International Journal of Geometry, Vol. 1, No. 1, 2012

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INT

V

Edited by

“Vasile Alecsandri

RNATIONAL

OURNAL

OF

EOMETRY

l. 1 No. 1 2012

Department of Mathematics

National College of Bacău, Romania



CONTENTS

VOL. 1 (2012) No. 1

GHEORGHE SZÖLL½OSY and OVIDIU T. POPA new proof of Neuberg’s theorem and one application .........................5-9

ION P ¼ATRASCU and C ¼AT ¼ALIN BARBUTwo new proofs of Goormaghtigh’s theorem ..........................................10-19

NICUSOR MINCULETESeveral geometric inequalities of Erdös - Mordell type in the

convex polygon ..............................................................................20-26

MIRCEA SELARIU, FLORENTIN SMARANDACHE and MARIAN NITU

Cardinal functions and integral functions ............................................27-40

PETRU BRAICA and ANDREI BUDA generalization of the isogonal point ..................................................41-45



INTERNATIONAL JOURNAL OF GEOMETRY

Vol. 1 (2012), No. 1, 5 - 9

A NEW PROOF OF NEUBERG’S THEOREM

AND ONE APPLICATION

GHEORGHE SZÖLL½OSY and OVIDIU T. POP

Abstract. In this paper, we will give a new proof of Neuberg’s Theorem.One application of this result is found in Theorem 2.2.

1. Introduction

In this section, we’ll recall some known results.

Theorem 1.1. Let ABC be a triangle, C (O; R) his circumscribed circle

and H the orthocentre of this triangle. If M is the midpoint of the side BC ,

then

(1) AH = 2OM

and

(2) OM ? BC:

For additional comments and proofs, see [1]-[4].In this paper, we will consider a convex quadrilateral ABCD and note

with H a, H b, H c, H d the orthocentres of triangles BC D, CDA, DAB,respectively ABC and with T [ABCD] the area of quadrilateral ABCD:

Theorem 1.2. If ABCD is a cyclic quadrilateral, then

(3) AB H aH b; ABkH aH b;

(4) AH b ? CD; BH a ? CD;

(5) AH b BH a

and analogues.

For comments and proofs see [3].

————————————–Keywords and phrases: Convex quadrilateral, area(2010)Mathematics Subject Classi…cation: 51M04, 51M25, 51M30



6 Gheorghe Szöll½osy and Ovidiu T. Pop

Corollary 1.3. If ABCD is a cyclic quadrilateral, then

(6) T [ABCD] = T [H aH bH cH d]:

Proof. It results from Theorem 1.2.

Corollary 1.4. Let ABCD be a convex quadrilateral. If AH b BH a,

BH c CH b, CH d DH c or DH a AH d, then ABCD is a cyclic quadri-

lateral.

Proof. Let’s consider the relation AH b BH a.Let C (O1) and C (O2) be thecircumscribed circles of the triangle ADC , respectively BC D and M themidpoint of DC . Taking Theorem 1.1 into account, we have that

AH b = 2O1M; BH a = 2O2M;

and

O1M ? DC;O2M ? DC;

from where, O1 coincide to O2. So, the quadrilateral ABCD is cyclic.

2. Main results

In this section, the main results are proved by using the analytic geometry.

The result from Theorem 2.1 is known in literature as Neuberg’s Theorem(see [4] or [6]) and Theorem 2.2 is an application of this.

Theorem 2.1. If M is a point in the same plan with the triangle ABC ,

M 62 AB[BC [CA and H a, H b, H c are the orthocentres of triangles M BC ,

M CA, respectively M AB, then

(7) T [ABC ] = T [H aH bH c]:

Proof. We consider a;b;c;; 2 R, a > 0, b < c and the points A(O; a),B(b; 0), C (c; 0) and M (; ) (Figure 1). From M 62 BC , it results that 6= 0. The equation of the line AB is

ax + by ab = 0;

and then, from M 62 AB it results that

a + b ab 6= 0:

Similarly, from M 62 AC it results that

a + c ac 6= 0:

Taking the remarks above, from M 62 AB [ BC [ CA we obtain that

(8) (a + b ab)(a + c ac) 6= 0:



A new pro of of N eub erg’s theo rem and one applicatio n 7

Writing the corresponding heights equations, after the calculations, we ob-tain

(9) H a

;2 + (b + c) bc

;

(10) H b

(c a + a2)

a + c ac;

(a b + b2)

a + b ab

and

(11) H c

(b a + a2)

a + b ab;

(a b + b2)

a + b ab

:

If = 0, then from M 62 AB [ CA it results that bc 6= 0. In this case

H a

0;

bc

, H b

a

c; 0

, H c

a

b; 0

and then T [H aH bH c] = jj,

where

=1

2

xH a yH a 1xH b

yH b1

xH c yH c 1

;

H a(xH a ; yH a) and analogues. We have that

=1

2a(b c);

so

T [H aH bH c] = T [ABC ]

and then, in this case the theorem was proved. If 6= 0, we note

E = (a + c ac)(a + b ab)

and then

(12) E = a2

2

+ a( a)(b + c) + bc( a)2

:



8 Gheorghe Szöll½osy and Ovidiu T. Pop

Taking (8) and (12) into account, we have that E 6= 0.Then

=1

2E

2 + (b + c) bc

(c a + a2) (a c + c2) a + c ac

(b a + a2) (a b + b2) a + b ab

and multiplying the …rst line with and dividing the second column with, we obtain

=1

2E

2 2 + (b + c) bc

c a + a2 a c + c2 a + c ac

b a + a2 a b + b2 a + b ab

:

Adding the …rst column to the second column, we obtain

= 12E

2 (b + c) bc

c a + a2 a2 + c2 a + c ac

b a + a2 a2 + b2 a + b ab

:

Deducting the third line from the second line, we get

=c b

2E

2 (b + c) bc

c + b a

b a + a2 a2 + b2 a + b ab

:

We multiply the second line with and adding it to the …rst line, and wemultiply the second line with b and adding it to the third line, then weobtain

=c b

2E

0 bc a

b + c aa( a) a2 bc a

:

Deducting the …rst line from the third line and taking that

T [ABC ] =a(c b)

2;

we have

=c b

2E

0 bc a

b + c aa( a) a2 0

=

1

E T [ABC ]

0 bc a

b + c a( a) a 0

=1

E T [ABC ]E = T [ABC ]:

Now, the relation (7) is proved.

Theorem 2.2. If ABCD is a convex quadrilateral, then

(13) T [ABCD] = T [H aH bH cH d]:

Proof. Any three from the points H a; H b; H c; H d are not collinear. Evenif H aH bH cH d is a concave quadrilateral, it has at least one diagonal situ-ated on the surface of the quadrilateral ABCD, for example [H bH d] (Figure

2).Applying Theorem 2.1 for the triangle ABD and for the point C , for the



A new pro of of N eub erg’s theo rem and one applicatio n 9

triangle CB D and for the point A respectively, we obtain that

T [ABD] = T [H aH bH d];

respectivelyT [CB D] = T [H cH bH d]:

From the equalities above, (13) follows.

Remark 2.3. Taking Theorem 2.2 into account, it results that (13) takesplace also in quadrilateral that are not cyclic quadrilaterals.

References

[1] Barbu, C., Fundamental Theorems of Triangle Geometry (Romanian), Ed. Unique,Bac¼au, 2008.

[2] F. G.-M., Exercices de géométrie , Huitième édition, Paris, 1931.[3] Golovina, L. I. and Iaglom, I. M., Induction in Geometry (Romanian), Ed. Tehnic¼a,

Bucharest, 1964.[4] Miculita, M. and Brânzei, D., Analogies Triangle-Tetrahedron (Romanian), Ed.

