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A `coordinate` command ignored

1

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 = ((sqrt(3)/4)(sqrt(3)-1), (1/4)(sqrt(3)-1))

and its radius is (1/4)(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 intersects leg AC at

Q = (sqrt(3)*(sqrt(3)-1), 3*(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*1},0) coordinate (B) ({8*(1/4)},{8*sqrt(3)/4}) 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={8*(sqrt(3)/4)*(sqrt(3)-1)}, n2={8*(1/4)*(sqrt(3)-1)} in coordinate (O) at (n1,n2);

draw[fill] (O) circle (1.5pt);

draw[blue] let n1={8*(sqrt(3)-1)/4} in (O) circle (n1);





%The points of tangency are located via the Angle-Bisector Theorem.

path let n1={8*(1/2+1/4)/(1/2+sqrt(3)/2+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={8*3*(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

asked 14 mins ago

A gal named DesireA gal named Desire

6781411

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If one adds the units, the problem disappears: path let n1={8*sqrt(3)*(sqrt(3)-1)*1pt}, n2={8*3*(sqrt(3)-1)*1pt} in coordinate (Q) at (n1,n2);.
  
  – marmot
  4 mins ago
  

  
  
  
  

add a comment | 

1

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 = ((sqrt(3)/4)(sqrt(3)-1), (1/4)(sqrt(3)-1))

and its radius is (1/4)(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 intersects leg AC at

Q = (sqrt(3)*(sqrt(3)-1), 3*(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*1},0) coordinate (B) ({8*(1/4)},{8*sqrt(3)/4}) 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={8*(sqrt(3)/4)*(sqrt(3)-1)}, n2={8*(1/4)*(sqrt(3)-1)} in coordinate (O) at (n1,n2);

draw[fill] (O) circle (1.5pt);

draw[blue] let n1={8*(sqrt(3)-1)/4} in (O) circle (n1);





%The points of tangency are located via the Angle-Bisector Theorem.

path let n1={8*(1/2+1/4)/(1/2+sqrt(3)/2+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={8*3*(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

asked 14 mins ago

A gal named DesireA gal named Desire

6781411

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If one adds the units, the problem disappears: path let n1={8*sqrt(3)*(sqrt(3)-1)*1pt}, n2={8*3*(sqrt(3)-1)*1pt} in coordinate (Q) at (n1,n2);.
  
  – marmot
  4 mins ago
  

  
  
  
  

add a comment | 

$triangle[ABC]$ 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 = ((sqrt(3)/4)(sqrt(3)-1), (1/4)(sqrt(3)-1))

and its radius is (1/4)(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 intersects leg AC at

Q = (sqrt(3)*(sqrt(3)-1), 3*(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*1},0) coordinate (B) ({8*(1/4)},{8*sqrt(3)/4}) 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={8*(sqrt(3)/4)*(sqrt(3)-1)}, n2={8*(1/4)*(sqrt(3)-1)} in coordinate (O) at (n1,n2);

draw[fill] (O) circle (1.5pt);

draw[blue] let n1={8*(sqrt(3)-1)/4} in (O) circle (n1);





%The points of tangency are located via the Angle-Bisector Theorem.

path let n1={8*(1/2+1/4)/(1/2+sqrt(3)/2+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={8*3*(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

asked 14 mins ago

A gal named DesireA gal named Desire

6781411

O = ((sqrt(3)/4)(sqrt(3)-1), (1/4)(sqrt(3)-1))

Q = (sqrt(3)*(sqrt(3)-1), 3*(sqrt(3)-1)) .

documentclass{amsart}

usepackage{amsmath}







usepackage{tikz}

usetikzlibrary{calc,intersections}





begin{document}



noindent hspace*{fill}

begin{tikzpicture}



path (0,0) coordinate (A) ({8*1},0) coordinate (B) ({8*(1/4)},{8*sqrt(3)/4}) 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={8*(sqrt(3)/4)*(sqrt(3)-1)}, n2={8*(1/4)*(sqrt(3)-1)} in coordinate (O) at (n1,n2);

draw[fill] (O) circle (1.5pt);

draw[blue] let n1={8*(sqrt(3)-1)/4} in (O) circle (n1);





%The points of tangency are located via the Angle-Bisector Theorem.

path let n1={8*(1/2+1/4)/(1/2+sqrt(3)/2+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={8*3*(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

asked 14 mins ago

A gal named DesireA gal named Desire

6781411

share|improve this question

asked 14 mins ago

A gal named DesireA gal named Desire

6781411

asked 14 mins ago

A gal named DesireA gal named Desire

6781411

asked 14 mins ago

A gal named DesireA gal named Desire

6781411