geometry for cosmos by h.e.pdf

Our Life of Geometry for Humankindby A researcher

2011-12-4 at A cafe shop

Circle is not only Circle, But It is a start of Theorem with lines

Geometry for Cosmos by H.E

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DOVAL

http://hoval.blogzine.jp/


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- 1 -

2011 9 29

DOVAL


DOVAL

DOVAL

2

Doval

DOVAL

X,Y

DOVAL

DOVALDOVAL


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Doval (12 http://aitoyume.de-blog.jp/

c

b d

i

j

k

a

e

f

g h

AB

CDE

F

G

H

I

JK

Doval Third DEF by H.E

P.56 Doval??? by?????

??????Now this is managed on http://hoval.blogzine.jp/


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P.2 Doval??? by?????



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2 (Dovd)

7400012 4 12 10Eelllail,1lII1111111,

7(pCl4Aea of her and Outer Part of Oval(Doval)which is double closed Cwes

2

Dovd

Doval

2

2

(Dovd)Zrl=

:=22

Doval31)2

3kmc/(mA2nA2), Icnc(mA2lllA2)mnc/(mA2nA2)

,(nA2c/(mA2-nA2),0)

[ly]1/) :! .

_2 cos)2s292222:::}rl=2_2

~

+

P.20 Doval??? by?????



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O O

O O

O O

O O

O O

O O

O O

O O

O O

O O

O O

O O

O O

O O

O O

# DOVAL CG by x-y STANDARD FORMULA transformed from Bipolar coordinates(m$R1GnR2=k$c) by Hirotaka Ebisui

with plots :#:#-------------------------------------------------------------------------------------------------------------

-----------------:# Doval(The Inner ond Outer Oval of Descartes is defined by Followings 4th Order x-y

Algeblic Equation.

# m2 Kn22$ y2 C x C n

2$cm2 Kn2

2

Kk2$m2 Ck2$n2 Cm2$n2

m2 Kn22 $c

22

=

K8$k2$m2$n2$c3

m2 Kn2$ x C n

2$cm2 Kn2

C4 k2$m2$n2$ k2 Cm2 Cn2 $c4

m2 Kn22 :

#k,m,n:Arbitraly constant with a condition k O m O n O 0 , c : the distance between Fisrt and Secand focus points :

#-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------:

# Example 1:m d 9 :n d 6 :k d 10 :c d 1 :#:

implicitplot m2 Kn22$ y2 C x C n

2$cm2 Kn2

2


m2 Kn22 $c

22

=

K8$k2$m2$n2$c3

m2 Kn2$ x C n

2$cm2 Kn2

C4 k2$m2$n2$ k2 Cm2 Cn2 $c4

m2 Kn22 , x =K20 ..10, y =

K10 ..10, numpoints = 100000 ;

P.22 Doval??? by?????



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O O

O O

O O

O O O O

O O O O

xK5 K4 K3 K2 K1 0 1

y

K3

K2

K1

1

2

3

# Example 2:m d 6 :n d 4 :k d 10 :c d 1 :#:

implicitplot m2 Kn22$ y2 C x C n

2$cm2 Kn2

2


m2 Kn22 $c

22

=

K8$k2$m2$n2$c3

m2 Kn2$ x C n

2$cm2 Kn2

C4 k2$m2$n2$ k2 Cm2 Cn2 $c4

m2 Kn22 , x =K20 ..10, y =

K10 ..10, numpoints = 100000 ;

P.23 Doval??? by?????



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O O

xK6 K4 K2 0 2

y

K4

K2

2

4

P.24 Doval??? by?????



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e(X,Y)

(0,0)(a,0)

(0,a)

(0,b)

rf.4

X=a cos(e)

Y=b sin(e)


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Doval

Y=a*cos(e)sin(h)t*sin(2h)

X=a*cos(e)cos(h)t*cos(2h)

2011-6-24

,

a

e,h

e

(X,Y)


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e

(X,Y)

a

t

(0,0)

X=t+a Cos(e)Cos(h)-2t Sin(h)^2=a Cos(e) Cos(h) +t Cos(2h)

Oo F F1

P

2O12O21

P,mR1+nR2=kC

F1F2C

a=C

mnC

e

Y=a Cos(e)Sin(h)+2Sin(h)Cos(h)=a Cos(e)Sin(h)+t Sin(2h)

Dovaleh

X=a Cos(e)Cos(h)+t Cos(2h)Y=a Cos(e)Sin(h)+t Sin(2h)

2011-4-10?????SPRING HI-TK-S-48

http://aitoyume.de-blog.jp/

??????