Paralela 45, 2000.[5] Mihalescu, C., The Geometry of Remarkable Elements (Romanian), Ed. Tehnic¼a,

Bucharest, 1957.[6] Neuberg, J., Mathesis (1881), 168.[7] Pop, O. T., Minculete, N. and Bencze, M., An Introduction to Quadrilateral Geometry

(to appear).

Received: December 16, 2011.

AVRAM IANCU 28E, 435500 SIGHETU MARMATIEI, ROMANIAE-mail address : [email protected]

NATIONAL COLLEGE "MIHAI EMINESCU",MIHAI EMINESCU 5, 440014 SATU MARE, ROMANIA

E-mail address : [email protected]




Vol. 1 (2012), No. 1, 10 - 19

TWO NEW PROOFS OF

GOORMAGHTIGH’S THEOREM

ION P ¼ATRASCU and C ¼AT ¼ALIN I. BARBU

Abstract. In this note we present two new demonstrations of the theo-rem of a Belgian mathematician René Goormaghtigh.

1. Introduction

In order to state our main results we need recall some important theoremsthat we need in proving the Goormaghtigh’s theorem. Consider a triangleABC is neither isosceles rectangular nor with circumcenter O. We presentbelow an interesting proposition given by Goormaghtigh.

Theorem 1.1. (Goormaghtigh [7;pp: 281 283]): Let A0

; B0

; C 0

be points on OA;OB;OC so that

OA0

OA=

OB0

OB=

OC 0

OC = k;

k 2 R

+; then the intersections of the perpendiculars to OA at A0, OB at B0,and OC at C 0 with the respective sidelines BC;CA;AB are collinear.

R. Musselman and R. Goormaghtigh are given in [7] a proof of this theo-rem using complex numbers. A synthetic demonstration is also given by K.Nguyen meet in [9].

Theorem 1.2. (Kariya [3; p: 109]): Let C a; C b; C c the points of tangency

of the incircle with the sides BC;CA;AB of triangle ABC and I center of the incircle. On the lines IC a; IC b; IC c the points A0; B0; C 0 are considered in the same direction so that IA0 = IB0 = IC 0. Then the lines AA0; BB 0;

and CC 0 are concurrent.

Theorem 1.3. (Desargues [5; p: 133]): Two triangles are in axial perspective if and only if they are in central perspective.

————————————–Keywords and phrases: Goormaghtigh’s theorem; Miquel’s point;

complete quadrilateral(2010) Mathematics Subject Classi…cation: 26D05; 26D15; 51N35



Two new dem onstrations of Go orm aghtigh’s theorem 11

Theorem 1.4. (Miquel [6; pp: 233 234]): The centers of the circles of the four triangles of a complete quadrilateral are on a circle. (Miquel’s Circle).

Theorem 1.5. (Steiner [6; p: 235]): Miquel’s point of the circles determined by the four triangles of a complete quadrilateral is situated on Miquel’s circle.

Theorem 1.6. (Sondat [11; p: 10]): If two triangles ABC and A0B0C 0 are perspective and orthologic, then the center of perspective P and the ortho-logic centers Q and Q0 are on the same line perpendicular to the axis of perspectivity d:

Theorem 1.7. (Thébault [12; pp: 2224]): If two triangles ABC and A0B0C 0

are perspective and orthologic, with the center of perspective P and the or-thologic centers Q and Q0; then the conics ABCPQ and A0B0C 0P Q0 are equilateral hyperbolas.

Theorem 1.8. (Brianchon - Poncelet [4; pp: 205220]): The centers of all equilateral hyperbolas passing through the vertices of a triangle ABC lie on the Euler circle of the triangle.

2. Main results

In this section we present two new demonstrations of the theorem of aBelgian mathematician René Goormaghtigh and some consequences deriving

from this theorem.

Solution 1. We noted with A00 the point of intersection of perpendicularstaken at B0 and C 0 on the OB;OC respectively. Similarly we de…ne thepoints B00 and C 00 (see Figure 1).



12 Ion P¼atrascu and C¼at¼alin I. Barbu

Since OA = OB = OC , from the relation

OA0

OA =OB0

OB =OC 0

OC = k;

we get OA0 = OB0 = OC 0: Because the lines OA0; OB0; OC 0 are perpendicu-lar on B00C 00; C 00A00; and A00B00 respectively, then the point O is the incenterof the triangle A00B00C 00: Applying theorem 2 in the triangle A00B00C 00 for thepoints A ; B ; C , it results that the lines AA00; BB 00 and CC 00 are concurrent(at one of Kariya’s points which corresponds to A00B00C 00 triangle), then tri-angles ABC and A00B00C 00 are homological. Thus, according to theorem 3,that the points of intersection of lines AB and A00B00; BC and B00C 00, andCA and C 00A00 are collinear.

Denote by X the intersection of the lines BC and B00C 00. Similarly wede…ne the points Y and Z:

Solution 2. Without restricting the generality suppose that \BCA >

\ABC . Let us designate by R the radius of the circle triangle ABC , by A1

intersection of the tangent in A to circumcircle of the triangle ABC withthe line CB , by T and X 0 the projections of the points B and X on thistangent, by M and M 0 the projections of points A0 and O; respectively, withthe line BT ; and by A0

1 the intersection of BC and OM 0 (see Figure 2).

We have: \CAA1 = \ABC; \ACA1 = \BAC + \ABC and

\AA1B = 180 \BAC 2 \ABC

= \BC A \ABC;\COA0

1 = 2 \ABC 90:

Applying the law of sines in the triangle OCA1; we have

A0

1C

sin(2B 90)=

OC

sin(C B);

so

A0

1

C =

R cos2B

sin(C B):




Because XX 0 = AA0 = OA OA0 = R(1 k); then

(1) XA1 = XX 0

sin(C B)= R(1 k)

sin(C B)

From OA0

OA=

A0

1X

A0

1A1

= k, we get

(2)A0

1X

XA1

=k

1 k

From relations (1) and (2) we get

(3) A0

1X =k

1 k

R(1 k)

sin(C B)=

kR

sin(C B)

Since,

(4) XC = XA0

1 + A0

1 =R(k cos2B)

sin(C B)

Because \M 0OB = 2\ACB90; BM 0 = BO sin(2C 90) = R cos2C;

M M 0 = OA0 = kR; then BM = BM 0 + M M 0 = R(k cos2C ): Since

(5) XB =BP

sin(C B)=

R(k cos2C )

sin(C B)

From relations (4) and (5) we get

XBXC

= k cos2C k cos2B

:

Similarly it is shown that

Y C

Y A=

k cos2A

k cos2C

andZA

ZB=

k cos2B

k cos2A:

We obtain thatXB

XC Y C

Y A ZA

ZB = 1

and from the converse of Menelaus’s theorem results that points X;Y; andZ are collinear.

Theorem 2.1. Let us consider C 1; C 2; C 3 and { the circumcircles of the triangles AY Z; BZX;CXY; and respectively ABC: The circles C 1; C 2; C 3;

and { pass through a common point.

The proof results from theorem 5.

Let P be the corresponding point of Miquel complete quadrilateral ABXY CZ

(see Figure 3).




Theorem 2.2. Centers of circles C 1; C 2; C 3; { and point P are on the same circle @:

The proof results from theorems 4 and 5.

Theorem 2.3. Let us consider C 01; C 02; and C 03 the circumcircles of the tri-angles A00Y Z; B00ZX; and respectively C 00XY . The circles C 01; C 02; and C 03pass through a common point.

