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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












2















$triangle[ABC]$ 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}









share|improve this question

























  • @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











  • 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






  • 2





    I multiplied both coordinates of Q by 8 errantly.

    – A gal named Desire
    1 hour ago
















2















$triangle[ABC]$ 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}









share|improve this question

























  • @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











  • 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






  • 2





    I multiplied both coordinates of Q by 8 errantly.

    – A gal named Desire
    1 hour ago














2












2








2


2






$triangle[ABC]$ 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}









share|improve this question
















$triangle[ABC]$ 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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








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 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











  • 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






  • 2





    I multiplied both coordinates of Q by 8 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











  • 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













  • TikZ is not putting Q on leg AC, though.

    – A gal named Desire
    2 hours ago






  • 2





    I multiplied both coordinates of Q by 8 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










2 Answers
2






active

oldest

votes


















2














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}


enter image description here






share|improve this answer


























  • 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













  • @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



















0














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}


enter image description here






share|improve this answer


























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    2 Answers
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    2 Answers
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    active

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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}


    enter image description here






    share|improve this answer


























    • 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













    • @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
















    2














    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}


    enter image description here






    share|improve this answer


























    • 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













    • @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














    2












    2








    2







    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}


    enter image description here






    share|improve this answer















    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}


    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 2 hours ago

























    answered 2 hours ago









    marmotmarmot

    120k6154290




    120k6154290













    • 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













    • @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



















    • 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













    • @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

















    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











    0














    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}


    enter image description here






    share|improve this answer






























      0














      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}


      enter image description here






      share|improve this answer




























        0












        0








        0







        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}


        enter image description here






        share|improve this answer















        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}


        enter image description here







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 33 mins ago

























        answered 1 hour ago









        ferahfezaferahfeza

        7,60912033




        7,60912033






























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