Mem.Fac.Educ.、Kagawa univ. II、58(2008)、31-45
Plane Algebraic Curves Drawn
by thelsotomie Conjugate 11)ra7[¥iangle
−Applicationsof a Drawing lool and ]Vlathematica−
by
Keisuke M:ATsUMOTo,Kazunori
FUJITA alidHiroo FUKAlsHI
(Received May 30, 2008)
Abstract
ln this paper we present a cQmputer tool for drawing a locus of the isotomic conjugate to a point for a triangle which moves along a distinguished curve. The drawing tool provides many plane algebraic curves witb simple expressions. <
§1.1ntroduction
ln the sequel to [4]'[9],[20]our
study aims
to develop
a drawing
tool on a
display for experimental
research on various curves using computers.
ln elementary
geometry
we have five significant notions for a triangle; that is,the
center of gravity,the center of an inScribed circle,the center of an escribed clrcle,the
circumcenter
and the orthocenter of i triangle,As a similar notion we have the isotomic
conjugate to a point for a triangle.
`Whichcurve is drawl! as a locus of l;heisotomic
conjugate
to a point for a triangle
which
moves
under
a certain condition? ln this study we limit ourselves
to the case
where
a triangle is fixed while the point moves
along a distinguished curve. Then
our
main
concem
is to find various unknown
curves with simple algebraic expressions
as a
The last author was paltiallysuppolted by Grant-in-Aid florScientificResearch(No. 19500761), the Ministry of Education, Culture,Sports,Science and Technology, Japan。
K.MATsUMOTO,K。FU,IITA and H. FUKAlsHI
locus of the isotomic conjugate to a point for a triangle.For terminology of geometry
throughout the paper, consult[2]−[31,[11],[13],[14],[18]and[22].
§2.A
program
for drawing a locus
On a plane let us consider a given triangle △Å召C andai)oint j)different from the vertices of tlle triangle.
Suppose lines Aj),召j),C7)intersect the opposite sides 6f a triangle △Å召C at points,£),£,F and Z:)・s・E19F・ be the reflections of に)・,E,Finthe midpoints ofthe corresponding sides. Then the lines λZ:)',召£',CLF' always intersect at a point ?', (or are
.pal‘allel,1.6ム,intersectin the point of infinity (x)),calledthe j加zθz?zjccθ司昭αた to」1)for 4AjC ; the map ψ :R\{Å,j,(フ)→RU{oo}defind by ψ(だ)=j="is dalled the 泌応 「心明/昭d回for△Å斑].
L6tA(s,r),召(−1,0),C(1,
0)be
the vertices of a triangle where
z≠O,and
?(M, 1')a
pOint different frOm
the vertlCeS. Then
the p01nt P’(XI,
y)iS
determined
aS
thejntersection of two straightlines
jS£'and
CF'
given by
・奉 ・・ y° y° 召バ雨+,4)−(1+4)v}
z(1+s)(1+j)+(3+y)(1−j)y
バz(zj−1)+(1−s)y}
z(1−s)(1−z,)+(3−s)(1+s)v
(い1),
/
.
. 回忿 ̄胴曰ビ= ̄屁J 、(レに1)
£ ) ぴ Å CPlane
AlgebraicCurvesDIJawn by thelsotomic Colljugateflolj
a Tfliangle
Proposition. The
ise7ie
Qf the point Qf 垣紅吻,‘φ ̄1(Qo),jりμwzゆy涜a面pg
z2λ12−2m:y十G2十3)y2−2り'−z2=O,
W11jdz芦z&s6法j'り叫冶哨Ewrrjc6Å,召,Cαz
「&z・μ/zEみαzでycazMrげ△Å召Cα・sj£scEnたz:
j)roo:f1 Let
7:)be apoint
such
that ψり))゜oo.
The
slope
s・・ of the linφjS£'is
equal to the s16pe j恥
レ
of the line F'C. 万
\ 犬
Find
the Groebner
bases (see[11,[12])by
applying
Mathematica
ver 6.0by
M/olfram Resea:rch lnc・,to the following code :
1
U:゜X
v:゜y
fl:=(t*(1+s)*(1+u)+v*(3+s)*(1−s))*mb-t*(tft*u-v-s*v)
f2:=(t*(1−s)*(1-u)+v*(3-s)*(1+s))*mc-t*(-t†t*u+v一s*v)
f3:=mb-mc
Gro9bne7`Basis[{f1,f2,f3}・,{!nc,mb,t,s,y,x}]
Factor[Z]
Then we obtain the term
z(−r2十&j2−2印−2my十3y2十jy2)
in thelistof the generating Groebner bases. Therefore, we have
(**) ゐ
;
2−
2m
E
y十(3十s2)y2−2印−z‰0
as anequation of ψ ̄1(oo)whichpresents the ellipse,because
r2(3+s2)−Gz)2>o.
