How to divide arc, line, and angle with a certain ratio?











up vote
5
down vote

favorite












A simple example:



documentclass[12pt,border=5pt]{standalone}

usepackage{newcent,pstricks,pst-eucl} 

usepackage{auto-pst-pdf}

begin{document}

begin{pspicture}

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

end{pspicture}

end{document}


The result of compiling:



enter image description here



Question 1:



How to get a point M such as MA=2/3AB, or more generally MA=(a/b)AB and a point M' belong to small arc CA such as arc M'A=2/3AB, or more generally M'A=(a/b)AB.



Question 2:



How to get bisector of angle A BUT are two bisectors or three bisectors.



Responding to AS'comment below.



enter image description here






pstricks pst-eucl







share|improve this question




















  • 1




    Can you explain your second question more clearly please? what do you mean by "BUT are ..."
    – Thruston
    Dec 10 at 9:16










  • @Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
    – chishimotoji
    Dec 10 at 9:19

















up vote
5
down vote

favorite












A simple example:



documentclass[12pt,border=5pt]{standalone}

usepackage{newcent,pstricks,pst-eucl} 

usepackage{auto-pst-pdf}

begin{document}

begin{pspicture}

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

end{pspicture}

end{document}


The result of compiling:



enter image description here



Question 1:



How to get a point M such as MA=2/3AB, or more generally MA=(a/b)AB and a point M' belong to small arc CA such as arc M'A=2/3AB, or more generally M'A=(a/b)AB.



Question 2:



How to get bisector of angle A BUT are two bisectors or three bisectors.



Responding to AS'comment below.



enter image description here






pstricks pst-eucl







share|improve this question




















  • 1




    Can you explain your second question more clearly please? what do you mean by "BUT are ..."
    – Thruston
    Dec 10 at 9:16










  • @Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
    – chishimotoji
    Dec 10 at 9:19















up vote
5
down vote

favorite









up vote
5
down vote

favorite











A simple example:



documentclass[12pt,border=5pt]{standalone}

usepackage{newcent,pstricks,pst-eucl} 

usepackage{auto-pst-pdf}

begin{document}

begin{pspicture}

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

end{pspicture}

end{document}


The result of compiling:



enter image description here



Question 1:



How to get a point M such as MA=2/3AB, or more generally MA=(a/b)AB and a point M' belong to small arc CA such as arc M'A=2/3AB, or more generally M'A=(a/b)AB.



Question 2:



How to get bisector of angle A BUT are two bisectors or three bisectors.



Responding to AS'comment below.



enter image description here






pstricks pst-eucl







share|improve this question















A simple example:



documentclass[12pt,border=5pt]{standalone}

usepackage{newcent,pstricks,pst-eucl} 

usepackage{auto-pst-pdf}

begin{document}

begin{pspicture}

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

end{pspicture}

end{document}


The result of compiling:



enter image description here



Question 1:



How to get a point M such as MA=2/3AB, or more generally MA=(a/b)AB and a point M' belong to small arc CA such as arc M'A=2/3AB, or more generally M'A=(a/b)AB.



Question 2:



How to get bisector of angle A BUT are two bisectors or three bisectors.



Responding to AS'comment below.



enter image description here






pstricks pst-eucl




pstricks pst-eucl






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 10 at 10:34









Artificial Stupidity

5,00611039




5,00611039










asked Dec 10 at 9:05









chishimotoji

599212




599212








  • 1




    Can you explain your second question more clearly please? what do you mean by "BUT are ..."
    – Thruston
    Dec 10 at 9:16










  • @Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
    – chishimotoji
    Dec 10 at 9:19
















  • 1




    Can you explain your second question more clearly please? what do you mean by "BUT are ..."
    – Thruston
    Dec 10 at 9:16










  • @Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
    – chishimotoji
    Dec 10 at 9:19










1




1




Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
Dec 10 at 9:16




Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
Dec 10 at 9:16












@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
Dec 10 at 9:19






@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
Dec 10 at 9:19












2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










Step 1



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

end{pspicture}

end{document}


Note: HomCoef cannot accept RPN 2 3 div so I have to insert 0.6666 (3 decimals places should suffice, I think).



enter image description here



Step 2



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

end{pspicture}

end{document}


Note: AngleCoef must come before RotAngle. It is not commutative!



enter image description here



Step 3 (Final)



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=0.6666,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}


Note: We have pstSegmentMark (ends with Mark) to mark a segment but we have pstMarkAngle (begins with Mark) to mark an angle. It seems the package author likes making inconsistent names.



enter image description here



Last Edit



pscalculate from pst-calculate package can make me possible to insert infix calculations fed to the HomCoef and AngleCoef. 



