Some Construction Problems Related to the Incircle of a Triangle

Home , Integer triangle, Triangle center

SOME CONSTRUCTION PROBLEMS

RELATED TO THE

INCIRCLE OF A TRIANGLE

Amy B. Bell

A Thesis Submitted to the Faculty of

The Charles E Schmidt College of Science

In Partial Fulfillment of the Requirements for the Degree of

Masters of Science in Teaching

Florida Atlantic University

Boca Raton, Florida

December 2006 Some Construction Problem s R elated to the Incircle of a Triangle by Amy B. B ell

This thesis was prepared under the direction of the candidate's thesis advisor, Dr. Paul Yiu, Department of Mathematical Sciences, and has been approved by the members of her supervisory committee. It was submitted to the faculty of the Charles E. Schmidt College of Science and was accepted in partial fulfi llment of the requirements for the degree of Master of Science in Teaching of Mathematics.

SUPERVISORY COMMITTEE:

I I i.. • I (C-v Thesis Advisor

Date

ll Acknowledgements

I would like to thank my graduate advisor, Professor Paul Yiu, for his

support, patience and encouragement throughout the process of this thesis. I

would also like to thank Professor Markus Schmidmeier for going beyond the

usual duties of a thesis committee member and giving valuable feedback that

'\,. made my thesis defense much more effective.

I would also like to acknowledge the support of my mother, Katherine

Gosney and my husband, Terry Bell. Both made it possible for me to spend the

additional hours I needed to work through the mathematical ideas presented in

this thesis.

111 ~------

Abstract

Author: Amy B. Bell

Title: Some Construction Problems Related to

the Incircle of a Triangle

Institution: Florida Atlantic University

Thesis Advisor: Dr. Paul Yiu

Degree: Masters of Science in Teaching

Year: 2006

This thesis explores several construction problems related to the incircle of

a triangle. Firstly, as a generalization of a theorem of D. W. Hansen, we find two

quadruples of quantities related to a triangle which have equal sums and equal

sums of squares. We also study the construction problems of triangles with

centroid on the incircle, and those with a specified cevian - a median, an angle

bisector, or an altitude- bisected by the incircle. Detailed analysis leads to dE>•:;i[.>' s

of animation pictures using the dynamic software Geometer's Sketchpad.

lV Contents

List of Tables ...... v11 List of Figures ...... vm 1 Preliminaries 1 1.1 Introduction ...... 1.2 Notations ...... 3 1.3 The Heron formula and Heron triangles ...... 4 1.3.1 The Heron formula...... 4 1.3 .2 Heron triangles ...... 4 1.4 Barycentric coordinates ...... 5 1.4.1 Homogeneous barycentric coordinates ...... 5 1.4.2 Absolute barycentric coordinates ...... 6 1.4.3 Equation ofthe incircle ...... 7 1.5 Geometric constructions ...... 7 1.5.1 Geometric solution of quadratic equations ...... 7 1.5.2 Non-constructibility ...... 8

2 Hansen's theorem and generalizations 9 2.1 Hansen's theorem ...... 9 2.2 Integer relations from Pythagorean triangles ...... 11 2.3 Generalization of Hansen's theorem ...... 13 2.4 Some triangle geometry ...... 15 2.5 ProofofTheorem 2.3 ...... 18 2.6 Converse ofHansen's Theorem ...... 19

v 2. 7 Integer relations ...... 21

3 Triangles with centroid on the incircle 24 3.1 From s- bands-c...... 25 3.2 From band c ...... 28 3.3 Triangle with a median trisected by the incircle ...... 32

';; 4 Triangle with midpoint of one cevian on incircle '•3 ' 4.1 Construction of triangle ABC with midpoint of A-median on incircle ...... 34 4.1.1 From s - b and s - c ...... 34 4.1 .2 From band c ...... 36 4.2 Construction of triangle ABC with midpoint of A-bisector on incircle ...... 39 4.2.1 From b and c ...... 40 4.2.2 An alternative construction ...... 42 4.2.3 From s- b and s - c ...... 43 4.3 Construction of triangle ABC with midpoint of A-altitude on incircle ...... 44 4.3.1 From band c ...... 45 4.3 .2 From s- bands-c...... 46

Bibliography

VI ------

List of Tables

Integers with equal sums and sums of squares

from Pythagorean triangles ...... 12 2 Integers with equal sums and sums of squares from Heron triangles ...... 21 3 More integers with equal sums and sums of squares from Heron triangles ...... 23

Vll ------· -

List of Figures

1.1 The incircle ...... 3 1.2 Homogeneous barycentric coordinates ...... 6 1.3 Solutions of quadratic equations ...... 8 2.1 The incircle of a right triangle ...... 9 2.2 The incircle and excircles of a right triangle ...... 10 2.3 Circumcenter and orthocenter ...... 13 'i'" 2.4 Two quadruples with equal sums and equal sums of squares . . . . . 14 2.5 Incenter and excenter ...... 15

2.6 ] 'f a= 2R ...... 16

2.7 ra + r b + rc = 4R + r...... 17 3.1 Isosceles with centroid on incircle ...... 25 3.2 Construction of u from v and w ...... 26 3.3 Triangle with centroid on incircle ...... 27 3.4 Triangle with centroid on incircle ...... 28

2 1 2 4 2 3.5 AP = -(b+ c) + -(b- c) ...... 29 5 5 3.6 Solution of (3 .3) ...... 30 3.7 Triangles with centroid on incircle ...... 31 3.8 Triangle with a median trisected by the incircle ...... 32 4.1 Isosceles triangle with median bisected by incircle ...... 34 4.2 Triangle with median bisected by incircle ...... 35 4.3 Triangle with median bisected by incircle ...... 36 4.4 Solution of(4.1) ...... 37

vm ------

4.5 Triangle with a median bisected by incircle ...... 38 4.6 Triangle with a median bisected by incircle ...... 38 4.7 Barycentric coordinates of midpoint of angle bisector ...... 39

4.8 Construction of A ...... 40 2 4.9 Triangle with an angle bisector bisected by incircle ...... 42 4.10 Triangle with an angle bisector bisected by incircle ...... 43 4.11 Triangle with two altitudes bisected by incircle ...... 45

~ .. 1 1 1 4.12 -+-=- ...... 46 p q r

1 1 1 4.13 -+-=- ...... 47 V W X

4.14 Construction of u from v and w ...... 48 4.15 Triangle with an altitude bisected by incircle ...... 48

IX Chapter 1

Preliminaries

1.1 Introduction

In thi s thesis we study several problems re lated to the inc irc le of a triang le. In the present chapter we sha ll ex pl a in some bac kground results in triangle geometry th at sha ll be freely used in the subsequent chapters, namely,

( i) the Heron formula for the area of a triang le and formulas for various radi i asso c iated with a triang le (s1 .3), (ii) the noti on of barycentric coordinates and the descriptio n of the inc irc le in terms of such ( ~ 1.4),

(iii ) a basic meth od o f solvin g quadratic equati ons as a ruler and compass construc ti on problem (§ 1.5).

