at a distanced apart. Animate a pointX on O(R) and construct a ray throughI oppositelyparallel to the rayOX to intersect the circleI(r) at a pointY . You will find thatthe lineXY always intersects the lineOI at the same pointT . This we call theinternal center of similitude, or simply theinsimilicenter, of the two circles. Itdivides the segmentOI in the ratioOT : TI = R : r. The absolute barycentriccoordinates ofP with respect toOI are

T =R · I + r ·O

R+ r.

O I

XY ′

Y

T T ′

If, on the other hand, we construct a ray throughI directly parallel to the rayOX to intersect the circleI(r) atY ′, the lineXY ′ always intersectsOI at anotherpointT ′. This is theexternal center of similitude, or simply theexsimilicenter, ofthe two circles. It divides the segmentOI in the ratioOT ′ : T ′I = R : −r, andhas absolute barycentric coordinates

T ′ =R · I − r ·O

R− r.

4.1.3 Tangent circles

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

4.1 Preliminaries 33

O IT O IT ′

4.1.4 Harmonic division

Two pointsX andY are said to divide two other pointsB andC harmonically if

BX

XC= −BY

Y C.

They areharmonic conjugatesof each other with respect to the segmentBC.

Examples

1. For two given circles, the two centers of similitude divide the centers har-monically.

2. Given triangleABC, let the internal bisector of angleA intersectBC atX. The harmonic conjugate ofX in BC is the intersection ofBC with theexternal bisector of angleA.

3. Let A andB be distinct points. IfM is the midpoint of the segmentAB,it is not possible to find afinite point N on the lineAB so thatM , NdivideA, B harmonically. This is becauseAN

NB= −AM

MB= −1, requiring

AN = −NB = BN , andAB = BN − AN = 0, a contradiction. Weshall agree to say that ifM andN divideA, B harmonically, thenN is theinfinite pointof the lineAB.

Exercise

1. If X, Y divideB, C harmonically, thenB, C divideX, Y harmonically.


A

B CX X′

2. Given a pointX on the lineBC, construct its harmonic associate with re-spect to the segmentBC. Distinguish between two cases whenX dividesBC internally and externally.1

3. The centersA andB of two circlesA(a) andB(b) are at a distanced apart.The lineAB intersect the circles atA′ andB′ respectively, so thatA,B arebetweenA′,B′.

A′A B

B′

4. Given two fixed pointsB andC and a positive constantk = 1, the locus ofthe pointsP 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 pointT and a nonzero constantk, the similarity transformationh(T, k)which carries a pointX to the pointX ′ on the lineTX satisfyingTX ′ : TX =k : 1 is called thehomothetywith centerT and ratiok. Explicitly,

h(T, k)(P ) = (1 − k)T + kP.

Any two circles are homothetic. LetP andQ be the internal and externalcenters of similitude of two circlesO(R) andI(r). Both the homothetiesh(Q, r

R)

andh(P,− rR) transform the circleO(R) into I(r).

4.1.6 The power of a point with respect to a circle

Thepowerof a pointP with respect to a circleC = O(R) is the quantity

C(P ) := OP 2 − R2.

This is positive, zero, or negative according asP is outside, on, or inside the circleC. If it is positive, it is the square of the length of a tangent fromP to the circle.

Theorem (Intersecting chords)

If a line L throughP intersects a circleC at two pointsX andY , the productPX · PY (of signed lengths) is equal to the power ofP with respect to the circle.

P

T

T ′

X

Y

O


4.2 Menelaus and Ceva theorems

4.2.1 Menelaus and Ceva Theorems

Consider a triangleABC with pointsX, Y , Z on the side linesBC, CA, ABrespectively.

Theorem 4.1 (Menelaus).The pointsX, Y , Z are collinear if and only if

BX

XC· CYY A

· AZZB

= −1.

A

B CX

Y

Z

Theorem 4.2 (Ceva).The linesAX,BY , CZ are concurrent if and only if

BX

XC· CYY A

· AZZB

= +1.