The proof results from theorem 5.

Let us designate by Oa; Ob; Oc; O0

a; O0

b; O0

c the circumcenters of the trian-gles AY Z; BZX; CXY; A00Y Z; B00ZX; C 00XY; respectively, by Q the point

of Miquel of A00

C 00

XZB00

Y complete quadrilateral (see Figure 4).




Open problems:

1) Point Q is on circle @:2) Point O is on Aubert’s line of complete quadrilaterals Y ABXCZ and

XZA00C 00B00Y:

Remark 2.1. Goormaghtigh’s theorem is true for k < 0, where !OA0 = k

!OA;

!OB0 = k

!OB;

!OC 0 = k

!OC; the demonstration is similar.

Remark 2.2. Points A00; B00 and C 00 are on the perpendicular bisectors of the sides of triangle ABC , therefore the triangles ABC and A00B00C 00 are bilogical, O is a common center of orthology.

Remark 2.3. If k = 0 Goormaghtigh’s theorem remains true as a special

case of Bobillier’s theorem [10; p:119]:




Remark 2.4. For k = 1

2we obtain Ayme’s theorem [2]:

Remark 2.5. For k = 1 we obtain Lemoine’s theorem and XY Z is Lemoine’s line of the triangle ABC [3; p:155]:

Remark 2.6. Theorem 4 is true for any transversal XY Z which cuts the sides of triangle ABC , the demonstration remains the same.

Remark 2.7. Because OaO0

a; ObO0

b; OcO

0

c are the perpendicular bisectors of

the segments Y Z; Z X ; and XY respectively, then OaO0

akObO0

bkOcO

0

c:

Remark 2.8. The triangles ABC and A00B00C 00 are perspective, XY Z being the axis of perspective. Let S be the perspective center of triangles ABC and A00B00C 00.

Theorem 2.4. The lines OS and XY Z are perpendicular.

The proof results by Sondat’s theorem (see Figure 5).

Theorem 2.5. The conics ABCSO and A0B0C 0SO are equilateral hyper-bolas.

Proof. Because the circumcenter O is the common center of orthology, byTheorem 1.7 we obtain the conclusion.

Corollary 2.6. The centers of the conics ABCSO and A0B0C 0SO lie on the Euler circles of the triangles ABC , respectively A0B0C 0 :

The proof results from Theorems 1.8 and 2.5 (see Figure 5).




Corollary 2.7. The points P ans Q are the focus of parabolas tangent to the sides of the complet quadrilaterals ABXY CZ and A00C 00XZB 00Y ,respectively :

(see [1, p. 109], Figure 5).




3. Dynamic properties

In this section we examine the dependence of considered con…guration onhomothety coe¢cient k. Firstly formulate

Lemma 3.1. Given two points A, B. The map f transforms the lines passing through A to the lines passing through B and conserve the cross-ratios of the lines. Then the locus of points l \ f (l) is a conic passing through A and B.

Indeed if X , Y , Z are three points of the thought locus, then lines l andf (l) intersect the conic ABXY Z in the same point.

Lemma 3.1 has also a dual formulating: if f is a projective map betweenlines a and b then the envelop of lines Af (A) is a conic touching a and b.

Using Lemma 3.1 we obtain that the envelop of lines XY Z from Theorem1.1 is a parabola touching the sidelines of ABC , and the locus of perspec-tivity centers from Theorem 1.2 is the Feuerbach hyperbola.

Theorem 3.1. Point P from Theorem 2.1 is …xed.

Proof. Immediately follows from Lemma 3.1 and Corollary 2.7.

Theorem 3.2. The locus of points Q is a line passing through O.

Proof. Using polar transformation with center O we obtain from Theorem2.5 that two parabolas from Corollary 2.7 are homothetic. Thus all pointsQ are the foci of homothetic parabolas.

Acknowledgement . The last section of the article was created by Mr.Alexey Zaslavsky, and the authors are grateful to him!.

References

[1] Akopyan, A. V. and Zaslavsky, A. A., Geometry of Conics , American MathematicalSociety, 2007, 109.

[2] Ayme, J. L. Geometry , http://pagesperso-orange.fr/jl.ayme/ [3] Barbu, C., Fundamental Theorems of Triangle Geometry (Romanian), Ed Unique,

Bac¼au, 2008, 109.[4] Brianchon, C. J. and Poncelet, J. V., Recherches sur la détermination d’une hyperbole

équiatère , Gergonne’s Annales de Math. 11(1820/21), 205-220.[5] Court, N., College Geometry , Johnson Publishing Company, New York, 1925, 133.[6] Mihalescu, C., Remarkable Elements of Geometry (Romanian), Ed. SSMR,

Bucharest, 2007, 233-235.[7] Musselman, J. R. and Goormaghtigh, R., Advanced Problem 3928, Amer. Math.

Monthly, 46(1939), 601; solution, 48(1941), 281-283.[8] Neuberg, V., Mathesis , 1922, 163.[9] Nguyen, K. L., A Synthetic Proof of Goormaghtigh’s and Generalization of Mussel-

man’s Theorem , Forum Geometricorum, 5(2005), 17-20.[10] Nicolescu, L. and Bosco¤, V., Practical Problems of Geometry (Romanian), Technical

Publishing House, Bucharest, 1990, 119.[11] Sondat, P., L’intermédiaire des mathématiciens , 1894, 10 [question 38, solved by

Sollerstinsky, 94].[12] Thébault, V., Perspective and Orthologic Triangles and Tetrahedrons , Amer. Math.

Monthly, 59(1952), 24-28.




Received: January 12, 2012.

FRATII BUZESTI COLLEGEION CANTACUZINO 15, S33, SC. 1, AP. 8, CRAIOVA, ROMANIAE-mail address : [email protected]

VASILE ALECSANDRI COLLEGEIOSIF COCEA 12, SC. A, AP. 13, BAC ¼AU, ROMANIAE-mail address : [email protected]




Vol. 1 (2012), No. 1, 20 - 26

SEVERAL GEOMETRIC INEQUALITIES OF

ERDÖS - MORDELL TYPE IN THE CONVEX POLYGON

NICUSOR MINCULETE

Abstract. In this paper we present the several geometric inequalities of Erdös-Mordell type in the convex polygon, using the Cauchy Inequality.

1. Introduction

In [6], in colaboration with A. Gobej, we present some geometric inequal-ities of Erdös-Mordell type in the convex polygon. Here, we found othersgeometric inequalities of Erdös-Mordell type, using several known inequali-ties, in the convex polygon.

Let A1,A2;...,An the vertices of the convex polygon, n 3; and M , apoint interior to the polygon. We note with Rk the distances from M tothe vertices Ak and we note with rk the distances from M to the sides[AkAk+1] of length AkAk+1 = ak; where k = 1; n and An+1 A1. For

all k 2 f1;:::;ng with An+1 A1 and m\ AkMAk+1

= k we have the

following property: 1 + 2 + ::: + n = 2:

L. Fejes Tóth conjectured a inequality which is refered to the convex polygon,recall in [1] si [3], thus

(1)n

Xk=1

rk

cos

n

n

Xk=1

Rk:

In 1961 H.-C. Lenhard proof the inequality (1), used the inequality

(2)nX

k=1

wk cosn

nXk=1

Rk;

which was established in [5], where wk the length of the bisector of theangle AkMAk+1, (8) k = 1; n with An+1 A1.