Setting the leftside of (**)byパyy),we
have the following:
・
1
)The
emPse
(**)passes through the vertices of △Å召C,
ii)The
ellipse (**)has the point
of the equations
が
一−
∂,χj
汗
ル)as
its center, which is given by the solution
O and
が
K.MATsUMOTo,K。FU,IITA and H. FUKAlsHI
Which
curve is drawn
as a locus of the isotomic
conjugate
7)’
to a point, 7)for
△AjSC
which
moves
along a distinguished curve
y
?
We
divide our operation of a drawing tool into the drawing part and the printing
part,due to the circumstances of our computer machines.
PART
ONE : To
draw
a locus of the isotorilicconjugate to a point for a given triangle.
A program
of our drawing
game
for Case
l of Theorem
ia § 3 is written in visua1
B asic Ver. 6.0 by Microsoft
Corporation
and consists of the following
five st6ps `(see
List l).
Step 1.
Step 2.
Step 3.
Step 4・
Step5.
Set the coordinates axes and triangle △Å召C in black.
Set the initialpositk)n of a point ?on
curve
ー(in blue),
Draw
the ellipse that pre帥ntsψ'1(邨)in
yellow ocher.
Plot the iiotomic conjugate j)'in red.
ニ
When
one moves
the point j)continuously
along the curve
ー by the mouse。
the point j)'continuously
draws a locus ,y with a solid line in red.
犬
PART Two
: 7ITo process a bitmap file(bmp)bypびIEX2ε'to
exhibitit on another
display,an4 to printit.
OUTLINE
of the PROGRAM. Let
A (0,2),召(−1,0),C(1,0)and
y:y=た.Where
a
pOint戸(Z4,1ノ)On
ーiS
SeleCted by the mOUSe,
then the iSOtOmiC
COnjUgate
7)'(XI,
y)to
■ I ・ ●
?is deter面ned
by the equations (*)for
G,r)=(0,2).
The program
for drawing
a locus y
of the point 7)'is given in List l. ln thiscase
we win have the curve
y: 4む2−8xy十3げ−4肪一計=0
on a display (Fig. 1).
Å(0,2),j(−1,0),C(1,0),
yレy=た(--?十1)
犬
(λ
;
≠O)レ彪卯
「・必z/
y: 4(た−2)x2−3(3た+2)y2=4(3た−2)y−4た+8=o:
i)the
straightline
ii),the ellipse
iii)the parabola
iv)the
hyperbola
忙ド
{J,た≠−1,(い
forん=2,
orん>2.
1・4(づ
for −でF<ん<2, forん=−{,Plane Algebraic Curves Drlawn by thむlsotomic Co哨ugate f6r a Triangle
§3. Algebraiccurvesobtained
as a locus of the isotomic conjugate
As a locus of the isotomic
coiljugate to a point for a given triangle we have varioui
curves. ln
this section
we exhibit some
of them
together
with
the cases of plane
algebraic curves
with simple expressions
WhoSe
graphs
are not given in the standard
texts of the subject [101,[121,[151,[171,[191,[21]and[251.
Theorem,£α油Jげがz・りW/θw訥gαな訪Γαjccμn.?Es 。ya,H,ε涌zαj,z 气ソHz勿csげ
the isotomic conjugate l)’Zθαjpθ訥Z?知r ・2 rrjα凧grZE△Å召C w/zjc/1謂θW・∫α/θzlgが1・E! czzr1.ぞ ー。 : 下
Case l (Fig.1)
Å(0,2),S(−1,0),C(1,0).
ー: χI=k,the stmight nne,
y:
4尨ヲー8,xy+3戻−4砂−4灸=o
thatis, '
i)the
straightline f()r匹−1, 0,1,
ii)the ellipse
iii)the parabola
)V)`the
hyp&bohi
Case 2 (Fig.2)
that is, forた</IJ
or
forλ
;
゜±{J,
f。fづニ{みdx
K.MAT`sUMOTo,K.FU,JITA and H. FUKAlsHI
Case
3 (Fig.3).
