Plane Algebraic Curves Drawn by the Orthocenter of a Pedal Triangle : Applications of a Drawing Tool and Mathematica-香川大学学術情報リポジトリ

肘･．.aむ.￡ゐむ.，x昭αｗαび,li,d7，57(2007)，51-72

Plane Algebraic Curves Drawn

by the Orthocenter of a PedaI Triangle

Applications of a Drawing

7n)oland Mathematica

by

Kazunori FUJITA and Hiroo FUKAlsHI

(Received May 31, 2007)

Abstract

§1.1ntroduction

   lnthis paper we present a computer too! for drawing a locus of the orthocenter of the pedal triangle for a triangle with two vertices fixed when one moves the third vertex along

a distinguished curve. The drawing tool provides many plane algebraic curves with simple expresslons.

   ln the sequel to [4]−[7]，[14]our study aims to develop a drawing tool on a display for experimanta1 research on various curves using computers.

   ln elementary geometry we have five significant notions for a triangle； thatis，the center of gravity，the center of an inscribed circle，the center of an escribed circle，the circumcenter and the orthocenter of a triangle.

   Whichcurveis draxvn as alocus of such a point of a pedal triangle for a triangle with two vertices fixed on the plane when the third vertex moves under a certain condition? ln this study we limit ourselves to the case where two vertices of a triangle

   The second author was partiallysupported by Grant-in-Aid for ScientificResearch (N0.19500761)， the Ministry of Education, Culture, Sports，Science and Technology, Japan･

K,FUJITA and H.FUKAlsHI

are fixed while the third vertex moves

along a distinguished curve. Then

our main

concem

is to find various unknown

curves with simple algebraic expressions as a locus

of the orthocenter of a pedal triangle.

   Forterminology of geometry throughout the paper, consult [2]−[3]and[16].

§2.A

program 蝕･drawing a locus

Let us consider a triangle △AjSC given on a plane.

LetA＝(貼,‰)，召＝(み,yﾑＣ＝(貼よ)，Then

the coordinates of the orthocenter

召’of△Å召C is given by (*) where

Ｒ

巧

-み

石

-λ7c， λc， qxl 92 一 一 -ぐ yか ‰ r1 乃 -92 − g2 y白 yo 一rlq1  P1r2 −巧rl −plqy p192 −鳶91 r1 r2 -じ恥 石 (心 (石 -う／ 恥)十知(yか 一刄 ）

石)十勤(知一刄)

ツ

  Let7)，￡，F be the feet of the perpendiculars from the vertices A，召，C to the opposite sides，respectively. The △￡)￡F is called a pedaHriangle tor△Å召C， Denote the orthocenter of △ＤＥＦby瓦｡

  Whichcurveis drawn as a locus of the orthocenter ￡of the pedal triangle △DEF tor △Å召C with the vertices 召，C fixed when the third vertex Amoves 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 : 7ITbdraw a locus of the orthocenter of a pedal triangle for a given triangle. A program of our drawing game for Case ll of Theorem in §3 is written in visual Basic Ver. 6.0 by Microsoft Corporation and consists of the following seven steps (see List l).

Step l. Set the coordinates axes and two fixed point j， C in black and a curve yin     blue.

Plane Algebraic Curves Drawn by the Orthocenter of a Pedal Triangle

     in black.

Step 3. Plot each foot ﾌ:)，Ｅ，Ｆof the perpendiculars from the vertices of △λSC to the      opposite sides in green.

Step 4. Draw the sides of the pedal triangle △ﾌ:)EF￡ヽor △Å召C in green and plot its      orthocenter瓦in red.

Step 5. Draw each of the perpendiculars from the vertices of △￡)￡F to the opposite      sides with a broken line in blue.

Step 6. NVhen △DEF is obtuse， extend each of two sides a(tjacent at the obtuse angle      with a broken line in blue.

Step 7. When one moves the vertex Acontinuously along the curve 貿by the mouse，      the orthocenter ￡of△ＤＥＦ continuously draws a locus ，y with a solid line      in red.

PART

Two

: lb process a bitmap file(bmp)bypLyIEX2ε・to

exhibit it on another

display，and to print it･

OUTLINE

of the PROGRAM.

Let召(−1，0)，C(1，O)be

the fixed vertices of a triangle

on the plane，and let

ｰbe

the straightline y °x. when

a poin A(･け)on

y

is

selected by the mouse， then the orthocenter jg'(ぷ2y)of the △Z)￡F is determined by

the formula (＊)with

7)＝(心，yα)，￡＝(為，yゐ)，Ｆ＝(恥，yc)｡

  The

program for drawing a locus y

of the point 瓦'is given in LiSt l. ln thiscase

we will have the curve

｡y: 4ぞ4十8緋ぞ十3)ソー4(6J2

− 1)y十緋2J2十2x−1)(2×2−2x−1)＝0

on a disPlay (Fig.11)

K.FUJITA and H.FUKAlsHI

§3.Algebraic

Curves obtained as locus of the orthocenter of a pedal triangle

   As a locus of the orthocenter of a pedal triangle for a given triangle we have various curves. ln this section we exhibit the cases of plane algebraic curves with simple expressions whose graphs are not given in the standard texts of the sub徊ct[8]イ9]， [11]−[131，[15]and[17]−[181.