————————————–Keywords and phrases: Inequality of Erdös-Mordell type, Cauchy’s

Inequality(2010) Mathematics Subject Classi…cation: 51 Mxx,26D15



Several geometric inequalities of Erdös-Mordell type in the convex polygon 21

M. Dinc¼a published other solution for inequality (1) in Gazeta Matematic ¼ a Seria B in 1998 (see [4]). Another inequality of Erdös-Mordell type for

convex polygon was given by N. Ozeki [9] in 1957, namely,

(3)nY

k=1

Rk

sec

n

n nYk=1

wk

which proved the inequality 16.8 from [3] due to L. Fejes-Tóth, so

(4)nY

k=1

Rk

sec

n

n nYk=1

rk:

R. R. Janic in [3], shows that in any convex polygon A1A2:::An, there is theinequality

(5)nX

k=1

Rk sinAk

2 nX

k=1

rk:

D. Busneag proposed in GMB no. 1/1971 the problem 10876, which is aninequality of Erdös-Mordell type for convex polygon, thus,

(6)nX

k=1

ak

rk 2 p2

;

where p is the semiperimeter of polygon A1A2:::An and is the area of polygon.

In connection with inequality (6), D. M. B¼atinetu established [2] the in-

equality(7)

nXk=1

ak

rk 2 p

r

if the polygon A1A2:::An is circumscribed about a circle of radius r.Among the relations established between the elements of polygon A1A2:::An we

can remark the following relation for - the area of convex polygon A1A2:::An:

(8) 2 = a1r1 + a2r2 + ::: + anrn:

We select several inequalities obtained from [6]:

(9) Rk rk1 + rk

2sin Ak

2

hold for all k 2 f1; 2;:::;ng ; with r0 = rn;

(10)

2cos

n

n nYk=1

Rk nY

k=1

(rk1 + rk) ; (r0 = rn)

and

(11)nX

k=1

rk1 + rk

Rk

2n cos

n

and

(12)n

Xk=1

R2k

rk sec

n

n

Xk=1

Rk:



22 Nicusor Minculete

2. Main results

First, we will follow some procedures used in paper [6], through whichwe will obtain some Erdös-Mordell-type inequalities for the convex polygon.Among these will apply the Cauchy Inequality

Theorem 2.1. In any convex polygon A1A2:::An, there is the inequality

(13)nX

k=1

rk

Rk + Rk+1

2n cos

n:

Proof. The inequality (11),nXk=1

rk1 + rk

Rk

2n cos

n;

with r0 = rn; is expanded in the following way,rn + r1

R1

+r1 + r2

R2

+ ::: +rn1 + rn

Rn

=

r1

1

R1

+1

R2

+ r2

1

R2

+1

R3

+ ::: + rn

1

Rn

+1

R1

2n cos

n

On the other hand, we have1

Rk1+

1

Rk

4

Rk1 + Rk; (8) k = 1; n

with R0 = Rn; from where we can deduce another inequality, of an Erdös-Mordell type, namely,

nXk=1

rk

Rk + Rk+1

n

2cos

n:

Theorem 2.2. For any convex polygon A1A2:::An,we have the inequality n

Xk=1

Rk + Rk+1

rk 2n

cosn

2

;

with Rn+1 = R1:

Proof. The inequality

nXk=1

xkyk nX

k=1

xk

yk

nXk=1

xk

!2

is well known, because it is a particulary case of Cauchy’s inequality. In thiswe will take xk = rk

Rk+Rk+1and yk = 1. Thus, the inequality becomes

n

Xk=1

rk

Rk + Rk+1

n

Xk=1

Rk + Rk+1

rk n2




and, if we use the inequality (13), we getnX

k=1

Rk

+Rk

+1rk

2n

cos n

2

;

with Rn+1 = R1:

Theorem 2.3. In any convex polygon A1A2:::An, there is the inequality

(15)nX

k=1

p rk1rk

Rk

n cos

n:

Proof. From Cauchy’s inequality, we haven

Xk=1xkyk n

Xk=1xk

yk

n

Xk=1 xk!2

:

Using the substitutions

xk =

p rk1rk

Rk

and yk = 1, we deduce that the inequalitynX

k=1

p rk1rk

Rk

nX

k=1

Rkp rk1rk

n2;

holds. However, from the relation (11), we obtainn

Xk=1p rk1rk

Rk

n cos

n

which implies the inequalitynX

k=1

Rkp rk1rk

n

cos

n

;

with r0 = rn:

Remark 1. The inequality (15) generalizes the problem 1045 of G. Tsinsifas from the magazine Crux Mathematicorum. This is also remarked in [7].

Theorem 2.4. In any convex polygon A1A2:::An there is the inequality

(16)

nXk=1

R2kp rk1rk

n

cos2

n;

with r0 = rn:

Proof. In the inequalitynX

k=1

xkyk nX

k=1

xk

yk

nXk=1

xk

!2

we replaced xk = yk = Rkp rk1rk

and the inequality becomes

n

n

Xk=1

R2k

p rk1rk n

Xk=1

Rk

p rk1rk!2

n2

cos2

n

:




This means that inequality (16) is true.

Theorem 2.5. In any convex polygon A1A2:::

An

there is the inequality nX

k=1

akrk

RkRk+1

n sin2

n;

with An+1 = A1:

Proof. We can be written the area of triangle AkMAk+1 in two ways, thus

akrk

2=

RkRk+1 sinAkMAk+1

2;

which implies the relationakrk

RkRk+1

= sinAkMAk+1

so, by passing to the sum, we get the relationnX

k=1

akrk

RkRk+1

=nX

k=1

sinAkMAk+1;

with An+1 = A1: Because the function f : (0;1) ! R, is de…ned as f (x) =sinx, is concave, we will apply the inequality Jensen, thus

1

n

nXk=1

sinAkMAk+1 sin1

n

nX

k=1

AkMAk+1

!= sin

2

n:

Therefore, we havenX

k=1

AkMAk+1 n sin 2n:

Consequently, we obtain the inequality of the statement.

Remark 2. The equality hold in the above mentioned theorems when the polygon is regular.

Remark 3. On the a hand, we have the equality (8),

2 = a1r1 + a2r2 + ::: + anrn =nX

k=1

akrk;

and on the other hand, we have the inequality Cauchy, where we will replace

xk = p akrk and yk =q

akrk

,then

2nX

k=1

ak

rk=

nXk=1

akrk

nXk=1

ak

rk

nXk=1

ak

!2

= 4 p2

which proves the inequality (6):


(18)nX

k=1

R2k sin

Ak

2 sec

n

n

nX

k=1

rk

!2

holds.




Proof. Applying inequality (9), we have the inequality

Rk = MAk rk1 + rk

2sin Ak

2(8) k = 1; n

with r0 = rn; and this, by squaring, becomes

4R2k sin

Ak

2 (rk1 + rk)2

sin Ak

2

and taking the sum, we deduce

4nX

k=1

R2k sin

Ak

2

nXk=1

(rk1 + rk)2

sin Ak

2

"nX

k=1

(rk1 + rk)2

#n

Xk=1

sin Ak

2

4

nX

k=1

rk

!2

n cos

n

so, we found inequality (18).


(19)nX

k=1

(Rk + Rk+1) rk 2sec

n

n

nX

k=1

rk

!2

holds.

Proof. From inequality (9), we have

Rk = MAk

rk1 + rk

2sinAk

2

; (

8) k = 1; n

with r0 = rn; and this, by multiply with (rk1 + rk) ,becomes

2 (rk1 + rk)Rk (rk1 + rk)2

sin Ak

2

and by passing to the sum, we obtain the relation

2nX

k=1

(rk1 + rk)Rk nX

k=1

(rk1 + rk)2

sin Ak

2

"nX

k=1

(rk1 + rk)

#2n

Xk=1 sin Ak

2

4

nX

k=1

rk

!2

n cos

n

:

Therefore, we havenX

k=1

(rk1 + rk)Rk 2sec

n

n

nX

k=1

rk

!2

:

But, it follows thatnX

k=1

(Rk + Rk+1) rk =nX

k=1

(rk1 + rk)Rk

which means that, we obtain the inequality of statement.