4.2.2 Desargues Theorem

As a simple illustration of the use of the Menelaus and Ceva theorems, we provethe following Desargues Theorem: Given three circles, the exsimilicenters of thethree pairs of circles are collinear. Likewise, the three lines each joining the in-similicenter of a pair of circles to the center of the remaining circle are concurrent.

We prove the second statement only. Given three circlesA(r1), B(r2) andC(r3), the insimilicentersX of (B) and(C), Y of (C), (A), andZ of (A), (B)are the points which divideBC, CA, AB in the ratios

BX

XC=r2r3,

CY

Y A=r3r1,

AZ

ZB=r1r2.

4.2 Menelaus and Ceva theorems 37

P

X

Y

Z

A

B C

A

B

C

X

Y

Z

P

X′

Y ′

Z′

It is clear that the product of these three ratio is+1, and it follows from the Cevatheorem thatAX,BY , CZ are concurrent.


4.2.3 The incircle and the Gergonne point

Theincircle is tangent to each of the three sidesBC,CA,AB (without extension).Its center, theincenterI, is the intersection of the bisectors of the three angles.The inradiusr is related to the area∆ by

S = (a+ b+ c)r.

I

A

B C

Ge

X

Y

Z

If the incircle is tangent to the sidesBC atX, CA atY , andAB atZ, then

AY = AZ =b+ c− a

2, BZ = BX =

c+ a− b

2, CX = CY =

a+ b− c

2.

These expressions are usually simplified by introducing thesemiperimeters =12(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 thatAX, BY , CZ are concurrent.The point of concurrencyGe is called theGergonne pointof triangleABC.

TriangleXY Z is called theintouch triangleof ABC. Clearly,

X =B + C

2, Y =

C + A

2, Z =

A+B

2.


It is always acute angled, and

Y Z = 2r cosA

2, ZX = 2r cos

B

2, XY = 2r cos

C

2.

Exercise

1. Given three pointsA, B, C not on the same line, construct three circles,with centers atA,B, C, mutually tangent to each otherexternally.

C A

B

Z

Y

X

2. Construct the three circles each passing through the Gergonne point andtangent to two sides of triangleABC. The 6 points of tangency lie on acircle.2

3. Two circles are orthogonal to each other if their tangents at an intersectionare perpendicular to each other. Given three pointsA, B, C not on a line,construct three circles with these as centers and orthogonal to each other.

(1) Construct the tangents fromA′ to the circleB(b), and the circle tangentto these two lines and toA(a) internally.

(2) Construct the tangents fromB ′ to the circleA(a), and the circle tangentto these two lines and toB(b) internally.

(3) The two circles in (1) and (2) are congruent.

2This is called the Adams circle. It is concentric with the incircle, and has radius√

(4R+r)2+s2

4R+r ·r.


I

A

B C

Ge

4. Given a pointZ on a line segmentAB, construct a right-angled triangleABC whose incircle touches the hypotenuseAB atZ. 3

5. LetABC be a triangle with incenterI.

(1a) Construct a tangent to the incircle at the point diametrically oppositeto its point of contact with the sideBC. Let this tangent intersectCA atY1

andAB atZ1.

(1b) Same in part (a), for the sideCA, and let the tangent intersectAB atZ2 andBC atX2.

(1c) Same in part (a), for the sideAB, and let the tangent intersectBC atX3 andCA atY3.

(2) Note thatAY3 = AZ2. Construct the circle tangent toAC andAB atY3

andZ2. How does this circle intersect the circumcircle of triangleABC?

6. The incircle of�ABC touches the sidesBC, CA, AB atD, E, F respec-tively. X is a point inside�ABC such that the incircle of�XBC touchesBC atD also, and touchesCX andXB atY andZ respectively.

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


(1) The four pointsE, F , Z, Y are concyclic.4

(2) What is thelocusof the center of the circleEFZY ? 5

7. Given triangleABC, construct a circle tangent toAC at Y andAB at Zsuch that the lineY Z passes through the centroidG. Show thatY G : GZ =c : b.