Å(0,−3),j(−1,0),C(1,0). .j
犬上
‘ぎ: y°.χ?−4,哨εραn2みθZα.
y
:
5y4十74y3十2(x2十5)y2十48(x2−1)y十9(x2−1)2=0
Case 4 (Fig.4). .・
Å(0,−1),j(−1,0),C(1,0).
y:\χ12−y2°−1,thehyperbola,
y : y3−y+ (3×2−1)y−(x+1)(x−1)=o
Case 5 (Fig.5).
Å(0,3),j(−1,0),C(1,0).
ー: ,x2十(y−1)2=1,向d
「ε.
y:
3y4−28y−2(7×2 − 45)y2十108(x2−1)y+27(x2−1)2=0
Case
6 (Fig.6).
Å(0,1),j(−1,0),C(1,0).
十
y
:,x2十(y十1)2=4,向d
「ε.
y: 2:y3 − 3y2− (x+1)(x−1)=o. ◇
j)n9げ Let れz4,V)aj j)″(XI,y)・
エ
Case l. Find
the Groebner bases by applying Matheinatica Ver 6.0 to the following code :
S:=0
u:=k
fl:=(t*(1+s)*(1+u)+v*(3+s)*(1−s))*mb-t*(t+t*u-v-s*v)
f2:=(t*(1−s)*(1-u)+v*(3−s)*(1+s))*mc-t*(-t+t*u+v-s*v)
f3:゜y-mb*(x゛1)
f4:゜y-mc*(x-1)
GroebnerBasis[{fl,f2,f3,f4}・,{mc,μ},v,k,y,x}・]
Factor[Z]
Then we obtain the term
r(一尨2十kty)cl− 2抑−4吋十3が)
Plane Algebraic Curves Drawn by the lsotomic Co哨ugate for a Trliangle
奴x2− 8巧十3砂2−4妙−4ん=0
as an equation of ,y.
K.MArsUMOTo、K.FU、IIIAand H. FUKAlsHI
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Plane Algebraic Curves Drawn by the lsotomic Cor!jugate for a Triangle
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Keisuke MATsUMOTO
9-121-2 lkeda-cho/lajimi-shi,Gifu,507-0048,JAPAN
Kazunori FUJITA
Depaltment of Mathematics, Faculty of Education, Kagawa university 1-1 Saiwai-ch0, 111kamatsu-shi,Kagawa,760-8522,JAPAN
£-z7zd 殍,as:fujita@ed。kagawa-u。ac.jp
Hiroo FUKAlsHI
Department of Mathematics, Faculty of Education・ Kagawa university 1-1 Saiwai-cho,Takamatsu-shi,Kagawa,760-8522,JAPAN
Case
Case
ん=1
ん=
且
K.MArsUMOTo,K.FUJITA and H. FUKAlsHI
Case Case 左=2.300 た=0.667 Fig. 1 ( ー y
λ7=ん
4旨七8ぞ十3似七4似一収=0
Case ん=2
Case ん=
Plane Algebraic CurvesDrawn by the lsotomic Conjugate for a Triangle
2 −3 Fig. 2 貿 y Case た=−0.300 Case 灸=−3.000 y 4 =ん( y十げ(ん≠o) (ん−2)y−3(3ん+2)ダー4(3ん−2)y−4ん+8=o
Fig. 3 貿 y
Fig. 5 貿 y
K.MxrsUMOTo,K.FUJITA and H. FUKAlsHI
jにぞ−4 5戸+24y3十2(ぞ十5)戸十48(x2−1)y+9(x2−1)2=0 Fig. 4 貿 y ぞ亘y−1)2=1 3ダー28ダー2(フズ2− 45)ダ十108ヅー1)y十27(y−1)2=0
ぞーゾニー1
ダーダ十(3y−1)バ(x十1)(x−1)=o
Fig. 6 貿 y ぞ十(y十1)2=4 y−3ダーG+1ぬ−1)=o- 一 4 ∼
Plane Algebraic Curves Drawn by the lsotomic Co哨ugate f6r a Triangle
List 1. A prggram
for drawing a locus
一 -ll● | -pse φ'1((x)) 一 一 -∼
that presents
Specifying statements
of the gineral viiriables
Dim
SX, sy,ex,ey
As Double
Dim
ax, ay, bx,by, cx,cy, dx, dy As Double
Dim
qx, qy As Double
Dim
px As Double
Drawing
