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Sejfriedian: existence, uniqueness, constructing and the proof of properties

Vladimir Shelomovskiie–mail: vvsss@rambler.ru

site http://deoma-cmd.ru/

Russia

Abstract

In this paper we investigate the properties of an amazingly beautiful construction made from five triangles and fourcircles, called together Sejfriedian in honor of its discoverer Michael Sejfried [1]. The definition of the Sejfriedian isgiven, the existence and uniqueness theorems are proved, two ways of constructing are given and many properties mostof which had previously been found numerically by Michael Sejfried [2] are proved. We have created the set of dynamicgeometric illustrations using freely distributed DGS GInMA which makes it convenient to introduce the Sejfriedianstudy in training courses on geometry. This study will enable to demonstrate the deep connections between the varioustransformations (eg, inversion and projective transformations), the possibilities of flat problems solving with the use ofa view from outside (from the space), to introduce students to little-known concepts of the simedians, the Lemoinepoints, Fermat points, Apollonius points and more. The richness of Sejfriedian properties and possibilities of their usein training courses may be comparable to the richness of the conic family.

1 Introduction

Ten years ago Polish researcher Michael Sejfried opened an exceptionally beautiful constructionwhich has a great potential in terms of geometry. Michael Sejfried was spellbound by its beauty anddevoted many years to numerical investigation of its properties with the help of computer programsGeometer's Sketchpad and Mathematica. In the early stages of research the construction which wename in honour of Sejfried as a Sejfriedian consisted of three triangles and a circle. In this paper thedefinition of the Sejfriedian is given, the existence and uniqueness theorems are proved, two waysof constructing are given and many properties are proved which had been previously foundnumerically by Michael Sejfried [1], [2]. We have created the set of dynamic geometric illustrationsusing freely distributed DGS GInMA which makes it convenient to introduce the Sejfriedian studyin training courses on geometry.

2 Definitions and denominations

We use the Sejfried notation for the Sejfriedian definition. Let an arbitrary triangle KLM and thepoint K1 be given. The triangle vertices are named in clockwise direction. The point K1 lies on thearc KL of the KLM circumcircle ω centered at the point O.We call K1L1M1 the Sejfried amicable triangle for KLM if it is inscribed into ω and there is aprojective transformation mapping ω into a circle, K1L1M1 and KLM into regular triangles. We callthe triangles KLM and K1L1M1 the Sejfried pair, ω is called the Sejfried circle.Let the point A be the intersection point of the straight lines KM and K1L1, B be the intersectionpoint of KL and M1L1, C be the intersection point of LM and K1M1. The triangle ABC have beencalled the Sejfried generating triangle.We name the tangent AAL “left” if the acute angle clockwise rotation map AAL onto AO. Let the lefttangents of ABC vertices touch with ω at the points AL, BL, and CL, and the right tangents touch withω at AR, BR, and CR. We call the triangles ALBLCL and ARBRCR the Sejfried generated triangles.Let the circles ωA, ωB, and ωC are perpendicular to ω and are centered at ABC vertices with radii RA =AАL, RB =BBL, RC =CCL. We name ωA, ωB, ωC the Sejfried satellite circles.The construction formed by five Sejfried triangles (the Sejfried pair, the generating triangle and twogenerated triangles) and four circles (the Sejfried circle and three satellite circles) is named the Sejfriedian

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We call the Sejfried number the ratio Se=KK 1⋅M 1 L1⋅ML

K 1 M 1⋅L1 M⋅LK.

Let the point S be the intersection point of the straight lines KM and BC, T be the intersection pointof KL and AC, U be the intersection point of LM and AB. The parameter stu, used by Sejfried, is

determined by the formula stu=AU⋅BS⋅CTBU⋅CS⋅AT

.

The Sejfried function is a function (in the general case it is two-dimensional function) allowing toidentify the position of the point O with respect to ABC.If the Sejfried pair consists of the regular triangles then the Sejfriedian have been called the regularSejfriedian. The example of the regular Sejfriedian is shown in Fig. 1,a.If the Sejfried circle ω is inscribed into the generating triangle ABC, the Sejfriedian have beencalled the base Sejfriedian. The example of the base Sejfriedian is shown in Fig. 1,b.The Sejfriedian in general form is shown in Fig. 1,c. The Sejfried pair is shown in red and blue, the generating triangle is highlighted by yellow, generated triangles are shown in black and green, the Sejfried circle is shown in crimson, the satellite circles are shown in black.

Fig.1,a. Regular Sejfriedian Fig.1,b. Base Sejfriedian

Fig.1,c. Sejfriedian in general form

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3 Main theorems3.1 Existence theorem

At least one Sejfriedian exists for an arbitrary triangle KLM and the point K1 on the arc KL of the KLM circumcircle.

At least one Sejfriedian exists for an arbitrary triangle KLM and the given Sejfried number Se(0,1]

3.2 Uniqueness theorem

One and only one Sejfriedian exists for an arbitrary triangle KLM and the point K1 on the arc KL of the KLM circumcircle if the Lemoine (Grabe) point Le of KLM does not coincide with the KLM circumcenter O.

One and only one Sejfriedian exists for an arbitrary triangle KLM and the number Se(0,1] if the Lemoine point Le of KLM does not coincide with the KLM circumcenter O.

3.3 Proofs and Corollaries3.3.1 Test of the regular triangle

Let an arbitrary triangle and the circumcircle be given, and the circumcircle center O coincides withthe Lemoine (Grabe) point Le of the triangle. Then the triangle is regular.

Proof If the points O and Le coincide, then each simedian lies on the circumcircle diameter. So thetangents to the circle at the ends of the simedian are perpendicular to simedians. Every tangent isparallel to the side which is crossed by the simedian.

We use the property of the tangents to a circumcircle at the points of intersection of the simedianand the circle. It is known that these tangents intersect at the point on the line containing the sidecrossed by the simedian or they are parallel to this side. We obtain that each simedian is thediameter perpendicular to the corresponding side and coincides with the median and the height.Thus the centroid, the orthocenter and the Lemoine point of the triangle coincide. It is known thatsuch triangle is regular.■

Corollary Each pair of regular triangles inscribed in one circle is the Sejfried pair.

3.3.1.a Test of the regular triangle

Let an arbitrary triangle and the incircle be given, and the incenter I coincides with the Gergonnepoint G. Then the triangle is regular.

Proof If the point I coincides with the point G, then the bisectors are perpendicular to the trianglesides. Therefore the bisectors coincide with the heights. The orthocenter H should coincide with theincenter I. It is possible only for a regular triangle.■

3.3.2 Sejfried pair of the regular triangles

Any two regular triangles inscribed in one circle form the Sejfried pair.

Proof When two regular triangles are inscribed in one circle, the Lemoine (Grabe) points coincidewith the circumcircle center. By the definition, it is the Sejfried pair.■

3.3.3 Projective transformation holding the incircle

Let the projective transformation map the incircle of the given triangle into a circle. Then it map thepreimage Gergonne point into the image Gergonne point, the Gergonne point is an invariant of this transformation.

Proof The Gergonne point is the intersection point of the segments joining the vertices of thetriangle and the points of tangency of the incircle with the sides. Points of tangency and points ofintersection of the straight lines are invariants of the projective transformation. Hence, the

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projective transformation map the preimage Gergonne point into the image Gergonne point.Gergonne point is an invariant of this transformation.■

3.3.3.a Projective transformation of the circumscribed triangle into regular triangle

Let the projective transformation map the incircle of the given triangle into a circle and map theGergonne point into the center of image circle. Then the given triangle is mapped into thecircumscribed regular triangle.

