Tick Marks in Geometry
Here is an excerpt from my code:
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (0.5,0.866);
draw (A)--(B)--(C)--(A);
end{tikzpicture}
I want to denote AB=BC using the tick mark notation. Also, point C isn't exactly where it should be. How can I fix that? I rounded sqrt(3)/2 to 0.866.
graphs
New contributor
add a comment |
Here is an excerpt from my code:
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (0.5,0.866);
draw (A)--(B)--(C)--(A);
end{tikzpicture}
I want to denote AB=BC using the tick mark notation. Also, point C isn't exactly where it should be. How can I fix that? I rounded sqrt(3)/2 to 0.866.
graphs
New contributor
2
Welcome to TeX.SE! Please show us -- as usual here --an short compilable code resulting in your issue ...
– Kurt
5 hours ago
2
It would also be helpful if you could include a sketch of how the output should look like.
– samcarter
5 hours ago
TikZ understands polar coordinates such ascoordinate[label=:$C$] (C) at (60:1);
It also understandscoordinate[label=:$C$] (C) at (0.5,{sqrt(3)/2});
.
– marmot
5 hours ago
add a comment |
Here is an excerpt from my code:
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (0.5,0.866);
draw (A)--(B)--(C)--(A);
end{tikzpicture}
I want to denote AB=BC using the tick mark notation. Also, point C isn't exactly where it should be. How can I fix that? I rounded sqrt(3)/2 to 0.866.
graphs
New contributor
Here is an excerpt from my code:
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (0.5,0.866);
draw (A)--(B)--(C)--(A);
end{tikzpicture}
I want to denote AB=BC using the tick mark notation. Also, point C isn't exactly where it should be. How can I fix that? I rounded sqrt(3)/2 to 0.866.
graphs
graphs
New contributor
New contributor
New contributor
asked 6 hours ago
M. C.M. C.
112
112
New contributor
New contributor
2
Welcome to TeX.SE! Please show us -- as usual here --an short compilable code resulting in your issue ...
– Kurt
5 hours ago
2
It would also be helpful if you could include a sketch of how the output should look like.
– samcarter
5 hours ago
TikZ understands polar coordinates such ascoordinate[label=:$C$] (C) at (60:1);
It also understandscoordinate[label=:$C$] (C) at (0.5,{sqrt(3)/2});
.
– marmot
5 hours ago
add a comment |
2
Welcome to TeX.SE! Please show us -- as usual here --an short compilable code resulting in your issue ...
– Kurt
5 hours ago
2
It would also be helpful if you could include a sketch of how the output should look like.
– samcarter
5 hours ago
TikZ understands polar coordinates such ascoordinate[label=:$C$] (C) at (60:1);
It also understandscoordinate[label=:$C$] (C) at (0.5,{sqrt(3)/2});
.
– marmot
5 hours ago
2
2
Welcome to TeX.SE! Please show us -- as usual here --an short compilable code resulting in your issue ...
– Kurt
5 hours ago
Welcome to TeX.SE! Please show us -- as usual here --an short compilable code resulting in your issue ...
– Kurt
5 hours ago
2
2
It would also be helpful if you could include a sketch of how the output should look like.
– samcarter
5 hours ago
It would also be helpful if you could include a sketch of how the output should look like.
– samcarter
5 hours ago
TikZ understands polar coordinates such as
coordinate[label=:$C$] (C) at (60:1);
It also understands coordinate[label=:$C$] (C) at (0.5,{sqrt(3)/2});
.– marmot
5 hours ago
TikZ understands polar coordinates such as
coordinate[label=:$C$] (C) at (60:1);
It also understands coordinate[label=:$C$] (C) at (0.5,{sqrt(3)/2});
.– marmot
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
A PSTricks solution just for fun purposes.
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;150){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture
end{document}
Bonus
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
foreach i in {90,100,...,170}{%
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;i){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture}
end{document}
add a comment |
Welcome to TeX.SE! You can add these marks with decorations.markings
. Since you want two of them, it is shorter to use the .list
key for that. Further, TikZ understands polar coordinates, and it also understands (0.5,{sqrt(3)/2})
, so there is no need to unbury your calculator. ;-)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (60:1);
draw[equal mark/.list={1/6,1/2}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
As you can see, this code starts with documentclass
and ends with end{document}
, and is compilable. Kurt asked you in his comment to add such a code.