the initial figure and the elPrivate Sub Form_Activate()
Forml.AutoRedraw=True
FormLLine(sx,0)-(ex,0) Forml.Line(0,sy)-(0,ey) h=0.06
For x =lnt(sx)To lnt(ex) Forml.Line(X,−h)-(X,h) Next x
For Y =lnt(sy)To lnt(ey) Forml.Line(-h,Y)-(h,Y) Next Y
Forml.DrawWidth=2 `Forml.Line(ax,ay)-(bx,by) Forml.Line(bx,by)-(cx,cy) Forml .Line (cx,cy)-(ax,ay)
pi=4 ・ Atn(1)
For t =O To 2 i pi Step 0.001 。・
レ d・=3・(ayべ2 ・ Cos(t)A 2 -・ 2 ・ ax t ay・1 Cos(t)・Sin(t)十(axA 2・ 十 3)1・ Sin(t)・A 2) r=Sqr((4・ayA2.)/d) ×=ax/3十r yCos(t) 。 = Y=ay/3十r ・ Sin・(t) Forml.PSet(×,Y),QBColor(6) Nextt Forml.AutoRedraw=False Forml.DrawWidth=2 Forml.Line(2,sy)-(2,ey),vbBlue Forml.DrawWidth=3 px=2 Textl.Text=px Locus End Sub
1 6 S 6 ax bx 1 - 一 一 4 g S S
-K.MAT`sUMOTo,K.FU,IITA and H. FUKAlsHI
- 一 一 - - 一 一
一 一 一 ・ 一 一
-Setting the form, the coordinate axes and the initial
values of the coordinates of each vertex of a trian9le
O: −1
Private Sub Form_Loado
Forml.T6p=500
SX=-4.8
ex=4.8
WX=ex
− SX
wy=wx
t Forryil.ScaleHeight / Forml .ScaleWidth
sy=-0.51wy
ey=0.5吟wy
づ
Forml .Scale(sx,ey)-(ex,sy) .・
Forml.BackColor=VbWhite
CX=1: Cy =OI
Textl.FontSize=16
End Sub
Calculating the intersection of two lines
qy=(rD・ t2 − r2 り:1)/d End Sub
-Private Su b.lntersection(xa,ya,xb,yb,xc,yc,xd,yd)
rl=ya
− ylb:sl =xb − xa: tl =ya゛xb−xa
゛yb
'
r2=yc
−yd: s2 =xd−xc:t2
°ylc゛xd−xc゛yd
d=rl
●s2 − r2・きsl
qx=(tl ・ S2 − t2・ SI)I/ d・
left button of mouse
Action corresponding
to the
Forml.Line(X,sy)-(X,ey),vbBlue Forml.DrawWidth=3 px=X Textl.Text=lnt(1000 1 px)/1000 Locus End Sub
Private Sub Form_MouseDown(Button
As lnteger, Shift As lnteger, X As Single,,YAs Sin9le)
Forml .Cls .
g
g
S
Plane Algebraic Curves Drawn by the lsotomic Coりjugate for a Triangle
- ・
Presentin9
一 一
-of
the isotomic conjugate
Private Sub Textl_DblClicko
px=val(TextLText)
Forml.Cls
Forml.DrawWidth=2
Forml.DrawWidth=3
LOCus
End Sub
Private Sub Disp(py)
lntersectionax,ay,px, py, bx,by,cx,cy dxl =qx: dyl =・qy dx2=cx − dxl 十bx dy2=cy−dyl +・by ・・ lntersection bx, by,px,py,cx,cy,ax,ay exl =qx:.eyl=qy ex2=ax−e)tl十cx ey2=ay − eyl 十cy
lntersection ax, ay,dx2,dy2, bx, by, ex2,ey2 pdx=qx: pdy ° qy
Forml.PSet(pdx,pdy),vbRed End Sub
Private Sub Locu50
For Y =50 t sy To 3 ゛sy Step O.1 Disp Y.
Next Y
For Y =3 ・ sy To 3 ・i ・ey Step 0.001
Disp Y ▽
Next Y
For Y = 3 ・ey To 50 1・ ey Step O.1
1 Disp Y ・,
Next Y
End Sub