Proof The Gergonne point and the property of “being incribed” are invariants under projectivetransformation. The mapped triangle is regular in accordance with 3.3.1.■

3.3.4 Projective transformation holding the circumcircle

Let the projective transformation map the circumcircle of the given triangle into a circle. Then thepreimage Lemoine point is mapped into the image Lemoine point, the Lemoine point is an invariantof such transformation.

Proof Let us construct the tangents to the circumcircle at the vertices of the triangle. It is knownthat the Gergonne point of the triangle formed by the tangents is the Lemoine point of the giventriangle. Under the projective transformation the property of tangency is invariant, the Gergonnepoint is invariant. Hence the preimage Lemoine point of the original triangle is mapped into theLemoine point of the image, that is the Lemoine point is an invariant of such transformation.■

3.3.4.a Projective transformation of the inscribed triangle into regular triangle

Let the projective transformation map the circumcircle of the given triangle into a circle and theLemoine point into the image circumcenter. Then the given triangle is mapped into the regulartriangle, inscribed into the circle image.

Proof The Lemoine point is an invariant of the transformation. The Lemoine point and thecircumcircle center coincide in the image triangle. This triangle is regular in accordance with3.3.1.■

3.3.5 The projective transformation of the Sejfried pair

Let the projective transformation map the Sejfried circle into a circle. Then the Sejfried pair ismapped into a Sejfried pair.

Proof The Lemoine point and the property of “being incribed” are invariant under suchtransformation. The Lemoine point remains the common Lemoine point for the image triangles. Thetriangles images correspond to the definition of the Sejfried pair.■

3.3.6 The projective transformation of the Sejfried pair into regular Sejfried pair

Let the projective transformation map the Sejfried circle into a circle and map the Lemoine pointinto the image circumcenter. Then the Sejfried pair is mapped into the Sejfried pair of regulartriangles.

Proof The statement follows from the invariance of the Lemoine point and the the property of“being inscribed”, 3.3.5 and 3.3.1.■

3.3.6.a Corollary: affinely equivalence of the Sejfriedians

Any Sejfriedian is affinely equivalent to the regular Sejfriedian under the projective transformation which map the Sejfried circle into a circle.

3.3.6.b Corollary: the polar line of the Lemoine point

The polar line of the common Lemoine point for the Sejfried pair with respect to the Sejfried circle is the preimage of the line of infinity of the regular Sejfriedian.

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3.3.6.c Corollary: property of the polar line of the Lemoine point

Preimages of parallel straight lines which may be constructed for any regular Sejfried pair intersectat the polar line of the common Lemoine point with respect to the Sejfried circle.

3.3.6.d Corollary: equidistant points property

Preimages of any group of points equidistant from the common Lemoine point of the regularSejfriedian lie on a conic.

3.3.7 Sejfried triangles vertices property

The straight line BC, the straight lines KK1, LM1, ML1 and the straight lines ARAL, BRCL, BLCR areconcurrent. The point of concurrence belongs to the polar line of the Lemoine point with respect tothe Sejfried circle.

Proof For any pair of regular Sejfried triangles the straight lines BC, KK1, LM1, ML1, ARAL, BRCL,and BLCR are parallel. Their images intersect at one point (point P in the figure 2) lying on the lineof infinity. For an arbitrary Sejfriedian a preimage of the line of infinity is the polar line of theLemoine point with respect to the Sejfried circle.■

3.3.7.a The Sejfried triangles vertices property

The straight lines AC, KM1, MK1, LL1, ARCL, BRBL, ALCR intersect at one point (point E in the figure 2). The straight lines AB, LK1, KL1, MM1, ARBL, BRAL, CLCR intersect at one point (point D in the figure 2). The points P, D and E belongs to the polar line of the Lemoine point with respect to the Sejfried circle.

Fig.2. Three groups of collinear lines

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4 Auxiliary constructions

4.1 Perspector construction for given circle

Let us construct the perspector S of the projective transformation, which maps the circle ω centered at О into the circle ω', and maps the given point L, different from О, into the center of ω'.

Construction Let the radius ОL be intersected by the circle ω at the point LS, the point L0 bediametrically opposite to LS, the segment SLS be perpendicular to the circle ω plane, and the ratioholds (see Fig.3,a):

SLS

OLS

=√ 2 L LS

OL.

In this case the perpendicular LSS1 from the vertex of the right angle LS to SL0 is halved by thesegment connecting L and S. Point O1 is the midpoint of LS S1.We construct the sphere with the diameter SLS . It is shown in Fig. 3,b by blue. The sphere with thediameter LSL0 shown by yellow. The circle ABC is the big circle of the constructed yellow sphere.The intersection curve of this spheres is the circle ω'. According to the definition, the circle ω is thestereographic projection of the circle ω' under the projection from the perspector S onto the planeALO.

Fig.3,a. Constructing of the center S of the projective transformation.

Fig.3,b. Stereographic projection mapping a circle into a circle.

4.2 Construction of a projective transformation for an isosceles triangle

Consider the projective transformation of the Sejfriedian isosceles triangle on the regular triangle.Figure 4 shows some of the possible situations.

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Fig.4. Projective transformation of the Sejfriedian isosceles triangle on the regular triangle

An equilateral triangle shown by green. Isosceles triangle shown by blue. It has vertex B with angleequal to the 2β. Perspector of the mapping is the point S. The line of intersection of planes passingthrough one of the ends of the axial diameter DE. If the angle B is less than 60°, the point D iscloser to the vertex B, than E. If the angle B is greater than 60°, the point E is closer to the vertex B,than D. Let us consider the view in the direction along the AC. Thus there can be two circuits shownin Figures at Fig.5. Red line corresponds to the axis of symmetry of the equilateral triangle A'B'C',the middle of its base point B'1. Blue line corresponds to the axis of symmetry of an isoscelestriangle ABC, the middle of its base point B1. Angle BEE' between the planes of the triangles

denote α. Parameter of the Sejfried is: τ=2 h '3 R '

=2 b '

R' √3=√ s+1

s−2, s=Se+

1Se

, where h'- height, R'

is the Sejfried circle radius, b' is half of the side of an equilateral triangle.

Figure 5. The cross section in the general plane of Sejfriedian symmetry

1. Let the apex B angle be less than 60°, β <30 °. This case is shown on the left figure 5. The ratio

of line B'B'1 segments is: B ' D' : D ' E : EB ' 1=2h '3

−R ' : 2 R' :h '3

+R ' . We write ratios:

tan D ' SB ' 1

tan D ' SE=

tan D' SB ' 1

tan α=

D ' B ' 1

D ' E=

h '6 R'

+12= τ

4+

12=τ1 .

tan D ' SB 'tan D ' SE

=tan D ' SB '

tan α=

D ' B 'D' E

=h '

3 R'−

12= τ

2−

12=τ2.

tan BSDtanα

=tan( BSE+α)

tanα=

τ2+1

1−τ2⋅tan2α

.tan B1 SE

tanα=

tan (B1 SD−α)

tanα=

τ1−1

1+ τ1⋅tan2α

.

The ratio of the altitude of an isosceles triangle and regular triangle is:

BB1

B' B ' 1

=tan BSE + tan ESB1

cos α⋅( tan B' SD' + tan B ' 1 SD ' )=

1cos3

α⋅(1+τ1⋅tan 2α)⋅(1−τ2⋅tan2

α).

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Let us consider the view in the plane which is perpendicular to the previous plane and find this ratio, using similar triangles SAB1 and SA'B'1:

BB1

B' B ' 1

=BB1

AB1

⋅AB1

A' B' 1

⋅A ' B ' 1

B ' B ' 1

=1

√3⋅tanβ⋅cos2α⋅(1+ τ1⋅tan2

α).

We obtain the equation for the angle α by using the equation between the same ratious.

cosα⋅(1−τ2⋅tan α)=√3⋅tan β . cos2α−2 √3⋅tanβ

τ+1cosα−

τ−1τ+1

=0.

