A `coordinate` command ignored Announcing the arrival of Valued Associate #679: Cesar Manara ...
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A `coordinate` command ignored
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizeConnecting an arced path in TikZTikZ/ERD: node (=Entity) label on the insideHow to draw points in TikZ?TikZ: Drawing an arc from an intersection to an intersectionRelative transparency in TikZ?Line up nested tikz enviroments or how to get rid of themRectanglar cloud shaped node in TikZCorrect linejoins at tikz to path II
$triangle[ABC]$ is a 30-60 right triangle, and its right angle is at C
. A
is at the origin. A circle is inscribed in it; its center is at
O = (2*sqrt(3)*(sqrt(3) - 1), 2*(sqrt(3) - 1))
and its radius is 12(sqrt(3) - 1)
. Leg AC
is the shorter leg. The equation of the line through it is y = sqrt(3)*x
. The line perpendicular to AC
has slope -sqrt(3)/3
, and the line through O
with slope -sqrt(3)/3
is
y = (-sqrt(3)/3)*(x - 2*(sqrt(3))*(sqrt(3)-1)) + 2*(sqrt(3)-1) .
The two lines intersect on leg AC
at
Q = (8*sqrt(3)*(sqrt(3)-1), 24*(sqrt(3)-1)) .
So, the command draw (O) -- (Q);
should draw a radius of the circle to leg AC
. On my computer, the command renders a line segment through the other leg and ridiculously long. It seems to me that the command locating point Q
has been ignored.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
path (0,0) coordinate (A) (8,0) coordinate (B) (2,{2*sqrt(3)}) coordinate (C);
node[anchor=north, inner sep=0, font=footnotesize] at (0,-0.15){textit{A}};
node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$){textit{B}};
node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$){textit{C}};
draw (A) -- (B) -- (C) -- cycle;
path let n1={2*(sqrt(3))*(sqrt(3)-1)}, n2={2*(sqrt(3)-1)} in coordinate (O) at (n1,n2);
draw[fill] (O) circle (1.5pt);
draw[blue] let n1={2*(sqrt(3)-1)} in (O) circle (n1);
path let n1={2*(sqrt(3))*(sqrt(3)-1)} in coordinate (P) at (n1,0);
node[anchor=north, inner sep=0, font=footnotesize] at ($(P) +(0,-0.15)$){textit{P}};
draw (O) -- (P);
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={24*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
tikz-pgf
|
show 4 more comments
$triangle[ABC]$ is a 30-60 right triangle, and its right angle is at C
. A
is at the origin. A circle is inscribed in it; its center is at
O = (2*sqrt(3)*(sqrt(3) - 1), 2*(sqrt(3) - 1))
and its radius is 12(sqrt(3) - 1)
. Leg AC
is the shorter leg. The equation of the line through it is y = sqrt(3)*x
. The line perpendicular to AC
has slope -sqrt(3)/3
, and the line through O
with slope -sqrt(3)/3
is
y = (-sqrt(3)/3)*(x - 2*(sqrt(3))*(sqrt(3)-1)) + 2*(sqrt(3)-1) .
The two lines intersect on leg AC
at
Q = (8*sqrt(3)*(sqrt(3)-1), 24*(sqrt(3)-1)) .
So, the command draw (O) -- (Q);
should draw a radius of the circle to leg AC
. On my computer, the command renders a line segment through the other leg and ridiculously long. It seems to me that the command locating point Q
has been ignored.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
path (0,0) coordinate (A) (8,0) coordinate (B) (2,{2*sqrt(3)}) coordinate (C);
node[anchor=north, inner sep=0, font=footnotesize] at (0,-0.15){textit{A}};
node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$){textit{B}};
node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$){textit{C}};
draw (A) -- (B) -- (C) -- cycle;
path let n1={2*(sqrt(3))*(sqrt(3)-1)}, n2={2*(sqrt(3)-1)} in coordinate (O) at (n1,n2);
draw[fill] (O) circle (1.5pt);
draw[blue] let n1={2*(sqrt(3)-1)} in (O) circle (n1);
path let n1={2*(sqrt(3))*(sqrt(3)-1)} in coordinate (P) at (n1,0);
node[anchor=north, inner sep=0, font=footnotesize] at ($(P) +(0,-0.15)$){textit{P}};
draw (O) -- (P);
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={24*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
tikz-pgf
@marmot Why didn't I have to include*1pt
in the commands locatingO
andP
?