And you may simplify/shorten the code using a foreach
loop.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
foreach X/Y in {210/A,-30/B,90/C}
{coordinate[label=X:$Y$] (Y) at (X:{1/sqrt(3)});}
draw[equal mark/.list={1/6,1/2,5/6}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
add a comment |
one more tikz
solution:
documentclass[tikz,border=3.141592mm]{standalone}
begin{document}
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=$C$] (C) at (0.5,0.866);
draw (A) -- node {$|$} (B) -- node[sloped] {$|$} (C) -- (A);
end{tikzpicture}
end{document}
(not tested since my tikz
is broken :-( )
The mark on BC is not perpendicular I think ( I compile with my heart).
– God Must Be Crazy
3 hours ago
1
@GodMustBeCrazy, of course not, i forgot to add optionsloped
... i correct this.
– Zarko
3 hours ago
add a comment |
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3 Answers
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3 Answers
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A PSTricks solution just for fun purposes.
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;150){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture
end{document}
Bonus
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
foreach i in {90,100,...,170}{%
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;i){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture}
end{document}
add a comment |
A PSTricks solution just for fun purposes.
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;150){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture
end{document}
Bonus
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
foreach i in {90,100,...,170}{%
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;i){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture}
end{document}
add a comment |
A PSTricks solution just for fun purposes.
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;150){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture
end{document}
Bonus
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
foreach i in {90,100,...,170}{%
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;i){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture}
end{document}
A PSTricks solution just for fun purposes.
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;150){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture
end{document}
Bonus
documentclass[pstricks,12pt,border=1cm]{standalone}
usepackage{pst-eucl}
begin{document}
foreach i in {90,100,...,170}{%
pspicture[MarkAngle=90](-4,4)
pstTriangle(4;i){C}(-4,0){A}(0,0){B}
pstSegmentMark{A}{B}
pstSegmentMark{B}{C}
endpspicture}
end{document}
edited 3 hours ago
answered 3 hours ago
God Must Be CrazyGod Must Be Crazy
5,89211039
5,89211039
add a comment |
add a comment |
Welcome to TeX.SE! You can add these marks with decorations.markings
. Since you want two of them, it is shorter to use the .list
key for that. Further, TikZ understands polar coordinates, and it also understands (0.5,{sqrt(3)/2})
, so there is no need to unbury your calculator. ;-)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (60:1);
draw[equal mark/.list={1/6,1/2}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
As you can see, this code starts with documentclass
and ends with end{document}
, and is compilable. Kurt asked you in his comment to add such a code.
And you may simplify/shorten the code using a foreach
loop.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
foreach X/Y in {210/A,-30/B,90/C}
{coordinate[label=X:$Y$] (Y) at (X:{1/sqrt(3)});}
draw[equal mark/.list={1/6,1/2,5/6}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
add a comment |
Welcome to TeX.SE! You can add these marks with decorations.markings
. Since you want two of them, it is shorter to use the .list
key for that. Further, TikZ understands polar coordinates, and it also understands (0.5,{sqrt(3)/2})
, so there is no need to unbury your calculator. ;-)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (60:1);
draw[equal mark/.list={1/6,1/2}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
As you can see, this code starts with documentclass
and ends with end{document}
, and is compilable. Kurt asked you in his comment to add such a code.
And you may simplify/shorten the code using a foreach
loop.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
foreach X/Y in {210/A,-30/B,90/C}
{coordinate[label=X:$Y$] (Y) at (X:{1/sqrt(3)});}
draw[equal mark/.list={1/6,1/2,5/6}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
add a comment |
Welcome to TeX.SE! You can add these marks with decorations.markings
. Since you want two of them, it is shorter to use the .list
key for that. Further, TikZ understands polar coordinates, and it also understands (0.5,{sqrt(3)/2})
, so there is no need to unbury your calculator. ;-)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (60:1);
draw[equal mark/.list={1/6,1/2}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
As you can see, this code starts with documentclass
and ends with end{document}
, and is compilable. Kurt asked you in his comment to add such a code.
And you may simplify/shorten the code using a foreach
loop.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
foreach X/Y in {210/A,-30/B,90/C}
{coordinate[label=X:$Y$] (Y) at (X:{1/sqrt(3)});}
draw[equal mark/.list={1/6,1/2,5/6}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
Welcome to TeX.SE! You can add these marks with decorations.markings
. Since you want two of them, it is shorter to use the .list
key for that. Further, TikZ understands polar coordinates, and it also understands (0.5,{sqrt(3)/2})
, so there is no need to unbury your calculator. ;-)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=:$C$] (C) at (60:1);
draw[equal mark/.list={1/6,1/2}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
As you can see, this code starts with documentclass
and ends with end{document}
, and is compilable. Kurt asked you in his comment to add such a code.