References

[1] Abi-Khuzam, F. F., A Trigonometric Inequality and its Geometric Applications ,Mathematical Inequalities & Applications, 3(2000), 437-442.

[2] B¼atinetu, D. M., An Inequality Between Weighted Average and Applications (Ro-manian), Gazeta Matematic¼a seria B, 7(1982).

[3] Botema, O., Djordjevic R. Z., Janic, R. R., Mitrinovic, D. S. and Vasic, P. M.,Geometric In equalities , Groningen,1969.

[4] Dinc¼a, M., Generalization of Erdös-Mordell Inequality (Romanian), Gazeta Matem-atic¼a seria B, 7-8(1998).

[5] Lenhard, H.-C., Verallgemeinerung und Verscharfung der Erdös-Mordellschen Ungle-ichung für Polygone , Arch. Math., 12(1961), 311-314.

[6] Minculete, N. and Gobej, A.,Geometric Inequalities of Erdös-Mordell Type in a Con-vex Polygon (Romanian), Gazeta Matematic¼a Seria A, nr. 1-2(2010).

[7] Minculete, N., Geometry Theorems and Speci…c Problems (Romanian), Editura Eu-rocarpatica, Sfântu Gheorghe, 2007.

[8] Mitrinovic, D. S., Peµcaric, J. E. and Volenec, V., Recent Advances in Geometric Inequalities , Kluwer Academic Publishers, Dordrecht, 1989.

[9] Ozeki, N., On P. Erdös Inequality for the Triangle , J. College Arts Sci. Chiba Univ.,2(1957), 247-250.

[10] Vod¼a, V. Gh., Spell Old Geometry (Romanian), Editura Albatros,1983.

"DIMITRIE CANTEMIR" UNIVERSITY107 BISERICII ROMÂNE, 500068 BRASOV, ROMANIAE-mail address : [email protected]




Vol. 1 (2012), No. 1, 27 - 40

CARDINAL FUNCTIONS AND

INTEGRAL FUNCTIONS

MIRCEA E. SELARIU, FLORENTIN SMARANDACHE andMARIAN NITU

Abstract. This paper presents the correspondences of the eccentricmathematics of cardinal and integral functions and centric mathematics,or ordinary mathematics. Centric functions will also be presented in theintroductory section, because they are, although widely used in undulatoryphysics, little known.

In centric mathematics, cardinal sine and cosine are de…ned as well as theintegrals. Both circular and hyperbolic ones. In eccentric mathematics, allthese central functions multiplies from one to in…nity, due to the in…nity of possible choices where to place a point. This point is called eccenter S (s; ")

which lies in the plane of unit circle UC(O; R = 1) or of the equilateral unityhyperbola HU(O; a = 1; b = 1). Additionally, in eccentric mathematics thereare series of other important special functions, as aex, bex, dex, rex,etc. If we divide them by the argument , they can also become cardinaleccentric circular functions, whose primitives automatically become integraleccentric circular functions.

All supermatematics eccentric circular functions (SFM-EC) can be of vari-able excentric , which are continuous functions in linear numerical eccen-tricity domain s 2 [1; 1], or of centric variable , which are continuous forany value of s. This means that s 2 [1; +1].

————————————–Keywords and phrases: C-Circular , CC- C centric, CE- C Eccentric,

CEL-C Elevated, CEX-C Exotic, F-Function, FMC-F Centric Mathemat-ics, M- Matemathics, MC-M Centric, ME-M Excentric, S-Super, SM- SMatematics, FSM-F Supermatematics FSM-CE- FSM Eccentric Circulars,FSM-CEL- FSM-C Elevated, FSM-CEC- FSM-CE- Cardinals, FSM-CELC-FSM-CEL Cardinals

(2010) Mathematics Subject Classi…cation: 32A17



28 M. Selariu, F . Sma randache and M. N itu

1. INTRODUCTION: CENTRIC CARDINAL SINE FUNCTION

According to any standard dictionary, the word "cardinal" is synonymouswith "principal", "essential", "fundamental".

In centric mathematics (CM), or ordinary mathematics, cardinal is, onthe one hand, a number equal to a number of …nite aggregate, called thepower of the aggregate, and on the other hand, known as the sine cardinalsinc(x) or cosine cardinal cosc(x), is a special function de…ned by the centriccircular function (CCF). sin(x) and cos(x) are commonly used in undulatoryphysics (see Figure 1) and whose graph, the graph of cardinal sine, which iscalled as "Mexican hat" (sombrero) because of its shape (see Figure 2).

Note that sinc(x) cardinal sine function is given in the speciality litera-ture, in three variants

sinc(x) =

(1; for x = 0

sin(x)

x; for x 2 [1; +1] n 0

(1)

=sin(x)

x= 1 x2

6+

x4

120 x76

5040+

x8

362880+ 0[x]11

=+1Xn=0

(1)nx2n

(2n + 1)!! sinc(

2) =

2

;

d(sinc(x))

dx=

cos(x)

x sin(x)

x2= cosc(x) sinc(x)

x;

(2) sinc(x) =sin(x)

x

(3) sinca(x) =sin(xa )

xa

It is a special function because its primitive, called sine integral and denotedSi(x)

Centric circular cardinalsine functions

Modi…ed centric circularcardinal sine functions

Figure 1: The graphs of centric circular functions cardinal sine, in 2D, as

known in literature



Cardinal functions and integral functions 29

Figure 2: Cardinal sine function in 3D Mexican hat (sombrero)

Si(x) = Z x

0

sin(t)

tdt = Z

x

0sinc(t) dt(4)

= x x3

18+

x5

600

x7

35280+

x9

3265920+ 0[x]11

= x x8

3:3!+

x5

5:5! x7

7:7!+ : : :

=+1Xn=0

(1)nx2n

(2n + 1)2(2n)!; 8x 2 R

can not be expressed exactly by elementary functions, but only by expansionof power series, as shown in equation (4). Therefore, its derivative is

(5) 8x 2 R; Si0

(x) =d(Si(x))

dx =sin(x)

x = sinc(x);

an integral sine function Si(x), that satis…es the di¤erential equation

(6) x f 000

(x) + 2f 00

(x) + x f 0

(x) = 0 ! f (x) = Si(x):

The Gibbs phenomenon appears at the approximation of the square witha continuous and di¤erentiable Fourier series (Figure 3 right I). This op-eration could be substitute with the circular eccentric supermathematicsfunctions (CE-SMF), because the eccentric derivative function of eccentricvariable can express exactly this rectangular function (Figure 3 N top) orsquare (Figure 3 H below) as shown on their graphs (Figure 3 J left).

1 cos x=2p 1sin(x=2)2

;

fx;; 2:01g

12 4xPSi nc[2(2k1)x];

fk; ngfn; 5gfx; 0; 1g




12dex[(;

2 ); S (1; 0)]

Gibbs phenomenon fora square wave with n = 5

and n = 10

Figure 3: Comparison between the square function, eccentric derivative

and its approximation by Fourier serial expansion

Integral sine function (4) can be approximated with su¢cient accuracy.The maximum di¤erence is less than 1%, except the area near the origin.By the CE-SMF eccentric amplitude of eccentric variable

(7) F () = 1:57 aex[; S (0:6; 0)];

as shown on the graph on Figure 5.