– A gal named Desire
2 hours ago
This was just a guess, and it was wrong. However, you ask TikZ to dopath let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent topath ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not needcalc
for that, and this is where TikZ places the point.
– marmot
2 hours ago
I want to manually locateQ
. You may not believe that the coordinates I give renderOQ
perpendicular to legAC
, but you should know that it will be a point on legAC
.
– A gal named Desire
2 hours ago
TikZ
is not puttingQ
on legAC
, though.
– A gal named Desire
2 hours ago
2
I multiplied both coordinates ofQ
by8
errantly.
– A gal named Desire
1 hour ago
|
show 4 more comments
$triangle[ABC]$ is a 30-60 right triangle, and its right angle is at C
. A
is at the origin. A circle is inscribed in it; its center is at
O = (2*sqrt(3)*(sqrt(3) - 1), 2*(sqrt(3) - 1))
and its radius is 12(sqrt(3) - 1)
. Leg AC
is the shorter leg. The equation of the line through it is y = sqrt(3)*x
. The line perpendicular to AC
has slope -sqrt(3)/3
, and the line through O
with slope -sqrt(3)/3
is
y = (-sqrt(3)/3)*(x - 2*(sqrt(3))*(sqrt(3)-1)) + 2*(sqrt(3)-1) .
The two lines intersect on leg AC
at
Q = (8*sqrt(3)*(sqrt(3)-1), 24*(sqrt(3)-1)) .
So, the command draw (O) -- (Q);
should draw a radius of the circle to leg AC
. On my computer, the command renders a line segment through the other leg and ridiculously long. It seems to me that the command locating point Q
has been ignored.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
path (0,0) coordinate (A) (8,0) coordinate (B) (2,{2*sqrt(3)}) coordinate (C);
node[anchor=north, inner sep=0, font=footnotesize] at (0,-0.15){textit{A}};
node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$){textit{B}};
node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$){textit{C}};
draw (A) -- (B) -- (C) -- cycle;
path let n1={2*(sqrt(3))*(sqrt(3)-1)}, n2={2*(sqrt(3)-1)} in coordinate (O) at (n1,n2);
draw[fill] (O) circle (1.5pt);
draw[blue] let n1={2*(sqrt(3)-1)} in (O) circle (n1);
path let n1={2*(sqrt(3))*(sqrt(3)-1)} in coordinate (P) at (n1,0);
node[anchor=north, inner sep=0, font=footnotesize] at ($(P) +(0,-0.15)$){textit{P}};
draw (O) -- (P);
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={24*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
tikz-pgf
$triangle[ABC]$ is a 30-60 right triangle, and its right angle is at C
. A
is at the origin. A circle is inscribed in it; its center is at
O = (2*sqrt(3)*(sqrt(3) - 1), 2*(sqrt(3) - 1))
and its radius is 12(sqrt(3) - 1)
. Leg AC
is the shorter leg. The equation of the line through it is y = sqrt(3)*x
. The line perpendicular to AC
has slope -sqrt(3)/3
, and the line through O
with slope -sqrt(3)/3
is
y = (-sqrt(3)/3)*(x - 2*(sqrt(3))*(sqrt(3)-1)) + 2*(sqrt(3)-1) .
The two lines intersect on leg AC
at
Q = (8*sqrt(3)*(sqrt(3)-1), 24*(sqrt(3)-1)) .