In Chapter 2, we generali ze a result of D. W. Hansen [I] and prove (Theorem

2.3) that for a given triang le with orthocentcr H , the two quadruples, namely,

(i) the inradius and th e three exradii, and

(ii ) the c ircumdiameter and the lengths of the segments AH, D H , C H , have equal sums and equal sums of squares.

In §2.6, we prove a strong converse of Hansen's theorem (Theorem 2.4).

In chapter 3, we study the constructi on problem of a triangle whose inc ircle contains the centro id . We analyze the problem w ith two sets of g iven data:

( i) g iven two le ngth s band c, to construct a third le ngth a so that they form a triang le w hose incirc le contains the centroid ( >3 3.2):

(ii ) g iven two po ints /3, C: together w ith a third po in t X on the segment I3C, to construct a third po in t A suc h that ABC has its inc irclc containing the centroid, and tangent to BC at X, (§3. 1).

A ft e r a deta il ed analysis, we incorporate the constructions with the use of the

Geometer's Sketchpad, a dy namic software for geometric construc tions.

F inally, in C hapter 4 , we study a variation of the the me of C hapter 3, na me ly the constructi on probl em o f a triang le w hose inc irclc b isects one specifi c cevian of the tri ang le. Recall that a ccvian is a line through a vertex of the tri ang le. T he cevians we consider in C hapte r 4 are a median, a n ang le bisector, and an a ltitude.

T hus, we analyze each proble m below to the exte nt that it is solvable as a rule r and compass constructi on pro ble m . In each suc h case, we produce a constructi on using the Geometer's Sketc hpad .

(i) Given two le ngth s band r, to construct a third le ngth a so that they form a triang le w hose incirc lc b isects

(a) th e median (§4. 1.2),

(b) the ang le bisector (§4.2. I),

(ii) Given two points B, C togethe r with a thi rd po in t X on the segment BC, to construct a triang le ABC w ith inc irc le tangent to BC at X and bisecting

(a) th e median (§4. 1. 1),

(b) the angle bisector (§4.2.3 ),

on the s ide BC.

2 ------.

1.2 Notations

We consider a triangle ABC with side length s B C = a, C A = !J , AJJ = c. It is common to adopt the foll ow ing standard notati ons.

semi perimeter area of tri angle cirumradius I' inradiu s rad iu s of exc ircles on the oppos it e sides of A, D, C respective ly (0) a circle with center 0 O(P) a circle with center 0, pass in g through a point]:>

The quantities s - a, s- !J, 8 - c are related to the incircle as indicated in Fi gure 1.1 below. They are fund amental in thi s thesis. In Chapte rs 3 and 4, the qu antities

8- a, s -- b, s -c are abbreviated as 11, v , w respecti ve ly.

s-c s-b

Figure 1. 1: The incircle

3 1.3 The Heron formula and Heron triangles

1.3.1 The Heron formula

The area of a triangle is given by the famous Heron formula

U. 2 = s(s- a)(s - b)(.s- c). (1.1)

Explicitly in terms of a, b, c, this can be written as

16U.2 =(a+ b + c)(b + c- a)(c +a- b)(a + b- c)

( 1.2)

The relations between the radii and the side lengths are as follows.

Proposition l.l. (a) R = ~~·

(b) T = ~·

1.3.2 Heron triangles

A Heron triangle is a triangle with integer sides and integer area. It is called prim

itive if its side lengths have no common divisor other than 1. It is well known that

the semiperimeter of a Heron triangle must be an integer. In other words, among

the sides of a primitive Heron triangle, exactly one has even lengths. We give a

simple proof here.

Lemma 1.1. In a primitive Heron triangle, exactly one side has even length.

Proqf Clearly at least one side length is an odd number. It is enough to eliminate

the following two cases.

(I) Suppose a, b, care all odd. Then u2 = 8a' + 1, b2 = 8b' + 1, c2 = 8c' + 1 for integers o', b', c'. Modulo 16, we have a 4 = b4 = c4 = 1. Also, 2b2c2 = 2c2a 2 =

4 2a2b2 =2 . It follows from (1.2) that 16.62 =2 + 2 + 2 - 1 - 1- 1 =3 mod 16, which is a contradiction. (2) Suppose a is odd and b, c are both even. Then, modulo 16, n:1 = 1 and b4 =c 4 =0. On the other hand, 2b2 c2 =0, 2c2 a 2 =2 a2b2 =8. It follows that 16.6 2 =0 + 8 + 8 - 1 - 0 - 0 =-1 mod 16, which is a contradiction. D

Corollary 1.1. The semiperimeter of a Heron triangle is an integer.

1.4 Barycentric coordinates

1.4.1 Homogeneous barycentric coordinates

In the geometry of the triangle, it is more convenient to work with barycentric co

ordinates than with Cartesian coordinates. Given a reference triangle ABC (with side lengths a, b, c), every point P in the plane of the triangle can be character ized by a triple of numbers, which are the magnitudes of forces at A, B, C with resultant at r . These numbers are clearly defined up to proportions, and are called the homogeneous barycentric coordinates of P with reference to triangle ABC.

They can also be defined in terms of areas. Let [ABC] denote the signed area of

triangle ABC , so that a triangle with opposite orientation has a negative area. The

homogeneous barycentric coordinates of a point P in the plane of triangle ABC

are [PBC] : [APC] : [ABP] .

See Figure 1.2.

For an introduction to the use of barycentric coordinates in triangle geometry,

see [5]. For the purpose of this thesis, we only need the coordinates of the following

triangle centers.

5 ~---~~~~

Figure 1.2: Homogeneous barycentric coordinates

Triangle center Notation Homogeneous barycentric coordinates ~~~~~~~~~=-~~~~~~~- centroid G 1 : 1 : 1 incenter I a : b :c circumcenter 0 a 2 ( b2 + c2 - a 2) : b2 ( c2 + a 2 - b2) : c2 (a 2 + b2 - c2) orthocenter H 1 . l . 1 ~ +~-~ . ~ + ~-~ . ~+~ - 2

lt is well known that the triangle centers G, 0, Hare collinear. The line con taining them is called the Euler line of the triangle.