Sin R x;

fx;

20; 20

g

Sin

R (x + Iy)

fx;

20; 20

g;

fy;

3; 3

g

Figure 4: The graph of integral sine function Si(x) N compared with thegraph CE-SMF Eccentric amplitude 1; 57aex[; S (0; 6; 0)] of eccentric

variable H




Figure 5: The di¤erence between integral sine and CE-SMF eccentricamplitude F () = 1; 5aex[; S (0; 6;0] of eccentric variable

2. ECCENTRIC CIRCULAR SUPERMATHEMATICS CARDINALFUNCTIONS, CARDINAL ECCENTRIC SINE (ECC-SMF)

Like all other supermathematics functions (SMF),they may be eccentric(ECC-SMF), elevated (ELC-SMF) and exotic (CEX-SMF), of eccentric vari-able , of centric variable 1;2 of main determination, of index 1, or secondarydetermination of index 2. At the passage from centric circular domain tothe eccentric one, by positioning of the eccenter S (s; ") in any point in the

plane of the unit circle, all supermathematics functions multiply from one toin…nity. It means that in CM there exists each unique function for a certaintype. In EM there are in…nitely many such functions, and for s = 0 one willget the centric function. In other words, any supermathematics functioncontains both the eccentric and the centric ones.

Notations sexc(x) and respectively, Sexc(x) are not standard in the lit-erature and thus will be de…ned in three variants by the relations:

(8) sexc(x) =sex(x)

x=

sex[; S (s; s)]

of eccentric variable and

(8’) Sexc(x) =Sex

(x

)x =Sex

[; S

(s; s

)]

of eccentric variable .

(9) sexc(x) =sex(x)

x;

of eccentric variable , noted also by sexc(x) and

(9’) Sexc(x) =sex(x)

x=

Sex[; S (s; s)]

;

of eccentric variable , noted also by Sexc(x)

(10) sexca(x) =sexx

a

xa

=sex

;




of eccentric variable , with the graphs from Figure 6 and Figure 7.

(10’) Sexa(x) =

Sex xa

xa

=

Sex aa

aa

Sin ArcSin [s Sin ()]

fs; 0; +1g; f;; 4gSin

ArcSin [s Sin ()]

fs;1; 0g; f;; g

Sin ArcSin [s Sin ()]

fs; 0; 1g; f;; gFigure 6: The ECCC-SMF graphs sexc1[; S (s; ")] of eccentric variable

Sin +ArcSin [s Sin ()]

fs; 0; 1g; f;4; 4gSin

ArcSin [0.1s Sin ()]

fs;10; 0g; f;; g

Sin ArcSin [0.1s Sin ()]

fs; 0:1; 0g; f;; gFigure 7: Graphs ECCC-SMF sexc2[; S (s; ")]; eccentric variable




3. ECCENTRIC CIRCULAR SUPERMATHEMATICSFUNCTIONS CARDINAL ELEVATED SINE AND COSINE

(ECC- SMF-CEL)

Supermathematical elevated circular functions (ELC-SMF), elevated sinesel() and elevated cosine cel(), is the projection of the fazor/vector

~r = rex() rad() = rex[; S (s; ")] rad()

on the two coordinate axis X S and Y S respectively, with the origin in the

eccenter S (s; "), the axis parallel with the axis x and y which originate inO(0; 0).

If the eccentric cosine and sine are the coordinates of the point W (x; y),by the origin O(0; 0) of the intersection of the straight line d = d + [ dae\,revolving around the point S (s; "), the elevated cosine and sine are the samecoordinates to the eccenter S (s; "); ie, considering the origin of the coordi-nate straight rectangular axes XSY /as landmark in S (s; "). Therefore, therelations between these functions are as follows:

(11)

x = cex() = X + s cos(") = cel() + s cos(")y = Y + s sin(") = sex() = sel() + s sin(")

Thus, for " = 0, ie S eccenter S located on the axis x > 0;sex() = sel(),and for " =

2 ;cex() = cel(), as shown on Figure 8.On Figure 8 were represented simultaneously the elevated cel() and the

sel() graphics functions, but also graphs of cex() functions, respectively,for comparison and revealing sex() elevation Eccentricity of the functionsis the same, of s = 0:4, with the attached drawing and sel() are " =

2 , andcel() has " = 0.

Figure 8: Comparison between elevated supermathematics function and

eccentric functions




Figure 9: Elevated supermathematics function and cardinal eccentricfunctions celc(x) J and selc(x) I of s = 0:4

Figure 10: Cardinal eccentric elevated supermathematics functioncelc(x) J and selc(x) I

Elevate functions (11) divided by become cosine functions and cardinalelevated sine, denoted celc() = [; S ] and selc() = [; S ], given by theequations

(12)

8><>:

X = celc() = celc[; S (s; ")] = cexc() s cos(s)

Y = selc() = selc[; S (s; ")] = sexc() s sin(s)

with the graphs on Figure 9 and Figure 10.




4. NEW SUPERMATHEMATICS CARDINAL ECCENTRICCIRCULAR FUNCTIONS (ECCC-SMF)

The functions that will be introduced in this section are unknown in math-ematics literature. These functions are centrics and cardinal functions orintegrals. They are supermathematics eccentric functions amplitude, beta,radial, eccentric derivative of eccentric variable [1], [2], [3], [4], [6], [7] cardi-nals and cardinal cvadrilobe functions [5].

Eccentric amplitude function cardinal aex(), denoted as

(x) = aex[; S (s; ")]; x ;

is expressed in

(13) aexc() = aex(

= aex[; S (s; s)]

= arcsin[s sin( s]

and the graphs from Figure 11.

Sin () ; fs; 0; 1g; f;4; +4g

Figure 11: The graph of cardinal eccentric circular supermathematicsfunction aexc()

The beta cardinal eccentric function will be

(14) bexc() =bex()

=

bex[S (s; s)]

=

arcsin[s sin( s)]

;

with the graphs from Figure 12.

Figure 12: The graph of cardinal eccentric circular supermathematicsfunction bexc() (

fs;

1; 1

g;

f;

4; 4

g)




The cardinal eccentric function of eccentric variable is expressed by

(15) rexc1;2() =rex()

=rex[; S (s; s)]

=s cos( s)

p 1 s2sin( s)

and the graphs from Figure 13, and the same function, but of centric variable is expressed by

(16) Rexc(1;2) =Rex(1;2)

1;2

=Rex[1;2S (s; s)]

1;2

=

p 1 + s2 2s cos(1;2 s)

1;2

sCos()+p 1[sSin()]2 ;

fs; 0; 1g; f;4; 4gsCos()p 1[s Sin()]2

;

fs;1; 0g; f;4; 4gsCos()p 1[sSin()]2

;

fs; 0; 1g; f;4; 4gFigure 13: The graph of cardinal eccentric circular supermathematical

function rexc1;2()

And the graphs for Rexc(1), from Figure 14.

Figure 14: The graph of cardinal eccentric radial circularsupermathematics function Rex c()

An eccentric circular supermathematics function with large applications,representing the function of transmitting speeds and/or the turning speeds of all known planar mechanisms is the derived eccentric dex1;2() and Dex(1;2),

functions that by dividing/reporting with arguments and, respectively,




lead to corresponding cardinal functions, denoted dexc1;2(), respectivelyDexc(1;2) and expressions

(17) dexc1;2() =dex1;2()

=

dex1;2[; S (s; s)]

=

1 scos(")1s2sin2(")

(18)

Dexc(1;2) =Dex(1;2)

1;2=

Dexf[1;2S (s; s)]g1;2

=

p 1 + s2 2s cos(1;2 s)

1;2

the graphs on Figure 15.