So, the command draw (O) -- (Q);
should draw a radius of the circle to leg AC
. On my computer, the command renders a line segment through the other leg and ridiculously long. It seems to me that the command locating point Q
has been ignored.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
path (0,0) coordinate (A) (8,0) coordinate (B) (2,{2*sqrt(3)}) coordinate (C);
node[anchor=north, inner sep=0, font=footnotesize] at (0,-0.15){textit{A}};
node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$){textit{B}};
node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$){textit{C}};
draw (A) -- (B) -- (C) -- cycle;
path let n1={2*(sqrt(3))*(sqrt(3)-1)}, n2={2*(sqrt(3)-1)} in coordinate (O) at (n1,n2);
draw[fill] (O) circle (1.5pt);
draw[blue] let n1={2*(sqrt(3)-1)} in (O) circle (n1);
path let n1={2*(sqrt(3))*(sqrt(3)-1)} in coordinate (P) at (n1,0);
node[anchor=north, inner sep=0, font=footnotesize] at ($(P) +(0,-0.15)$){textit{P}};
draw (O) -- (P);
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={24*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
tikz-pgf
tikz-pgf
edited 1 hour ago
A gal named Desire
asked 2 hours ago
A gal named DesireA gal named Desire
6831411
6831411
@marmot Why didn't I have to include*1pt
in the commands locatingO
andP
?
– A gal named Desire
2 hours ago
This was just a guess, and it was wrong. However, you ask TikZ to dopath let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent topath ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not needcalc
for that, and this is where TikZ places the point.
– marmot
2 hours ago
I want to manually locateQ
. You may not believe that the coordinates I give renderOQ
perpendicular to legAC
, but you should know that it will be a point on legAC
.
– A gal named Desire
2 hours ago
TikZ
is not puttingQ
on legAC
, though.
– A gal named Desire
2 hours ago
2
I multiplied both coordinates ofQ
by8
errantly.
– A gal named Desire
1 hour ago
|
show 4 more comments
@marmot Why didn't I have to include*1pt
in the commands locatingO
andP
?
– A gal named Desire
2 hours ago
This was just a guess, and it was wrong. However, you ask TikZ to dopath let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent topath ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not needcalc
for that, and this is where TikZ places the point.
– marmot
2 hours ago
I want to manually locateQ
. You may not believe that the coordinates I give renderOQ
perpendicular to legAC
, but you should know that it will be a point on legAC
.
– A gal named Desire
2 hours ago
TikZ
is not puttingQ
on legAC
, though.
– A gal named Desire
2 hours ago
2
I multiplied both coordinates ofQ
by8
errantly.
– A gal named Desire
1 hour ago
@marmot Why didn't I have to include
*1pt
in the commands locating O
and P
?– A gal named Desire
2 hours ago
@marmot Why didn't I have to include
*1pt
in the commands locating O
and P
?– A gal named Desire
2 hours ago
This was just a guess, and it was wrong. However, you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent to path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not need calc
for that, and this is where TikZ places the point.– marmot
2 hours ago
This was just a guess, and it was wrong. However, you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent to path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not need calc
for that, and this is where TikZ places the point.– marmot
2 hours ago
I want to manually locate
Q
. You may not believe that the coordinates I give render OQ
perpendicular to leg AC
, but you should know that it will be a point on leg AC
.– A gal named Desire
2 hours ago
I want to manually locate
Q
. You may not believe that the coordinates I give render OQ
perpendicular to leg AC
, but you should know that it will be a point on leg AC
.– A gal named Desire
2 hours ago
TikZ
is not putting Q
on leg AC
, though.– A gal named Desire
2 hours ago
TikZ
is not putting Q
on leg AC
, though.– A gal named Desire
2 hours ago
2
2
I multiplied both coordinates of
Q
by 8
errantly.– A gal named Desire
1 hour ago
I multiplied both coordinates of
Q
by 8
errantly.– A gal named Desire
1 hour ago
|
show 4 more comments
2 Answers
2
active
oldest
votes
I am sorry, I cannot follow your equations at all. you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent to
path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not need calc for that), and this is where TikZ places the point. I cannot tell you everything that went wrong in your computation of Q
, but here is one point: how is it possible that you do not need the coordinates of O
in your way of doing things? You should be solving
alpha * 1 = O_x + beta
alpha * sqrt(3) = O_y - beta * sqrt(3)/3
if you want to find the point where AC
intersects with the line that is perpendicular and runs through O
, but I cannot see you doing this. (BTW, there is intersection cs:
specifically for that, you do not need to do such things by hand.)