1.4.2 Absolute barycentric coordinates

The absolute coordinate of a (finite) point in the plane of triangle ABC is obtained by normalization, namely, adjusting the sum of the coordinates to 1. Thus, if x + y+ z i= 0, the point (x, y, z) has absolute coordinate x · A;!:~~z - C . (If x+y+z = 0,

then x : y : z is an infinite point).

Lemma 1.2. Given two finite points P and Q in absolute barycentric coordinates,

the point X dividing PQ in the ratio P X : X Q = >. : fJ, has absolute coordinate ~~ : !~Q. In particular, the midpoint of the segment PQ has absolute barycentric coordinate P ~Q .

6 1.4.3 Equation of the incircle

Lines and circles in the plane of triangle ABC are represented by linear and quadratic equations in homogeneous barycentric coordinates. For example, the equation of the Euler I ine is 1

(b2- c2)(b2 + c2- a2)x + (c2- a.2)(c2 + a.2 - l})y+ (a2- b2)(a.2 + b2- c2) z = 0. ( 1.3)

The equation of the circumcircle is 2

( 1.4)

We shall need the equation of the incircle. For details, see [5, §6.1].

Lemma 1.3. The equation of the incircle is given by

2(s - b)( s- c) yz + 2(s- c)(s- o.)zx + 2(s- a.)(s- b).Ty

-(s - a.) 2 x 2 - (s- b) 2 y 2 - (s- c) 2 z 2 = 0. ( 1.5)

1.5 Geometric constructions

1.5.1 Geometric solution of quadratic equations

In Chapters 3 and 4 we study a group of geometric construction problems using ruler and compass. We shall make use of the following intersecting chords theorem to handle the solutions of quadratic equations.

Theorem 1.1. Let P be a .fixed point in the plane of a circle e. !fa line through P intersects the circle at X and Y, then the product P X · PY is a constant (called the power) of p with respect to e.

I (5 , §4.1]. 2[5, §5.2].

7 ------·--·- - ·

Proposition 1.2. Consider a circle ( 0), diameter p, and a segment of length q.

(a) Let P be a point inside the circle such that the chord perpendicular to O.P at P has length 2q. If X X' is the diameter containing P, then the roots of the quadratic equation 2 2 .r - p · x + q = 0

are the lengths of P X and P X'.

Figure 1.3: Solutions of quadratic equations

(b) Let P be a point outside the circle such that the tangents from P have length q. If X X' is the diameter containing P, then the roots of the quadratic equation

2 2 :r - p · x - lj = 0

are the lengths of P X and P X' with appropriate signs.

1.5.2 Non-constructibility

It is well known that not every constmction problem in plane euclidean geometry is

solvable using ruler and compass. We shall make use of the following well known

non-constructibility theorem. See, for example, [3, Theorem 5. 15].

Theorem 1.2. The roots of an irreducible cubic polynomial are not constructible using ruler and compass.

8 Chapter 2

Hansen's theorem and generalizations

2.1 Hansen's theorem

In an interesting article in Mathematics Teacher, D. W. Hansen [I] has found some remarkable identities associated with a right triangle. We shall extend these for a general triangle. Specifically, given a triangle, we shall find two quadruples with equal sums and equal sums of squares. Let ABC be a triangle with a right angle at C , along with its incircle. The

inradius is clearly T = s - c. B

/ I ' r

s-c :.-.--, . _[ )

C s-c -~------" A

Figure 2.1: The incircle of a ri ght triangle

9 ------

On each side of the triangle, there is an excircle tangent to the side, and the

extensions of the remaining two sides. The radii of these circles are given in Propo

sition 1. I (c).

Figure 2.2: The incircle and excircles of a right triangle

Hansen [1] has proved the following interesting theorem.

Theorem 2.1. (I) The sum of the four radii is equal to the perimeter of the right

triangle:

T a + Tb + T c + T =a+ b +C. (2.1)

(2) The sum of the squares of the four radii is equal to the sum of the squares of the sides of the right triangle:

(2.2)

10 Instead of proving Theorem 2.1 here, we prove a simple characterization of the right triangles in terms of the inradius and exradii.

Proposition 2.1. The following statements for a triangle ABC are equivalent.

(l)rc= S.

(2) Ta = S- b.

(3) rb = 8- a.

(4) r = 8 - c.

(5) C is a right angle.

Proo.f By the formulas for the exradii and the Heron formula, each of (1 ), (2), (3),

( 4) is equivalent to the condition

(s- a)(8- b) = 8(8 - c). (2.3)

2 2 Assuming (2.3), we have 8 - (a+ b)s + ab = s - cs, (a+ b- c)s = ab, 2 2 2 2 2 (a+ b- c)(a + b +c) = 2ab, (a+ b) - c = 2ab, a + b = c . This shows that each of( l), (2), (3), (4) implies (5). The converse is clear. D

2.2 Integer relations from Pythagorean triangles

Recall that every integer right triangle has side lengths

a= k(p 2 - q 2 ), b = k · 2pq , c = k(p 2 + q 2 ),

where p and q are relatively prime positive integers of different parity. Note that with k = 1, we obtain the primitive Pythagorean triangles. With these, we have

ra = kp(p- q) , rb = kq(p + q), rc = kp(p + q), r = kq(p- q).

It is clear that the relations (2.1) and (2.2) can be realized by primitive Pythagorean

triangles.

11 Theorem 2.2. If p and q are positive integers, then the two sets of numbers

(i) a= p2 - q2, b = 2pq, c = p 2 + q2, and

(ii) Ta = p(p- q), rb = q(p + q), Tc = p(p + q), r = q(p- q) satL\fy (2.1) and (2.2).

Here are the data generated by taking relati ve ly prime integers ]J, q ::; 10 w ith different parity.

Table I : Integers with equal sums and sums of squares from Pythagorean triangles

2 1 3 4 5 2 3 6 1 12 50 3 2 5 12 13 3 10 15 2 30 338 4 1 15 8 17 12 5 20 3 40 578 4 3 7 24 25 4 21 28 3 56 1250 5 2 21 20 29 15 14 35 6 70 1682 5 4 9 40 41 5 36 45 4 90 3362 6 1 35 12 37 30 7 42 5 84 2738 6 5 11 60 61 6 55 66 5 132 7442 7 2 45 28 53 J5 18 63 10 126 5618 7 4 33 56 65 21 44 77 12 154 8450 7 6 13 84 85 7 78 91 6 182 144 50 8 1 63 16 65 56 9 72 7 144 8450 8 3 55 48 73 40 33 88 15 176 10658 8 5 39 80 89 24 65 104 15 208 15842 8 7 15 112 113 8 10.'5 120 7 240 25538 9 2 77 36 85 63 22 99 14 198 14450 9 4 65 72 97 45 52 117 20 234 18818 t--9 8 17 144 145 9 136 153 8 306 42050 10 1 99 20 101 90 11 110 9 220 20402 10 3 91 60 109 70 39 130 21 260 23762 10 7 51 1.40 149 30 11 9 170 21 340 44402 10 9 19 180 181 10 171 190 9 380 65522

For example, from the primitive Pythagorean triangle (:3 , 4, 5), we have

2 + 3 + 6 + 1 =3 + 4 + 5, 22 + 32 + 62 + 12 =32 + 42 +52.