?61


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risou

Doval

Y=b*cos(e)sin(h)-t*sin(2h)

X=b*cos(e)cos(h)-t*cos(2h)

(X,Y)

b

h

e

2011-6-24

e,h

a

,

2011-6-24

e

(X,Y)

X=a*cos(e)cos(h)t*cos(2h)

Y=a*cos(e)sin(h)t*sin(2h)

Doval

P.59 Doval??? by?????



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(=

a

E,H

Ee E1EE2ePiH KHPi

X,Y)

2 (

x=a*cos(e)cos(h)t*cos(2h)

y=a*cos(e)sin(h)t*sin(2h)

E

e

H

X=b*cos(E)cos(H)-t*cos(2H)

Y=*cos(E)sin(H)-t*sin(2H)

Doval 2011-6-24

e,h

P.62 Doval??? by?????



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2011

3

Pr.1-1 Pr.1-2

Pr.1-a


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Pr.2

2011

Pr.2-1

Pr.2-b

Pr.2-PB


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2011

3 i

:Pr.3

Pr.3-a


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2011

Pr.1

Pr.2 Pr.3

Pr.4 Pr.5Pr.6


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42

Pr.4-0

2011

43

Pr.4-1


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by H.EBISUI

54

3

2

1

by H.E

4

2011

Pr.5-0-a


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' 8 (ABCD

Pr.5-0-b


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2009-3-15

5-2

:

Pr.5-0-c


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6

HEXAGON THEOREM

423

GGJFGJ-001)-01

2011

6

Pr.6-0-a


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2011-9-66' .HEXAGON 5 ten teiri

2011

Pr.6-0-a-1


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2011-9-67.HEX 10 ten teiri

2011

Pr.6-0-a-2

:


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Projective Geometry Theorem named as ROSE Theorem by Hirotaka EbisuiHirotaka Ebisui

(10.97, 15.42)

(21.17, 7.42)

AB

C

D

E

FG

H

I

K

L

M

N

O

PQ

R

S

T

U

V

W


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Projective Geometry Theoremon Rose theorem with it's Proof of RED ROSE and Line only Figure

Hirotaka EbisuiOval Research Center

[email protected]

Abstract:We found ROSE THeorem 5 year ago. And, We found Appending Theorem which areBlue and Mixed Rose theorem derived from red and blue rose. Red Rose theorem are proved.And We can constract Rose thereom using Two lines on which 4points each are given insteadof Circle on which 8points are given.From this view point, we can say," Rose theorem are projective theorem that keep line and circlecombination like Papus and Pascal Theorem".In this paper, We show 7 sheets where Compositionof RED ROSE theorem , it's PROOF figure with brief explanation,and Line only figure are drawn.We show them on following pages.

PROOF-STEP Of RED ROSE THeorem 2010-4-(1213)Diagram-1

Construction Order

Composition Expect figure Conformation figure

Projective Geometry Theorem on Rose Theorem ATCM2011 Taiwan Dec17-19

BYHirotakaEbisui P.1


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RED Rose Line Theorem

2006-8-7

2011-10-26

Projective ROSE Theorem




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ABC

D

The Proof Principale Of RED ROSE THEOREM is Here.

1

4

PASCAL Line

A,B,C,D construct Pascal Collinear line 1,4,7

Fig.1

Hirotaka Ebisui

Elementaly Pascal Theorem Figure




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6 points PASCAL-line reduse 4points pascal-line




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ABC

D

E

F


1

2

3

4 5

6

Hirotaka Ebisui

Brianshon Point

Tangent Hexagon of makes Brianshon Point of A,B,C,D,E,F

Fig.2




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ABC

D


G

H

1

9

10

11 4 8

19104118Hexagon construct Brianshon Poit of A,D,G,C,B,H

Brianshon point

At the same Time 3 Pascal lines are concurent on Brianshin Point

Fig.3




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by H.E

A

BC

D

E

F


Brianshon Point of A,B,C,D,E,F is on Pascal Line of A,B,C,D

Tangent Hexagon of makes Brianshon Point of A,B,C,D,E,F

These reason set establish the proof of RED ROSE THEOREM

1

2

3

4 5

6

Hirotaka Ebisui

http://aitoyume.de-blog.jp/

2010-12-19 Finished

2010-4-12Proof days

found

PASCAL Line

Brianshon Point




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RED nad Blue Mixed Rose Theorem and it's Line only Theorem

Blue Rose Theorem and it's Line only Theorem




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Projective lines Theorem

Appendix Projective 11 lines Theorem by Hirotaka EbisuiGeometry for Cosmos by H.E

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42

24

4

2011-10-20

A deffinition on Incenter and Orthocenter of Quadrangle by Hirotaka Ebisui

ATCM2011 Taiwan on 17?19th DEC


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Orthocenter of Quadrangle on Circle

Orthocenter of Quadrangle on Plane

2011-11-7

for ATCM2011 Taiwan

A deffinitionon Incenter and Orthocenter of Quadrangle by Hirotaka Ebisui

ATCM2011 Taiwan on 17?19th DEC


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Orthocenter of Quadrangle on Circle