Let 召(−1,0)，C(1，0)be the vertices fixed throughout this section.

Theorem. ￡･2湧げ治りＷ／ｎﾉ㈲μ面訪7㎡cantg貿c回＆涌屈zz

歛α/Qcsげがz･g

orthocenter of tk Pedal triangle△DEF

Jor△A召Ｃｗ加z2が1εがzj

｢wr砥x:Å謂∂v6

｢θz7g

z/zEcθ㎡c認cljθηy.

Casd-1(Fig.1-1)

ｙｙ

}に1，the stmight nne

.

λ勺4十2×2y3十2×2(ぞ十4)ダ十2(x4−4y−4)y十(ぞ十2)(ぞ−2)2＝0

Casd-2(Fig.1-2)

Case 2 (F

゜恥貿

ｙ

ｊ ２ ・ 争   ・ ︱

Case 3 (Fig.3)

柘貿ｙ

Case 4 (Fig.

    貿

    ｙ

ｊ ４

ｙニーぞ十1，the Pambola

λ勺4十2×2y3十2y(ぞ十4)ゾ十2(x4−4×2−4)y十(ぞ十2)(ぞ−2)2＝0

λ ;

ニゾ十1，thePambota

4yy4十ぞ(5ぞ十24が十68λ;十88)ゾ十(λ7-2)(ぞ十3y−8)2ニ０

Ｘ

_{ニゾー1，the Parabola}

ぞ戸十x2(3λ73十2j十9x十18)ダ

十(3y十4×6十ﾌぞ−14y

− 16y − 16y 十48x − 32)ダ

十(x十2)(λ7−1)2(ぞ十ぞ−4)2＝0.

ゾーぞ＝恥the

hでyPerbota

4yy8十x2(16y

− 15)戸十2(12×6十5×4十143y十56)ダ

Case 5 (Fig.

    貿

    ｙ

Case 6 (Fig.

    貿

    ｙ

Case 7 (Fig.

    貿

    ｙ

Case 8 (Fig.

     貿

     ｙ

Case 9 (Fig.

    貿

    y

５

Plane Algebraic Curves Drawn by the Orthocenter of a Peda1 Tri皿gle

）  ゾー ？＝2、治f哺ypど油 ｢α． 6） 7） ・ Φ ８ ・ ・ Ｉ ・ ９ ・ ・ Φ ・

144yy8十6∠Lj(9ぞ十10)ゾ十8(108×6十160λ74十257が十144)ダ

十16(36が十40y−3ダー114λソー49)ｿﾞ十9ぞ(4ﾔｰﾌ)2ニO，

ぞ−3ｿﾞ＝1，the肪Perbola．

１

6粧4＋4(8×4十7oj−3)ｿﾞﾊﾟｽﾞ十2)(x−2)(牡七7)2＝o

(x十2)2−ゾニ1，治ど/7yμど油θ/ふ

yy6十(3ダ十8が十6x十6)ダ十(3y十16×4十３１ｙ十20y

− 15λ7−

26)ゾ

十(λ7十2)(ぞ十3y十2x−2)2＝0.

）

 ぞーゾ＝ｌ、出c胎Perbota

16yy8十2×2(32×2十384)y6十4(24×6十352×4十1762×2− 32)ソ 十16(心78十32×6− 211×4十400×2 − 289)ｿﾞ十J2(4×4− 16ぞ十17)2＝0

)

 (2x十1)2−4ゾニ1，the肪Perbola.

 16原6十8バ6×2十4･v十15)y4

 十(48×5十64が十64y−42心ﾌ十８ｈ−８)ソ

 十(x十2)(4y

−7x十1)2＝0.

Case 10.1(Fig.10.1) 貿： ｙ： y°2x十2，z尨･szrαjがzz /訥ε．

125ぞ2＋1202y＋(5x＋6)(5x＋7)(5x−1ト０

Case 10-2 (Fig.10･2)

貿

ｙ

ｌ ｌ _ｙ Ｉ -１ − ２

25ぞ2− 120y＋(5x−6)(5x−7)(5x＋1)＝0

K.FUJITA and H.FUKAlsHI Case 11 (Fig.11)      ｰ: 2)にχ，治ε幼7zjがzl /訥ど

4･yy4十8x(ぞ十3)ソー4(6y−1)y十J(2λｦ十2J−1)(2×2−2x−1)＝0

Case 12 (Fig.12)

    

ｰ： ？十y2°

3，

Z＆dydE

ｙ：収ｙ十ぞ(敗2十56)ダ十(6?

− 27? 十52×2− 48)ソ十(ぞ−2)4＝0

Case 13 (Fig.13)     貿:(jv−1)2十ゾ＝1.the circle.     jﾀﾞ: 6もう6十64x(3y十3j十1&v−4)ダ        十16x(12λ75十32? − 90×3− 30λ72十133λフー120)ゾ        十(ぞ十壮一8)2(8j − 12x 十1)2＝0.