The equation has a unique solution for the allowable angle α[0,90°):

cosα=√3⋅tan β

τ+1+√ 3⋅tan2

β

(τ+1)2 +τ−1τ+1

.

2. Let the apex B angle be more than 60°, β > 30°. This case is shown on the right figure 5. Wewrite ratios for angles in the form of:

tan D ' SB ' 1

tan D' SE=

tan D ' SB ' 1

tanα=

D ' B ' 1

D' E=

h '6 R '

−12= τ

4−

12=τ1−1.

tan D ' SB 'tan D ' SE

=tan D ' SB '

tan α=

D ' B 'D' E

=h '

3 R'+

12= τ

2+

12=τ2+1.

tan BSEtan α

=tan(BSD '−α)

tan α=

(τ2+1)−1

1+( τ2−1)⋅tan2α

.

tan B1 SE

tanα=

tan (B1 SD ' +α)

tanα=

(τ1−1)+1

1−( τ1+1)⋅tan 2α

.

The ratios of the altitude of an isosceles triangle and regular triangle are:

BB1

B' B ' 1

=tan BSE + tan ESB1

cos α⋅( tan B' SD' + tan B ' 1 SD ' )=

1cos3

α⋅(1−(τ1+1)⋅tan2α)⋅(1+(τ2−1)⋅tan 2

α).

BB1

B' B ' 1

=BB1

AB1

⋅AB1

A' B' 1

⋅A ' B ' 1

B ' B ' 1

=1

√3⋅tanβ⋅cos2α⋅(1−(τ1−1)⋅tan2

α).

The equation for the angle α is:

cosα⋅(1+(τ2−1)⋅tan 2α)=√3⋅tanβ , cos2α+2 √3⋅tanβ

τ−1cosα−

τ+1τ−1

=0.

The equation has a unique solution for the allowable angle α[0,90°):

cosα=−√3⋅tanβ

τ−1+√ 3⋅tan2

β

( τ−1)2 +τ+1τ−1

.

We know the angle α and the ratio of BD: DE: EB1, so we can build Sejfriedian.

4.3 Triangle with given circumcenter and the Lemoine point construction

Let construct the triangle АВС such that the vertex А, circumcenter О and the Lemoine point L be given. If the points A, O, and L are concurrent, we decide that the plane АВС is known.

Construction We construct step by step the circumference ω(О, ОА), the Brocar axis LO, the pointsof the intersection of the circle ω and LO (the points L1 and L0), the perspector S and the circle-image ω'.

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We construct the point А1 as a result of intersection of the straight line AS and the circle ω'. We construct the regular triangle А1В1С1 inscribed into the circle ω'. It is shown in Fig 6,a. The bluetriangle А1В1С1 is inscribed into the green circle ω'.

We fulfill the projective transformation of the triangle А1В1С1 with the perspector S onto the planeAOL. This is shown in Fig 6,b. Purple rays from the perspector S cross the red circle ω at the pointsA, B and C. The triangle АВС is constructed.

Proof We have choose the projective transformation which maps the center of the regular triangleА1В1С1 (its Lemoine point) into the point L. The point L is the Lemoine point preimage inaccordance with 3.3.4.■

Fig.6,a. Image А1В1С1 is shown. Fig.6,b. Preimage АВС is shown.

4.4 Amicable Sejfried triangle construction

Let an arbitrary triangle ABC and the point A1 belonging to the circumcircle ABC be given. Use the projection transformation to construct a Sejfried triangle.

Construction We construct the perspector S and the circle ω'. We construct the image of the triangle KLM. In Fig. 7,a it is noted as K'L'M'.We construct the point K1' on the circle ω' which is the image of the point K1 (Fig.7,a).We construct the regular triangle K1'L1'M1' inscribed in a the circle ω' (Fig.7,b).We find the vertices of the regular generating triangle A'B'C' at the intersections of the straight lineswhich contain the sides of the triangle (Fig. 7,c).We construct the Sejfriedian preimages K1L1M1 и АВС using the corresponding points of theSejfriedian images K1'L1'M1' and A'B'C' (Fig.7,d).

Fig.7,a. Constructing of the images K'L'M' and K1'.

Fig.7,b. Constructing of the images K1'L1'M1' and A'B'C'

Fig.7,c. Constructing of the K1L1M1

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Fig.7,d. Sejfriedian is built.

4.5 Construction of Sejfriedian by one vertex of the amicable triangle

Let an arbitrary triangle KLM and the point K1 belonging to the arc KL of the circle ω (KLM) be given. Construct a Sejfriedian.

Construction We construct the Brocar axis OLe and the polar line of the Lemoine point Le. Thepolar is perpendicular to the Brocar axis and contains the point L' inverted to Le with respect to thecircle ω. OLe OL' = OK⋅ ² (see left Fig.8).We construct the point Р as a result of intersection of KK1 and the polar.We construct the vertices L1 and M1 of the Sejfried triangle as the second points of intersection ofthe straight lines PL and PM with the circle ω (see left Fig.8). We construct the generating triangle ABC using the intersection points of the straight linescontaining the sides of the Sejfried triangles (see right Fig.8). We construct the generated Sejfried pair by using the tangents. The values of the Sejfried numberSe and parameter stu are shown.Proof See the Corollary 3.3.6.1.

Fig.8. Construction of Sejfriedian by one vertex of the amicable triangle

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4.6 Construction of Sejfriedian by Sejfried number

Let an arbitrary triangle KLM and the real number Se,1 Se > 0 be given. Construct a Sejfriedian.

Construction We construct the point K1 according to the conditionKK 1

LK 1

=SeKMLM

. We use the

Apollonius circle and the point of intersection of this circle and the circumcircle KLM (see Fig.9).Then we use 4.5 construction.

Fig. 9. Point K1 construction.

5 Special types Sejfriedian properties5.1 Regular Sejfriedian properties

The base regular Sejfriedian is shown in Figure 10,a, the regular Sejfriedian is shown in Figure10.b. The point Q is the center of ABC. The lines described by Sejfried, the family of concentriccircles uniting the points equidistant from the center Q, the family of straight lines parallel to thebase АС, the Sejfried number Se and the parameter stu are shown.

Figure10,a. Base regular Sejfriedian. Figure10.b. Regular Sejfriedian.

For the base regular Sejfriedian we have AC 1=KL=K1 L1=AB2

, and АС1 touches the incircle.

By the property of a tangent and a secant, several segments meet the conditions of the goldensection:

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A K1

AC1

=AC 1

AL1

=AC 1

AK 1+ AC 1

=K 1C1

C1 L1

. (1)

For an arbitrary regular Sejfriedian we take as a unit the generating triangle side length and denote

AT =AU 1=BU =BU 1=CS=CT1=ξ ≤12

. (2)

We have

AВ1=АС1=BC 1=AC 1=BA1=CA1=12

, AQ=BQ=CQ=1

√3,QA1=QB1=QC1=

1

2√3, (3)

AS=АS1=BT=BT 1=CU=CU 1=η=√1−ξ +ξ 2=

√1+Se+Se2

1+Se. (4)

We use the Sejfried number

Se=ξ

1−ξ,ξ =

Se1+Se

. (5)

and the parameter

stu=1

Se3. (6)

We use the ratios for skew segments and obtain the length of Sejfried triangles:

KL=LM =MK=K 1 L1=L1 K1=K1 M 1=1−2ξ

η =ζ =1−Se

√1+Se+Se2. (7)

RSe=QK=QL=QM =QK 1=QAR=ζ

√3=

1−Se

√3(1+Se+Se2)

. (8)

The vertices of the Sejfried triangles form the segments with only four different lengths.

