Mdthbs

$triangle[ABC]$ 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}

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}

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}

deffangle{60}

defsangle{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}