Case 14 (Fig.14)

貿 ・ ・

ｙ

ｙ：

１ う／ 十戸 -ぐ Ｉ ２ う／ .thg circle

(89y＋4oox＋464)ゾ＋り−2)(x＋4)2(5xづ)2＝o

the circle 64ぞ4 ２ ＋Ｘ

Case 15 (Fig.15)

    y: G−1)2十(y−2)2＝4，函ｄ

｢ε．

    ｙ：弓4十2x(2x−1)y3＋2(3×3十3x−1)ソ十2(2?＋3y十＆しx−2)y

       ＋G＋1)(5x十1)(x3＋212十x−2)＝0.

Case 16 (Fig.16)

    

ｰ:(x−3)≒y2＝4，心ｄ

｢ε．

    ｙ：．ｙ＋2(13x七3占べｎに駄白べに寂ご庁ekご庁ぺ.

Case 17 (Fig.17)

    

ｰ:(x−1)2十y＝2，r/zg

d

｢ε.

    ,y: jy十む(3χ3−2.x:2十14χ−4)y4十4(3? −18？ 十24?−2j−8x−4)y2

      十(x2十2x−4)2(2J2−2x−1)2＝0.

Case 18 (Fig.18)

    y: G−2)2十y＝9，

36jvy4十(261y − 1200×2 十1216λ7− 360)ソ パx＋2)(3x−2)2(5xぺ)2＝0.

Plane Algebraic Curves Drawn by the Orthocenter of a Peda1 Triangle

Case 19-1 (Fig. 19-1)

    貿:χ2十5y≒

ｙ _144舒y4十

6se

19-2 (Fig.19-2)

貿

ｙ

・・ λﾄﾞ＋ 144X Ｉ 5 2y4 ｙ 1， (2 ｆ Ｈ he dipse， 88×4＋2200×2＋1620)ｿ＋9(x＋2)(x−2)(4×2＋     ﹃ ／ Ｉ   一 一 １ 乙

lhe empse.

２ ｊ ｌ ０

＋(288×4＋2200×2＋1620)ｿ＋9(x＋2)(x−2)(4×2＋1)2＝0

Case 20-1 (Fig. 20-1)

    

ｰ:χ2十ﾌﾀﾞﾆ

y : 324×2y4 十 Case 20-2 (Fig.20-2)

貿

.が＋ １ ７ ｙ

X、lhe dipse、

(648×4＋4185y＋1792)ｿﾞ＋4(x＋2)(x−2)(9J2−

2)2＝0

    Ｉ ／ Ｉ   ＝ ︷ Ｚ the dnpse，

y:

324xy＋(648×4＋4185×2＋1792)ゾ＋4(x＋2)(x−2)(9×2− 2)2＝0.

Proor ＬｅｔＡ(ａげ)and f収,y)，whileβ＝

are nxed. Then we have the following :

皿d Ｈｅｎｃｅ ｙ十β＝ ｙ十β＝ ぐ  ＝  ￡ Ｆ -ぐ ｙ十β μ−α

(-

Ｉ，0） -(ｘ−α)， μ−(l slnCe ａ     角 ／ Ｉ   ＝ β -０ ｖ十β −(M−1)2十172  (14−1)2十v2'  (M十1)2−y2  GI＋1)2十ｙ' (x＋α) ＝(―貼−β)，Ｃ＝ −2v匯 (zj−1)2 -1） ＋F2 う／  2vφ＋1) (M十1)2十ジ う Ｉ ぐ ，0)＝(貼−β)

Case 9.

WOlfram

(4，yc):

fo ｆ ｆ ・ ・ １： ２：

f3

f4

f5

f6

f7

f8

f9

・ ・ Ｓ Ｉ ・ ・ ・ ・ 争 ・ ・ ・ -K.FUJITAand H.FUKAlsHI

Find the Groebner bases (see[1]、[10])by

Research lnc･ ，to the following code， where

(2*u+1)^2-4*v^2-1

゛ｘａ一Ｕ ゜ya ＝（ ３２ 作 Ｑ ｊｌ に ぐ ３ ２ ︷ Ｖ μ ド ︲1 叫 べ

５２

Ｗ

Ｑ

３

１皿

ぐべ

３ ２ Ｗ つ‘ り １ ｊ ぐ べ

゜P1−(xc-xb)

゜q1−(yc-yb)

applying Mathematica

Ver 3.0 by

Ｄこ

(心，‰)，石＝吊ふ)，Ｆ＝

*xb一(v^2-(u-1)^2)

*yb十2*(u-1)*v

*xc−((u+1)^2-v^2)

*yc-2*(u十1)*v

:゜r1−((xc-xb)*xa+(yc-yb)*ya)

f10:=P2-(xa-xc)

ｊ ｙ 叫 四 ・争 ｍ f12:=r2−

_{((xa-xc)*xb4･(ya-yc)*yb)}

３ １  ｑ  ＊  ２  ｒ  一 ２  ｑ  ＊  １  ｒ ぐ  一  Ｘ  ＊ ｊ ｌ  ｑ  ＊  り心  Ｐ  一 り ＊ １  ｐ ぐ  一一  ｅｓ ３ １ ｆ

f14:＝(P1*q2-p2*q1)*y-(P1*r2-P2*r1

ｊ

GroebnerBasis[{f0,f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12,f13,f14}，