There are six sides of the triangles KL=LM =MK=K 1 L1=L1 K1=K1 M 1=1−2ξ

η =ζ . (9)

There are three short segments L L1=KK 1=MM 1=ξ⋅ζη , (10)

There are three middle segments L K 1=KM 1=ML1=(1−ξ )⋅ζ

η , (11)

There are three long segments L M 1=KL1=MK 1=ζη . (12)

We combine these segments and found the products:

KK 1⋅ML1⋅LM 1=L L1⋅MK 1⋅KM 1=MM 1⋅LK 1⋅KL1=ξ⋅(1−ξ )ζ 3

η3 . (13)

KM 1⋅LK 1⋅ML1=(1−ξ )

3⋅ζ 3

η3 , (14)

The product of the short segments is KK 1⋅L L1⋅MM 1=ξ 3

⋅ζ 3

η3 , (15)

The product of the long segments is KL1⋅LM 1⋅MK 1=ζ 3

η3 . (16)

The ratio of the lengths is: KK 1

KM 1

=KK 1

LK1

=KK 1

ML1

= ...=ξ

1−ξ=Se. (17)

According to Sejfried, we denote the points of contact of the lines passed from the generatingtriangle vertices with Sejfried circle as AL, AR, BL, BR, CL, CR. We get:

AAL=AAR=...=√AK⋅AM =√ξ (1−ξ )

η , (18)

KAR

MAR

=MC R

LCR

=LAR

KAR

=K1 AL

L1 AL

=L1 BL

M 1 BL

=M 1 C L

K 1C L

=√ AKAM

=√ ξ

1−ξ=√Se=

16√ stu

. (19)

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We use similarity of the triangles AQAR and AS2A1 and get:

A1 S 2=RSe

AA1

AAR

=1−Se2√Se

=A1 S 3=B1 T 2=B1T 3=C 1U 2=C1 U 3 . (20)

We denote the points of intersection of the Sejfried circle and АА1 as А2 and А3:

AA2=BB2=CC 2=1−ζ

√3, AA3=BB 3=CC 3=

1+ζ

√3, A1 A2=B1 B2=C1 C 2=

0.5+ζ

√3. (21)

We obtain:

AK⋅SMAM⋅KS

=AK 1⋅S 1 L1

AL1⋅K1 S 1

=BL⋅KTBK⋅L T

=

ξη⋅

ξ 2

η

1−ξη ⋅

(1−ξ )2

η

=ξ 3

(1−ξ )3=Se3

=1

stu. (22)

(BC , S1 S )=BS 1⋅CS

CS1⋅BS=( AB ,U 1U )=

AU 1⋅BU

BU 1⋅AU=(CA ,T 1T )=

CT 1⋅AT

CT⋅AT 1

=ξ 2

(1−ξ )2=Se2. (23)

(BC , S1 A1)=BS 1⋅CA1

CS1⋅BA1

=( AB ,U 1C1)=AU 1⋅BC 1

BU 1⋅AC 1

=(CA ,T 1 B1)=CT 1⋅AB1

CB1⋅AT 1

=ξ

1−ξ=Se . (24)

(BA1 , S 3 S 2)=BS3⋅A1 S2

A1 S 3⋅BS2

=(CA1, S2 S3)=CS 2⋅A1 S 3

CS 3⋅A1 S 2

=|√ Se−1+Se |√Se+1−Se

= S̃e . (25)

(BC , S3 S 2)=BS3⋅CS2

CS 3⋅BS2

=(AB ,U 3 U 2)=AU 3⋅BU 2

BU 3⋅AU 2

=(CA ,T 3 T 2)=CT 3⋅AT 2

CT 2⋅AT 3

= ̃Se2. (26)

Let us denote the point of intersection of the straight lines L1K and AC as T4 Then:

CT 4

AT 4

=Se+1Se

,(CT 4 , B1 A)=CT 4

AT 4

⋅AB1

CB1

=Se+1Se

. (27)

The straight line L1K is parallel to the side AB and intersects the axis of symmetry ВВ1 at the point Bi, inverted for B with respect to Sejfried circle.

5.1.1 Straight lines containing vertices of generated triangles

We have the following statement for an arbitrary regular Sejfriedian. For each group of the triples ofpoints A1BLCL , A1CRBR , B1CLAL , B1ARCR , B1ALBL , C1BRAR the points in the group are concurrent.

Proof Sejfried generated triangles are regular, since all their defining points are mapped into eachother by turning the triangle ABC by 120° around the point Q. Let us consider the triple of thepoints A1BLCL. The radius QBL divides the angle ALBLCL in half, the angle QBLCL = 30°. The points A1 and BL belong to the circle with the diameter BQ. So the sum of the angles isQBLA1 + QBA1 = 180°, where QBA1 = 30°. Hence the angle A1BLCL = QBLCL + QBLA1 =180° is a flat angle, and the points A1 and BL belong to the straight line A1CL. We have a similarsituation for the other triples of points.■

5.1.2 The collinear points

Let the straight lines for an arbitrary regular Sejfriedian A1СR and С1BL intersect at the point F1 , the straight lines B1СR and С1BL intersect at the point G1. Then F1 belongs to the median BQ, G1 belongs to the median AQ.

Proof All defining points are mapped into each other under the symmetry about ABC medians.■

5.1.2.a Corollary

Let the point A1 is the point of intersection of the straight lines AQ and BС, the point В1 is the pointof intersection of the straight lines ВQ and АС, the point С1 is the point of intersection of thestraight lines СQ and АВ. Then the triangle A1B1C1 is the generating triangle for the Sejfriedgenerated triangles.

14

5.1.3 Formula

For an arbitrary regular Sejfriedian we have:

BLCR

BR C L

=|√ξ (1−ξ )−1+2ξ |

√ξ (1−ξ )+1−2ξ=

|√ ξ

1−ξ−1+

ξ

1−ξ|

√ ξ1−ξ

+1−ξ

1−ξ

=|√Se−1+Se |√Se+1−Se

. (28)

Proof We use the symmetry about the medians of the triangle ABC, the sine theorem and equations:A1QBL = QВBL – 30°, С1QBL = QВBL + 30°.■

5.1.4 Ellipses of the Sejfriedian

There are the ellipses which are tangent to the sides of generated triangle ABC at the points U, S1,

and B1, S, T1, and C1, T, U1, and A1. In the case of Se=12

, stu=8 these ellipses are tangent to the

Sejfried circle at the vertices of the Sejfried triangles. One of the ellipses is shown in Fig.11.

Proof It is trivial for the regular Sejfriedian and is invariant under the transformations.■

Fig.11. Ellipse in special case stu = 8

5.2 Sejfriedian for isosceles triangle

5.2.1 Radii of the accompanying circles

Let the isosceles triangle ABC (AB = BC) be given. We want to find the radii of the accompanying circles RA = RC and RВ .

5.2.1.a The collinear points

Let Bi be the inverse point for B with respect to inversion for ω. Let G be the mutual inversioncenter (dilation) of the circles with centers A and B. The points G, Bi, K, L1 and Т4 are collinear.

Proof. Consider the straight line KL1.

We prove that point G belongs to KL1. When a projective transformation in the regular Sejfriediangroup of the straight lines АВ, M1L, KL1, MK1 mapped into parallel lines, so they have a commonpoint at infinity, which corresponds to the center of homothety circles with centers A and Bperpendicular for ω.

We prove that point Т4 belongs to KL1. A projective transformation of the Sejfriedian retained the

double ratio (27) (CT 4 ,B1 A)=CT 4

AT 4

⋅AB1

CB1. For the isosceles triangle the point B1 is midpoint of

BC. It means the ratio of any two segments that make up the double relationship with segments AB1

15

and CB1 is saved. In particular CT 4

AT 4

=Se+1Se

.