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 

usepackage{pst-calculate}

begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=pscalculate{2/3}]{A}{B}[M]% Now without hard coded 0.6666

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=pscalculate{2/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}





share|improve this answer



















  • 1




    Very good. This is a exciting trick.
    – chishimotoji
    Dec 10 at 13:44


















up vote
4
down vote













documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

  pstTriangle(2,4){A}(0,0){B}(6,0){C}

  pstCircleABC[PosAngle=60]{A}{B}{C}{O}

  pstHomO[HomCoef=0.667]{A}{B}[M]

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

  psset{PointSymbol=none,PointName=none}

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M1]

  pstRotation[AngleCoef=0.667,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M2]

  pcline[linestyle=dashed](A)(M1)

  pcline[linestyle=dashed](A)(M2)

end{pspicture}

end{document}


enter image description here






share|improve this answer























  • Mr. Herbert, could you modify the HomCoef and AngleCoef to be able to accept postfix notation? Defining a new command named pstAngleMark as an alias of pstMarkAngle. Modifying the core such that AngleCoef and RotAngle can be interchanged. Thank you!
    – Artificial Stupidity
    Dec 10 at 10:41












  • I found a "bug" in psCircleTangents. See my answer. Try change x from 3 to 5 with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
    – Artificial Stupidity
    Dec 10 at 17:41













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Step 1



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

end{pspicture}

end{document}


Note: HomCoef cannot accept RPN 2 3 div so I have to insert 0.6666 (3 decimals places should suffice, I think).



enter image description here



Step 2



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

end{pspicture}

end{document}


Note: AngleCoef must come before RotAngle. It is not commutative!



enter image description here



Step 3 (Final)



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=0.6666,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}


Note: We have pstSegmentMark (ends with Mark) to mark a segment but we have pstMarkAngle (begins with Mark) to mark an angle. It seems the package author likes making inconsistent names.



enter image description here



Last Edit



pscalculate from pst-calculate package can make me possible to insert infix calculations fed to the HomCoef and AngleCoef. 



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 

usepackage{pst-calculate}

begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=pscalculate{2/3}]{A}{B}[M]% Now without hard coded 0.6666

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=pscalculate{2/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}





share|improve this answer



















  • 1




    Very good. This is a exciting trick.
    – chishimotoji
    Dec 10 at 13:44















up vote
5
down vote



accepted










Step 1



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

end{pspicture}

end{document}


Note: HomCoef cannot accept RPN 2 3 div so I have to insert 0.6666 (3 decimals places should suffice, I think).



enter image description here



Step 2



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

end{pspicture}

end{document}


Note: AngleCoef must come before RotAngle. It is not commutative!



enter image description here



Step 3 (Final)



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=0.6666,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}


Note: We have pstSegmentMark (ends with Mark) to mark a segment but we have pstMarkAngle (begins with Mark) to mark an angle. It seems the package author likes making inconsistent names.



enter image description here



Last Edit



pscalculate from pst-calculate package can make me possible to insert infix calculations fed to the HomCoef and AngleCoef. 



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 

usepackage{pst-calculate}

begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=pscalculate{2/3}]{A}{B}[M]% Now without hard coded 0.6666

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=pscalculate{2/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}





share|improve this answer



















  • 1




    Very good. This is a exciting trick.
    – chishimotoji
    Dec 10 at 13:44













up vote
5
down vote



accepted







up vote
5
down vote



accepted






Step 1



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

end{pspicture}

end{document}


Note: HomCoef cannot accept RPN 2 3 div so I have to insert 0.6666 (3 decimals places should suffice, I think).



enter image description here



Step 2



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

end{pspicture}

end{document}


Note: AngleCoef must come before RotAngle. It is not commutative!



enter image description here



Step 3 (Final)



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=0.6666,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}


Note: We have pstSegmentMark (ends with Mark) to mark a segment but we have pstMarkAngle (begins with Mark) to mark an angle. It seems the package author likes making inconsistent names.



enter image description here



Last Edit



pscalculate from pst-calculate package can make me possible to insert infix calculations fed to the HomCoef and AngleCoef. 



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 

usepackage{pst-calculate}

begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=pscalculate{2/3}]{A}{B}[M]% Now without hard coded 0.6666

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=pscalculate{2/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}





share|improve this answer














Step 1



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

end{pspicture}

end{document}


Note: HomCoef cannot accept RPN 2 3 div so I have to insert 0.6666 (3 decimals places should suffice, I think).



enter image description here



Step 2



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

end{pspicture}

end{document}


Note: AngleCoef must come before RotAngle. It is not commutative!



enter image description here



Step 3 (Final)



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=0.6666,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}


Note: We have pstSegmentMark (ends with Mark) to mark a segment but we have pstMarkAngle (begins with Mark) to mark an angle. It seems the package author likes making inconsistent names.



enter image description here



Last Edit



pscalculate from pst-calculate package can make me possible to insert infix calculations fed to the HomCoef and AngleCoef. 



documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 

usepackage{pst-calculate}

begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=pscalculate{2/3}]{A}{B}[M]% Now without hard coded 0.6666

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=pscalculate{1/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=pscalculate{2/3},RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}






share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 10 at 13:41

























answered Dec 10 at 9:40









Artificial Stupidity

5,00611039




5,00611039








  • 1




    Very good. This is a exciting trick.
    – chishimotoji
    Dec 10 at 13:44














  • 1




    Very good. This is a exciting trick.
    – chishimotoji
    Dec 10 at 13:44








1




1




Very good. This is a exciting trick.
– chishimotoji
Dec 10 at 13:44




Very good. This is a exciting trick.
– chishimotoji
Dec 10 at 13:44










up vote
4
down vote













documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

  pstTriangle(2,4){A}(0,0){B}(6,0){C}

  pstCircleABC[PosAngle=60]{A}{B}{C}{O}

  pstHomO[HomCoef=0.667]{A}{B}[M]