        {xa,ya,xb,yb,xc,y{;,P1,P2,q1,q2,r1,r2,v,u,y,x}]

Factor[Z]

ぐ

Then we obtain the term

−１十a)(1十ｇ)２(−１十2M)(2−27λﾌ十84j十65λソー104λ･4 − 56×5 十32×6  十1＆7−8戸十81ぞ2 − 424舒y2十64λ勺2十64×4y2十48×5ｿﾞ十120原4

 十32λ72y4十48ぞﾀﾞ十16ぞ6)

in the list of the generating Groebner bases. From the last factor we have

Ｉ

6ぞ6十8x(6j十4j十15)ダ

十(48y十64λ74十64y

− 424y

十８ｈ−８)ｿﾞ十(x十2)(4λｿｰﾌx十1)2＝0

as an equation of jf.

Case 11.

Find the Groebner basesby

applying Mathematica

Ver 3.0 to the following

code. At firstwe put

fo:゜u^2-7^2，because the calculation by Mathematica

stopped

fo

f1

f2

f3

f4

f5

f6

f7

f8

：＝Ｕ＾２−Ｖ＾２ ● ● ・ ・ ゛ｘａ−Ｕ °ｙａ ぐ φ ・ ぐ ・・ ぐ ＠ ・ ぐ ・ ＠ ・ ・ ・ ・

Plane Algebraic CurvesDrawn by the Orthocenter of a Peda1 Triangle

(u-1)^2+v^2)*xb-(v^2-(u-1)^2)

(u-1)^2+v^2)*yb+2*(u-1)*v

(u+1)^2+v^2)*xc-((u+1)^2-v^2)

(u十1)^2十v^2)*yc−2*(u+1)*V

ﾆp1−(xc-xb)

゜q1−(yc-yb)

f9:=r1−((xc-xb)*xa+(yc-yb)*ya)

f10:゜P2-(xa-xc)

f11:'q2-(ya-yc)

f12:=r2-((xa一xc)*xb+(ya-yc)*yb)

f13:゜(P1*q2-P2*q1)*x-(r1*q2-r2*q1)

f14:゜(P1*q2-P2*q1)*y-(P1*r2-P2*r1)

GroebnerBasis[{f0,f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12,f13,f14}。

         {xa,ya,xb,yb,xc,yc,P1,p2,q1,q2,r1,r2,v,u,y,x}]

Factor[Z]

Then we obtain the term

(−１十μ)(1十μ)(−１十2j)(x−8y十4y十4Jy

− 24yy

十24乃八十8yｿﾞ十4λly4)

 ×(J−8ぞ十4y

− 4y 十24λう十24xy2十8×3戸十4Jy4)

in the list of the generating Groebner bases

Ｏ「

From the last two factors we have

λ;−8×3十4×5十牡−24頒y十2∠凧y2十8試y2十4yy4ニO･

J−8y十4y − 4y 十24舒y十24ぞ2十8yｿﾞ十疫yy4ニ０

as two candidates for an equation of the former gives a required equation ；

y.

By checking theirgraphs we conclude that

that is、

牡y4十8緋ぞ十3)ソー4(6×2 − 1 )y十亘2λ72十2λ7−1)(2ぞ−2λ7−1)＝0

K.FUJITA and H，FUKAlsHI

§4.Questions

Question l.

The two curves y

in the following cases coincide with each other :

i)Cases

1-1 and 1-2.

ii)Cases 19-1 and 19-2

iii)Cases 20-l and 20-2

Whyisitso ？

Question

2.

The two curves y

in Case 10-l and 10-2 are renexive in the y-axis.

Whyisitso

？

References

［1］XV.NV.Adams and P. Loustaunau : Azz7/lzmjM￡zj∂ylz∂Gnijゐzlgr＆M,11s,Graduation Studies    in Math.，VOI.3，Amer.Math.Soc.，Providence，1994.

［2］H. S. M.Coxeter : /y7zmj￡,czj∂72記,GE∂z?2Ezry，2nded.，John M/iley and Sons lnc.，New    York，1980.〔コクセター（銀林 浩・訳）： 幾何学入門 第２版，明治図書，東京，

   1982.〕

F3］H.S.M.Coxeter and S. L.Greitzer : GEθ謂どzり,＆?1心Ed，New Mathematical Library，    Number6， School Mathematics Study Group， Random House， lnc･， New York， 1967.    〔コークスター，グレイツァー（寺阪英孝・訳）： 幾何学再入門，SMSG双書，河

   出書房新社，東京，1970.〕

［4］K.F1!jita,A.lviatsushima and H. Fukaishi : A Locus of the Orthocenter of a Triangle −    lnstruction in Geometry by a Moving Locus on a Computer， A右?誂.jic.￡dzjc･，瓦αgαwα    び,ljwrsi印弑 55（2005），1-13.，

［5］K.Fujita，A.Matsushima and H. Fukaishi : A Triangle with Three Distinguished    ComnearPoints￣lnstruction of Geometry by use of a Drawing Game on a Display，    A＆。.ac.￡jMc.，瓦昭αｗαび ｢ｗだ匈瓦55（2005），25-41.