Fig. 12. The isosceles triangle Sejfriedian and straight line KL1.We prove that point Bi belongs to KL1. Circle ω and ωB are perpendicular, point Bi belongs to theintersection point of its radical axis and the center line. So under the inversion with respect ω itmapped into B, and under the inversion with respect ωB it mapped into O. Under the inversion withrespect to ω the straight line KL1 mapped into the circle IKL1, containing point B. This follows fromthe symmetry points K and K1 with respect to the axis of symmetry of the triangle and points K1 andL1 with respect to ωB . The straight line KL1 is the radical axis of the circles and IKL1 and ωВ. Itfollows that the angles L1BI and L1KI are equal one to another.

5.2.2 The values of the radii of the accompanying circles

Menelaus condition for the straight line GKBi, intersecting sides of the triangle АВВ1, leads to theequation:

RB⋅RC=BT 2⋅RC+RC2−

AC 2

2. (29)

The condition that the center O lies on the radical axis of the circles ωB and ωC is:

2 RB⋅BT 2=RB2−RC

2+ BC 2. (30)

Solution of the system has the form of:

RC=AC 2

2√BT 22+ AC 2

−BC 2. (31)

RB=BT 2−b2

+d 2

√d 2+3b2

=√h2+d 2

−b(s−1)

√s+1. (32)

5.2.3 Sejfriedian for isosceles triangle constructionLet point В1 be midpoint AC, b = AB1, h = BB1. We use points Т and Т2 belonging to АС.

ATCT

=r A Se

rC

=Se ,

AT 2

CT 2

=r A S̃e

rC

, S̃e=√Se−1+Se√Se+1−Se

.(33)

16

Let d be d = B1T2: B1T 2

AB1

=db=

1−Se√Se

, BT 22=h2

+d 2 . (34)

We express the radii by b, d and Se:

RA=RC=2b2

√d 2+3b2=

2b

√ 1Se

+Se+1

=2b

√s+1, s=

1Se

+Se .(35)

RB=BT 2−b2

+d 2

√d 2+3b2

=√h2+d 2

−b(s−1)

√s+1. (36)

We consistently construct next lines and points (see Fig.12):circles ωА and ωB , point О is the intersection point of the radical axis of the circles ωА and ωB and straight line ВВ1,straight line BT2,perpendicular OBr on BT2 , find the radius OBr of the circle ω, straight line BT and points K and L of the intersection BT and ω.

5.2.4 Sejfried function

In the case of the isosceles triangle the point O belong the axis of symmetry of the triangle ABC.Hence it is scalar function. For example, we can use the distance BO between the center O and thevertex B. It is evidence that

BO=RB

sin BIBR

=RB

sin BT 2 C=

RB⋅BT 2

h, d 2

h2 =d 2

b2

b2

h2=(s−2)b2

h2 .

BOh

=1+d 2

h2−

bh √1+

d 2

h2

s−1

√s+1. (37)

Let us verify the formula in trivial cases.

1. Let the radius of the Sejfried circle be near zero. In this case Sejfried number Se → 1, s → 2,

lims→2

BOh

=1−b

h √3. It is Fermat point.

2. Let the Sejfried circle be incircle. In this case Sejfried number Se=3−√5

2≈0.3819 , s=3.

lims→3

BOh

=1+b2

h2−

bh √1+

b2

h2. It is incenter.

3. Let the Sejfried circle be near circumcircle. In this case Sejfried number Se → 0, s → ,

lims→∞

BOh

=12+

b2

2h2 . It is circumcenter.

At the Fig. 13 Sejfried functionBOh

(Se) for seven values ofbh={3

2,54

,1,1

√3,12

,14

,17} is

shown. For the regular triangle bh=

1

√3, the point О is stationary and located at its center,

BOh

=23

. For an arbitrary triangle function changes monotonically from F (0)=1−b

h√3till

F (1)=1+b2

h2−

bh √1+

b2

h2. If the angle in vertex B is less than 60°, β <30°,

bh<

1

√3. In this case,

17

the function increases monotonically with increasing values of the number of Se from

F (0)=1−b

h√3<

23 till F(1). Point O is move away from the vertex B and approach to the base

AC. If the angle in vertex B is more than 60°, β >30°, bh>

1

√3. In this case, the function decreases

monotonically with increasing values of the number of Se from F (0)=1−b

h√3>

23 till F(1).

Fig.13. Sejfried function for isosceles triangle

5.3 The base Sejfriedian, building

Let a triangle ABC with AB ≠ BC be given. Let us build the base Sejfriedian.Constructing. Let ω = A1B1C1 be the inscribed circle ABC (A1B1C1 are the points of contact). Weconstruct a point D as the point of intersection of lines A1C1 and AC, point E as the point ofintersection of lines A1B1 and CC1 (see Fig.14, left). Points M and K1 are the points of intersectionof DE and ω. These points are the vertices of two Sejfried triangles. Construction of the remainingvertices is obvious.

Fig. 14. Building of the base SejfriedianProof. In the regular Sejfriedian АС||A1C1||K1M, the segments A1В1, LM1 and K1M intersects at onepoint (E). As a result of the projective transformation, preserving the incircle and transformingSejfriedian in the regular Sejfriedian, point D of intersection of AC and A1C1 would be on thestraight line K1MЕ.

5.4 Sejfriedian in the case of Se = 0,5

Let the Sejfried number be equal to 0,5. It is trivial to prove that for regular Sejfriedian the points A1, L1 and U are collinear, at that the straight line A1L1 is tangent for Sejfried circle. Similarly such properties have triples of points (A1, T1 and M), (B1, M1 and S), (B1, U1 and K), (C1, K1 and T), (C1, S1

and L). It is shown on Fig.15 at left for regular Sejfriedian and at right for arbitrary one.

18

The straight lines S1T, T1U, U1S contain the point O of the crossing straight lines A1A, B1B, C1C.

The straight lines A1C1, BT1, TS1 are concurrent. Similarly the triples of lines (A1C1, BT, T1U), (A1B1, SU1, CU), (A1B1, СU1, ST), (B1C1, T1U1, AS), (B1C1, SU1, AS1) are concurrent.

The straight lines A1C1, BB1, CU, AS1 are concurrent. Similarly lines A1B1, CC1, AS, BT1 and lines B1C1, AA1, BT, CU1 are concurrent. These properties are invariant under projective transformations so every Sejfriedian has them if Se = 0,5.

Fig.15. Regular and arbitrary Sejfriedian for Se = 0,5

5.5 Sejfriedian in the case of Se = 0,618...

Let the Sejfried number be equal to Se=√5−12

≈0.618. It is trivial to prove that for regular

Sejfriedian the points S1, U1 and L, at that the straight line U1L is tangent for Sejfried circle. Similarly such properties have triples of points (U, L1, S), (S1, T1, M), (S, M1, T), (T1, U1, K), (U, K1, T). It is shown on Fig.16 at left for regular Sejfriedian and at right for arbitrary one.

The polar ALAR of the vertex А contain the points T1, U and crossing points of the lines AS1 , C1C andBT, of lines AS, CU1 и BB1. Similarly the polar BLBR contain the points U1, S and crossing points of the lines AS1 , C1C и BT, and lines CU, BT1 and AA1. The polar CLCR contain the points S1, T and crossing points of the lines AS1 , C1C и BT, and lines CU, BT1 and AA1.

The points A1, C1, L1 and L are collinear. Similarly points (A1, B1, M1 and M), (B1, C1, K1 and K) are collinear.

These properties are invariant under projective transformations so every Sejfriedian has them.

Fig.16. Regular and arbitrary Sejfriedian for Se = 0,618...