Luckily these projections can be done with calc
out of the box.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
draw (0,0) coordinate[label=below:$scriptstyle A$] (A) --
({8*1},0) coordinate[label=below:$scriptstyle B$] (B) --
({8*(1/4)},{8*sqrt(3)/4}) coordinate[label=above:$scriptstyle A$] (C) -- cycle;
draw[fill] ({8*(sqrt(3)/4)*(sqrt(3)-1)},{8*(1/4)*(sqrt(3)-1)})
coordinate (O) circle (1.5pt);
draw[blue] (O) circle({8*(sqrt(3)-1)/4});
path ($(A)!(O)!(C)$) coordinate[label=left:$scriptstyle Q$] (Q)
($(A)!(O)!(B)$) coordinate[label=below:$scriptstyle P$] (P);
draw (O) -- (P);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
Do you agree that the coordinate that I give forQ
are the coordinates for a point on the liney = sqrt(3)*x
?
– A gal named Desire
2 hours ago
A
andC
are points on the line. The y-coordinate issqrt(3)
times bigger than the x-coordinate for these points. Same is true forQ
. Why doesTikZ
not plot Q onAC
?
– A gal named Desire
2 hours ago
@AgalnamedDesire How is that important? This answer provides a way to do the projection independently of these computations and irrespective of whether or not you set the origin atA
. If your comment is to ask whether the fact that in your Q does not appear where you want it to be due to an error in TikZ or in your code, my bet is that it is not TikZ. As I said, I could not follow your logic.
– marmot
2 hours ago
add a comment |
Just for fun with tkz-euclide
.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz,tkz-euclide}
usetikzlibrary{calc}
usetkzobj{all}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (8,0);
coordinate (C) at (2,{2*sqrt(2.99)});
tkzDefCircle[in](A,B,C)
tkzGetPoint{O} tkzGetLength{rIN}
tkzDrawPoints(O)
tkzDrawCircle[R,color=blue](O,rIN pt)
tkzLabelPoints[below](B)
tkzLabelPoints[above left](A,C)
tkzDrawPolygon(A,B,C)
tkzDefPointBy[projection= onto A--C](O) tkzGetPoint{Q}
tkzDefPointBy[projection= onto A--B](O) tkzGetPoint{P}
draw (O)--(Q) (O)--(P)node[below]{$P$};
filldraw [green](Q) circle (1.5pt);
node at (Q)[left]{$Q$};
end{tikzpicture}
end{document}
add a comment |
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I am sorry, I cannot follow your equations at all. you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent to
path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not need calc for that), and this is where TikZ places the point. I cannot tell you everything that went wrong in your computation of Q
, but here is one point: how is it possible that you do not need the coordinates of O
in your way of doing things? You should be solving
alpha * 1 = O_x + beta
alpha * sqrt(3) = O_y - beta * sqrt(3)/3
if you want to find the point where AC
intersects with the line that is perpendicular and runs through O
, but I cannot see you doing this. (BTW, there is intersection cs:
specifically for that, you do not need to do such things by hand.)
Luckily these projections can be done with calc
out of the box.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
draw (0,0) coordinate[label=below:$scriptstyle A$] (A) --
({8*1},0) coordinate[label=below:$scriptstyle B$] (B) --
({8*(1/4)},{8*sqrt(3)/4}) coordinate[label=above:$scriptstyle A$] (C) -- cycle;
draw[fill] ({8*(sqrt(3)/4)*(sqrt(3)-1)},{8*(1/4)*(sqrt(3)-1)})
coordinate (O) circle (1.5pt);
draw[blue] (O) circle({8*(sqrt(3)-1)/4});
path ($(A)!(O)!(C)$) coordinate[label=left:$scriptstyle Q$] (Q)
($(A)!(O)!(B)$) coordinate[label=below:$scriptstyle P$] (P);
draw (O) -- (P);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
Do you agree that the coordinate that I give forQ
are the coordinates for a point on the liney = sqrt(3)*x
?
– A gal named Desire
2 hours ago
A
andC
are points on the line. The y-coordinate issqrt(3)
times bigger than the x-coordinate for these points. Same is true forQ
. Why doesTikZ
not plot Q onAC
?
– A gal named Desire
2 hours ago
@AgalnamedDesire How is that important? This answer provides a way to do the projection independently of these computations and irrespective of whether or not you set the origin atA
. If your comment is to ask whether the fact that in your Q does not appear where you want it to be due to an error in TikZ or in your code, my bet is that it is not TikZ. As I said, I could not follow your logic.
