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:
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.
pstricks pst-eucl
edited yesterday
Artificial Stupidity
4,82111039
asked yesterday
chishimotoji
543212
add a comment |
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:
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.
pstricks pst-eucl
edited yesterday
Artificial Stupidity
4,82111039
asked yesterday
chishimotoji
543212
1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
yesterday
@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
yesterday
add a comment |
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:
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.
pstricks pst-eucl
edited yesterday
Artificial Stupidity
4,82111039
asked yesterday
chishimotoji
543212
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:
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.
pstricks pst-eucl
pstricks pst-eucl
edited yesterday
Artificial Stupidity
4,82111039
asked yesterday
chishimotoji
543212
edited yesterday
Artificial Stupidity
4,82111039
asked yesterday
chishimotoji
543212
edited yesterday
Artificial Stupidity
4,82111039
edited yesterday
Artificial Stupidity
4,82111039
edited yesterday
Artificial Stupidity
4,82111039
4,82111039
asked yesterday
chishimotoji
543212
asked yesterday
chishimotoji
543212
asked yesterday
chishimotoji
543212
543212
1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
yesterday
@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
yesterday
add a comment |
1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
yesterday
@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
yesterday
1
1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
yesterday
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
yesterday
@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
yesterday
@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
yesterday
add a comment |
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).
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!
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.
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}
answered yesterday
Artificial Stupidity
4,82111039
1
Very good. This is a exciting trick.
– chishimotoji
yesterday
add a comment |
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}
answered yesterday
Herbert
267k23406716
Mr. Herbert, could you modify theHomCoefandAngleCoefto be able to accept postfix notation? Defining a new command namedpstAngleMarkas an alias ofpstMarkAngle. Modifying the core such thatAngleCoefandRotAnglecan be interchanged. Thank you!
– Artificial Stupidity
yesterday
I found a "bug" inpsCircleTangents. See my answer. Try changexfrom3to5with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
– Artificial Stupidity
yesterday
add a comment |
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).
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!
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.
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}
answered yesterday
Artificial Stupidity
4,82111039
1
Very good. This is a exciting trick.
– chishimotoji
yesterday
add a comment |
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).
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!
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.
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}
answered yesterday
Artificial Stupidity
4,82111039
1
Very good. This is a exciting trick.
– chishimotoji
yesterday
add a comment |
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).
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!
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.
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}
answered yesterday
Artificial Stupidity
4,82111039
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).
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!
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.
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}
answered yesterday
Artificial Stupidity
4,82111039
edited yesterday
answered yesterday
Artificial Stupidity
4,82111039
answered yesterday
Artificial Stupidity
4,82111039
answered yesterday
Artificial Stupidity
4,82111039
4,82111039
1
Very good. This is a exciting trick.
– chishimotoji
yesterday
add a comment |
1
Very good. This is a exciting trick.
– chishimotoji
yesterday
1
1
Very good. This is a exciting trick.
– chishimotoji
yesterday
Very good. This is a exciting trick.
– chishimotoji
yesterday
add a comment |
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}
answered yesterday
Herbert
267k23406716
Mr. Herbert, could you modify theHomCoefandAngleCoefto be able to accept postfix notation? Defining a new command namedpstAngleMarkas an alias ofpstMarkAngle. Modifying the core such thatAngleCoefandRotAnglecan be interchanged. Thank you!
– Artificial Stupidity
yesterday
I found a "bug" inpsCircleTangents. See my answer. Try changexfrom3to5with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
– Artificial Stupidity
yesterday
add a comment |
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}
answered yesterday
Herbert
267k23406716
Mr. Herbert, could you modify theHomCoefandAngleCoefto be able to accept postfix notation? Defining a new command namedpstAngleMarkas an alias ofpstMarkAngle. Modifying the core such thatAngleCoefandRotAnglecan be interchanged. Thank you!
– Artificial Stupidity
yesterday
I found a "bug" inpsCircleTangents. See my answer. Try changexfrom3to5with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
– Artificial Stupidity
yesterday
add a comment |
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}
answered yesterday
Herbert
267k23406716
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}
answered yesterday
Herbert
267k23406716
edited yesterday
answered yesterday
Herbert
267k23406716
answered yesterday
Herbert
267k23406716
answered yesterday
Herbert
267k23406716
267k23406716
Mr. Herbert, could you modify theHomCoefandAngleCoefto be able to accept postfix notation? Defining a new command namedpstAngleMarkas an alias ofpstMarkAngle. Modifying the core such thatAngleCoefandRotAnglecan be interchanged. Thank you!
– Artificial Stupidity
yesterday
I found a "bug" inpsCircleTangents. See my answer. Try changexfrom3to5with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
– Artificial Stupidity
yesterday
add a comment |
Mr. Herbert, could you modify theHomCoefandAngleCoefto be able to accept postfix notation? Defining a new command namedpstAngleMarkas an alias ofpstMarkAngle. Modifying the core such thatAngleCoefandRotAnglecan be interchanged. Thank you!
– Artificial Stupidity
yesterday
I found a "bug" inpsCircleTangents. See my answer. Try changexfrom3to5with step 1. The tangent line moves from one side to opposite side when the radii change from smaller to equal.
– Artificial Stupidity
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
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1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
yesterday
@Thruston In pst-eucl documentation, only to draw one bisector, " BUT are " is equivalent " to get". My English is not so good. :-))
– chishimotoji
yesterday