［6］K.Fujita，A. Matsushima and H. Fukaishi : A Triangle with Distinguished Concyclic Points lnstruction of Geometry by use of a Drawing Game on a Display.Mem.Fac，

Plane Algebraic Curves Drawn by the Orthocenter of a Peda1 Triangle

［7］H.Fukaishi，K.Fujita and A. Matsushima : Alternative Geometric Proofs of Theorems    for Concyclic Points for a Triangle，A＆z7z.Rzc.￡dgc･， jQzgαwa lﾉＭｗａｊ印77，56    （2006），51-59.

［

［

［

ΓＬ

8］XV.Fulton : Åなd?πzjca4ry6，1V.A.Benjamin，New ybrk， 1969.

9］一松 信・他：新数学事典，大阪書籍，大阪，1979.〔＝S.Hitotumatu et al.:Shinl-  Saga㎞-JjMzz，（2）saka-shoseki，1979.〕 10］一松 信： 代数学入門 第三課，近代科学社，束京，1994.〔＝S.Hitotumatu :   佃zrり面dθyz zθÅな訪rどz，772gnj ｢む∬∂yz，Kindai-kagaku-sha，7n〕ky0，1994.〕 11］飯高 茂：平面曲線の幾何，共立出版，東京，2001.（＝S.litaka : Gg∂772gzり･げ   j）jαylECi4rw?s，Kyoritsu-Shuppan，’Tbkyo，2001.〕

[12]河田敬義：代数曲線論入門，至文堂，東京，1975.〔＝YI Kawada l lnlroduction to    Åなεゐr ｢c CMrw 777ど∂り，Shibundo/Tokyo，1975.〕

[13]E.Kunz: 佃zyり＆4czjθzl zθj)jαzzE/哨μ泌jtzjc Cz4rws， Birkhiiuser， Boston， 2005. (Translated    from the originaI German by R. G.Belshoff.)

[14]A.Matsushima，K.Fujita and H. Fukaishi : A Locus of the Orthocenter of a Pedal    Ti‘iangle￣lnstruction of Geometry by use of a Drawing Galne on aDisplay， A＆謂.刄zc.    JEj￡,c.，瓦昭αｗα防ljw肖吟瓦 57(2007)，1-15.

[15] [16]

G.0rzech and M. 0rzech : 戸1α7zg/哨μ泌r㎡cCMnﾉ6，Marcel Dekker， New York， 1981 D.Pedoe: Ga琲zErり，Dover Publ. lnc･，New York， 1 970.

［17］坂井忠次： グラフと追跡，培風館，束京，1963.〔＝C.Sakai

: G四涵ω乃心dj，

   Baifukan/T〔jkyo，

1963.〕

［18］R.J.iValker : Åなεわn21

Cgrws， Springer-Verlag，NewYork，1950，

7

Kazunori FUJITA

Department of Mathematics， Faculty of Education， Kagawa university l-1 Saiwai-cho/lakamatsu-shi， Kagawa，760-8522，JAPAN

￡-majla＆1,。g: fujita＠ed.kagawa-u.ac.jp

Hiroo FUKAlsHI

Department of Mathematics， Faculty of Education， Kagawa university 1-1 Saiwai-cho/nlkamatsu-shi， Kagawa， 760-8522， JAPAN

Fig.1-1貿   ｙ Fig.2（が   ｙ Ｘ ｙ°１

K.FUJITA and H.FUKAlsHI

Ｆ_ig.1-2 yy4十2χ2y3十2×2(ぞ十4)ダ  十2(x4−収2−4)六(x2＋2)(x2−2)2＝0 一- ダ十１ 4ぞy4十ぞ(5×3十24ぞ十68x十  十(χ−2)(ぞ十3×2− 8)2＝0 88)ダ Fig. 3 貿   ｙ ｙ： ｙ＝−ｘ２十1 jf:斌y4十2χ2J3十2×2(χ2＋4)y2    ＋2(x4−4×2−4)y＋(x2＋2)(x2−2)2＝0 ｘニダー１ 舒y6十x2(3×3十2×2こﾄ9x十18)ダ  十(3ぞ十4λJ6十万5−14y−IG3−16×2十4沿−32)y2

Fig. 4 貿   ｙ   − − Ｒ g.6

貿

ｙ

Plane Algebraic Curves Drawn by the Orthocenter of a Peda1 Triangle

列−x2ニフ 4ぞy8十ぞ(16λ;2−15)y6  十2(12χ6十5？十143が十56)y4  十(16×8十65×6十2×4 − 479×2 − 4)y2  十4λ;2(x4十5J2十2)2＝0 ぞ−3y2ニ１

Fig.5貿

  ｙ

Fig. ７ 戸−ぞ゜２ 144×2y8十64×2(9×2十10)y6  十8(108χ6十160y4十257ぞ十144)ダ  十16(36×8十40×6− 3ぞ−114ぞ−49)ダ  十9jxr2(4×4−7)2＝0

ｙｙ

収＋2)2づ2＝1 刄6十(3が十8×2十6x十6)ダ

FjR.8 ‘r: χ2−y2＝2

K.FUJITA and H.FUKAlsHI

jﾀﾞ: 16×2y8十2×2(32が十384)y6

    十4(2収6十352×4十1762×2− 32)ダ     十16(収8十32×6−211×4十400×2−289)y2     十x2(叙4− 16×2十17)2＝0