A

B C

GY

Z

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 thecorresponding sides of triangleABC. The linesAX ′, BY ′, CZ ′ are concurrent.The common pointNa is called theNagel pointof triangleABC.

Exercise

1. Construct thetritangent circlesof a triangleABC.

(1) Join each excenter to the midpoint of the corresponding side ofABC.These three lines intersect at a pointMi. (This is called theMittenpunktofthe triangle).

4International Mathematical Olympiad 1996.5IMO 1996.


Ic

Ib

Na Y ′

Z′

Ia

X′

I

A

B

C

X

Y

Z

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

(3) The linesAMi andAT are symmetric with respect to the bisector of an-gleA; so are the linesBMi,BT andCMi,CT (with respect to the bisectorsof anglesB andC).

2. Construct the excircles of a triangleABC.

(1) LetD, E, F be the midpoints of the sidesBC, CA, AB. Construct theincenterSp of triangleDEF , 6 and the tangents fromS to each of the threeexcircles.

(2) The 6 points of tangency are on a circle, which isorthogonalto each ofthe excircles.

6This is called the Spieker point of triangleABC.

4.3 The nine-point circle 43

Ic

Ib

Y ′

Z′

Ia

X′

I

A

B

C

X

Y

Z Mi

T

3. LetD,E, F be the midpoints of the sidesBC,CA,AB, and let the incircletouch these sides atX, Y , Z respectively. The lines throughX parallel toID, throughY to IE and throughZ 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 triangleABC. The midpoints of the segmentsAP , BP , CP form themidway triangleof P . It is the image ofABC under thehomothetyh(P, 1

2). The midway triangle of the orthocenterH is called theEuler

7Crux Math. Problem 2250. The reflection of the Nagel pointN a in the incenter. This isX145

of ETC.


Ia

Ic

Ib

SpY ′

Z′

X′

A

B

C

triangle. The circumcenter of the midway triangle ofP is the midpoint ofOP . Inparticular, the circumcenter of the Euler triangle is the midpoint ofOH, which isthe same asN , the circumcenter of the medial triangle. (See Exercise??.). Themedial triangle and the Euler triangle have the same circumcircle.

4.3.2 The orthic triangle as a pedal triangle

Thepedalsof a point are the intersections of the sidelines with the correspondingperpendiculars throughP . They form thepedal triangleof P . The pedal triangleof the orthocenterH is called theorthic triangleof ABC.

The pedalX of the orthocenterH on the sideBC is also the pedal ofA onthe same line, and can be regarded as thereflectionof A in the lineEF . It followsthat

∠EXF = ∠EAF = ∠EDF,


D

EF

I

A

B CX

Y

Z

P

O

A

B C

A′

B′ C′

P

O′

sinceAEDF is a parallelogram. From this, the pointX lies on the circumcircleof the medial triangleDEF ; similarly for the pedalsY andZ of H on the othertwo sidesCA andAB.


A

B CX

Y

Z

P

4.3.3 The nine-point circle

From§§4.3.1, 4.3.2 above, the medial triangle, the Euler triangle, and the orthictriangle all have the same circumcircle. This is called thenine-point circleoftriangleABC. Its centerN , the midpoint ofOH, is called thenine-point centerof triangleABC.

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) thecircumcenter of the excentral triangle is the reflection ofI in O.

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

3. If the midpoints ofAP ,BP ,CP are all on the nine-point circle, mustP bethe orthocenter of triangleABC? 8

8P. Yiu and J. Young, Problem 2437 and solution,Crux Math.25 (1999) 173; 26 (2000) 192.