Figure 15: The graph of supermathematical cardinal eccentric radialcircular function dex c1()

Because Dex(1;2) = 1dex1;2()

results that Dexc(1;2) = 1dexc1;2()

siq ()

and coq () are also cvadrilobe functions, dividing by their arguments leadto cardinal cvadrilobe functions siqc() and coqc() obtaining with the ex-pressions

(19) coqc() =coq ()

=

coq [S (s; s)]

=

cos( s)

p

1 s2sin2( s)

(20) siqc() =siq ()

=

siq [S (s; s)]

=

sin( s)

p

1 s2cos2( s)

the graphs on Figure 16.

Figure 16: The graph of supermathematics cardinal cvadrilobe functionceqc() J and siqc() I




It is known that, by de…nite integrating of cardinal centric and eccentricfunctions in the …eld of supermathematics, we obtain the corresponding

integral functions.Such integral supermathematics functions are presented below. For zero

eccentricity, they degenerate into the centric integral functions. Otherwisethey belong to the new eccentric mathematics.

5. ECCENTRIC SINE INTEGRAL FUNCTIONS

Are obtained by integrating eccentric cardinal sine functions (13) and are

(21) sie(x) = Z x

0

sexc()

d

with the graphs on Figure 17 for the ones with the eccentric variable x .

Figure 17: The graph of eccentric integral sine function sie 1(x)N andsie 2(x)H

Unlike the corresponding centric functions, which is denoted Sie(x), theeccentric integral sine of eccentric variable was noted sie(x), without thecapital S , which will be assigned according to the convention only for theECCC-SMF of centric variable. The eccentric integral sine function of cen-tric variable, noted Sie(x) is obtained by integrating the cardinal eccentricsine of the eccentric circular supermathematics function, with centric vari-able

(22) Sexc(x) = Sexc[; S (s; ")];




thus

(23) Sie(x) =Z x0

Sex[; S (s; "]

;

with the graphs from Figure 18.

Figure 18: The graph of eccentric integral sine function sie 2(x)

6. C O N C L U S I O N

The paper highlighted the possibility of inde…nite multiplication of car-

dinal and integral functions from the centric mathematics domain in theeccentric mathematics’s or of supermathematics’s which is a reunion of thetwo mathematics. Supermathematics, cardinal and integral functions werealso introduced with correspondences in centric mathematics, a series newcardinal functions that have no corresponding centric mathematics.

The applications of the new supermathematics cardinal and eccentricfunctions certainly will not leave themselves too much expected.

References

[1] Selariu, M. E.,Eccentric circular functions

, Com. I C onferinta National¼a de Vibratiiîn Constructia de Masini, Timisoara , 1978, 101-108.[2] Selariu, M. E., Eccentric circular functions and their extension , Bul .St. Tehn. al I.P.T.,

Seria Mecanic¼a, 25(1980), 189-196.

[3] Selariu, M. E., Supermathematica , C om.VII Conf. International¼a. De Ing. Manag. siTehn.,TEHNO’95 Timisoara, 9(1995), 41-64.

[4] Selariu, M. E., Eccentric circular supermathematic functions of centric variable,

Com.VII Conf. International¼a. De Ing. Manag. si Tehn.,TEHNO’98 Timisoara, 531-548.

[5] Selariu, M. E., Quadrilobic vibration systems , The 11th International Conference on

Vibration Engineering, Timisoara, Sept. 27-30, 2005, 77-82.[6] Selariu, M. E., Supermathematica. Fundaments, Vol.II , Ed. Politehnica, Timisoara,

2007.[7] Selariu, M. E., Supermathematica. Fundaments, Vol.II , Ed. Politehnica, Timisoara,

2011 (forthcoming).




Received: December, 2011

POLYTEHNIC UNIVERSITY OF TIMISOARA, ROMANIAE-mail address : [email protected]

UNIVERSITY OF NEW MEXICO-GALLUP, USAE-mail address : [email protected]

INSTITUTUL NATIONAL DE CERCETARE - DEZVOLTAREPENTRU ELECTROCHIMIE SI MATERIE CONDENSAT ¼ATIMISOARA, ROMANIAE-mail address : [email protected]




Vol. 1 (2012), No. 1, 41 - 45

A GENERALIZATION OF THE ISOGONAL POINT

PETRU I. BRAICA and ANDREI BUD

Abstract. In this paper we give a generalization of the isogonal point interms of concurrency, starting with the ideea of "breaking" the equilateraltriangles constructed on the sides of one triangle, rotating them with thesame acute angle and compressing with the same ratio the broken sides.

1. Introduction

We will denote by E 2 the euclidean space.

De…nition 1.1. For X 2 E 2 …xed and 2 (; ), we call rotation of

center X and angle of Y 2 E 2, the point z = Rx;(Y ), having the followingproperties:

(i) m(]Y XZ ) = ;

(ii)!XY =

!XZ .

De…nition 1.2. Let O 2 E 2 a …xed point and k 2 Rnf0g. We call homothety

with center O and ratio k an application:

H 0;k : E 2 ! E 2 : H 0;k(M ) = M 0

with the following properties:

(1) H 0;k(O) = O;(2) If M 6= 0, then the points O;M ;M 0 are collinear;(3) If k > 0, then M 0 2 [OM , and if k < 0, then O 2 [M M 0];(4) OM 0 = jkj OM .

————————————–Keywords and phrases: Triangle, isogonal point(2010)Mathematics Subject Classi…cation: 51M04, 51M25, 51M30



42 Petru I. Braica and Andrei Bud

2. Main Results

If we consider an arbitrary triangle ABC and the points C a = (H A;k RA;)(B), C b = (H B;k RB;)(A), Ab = (H B;k RB;)(C ), Ac = (H C;k RC;)(B), Bc = (H C;k RC;)(A), Ba = (H A;k RA;)(C ) with k 2 R

and 2 (; ) …xed.

Lemma 2.1. If C 0; B0 are points in the interior of the triangle ABC with

4ABC 0 4ACB 0 and CC 0 \ AB = fC 1g, also BB 0 \ AC = fB1g, then

the following identity is true:

BA0

A0C =

sin C sin(B + y)

sin B sin(C + y);

with ynot= m(]ABC 0) = m(]ACB0).

Proof. We have

B1A

B1C =

S 4BB 0C

S 4BAB 0

=BC B0C sin(C + y)

AB AB0 sin(A + x)

=sin C sin y sin(A + x)

sin A sin x sin(C + y);

where xnot= m(]BAC 0) = m(]CAB0) (see Figure 1):

Similarly, we have

C 1A

C 1B=

S 4ACC 0

S 4CC 0B

=AC 0 AC sin(A + x)

BC 0 BC sin(B + y)=

sin y sin B sin(A + x)

sin x sin A sin(B + y):

Now, using the Ceva’s Theorem, in the triangle ABC , one obtains

A0C

BA0=

B1C

B1A

C 1A

C 1B=

sin B sin(C + y)

sin C sin(B + y):

Theorem 2.1. Let us consider now an arbitrary triangle ABC and the

points C a = (H A;k RA;)(B), C b = (H B;k RB;k)(A), Ab = (H B;k RB;)(C ), Ac = (H C;k RC;)(B), Bc = (H C;k RC;)(A), Ba = (H A;k RA;)(C ), with k 2 R, 2 (; ) …xed. Using this points, de…ned above,

we consider the following intersections: BBc \ CC b = fP ag, AP a \ BC =

fP Ag, BBa \ AAb = fP cg, CP c \ AB = fP C g, CC a \ AAc = fP bg, BP b \



A generalization of the isogonal point 43

AC = fP Bg. The concurrency of the cevians AP A, BP B, CP C takes place

in the point T ;k.