Orthocenter of Quadrangle on Plane

2011-11-7for ATCM2011 Taiwan

General Orthocenter is same as Special Orthocenter

General Orthocenter is same as Special Orthocenter

2011-11-19

General Orthocenter and Special Orthocenter By Hirotaka Ebisui

ATCM2011 Taiwan on 17th?19th DEC


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- 1 -

456

456

4 2 O4 4 2 4 4 1 O4 4 4

5 5 5

6 6 3 5 63 4 35 1 44 1 4 2


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- 2 -

2011-11-20


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- 3 -

2011-11-20

Orthocenter of Hexagon


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>>

>>

(1)(1)

>>

>>

>>

>>

(4)(4)

>>

(2)(2)

(3)(3)

# sakna1 by H.E:

with(plots):

P:=sin(5*sin(x2)^3*cos(x1)+6)+2*sin(x2)*cos(x1)^2+3;

P := sin 5 sin x2 3 cos x1 C6 C2 sin x2 cos x1 2 C3

Y1:=simplify(P*sin(x2)*sin(x1)+diff(P,x2)*cos(x2)*sin(x1)+diff(P,x1)

*cos(x1)/sin(x2));

Y1 := Ksin x1 sin x2 sin K5 cos x1 sin x2 C5 cos x1 sin x2 cos x2 2 K6

C2 cos x1 2 K3 sin x2 K20 cos x1 cos x2 2 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6 C15 cos x1 cos x2 4 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6 C5 cos x1 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6

Y2:=simplify(P*sin(x2)*cos(x1)+diff(P,x2)*cos(x2)*cos(x1)-diff(P,x1)

*sin(x1)/sin(x2));

Y2 := Kcos x1 sin x2 sin K5 cos x1 sin x2 C5 cos x1 sin x2 cos x2 2 K6

K2 cos x1 3 C3 cos x1 sin x2 C20 cos x1 2 cos x2 2 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6 K15 cos x1 2 cos x2 4 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6 C5 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6 K5 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6 cos x2 2 K5 cos K5 cos x1 sin x2

C5 cos x1 sin x2 cos x2 2 K6 cos x1 2 C4 cos x1

Y3:=simplify(P*cos(x2)-diff(P,x2)*sin(x2));

Y3 := cos x2 Ksin K5 cos x1 sin x2 C5 cos x1 sin x2 cos x2 2 K6 C3 K15 cos

K5 cos x1 sin x2 C5 cos x1 sin x2 cos x2 2 K6 sin x2 cos x1 C15 cos

K5 cos x1 sin x2 C5 cos x1 sin x2 cos x2 2 K6 sin x2 cos x1 cos x2 2

plot3d([Y1,Y2,Y3],x2=0..Pi,x1=0..2*Pi,scaling=CONSTRAINED);


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> # ovaloid-3d sakana 1 kuchiakeanime by H.E:> with(plots):Warning, the name changecoords has been redefined

> n1:=sin(x2)*sin(x1):> n2:=sin(x2)*cos(x1):> n3:=cos(x2):> P:=sin((4+s)*sin(x2)^3*cos(x1)+5+s)+2*sin(x2)*cos(x1)^2+3;

:=P ( )sin ( )4 s ( )sin x2 3 ( )cos x1 5 s 2 ( )sin x2 ( )cos x1 2 3> Y1:=simplify(P*sin(x2)*sin(x1)+diff(P,x2)*cos(x2)*sin(x1)+diff(P,x1)*cos(x1)/sin(x2));

Y1 ( )sin x1 ( )sin x2 ( )sin 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s(:=2 ( )cos x1 2 3 ( )sin x2 16 ( )cos x1 ( )cos x2 2 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s12 ( )cos x1 ( )cos x2 4 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s4 ( )cos x1 ( )cos x2 2 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s3 ( )cos x1 ( )cos x2 4 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s4 ( )cos x1 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s

( )cos x1 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s )> Y2:=simplify(P*sin(x2)*cos(x1)+diff(P,x2)*cos(x2)*cos(x1)-diff(P,x1)*sin(x1)/sin(x2));

Y2 ( )sin x2 ( )cos x1 ( )sin 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s:=2 ( )cos x1 3 3 ( )sin x2 ( )cos x1 16 ( )cos x1 2 ( )cos x2 2 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s12 ( )cos x1 2 ( )cos x2 4 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s4 ( )cos x1 2 ( )cos x2 2 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s


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3 ( )cos x1 2 ( )cos x2 4 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s4 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s4 ( )cos x2 2 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s4 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s ( )cos x1 2

( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s ( )cos x2 2( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s s ( )cos x1 2 4 ( )cos x1