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

  psset{PointSymbol=none,PointName=none}

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M1]

  pstRotation[AngleCoef=0.667,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M2]

  pcline[linestyle=dashed](A)(M1)

  pcline[linestyle=dashed](A)(M2)

end{pspicture}

end{document}


enter image description here






share|improve this answer























  • Mr. Herbert, could you modify the HomCoef and AngleCoef to be able to accept postfix notation? Defining a new command named pstAngleMark as an alias of pstMarkAngle. Modifying the core such that AngleCoef and RotAngle can be interchanged. Thank you!
    – Artificial Stupidity
    Dec 10 at 10:41












  • I found a "bug" in psCircleTangents. See my answer. Try change x from 3 to 5 with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
    – Artificial Stupidity
    Dec 10 at 17:41

















up vote
4
down vote













documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

  pstTriangle(2,4){A}(0,0){B}(6,0){C}

  pstCircleABC[PosAngle=60]{A}{B}{C}{O}

  pstHomO[HomCoef=0.667]{A}{B}[M]

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

  psset{PointSymbol=none,PointName=none}

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M1]

  pstRotation[AngleCoef=0.667,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M2]

  pcline[linestyle=dashed](A)(M1)

  pcline[linestyle=dashed](A)(M2)

end{pspicture}

end{document}


enter image description here






share|improve this answer























  • Mr. Herbert, could you modify the HomCoef and AngleCoef to be able to accept postfix notation? Defining a new command named pstAngleMark as an alias of pstMarkAngle. Modifying the core such that AngleCoef and RotAngle can be interchanged. Thank you!
    – Artificial Stupidity
    Dec 10 at 10:41












  • I found a "bug" in psCircleTangents. See my answer. Try change x from 3 to 5 with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
    – Artificial Stupidity
    Dec 10 at 17:41















up vote
4
down vote










up vote
4
down vote









documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

  pstTriangle(2,4){A}(0,0){B}(6,0){C}

  pstCircleABC[PosAngle=60]{A}{B}{C}{O}

  pstHomO[HomCoef=0.667]{A}{B}[M]

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

  psset{PointSymbol=none,PointName=none}

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M1]

  pstRotation[AngleCoef=0.667,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M2]

  pcline[linestyle=dashed](A)(M1)

  pcline[linestyle=dashed](A)(M2)

end{pspicture}

end{document}


enter image description here






share|improve this answer














documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

  pstTriangle(2,4){A}(0,0){B}(6,0){C}

  pstCircleABC[PosAngle=60]{A}{B}{C}{O}

  pstHomO[HomCoef=0.667]{A}{B}[M]

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

  psset{PointSymbol=none,PointName=none}

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M1]

  pstRotation[AngleCoef=0.667,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M2]

  pcline[linestyle=dashed](A)(M1)

  pcline[linestyle=dashed](A)(M2)

end{pspicture}

end{document}


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 10 at 10:09

























answered Dec 10 at 10:04









Herbert

267k23407716




267k23407716












  • Mr. Herbert, could you modify the HomCoef and AngleCoef to be able to accept postfix notation? Defining a new command named pstAngleMark as an alias of pstMarkAngle. Modifying the core such that AngleCoef and RotAngle can be interchanged. Thank you!
    – Artificial Stupidity
    Dec 10 at 10:41












  • I found a "bug" in psCircleTangents. See my answer. Try change x from 3 to 5 with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
    – Artificial Stupidity
    Dec 10 at 17:41




















  • Mr. Herbert, could you modify the HomCoef and AngleCoef to be able to accept postfix notation? Defining a new command named pstAngleMark as an alias of pstMarkAngle. Modifying the core such that AngleCoef and RotAngle can be interchanged. Thank you!
    – Artificial Stupidity
    Dec 10 at 10:41












  • I found a "bug" in psCircleTangents. See my answer. Try change x from 3 to 5 with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
    – Artificial Stupidity
    Dec 10 at 17:41


















Mr. Herbert, could you modify the HomCoef and AngleCoef to be able to accept postfix notation? Defining a new command named pstAngleMark as an alias of pstMarkAngle. Modifying the core such that AngleCoef and RotAngle can be interchanged. Thank you!
– Artificial Stupidity
Dec 10 at 10:41






Mr. Herbert, could you modify the HomCoef and AngleCoef to be able to accept postfix notation? Defining a new command named pstAngleMark as an alias of pstMarkAngle. Modifying the core such that AngleCoef and RotAngle can be interchanged. Thank you!
– Artificial Stupidity
Dec 10 at 10:41














I found a "bug" in psCircleTangents. See my answer. Try change x from 3 to 5 with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
– Artificial Stupidity
Dec 10 at 17:41






I found a "bug" in psCircleTangents. See my answer. Try change x from 3 to 5 with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
– Artificial Stupidity
Dec 10 at 17:41




















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