– marmot
2 hours ago
add a comment |
I am sorry, I cannot follow your equations at all. you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent to
path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not need calc for that), and this is where TikZ places the point. I cannot tell you everything that went wrong in your computation of Q
, but here is one point: how is it possible that you do not need the coordinates of O
in your way of doing things? You should be solving
alpha * 1 = O_x + beta
alpha * sqrt(3) = O_y - beta * sqrt(3)/3
if you want to find the point where AC
intersects with the line that is perpendicular and runs through O
, but I cannot see you doing this. (BTW, there is intersection cs:
specifically for that, you do not need to do such things by hand.)
Luckily these projections can be done with calc
out of the box.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
draw (0,0) coordinate[label=below:$scriptstyle A$] (A) --
({8*1},0) coordinate[label=below:$scriptstyle B$] (B) --
({8*(1/4)},{8*sqrt(3)/4}) coordinate[label=above:$scriptstyle A$] (C) -- cycle;
draw[fill] ({8*(sqrt(3)/4)*(sqrt(3)-1)},{8*(1/4)*(sqrt(3)-1)})
coordinate (O) circle (1.5pt);
draw[blue] (O) circle({8*(sqrt(3)-1)/4});
path ($(A)!(O)!(C)$) coordinate[label=left:$scriptstyle Q$] (Q)
($(A)!(O)!(B)$) coordinate[label=below:$scriptstyle P$] (P);
draw (O) -- (P);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
Do you agree that the coordinate that I give forQ
are the coordinates for a point on the liney = sqrt(3)*x
?
– A gal named Desire
2 hours ago
A
andC
are points on the line. The y-coordinate issqrt(3)
times bigger than the x-coordinate for these points. Same is true forQ
. Why doesTikZ
not plot Q onAC
?
– A gal named Desire
2 hours ago
@AgalnamedDesire How is that important? This answer provides a way to do the projection independently of these computations and irrespective of whether or not you set the origin atA
. If your comment is to ask whether the fact that in your Q does not appear where you want it to be due to an error in TikZ or in your code, my bet is that it is not TikZ. As I said, I could not follow your logic.
– marmot
2 hours ago
add a comment |
I am sorry, I cannot follow your equations at all. you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent to
path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not need calc for that), and this is where TikZ places the point. I cannot tell you everything that went wrong in your computation of Q
, but here is one point: how is it possible that you do not need the coordinates of O
in your way of doing things? You should be solving
alpha * 1 = O_x + beta
alpha * sqrt(3) = O_y - beta * sqrt(3)/3
if you want to find the point where AC
intersects with the line that is perpendicular and runs through O
, but I cannot see you doing this. (BTW, there is intersection cs:
specifically for that, you do not need to do such things by hand.)
Luckily these projections can be done with calc
out of the box.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
draw (0,0) coordinate[label=below:$scriptstyle A$] (A) --
({8*1},0) coordinate[label=below:$scriptstyle B$] (B) --
({8*(1/4)},{8*sqrt(3)/4}) coordinate[label=above:$scriptstyle A$] (C) -- cycle;
draw[fill] ({8*(sqrt(3)/4)*(sqrt(3)-1)},{8*(1/4)*(sqrt(3)-1)})
coordinate (O) circle (1.5pt);
draw[blue] (O) circle({8*(sqrt(3)-1)/4});
path ($(A)!(O)!(C)$) coordinate[label=left:$scriptstyle Q$] (Q)
($(A)!(O)!(B)$) coordinate[label=below:$scriptstyle P$] (P);
draw (O) -- (P);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
I am sorry, I cannot follow your equations at all. you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent to
path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not need calc for that), and this is where TikZ places the point. I cannot tell you everything that went wrong in your computation of Q
, but here is one point: how is it possible that you do not need the coordinates of O
in your way of doing things? You should be solving
alpha * 1 = O_x + beta
alpha * sqrt(3) = O_y - beta * sqrt(3)/3
if you want to find the point where AC
intersects with the line that is perpendicular and runs through O
, but I cannot see you doing this. (BTW, there is intersection cs:
specifically for that, you do not need to do such things by hand.)