> Y3:=simplify(P*cos(x2)-diff(P,x2)*sin(x2));

Y3 ( )cos x2 ( )sin 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s 3 (:=12 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s ( )sin x2 ( )cos x1

12 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s ( )sin x2 ( )cos x1 ( )cos x2 2

3 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s ( )sin x2 ( )cos x1 s 3 ( )cos 4 ( )sin x2 ( )cos x1 4 ( )sin x2 ( )cos x1 ( )cos x2 2 ( )sin x2 ( )cos x1 s ( )sin x2 ( )cos x1 s ( )cos x2 2 5 s ( )sin x2 ( )cos x1 s ( )cos x2 2 )

>> animate3d([Y1,Y2,Y3],x2=0..Pi,x1=0..2*Pi,s=0.5..1,scaling=CONSTRAINED);


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-47-

- 1 -

740-0012 4 12-10

E-mail [email protected]

On -Dim rectangle devided equally and its generalization

Hirotaka Ebisui

Oval Research Center, Motomachi 4-12-10 Iwakunisi 740-0012 , JAPAN

AbstractThe 2-dimensional rectangle whose edges have golden section 1:(1+sqrt(5))/2, are

extended to a n-dimensional rectangle by Ebisui. Another problem is to extend the rectangle, whose

edges have the ratio 1:sqrt(2), to n-dimensional rectangle, whose edges satisfy the equation

The length of the k th edge is ( k=0...n-1) .

This result is generalized to the case with edge ratio :

Values of (k=1... n-1)is obtained. And some figures of 4-dim case are shown.

Keywords: Hyper rectangle, A4 form, Golden section, Similarity

1

[1]

1

3

2

1:2

:1 11xx 21 x1x

21 ,,1 xx 21 , xx

1:2

:2

::1 2121xxxx

1::...:::...:::11

1

2

2

1

1121

n

nn b

xbx

bxxxx

1:2

:...:2

:2

:...:::1 121121

nn

xxxxxx

)(2 n

k

kx


-48-

- 2 -

2

2 2 2 3

3

(2)

3 ,

3

4 2 12

1:x=x/4:1 3 1 2

21

21 122 xxxx

232

31 )2(,2 xx

21 , xx

1:3

:3

::1 2121xxxx

3 4

232

31 )3(,3 xx

23 )4(,

2x


-49-

- 3 -

3

3

3

1 b

n-1

3 4 3

(a) (b) (c)

3

1::...:::...:::1 121121 bx

bx

bxxxx nn

nkk bx

/

nnb /)1(

1nx


-50-

- 4 -

..624.3:..627.2:..904.1:..380.1:15:5:5:5:1

..344.3:..236.2:..495.1:15:5:5:1

..924.2..710.1:15:5:1

..236.2:15:1

..031.3:..297.2:..741.1:..320.1:14:4:4:4:1

..828.2:2:..414.1:14:4:4:1

..520.2:..587.1:14:4:1

2:14:1

..465.2:..933.1:..552.1:..246.1:13:3:3:3:1

..280.2:..732.1:..316.1:13:3:3:1

..080.2:..442.1:13:3:1

..732.1:13:1

..741.1:..516.1.1:..320.1:..149.1:12:2:2:2:1

..682.1:..414.1:..189.1:12:2:2:1

..587.1:..260.1:12:2:1

..414.1:12:1

5 45 35 254 34 243 232

5 45 35 254 34 243 232

5 45 35 254 34 243 232

5 45 35 254 34 243 232

1 , n-1

(k=1... n-1)

b1=4, b2=3, b3=2

Maple

.

( 3 4 4

5 )b=2 5

2 3 4 5

2

3

4

5

3 4

A4

1

1::...:::...:::11

1

2

2

1

1121

n

nn b

xbx

bxxxx

]}[{1

1

1

1

j

k

j

n

i

ni bb

kX

)4/1()2/1()4/1( )24(2/)24()24(41

2/32/14/34/24/1 2:2:2:14:4:4:1


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- 5 -

1/2

3 3

, b 3

1

[1]

Hyper SpaceVol 2, No 3, p.18-231993

x=x 1:(1+5^(1/2))/2

1/(x-1)=x/(y-1)=y/(z-1)=z

z=1/(x-1)

1.927561975

1.787933192

1.518790064

4

1X X X =X X X ....Xn 1

Xn Xn^n =X +Xn^(n-1) ....+Xn+

xn-1,xn-2x1


-52-

For HumankindOur life of Geometry for Humankind 2011-12-4.pdfAdeffinitiononincenter and Orthocenter of quadrangleHI-ORTHOCENTER of QUADRANGLEHI-ORTHOCENTER of QUADRANGLE 2 type

sakana1sakana1-kuchiake anime

geometry for cosmos by h.e.pdf

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