How to divide arc, line, and angle with a certain ratio?
5
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 '18 at 10:34
God Must Be Crazy
5,60511039
asked Dec 10 '18 at 9:05
chishimotoji
820316
add a comment |
5
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 '18 at 10:34
God Must Be Crazy
5,60511039
asked Dec 10 '18 at 9:05
chishimotoji
820316
1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
Dec 10 '18 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 '18 at 9:19
add a comment |
5
5
5
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 '18 at 10:34
God Must Be Crazy
5,60511039
asked Dec 10 '18 at 9:05
chishimotoji
820316
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 '18 at 10:34
God Must Be Crazy
5,60511039
asked Dec 10 '18 at 9:05
chishimotoji
820316
edited Dec 10 '18 at 10:34
God Must Be Crazy
5,60511039
asked Dec 10 '18 at 9:05
chishimotoji
820316
edited Dec 10 '18 at 10:34
God Must Be Crazy
5,60511039
edited Dec 10 '18 at 10:34
God Must Be Crazy
5,60511039
edited Dec 10 '18 at 10:34
God Must Be Crazy
5,60511039
5,60511039
asked Dec 10 '18 at 9:05
chishimotoji
820316
asked Dec 10 '18 at 9:05
chishimotoji
820316
asked Dec 10 '18 at 9:05
chishimotoji
820316
820316
1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
Dec 10 '18 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 '18 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 '18 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 '18 at 9:19
1
1
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
Dec 10 '18 at 9:16
Can you explain your second question more clearly please? what do you mean by "BUT are ..."
– Thruston
Dec 10 '18 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 '18 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 '18 at 9:19
add a comment |
2 Answers
2
active
oldest
votes
5
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 '18 at 9:40
God Must Be Crazy
5,60511039
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 '18 at 13:44
add a comment |
4
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 '18 at 10:04
Herbert
269k24408717
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!
– God Must Be Crazy
Dec 10 '18 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.
– God Must Be Crazy
Dec 10 '18 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
5
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 '18 at 9:40
God Must Be Crazy
5,60511039
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 '18 at 13:44
add a comment |
5
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 '18 at 9:40
God Must Be Crazy
5,60511039
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 '18 at 13:44
add a comment |
5
5
5
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 '18 at 9:40
God Must Be Crazy
5,60511039
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 '18 at 9:40
God Must Be Crazy
5,60511039
edited Dec 10 '18 at 13:41
answered Dec 10 '18 at 9:40
God Must Be Crazy
5,60511039
answered Dec 10 '18 at 9:40
God Must Be Crazy
5,60511039
answered Dec 10 '18 at 9:40
God Must Be Crazy
5,60511039
5,60511039
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 '18 at 13:44
add a comment |
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 '18 at 13:44
1
1
Very good. This is a exciting trick.
– chishimotoji
Dec 10 '18 at 13:44
Very good. This is a exciting trick.
– chishimotoji
Dec 10 '18 at 13:44
add a comment |
4
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 '18 at 10:04
Herbert
269k24408717
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!
– God Must Be Crazy
Dec 10 '18 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.
– God Must Be Crazy
Dec 10 '18 at 17:41
add a comment |
4
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 '18 at 10:04
Herbert
269k24408717
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!
– God Must Be Crazy
Dec 10 '18 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.
– God Must Be Crazy
Dec 10 '18 at 17:41
add a comment |
4
4
4
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 '18 at 10:04
Herbert
269k24408717
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 '18 at 10:04
Herbert
269k24408717
edited Dec 10 '18 at 10:09
answered Dec 10 '18 at 10:04
Herbert
269k24408717
answered Dec 10 '18 at 10:04
Herbert
269k24408717
answered Dec 10 '18 at 10:04
Herbert
269k24408717
269k24408717
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!
– God Must Be Crazy
Dec 10 '18 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.
– God Must Be Crazy
Dec 10 '18 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!
– God Must Be Crazy
Dec 10 '18 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.
– God Must Be Crazy
Dec 10 '18 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!– God Must Be Crazy
Dec 10 '18 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!– God Must Be Crazy
Dec 10 '18 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.– God Must Be Crazy
Dec 10 '18 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.– God Must Be Crazy
Dec 10 '18 at 17:41
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
Dec 10 '18 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 '18 at 9:19