12 2.3 Generalization of Hansen's theorem

We seek to generalize Hansen's theorem to an arbitrary triangle, by repl acing a, b, c by appropriate quantities whose sum and sum of squares are respectively equal to those of ra , rb , rc and r . Now, for a ri ght triangle ABC with ri ght angle vertex

C, this latter vertex is the orthocenter of the triangle, which we genericall y denote by H . Note that a = B H and b = AH.

On the other hand, the hypotenuse being a diameter of the circumcircle, c = 2R.

Note also that CH = 0 since C and H coincide. This suggests that a proper generali zation of Hansen's theorem is to replace the triple a, b, c by the quadruple AH, B H , C H and 2R. The quantities AH , B H , C H should be regarded as signed lengths.

Lemma 2.1. AH = 2 · OD = 2Rcos A.

Figure 2.3: Circumcenter and orthocenter

We shall establish the following theorem.

13 ------

Theorem 2.3. Let ABC be a triangle with orthocenter Hand circumradius R.

(I) Ta + Tb + Tc + r = AI-I+ BH + CH + 2R; 2 2 2 2 2 (2) r~ + r~ + r~ + r = AJ-! + BJ-! + CI-! + (2R) .

Figure 2.4: Two qu adruples with equal sums and equal sums of squares

Thus, we have two quadruples, namely, Ta, Tb, Tc, r, and AI-I, BH, CH, 2R,

with equal sums and equal sums of squares, given in the multigrade equations (1)

and (2) above. See Figure 2.4. The proof of Theorem 2.3 will be given in §2.5

below.

14 2.4 Some triangle geometry

As a preparation for the proof of Theorem 2.3, we study the excircles in relation to the circumcircle and the incircle. We establish a basic result, Proposition 2.4, below. 1 Consider triangle ABC with its circumcircle ( 0). Let the bisector of angle A intersect the circumcircle atM. Clearly, M is the midpoint of the arc BMC. The

line BNI clearly contains the incenter I and the excenter Ia.

Proposition 2.2. Af B = M I = MIa = l\fC.

Figure 2.5: Incenter and excenter

1 Proposition 2.2 and the statement of Proposition 2.4 can be found in [2, pp.l85- 193]. An outline proof of Proposition 2.3 can be found in [4, §2.4. 1]. Propositions 2.3 and 2.4 can also be found in [5, §4.6. 1]. We present a unified detailed proof of these propositions here, simpler and more geometric than the trigonometric proofs outlined in [2].

15 -~------~------

Proof It is enough to prove that M B = MI. This follows by an easy calculation

of angles. (i) L. BI M = ~(A+ B), (ii) L.BM I = L. BM A = L.BCA =C. It follows that

(iii) L. IB/11 = 180° - ~( A + B) - C =~(A+ B )= L. BIA1, and

(iv) ME= ~MI . Note that

(v) L. I Bia = 90° since the two bisectors of angle B are perpendicular to each other;

(vi) L. MBia = 90° - ~(A+ B)= ~C ; (vii) L.Mi aB = L. IM B - L.NIBia = C- ~C =~C .

It follows that M Ia = M B = Af I, and M is the midpoint of I I a.

The same reasoning shows that JvfC = M I = 111 Ia as well. D

Figure 2.6: I' I a = 2R

16 ------

Now, let I' be the intersection of the line IO and the perpendicular from I a to

IJC. See Figure 2.6. Note that this latter line is parallel to OM. Since 1M is the

midpoint of I I a, 0 is the midpoint of I I'. It follows that I' is the reflection of I in

0. Also, I' I a = 2 ·OM= 2R. Similarly, I' Ib = I' Ic = 2R. We summarize this

in the following proposition.

Proposition 2.3. The circle through the three excenters has radius 2R and center I', the reflection of I in 0.

Proposition 2.4. Ta + rb + r c = 4R + T.

Figure 2.7: Ta + Tb + T c = 4R + T

Proof The line Iai' intersects BC at the point X' of tangency with the excircle.

Note that I' X'= 2R- Ta. Since 0 is the midpoint of II', we have IX+ I' X'=

17 2 · OD. From this, we have

2 · OD = T + (2R- Ta)· (2.4)

Consider the excenters hand Ic. Since the angles hBic and IbCic are both right angles, the four points h. Ic, B, Care on a circle, whose center is the mid point N of hie. The center N must lie on the perpendicular bisector of BC, which is the line 0111. Therefore N is the antipodal point of 111 on the circumcircle, and we have 2ND = rb + Tc. Thus, 2(R + OD) = Tb + Tc . From (2.4), we have r 0. +rb+rc =4R+r. D

2.5 Proof of Theorem 2.3

(I) Since AH = 2 · OD by Lemma 2.1, we express it in tenns of R, rand ra by using (2.4); similarly for BH and C H :

AH = 2R + T - Ta , BH = 2R + T- Tb, C H =2R+r-rc·

From these,

AH + BH + CH + 2R =8R + 3r- (ra + rb + rc)

=2(4R + r) + T- (ra + rb + rc)

=2(ra + Tb + rc) + r - (ra + Tb + rc)

(2) This follows from simple calculation making use of Proposition 2.4.

AH2 + BH2 + CH 2 + (2R)2

=(2R + r- ra) 2 + (2R + r- rb) 2 + (2R + r- rc) 2 + 4R2

2 2 =3(2R + r) - 2(2R + r)(ra + rb + rc) + r~ + r~ + r~ + 4R

2 2 =3(2R + r) - 2(2R + r )(4R + r) + 4R + r~ + r~ + r~

=r2 + r~ + r~ + r~.

18 ------

2.6 Converse of Hansen's theorem

In this section we prove a strong converse of Hansen's theorem (Theorem 2.4 be-

low ).

Proposition 2.5. A triangle ABC satisfies

(2.5)

if and only 1f it contains a right angle.