19

6 Some geometric facts6.1 Invariants of the projective transforms

6.1.1 Four points ratio

Let a circle and belonging to it points А, B, C, D be given. We call a d-ratio of four points the ratio

τ=| A−B |⋅|C−D || B−D |⋅|C−A |

. (38)

Let the projective transformation maps the circle ABC into a circle and the given points are mappedinto the points А', B', C', D', respectively. Then d-ratio is invariant under this transformation:

τ=| A−B |⋅|C−D || B−D |⋅|C−A |

=| A' −B ' |⋅|C '−D' || B' −D ' |⋅|C '−A' |

. (39)

Proof This transformation with respect to the points of the circle ABC can be considered as theinversion (S0,R). We use the formula for the distance between the image points and the preimagepoints in the form of:

| A−B |=| A '−B ' |⋅R2

|S 0−A' |⋅|S 0−B' |.

We get: | A−B |⋅|C−D |=| A'−B' |⋅|C '−D ' |⋅R4

| S0−A ' |⋅|S 0−B ' |⋅|S 0−C ' |⋅|S 0−D ' |,

| B−D |⋅|C−A |=| B '−D' |⋅|C '−A' |⋅R4

| S0−A ' |⋅|S 0−B ' |⋅|S 0−C ' |⋅|S 0−D ' |. We fulfill cancellation and get (38).■

6.1.2 Six points ratio

Let a circle and belonging to it points А, B, C, D, E, F be given. We call a triple ratio of six points the ratio

t=| A−B |⋅|C−D |⋅| E−F || B−D |⋅|C−F |⋅| E−A |

. (40)

Let the projective transformation maps the circle ABC into a circle and the given points are mappedinto the points А', B', C', D', E', F', respectively. Then the triple ratio is invariant under thistransformation:

| A−B |⋅|C −D |⋅| E−F || B−D |⋅|C −F |⋅|E−A |

=| A'−B' |⋅|C '−D ' |⋅| E '−F ' || B '−D' |⋅|C '−F ' |⋅| E '−A' |

. (41)

6.1.3 2n points ratio

Let a circle and belonging to it 2n points А, B, C, D...Z be given. We call a n-ratio of 2n points the ratio

t n=| A−B |⋅|C−D |⋅...| B−D |⋅...⋅| Z−A |

. (42)

Let the projective transformation maps the circle ABC into a circle and the given points are mappedinto the points А', B', C', D',...,Z', respectively. Then n-ratio is invariant under this transformation:

t n=| A−B |⋅|C−D |⋅...| B−D |⋅...⋅| Z−A |

, t ' n=| A'−B ' |⋅|C '−D' |⋅...| B '−D ' |⋅...⋅| Z '−A' |

, t n=t ' n. (43)

6.1.4 Clearly about the ratios

We may formulate the statement 6.1 in another way. Let us take a circle and inscribe in it a closed broken line, passing once through each given point. Then we paint the segments of broken line

20

through one in two colors, such as red and blue as it is shown in Fig. 14,a. We find the product of the lengths of blue segments and the product the lengths of the red segments. The ratio of this products does not change under the projective transformation mapping the given circle into any other circle. We see the calculated value of t – t' in the up of figure 14,b.

Fig.17,a. Projective transformation. Fig.17,b. Illustration of invariance.

6.2 Tangents proportion

Let the circle, the points А and B lying on it and the tangents AAR, AAL, BBR, BBL be given. We usethe subscript «R» for the tangent AAR if the ray АО is mapped into the ray AAR under the acute angleclockwise rotation. Let the straight line ARBR intersects the straight line AB at the point E. Then:

AAR

BBR

=AEBE

. (44)

Proof Let the straight line AАR intersects the lineВBR at the point D, DBR = DAR. In accordancewith the Menelaus' theorem for the triangle ABD,we have:

DAR⋅AE⋅BBR=DBR⋅BE⋅AAR ⇔AAR

BBR

=AEBE

. ■

Fig.18. Tangents proportion

6.3 The Newton-Gauss line

The Newton-Gauss line of the cyclic quadrangle contains the center of the inscribed circle.

Proof It is the corollary for the Newton theorem for a cyclic quadrangle. ■

6.4 The property of the intersecting tangents

Let the circle, the points А and B lying on it and the tangents AAR, AAL, BBR, BBL be given. Thepoints of intersection of the tangents are C, C', D, D', as it is shown in figures 16,a and 16,b. Then:

– the straight lines CC', ALBL and ARBR are concurrent, the point E of intersection belongs to АВ,– the straight lines DD', ALBR and ARBL are concurrent, the point E' of intersection belongs to АВ,

21

– the tangents from the point E touch the circle at the points lying on DD',– the tangents from the point E' touch the circle at the points lying on СC',– the straight lines ABR and BAL intersect at the point C1 belonging to СC',– the straight lines ABL and BAL intersect at the point D1 belonging to DD',– the straight lines ABR and BAR intersect at the point D2 belonging to DD',– the straight lines ABL and BAR intersect at the point C0 belonging to СC'.

Proof It is known that the straight lines CC', DD', ALAR and BRBL intersect at one point (the point G).We perform projective transformations which maps the given circle into a circle, and the point Ginto the image circle center. We use the invariance properties of tangency and find that the polar ofthe point G (the line AB) is mapped into the line at infinity, and the tangents are pairwise parallel.■

Fig.19,a. The property of the intersecting tangents

6.4.1 The property of the self-polar triangle

Under the conditions 6.4, DD'OE, CC'OE', ABOG, the triangle EGE' is self-polar triangle, the point О is the orthocenter of the triangle EE'G, the point G is the orthocenter of the triangle EE'O.

Fig.19,b. The property of the intersecting tangents

Proof The tangents from the point E touch the circle at the points lying on DD'. The tangents fromthe point E' touch the circle at the points lying on CC' (6.2).■

6.4.2 The radical axes property

Under the conditions 6.4 the radical axis of the circles (A, AAR) and (B, BBR) is perpendicular to AB. The radical axis contains the center O and the point G. If the circles intersect, the radical axis contains the points H and H' of their intersection.

Proof The radical axis HH' of the circles (A, AAR) and (B, BBR) is perpendicular to the line of thecircumcenters and contains the point О, power of which with respect to this circles is equal to the

22

square of radius of the circle centered at O. ABOG.■

6.4.3 The Newton-Gauss line

Under the conditions 6.4 the Newton-Gauss line of the cyclic quadrangle СDC'D' contains the center O of the inscribed circle (AM = BM, DM1 = D'M1, CM2 = C'M2).

6.4.4 Under the conditions 6.4 the following equalities hold:

AEAAR

=BEBBR

,AE 'AAR

=BE 'BBR

. (45)

Proof Equalities follow from 6.2.■

7 Sejfriedian properties

7.1 Collinearity

7.1.1 The image of the center of the regular Sejfriedian

Let Q be the intersection point of straight lines AA1 and ВВ1. Then the straight line CC1 contains this point. Point Q is the Lemoine point of the triangles Q with respect to the Sejfried circle. Proof. Point Q is the preimage of the center of the regular triangle under mapping Sejfriedian into regular Sejfriedian. Lemoine point is mapping invariant. ■

7.1.2 The set of points

The set of points, collinearity of which have been indicated by Sejfried [1],[2], for example, A, D, Q, G, G1 , A1 , are really collinear.

Proof The collinearity is trivial for the regular Sejfriedian. All Sejfriedians are affine equivalent(3.3.6.1). Collinearity is the affine property.■

7.1.3 The concurrence

The straight lines LL1, A1C1, ALCR, EG, SU1, K1M, DH, …, AC are parallel or intersect at one point belonging to the polar line of the Lemoine point Q with respect to the Sejfried circle.