ON

D

EF

A

B C

H

X

Y

Z

A′

B′ C′

4. LetABC be a triangle andP a point. The perpendiculars atP toPA, PB,PC intersectBC, CA, AB respectively atA′,B′, C ′.

(1)A′,B′, C ′ are collinear.9

(2) The nine-point circles of the (right-angled) trianglesPAA′, PBB′,PCC ′ are concurrent atP and another pointP ′. Equivalently, their cen-ters are collinear.10

5. (Triangles with nine-point center on the circumcircle)

Begin with a circle, centerO and a pointN on it, and construct a family oftriangles with(O) as circumcircle andN as nine-point center.

(1) Construct the nine-point circle, which has centerN , and passes through

9B. Gibert, Hyacinthos 1158, 8/5/00.10A.P. Hatzipolakis, Hyacinthos 3166, 6/27/01. The three midpoints ofAA ′, BB′, CC ′ are

collinear. The three nine-point circles intersect atP and its reflection in this line.


0

Ic

Ib

Ia

I

A

B

C

I′

A

BC

P

B′

C′

A′

P ′

4.4 TheOI-line 49

the midpointM of ON .

(2) Animate a pointD on the minor arc of the nine-point circleinsidethecircumcircle.

(3) Construct the chordBC of the circumcircle withD as midpoint. (Thisis simply the perpendicular toOD atD).

(4) LetX be the point on the nine-point circle antipodal toD. Complete theparallelogramODXA (by translating the vectorDO toX).

The pointA lies on the circumcircle and the triangleABC has nine-pointcenterN on the circumcircle.

Here is a curious property of triangles constructed in this way: letA′, B′,C ′ be the reflections ofA, B, C in their own opposite sides. The reflectiontriangleA′B′C ′ degenerates,i.e., the three pointsA′,B′,C ′ are collinear.11

4.4 TheOI-line

4.4.1 The homothetic center of the intouch and excentral tri-angles

TheOI-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 thehomothety of the intouch and excentral triangles. These triangles are homotheticsince their corresponding sides are parallel, being perpendicular to the same anglebisector of the reference triangle.

The ratio of homothetic is clearly2Rr

. Their circumcenters areI ′ = 2O − IandI. The homothetic center is the pointT such that

2O − I = h

(T,

2R

r

)(I) =

(1 − 2R

r

)T +

2R

r· I.

This gives

T =(2R+ r)I − 2r ·O

2R− r.

It is the point which dividesOI externally in the ratio2R+ r : −2r. 12

11O. Bottema,Hoofdstukken uit de Elementaire Meetkunde, Chapter 16.12The pointT is X57 in ETC.


Ic

Ib

Ia

I

A

B

C

X

Y

ZT

4.4.2 The centers of similitude of the circumcircle and the in-circle

These are the pointsT+ andT− which divide the segmentOI harmonically in theratio of the circumradius and the inradius.13

T+ =1

R+ r(r ·O +R · I),

T− =1

R− r(−r ·O +R · I).

13T+ andT− are respectivelyX55 andX56 in ETC.

4.4 TheOI-line 51

OI

A

B CX

Y

Z

T+T−

M

M′

4.4.3 Reflection ofI in O

The reflectionI ′ of I in O is the circumcenter of the excentral triangle. It is theintersections of the perpendiculars from the excenters to the sidelines.14

The midpointM of the arcBC is also the midpoint ofIIa. SinceIM andI ′Ia are parallel,I ′Ia = 2R. Similarly, I ′Ib = Pc = 2R. This shows that theexcentral triangle has circumcenterI ′ = 2O − I and circumradius2R. SinceIis the orthocenter ofIaIbIc, its follows thatO, the midpoint ofI andI ′, is thenine-point center of the excentral triangle. In other words, the circumcircle oftriangleABC is the nine-point circle of the excentral triangle. Apart fromA, thesecond intersection ofIbIc with the circumcircle ofABC is the midpointM ′ ofIbIc. From this we deduce the following interesting formula:15

ra + rb + rc = 4R+ r.

14I ′ appears asX40 in ETC.15Proof. ra+rb+rc = 2MD+(I ′Ia−I ′X ′) = 2R+2OD+I ′Ia−I ′X ′ = 4R+IX = 4R+r.