Proof Since AH = 2R cos A and a = 2R sin A (law of sines) etc., we have

2 =4R2(cos2 A+ cos2 B + cos2 C + 1 - sin2 A- sin2 B- sin C)

=4R2 (2 cos2 A+ cos 2B +cos 2C)

=8R2(cos2 A+ cos(B +C) cos(B- C))

=- 8R2 cos A(cos(B +C)+ cos(B- C))

=- 16R2 cos A cos B cos C.

By Theorem 2.3(2), the condition (2.5) holds if and only if AH2 + BH2 + C H 2 + (2R)2 = a2 + b2 + c2. One of cos A, cos B, cos C must be zero from above. This means that triangle ABC contains a right angle. 0

2 Lemma 2.2. (1) rarb + rbrc + rcra = s . 2 2 2 2 (2) r~ + r~ + r ~ + r = (4R + r) - 2s + r .

(3) ab + be + ca = s 2 + ( 4R + r )r. 2 (4) a2 + b2 + c2 = 2s - 2(4R + r)r.

19 ------

Proof (l) follows from the formulas for the exradii and the Heron formula.

62 62 62 TaTb + TbTc + T cTa = · + + ----- (8- a)(s- b) (s- b)(s- c) (s- c)(s- a) =s((s- c)+ (s-a)+ (s- b))

From this (2) easily follows.

Again, by Proposition 2.4,

4R+r

=Ta + Tb + Tc 6 6 6 =--+--+-- s-a s-b s-c 6 (s _ a)(s _ b)(s _c) ((s- b)(s- c)+ (s -- e)(s- a)+ (s- a)(s- b))

2 =~ (3s - 2(a + b + e)s + (ab +be+ ca)) r 1 2 =- ( ( ab + be + ea) - s ) . r

An easy rearrangement gives (3). 2 2 (4) follows from (3) since a 2 + b2 + c = (a+ b + c) - 2(ab +be+ ea) =-~ 2 4s 2 - 2(s2 + (4R + r)r) = 2s - 2(4R + r)r. 0

Proposition 2.6. r~ + r~ + r~ + r 2 = a2 + b2 + c2 if and only ij2R + r = s.

Proof By Lemma 2.2(2) and (4), r~ + rl + r ~ + r 2 = a 2 + b2 + c2 if and only if (4R+r)2 -2s2 +r2 = 2s2 -2(4R+r)r;4s2 = (4R+r?+2(4R+r)r+r2 =

2 2 (4R + 2r) = 4(2R + r) ; s = 2R + r. 0

20 Theorem 2.4. The following statements for a triangle ABC are equivalent.

( I) r a + Tb + T c + r = a+ b + c.

(2) r~ + T~ + T~ + r2 = a2 + b2 + c2.

(3) R + 2r = s. ( 4) One of the angles is a right angle.

Proof ( 1) ==> (3): This follows easily from Proposition 2.4.

(3) -¢=:=;> (2): Proposition 2.6 above.

(2) -¢=:=;> ( 4 ): Proposition 2.5 above.

( 4) ==> ( I ): Theorem 2.1 (l ). 0

2.7 Integer relations from Heron triangles

We find triangles with integer sides for which ra, rb, 'l "c , r, AH, BH, CH, and 2R are integers. Since the area of the triangle is given by 6 = rs, this is an integer. 2

For a Heron triangle, the radii Ta,, rb, rc, rand the lengths AH, BH, CH, 2R are not necessaril y integers. We li st those primitive Heron triangles (a , b, c) with sides ::=; 100 whose associated length s are all integers.

Table 2: Integers with equal sum s and sums of squares from Heron triangles

b T c r J AH BH CH 2R I 25 33 .52 11 15 110 6 60 56 - 39 65 25 39 56 12 20 ] 05 7 60 52 -33 65 13 40 51 4 13 156 3 84 75 - 68 85 35 44 75 11 14 231 6 120 117 - 100 125 68 75 77 55 66 70 21 51 40 36 85 40 51 77 21 28 132 11 75 68 -36 85 25 52 63 14 35 90 9 60 39 ·-16 65 13 68 75 6 39 130 5 84 51 - 40 85 17 87 100 6 34 255 5 144 116 - 105 145

2 Assuming a. b. c and r integer, it is clear that s is a multiple of a half-integer. Thus. 6 is rational. It is wel l known that if an integer triangle has rational area, it must have integer area.

21 For example, from the Heron triangle (23, 33, 52), we have

11 + 15 + 11 0 + 6 = 60 + 56+ ( - 39) + 65,

2 2 2 2 2 2 2 11 2 + 15 + 110 + 6 =60 + 56 + ( - 39) + 65 .

Most of these triangles are obtuse angled, except (68, 75, 77) from whi ch we have the identities

55 + 66 + 70 + 21 = 51 + 40 + 36 + 85,

2 2 2 2 2 2 2 2 55 + 66 + 70 + 21 = 51 + 40 + 36 + 85 .

A routine computer search gives all acute Heron triangles with sides ::; 1000 whose associated length s are all integers. We li st these in Table 3.

22 Table 3: Integers with equal sums and sums of squares from acute Heron triangles

I a b T c T I AH B H c H 2R I 68 75 77 55 66 70 21 51 40 36 85 105 116 141 78 91 154 33 100 87 24 145 148 153 175 119 126 170 45 111 104 60 185 111 175 176 77 165 168 40 148 60 57 185 123 187 200 8.5 16.5 204 44 164 84 45 20.5 212 225 247 171 190 234 65 159 140 96 26.5 253 260 31.5 198 207 322 77 204 195 80 32S 195 280 323 133 228 :357 68 260 165 36 325 275 292 3.57 210 231 374 85 240 219 76 365 125 312 323 76 285 340 .51 300 91 36 325 255 385 416 176 336 429 91 340 180 87 425 325 388 483 230 299 546 105 360 291 44 485 267 400 437 184 345 456 95 356 195 84 445 377 404 4!:!5 286 319 522 117 336 303 100 505 319 444 455 231 406 435 110 360 185 1f)6 481 327 455 544 221 357 624 112 436 300 33 545 436 455 513 351 378 494 133 327 300 184 545 339 493 560 232 408 G09 119 452 276 75 565 .soo 527 585 403 442 558 153 375 336 220 625 525 548 667 406 435 690 161 440 411 156 685 296 595 621 189 540 644 115 .555 204 100 629 435 644 715 299 546 759 154 580 333 120 725 - 595 676 837 434 527 918 180 600 507 116 845 447 700 713 310 651 690 161 596 255 216 745 692 703 825 555 570 814 209 519 504 260 865 203 715 720 117 693 728 88 696 120 85 725 697 788 975 510 615 1066 221 696 591 140 985 ~· 795 899 310 651 1015 174 765 424 60 901 325 819 836 198 770 855 B3 780 208 123 845 715 851 876 555 759 814 230 624 420 36.') 949 533 875 888 364 820 861 195 756 300 259 925 820 897 935 663 782 858 2.53 615 496 420 1025 665 935 984 476 836 969 23 1 780 420 287 1025 943 952 975 805 820 861 276 576 561 520 1105 ··--

23 Chapter 3

Triangles with centroid on the in circle

In this chapter we study triangles with centroid on the incircle. Specifi call y, we study the fo ll owing two consttuction problems.