Proof The statements are trivial for the regular Sejfriedian. All Sejfriedians are affine equivalent(3.3.6.1). The statements are affine properties The polar line of the point Q with respect to theSejfried circle is on the line of infinity.■

7.1.4 The straight lines containing pairs of vertices of Sejfried triangles

The straight lines BC, KK1, LM1, ML1, ALAR, BLCR, CLBR are parallel or intersect at one point belonging to the polar line of the Lemoine point with respect to the Sejfried circle.

Proof The statements are trivial for the regular Sejfriedian. All Sejfriedians are affine equivalent(3.3.6.1).■

7.1.5 Tangents

The straight lines SM1, TK1, UL1, LS1, MT1, KU1 are tangent to the Sejfried circle.

7.1.6 The tangents passed from the vertices of the generating triangle

The tangents to the Sejfried circle passed from the vertices of the generating triangle areintersecting at the points belonging to the straight lines which contain the point Q. The tangents tothe Sejfried circle passed from the vertices А and В are intersecting at the point belonging to CQC1,the tangents passed from the vertices B and C are intersecting at the point belonging to AQA1, thetangents passed from the vertices A and C are intersecting at the point belonging to BQB1.

7.1.7 The point of concurrence

The tangent to the Sejfried circle DDR is concurrent with the straight lines EFR and FER. Similarly the straight lines DFR, EER and FDR, DER , EDR and FFR , DDL, EFL and FEL , DFL, EEL and FDL,

23

DEL , EDL and FFL are concurrent.

Proof The statements are trivial for the regular Sejfriedian. All Sejfriedians are affine equivalent(3.3.6.1). The statements are affine properties. The polar line of the point Q with respect to theSejfried circle is on the line of infinity.■

Fig. 20. Common point of the straight lines joining the Sejfriedian corresponding vertices.

7.2 Invariants

7.2.1 Sejfried number and the parameter stu

The Sejfried number Se and the parameter stu are invariant under the projective transformationmapping the Sejfriedian into the regular Sejfriedian. The following ratios hold for any Sejfriedian:

KM⋅L L1⋅K 1 M 1

KM 1⋅L1 K 1⋅ML=

ξ

1−ξ=Se=

13√stu

, (46)

KK 1⋅ML1⋅LM 1

L L1⋅MK1⋅KM 1

=MM 1⋅LK1⋅KL1

L L1⋅MK 1⋅KM 1

=1 , (47)

KK 1⋅L L1⋅MM 1

KM 1⋅LK 1⋅ML1

=( ξ1−ξ )

3

=Se3=

1stu

. (48)

KAR⋅MC R⋅LAR

MAR⋅LC R⋅KAR

=K 1 AL⋅L1 BL⋅M 1C L

L1 AL⋅M 1 BL⋅K 1C L

=( ξ1−ξ )

1.5

=Se1.5=

1√stu

. (49)

Proof We use the triple ratios for the sets of following points: K ,M , L , L1 , K 1 ,M 1 for (46),K , K1 ,M , L1 , L ,M 1 and M , M 1 , L , K 1 ,K , L1 for (47), K , K 1 , L , L1 ,M ,M 1 for (48),K , AR , M ,C R , L , AR and K 1, AL , L1 , BL ,M 1 ,C L for (49). All these points belong to the

Sejfried circle. Hence the triple ratios are invariant under projective transformation mapping theSejfriedian into the regular Sejfriedian. The equations are trivial for regular Sejfriedian.■

7.2.2 Sejfriedian's ternary sets

The following ratio holds for any Sejfriedian:

AK⋅BL⋅CMAAR⋅BBR⋅CC R

=AK 1⋅BL1⋅CM 1

AAL⋅BBL⋅CC L

=( ξ1−ξ )

1.5

=Se1.5=

1√ stu

. (50)

AK⋅BL⋅CM =AK 1⋅BL1⋅CM 1. (51)

Proof According to the properties of a tangent and a secant we get six equations:

24

AKAAR

=KAR

MAR

,BLBBR

=MC R

LC R

, ... (52)

We multiply them by three and find the expression for the regular Sejfriedian for the segments withendpoints belonging to the Sejfried circle. Thus we get (40). Since the left and the right tangentsfrom one point are equal, we obtain (41).■

7.2.3 After the Carnot's theorem

The following ratios hold for any Sejfriedian:

AT⋅BU⋅CSAU⋅BS⋅CT

=AU 1⋅BS1⋅CT 1

AT 1⋅BU 1⋅CS1

=( ξ1−ξ )

3

=Se3=

1stu

,(53)

AT 2⋅BU 2⋅CS 2

AU 2⋅BS2⋅CT 2

=AU 3⋅BS 3⋅CT 3

AT 3⋅BU 3⋅CS 3

= ̃Se3 ,(54)

Proof The points S ,S1 ,T , T 1 , U , U 1 (and the points S 2, S3 ,T 2,T 3 , U 2,U 3 ) are concyclic in aregular Sejfriedian, hence they belong to the same quadric in an arbitrary Sejfriedian. According tothe Carnot's theorem we get:

ATCT

⋅AT 1

CT 1

⋅CSBS

⋅CS1

BS1

⋅BUAU

⋅BU 1

AU 1

=1,

AT 2

CT 2

⋅AT 3

CT 3

⋅CS 2

BS 2

⋅CS3

BS3

⋅BU 2

AU 2

⋅BU 3

AU 3

=1. (55)

Then we use the double cross-ratios from 7.3.■7.2.4 The base SejfriedianThe following ratios hold for any base Sejfriedian:

AT⋅BU⋅CSAU⋅BS⋅CT

=AU 1⋅BS1⋅CT 1

AT 1⋅BU 1⋅CS1

=Se3=(

√5−12

)6

=(3−√5

2)

3

.

7.3 Sejfriedian's double sets

It is known that double cross-ratios are invariant under projective transformation. So we find thedouble cross-ratios for the regular Sejfriedian and use them for any Sejfriedian. For example, wewrite the ratio of a pair of double cross-ratios, which is also an invariant:

(AT ,T 1C )

(AT 1 ,TC )=

AT 1⋅CTAC⋅TT 1

:AT⋅CT 1

AC⋅TT 1

=AT⋅CT 1

AT 1⋅CT=( ξ

1−ξ )2

=Se2=stu

−23 . (56)

Similarly CS⋅BS1

CS 1⋅BS=

BU⋅AU 1

BU 1⋅AU=Se2=stu

−23 . (57)

(AS , KM )=AK⋅MSAM⋅KS

=(BT , LK )=BL⋅KTBK⋅TL

=(CU , LM )=CM⋅LUCL⋅MU

=Se3=

1stu

.(58)

7.3.1 Formulas for radii of satellite circles

The following ratios hold for any Sejfriedian:

AK⋅MSr A⋅SM 1

=BL⋅KTr B⋅TK 1

=CM⋅LUrC⋅UL1

=Se3=√ stu .

(59)A1 B

r B

=A1 C

rC

,B1 A

r A

=B1 C

rC

,C1 A

r A

=C 1 B

r B

.(60)

Proof We use (22), (18), and (44).■

7.3.2 Segments on the sides of the generator triangle

The following ratios hold for any Sejfriedian:

25

CS⋅BA1

BS⋅CA1

=BU⋅AC 1

AU⋅BC 1

=AT⋅CB1

CT⋅AB1

=Se=1

3√ stu

,(61)

BS1⋅CA1

CS 1⋅BA1

=AU 1⋅BC 1

BU 1⋅AC 1

=CT 1⋅AB1

AT1⋅CB1

=Se=1

3√stu

. (62)

CSBS

=CA1

BA1

⋅Se=Ser C

r B

=rC

r B⋅3√ stu

,BUAU

=Ser B

r A

,ATCT

=Ser A

r C

,(63)

CS1

BS1

=CA1

BA1⋅Se=

rC

Se⋅r B

=rC

r B

⋅3√ stu ,

BU 1

AU 1

=r B

Se⋅r A

,AT 1

CT 1

=r A

Se⋅r C

.(64)

KK 1

LK 1

=Se⋅KMLM

.(65)

Proof We use the invariance of the double cross-ratio (BC, A1S) = (AB, C1U) = (CA, B1T) = Se, (BC, S1A1) = (AB, U1C1) = (CA, T1B1) = Se.■

7.3.3 Sejfriedian points coordinates

The point Q of a Sejfriedian has the barycentric coordinates with respect to the triangle АВС equal to:

Q=(

1r A

,1r B

,1rC

) .