And you may simplify/shorten the code using a foreach
loop.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.markings}
begin{document}
begin{tikzpicture}[scale=5.5,equal mark/.style={postaction={decorate,
decoration={markings,mark=at position #1 with {draw (-2pt,-4pt) -- (-2pt,4pt);
draw (2pt,-4pt) -- (2pt,4pt);}}}}]
foreach X/Y in {210/A,-30/B,90/C}
{coordinate[label=X:$Y$] (Y) at (X:{1/sqrt(3)});}
draw[equal mark/.list={1/6,1/2,5/6}] (A)--(B)--(C)--cycle;
end{tikzpicture}
end{document}
edited 3 hours ago
answered 4 hours ago
marmotmarmot
89.4k4103194
89.4k4103194
add a comment |
add a comment |
one more tikz
solution:
documentclass[tikz,border=3.141592mm]{standalone}
begin{document}
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=$C$] (C) at (0.5,0.866);
draw (A) -- node {$|$} (B) -- node[sloped] {$|$} (C) -- (A);
end{tikzpicture}
end{document}
(not tested since my tikz
is broken :-( )
The mark on BC is not perpendicular I think ( I compile with my heart).
– God Must Be Crazy
3 hours ago
1
@GodMustBeCrazy, of course not, i forgot to add optionsloped
... i correct this.
– Zarko
3 hours ago
add a comment |
one more tikz
solution:
documentclass[tikz,border=3.141592mm]{standalone}
begin{document}
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=$C$] (C) at (0.5,0.866);
draw (A) -- node {$|$} (B) -- node[sloped] {$|$} (C) -- (A);
end{tikzpicture}
end{document}
(not tested since my tikz
is broken :-( )
The mark on BC is not perpendicular I think ( I compile with my heart).
– God Must Be Crazy
3 hours ago
1
@GodMustBeCrazy, of course not, i forgot to add optionsloped
... i correct this.
– Zarko
3 hours ago
add a comment |
one more tikz
solution:
documentclass[tikz,border=3.141592mm]{standalone}
begin{document}
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=$C$] (C) at (0.5,0.866);
draw (A) -- node {$|$} (B) -- node[sloped] {$|$} (C) -- (A);
end{tikzpicture}
end{document}
(not tested since my tikz
is broken :-( )
one more tikz
solution:
documentclass[tikz,border=3.141592mm]{standalone}
begin{document}
begin{tikzpicture}[scale=5.5]
coordinate[label=left:$A$] (A) at (0,0);
coordinate[label=right:$B$] (B) at (1,0);
coordinate[label=$C$] (C) at (0.5,0.866);
draw (A) -- node {$|$} (B) -- node[sloped] {$|$} (C) -- (A);
end{tikzpicture}
end{document}
(not tested since my tikz
is broken :-( )
edited 3 hours ago
answered 3 hours ago
ZarkoZarko
121k865158
121k865158
The mark on BC is not perpendicular I think ( I compile with my heart).
– God Must Be Crazy
3 hours ago
1
@GodMustBeCrazy, of course not, i forgot to add optionsloped
... i correct this.
– Zarko
3 hours ago
add a comment |
The mark on BC is not perpendicular I think ( I compile with my heart).
– God Must Be Crazy
3 hours ago
1
@GodMustBeCrazy, of course not, i forgot to add optionsloped
... i correct this.
– Zarko
3 hours ago
The mark on BC is not perpendicular I think ( I compile with my heart).
– God Must Be Crazy
3 hours ago
The mark on BC is not perpendicular I think ( I compile with my heart).
– God Must Be Crazy
3 hours ago
1
1
@GodMustBeCrazy, of course not, i forgot to add option
sloped
... i correct this.– Zarko
3 hours ago
@GodMustBeCrazy, of course not, i forgot to add option
sloped
... i correct this.– Zarko
3 hours ago
add a comment |
M. C. is a new contributor. Be nice, and check out our Code of Conduct.
M. C. is a new contributor. Be nice, and check out our Code of Conduct.
M. C. is a new contributor. Be nice, and check out our Code of Conduct.
M. C. is a new contributor. Be nice, and check out our Code of Conduct.
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2
Welcome to TeX.SE! Please show us -- as usual here --an short compilable code resulting in your issue ...
– Kurt
5 hours ago
2
It would also be helpful if you could include a sketch of how the output should look like.
– samcarter
5 hours ago
TikZ understands polar coordinates such as
coordinate[label=:$C$] (C) at (60:1);
It also understandscoordinate[label=:$C$] (C) at (0.5,{sqrt(3)/2});
.– marmot
5 hours ago