(A) Given a point X on a segment BC, to construct a triangle AB C whose incircle touches BC at X and contain s the centroid.

(B) Given two length s band c, to construct a triangle with sides a, b, c whose incirc le contain s the centroid.

From the equation of the incircle, we easil y establish the following necessary and suffi cient condition.

Proposition 3.1. The centroid of a triangle lies on its incircle if and only if

2 2 2 5(a + b + c )- 6(ab +be+ ca) = 0. (3.1 )

An isosceles triangle with b = c sati sfi es this condition if and only if (a -

2/J)(5a. - 2b) = 0, i.e., a : b = 2 : 1 or 2 : G. Since a : b = 2 : 1 gives a degenerate triangle, the only isosceles triangle with centroid on the incircle is a : b : c = 2 : 5 : 5.

24 ------

Figure 3. L: Isosceles with centroid on incircle

3.1 From s- bands- c

In the remainder of this thesis, we shall write

u =s-a, v = s- b, ·w=s-c

for convenience. In terms of v, v, w, the equation of the incircle becomes

2 2 2uv + 2vw + 2wu - v? - 1; - w = 0.

From (1.2), this condition is equivalent to

(.;u + vv + vw)( -JU + vv + vw)( JU- JV + JW)( JU + vv- vw) = o,

i. e., JU± v'v± Vw = 0. (3.2)

25 ------· -.

This gives the following simple characterization of triangles with centroid on

the incircle.

Proposition 3.2. The centroid of triangle (a , b, c) lies on its incircle if and only if

JS-=0: ± Vs- b ± vs=c = 0.

More explicitly, from (3.2) above, we have

2 1J, = ( Fu ± vw) = v + 'W ± 2/VW.

Construction 3.1. Let X be a point on a segment BC with midpoint 0.

Figure 3.2: Construction of u from v and w

(I) Construct the circle with diameter BC. Let H H' he the diameter perpendicular toBC. (2) Let X' be the reflection of X in 0.

26 ------..

(3) Construct the perpendicular to BC at X , intersecting the circle at P on the same side of H .

(4) Construct the line through X' parallel to PH (respectively PH', to intersect

the perpendicular in (3) at J( (respectively K' ).

Lemma 3.1. ff B X = s - b and C X = s - c, then X K = a - 2 J (s - b) ( s - c) and XK' = a+ 2y1(s- b)(s- c).

Proof Since X' I< and H P are parallel, and X X' = 2 · X 0,

XK = 2(} - X P) = a- 2y'(S= b)(s - c). The proof for X J(' is similar. v· D H

Figure 3.3: Triangle with centroid on incircle

Construction 3.2. (I) Construct the circle, center X (K ) to intersect D C at Y' and Z' (so that Y' is bet1veen X and C, and Z' is between X and B).

(2) Construct the circles B (Y') and C ( Z') to intersect at a point A. Then ABC is a triangle whose centroid lies on the incircle.

27 Repeating the above constmction with I< replaced by I<', we obtain another triangle with centroid on the incircle.

Figure 3.4: Triangle with centroid on incircle

3.2 From b and c

Given equation (3.1) we solve for a in terms of band c:

2 2 5a - 6(b + c)a + (5li- 6bc + 5c ) = 0. (3.3)

This equation has real solutions if and only if

This requires 3 - V5 b 3+V5 --2 - <-<----c- 2 ·

28 In other words, ~ :S ~ :S v/, where tp = 1+2v'5 is the golden ratio. To facilitate constructions, we rewrite (3.3) as

a2 - -(6h+ c)o. + (1-( b +c)2 + -(4b- c)2 ) =0. 5 5 5

Construction 3.3. Given a segment AB, let C0 be a point on the lin e AB such 2 that AB = c and AGo = b satisfy ~ 2 :S ~ :S tp . (I) Construct a square on BCo and let J\1 be the midpoint of the side opposite to B Co. (2) Construct a square on Alvf and a semicicle inside this square with AA1 as diameta

(3) Join A to the midpoint of the side perpendicular to AM at 111, to intersect the semicircle at P.

Figure 3.5: AP2 = -1 (b + c) 2 + ~-( b - c) 2 5 ,)

29 Proof Since AGo b, AB = c, and M is "vertically" above the midpoint of EGo, AM2 = cJ ;c)2+(b-c)2.

In the ri ght triangle AP M, AP : PM = 2 : 1, it follows that AP : AM = 2 :

15, and

Construction 3.4. ( I) Construct a circle through P with center on the perpendic ular to AP at A, and radius ~ ( b + c). (2) Construct the diameter XY through A.

Lemma 3.3. The lengths of AX and AY satisfy the equation ( 3.1).

Figure 3.6: Solution of (3.3)

30 Construction 3.5. ( 1) Construct the circle A ( C'o)- (2) Construct a circle, center B. radius AX to intersect the circle in (I) at C. (3) Construct a circle, center B, radius AY to intersect the circle in ( 1) at C'. ABC and ABC' are triangles with centroid on the incircle.

Figure 3.7: Triangles with centroid on the incircle

31 3.3 Triangle with a tnedian trisected by the incircle

In the next chapter we shall study, among other things, triangles with a median bisected by its incircle. The analogous problem of the incircle trisecting a median has a simple and unique solution which we can easily work out here.

If triangle ABC has one of its medians, say the A-median, trisected by its incircle, then its centroid must lie on the incircle. The other trisection, by Lemma

1.2, has homogeneous barycentric coordinates ( 4, 1, 1). It follows that

2 2 2 5a - 6a(b +c) + 5b - 6bc + 5c =0,

2 2 2 8a - 12a(b + c) + 5b + 6bc + 5c =0.

2 Upon subtraction, 3a - 6a(b +c)+ 12bc = 0; 3(a- 2b)(a- 2c) = 0. This

2 2 2 means that a= 2b, or 2c. If a= 2b, then from 5(a + b + c ) = 6(ab +be+ ca), we have (b - c)(13b - 5c) = 0. While b - c = 0 gives a degenerate triangle with a : b : c = 2 : 1 : 1, from 13b - 5c = 0, we have a : b : c = 10 : 5 : 13.

