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 Dec 10 at 10:34
Artificial Stupidity
5,00611039
asked Dec 10 at 9:05
chishimotoji
599212
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 Dec 10 at 10:34
Artificial Stupidity
5,00611039
asked Dec 10 at 9:05
chishimotoji
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
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 Dec 10 at 10:34
Artificial Stupidity
5,00611039
asked Dec 10 at 9:05
chishimotoji
599212
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 Dec 10 at 10:34
Artificial Stupidity
5,00611039
asked Dec 10 at 9:05
chishimotoji
599212
edited Dec 10 at 10:34
Artificial Stupidity
5,00611039
asked Dec 10 at 9:05
chishimotoji
599212
edited Dec 10 at 10:34
Artificial Stupidity
5,00611039
edited Dec 10 at 10:34
Artificial Stupidity
5,00611039
edited Dec 10 at 10:34
Artificial Stupidity
5,00611039
5,00611039
asked Dec 10 at 9:05
chishimotoji
599212
asked Dec 10 at 9:05
chishimotoji
599212
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
add a comment |
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
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 Dec 10 at 9:40
Artificial Stupidity
5,00611039
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 at 13:44
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 Dec 10 at 10:04
Herbert
267k23407716
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
Dec 10 at 10:41
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
Dec 10 at 17:41
add a comment |
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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).
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 Dec 10 at 9:40
Artificial Stupidity
5,00611039
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 at 13:44
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 Dec 10 at 9:40
Artificial Stupidity
5,00611039
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 at 13:44
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 Dec 10 at 9:40
Artificial Stupidity
5,00611039
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 Dec 10 at 9:40
Artificial Stupidity
5,00611039
edited Dec 10 at 13:41
answered Dec 10 at 9:40
Artificial Stupidity
5,00611039
answered Dec 10 at 9:40
Artificial Stupidity
5,00611039
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
add a comment |
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
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 Dec 10 at 10:04
Herbert
267k23407716
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
Dec 10 at 10:41
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
Dec 10 at 17:41
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 Dec 10 at 10:04
Herbert
267k23407716
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
Dec 10 at 10:41
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
Dec 10 at 17:41
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 Dec 10 at 10:04
Herbert
267k23407716
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 Dec 10 at 10:04
Herbert
267k23407716
edited Dec 10 at 10:09
answered Dec 10 at 10:04
Herbert
267k23407716
answered Dec 10 at 10:04
Herbert
267k23407716
answered Dec 10 at 10:04
Herbert
267k23407716
267k23407716
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
Dec 10 at 10:41
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
Dec 10 at 17:41
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
Dec 10 at 10:41
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
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
add a comment |
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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