Fig. 10-1‘i : y ＝2x十2

Fig.9貿

  ｙ

Fig. 10-2 貿 巾＋1)2−4y2＝1 16が＋&v(6×2＋載＋15)y4  十(48y十64×4十64y − 424が十８ｈ−８)ダ  ＋収十2)(4y −7x十1)2＝o ｙ° ・＋ １ − ２ １ − ２

Fig.11貿

  ｙ

Fig.13貿

  ｙ

ｙニj『

Plane Algebraic Curves Drawn by the Orthocenter of a Peda1 Triangle

瓶y4十8x(x2十3)ダー4(6y−1)y  ＋x(2×2＋2x−1)(2×2−2x−1)＝0

Fig.12貿

  ｙ

(x−1)2十y2＝1 64×2y6十64x(3y十3ぞ＋16ぶ−4)y4   十16x(12y十32×4−90y − 30×2 十133x−120)ダ x2十ｙ２°３ 収2y6十ぞ(9が十56)ダ  ＋(6ぞ−27×4＋52×2−48)ソ＋(x2−2)4＝ Fig. 14 貿   ｙ

レ

ﾚｱ 十y2=(寸 ア 64ぞ4十到8敗2十40(h十464)ダ ０

K.FUJITA and H.FUKAlsHI Fig.15 y:(ズー1)2＋(y−2)2＝4    jr : xy4 ＋2x(2jr − 1)y3十2(3×3十3x−I)y2 十2(2×4十3×3十6×2−x−2)y ＋(x＋1)(5x＋I)(y＋y＋x−2)＝ ０ Fig.17 ？:(x−1)2十戸＝2     jf: x2y6＋壮(3×3 − 2×2十1壮−4)ダ Fig. 16 貿   ｙ ４ (x−3)2＋y2＝4 ぞ4十2(13×3−25×2十23x −5)ダ  パx−2)(x−1)2(5x−7)2＝0 18 ｰ：(戈−2)2＋y2＝9   y: 36司,4十(261×3 − 1200×2 十1216x−360)y

Fig. 19-1 貿   ｙ

Fig.20-l y:     y:

Plane Algebraic Curves Drawn by the Orthocenter of a Pedal Triangle

χ2＋5y2＝1 144λ72y4十(288χ4十2200×2十1620)y2   ＋9(x＋2)(x−2)(4J2＋↓)2＝0 x2十7y2＝1 32収2y4十(648×4十4185ぞ十1792)y2 ＋4(x十2)(x−2)(9×2 − 2)2＝0 Fig. 19-2 y : x2 ＋ Fig.20-2 y    y X2 ＋ ダニ１ y2 一 一 １ １ − ５ ﾀﾞ:1441y十(288×4十220(h24･1620):y2    ＋9G＋2)(･一2)(4×2＋1)2＝o １ − ７ 324×2y4十(648？十4185×2十1792)y2  ＋4(x＋2)(x−2)(9×2 − 2)2＝0

ｉ ｌ ｌ Ｓ 皿 d p ecifyi

Ｋ．ＦＵ訂TA and H.FUKAlsHI

List l.

A program

for dllawing a locus

ing statements of t − h･

Dim sx As Double， ex As Double Dim sy As Dou ble，ey As Double Dim ax As Double， ay As Double Dim bx As Double， by As Double Dim cx As Double， cy As Double Dimxd As Double， yd As Double Dim xe As Dou ble，ye As Double Dim xf As Double, yfAs Double Dim k As lnteger

Dim ra(4)As Double

｜ ｜ ｜ Ｄｒａｗｉｎｇ ｔｈｅ ｉ ’け 個︲ ｌ ｎ al _figure e general variables

Private Sub Form_Activateo  Forml.Line(sx，0)-(ex，0)  Forml.Line(0，sy)-(0，ey)

 For Y ＝lnt(sy)To lnt(ey)   Forml.Line(-0.02，Y)-(O.02，Y) NextY Forml.DrawWidth＝2 Forml.Line(sx，sx)-(ex，ex)，QBColor(9) Forml.Line(bx，by)-(cx，cy) Forml.DrawMode＝vbNotxorPen Forml.Line(ax，ay)-(bx，by) ㎜   ㎜ ㎜㎜  j    ㎜ a Forml.Line((:x，cy)-(ax， Forml.DrawWidth＝1 ay）

 Anth_trian9le ax， ay， bx， by， cx， cy， 2 End Sub

●

・

ｌ

ｌ

Setting the form， the coordinate axes and the initial values of the coordinates of each vertex of the triangle Private Sub Form_Loado

 FOrml.Top＝100  Forml.Left＝2000  Forml.Width＝0.5 ゛Screen.Width  Forml.Height＝0.9

_{こ函卿ｗ４}

゛Screen.Height ＝ ＝ -フ 一   一 つ ／ ﹄ ｅｘ 一 SX

＝wx ゛Forml.ScaleHeight / Forml .ScaleWidth

-−０．５゛ｗｙ ０．５゛ｗｙ

Plane Algebraic CurvesDrawn by the Orthocenter of a PedaI Triangle  Forml.Scale(sx，ey)-(ex，sy)  Forml.BackColor＝vbWhite  Forml.DrawWidth＝1  ax＝0.6: ay ＝0.6  bx＝−1  by＝0  CX＝1  cy＝0  k＝1 End Sub １ １