B D c

Figure 3.8: Triangle with a median trisected by the incircle

Pt·oposition 3.3. If the A-median of triangle ABC is trisected by the incircle, then

a : b : c = 10 : 5 : 13 or 10 : 13 : 5.

32 Chapter 4

Triattgles with midpoi11t of 011e cevian on incircle

In this chapter we study triangles ABC with its incircle containing the midpoint of one of the following cevians:

(A) a median,

(B) an angle bisector,

For example, if triangle ABC is isosceles with AC = AB, then its A-median,

A-bisector, and A-altitude are the same. It is easy to see that such a triangle must

sati sfy a : b : c = 2 : 3 : 3. See Figure 4. I.

We shall study each of these cases as a construction problem.

( I) Given band c, to construct, if possible, a triangle ABC whose incircle

contains the midpoint of its A-median, A-bisector, or A-altitude. (2) Given a segment BC with a point X in it, to construct, if possible, a tri angle ABC whose incircle touches BC at X, and contains the midpoints of its A-median, A-bisector, or A-altitude.

We shall also study integer triangles with these properties.

33 A

Figure 4. I : Isosceles triangle with median bi sected by incircle

4.1 Construction of triangle ABC with midpoint of A median on inch·cle

The midpoint of the A-median has coordinates (2 : 1 : 1). It lies on the incircle if and only if 2 2 3a? - 4a(b +c) + 2(b + c ) = 0. (4.1)

In term s of u , v , w, this condition becomes

2 2 4u - 4'U(v + w) + (v- w) = 0. (4.2)

4.1.1 From s - b and s - c

The equation (4.2) can be easily solved:

1 2 1 a 11, = -(JV ± vw) =- (v + w ± 2,/UW) =- + ..;vw. 2 2 2

34 Construction 4.1. Let X be a point on a segment DC with midpoint D.

( 1) construct a semicircle with diameter BC.

(2) Construct the perpendicular to BC at X, intersecting the semicircle at P.

(3) On this perpendicular, choose two points f{ and K' each at a distance ~fro m P.

K Figure 4.2: Triangle with median bisected by incircle

Construction 4.2. (I) Construct the circle X (K) to intersect BC at Y' and Z' (so that Y' is between X and C, and Z' is between X and B).

(2) Construct the circles B(Y') and C(Z') to intersect at a point A. Then ABC is a triangle whose incircle contains the midpoint of the A-median.

See Figure 4.2.

Repeating the above construction with K replaced by K', we obtain another

triangle with the same prope1ty. See Figure 4.3.

35 Y'

Figure 4.3: Triangle with median bisected by incircle

4.1.2 From band c

Given band c, the equation ( 4. 1) has real solutions in a if and only if

This is equivalent to

Construction 4.3. Given a segment AB of length c and a point Co on the half line

AB such that AGo = b satisfying the above condition, (1) let D be the midpoint of BCo, and construct a square AD EF: (2) construct the segment BE and a line through B making an angle 30° with BE;

(3) construct the perpendicular to BE atE to intersect the line in (2) at K.

36 Construction 4.4. (I) Construct the perpendicular to B J( at B.

(2) Construct a circle of radius ~ (b + c), with center on the lin e in (1), and passing thro11gh K. Let the circle and the line intersect at two points X and X'.

Then the lengths of BX and BX' sati:;fy equation (4.1). See Figure 4.4.

Figure 4.4: Solution of (4.1)

Construction 4.5. ( I ) Construct the circle A( Co) .

(2) Construct the circle B(X) to intersect the circle in ( 1) at a point C. Triangle ABC has the midpoint of its A-median on the incircle. See Figure 4.5.

37 Figure 4.5: Triangle with a median bisected by incircle

Repeat the same construction with X replaced by X'. This gives another trian gle ABC' with the same property. See Figure 4.6.

Figure 4.6: Triangle with a median bisected by incircle

38 4.2 Construction of triangle ABC with midpoint of A bisector on incircle

We consider the construction of triangle ABC whose incircle contains the mid point of the A-angle bisector.

The bisector of angle A divides BC at a point D such that flD : DC = c : b. The point D has homogeneous barycentric coordinates 0 : b : c with reference to triangle ABC. In absolute barycentric coordinates, D = b~ !~·C . The midpoint 111 of AD has absolute barycentric coordinate A+D (b + c)A + b · B + c · C 2 2(b +c) Its homogeneous barycentric coordinates are b + c : b : c.

A(b+c) \ I \ \ M \ \ \ \ B (b) D (b+c) C (c)

Figure 4.7: Barycentric coordinates of midpoint of angle bisector

Proposition 4.1. The incircle of triangle ABC contains the midpoint of the A angle bisector if and unly if

(4.3)

Proof Substituting the homogeneous barycentric of M into the equation of the incircle, and simplifying, we obtain

Since b + c- a > 0, we must have (4.3) above. D

39 4.2.1 From b and c

Given b and c, the triangle with incircle containing the midpoint of the A-angle bisector has its third side a given by

3 3 2 2 b + c . b - be + c a= = (b +c)· 2 b2 + be + c2 b2 + be + c · By the law of cosines, we have

/} + c2 _ a 2 b4 + 2b3c + b2c2 + 2bc3 + c4 2b2c2 cos A = bc 1 2 ( b2 + be + c2 ) 2 = - ( b2 + be + c2 ) 2 ·

Since cos A = 1 - 2 sin2 4, we conclude that

A be sin -= . . (4.4) 2 b2 + be+ c2

Figure 4.8: Construction of 4

40 Construction 4.6. Let A. Co, Ao be collinear points with AGo = b, CoAo = c, and AAo = b + c. (I) Construct an equilateral triangle CoAoX.

(2) Let Bo be a point on the segment AAo such that ABo = c.

(3) Construct an equilateral triangle AoBoY such that X andY are on opposite sides of AAo.

2 2 2 PI·oposition 4.2. ( l) XY = IJ +be+ c .

(2) AoXY, BoAY, and CoX A are congruent triangles. (3) AXY is an equilateral triangle. (4) A, X, Ao, andY are concyclic. See Figure 4.8.

Proof (I) It follows from the law of cosines since L X AoY = 120°.

(2) The three triangles in question each has a pair of sides band c, with included angle 120°. (3) From the congruence of these triangles we have XY = AY = X A.

