How to node midpoint of an arc automatically? - TeX

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user 3.14159

An arc between two points on a circle is not uniquely defined, there are two options. In the case at hand, there is also a midpoint close to `C` of an arc between `A` and `B` on a circle with center `O`. This is why the following code has two angles, `\n1` and `\n2`, which differ by `180`. While in this example the angles of the points are known, in general they may not be, which is why they get determined with the `calc` library. In more detail, after saying inside the `let ... in` block
```
\p{A}=(A)
```
the two lengths `\x{A}` and `\y{A}` contain the x and y coordinates on the screen. Then 
```
\n{A}={scalar(atan2(\y{A},\x{A}))}
``` 
contains the polar angle of `A`, and likewise for `B`. Note also that using polar coordinates is convenient in this case.
```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[declare function={R=2;},c/.style={circle,fill,inner sep=1pt}]
	\path 
	(0,0)	coordinate (O)
	(60:R)	coordinate (A)
	(-30:R)	coordinate (C)
	(210:R)	coordinate (B)
	let \p{A}=(A),\p{B}=(B),
	\n{A}={scalar(atan2(\y{A},\x{A}))},\n{B}={scalar(atan2(\y{B},\x{B}))},
	\n1={0.5*\n{A}+0.5*\n{B}},\n2={\n1+180}
	in 
	(\n2:R) coordinate[label=left:{$m$}] (M)	
	;
\draw (O) circle[radius = R];		
\path foreach \p/\g in {O/0,A/90,B/0,C/0}
{(\p)node[c]{}+(\g:2.5mm) node{$\p$}};
\end{tikzpicture}
\end{document}  
```
![Screen Shot 2021-06-19 at 7.59.12 PM.png](/image?hash=b4e3da90e9751483f1bda23390044564605be812d0ff01c576b64580f3eab9f4)

![adfd.png](/image?hash=492d78e43a0e14124d0d40389ef8f23b8a0752a5ac227f99812649087783231b) ``` settings.outformat="png"; settings.render=6; size(5cm); real r=1.2; path testarc=circle((0,0),r); pair A=r*dir(60),B=r*dir(210); draw(testarc); dot("$A$",A,dir(A),Fill); dot("$B$",B,dir(B),Fill); dot("$O$",(0,0),E,Fill); path m=arc((0,0),A,B); path C=arc((0,0),A,B,CW); dot("$m$",midpoint(m),dir(135),Fill); dot("$C$",midpoint(C),dir(-45),Fill); shipout(bbox(mm,invisible)); ```

It has `m=\pgfmathprintnumber{\n1}^\circ`, which shows `-45`, and `-45+180=135`, so everything is consistent, I think. You can use `Mod` if you find this confusing. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture}[declare function={R=2;},c/.style={circle,fill,inner sep=1pt}] \path (0,0) coordinate (O) (60:R) coordinate (A) %(-30:R) coordinate (C) (210:R) coordinate (B) let \p{A}=(A),\p{B}=(B), \n{A}={scalar(atan2(\y{A},\x{A}))},\n{B}={scalar(atan2(\y{B},\x{B}))}, \n1={Mod(0.5*\n{A}+0.5*\n{B},360)},\n2={Mod(\n1+180,360)} in (\n1:R) coordinate[label=right:{$m=\pgfmathprintnumber{\n1}^\circ$}] (M) (\n2:R) coordinate[label=left:{$n=\pgfmathprintnumber{\n2}^\circ$}] (N) ; \draw (O) circle[radius = R]; \draw[dashed] (M) -- (N); \path foreach \p/\g in {O/0,A/90,B/0} {(\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ```

Yes, I think so. One statement I wrote in the answer is incorrect, though, `C` is **not** the other midpoint. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture}[declare function={R=2;},c/.style={circle,fill,inner sep=1pt}] \path (0,0) coordinate (O) (60:R) coordinate (A) %(-30:R) coordinate (C) (210:R) coordinate (B) let \p{A}=(A),\p{B}=(B), \n{A}={scalar(atan2(\y{A},\x{A}))},\n{B}={scalar(atan2(\y{B},\x{B}))}, \n1={0.5*\n{A}+0.5*\n{B}},\n2={\n1+180} in (\n1:R) coordinate[label=right:{$m=\pgfmathprintnumber{\n1}^\circ$}] (M) (\n2:R) coordinate[label=left:{$n=\pgfmathprintnumber{\n2}^\circ$}] (N) ; \draw (O) circle[radius = R]; \draw[dashed] (M) -- (N); \path foreach \p/\g in {O/0,A/90,B/0} {(\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-06-19 at 8.29.26 PM.png](/image?hash=95eccc7fe6de9bc654f20adda0289438a0e2727afd2a30fec7ea921eb811c0c8)

@marmot, re: [your answer](#a2068), Is this true? ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture}[declare function={R=2;},c/.style={circle,fill,inner sep=1pt}] \path (0,0) coordinate (O) (60:R) coordinate (A) %(-30:R) coordinate (C) (210:R) coordinate (B) let \p{A}=(A),\p{B}=(B), \n{A}={scalar(atan2(\y{A},\x{A}))},\n{B}={scalar(atan2(\y{B},\x{B}))}, \n1={0.5*\n{A}+0.5*\n{B}},\n2={\n1+180} in (\n1:R) coordinate[label=right:{$m$}] (M) (\n2:R) coordinate[label=left:{$n$}] (N) ; \draw (O) circle[radius = R]; \path foreach \p/\g in {O/0,A/90,B/0} {(\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ```

1 Answer