Q⃗=

A⃗r A

+B⃗r B

+C⃗r C

1r A

+1r B

+1r C

.

(66)

The points K,L,... of a Sejfriedian have the barycentric coordinates with respect to the triangle АВСequal to:

K=(1

r A Se,Ser B

,1rC

) , L=(1r A

,1

r B Se,Ser C

) , M =(Ser A

,1r B

,1

r C Se) ,

K1=(1

r A Se,

1r B

,SerC

) , L1=(Ser A

,1

r B Se,

1rC

) ,M 1=(1r A

,Ser B

,1

rC Se) .

A1=(0,

1r B

,1rC

) , S=(0,1r B

,1

rC Se) , S1=(0,

1r B Se

,1r C

) ,... (67)

Respectively, in the Cartesian coordinates we have:

K⃗=

A⃗r A Se

+B⃗ Ser B

+C⃗rC

1r A Se

+Ser B

+1rC

, L⃗1=

A⃗ Ser A

+B⃗

r B Se+

C⃗rC

Ser A

+1

r B Se+

1r C

.

(68)Proof The segments AS1 and BT1 are intersecting at the point L1. We use (32) in the form ofBS1

CS1

=r B Se

r C and

AT 1

CT 1

=r A

rC Se and the Cheva's formulas for intersecting segments.

7.4 Sejfriedian inverse properties

Let the circles centered at the vertices of the generating triangle and perpendicular to the Sejfriedcircle be given. Then the vertices of the triangles of the Sejfried pair are mutually inverse withrespect to the circles.

Proof The radius of the circle centered at A and perpendicular to the Sejfried circle is equal tor A=AAE . According to the property of a tangent and a secant we obtain AK⋅AM =r A

2 . ■

26

8 Conclusions In this paper we have proved all Sejfriedian properties described by M. Sejfried in the paper [2]. Todate, only Sejfried function is not found, which describes the position of the Sejfried circle centerrelative to the generating triangle. Beautiful construction with the variety of easily provableproperties must enter the "golden fund" of the world geometry and be used at in-depth study of thegeometry for the development of geometric vision of students. Using geometric freeware DGSGinMA we have created interactive supporting files that illustrate Sejfriedian properties.

9 Literature

[1] M.Sejfried. Amicable triangles and perfect circles., at the Proceedings of the 15th International Conference on Geometry and Grafics (ICGG 2012), pp 682-687, 2012. ISBN 978-0-7717-0717-9.

[2] M.Sejfried. Amicable triangles and perfect circles., Journal for Geometry and Graphics, Volume 17 (2013), No. 1, 53-67.

[3] M.Sejfried. Trójkąty zaprzyjaźnione i niektóre ich własności geometryczne., Częstochowa, 2010.

[4] M. Sejfried, V.Shelomovskii. Elementary Proof of Sejfriedian Properties.,at the Proceedings ofthe 17th Asian Technology Conference in Mathematics (ATCM 2012), pp 342-352, 2012. ISBN978-0-9821164-4-9.

[5] A. Акопян, A. Заславский. Геометрические свойства кривых второго порядка. М.:МЦНМО, 2007.

[6] Yiu P. Introduction to the Geometry of the Triangle. Chapter 10. General conics. р.117.[http://www.math.fau.edu/yiu/geometry.html]

[7] Я. П. Понарин. Элементарная геометрия. Т.1. – М.: МЦНМО, 2004. – 312 с.

Software Packages

[8] GinMA Software, http://deoma-cmd.ru/en/Products/Geometry/GInMA.aspx

27

Constructions

9.1 The base Sejfriedian construction

Let triangle ABC be not regular and AB ≠ BC.We want to find the base Sejfriedian.

Construction. Let A1B1C1 be incircle ABC (A1B1C1 are the points of contact). We find point D ascrossing point of the straight lines A1C1 and AC, E is crossing point of the straight lines A1B1 andCC1,(see Fig.24,a) Points M and K1 are crossing points of the straight line DE and ABC incircle.These points are vertices of two Sejfried triangles.

Fig. 18. The base Sejfriedian construction

9.2 The isosceles triangle Sejfriedian construction

Let isosceles triangle ABC (AB = BC) and Sejfried number be given. We want to find Sejfriedian.

Construction. We find points T:

ATCT

=r A Se

rC

=Se and Т2:

AT 2

CT 2

=r A S̃e

rC

, S̃e=| √Se−1+Se |√Se+1−Se

.

Let В1 be midpoint AC, b = AB1, h = BB1, d = B1T2: B1T 2

AB1

=db=

1−Se√Se

, BT 22=h2

+d 2 .

We construct a circle with center A and radius

RA=RC=2b2

√d 2+3b2=

2b

√ 1Se

+Se+1

=2b

√s+1, s=

1Se

+Se .

We construct a circle with center B and radius RB=BT 2−b2

+d 2

√d 2+3b2

=√h2+d 2

−b(s−1)

√s+1.

Let Br be the point of intersection of the circle and straight line BT2. The center I of the Sejfriedcircle is the point of the intersection of the perpendicular IBr to the straight line BT2 and the axis ofsymmetry of the triangle BB1.

The Sejfried function defined by the distance BI between fixed point B and center I:

BI=RB

sin BIBR

=RB

sin BT 2 C=

RB⋅BT 2

h,BIh

=1+d 2

h2−

bh √1+

d 2

h2

s−1

√ s+1,

d 2

h2 =d 2

b2

b2

h2=(s−2)b2

h2 .

Let us check this formula for trivial cases.

1. Let the radius of the Sejfried circle be near zero. In this case Sejfried number Se → 1, s → 2,

lims→2

BIh

=1−b

h √3. It is Fermat point.

2. Let the Sejfried circle be incircle. In this case Sejfried number Se=3−√5

2, s=3.

lims→3

BIh

=1+b2

h2−

bh √1+

b2

h2. It is incenter.

28

3. Let the Sejfried circle be circumcircle. In this case Sejfried number Se → 0, s → ,

lims→∞

BIh

=12+

b2

2h2 . It is circumcenter.

We show Sejfried function for three value of h/b: 3,4 and 7.

Points K and L belong to theintersection of the straight line BT andline and the Sejfried circle. Point M isthe point of the intersection of thelines AK and CL. Amicable Sejfriedtriangle be symmetric for KLM withrespect to the axis of symmetry of thegiven triangle.

Proof: Let Ci be the point inverse forC with respect to the circle (A, RA),RA = RC. Let Bi be the point inversefor B with respect to the Sejfriedcircle. Let D be the mutual inversioncenter (and dilation center) of thecircles with centers A and B. Then thepoints D, Bi, K and Ci by concurrent.

Menelay's condition for points D, K,and Bi concurrenсу with respect tothe triangle АВВ1 is:

RB⋅RC=BT 2⋅RC+ RC2−

AC 2

2.

The condition that the center I belong to the radical axis of the circles (B, RB) and (C, RC) is:

2 RB⋅BT 2=RB2−RC

2+ BC 2.

The solution of the system of equation is: RC=AC 2

2√BT 22+ AC 2

−BC 2.

Fig. 25. The Sejfriedian of the isosceles triangle construction

29

Fig. 26 The the points D, Bi, K and Ci concurrence