Action corresponding to the left button of mouse

Private Sub Form_MouseDown(Button As lnteger, Shift As lnteger, X As Sin9le，Y As Sin9le)  lf k ＝2 Then   k＝1   Exit Sub  End lf ｋ End ｜ ・ ｜ -２ ｘｂ

Action corresponding to the movement

of mouse

Private Sub Form_MouseMove(Button As lnteger，  lfk＝I Then Exit Sub

・ − Ｄ End ｉ ｌ ｘｂ

Presentin9 of the orthocenter of a trian9le Private Sub Orthocenter(jxa

 xa＝jxa:ya °jya  xb＝jxb:yb＝jyb

ＸＣ＝ _{Jxc: yc ＝Jyc}

Shift As lnteger, XAs Sin9le， Y As Sin9le）

，jya,jxb,Jyb,jxc,Jy(，pcl)

ｕｘ＝ｘｂ − ｘａ：ｕｙ ＝ｙｂ − ｙａ ｖｘ＝ｘａ − ｘｃ：ｖｙ ＝ｙａ− yc

 lfah＞＝O And bh ＞＝O And ch ＞＝O And a ＞10ハ(-４)Ａｎｄ ｂ＞１０∧(-４)Ａｎｄｃ ＞１０ ハ(-4)Then

  pl＝xb゛yc−xc゛yb:ql＝-wy゛ya−wx゛xa   p2＝xc゛ya−xa゛yc:q2＝-vy゛yb−vx゛xb   p3＝xa゛yb−xb゛ya:q3＝−uy゛yc−ux゛xc

xd yd xe

K.FUJITA and H.FUKAlsHI

＝(ｐｌ ゛‘ｗｙ−ｑｌ ｉｗｘ)／ａ∧２ ＝(-ｐｌ゛ｗｘ−ｑｌ ゛ ｗｙ)／ａハ２ ＝(ｐ２ ゛ｖｙ−ｑ２゛ｖｘ)／ｂＡ２  』 − ．    ＝   ■  ㎜  ＝ ye°(-p2゛vx−q2'vy)/bハ2 xf＝(p3゛uy−q3りＪＸ)／ｃＡ２ ｙｆ'(Ｔｐ)３゛ｕｘ−ｑ３ ゛ｕｙ)／ｃＡ２ xh＝(q2゛wy−q￣1 ゛vy) yh＝(ql ゛vx−q2゛wx) Forml.DrawStyle ° 2 Forml.Line(xa,ya)-(xd， ／(ｗｘ ゛ ｖｙ −ｗｙ゛ vx） yh＝(qPvx−q2゛wx)/(wx゛vy−wy゛vx) 一  −−  − ・   − yd)， Forml.Line(xb，yb)-(xe，ye)， ㎜   ㎜■ ㎜㎜  −    ■ J − − Forml.Line(xc,yc)-(xf,yf)， Forml.DrawStyle＝0 Forml.DrawWidth＝ Forml.DrawMode＝ Forml.PSet(xh，yh)， Forml.DrawMode＝  Forml.DrawWidth Else lfbh＜O Then ｄｍ＝ｘａ： ｘａ ＝  dm＝ End lf ya: ya ° lfch＜O Then ｄｍ＝ｘａ：ｘａ＝  dm End lf ＵＸ ＶＸ -- ya: ya ＝ -４ Ｖ Ｖ vb bBlue bBlue Blue

vbCopyPen

QBColor(pcl)

vbNotxorPen l

xb:xb＝dm

yb: yb ＝dm

ｘｃ：ｘｃ＝ｄｍ ｙｃ：ｙｃ ＝ｄｍ 一 -xb − xa･: uy ＝yb − ya ｘａ一X(:: ｗｘ＝ｘｃ − ｘｂ vy＝ya − yc wy＝yc −yb d＝wx゛vy−wy゛vx lfAbs(d)＞0.001 Then ａ＝ ｂ＝ Ｃ＝_{ｐｑｘ １｀／﹄ｄ} yd xh Ⅶ’ＨＨ Sqr(wxA2＋wyA2) Sqr『vxA2十vyAり 皿 』 ・ − Sqr(uxA2 -一 -一 -xb゛yc− ＋uyA2） xc ゛yb: ql 一 一 −ｗｙ゛ｙａ−ｗｘ゛ｘａ -vy゛yb−VX゛xb (pl゛wy−qlりvx)/aハ２ (-ｐｌ ゛ｗｘ−ｑｌ ｉｗｙ)／ａハ２ くｑ２゛ｗｙ−ｑいvy)/d       ＝    ■  − (ql゛vx− -Prolong xb， y q2 ゛wx)/d b，xa, ya