Proof. Applying Lema 2.1 one obtains:

BP AP AC

=sin C sin(B + )

sin B sin(C + )

CP BP BA

=sin A sin(C + )

sin C sin(A + )

AP C P C B

=sin B sin(A + )

sin A sin(B + ):

Now, evaluating the product

BP AP AC

CP BP BA

AP C P C B

= 1

and from the reciprocal of Ceva’s Theorem, we have the concurrency in T ;k

(see Figure 2).

Corrolary 2.1. For = 60 one obtains the …rst Torricelli point, and for

= 60 one obtains the second Torricelli point, replacing k = 1.

Corrolary 2.2. For k = 1=(2cos ), one obtains: C a = C b; Ba = Bc;

Ac = Ab and the ABC a, BAbC , CBaA are similar isosceles triangles and

the Theorem of Kiepert takes place.[1]:

Corrolary 2.3. From Lemma 2:2 one obtaines that the geometrical locus

of the intersection point CC 0 \ BB 0 is the line AA0, where

BA0=A0C =sin C sin(B + y)

sin B sin(C + y)

:



44 Petru I. Braica and Andrei Bud

Theorem 2.2. Let us consider now an arbitrary triangle ABC and the

following points C a = (H A;k RA;)(B), C b = (H B;k RB;)(A), Ab =

(H B;k RB;)(C ),Ac = (H C;k RC;)(B), Bc = (H C;k RC;)(A), Ba =(H A;k RA;)(C ) with k 2 R

, 2 (; ) …xed. Using this points, we

consider the following intersections: fQag = CC a\BBa, fQbg = AAb\CC b,

fQcg = BBc \ AAc, AQa \ BC = fQAg, BQb \ AC = fQBg, CQc \ AB =fQC g. The concurrency of lines AQA, BQB, CQC takes place in the point

T ;k.

Proof. We make the following notations: = m(]C aBA) = m(]C bAB) =m(]BaCA) = m(]BcAC ) = m(]AcBC ) = m(]AbCB).

Applying Lemma 2.1 one obtains:

BQA

QAC =

sin C sin(B + )

sin B sin(C + )

CQB

QBA=

sin A sin(C + )

sin C sin(A + )

AQC

QC B=

sin B sin(A + )

sin A sin(B + ):

Now, applying again the reciprocal of the Ceva’s Theorem, one obtains therequired concurrency in the point T ;k (See Figure 3).

Remark 2.1. The corrolary 2.1, 2.2 and 2.3 takes place also for Theorem2.2.

Remark 2.2. The barycentric coordinates for the points T k; and T k; canremain an open problem.

Remark 2.3. For k = 1, Theorem 2.1 become Theorem 1.1 from [4].



A generalization of the isogonal point 45

References

[1] Barbu, C., Fundamental Theorems of Triangle Geometry (Romanian), Ed. Unique,Bac¼au, 2008.

[2] Nicolescu, L. and Bosko¤, V., Practical Problems of Geometry (Romanian), Ed.Tehnic¼a, Bucuresti, 1990.

[3] Mathematical Gazette (Romanian), Bucharest.[4] Braica, P. and Pop, O. T., An Extension of Torricelli’s Theorem (Romanian).

Received: February 2, 2012.

ELEMENTARY SCHOOL "GRIGORE MOISIL"MILENIULUI 1, 440037 SATU MARE, ROMANIAE-mail address : [email protected]

ICHB BUCURESTI, ROMANIAE-mail address : [email protected]



AIMS AND SCOPE

International Journal of Geometry publishes high quality original re-search papers and survey articles in areas of euclidean geometry, non - euclid-ean geometry and combinatorial geometry. It will also occasionally publish,as special issues, proceedings of international conferences (co)-organized by theDepartment of Mathematics and Computer Science, Vasile Alecsandri NationalCollege of Bacau and Vasile Alecsandri University of Bacau.

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Manuscripts should be written in English, following the style of our journalin what concerns the technical preparation of the papers. The manuscripts mustbe prepared electronically in LATEX macro package, and should be submittedeither in two paper copies and .tex …le on a ‡oppy diskette, or by E-mail (the

.tex-…le will be accompanied by .pdf …le). The style "ijgeometry.sty" and alsoan example of its usage may be found on the web site of the journal (addressbelow). On the title page author should include the title of the article, au-thor’s name (no degrees), author’s a¢liation, e-mail addresses, mailing addressof the corresponding author and running head less than 60 characters. Themanuscript must be accompanied by a brief abstract, no longer than 200-250words. It should make minimal use of mathematical symbols and displayedformulas. Mathematics Subject Classi…cation with primary (and secondary)subject classi…cation codes and a list of 4-5 key words must be given. Biblio-graphic references should be listed alphabetically at the end of the article. Theauthor should consult Mathematical Reviews for the standard abbreviations of

journal names. References should be listed in alphabetical order; the followingreference style should be used:

[1] Andreescu, T. and Andrica, D., Complex Numbers from A to . . . Z ,Birkhauser Verlag, Boston-Berlin-Basel, 2005.

[2] Andrica, D. and Barbu, C., A Geometric Proof to Blundon’s Inequalities ,Mathematical Inequalities & Applications, 15 (2011), 180-192.

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The (…rst) author of each published paper will receive the .pdf reprintof the paper.

Further information about the journal can be found at:http://ijgeometry.com/

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AIMS AND SCOPE

International Journal of Geometry publishes high quality original re-search papers and survey articles in areas of euclidean geometry, non - euclid-ean geometry and combinatorial geometry. It will also occasionally publish,as special issues, proceedings of international conferences (co)-organized by theDepartment of Mathematics and Computer Science, Vasile Alecsandri NationalCollege of Bacau and Vasile Alecsandri University of Bacau.

MANUSCRIPT SUBMISSION

Manuscripts should be written in English, following the style of our journalin what concerns the technical preparation of the papers. The manuscripts mustbe prepared electronically in LATEX macro package, and should be submittedeither in two paper copies and .tex …le on a ‡oppy diskette, or by E-mail (the

.tex-…le will be accompanied by .pdf …le). The style "ijgeometry.sty" and alsoan example of its usage may be found on the web site of the journal (addressbelow). On the title page author should include the title of the article, au-thor’s name (no degrees), author’s a¢liation, e-mail addresses, mailing addressof the corresponding author and running head less than 60 characters. Themanuscript must be accompanied by a brief abstract, no longer than 200-250words. It should make minimal use of mathematical symbols and displayedformulas. Mathematics Subject Classi…cation with primary (and secondary)subject classi…cation codes and a list of 4-5 key words must be given. Biblio-graphic references should be listed alphabetically at the end of the article. Theauthor should consult Mathematical Reviews for the standard abbreviations of

journal names. References should be listed in alphabetical order; the followingreference style should be used:

[1] Andreescu, T. and Andrica, D., Complex Numbers from A to . . . Z ,Birkhauser Verlag, Boston-Berlin-Basel, 2005.

[2] Andrica, D. and Barbu, C., A Geometric Proof to Blundon’s Inequalities ,Mathematical Inequalities & Applications, 15 (2011), 180-192.

Reprints

The (…rst) author of each published paper will receive the .pdf reprintof the paper.

Further information about the journal can be found at:http://ijgeometry.com/

Technical Editors

EVELIN BARBUCONSTANTIN CIOFU

Cover Design

CONSTANTIN CIOFU

International Journal of Geometry, Vol. 1, No. 1, 2012

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