Luckily these projections can be done with calc
out of the box.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
noindent hspace*{fill}
begin{tikzpicture}
draw (0,0) coordinate[label=below:$scriptstyle A$] (A) --
({8*1},0) coordinate[label=below:$scriptstyle B$] (B) --
({8*(1/4)},{8*sqrt(3)/4}) coordinate[label=above:$scriptstyle A$] (C) -- cycle;
draw[fill] ({8*(sqrt(3)/4)*(sqrt(3)-1)},{8*(1/4)*(sqrt(3)-1)})
coordinate (O) circle (1.5pt);
draw[blue] (O) circle({8*(sqrt(3)-1)/4});
path ($(A)!(O)!(C)$) coordinate[label=left:$scriptstyle Q$] (Q)
($(A)!(O)!(B)$) coordinate[label=below:$scriptstyle P$] (P);
draw (O) -- (P);
draw[fill=green] (Q) circle (1.5pt);
draw[green] (O) -- (Q);
end{tikzpicture}
end{document}
edited 2 hours ago
answered 2 hours ago
marmotmarmot
120k6154290
120k6154290
Do you agree that the coordinate that I give forQ
are the coordinates for a point on the liney = sqrt(3)*x
?
– A gal named Desire
2 hours ago
A
andC
are points on the line. The y-coordinate issqrt(3)
times bigger than the x-coordinate for these points. Same is true forQ
. Why doesTikZ
not plot Q onAC
?
– A gal named Desire
2 hours ago
@AgalnamedDesire How is that important? This answer provides a way to do the projection independently of these computations and irrespective of whether or not you set the origin atA
. If your comment is to ask whether the fact that in your Q does not appear where you want it to be due to an error in TikZ or in your code, my bet is that it is not TikZ. As I said, I could not follow your logic.
– marmot
2 hours ago
add a comment |
Do you agree that the coordinate that I give forQ
are the coordinates for a point on the liney = sqrt(3)*x
?
– A gal named Desire
2 hours ago
A
andC
are points on the line. The y-coordinate issqrt(3)
times bigger than the x-coordinate for these points. Same is true forQ
. Why doesTikZ
not plot Q onAC
?
– A gal named Desire
2 hours ago
@AgalnamedDesire How is that important? This answer provides a way to do the projection independently of these computations and irrespective of whether or not you set the origin atA
. If your comment is to ask whether the fact that in your Q does not appear where you want it to be due to an error in TikZ or in your code, my bet is that it is not TikZ. As I said, I could not follow your logic.
– marmot
2 hours ago
Do you agree that the coordinate that I give for
Q
are the coordinates for a point on the line y = sqrt(3)*x
?– A gal named Desire
2 hours ago
Do you agree that the coordinate that I give for
Q
are the coordinates for a point on the line y = sqrt(3)*x
?– A gal named Desire
2 hours ago
A
and C
are points on the line. The y-coordinate is sqrt(3)
times bigger than the x-coordinate for these points. Same is true for Q
. Why does TikZ
not plot Q on AC
?– A gal named Desire
2 hours ago
A
and C
are points on the line. The y-coordinate is sqrt(3)
times bigger than the x-coordinate for these points. Same is true for Q
. Why does TikZ
not plot Q on AC
?– A gal named Desire
2 hours ago
@AgalnamedDesire How is that important? This answer provides a way to do the projection independently of these computations and irrespective of whether or not you set the origin at
A
. If your comment is to ask whether the fact that in your Q does not appear where you want it to be due to an error in TikZ or in your code, my bet is that it is not TikZ. As I said, I could not follow your logic.– marmot
2 hours ago
@AgalnamedDesire How is that important? This answer provides a way to do the projection independently of these computations and irrespective of whether or not you set the origin at
A
. If your comment is to ask whether the fact that in your Q does not appear where you want it to be due to an error in TikZ or in your code, my bet is that it is not TikZ. As I said, I could not follow your logic.– marmot