O

Ic

Ib

XbXc

Ia

I

A

B

C

I′

M

M′

D

4.4.4 Orthocenter of intouch triangle

TheOI-line is the Euler line of the excentral triangle, sinceO and I are thenine-point center and orthocenter respectively. The corresponding sides of theintouch triangle and the excentral triangle are parallel, being perpendicular to therespective angle bisectors. Their Euler lines are parallel. Since the intouch trianglehas circumcenterI, its Euler line is actually the lineOI, which therefore containsits orthocenter. This is the point which dividesOI in the ratioR + r : −r. 16 Italso lies on the line joining the Gergonne and Nagel points.

16This is the pointX65 in ETC.

4.4 TheOI-line 53

OI

A

B CX

Y

Z

GeNa

H′

4.4.5 Centroids of the excentral and intouch triangles

The centroid of the excentral triangle is the point which dividesOI in the ratio−1 : 4. 17 From the homothetyh(T, r

2R), it is easy to see that the centroid of the

intouch triangle is the point which dividesOI in the ratio3R+ r : −r. 18

Exercises

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

17This is the point which divides the segmentI ′I in to the ratio1 : 2. It is X165 of ETC.18The centroid of the intouch triangle isX354 of ETC.


O

Ic

Ib

Ia

I

A

B

C

X

Y

ZT

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

3. Identify the midpoint of the Nagel point and the deLongchamps point as apoint on theOI-line. 20

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

19Feuerbach’s theorem.20Reflection ofI in O. This is because the pedal triangle ofX40 is the cevian triangle of the

Nagel point, and the reflections of the pedals of the Nagel point in the respective traces form thepedals of the de Longchamps point.

4.4 TheOI-line 55

ABC. Identify the homothetic center.21

O

I

A

B CX

Y

Z

H′

5. Construct the external common tangent of each pair of excircles. Thesethree external common tangents bound a triangle.

(i) This triangle is perpsective withABC. Identify the perspector.22

(ii) Identify the incenter of this triangle.23

6. ExtendAB andAC by lengtha and join the two points by a line. Similarlydefine the two other lines. The three lines bound a triangle with perspectorX65

24

7. Let H ′ be the orthocenter of the intouch triangleXY Z, andX ′, Y ′, Z ′ itspedals on the sidesY Z, ZX, XY respectively. Identify the common pointof the three linesAX ′, BY ′, CZ ′ as a point on theOI-line. Homotheticcenter of intouch and excentral triangles.

21T .22Orthocenter of intouch triangle.23Reflection ofI in O.242/18/03.


8. Let P be the point which dividesOI in the ratioOP : PI = R : 2r. Thereis a circle, centerP , radius Rr

R+2r, which is tangent to three congruent circles

of the same radius, each tangent to two sides of the triangle. Construct thesecircles.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 thepoints of tangency of each circle with(O) and(I) pass through the internalcenterT+ of similitude of(O) and(I). Construct these three circles.26

OI

A

B C

T+

10. LetT+ be the insimilicenter of(O) and(I), with pedalsY andZ onCA andAB respectively. IfY ′ andZ ′ are the pedals ofY andZ onBC, calculatethe length ofY ′Z ′. 27

11. Let P be the centroid of the excentral triangle, with pedalsX, 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 TheOI-line 57

O

I

A

B C

T+

X

Y

Z

Y ′Z′

sidesBC, CA, AB respectively. Show that28

AY + AZ = BZ +BX = CX + CY =1

3(a+ b+ c).

4.4.6 Mixtilinear incircles

A mixtilinear incircleof triangleABC is one that is tangent to two sides of thetriangle and to the circumcircle internally. Denote byA′ the point of tangencyof the mixtilinear incircleK(ρ) in angleA with the circumcircle. The centerKclearly lies on the bisector of angleA, andAK : KI = ρ : −(ρ− r). In terms ofbarycentric coordinates,

K =1

r(−(ρ− r)A+ ρI) .

