A `coordinate` command ignored
$triangle[ABC]$ 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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2 Answers
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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 27 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