2 hours ago
add a comment |
Just for fun with tkz-euclide
.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz,tkz-euclide}
usetikzlibrary{calc}
usetkzobj{all}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (8,0);
coordinate (C) at (2,{2*sqrt(2.99)});
tkzDefCircle[in](A,B,C)
tkzGetPoint{O} tkzGetLength{rIN}
tkzDrawPoints(O)
tkzDrawCircle[R,color=blue](O,rIN pt)
tkzLabelPoints[below](B)
tkzLabelPoints[above left](A,C)
tkzDrawPolygon(A,B,C)
tkzDefPointBy[projection= onto A--C](O) tkzGetPoint{Q}
tkzDefPointBy[projection= onto A--B](O) tkzGetPoint{P}
draw (O)--(Q) (O)--(P)node[below]{$P$};
filldraw [green](Q) circle (1.5pt);
node at (Q)[left]{$Q$};
end{tikzpicture}
end{document}
add a comment |
Just for fun with tkz-euclide
.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz,tkz-euclide}
usetikzlibrary{calc}
usetkzobj{all}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (8,0);
coordinate (C) at (2,{2*sqrt(2.99)});
tkzDefCircle[in](A,B,C)
tkzGetPoint{O} tkzGetLength{rIN}
tkzDrawPoints(O)
tkzDrawCircle[R,color=blue](O,rIN pt)
tkzLabelPoints[below](B)
tkzLabelPoints[above left](A,C)
tkzDrawPolygon(A,B,C)
tkzDefPointBy[projection= onto A--C](O) tkzGetPoint{Q}
tkzDefPointBy[projection= onto A--B](O) tkzGetPoint{P}
draw (O)--(Q) (O)--(P)node[below]{$P$};
filldraw [green](Q) circle (1.5pt);
node at (Q)[left]{$Q$};
end{tikzpicture}
end{document}
add a comment |
Just for fun with tkz-euclide
.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz,tkz-euclide}
usetikzlibrary{calc}
usetkzobj{all}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (8,0);
coordinate (C) at (2,{2*sqrt(2.99)});
tkzDefCircle[in](A,B,C)
tkzGetPoint{O} tkzGetLength{rIN}
tkzDrawPoints(O)
tkzDrawCircle[R,color=blue](O,rIN pt)
tkzLabelPoints[below](B)
tkzLabelPoints[above left](A,C)
tkzDrawPolygon(A,B,C)
tkzDefPointBy[projection= onto A--C](O) tkzGetPoint{Q}
tkzDefPointBy[projection= onto A--B](O) tkzGetPoint{P}
draw (O)--(Q) (O)--(P)node[below]{$P$};
filldraw [green](Q) circle (1.5pt);
node at (Q)[left]{$Q$};
end{tikzpicture}
end{document}
Just for fun with tkz-euclide
.
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz,tkz-euclide}
usetikzlibrary{calc}
usetkzobj{all}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (8,0);
coordinate (C) at (2,{2*sqrt(2.99)});
tkzDefCircle[in](A,B,C)
tkzGetPoint{O} tkzGetLength{rIN}
tkzDrawPoints(O)
tkzDrawCircle[R,color=blue](O,rIN pt)
tkzLabelPoints[below](B)
tkzLabelPoints[above left](A,C)
tkzDrawPolygon(A,B,C)
tkzDefPointBy[projection= onto A--C](O) tkzGetPoint{Q}
tkzDefPointBy[projection= onto A--B](O) tkzGetPoint{P}
draw (O)--(Q) (O)--(P)node[below]{$P$};
filldraw [green](Q) circle (1.5pt);
node at (Q)[left]{$Q$};
end{tikzpicture}
end{document}
edited 29 mins ago
answered 1 hour ago
ferahfezaferahfeza
7,60912033
7,60912033
add a comment |
add a comment |
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@marmot Why didn't I have to include
*1pt
in the commands locatingO
andP
?– A gal named Desire
2 hours ago
This was just a guess, and it was wrong. However, you ask TikZ to do
path let n1={8*sqrt(3)*(sqrt(3)-1)}, n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (n1,n2);
which is equivalent topath ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q);
(meaning you do not needcalc
for that, and this is where TikZ places the point.– marmot
2 hours ago
I want to manually locate
Q
. You may not believe that the coordinates I give renderOQ
perpendicular to legAC
, but you should know that it will be a point on legAC
.– A gal named Desire
2 hours ago
TikZ
is not puttingQ
on legAC
, though.– A gal named Desire
2 hours ago
2
I multiplied both coordinates of
Q
by8
errantly.– A gal named Desire
1 hour ago