(4) AXA0Y is a cyclic quadrilateral since L XAY + L XA0Y = 180°. D

Corollat·y 4.1. Let P be the intersection of AAo and XY. Then, sin 4 = ~~0::.

Proof Let Q and N be the orthogonal projections of A and Ao on XY respec tively. SincesiuXAoY = sinXAY =~ .from (4.4),

A [XA0 Y] AoN AoP siu 2 = [Y AX] AQ AP. 0

Construction 4.7. (1) Construct the circle with diameter AP.

(2) Construct the circle P (Ao) to intersect the circle in (l) at Hand K. Angle H AK satisfies (4.4) and is bisected by the line AAo. See Figure 4.8.

41 Figure 4.9: Triangle with an angle bisector bisected by incircle

Construction 4.8. ( 1) Construct a point B on Ali such that AB = AGo = c and a point C on Af( such that AC = ABo = b.

The incircle of triangle ABC contains the midpoint of its A-angle bisector. See Figure 4.9.

4.2.2 An alternative construction

If we write tan() = __ r..fijC then 2 2 vb +c '

1 - tan2 fJ a = ( b + c) · --- - = ( b + c) cos 2(). 1 +ta2n ()

This leads to the following construction.

42 B A

Figure 4. I 0: Triangle with an angle bisector bisected by incircle

Construction 4.9. Let B. A, Co be three collinear points such that ACo = band AB =c. (I) Construct a semicircle with diameter BCo. (2) Construct the perpendicular to BCo at A to intersect the semicircle at P.

(3) On the ray GoA choose a point Q such that AQ = .JFJ2+ c2. (4) Join P, Q and construct a parallel through B to PQ.

(5) Reflect the ray AB in the line constructed in (4), to intersect the semicircle at X.

(6) Construct the circles B (X) and A(Co) to intersect at a point C.

Triangle ABC has the property that its incircle contains the midpoint of the bisector of angle A. See Figure 4.1 0.

4.2.3 ·From s - b and s - c

The condition (4.3) becomes

3 u - 3uv'W- V'W(v + w) = 0. (4.5)

43 if we put a = v + w, b = w + u, and c = u + v. It is clear that 11. cannot be constructed from v and w using ruler and compass. See Theorem 1.2. However, writing v = p3 and w = q3 , we have

which decomposes into

From thi s we see that the cubic equation has exactly one real root

l 1 1 I u = pq (p + q) = v j w 3 ( v 3 + w 3 ) .

Proposition 4.3. Let ]J and q be positive integers. The triangle with sides

has its incircle containing the midpoint of the A-angle bisect01:

The length of the A-bisector is ;!~ cos 4 = 2pq }P2+q 2 . By choosing p = 3, q = 4, we obtain the triangle

a = 91 , b = 111 , c = 148 with length of A-angle bisector 120.

4.3 Construction of triangle ABC with midpoint of A altitude on incircle

The homogeneous barycentric coordinates of the otthocenter H are P"+c\ _ 2 0 c2 + a\_62 : a2 + ~Lc2 as given in § 1.4. 1. From this, we obtain the midpoint of the A-altitude as

44 Lemma 4.2. The midpoint of the A-altitude lies on the incircle if and only if

3 2 2 3a - a?(b +c)+ a(b - c) - 3(b- c) (b +c) = 0. (4.6)

Proof Substituting the coordinates of 111 in the equation of the incircle, simplifying, we have

Since b + c- a is nonzero, the result follows. 0

4.3.1 From b and c

Proposition 4.4. The construction of triangle ABC with given band c, and mid point of A-altitude on incircle is in general impossible with ruler and compass.

Figure 4.11: Triangle with two altitudes bisected by incircle

45 Apart from the isosceles triangle with a : b : c = 2 : 3 : 3 in Figure 4.1, we

seek an isosceles triangle with a = b satisfying (4.6). With a = b, this becomes

(4a - 3c)c2 = 0, and we have a : b: c = 3 : 3 : 4. Naturally, the midpoint of the B-altitude is also on the incircle. With reference to the triangle, the midpoints of

the A- and B-altitudes are 111 = 9 : 1 : 8 and N = 1 : 9 : 8, and the equation of the incircle is 2 2 2 4yz + 4zx + 8x y - 4.T - 4y - z = 0.

lt is interesting to note that the orthocenter H = 1 : 1 : 8 also lies on the incircle.

See Figure 4.11 . •• 4.3.2 From s - b and s - c

In terms of v = s- band w = .s - c, equation (4.6) becomes

vw(v + w) u- (4.7) - v 2 - vw + w 2 ·

Lemma 4.3. In each of the following cases, ~ + i = ~ (a) The segments of lengths p, q, r are parallel.

(b) In a right triangle with shorter sides p and q, the inscribed square has side length r.

(b) Figure 4.12: 1 + 1. = 1 p q T

In particular, if squares are constructed on the same side of two consecutive

46 segments as in Figure 4. 13 below, then

1 1 1 - +- = -. V W X

v 0 w

Figure 4. 13: ~ + t, = ~

Given v and w, to construct a length u satisfying ( 4 . 7), we rewrite the condition

as 2 2 1 v - vw + w 1 1 3 =-+ ----. (4.8) u vw(v+w) v w v+w Repeated applications of Lemma 4.3(b) give, as in F igure 4. 14,

1 1 1 - + -- =-, y v+w x 1 1 1 -+-- =-, z v+w y 1 1 1 - + -- =-. u v+w z

From these we have

1 3 1 1 1 - + --=- =- + -, ·u v+w :c v w

which is equivalent to (4.8) above.

47 u

Figure 4. 14: Construction of u from v and w ,. Construction 4.10. Let X be a point on a segment BC such that BX v and CX=w.

(1) Construct points Bo and Co on the line BC such that X Bo = X Co = 'U given

by (4.8). (2) Construct an intersection A on C(Bo) and B(Co). Triangle ABC has its A -altitude bisected by the incircle. See Figure 4.15.

Figure 4.15: Triangle with an altitude bisected by incircle

48 Bibliography

[I] D. W. Hansen, On inscribed and escribed circles of right triangles, circum scribed triangles, and the four-square, three-square problem, Mathematics •• Teacher, 96 (2003) 358- 364 .

[2] R. A. Johnson, Advanced Euclidean Geometry, Dover reprint, 1960.

[3] N. D. Kazarinoff, Ruler and the Round, Classic Problems in Geometric Con structions, Dover reprint, 2003.

[41 P. Yiu, Euclidean Geometry, Florida Atlantic University Lecture Notes, 1998.

[5] P. Yiu, Introduction to the Geometry of Triangle, Florida Atlantic University Lecture Notes, 2001.