3

$\triangle[ABC]$ 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}
  • @marmot Why didn't I have to include *1pt in the commands locating O and P? – A gal named Desire Apr 20 at 21:38
  • 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 Apr 20 at 21:52
  • 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 Apr 20 at 21:59
  • TikZ is not putting Q on leg AC, though. – A gal named Desire Apr 20 at 22:00
  • 2
    I multiplied both coordinates of Q by 8 errantly. – A gal named Desire Apr 20 at 22:48
3

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

  • 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 Apr 20 at 21:45
  • 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 Apr 20 at 21:48
  • @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 Apr 20 at 21:48
1

To draw the circumscribed circle: Draw the perpendicular bisectors of AB and AC; their intersection is the center O of the circle.

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,through,intersections}

\begin{document}
\begin{tikzpicture}
\def\fangle{60}
\def\sangle{30}
\coordinate (A) at (0,0);
\coordinate (B) at (8,0);
\coordinate (C) at (2,{2*tan(\fangle)});
\path [draw,name path=AB](A)node[left]{$A$}--(B);
\path [draw,name path=BC](B)node[right]{$B$}--(C);
\path [draw,name path=CA](C)node[above]{$C$}--(A);
\path [name path=A-bisector] (A)--++(\fangle/2:8);
\path [name path=B-bisector] (B)--++(180-\sangle/2:8);
\path [name intersections={of=A-bisector and B-bisector, by={O}}];
\path [name path=radius] (O)--++(-90:8);
\path [name intersections={of=AB and radius, by={P}}];
\node [draw,name path=circle,blue] at (O) [circle through={(P)}] {};
\path [name intersections={of=circle and CA, by={Q}}];
\filldraw (O) circle (1.5pt);
\draw (O)--(P)node[below]{$P$};
\draw[green] (O)--(Q)node[left,color=black]{$Q$};
\filldraw (Q) circle (1.5pt);
\end{tikzpicture}

\end{document}

enter image description here

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