-Prolong xc,yc

Forml.DrawStyle ，ｘａ， ｙａ  ＝２ Forml.Line(xh,yh)-(xd,yd),VbBlue Forml.Line(xb,yb)-(xh,yh),VbBlue ㎜㎜       ふ   ■ ・ 晶  還･ ａ ■ Forml.Ljne(xc,yc)-(xh，yh),vbBlue Forml.DrawStyle＝0 Forml.DrawWidth＝4 Forml.DrawMode＝vbCopyPen F(jrml.PSet(xh，yh)，QBColor(pcl) Forml.DrawMode＝vbNotxorPen

yf＝(-p3゛ux−q3゛uy)/cA2 Forml.DrawWidth＝5

Plane Algebraic Curves Drawn by the Orthocenter of a Peda1 Trianglc

   Forml.DrawWidth＝1   End lf  End lf End Sub ｌ ・

Extendin9 of two sides adjacent at the vertex with an obtuse angle Private Sub H_Prolong(xl

 Forml.DrawWidth＝1 ，yl，x2,y2） ｜＝Sqr((xl − x2)∧2十(yl − y2)∧2) ra(1) ra(2) ｢a(3) ra(4) ＝Sqr((xl − sx)ハ ＝Sqr((xl − ex)∧ ＝Sqr((xl − ex)ハ ＝Sqr((xl − sx)ハ r＝ra(1) FOr i ＝ 2To4 一 一 一 一 １１−１−１ ’Ｗ’Ｗ’Ｗ‘Ｗ ＋十十十 NNNN lfra(i)＞rThen r ＝ra(i) Nexti p＝r/｜ tx＝(1−p)゛xl＋p ty＝(1−p)・ yl ＋p       − −   − 女 女 X2 y2 sy） sy） ey） ey） ∧ ∧ ∧ ∧ 2） 2） 2） 2）  Forml.DrawStyle＝2  Forml.Line(x2，y2)-(tx，ty)，QBColor(9)  Forml.DrawStyle＝0 End Sub ｌ Ｗ Ｗ Presenting of a

Private Sub Anth_t「  ux＝xb−xa: uy pedal triangle

iangle(xa,ya,xb,yb

＝yb − ya ｖｘ＝ｘａ−ｘｃ：ｖｙ＝ｙａ − ｙｃ ｗｘ＝ｘｃ − ｘｂ： ｗｙ＝ｙｃ−ｙ    ｉ  ａ   −      -･ bh＝ ch＝ pl＝ p2＝ p3＝ Ｃ∧ (-p2゛ ■ − -一 一 -一 (-ｐｌ ゛ｗｘ−ｑｌ ゛ｗｙ)／ａＡ２ (ｐ２゛ｖｙ−ｑ２゛ｖｘ)／ｂＡ２ 還 − ・    ふ ・  ・ −●  ミ ＝(ｐｌ ゛ｗｙ−ｑｌ ゛ｗｘ)／ａＡ２

xd

yd

xe

y9

ｘａ _{゛yb − xb ゛ya: q3 ＝-uy゛yc−ux゛xc}

xc ゛ya 一xa゛yc:q2 xb゛yc一ｘｃ゛ ａＡ２十 ２十 (一４)Ａｎｄ ｂ＞１０∧(-４)Ａｎｄ ｃ ＞１０ハ(-４)Ｔｈｅｎ ２十ｃＡ２−ａ∧２ Ａ２十ｗｙＡ２) ∧２十vy∧2) A2十uyA2) (一4)And b＞ xf＝(p3゛  一  一 一 yb: ql ｃＡ２−ａ∧２ ａ∧２−ｂ∧２ ｂ∧２−ｃ∧２ ｂ ，xc，yc， col） ゜−ｗｙ‘゛ｙａ − ｗｘ ゛ ｘａ °−ｖｙ゛ｙｂ−ｖｘ゛ｘｂ ゛ｖｘ−ｑ２゛ｖｙ)／ｂＡ２ ｕｙ−ｑ３゛ｕｘ)／ｃＡ２

K.FUJITA and H.FUKAlsHI   Forml.PSet(xd，yd)，QBColor(col)   Forml.PSet(xe，ye)，QBColor(col)   Forml.PSet(xf,yf)，QBColor(col)   Forml.DrawWidth＝2   Forml.Line(xd，yd)-(xe，ye)，qBColor(col)   Forml.Line(xe,ye)-(xf,yf)，qBColor(col)   Forml.Line(xf,yf)-(xd，yd)，QBColor(col)   Forml.DrawWidth＝1

  0rthocenterxd，yd， xe，ye，xf, yf，12  End lf

End Sub

Private Sub Ext_Click()

 Forx＝ex To 40 ゛ex Step 0.1   Disp x  Nextx For x ＝40 ゛ sx To sx Step 0.1  Disp x Nextx For x ＝2  Disp x Next x End ｌｂ ９ 感１ ５

・sxTo2゛exStep

0.001

Private Sub Disp(x)  Forml.DrawWidth＝2  Forml.Line(ax，ay)-(bx，by)  Forml.Line(cx，cy)-(ax，ay)  Y＝.×  Forml.Line(x,Y)-(bx，by)  Forml.Line(cx，cy)-(x，Y)  Forml.DrawWidth＝1

 Anth_triangle ax， ay， bx，by,cx. cy, 2  Anth_triangle x,Y， bx， by, cx， cy， 2  ax＝x: ay ＝Y