# LaTeX command for unknown symbol

I am trying to type the shown symbol in LaTeX and I am hoping to find a command producing this symbol. This is just a very crude sketch. The three lines are supposed to be equal length with 120 degrees between them. It is quite large here, but I would like it to be of "normal character size", comparable to a + or so. Is there any direct way of obtaining this without including it as a graphic or perhaps drawing it in TikZ?

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did you check on detexify.kirelabs.org/classify.html ? –  Dominic Michaelis Mar 20 '13 at 14:29
I tried some sketches, but DeTeXify didn't gave my a valuable answer. –  CommuSoft Mar 20 '13 at 14:31
i don't believe this symbol is in unicode. what is its meaning, and can you provide a reference to where it is used in a published article or book? (if good documentation is available, i can convey it to the unicode technical committee with a request that it be encoded.) –  barbara beeton Mar 20 '13 at 14:41
@barbara In my context it does not have a special meaning and I do not know of any particular meaning of the symbol. I simply have a plot of some data where this symbol marks the data points. I would like to use the symbol in the caption text to refer to the plotted data. –  Thomas Arildsen Mar 20 '13 at 17:03
@barbarabeeton Unicode has a symbol for a pile of poo. I'd be surprised if this wasn't there... –  millimoose Mar 21 '13 at 0:23

Within the package \usepackage{MnSymbol} you can find

$A \downY B \upY C$


which renders like:

I took that from comprehensive symbols, page 23f.

Unfortunately (thanks for the hint @egreg) the use of MnSymbol does not only add additional commands but altering some as well. This might produce ugly side-effects. Good for us there is a step by step tutorial How can I use a symbol provided by a package without changing the entire mathematics font?

Following this is fairly easy (give Leo Liu as much credit as you can) and so you can circumvent the usage of the hole package as follows:

\documentclass{article}

\DeclareFontFamily{U}{MnSymbolC}{}
\DeclareFontShape{U}{MnSymbolC}{m}{n}{
<-6>  MnSymbolC5
<6-7>  MnSymbolC6
<7-8>  MnSymbolC7
<8-9>  MnSymbolC8
<9-10> MnSymbolC9
<10-12> MnSymbolC10
<12->   MnSymbolC12}{}
\DeclareFontShape{U}{MnSymbolC}{b}{n}{
<-6>  MnSymbolC-Bold5
<6-7>  MnSymbolC-Bold6
<7-8>  MnSymbolC-Bold7
<8-9>  MnSymbolC-Bold8
<9-10> MnSymbolC-Bold9
<10-12> MnSymbolC-Bold10
<12->   MnSymbolC-Bold12}{}

\DeclareSymbolFont{MnSyC}{U}{MnSymbolC}{m}{n}
\SetSymbolFont{MnSyC}{bold}{U}{MnSymbolC}{b}{n}
\DeclareMathSymbol{\upY}{\mathbin}{MnSyC}{41}
\DeclareMathSymbol{\downY}{\mathbin}{MnSyC}{43}

\begin{document}
$A \downY B \upY C$
\boldmath
$A \downY B \upY C$
\end{document}


Which (thanks to egreg again) renders even in \bold

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The package changes all math symbols, though. –  egreg Mar 20 '13 at 15:10
Hmm - that is bad. But there must be a way to circumvent this... –  Dominikus K. Mar 20 '13 at 15:40
–  egreg Mar 20 '13 at 15:52
Well done; you can also add \SetSymbolFont{MnSyC}{bold}{U}{MnSymbolC}{b}{n} to get a bold symbol when \mathversion{bold} (or \boldmath) is in force. Please, change also \documentclass{minimal} into \documentclass{article}, the minimal class should never be used for these examples. –  egreg Mar 20 '13 at 16:15

You can use the picture environment:

\documentclass{article}
\usepackage{pict2e,picture}
\newcommand{\triplus}{\mathbin{
\sbox0{$+$}
\begin{picture}(\wd0,\ht0)
\put(0,0){\line(1,.577){.5\wd0}}
\put(.5\wd0,.289\wd0){\line(-1,.577){-.5\wd0}}
\put(.5\wd0,.289\wd0){\line(0,1){\dimexpr\ht0-.289\wd0\relax}}
\end{picture}
}
}

\begin{document}
$a+b\triplus c$
\end{document}


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Can I use the box 0 without any risk? –  Marco Daniel Mar 20 '13 at 14:50
@MarcoDaniel Yes: the contents of \mathbin is processed in a group and the picture mode doesn't use box 0, as far as I know. –  egreg Mar 20 '13 at 14:58
Thanks. I wasn't sure. –  Marco Daniel Mar 20 '13 at 15:01
I suppose a picture environment is less computationally demanding than firing up TikZ? –  Thomas Arildsen Mar 21 '13 at 7:56

Second cut, trying to preserve the 120 degree angles, as well as getting proper alignment.

\documentclass{article}
\usepackage{scalerel}

\let\savearraystretch\arraystretch
\renewcommand\arraystretch{0}
\begin{array}{c}
\rule{-.01ex}{.01ex}\rule{.05ex}{.8ex}\\
\stretchto{
\scalerel*[\widthof{=}]{\wedge}
{\rule{1ex}{3ex}}%
}{0.42ex}\\
\vphantom{\rule{1ex}{.01ex}}
\end{array}
\let\arraystretch\savearraystretch
}

\begin{document}

$x^2 \triad \begin{array}{c}a\\b\\c\end{array}$

\end{document}


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You could also play around with the Mercedes star mark in TikZ:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{plotmarks}

\newcommand{\unkwn}{\tikz[scale=2] \draw plot [mark=Mercedes star] coordinates {(0,0)};}

\begin{document}

$a \unkwn b$

\end{document}


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Here's my TikZ shot. The length of the line was chosen to be as close to the length of the line in a + symbol as I could get it (interestingly, the ends of the lines on the + are not perfect half circles). And the height was so that the intersection was the same as the centre of the cross of the +.

Edit Looking at egreg's answer, I see that I can compute the dimensions automatically.

\documentclass{article}
%\url{http://tex.stackexchange.com/q/103427/86}
\usepackage{tikz}

\mathbin{%
\tikz[overlay,baseline=0 pt] {
\pgfmathparse{(height("$+$") + depth("$+$") -
\the\pgflinewidth)/2}%
\let\triht=\pgfmathresult
\pgfmathparse{(height("$+$") - depth("$+$"))/2}%
\let\tribs=\pgfmathresult
\draw[yshift=\tribs pt,xshift=-6pt,red,line cap=round] (0,0) --
(0,\triht pt) (0,0) -- (-30:\triht pt) (0,0) -- (210:\triht pt);%
}
}%
}

\begin{document}

$a + b a +\triad b$
\end{document}


Close up (overlaid on a +):

In "normal" use:

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Since you want a normal character, an alternative would be to use the similar symbol \Yup of the package stmaryrd. The angles are about 135°, 135°, 90°. The picture shows $a \Yup b$.

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You could use the scalerel package to compress that to 120/120/120 :^) –  Steven B. Segletes Mar 20 '13 at 14:52