28The projections ofO and I on the sideBC are the midpointD of BC, and the point oftangencyD′ of the incircle with this side. Clearly,BD = a

2 andBD′ = 12 (c + a − b). It follows

that

BX = BD′ +43D′D =

43BD − 1

3BD′ =

16(3a + b − c).

Similarly,BZ = 16 (3c + b− a), andBX + BZ = 1

3 (a + b + c). A similar calculation shows thatAY + AZ = CX + CY = 1

3 (a + b + c).


Also, since the circumcircleO(A′) and the mixtilinear incircleK(A′) touch eachother atA′, we haveOK : KA′ = R − ρ : ρ, whereR is the circumradius. Fromthis,

K =1

R(ρO + (R − ρ)A′) .

I

O

K

A′

A

B C

Comparing these two equations, we obtain, by rearranging terms,

RI − rO

R− r=R(ρ− r)A+ r(R− ρ)A′

ρ(R− r).

We note some interesting consequences of this formula. First of all, it gives theintersection of the lines joiningAA′ andOI. Note that the point on the lineOIrepresented by the left hand side isT−, the external center of similitude of thecircumcircle and the incircle.

This leads to a simple construction of the mixtilinear incircle. Given a triangleABC, letP be the external center of similitude of the circumcircle(O) and incir-cle (I). ExtendAP to intersect the circumcircle atA′. The intersection ofAI andA′O is the centerKA of the mixtilinear incircle in angleA.

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

KA

A′

A

B C

T−

M

I

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 angleA intersect the circumcircle atM . Construct thecircleM(B) to intersect this bisector at a pointI. This is the incenter since

∠IBC =1

2∠IMC =

1

2∠AMC =

1

2∠ABC,

and for the same reason∠ICB = 12∠ACB. Note that

(1) IM = MB = MC = 2R sin A2,

(2) IA = rsin A

2

, and

(3) by the theorem of intersecting chords,OI 2−R2 = thepowerof I with respectto the circumcircle =IA · IM = −2Rr.

4.5.2 Steiner’s porism

Construct the circumcircle(O) and the incircle(I) of triangleABC. Animate apointA′ on the circumcircle, and construct the tangents fromA′ to the incircle(I). Extend these tangents to intersect the circumcircle again atB ′ andC ′. ThelinesB′C ′ is always tangent to the incircle. This is the famous theorem on Steiner


OI

A

B C

M

porism: if two given circles are the circumcircle and incircle of one triangle, thenthey are the circumcircle and incircle of a continuous family ofporistic triangles.

Exercises

1. r ≤ 12R. When does equality hold?

2. SupposeOI = d. Show that there is a right-angled triangle whose sides ared, r andR − r. Which one of these is the hypotenuse?

3. Given a pointI inside a circleO(R), construct a circleI(r) so thatO(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 thetriangle.

5. Construct an animation picture of a triangle whose circumcenter lies on theincircle.29

29Hint: OI = r.

4.5 Euler’s formula and Steiner’s porism 61

OI

A

B CX

Y

Z

B′

A′

C′

6. What is the locus of the centroids of the poristic triangles with the samecircumcircle and incircle of triangleABC? How about the orthocenter?

7. LetA′B′C ′ be a poristic triangle with the same circumcircle and incircle oftriangleABC, and let the sides ofB ′C ′,C ′A′,A′B′ touch the incircle atX,Y , Z.

(i) What is the locus of the centroid ofXY Z?

(ii) What is the locus of the orthocenter ofXY Z?

(iii) What can you say about the Euler line of the triangleXY Z?

Chapter 4 Introduction to Triangle Geometrymath.fau.edu/yiu/PSRM2015/yiu/New Folder (4)/Downloaded...

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Transcript of Chapter 4 Introduction to Triangle Geometrymath.fau.edu/yiu/PSRM2015/yiu/New Folder (4)/Downloaded...